cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 11-20 of 134 results. Next

A382404 a(n) = -Sum_{k=0..n} (-1)^k * A039599(n,k)^3.

Original entry on oeis.org

-1, 0, 18, 480, 11550, 275184, 6597360, 159629184, 3897563670, 95946708000, 2378998624860, 59359563244800, 1489281975509328, 37545821365718400, 950601539891016000, 24159023128878865920, 616066120184552310150, 15757649689979967739200
Offset: 0

Views

Author

Seiichi Manyama, Mar 24 2025

Keywords

Links

Crossrefs

Cf. A039599.

Programs

  • PARI
    a(n) = (n-1)/(2*n+1)*binomial(2*n, n)*binomial(3*n, n);

Formula

a(n) = ((n-1)/(2*n+1)) * binomial(2*n,n) * binomial(3*n,n).

A000217 Triangular numbers: a(n) = binomial(n+1,2) = n*(n+1)/2 = 0 + 1 + 2 + ... + n.

Original entry on oeis.org

0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also referred to as T(n) or C(n+1, 2) or binomial(n+1, 2) (preferred).
Also generalized hexagonal numbers: n*(2*n-1), n=0, +-1, +-2, +-3, ... Generalized k-gonal numbers are second k-gonal numbers and positive terms of k-gonal numbers interleaved, k >= 5. In this case k = 6. - Omar E. Pol, Sep 13 2011 and Aug 04 2012
Number of edges in complete graph of order n+1, K_{n+1}.
Number of legal ways to insert a pair of parentheses in a string of n letters. E.g., there are 6 ways for three letters: (a)bc, (ab)c, (abc), a(b)c, a(bc), ab(c). Proof: there are C(n+2,2) ways to choose where the parentheses might go, but n + 1 of them are illegal because the parentheses are adjacent. Cf. A002415.
For n >= 1, a(n) is also the genus of a nonsingular curve of degree n+2, such as the Fermat curve x^(n+2) + y^(n+2) = 1. - Ahmed Fares (ahmedfares(AT)my_deja.com), Feb 21 2001
From Harnack's theorem (1876), the number of branches of a nonsingular curve of order n is bounded by a(n-1)+1, and the bound can be achieved. See also A152947. - Benoit Cloitre, Aug 29 2002. Corrected by Robert McLachlan, Aug 19 2024
Number of tiles in the set of double-n dominoes. - Scott A. Brown, Sep 24 2002
Number of ways a chain of n non-identical links can be broken up. This is based on a similar problem in the field of proteomics: the number of ways a peptide of n amino acid residues can be broken up in a mass spectrometer. In general, each amino acid has a different mass, so AB and BC would have different masses. - James A. Raymond, Apr 08 2003
Triangular numbers - odd numbers = shifted triangular numbers; 1, 3, 6, 10, 15, 21, ... - 1, 3, 5, 7, 9, 11, ... = 0, 0, 1, 3, 6, 10, ... - Xavier Acloque, Oct 31 2003 [Corrected by Derek Orr, May 05 2015]
Centered polygonal numbers are the result of [number of sides * A000217 + 1]. E.g., centered pentagonal numbers (1,6,16,31,...) = 5 * (0,1,3,6,...) + 1. Centered heptagonal numbers (1,8,22,43,...) = 7 * (0,1,3,6,...) + 1. - Xavier Acloque, Oct 31 2003
Maximum number of lines formed by the intersection of n+1 planes. - Ron R. King, Mar 29 2004
Number of permutations of [n] which avoid the pattern 132 and have exactly 1 descent. - Mike Zabrocki, Aug 26 2004
Number of ternary words of length n-1 with subwords (0,1), (0,2) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012
Number of ways two different numbers can be selected from the set {0,1,2,...,n} without repetition, or, number of ways two different numbers can be selected from the set {1,2,...,n} with repetition.
Conjecturally, 1, 6, 120 are the only numbers that are both triangular and factorial. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Mar 30 2005
Binomial transform is {0, 1, 5, 18, 56, 160, 432, ...}, A001793 with one leading zero. - Philippe Deléham, Aug 02 2005
Each pair of neighboring terms adds to a perfect square. - Zak Seidov, Mar 21 2006
Number of transpositions in the symmetric group of n+1 letters, i.e., the number of permutations that leave all but two elements fixed. - Geoffrey Critzer, Jun 23 2006
With rho(n):=exp(i*2*Pi/n) (an n-th root of 1) one has, for n >= 1, rho(n)^a(n) = (-1)^(n+1). Just use the triviality a(2*k+1) == 0 (mod (2*k+1)) and a(2*k) == k (mod (2*k)).
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3)^(n-1). - Sergio Falcon, Feb 12 2007
a(n+1) is the number of terms in the complete homogeneous symmetric polynomial of degree n in 2 variables. - Richard Barnes, Sep 06 2017
The number of distinct handshakes in a room with n+1 people. - Mohammad K. Azarian, Apr 12 2007 [corrected, Joerg Arndt, Jan 18 2016]
Equal to the rank (minimal cardinality of a generating set) of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, May 03 2007
a(n) gives the total number of triangles found when cevians are drawn from a single vertex on a triangle to the side opposite that vertex, where n = the number of cevians drawn+1. For instance, with 1 cevian drawn, n = 1+1 = 2 and a(n)= 2*(2+1)/2 = 3 so there is a total of 3 triangles in the figure. If 2 cevians are drawn from one point to the opposite side, then n = 1+2 = 3 and a(n) = 3*(3+1)/2 = 6 so there is a total of 6 triangles in the figure. - Noah Priluck (npriluck(AT)gmail.com), Apr 30 2007
For n >= 1, a(n) is the number of ways in which n-1 can be written as a sum of three nonnegative integers if representations differing in the order of the terms are considered to be different. In other words, for n >= 1, a(n) is the number of nonnegative integral solutions of the equation x + y + z = n-1. - Amarnath Murthy, Apr 22 2001 (edited by Robert A. Beeler)
a(n) is the number of levels with energy n + 3/2 (in units of h*f0, with Planck's constant h and the oscillator frequency f0) of the three-dimensional isotropic harmonic quantum oscillator. See the comment by A. Murthy above: n = n1 + n2 + n3 with positive integers and ordered. Proof from the o.g.f. See the A. Messiah reference. - Wolfdieter Lang, Jun 29 2007
From Hieronymus Fischer, Aug 06 2007: (Start)
Numbers m >= 0 such that round(sqrt(2m+1)) - round(sqrt(2m)) = 1.
Numbers m >= 0 such that ceiling(2*sqrt(2m+1)) - 1 = 1 + floor(2*sqrt(2m)).
Numbers m >= 0 such that fract(sqrt(2m+1)) > 1/2 and fract(sqrt(2m)) < 1/2, where fract(x) is the fractional part of x (i.e., x - floor(x), x >= 0). (End)
If Y and Z are 3-blocks of an n-set X, then, for n >= 6, a(n-1) is the number of (n-2)-subsets of X intersecting both Y and Z. - Milan Janjic, Nov 09 2007
Equals row sums of triangle A143320, n > 0. - Gary W. Adamson, Aug 07 2008
a(n) is also an even perfect number in A000396 iff n is a Mersenne prime A000668. - Omar E. Pol, Sep 05 2008. Unnecessary assumption removed and clarified by Rick L. Shepherd, Apr 14 2025
Equals row sums of triangle A152204. - Gary W. Adamson, Nov 29 2008
The number of matches played in a round robin tournament: n*(n-1)/2 gives the number of matches needed for n players. Everyone plays against everyone else exactly once. - Georg Wrede (georg(AT)iki.fi), Dec 18 2008
-a(n+1) = E(2)*binomial(n+2,2) (n >= 0) where E(n) are the Euler numbers in the enumeration A122045. Viewed this way, a(n) is the special case k=2 in the sequence of diagonals in the triangle A153641. - Peter Luschny, Jan 06 2009
Equivalent to the first differences of successive tetrahedral numbers. See A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus a(k) = |2^(-3)(P(2,1)-(-1)^k P(2,2k+1))|. - Peter Luschny, Jul 12 2009
a(n) is the smallest number > a(n-1) such that gcd(n,a(n)) = gcd(n,a(n-1)). If n is odd this gcd is n; if n is even it is n/2. - Franklin T. Adams-Watters, Aug 06 2009
Partial sums of A001477. - Juri-Stepan Gerasimov, Jan 25 2010. [A-number corrected by Omar E. Pol, Jun 05 2012]
The numbers along the right edge of Floyd's triangle are 1, 3, 6, 10, 15, .... - Paul Muljadi, Jan 25 2010
From Charlie Marion, Dec 03 2010: (Start)
More generally, a(2k+1) == j*(2j-1) (mod 2k+2j+1) and
a(2k) == [-k + 2j*(j-1)] (mod 2k+2j).
Column sums of:
1 3 5 7 9 ...
1 3 5 ...
1 ...
...............
---------------
1 3 6 10 15 ...
Sum_{n>=1} 1/a(n)^2 = 4*Pi^2/3-12 = 12 less than the volume of a sphere with radius Pi^(1/3).
(End)
A004201(a(n)) = A000290(n); A004202(a(n)) = A002378(n). - Reinhard Zumkeller, Feb 12 2011
1/a(n+1), n >= 0, has e.g.f. -2*(1+x-exp(x))/x^2, and o.g.f. 2*(x+(1-x)*log(1-x))/x^2 (see the Stephen Crowley formula line). -1/(2*a(n+1)) is the z-sequence for the Sheffer triangle of the coefficients of the Bernoulli polynomials A196838/A196839. - Wolfdieter Lang, Oct 26 2011
From Charlie Marion, Feb 23 2012: (Start)
a(n) + a(A002315(k)*n + A001108(k+1)) = (A001653(k+1)*n + A001109(k+1))^2. For k=0 we obtain a(n) + a(n+1) = (n+1)^2 (identity added by N. J. A. Sloane on Feb 19 2004).
a(n) + a(A002315(k)*n - A055997(k+1)) = (A001653(k+1)*n - A001109(k))^2.
(End)
Plot the three points (0,0), (a(n), a(n+1)), (a(n+1), a(n+2)) to form a triangle. The area will be a(n+1)/2. - J. M. Bergot, May 04 2012
The sum of four consecutive triangular numbers, beginning with a(n)=n*(n+1)/2, minus 2 is 2*(n+2)^2. a(n)*a(n+2)/2 = a(a(n+1)-1). - J. M. Bergot, May 17 2012
(a(n)*a(n+3) - a(n+1)*a(n+2))*(a(n+1)*a(n+4) - a(n+2)*a(n+3))/8 = a((n^2+5*n+4)/2). - J. M. Bergot, May 18 2012
a(n)*a(n+1) + a(n+2)*a(n+3) + 3 = a(n^2 + 4*n + 6). - J. M. Bergot, May 22 2012
In general, a(n)*a(n+1) + a(n+k)*a(n+k+1) + a(k-1)*a(k) = a(n^2 + (k+2)*n + k*(k+1)). - Charlie Marion, Sep 11 2012
a(n)*a(n+3) + a(n+1)*a(n+2) = a(n^2 + 4*n + 2). - J. M. Bergot, May 22 2012
In general, a(n)*a(n+k) + a(n+1)*a(n+k-1) = a(n^2 + (k+1)*n + k-1). - Charlie Marion, Sep 11 2012
a(n)*a(n+2) + a(n+1)*a(n+3) = a(n^2 + 4*n + 3). - J. M. Bergot, May 22 2012
Three points (a(n),a(n+1)), (a(n+1),a(n)) and (a(n+2),a(n+3)) form a triangle with area 4*a(n+1). - J. M. Bergot, May 23 2012
a(n) + a(n+k) = (n+k)^2 - (k^2 + (2n-1)*k -2n)/2. For k=1 we obtain a(n) + a(n+1) = (n+1)^2 (see below). - Charlie Marion, Oct 02 2012
In n-space we can define a(n-1) nontrivial orthogonal projections. For example, in 3-space there are a(2)=3 (namely point onto line, point onto plane, line onto plane). - Douglas Latimer, Dec 17 2012
From James East, Jan 08 2013: (Start)
For n >= 1, a(n) is equal to the rank (minimal cardinality of a generating set) and idempotent rank (minimal cardinality of an idempotent generating set) of the semigroup P_n\S_n, where P_n and S_n denote the partition monoid and symmetric group on [n].
For n >= 3, a(n-1) is equal to the rank and idempotent rank of the semigroup T_n\S_n, where T_n and S_n denote the full transformation semigroup and symmetric group on [n].
(End)
For n >= 3, a(n) is equal to the rank and idempotent rank of the semigroup PT_n\S_n, where PT_n and S_n denote the partial transformation semigroup and symmetric group on [n]. - James East, Jan 15 2013
Conjecture: For n > 0, there is always a prime between A000217(n) and A000217(n+1). Sequence A065383 has the first 1000 of these primes. - Ivan N. Ianakiev, Mar 11 2013
The formula, a(n)*a(n+4k+2)/2 + a(k) = a(a(n+2k+1) - (k^2+(k+1)^2)), is a generalization of the formula a(n)*a(n+2)/2 = a(a(n+1)-1) in Bergot's comment dated May 17 2012. - Charlie Marion, Mar 28 2013
The series Sum_{k>=1} 1/a(k) = 2, given in a formula below by Jon Perry, Jul 13 2003, has partial sums 2*n/(n+1) (telescopic sum) = A022998(n)/A026741(n+1). - Wolfdieter Lang, Apr 09 2013
For odd m = 2k+1, we have the recurrence a(m*n + k) = m^2*a(n) + a(k). Corollary: If number T is in the sequence then so is 9*T+1. - Lekraj Beedassy, May 29 2013
Euler, in Section 87 of the Opera Postuma, shows that whenever T is a triangular number then 9*T + 1, 25*T + 3, 49*T + 6 and 81*T + 10 are also triangular numbers. In general, if T is a triangular number then (2*k + 1)^2*T + k*(k + 1)/2 is also a triangular number. - Peter Bala, Jan 05 2015
Using 1/b and 1/(b+2) will give a Pythagorean triangle with sides 2*b + 2, b^2 + 2*b, and b^2 + 2*b + 2. Set b=n-1 to give a triangle with sides of lengths 2*n,n^2-1, and n^2 + 1. One-fourth the perimeter = a(n) for n > 1. - J. M. Bergot, Jul 24 2013
a(n) = A028896(n)/6, where A028896(n) = s(n) - s(n-1) are the first differences of s(n) = n^3 + 3*n^2 + 2*n - 8. s(n) can be interpreted as the sum of the 12 edge lengths plus the sum of the 6 face areas plus the volume of an n X (n-1) X (n-2) rectangular prism. - J. M. Bergot, Aug 13 2013
Dimension of orthogonal group O(n+1). - Eric M. Schmidt, Sep 08 2013
Number of positive roots in the root system of type A_n (for n > 0). - Tom Edgar, Nov 05 2013
A formula for the r-th successive summation of k, for k = 1 to n, is binomial(n+r,r+1) [H. W. Gould]. - Gary Detlefs, Jan 02 2014
Also the alternating row sums of A095831. Also the alternating row sums of A055461, for n >= 1. - Omar E. Pol, Jan 26 2014
For n >= 3, a(n-2) is the number of permutations of 1,2,...,n with the distribution of up (1) - down (0) elements 0...011 (n-3 zeros), or, the same, a(n-2) is up-down coefficient {n,3} (see comment in A060351). - Vladimir Shevelev, Feb 14 2014
a(n) is the dimension of the vector space of symmetric n X n matrices. - Derek Orr, Mar 29 2014
Non-vanishing subdiagonal of A132440^2/2, aside from the initial zero. First subdiagonal of unsigned A238363. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of complete graphs. - Tom Copeland, Apr 05 2014
The number of Sidon subsets of {1,...,n+1} of size 2. - Carl Najafi, Apr 27 2014
Number of factors in the definition of the Vandermonde determinant V(x_1,x_2,...,x_n) = Product_{1 <= i < k <= n} x_i - x_k. - Tom Copeland, Apr 27 2014
Number of weak compositions of n into three parts. - Robert A. Beeler, May 20 2014
Suppose a bag contains a(n) red marbles and a(n+1) blue marbles, where a(n), a(n+1) are consecutive triangular numbers. Then, for n > 0, the probability of choosing two marbles at random and getting two red or two blue is 1/2. In general, for k > 2, let b(0) = 0, b(1) = 1 and, for n > 1, b(n) = (k-1)*b(n-1) - b(n-2) + 1. Suppose, for n > 0, a bag contains b(n) red marbles and b(n+1) blue marbles. Then the probability of choosing two marbles at random and getting two red or two blue is (k-1)/(k+1). See also A027941, A061278, A089817, A053142, A092521. - Charlie Marion, Nov 03 2014
Let O(n) be the oblong number n(n+1) = A002378 and S(n) the square number n^2 = A000290(n). Then a(4n) = O(3n) - O(n), a(4n+1) = S(3n+1) - S(n), a(4n+2) = S(3n+2) - S(n+1) and a(4n+3) = O(3n+2) - O(n). - Charlie Marion, Feb 21 2015
Consider the partition of the natural numbers into parts from the set S=(1,2,3,...,n). The length (order) of the signature of the resulting sequence is given by the triangular numbers. E.g., for n=10, the signature length is 55. - David Neil McGrath, May 05 2015
a(n) counts the partitions of (n-1) unlabeled objects into three (3) parts (labeled a,b,c), e.g., a(5)=15 for (n-1)=4. These are (aaaa),(bbbb),(cccc),(aaab),(aaac),(aabb),(aacc),(aabc),(abbc),(abcc),(abbb),(accc),(bbcc),(bccc),(bbbc). - David Neil McGrath, May 21 2015
Conjecture: the sequence is the genus/deficiency of the sinusoidal spirals of index n which are algebraic curves. The value 0 corresponds to the case of the Bernoulli Lemniscate n=2. So the formula conjectured is (n-1)(n-2)/2. - Wolfgang Tintemann, Aug 02 2015
Conjecture: Let m be any positive integer. Then, for each n = 1,2,3,... the set {Sum_{k=s..t} 1/k^m: 1 <= s <= t <= n} has cardinality a(n) = n*(n+1)/2; in other words, all the sums Sum_{k=s..t} 1/k^m with 1 <= s <= t are pairwise distinct. (I have checked this conjecture via a computer and found no counterexample.) - Zhi-Wei Sun, Sep 09 2015
The Pisano period lengths of reading the sequence modulo m seem to be A022998(m). - R. J. Mathar, Nov 29 2015
For n >= 1, a(n) is the number of compositions of n+4 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
In this sequence only 3 is prime. - Fabian Kopp, Jan 09 2016
Suppose you are playing Bulgarian Solitaire (see A242424 and Chamberland's and Gardner's books) and, for n > 0, you are starting with a single pile of a(n) cards. Then the number of operations needed to reach the fixed state {n, n-1,...,1} is a(n-1). For example, {6}->{5,1}->{4,2}->{3,2,1}. - Charlie Marion, Jan 14 2016
Numbers k such that 8k + 1 is a square. - Juri-Stepan Gerasimov, Apr 09 2016
Every perfect cube is the difference of the squares of two consecutive triangular numbers. 1^2-0^2 = 1^3, 3^2-1^2 = 2^3, 6^2-3^2 = 3^3. - Miquel Cerda, Jun 26 2016
For n > 1, a(n) = tau_n(k*) where tau_n(k) is the number of ordered n-factorizations of k and k* is the square of a prime. For example, tau_3(4) = tau_3(9) = tau_3(25) = tau_3(49) = 6 (see A007425) since the number of divisors of 4, 9, 25, and 49's divisors is 6, and a(3) = 6. - Melvin Peralta, Aug 29 2016
In an (n+1)-dimensional hypercube, number of two-dimensional faces congruent with a vertex (see also A001788). - Stanislav Sykora, Oct 23 2016
Generalizations of the familiar formulas, a(n) + a(n+1) = (n+1)^2 (Feb 19 2004) and a(n)^2 + a(n+1)^2 = a((n+1)^2) (Nov 22 2006), follow: a(n) + a(n+2k-1) + 4a(k-1) = (n+k)^2 + 6a(k-1) and a(n)^2 + a(n+2k-1)^2 + (4a(k-1))^2 + 3a(k-1) = a((n+k)^2 + 6a(k-1)). - Charlie Marion, Nov 27 2016
a(n) is also the greatest possible number of diagonals in a polyhedron with n+4 vertices. - Vladimir Letsko, Dec 19 2016
For n > 0, 2^5 * (binomial(n+1,2))^2 represents the first integer in a sum of 2*(2*n + 1)^2 consecutive integers that equals (2*n + 1)^6. - Patrick J. McNab, Dec 25 2016
Does not satisfy Benford's law (cf. Ross, 2012). - N. J. A. Sloane, Feb 12 2017
Number of ordered triples (a,b,c) of positive integers not larger than n such that a+b+c = 2n+1. - Aviel Livay, Feb 13 2017
Number of inequivalent tetrahedral face colorings using at most n colors so that no color appears only once. - David Nacin, Feb 22 2017
Also the Wiener index of the complete graph K_{n+1}. - Eric W. Weisstein, Sep 07 2017
Number of intersections between the Bernstein polynomials of degree n. - Eric Desbiaux, Apr 01 2018
a(n) is the area of a triangle with vertices at (1,1), (n+1,n+2), and ((n+1)^2, (n+2)^2). - Art Baker, Dec 06 2018
For n > 0, a(n) is the smallest k > 0 such that n divides numerator of (1/a(1) + 1/a(2) + ... + 1/a(n-1) + 1/k). It should be noted that 1/1 + 1/3 + 1/6 + ... + 2/(n(n+1)) = 2n/(n+1). - Thomas Ordowski, Aug 04 2019
Upper bound of the number of lines in an n-homogeneous supersolvable line arrangement (see Theorem 1.1 in Dimca). - Stefano Spezia, Oct 04 2019
For n > 0, a(n+1) is the number of lattice points on a triangular grid with side length n. - Wesley Ivan Hurt, Aug 12 2020
From Michael Chu, May 04 2022: (Start)
Maximum number of distinct nonempty substrings of a string of length n.
Maximum cardinality of the sumset A+A, where A is a set of n numbers. (End)
a(n) is the number of parking functions of size n avoiding the patterns 123, 132, and 312. - Lara Pudwell, Apr 10 2023
Suppose two rows, each consisting of n evenly spaced dots, are drawn in parallel. Suppose we bijectively draw lines between the dots of the two rows. For n >= 1, a(n - 1) is the maximal possible number of intersections between the lines. Equivalently, the maximal number of inversions in a permutation of [n]. - Sela Fried, Apr 18 2023
The following equation complements the generalization in Bala's Comment (Jan 05 2015). (2k + 1)^2*a(n) + a(k) = a((2k + 1)*n + k). - Charlie Marion, Aug 28 2023
a(n) + a(n+k) + a(k-1) + (k-1)*n = (n+k)^2. For k = 1, we have a(n) + a(n+1) = (n+1)^2. - Charlie Marion, Nov 17 2023
a(n+1)/3 is the expected number of steps to escape from a linear row of n positions starting at a random location and randomly performing steps -1 or +1 with equal probability. - Hugo Pfoertner, Jul 22 2025
a(n+1) is the number of nonnegative integer solutions to p + q + r = n. By Sylvester's law of inertia, it is also the number of congruence classes of real symmetric n-by-n matrices or equivalently, the number of symmetric bilinear forms on a real n-dimensional vector space. - Paawan Jethva, Jul 24 2025

Examples

			G.f.: x + 3*x^2 + 6*x^3 + 10*x^4 + 15*x^5 + 21*x^6 + 28*x^7 + 36*x^8 + 45*x^9 + ...
When n=3, a(3) = 4*3/2 = 6.
Example(a(4)=10): ABCD where A, B, C and D are different links in a chain or different amino acids in a peptide possible fragments: A, B, C, D, AB, ABC, ABCD, BC, BCD, CD = 10.
a(2): hollyhock leaves on the Tokugawa Mon, a(4): points in Pythagorean tetractys, a(5): object balls in eight-ball billiards. - _Bradley Klee_, Aug 24 2015
From _Gus Wiseman_, Oct 28 2020: (Start)
The a(1) = 1 through a(5) = 15 ordered triples of positive integers summing to n + 2 [Beeler, McGrath above] are the following. These compositions are ranked by A014311.
  (111)  (112)  (113)  (114)  (115)
         (121)  (122)  (123)  (124)
         (211)  (131)  (132)  (133)
                (212)  (141)  (142)
                (221)  (213)  (151)
                (311)  (222)  (214)
                       (231)  (223)
                       (312)  (232)
                       (321)  (241)
                       (411)  (313)
                              (322)
                              (331)
                              (412)
                              (421)
                              (511)
The unordered version is A001399(n-3) = A069905(n), with Heinz numbers A014612.
The strict case is A001399(n-6)*6, ranked by A337453.
The unordered strict case is A001399(n-6), with Heinz numbers A007304.
(End)

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See Chapter 1.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 109ff.
  • Marc Chamberland, Single Digits: In Praise of Small Numbers, Chapter 3, The Number Three, p. 72, Princeton University Press, 2015.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 33, 38, 40, 70.
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 309 pp 46-196, Ellipses, Paris, 2004
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.
  • Martin Gardner, Colossal Book of Mathematics, Chapter 34, Bulgarian Solitaire and Other Seemingly Endless Tasks, pp. 455-467, W. W. Norton & Company, 2001.
  • James Gleick, The Information: A History, A Theory, A Flood, Pantheon, 2011. [On page 82 mentions a table of the first 19999 triangular numbers published by E. de Joncort in 1762.]
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.6 Mathematical Proof and §8.6 Figurate Numbers, pp. 158-159, 289-290.
  • Cay S. Horstmann, Scala for the Impatient. Upper Saddle River, New Jersey: Addison-Wesley (2012): 171.
  • Elemer Labos, On the number of RGB-colors we can distinguish. Partition Spectra. Lecture at 7th Hungarian Conference on Biometry and Biomathematics. Budapest. Jul 06 2005.
  • A. Messiah, Quantum Mechanics, Vol.1, North Holland, Amsterdam, 1965, p. 457.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 52-53, 129-132, 274.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 2-6, 13.
  • T. Trotter, Some Identities for the Triangular Numbers, Journal of Recreational Mathematics, Spring 1973, 6(2).
  • D. Wells, The Penguin Dictionary of Curious and Interesting Numbers, pp. 91-93 Penguin Books 1987.

Links

Crossrefs

The figurate numbers, with parameter k as in the second Python program: A001477 (k=0), this sequence (k=1), A000290 (k=2), A000326 (k=3), A000384 (k=4), A000566 (k=5), A000567 (k=6), A001106 (k=7), A001107 (k=8).
Cf. A000096, A000124, A000292, A000330, A000396, A000668, A001082, A001788, A002024, A002378, A002415, A003056 (inverse function), A004526, A006011, A007318, A008953, A008954, A010054 (characteristic function), A028347, A036666, A046092, A051942, A055998, A055999, A056000, A056115, A056119, A056121, A056126, A062717, A087475, A101859, A109613, A143320, A210569, A245031, A245300, A060544, A016754.
a(n) = A110449(n, 0).
a(n) = A110555(n+2, 2).
A diagonal of A008291.
Column 2 of A195152.
Numbers of the form n*t(n+k,h)-(n+k)*t(n,h), where t(i,h) = i*(i+2*h+1)/2 for any h (for A000217 is k=1): A005563, A067728, A140091, A140681, A212331.
Boustrophedon transforms: A000718, A000746.
Iterations: A007501 (start=2), A013589 (start=4), A050542 (start=5), A050548 (start=7), A050536 (start=8), A050909 (start=9).
Cf. A002817 (doubly triangular numbers), A075528 (solutions of a(n)=a(m)/2).
Cf. A104712 (first column, starting with a(1)).
Some generalized k-gonal numbers are A001318 (k=5), this sequence (k=6), A085787 (k=7), etc.
A001399(n-3) = A069905(n) = A211540(n+2) counts 3-part partitions.
A001399(n-6) = A069905(n-3) = A211540(n-1) counts 3-part strict partitions.
A011782 counts compositions of any length.
A337461 counts pairwise coprime triples, with unordered version A307719.
Cf. A000212, A001840, A007304, A156040, A220377, A337483, A337604.
Cf. A099174, A130534, A132440, A238363.

Programs

  • Haskell
    a000217 n = a000217_list !! n
a000217_list = scanl1 (+) [0..] -- Reinhard Zumkeller, Sep 23 2011
  • J
    a000217=: *-:@>: NB. Stephen Makdisi, May 02 2018
  • Magma
    [n*(n+1)/2: n in [0..60]]; // Bruno Berselli, Jul 11 2014
  • Magma
    [n: n in [0..1500] | IsSquare(8*n+1)]; // Juri-Stepan Gerasimov, Apr 09 2016
  • Maple
    A000217 := proc(n) n*(n+1)/2; end;
istriangular:=proc(n) local t1; t1:=floor(sqrt(2*n)); if n = t1*(t1+1)/2 then return true else return false; end if; end proc; # N. J. A. Sloane, May 25 2008
ZL := [S, {S=Prod(B, B, B), B=Set(Z, 1 <= card)}, unlabeled]:
seq(combstruct[count](ZL, size=n), n=2..55); # Zerinvary Lajos, Mar 24 2007
isA000217 := proc(n)
    issqr(1+8*n) ;
end proc: # R. J. Mathar, Nov 29 2015 [This is the recipe Leonhard Euler proposes in chapter VII of his "Vollständige Anleitung zur Algebra", 1765. Peter Luschny, Sep 02 2022]
  • Mathematica
    Array[ #*(# - 1)/2 &, 54] (* Zerinvary Lajos, Jul 10 2009 *)
FoldList[#1 + #2 &, 0, Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
Accumulate[Range[0,70]] (* Harvey P. Dale, Sep 09 2012 *)
CoefficientList[Series[x / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 30 2014 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[n], {n, 0, 53}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
LinearRecurrence[{3, -3, 1}, {0, 1, 3}, 54] (* Robert G. Wilson v, Dec 04 2016 *)
(* The following Mathematica program, courtesy of Steven J. Miller, is useful for testing if a sequence is Benford. To test a different sequence only one line needs to be changed. This strongly suggests that the triangular numbers are not Benford, since the second and third columns of the output disagree. - N. J. A. Sloane, Feb 12 2017 *)
fd[x_] := Floor[10^Mod[Log[10, x], 1]]
benfordtest[num_] := Module[{},
   For[d = 1, d <= 9, d++, digit[d] = 0];
   For[n = 1, n <= num, n++,
    {
     d = fd[n(n+1)/2];
     If[d != 0, digit[d] = digit[d] + 1];
     }];
   For[d = 1, d <= 9, d++, digit[d] = 1.0 digit[d]/num];
   For[d = 1, d <= 9, d++,
    Print[d, " ", 100.0 digit[d], " ", 100.0 Log[10, (d + 1)/d]]];
   ];
benfordtest[20000]
Table[Length[Join@@Permutations/@IntegerPartitions[n,{3}]],{n,0,15}] (* Gus Wiseman, Oct 28 2020 *)
  • PARI
    A000217(n) = n * (n + 1) / 2;
  • PARI
    is_A000217(n)=n*2==(1+n=sqrtint(2*n))*n \\ M. F. Hasler, May 24 2012
  • PARI
    is(n)=ispolygonal(n,3) \\ Charles R Greathouse IV, Feb 28 2014
  • PARI
    list(lim)=my(v=List(),n,t); while((t=n*n++/2)<=lim,listput(v,t)); Vec(v) \\ Charles R Greathouse IV, Jun 18 2021
  • Python
    for n in range(0,60): print(n*(n+1)//2, end=', ') # Stefano Spezia, Dec 06 2018
  • Python
    # Intended to compute the initial segment of the sequence, not
# isolated terms. If in the iteration the line "x, y = x + y + 1, y + 1"
# is replaced by "x, y = x + y + k, y + k" then the figurate numbers are obtained,
# for k = 0 (natural A001477), k = 1 (triangular), k = 2 (squares), k = 3 (pentagonal), k = 4 (hexagonal), k = 5 (heptagonal), k = 6 (octagonal), etc.
def aList():
    x, y = 1, 1
    yield 0
    while True:
        yield x
        x, y = x + y + 1, y + 1
A000217 = aList()
print([next(A000217) for i in range(54)]) # Peter Luschny, Aug 03 2019
  • SageMath
    [n*(n+1)/2 for n in (0..60)] # Bruno Berselli, Jul 11 2014
  • Scala
    (1 to 53).scanLeft(0)( + ) // Horstmann (2012), p. 171
  • Scheme
    (define (A000217 n) (/ (* n (+ n 1)) 2)) ;; Antti Karttunen, Jul 08 2017

Formula

G.f.: x/(1-x)^3. - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(x)*(x+x^2/2).
a(n) = a(-1-n).
a(n) + a(n-1)*a(n+1) = a(n)^2. - Terrel Trotter, Jr., Apr 08 2002
a(n) = (-1)^n*Sum_{k=1..n} (-1)^k*k^2. - Benoit Cloitre, Aug 29 2002
a(n+1) = ((n+2)/n)*a(n), Sum_{n>=1} 1/a(n) = 2. - Jon Perry, Jul 13 2003
For n > 0, a(n) = A001109(n) - Sum_{k=0..n-1} (2*k+1)*A001652(n-1-k); e.g., 10 = 204 - (1*119 + 3*20 + 5*3 + 7*0). - Charlie Marion, Jul 18 2003
With interpolated zeros, this is n*(n+2)*(1+(-1)^n)/16. - Benoit Cloitre, Aug 19 2003
a(n+1) is the determinant of the n X n symmetric Pascal matrix M_(i, j) = binomial(i+j+1, i). - Benoit Cloitre, Aug 19 2003
a(n) = ((n+1)^3 - n^3 - 1)/6. - Xavier Acloque, Oct 24 2003
a(n) = a(n-1) + (1 + sqrt(1 + 8*a(n-1)))/2. This recursive relation is inverted when taking the negative branch of the square root, i.e., a(n) is transformed into a(n-1) rather than a(n+1). - Carl R. White, Nov 04 2003
a(n) = Sum_{k=1..n} phi(k)*floor(n/k) = Sum_{k=1..n} A000010(k)*A010766(n, k) (R. Dedekind). - Vladeta Jovovic, Feb 05 2004
a(n) + a(n+1) = (n+1)^2. - N. J. A. Sloane, Feb 19 2004
a(n) = a(n-2) + 2*n - 1. - Paul Barry, Jul 17 2004
a(n) = sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)) = sqrt(A000537(n)). - Alexander Adamchuk, Oct 24 2004
a(n) = sqrt(sqrt(Sum_{i=1..n} Sum_{j=1..n} (i*j)^3)) = (Sum_{i=1..n} Sum_{j=1..n} Sum_{k=1..n} (i*j*k)^3)^(1/6). - Alexander Adamchuk, Oct 26 2004
a(n) == 1 (mod n+2) if n is odd and a(n) == n/2+2 (mod n+2) if n is even. - Jon Perry, Dec 16 2004
a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 1. - Miklos Kristof, Mar 09 2005
a(n) = a(n-1) + n. - Zak Seidov, Mar 06 2005
a(n) = A108299(n+3,4) = -A108299(n+4,5). - Reinhard Zumkeller, Jun 01 2005
a(n) = A111808(n,2) for n > 1. - Reinhard Zumkeller, Aug 17 2005
a(n)*a(n+1) = A006011(n+1) = (n+1)^2*(n^2+2)/4 = 3*A002415(n+1) = 1/2*a(n^2+2*n). a(n-1)*a(n) = (1/2)*a(n^2-1). - Alexander Adamchuk, Apr 13 2006 [Corrected and edited by Charlie Marion, Nov 26 2010]
a(n) = floor((2*n+1)^2/8). - Paul Barry, May 29 2006
For positive n, we have a(8*a(n))/a(n) = 4*(2*n+1)^2 = (4*n+2)^2, i.e., a(A033996(n))/a(n) = 4*A016754(n) = (A016825(n))^2 = A016826(n). - Lekraj Beedassy, Jul 29 2006
a(n)^2 + a(n+1)^2 = a((n+1)^2) [R B Nelsen, Math Mag 70 (2) (1997), p. 130]. - R. J. Mathar, Nov 22 2006
a(n) = A126890(n,0). - Reinhard Zumkeller, Dec 30 2006
a(n)*a(n+k)+a(n+1)*a(n+1+k) = a((n+1)*(n+1+k)). Generalizes previous formula dated Nov 22 2006 [and comments by J. M. Bergot dated May 22 2012]. - Charlie Marion, Feb 04 2011
(sqrt(8*a(n)+1)-1)/2 = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
a(n) = A023896(n) + A067392(n). - Lekraj Beedassy, Mar 02 2007
Sum_{k=0..n} a(k)*A039599(n,k) = A002457(n-1), for n >= 1. - Philippe Deléham, Jun 10 2007
8*a(n)^3 + a(n)^2 = Y(n)^2, where Y(n) = n*(n+1)*(2*n+1)/2 = 3*A000330(n). - Mohamed Bouhamida, Nov 06 2007 [Edited by Derek Orr, May 05 2015]
A general formula for polygonal numbers is P(k,n) = (k-2)*(n-1)n/2 + n = n + (k-2)*A000217(n-1), for n >= 1, k >= 3. - Omar E. Pol, Apr 28 2008 and Mar 31 2013
a(3*n) = A081266(n), a(4*n) = A033585(n), a(5*n) = A144312(n), a(6*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = A022264(n) - A049450(n). - Reinhard Zumkeller, Oct 09 2008
If we define f(n,i,a) = Sum_{j=0..k-1} (binomial(n,k)*Stirling1(n-k,i)*Product_{j=0..k-1} (-a-j)), then a(n) = -f(n,n-1,1), for n >= 1. - Milan Janjic, Dec 20 2008
4*a(x) + 4*a(y) + 1 = (x+y+1)^2 + (x-y)^2. - Vladimir Shevelev, Jan 21 2009
a(n) = A000124(n-1) + n-1 for n >= 2. a(n) = A000124(n) - 1. - Jaroslav Krizek, Jun 16 2009
An exponential generating function for the inverse of this sequence is given by Sum_{m>=0} ((Pochhammer(1, m)*Pochhammer(1, m))*x^m/(Pochhammer(3, m)*factorial(m))) = ((2-2*x)*log(1-x)+2*x)/x^2, the n-th derivative of which has a closed form which must be evaluated by taking the limit as x->0. A000217(n+1) = (lim_{x->0} d^n/dx^n (((2-2*x)*log(1-x)+2*x)/x^2))^-1 = (lim_{x->0} (2*Gamma(n)*(-1/x)^n*(n*(x/(-1+x))^n*(-x+1+n)*LerchPhi(x/(-1+x), 1, n) + (-1+x)*(n+1)*(x/(-1+x))^n + n*(log(1-x)+log(-1/(-1+x)))*(-x+1+n))/x^2))^-1. - Stephen Crowley, Jun 28 2009
a(n) = A034856(n+1) - A005408(n) = A005843(n) + A000124(n) - A005408(n). - Jaroslav Krizek, Sep 05 2009
a(A006894(n)) = a(A072638(n-1)+1) = A072638(n) = A006894(n+1)-1 for n >= 1. For n=4, a(11) = 66. - Jaroslav Krizek, Sep 12 2009
With offset 1, a(n) = floor(n^3/(n+1))/2. - Gary Detlefs, Feb 14 2010
a(n) = 4*a(floor(n/2)) + (-1)^(n+1)*floor((n+1)/2). - Bruno Berselli, May 23 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=1. - Mark Dols, Aug 20 2010
From Charlie Marion, Oct 15 2010: (Start)
a(n) + 2*a(n-1) + a(n-2) = n^2 + (n-1)^2; and
a(n) + 3*a(n-1) + 3*a(n-2) + a(n-3) = n^2 + 2*(n-1)^2 + (n-2)^2.
In general, for n >= m > 2, Sum_{k=0..m} binomial(m,m-k)*a(n-k) = Sum_{k=0..m-1} binomial(m-1,m-1-k)*(n-k)^2.
a(n) - 2*a(n-1) + a(n-2) = 1, a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0 and a(n) - 4*a(n-1) + 6*a(n-2) - 4*(a-3) + a(n-4) = 0.
In general, for n >= m > 2, Sum_{k=0..m} (-1)^k*binomial(m,m-k)*a(n-k) = 0.
(End)
a(n) = sqrt(A000537(n)). - Zak Seidov, Dec 07 2010
For n > 0, a(n) = 1/(Integral_{x=0..Pi/2} 4*(sin(x))^(2*n-1)*(cos(x))^3). - Francesco Daddi, Aug 02 2011
a(n) = A110654(n)*A008619(n). - Reinhard Zumkeller, Aug 24 2011
a(2*k-1) = A000384(k), a(2*k) = A014105(k), k > 0. - Omar E. Pol, Sep 13 2011
a(n) = A026741(n)*A026741(n+1). - Charles R Greathouse IV, Apr 01 2012
a(n) + a(a(n)) + 1 = a(a(n)+1). - J. M. Bergot, Apr 27 2012
a(n) = -s(n+1,n), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n)*a(n+1) = a(Sum_{m=1..n} A005408(m))/2, for n >= 1. For example, if n=8, then a(8)*a(9) = a(80)/2 = 1620. - Ivan N. Ianakiev, May 27 2012
a(n) = A002378(n)/2 = (A001318(n) + A085787(n))/2. - Omar E. Pol, Jan 11 2013
G.f.: x * (1 + 3x + 6x^2 + ...) = x * Product_{j>=0} (1+x^(2^j))^3 = x * A(x) * A(x^2) * A(x^4) * ..., where A(x) = (1 + 3x + 3x^2 + x^3). - Gary W. Adamson, Jun 26 2012
G.f.: G(0) where G(k) = 1 + (2*k+3)*x/(2*k+1 - x*(k+2)*(2*k+1)/(x*(k+2) + (k+1)/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 23 2012
a(n) = A002088(n) + A063985(n). - Reinhard Zumkeller, Jan 21 2013
G.f.: x + 3*x^2/(Q(0)-3*x) where Q(k) = 1 + k*(x+1) + 3*x - x*(k+1)*(k+4)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Mar 14 2013
a(n) + a(n+1) + a(n+2) + a(n+3) + n = a(2*n+4). - Ivan N. Ianakiev, Mar 16 2013
a(n) + a(n+1) + ... + a(n+8) + 6*n = a(3*n+15). - Charlie Marion, Mar 18 2013
a(n) + a(n+1) + ... + a(n+20) + 2*n^2 + 57*n = a(5*n+55). - Charlie Marion, Mar 18 2013
3*a(n) + a(n-1) = a(2*n), for n > 0. - Ivan N. Ianakiev, Apr 05 2013
In general, a(k*n) = (2*k-1)*a(n) + a((k-1)*n-1). - Charlie Marion, Apr 20 2015
Also, a(k*n) = a(k)*a(n) + a(k-1)*a(n-1). - Robert Israel, Apr 20 2015
a(n+1) = det(binomial(i+2,j+1), 1 <= i,j <= n). - Mircea Merca, Apr 06 2013
a(n) = floor(n/2) + ceiling(n^2/2) = n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = floor((n+1)/(exp(2/(n+1))-1)). - Richard R. Forberg, Jun 22 2013
Sum_{n>=1} a(n)/n! = 3*exp(1)/2 by the e.g.f. Also see A067764 regarding ratios calculated this way for binomial coefficients in general. - Richard R. Forberg, Jul 15 2013
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2) - 2 = 0.7725887... . - Richard R. Forberg, Aug 11 2014
2/(Sum_{n>=m} 1/a(n)) = m, for m > 0. - Richard R. Forberg, Aug 12 2014
A228474(a(n))=n; A248952(a(n))=0; A248953(a(n))=a(n); A248961(a(n))=A000330(n). - Reinhard Zumkeller, Oct 20 2014
a(a(n)-1) + a(a(n+2)-1) + 1 = A000124(n+1)^2. - Charlie Marion, Nov 04 2014
a(n) = 2*A000292(n) - A000330(n). - Luciano Ancora, Mar 14 2015
a(n) = A007494(n-1) + A099392(n) for n > 0. - Bui Quang Tuan, Mar 27 2015
Sum_{k=0..n} k*a(k+1) = a(A000096(n+1)). - Charlie Marion, Jul 15 2015
Let O(n) be the oblong number n(n+1) = A002378(n) and S(n) the square number n^2 = A000290(n). Then a(n) + a(n+2k) = O(n+k) + S(k) and a(n) + a(n+2k+1) = S(n+k+1) + O(k). - Charlie Marion, Jul 16 2015
A generalization of the Nov 22 2006 formula, a(n)^2 + a(n+1)^2 = a((n+1)^2), follows. Let T(k,n) = a(n) + k. Then for all k, T(k,n)^2 + T(k,n+1)^2 = T(k,(n+1)^2 + 2*k) - 2*k. - Charlie Marion, Dec 10 2015
a(n)^2 + a(n+1)^2 = a(a(n) + a(n+1)). Deducible from N. J. A. Sloane's a(n) + a(n+1) = (n+1)^2 and R. B. Nelson's a(n)^2 + a(n+1)^2 = a((n+1)^2). - Ben Paul Thurston, Dec 28 2015
Dirichlet g.f.: (zeta(s-2) + zeta(s-1))/2. - Ilya Gutkovskiy, Jun 26 2016
a(n)^2 - a(n-1)^2 = n^3. - Miquel Cerda, Jun 29 2016
a(n) = A080851(0,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A000290(n-1) - A034856(n-4). - Peter M. Chema, Sep 25 2016
a(n)^2 + a(n+3)^2 + 19 = a(n^2 + 4*n + 10). - Charlie Marion, Nov 23 2016
2*a(n)^2 + a(n) = a(n^2+n). - Charlie Marion, Nov 29 2016
G.f.: x/(1-x)^3 = (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...), where r(x) = (1 + x + x^2)^3 = (1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6). - Gary W. Adamson, Dec 03 2016
a(n) = sum of the elements of inverse of matrix Q(n), where Q(n) has elements q_i,j = 1/(1-4*(i-j)^2). So if e = appropriately sized vector consisting of 1's, then a(n) = e'.Q(n)^-1.e. - Michael Yukish, Mar 20 2017
a(n) = Sum_{k=1..n} ((2*k-1)!!*(2*n-2*k-1)!!)/((2*k-2)!!*(2*n-2*k)!!). - Michael Yukish, Mar 20 2017
Sum_{i=0..k-1} a(n+i) = (3*k*n^2 + 3*n*k^2 + k^3 - k)/6. - Christopher Hohl, Feb 23 2019
a(n) = A060544(n + 1) - A016754(n). - Ralf Steiner, Nov 09 2019
a(n) == 0 (mod n) iff n is odd (see De Koninck reference). - Bernard Schott, Jan 10 2020
8*a(k)*a(n) + ((a(k)-1)*n + a(k))^2 = ((a(k)+1)*n + a(k))^2. This formula reduces to the well-known formula, 8*a(n) + 1 = (2*n+1)^2, when k = 1. - Charlie Marion, Jul 23 2020
a(k)*a(n) = Sum_{i = 0..k-1} (-1)^i*a((k-i)*(n-i)). - Charlie Marion, Dec 04 2020
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)/(2*Pi).
Product_{n>=2} (1 - 1/a(n)) = 1/3. (End)
a(n) = Sum_{k=1..2*n-1} (-1)^(k+1)*a(k)*a(2*n-k). For example, for n = 4, 1*28 - 3*21 + 6*15 - 10*10 + 15*6 - 21*3 + 28*1 = 10. - Charlie Marion, Mar 23 2022
2*a(n) = A000384(n) - n^2 + 2*n. In general, if P(k,n) = the n-th k-gonal number, then (j+1)*a(n) = P(5 + j, n) - n^2 + (j+1)*n. More generally, (j+1)*P(k,n) = P(2*k + (k-2)*(j-1),n) - n^2 + (j+1)*n. - Charlie Marion, Mar 14 2023
a(n) = A109613(n) * A004526(n+1). - Torlach Rush, Nov 10 2023
a(n) = (1/6)* Sum_{k = 0..3*n} (-1)^(n+k+1) * k*(k + 1) * binomial(3*n+k, 2*k). - Peter Bala, Nov 03 2024
From Peter Bala, Jul 05 2025: (Start)
The following series telescope: for k >= 0,
Sum_{n >= 1} a(n)*a(n+2)*...*a(n+2*k)/(a(n+1)*a(n+3)*...*a(n+2*k+3)) = 1/(2*k + 3);
Sum_{n >= 1} a(n+1)*a(n+3)*...*a(n+2*k+1)/(a(n)*a(n+2)*...*a(n+2*k+2)) = 2/(2*k + 3) * Sum_{i = 1..2*k+3} 1/i. (End)

Extensions

Edited by Derek Orr, May 05 2015

A000108 Catalan numbers: C(n) = binomial(2n,n)/(n+1) = (2n)!/(n!(n+1)!).

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, 18367353072152, 69533550916004, 263747951750360, 1002242216651368, 3814986502092304
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

These were formerly sometimes called Segner numbers.
A very large number of combinatorial interpretations are known - see references, esp. R. P. Stanley, "Catalan Numbers", Cambridge University Press, 2015. This is probably the longest entry in the OEIS, and rightly so.
The solution to Schröder's first problem: number of ways to insert n pairs of parentheses in a word of n+1 letters. E.g., for n=2 there are 2 ways: ((ab)c) or (a(bc)); for n=3 there are 5 ways: ((ab)(cd)), (((ab)c)d), ((a(bc))d), (a((bc)d)), (a(b(cd))).
Consider all the binomial(2n,n) paths on squared paper that (i) start at (0, 0), (ii) end at (2n, 0) and (iii) at each step, either make a (+1,+1) step or a (+1,-1) step. Then the number of such paths that never go below the x-axis (Dyck paths) is C(n). [Chung-Feller]
Number of noncrossing partitions of the n-set. For example, of the 15 set partitions of the 4-set, only [{13},{24}] is crossing, so there are a(4)=14 noncrossing partitions of 4 elements. - Joerg Arndt, Jul 11 2011
Noncrossing partitions are partitions of genus 0. - Robert Coquereaux, Feb 13 2024
a(n-1) is the number of ways of expressing an n-cycle (123...n) in the symmetric group S_n as a product of n-1 transpositions (u_1,v_1)*(u_2,v_2)*...*(u_{n-1},v_{n-1}) where u_iA000272. - Joerg Arndt and Greg Stevenson, Jul 11 2011
a(n) is the number of ordered rooted trees with n nodes, not including the root. See the Conway-Guy reference where these rooted ordered trees are called plane bushes. See also the Bergeron et al. reference, Example 4, p. 167. - Wolfdieter Lang, Aug 07 2007
As shown in the paper from Beineke and Pippert (1971), a(n-2)=D(n) is the number of labeled dissections of a disk, related to the number R(n)=A001761(n-2) of labeled planar 2-trees having n vertices and rooted at a given exterior edge, by the formula D(n)=R(n)/(n-2)!. - M. F. Hasler, Feb 22 2012
Shifts one place left when convolved with itself.
For n >= 1, a(n) is also the number of rooted bicolored unicellular maps of genus 0 on n edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 15 2001
Number of ways of joining 2n points on a circle to form n nonintersecting chords. (If no such restriction imposed, then the number of ways of forming n chords is given by (2n-1)!! = (2n)!/(n!*2^n) = A001147(n).)
Arises in Schubert calculus - see Sottile reference.
Inverse Euler transform of sequence is A022553.
With interpolated zeros, the inverse binomial transform of the Motzkin numbers A001006. - Paul Barry, Jul 18 2003
The Hankel transforms of this sequence or of this sequence with the first term omitted give A000012 = 1, 1, 1, 1, 1, 1, ...; example: Det([1, 1, 2, 5; 1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132]) = 1 and Det([1, 2, 5, 14; 2, 5, 14, 42; 5, 14, 42, 132; 14, 42, 132, 429]) = 1. - Philippe Deléham, Mar 04 2004
a(n) equals the sum of squares of terms in row n of triangle A053121, which is formed from successive self-convolutions of the Catalan sequence. - Paul D. Hanna, Apr 23 2005
Also coefficients of the Mandelbrot polynomial M iterated an infinite number of times. Examples: M(0) = 0 = 0*c^0 = [0], M(1) = c = c^1 + 0*c^0 = [1 0], M(2) = c^2 + c = c^2 + c^1 + 0*c^0 = [1 1 0], M(3) = (c^2 + c)^2 + c = [0 1 1 2 1], ... ... M(5) = [0 1 1 2 5 14 26 44 69 94 114 116 94 60 28 8 1], ... - Donald D. Cross (cosinekitty(AT)hotmail.com), Feb 04 2005
The multiplicity with which a prime p divides C_n can be determined by first expressing n+1 in base p. For p=2, the multiplicity is the number of 1 digits minus 1. For p an odd prime, count all digits greater than (p+1)/2; also count digits equal to (p+1)/2 unless final; and count digits equal to (p-1)/2 if not final and the next digit is counted. For example, n=62, n+1 = 223_5, so C_62 is not divisible by 5. n=63, n+1 = 224_5, so 5^3 | C_63. - Franklin T. Adams-Watters, Feb 08 2006
Koshy and Salmassi give an elementary proof that the only prime Catalan numbers are a(2) = 2 and a(3) = 5. Is the only semiprime Catalan number a(4) = 14? - Jonathan Vos Post, Mar 06 2006
The answer is yes. Using the formula C_n = binomial(2n,n)/(n+1), it is immediately clear that C_n can have no prime factor greater than 2n. For n >= 7, C_n > (2n)^2, so it cannot be a semiprime. Given that the Catalan numbers grow exponentially, the above consideration implies that the number of prime divisors of C_n, counted with multiplicity, must grow without limit. The number of distinct prime divisors must also grow without limit, but this is more difficult. Any prime between n+1 and 2n (exclusive) must divide C_n. That the number of such primes grows without limit follows from the prime number theorem. - Franklin T. Adams-Watters, Apr 14 2006
The number of ways to place n indistinguishable balls in n numbered boxes B1,...,Bn such that at most a total of k balls are placed in boxes B1,...,Bk for k=1,...,n. For example, a(3)=5 since there are 5 ways to distribute 3 balls among 3 boxes such that (i) box 1 gets at most 1 ball and (ii) box 1 and box 2 together get at most 2 balls:(O)(O)(O), (O)()(OO), ()(OO)(O), ()(O)(OO), ()()(OOO). - Dennis P. Walsh, Dec 04 2006
a(n) is also the order of the semigroup of order-decreasing and order-preserving full transformations (of an n-element chain) - now known as the Catalan monoid. - Abdullahi Umar, Aug 25 2008
a(n) is the number of trivial representations in the direct product of 2n spinor (the smallest) representations of the group SU(2) (A(1)). - Rutger Boels (boels(AT)nbi.dk), Aug 26 2008
The invert transform appears to converge to the Catalan numbers when applied infinitely many times to any starting sequence. - Mats Granvik, Gary W. Adamson and Roger L. Bagula, Sep 09 2008, Sep 12 2008
Limit_{n->oo} a(n)/a(n-1) = 4. - Francesco Antoni (francesco_antoni(AT)yahoo.com), Nov 24 2008
Starting with offset 1 = row sums of triangle A154559. - Gary W. Adamson, Jan 11 2009
C(n) is the degree of the Grassmannian G(1,n+1): the set of lines in (n+1)-dimensional projective space, or the set of planes through the origin in (n+2)-dimensional affine space. The Grassmannian is considered a subset of N-dimensional projective space, N = binomial(n+2,2) - 1. If we choose 2n general (n-1)-planes in projective (n+1)-space, then there are C(n) lines that meet all of them. - Benji Fisher (benji(AT)FisherFam.org), Mar 05 2009
Starting with offset 1 = A068875: (1, 2, 4, 10, 18, 84, ...) convolved with Fine numbers, A000957: (1, 0, 1, 2, 6, 18, ...). a(6) = 132 = (1, 2, 4, 10, 28, 84) dot (18, 6, 2, 1, 0, 1) = (18 + 12 + 8 + 10 + 0 + 84) = 132. - Gary W. Adamson, May 01 2009
Convolved with A032443: (1, 3, 11, 42, 163, ...) = powers of 4, A000302: (1, 4, 16, ...). - Gary W. Adamson, May 15 2009
Sum_{k>=1} C(k-1)/2^(2k-1) = 1. The k-th term in the summation is the probability that a random walk on the integers (beginning at the origin) will arrive at positive one (for the first time) in exactly (2k-1) steps. - Geoffrey Critzer, Sep 12 2009
C(p+q)-C(p)*C(q) = Sum_{i=0..p-1, j=0..q-1} C(i)*C(j)*C(p+q-i-j-1). - Groux Roland, Nov 13 2009
Leonhard Euler used the formula C(n) = Product_{i=3..n} (4*i-10)/(i-1) in his 'Betrachtungen, auf wie vielerley Arten ein gegebenes polygonum durch Diagonallinien in triangula zerschnitten werden könne' and computes by recursion C(n+2) for n = 1..8. (Berlin, 4th September 1751, in a letter to Goldbach.) - Peter Luschny, Mar 13 2010
Let A179277 = A(x). Then C(x) is satisfied by A(x)/A(x^2). - Gary W. Adamson, Jul 07 2010
a(n) is also the number of quivers in the mutation class of type B_n or of type C_n. - Christian Stump, Nov 02 2010
From Matthew Vandermast, Nov 22 2010: (Start)
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color while satisfying the following conditions: 1. No two colors are chosen the same positive number of times. 2. For any two colors (c, d) that are chosen at least once, color c is chosen more times than color d iff color c appears more times in the original set than color d.
If the second requirement is lifted, the number of acceptable ways equals A000110(n+1). See related comments for A016098, A085082. (End)
Deutsch and Sagan prove the Catalan number C_n is odd if and only if n = 2^a - 1 for some nonnegative integer a. Lin proves for every odd Catalan number C_n, we have C_n == 1 (mod 4). - Jonathan Vos Post, Dec 09 2010
a(n) is the number of functions f:{1,2,...,n}->{1,2,...,n} such that f(1)=1 and for all n >= 1 f(n+1) <= f(n)+1. For a nice bijection between this set of functions and the set of length 2n Dyck words, see page 333 of the Fxtbook (see link below). - Geoffrey Critzer, Dec 16 2010
Postnikov (2005) defines "generalized Catalan numbers" associated with buildings (e.g., Catalan numbers of Type B, see A000984). - N. J. A. Sloane, Dec 10 2011
Number of permutations in S(n) for which length equals depth. - Bridget Tenner, Feb 22 2012
a(n) is also the number of standard Young tableau of shape (n,n). - Thotsaporn Thanatipanonda, Feb 25 2012
a(n) is the number of binary sequences of length 2n+1 in which the number of ones first exceed the number of zeros at entry 2n+1. See the example below in the example section. - Dennis P. Walsh, Apr 11 2012
Number of binary necklaces of length 2*n+1 containing n 1's (or, by symmetry, 0's). All these are Lyndon words and their representatives (as cyclic maxima) are the binary Dyck words. - Joerg Arndt, Nov 12 2012
Number of sequences consisting of n 'x' letters and n 'y' letters such that (counting from the left) the 'x' count >= 'y' count. For example, for n=3 we have xxxyyy, xxyxyy, xxyyxy, xyxxyy and xyxyxy. - Jon Perry, Nov 16 2012
a(n) is the number of Motzkin paths of length n-1 in which the (1,0)-steps come in 2 colors. Example: a(4)=14 because, denoting U=(1,1), H=(1,0), and D=(1,-1), we have 8 paths of shape HHH, 2 paths of shape UHD, 2 paths of shape UDH, and 2 paths of shape HUD. - José Luis Ramírez Ramírez, Jan 16 2013
If p is an odd prime, then (-1)^((p-1)/2)*a((p-1)/2) mod p = 2. - Gary Detlefs, Feb 20 2013
Conjecture: For any positive integer n, the polynomial Sum_{k=0..n} a(k)*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
a(n) is the size of the Jones monoid on 2n points (cf. A225798). - James Mitchell, Jul 28 2013
For 0 < p < 1, define f(p) = Sum_{n>=0} a(n)*(p*(1-p))^n, then f(p) = min{1/p, 1/(1-p)}, so f(p) reaches its maximum value 2 at p = 0.5, and p*f(p) is constant 1 for 0.5 <= p < 1. - Bob Selcoe, Nov 16 2013 [Corrected by Jianing Song, May 21 2021]
No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
From Alexander Adamchuk, Dec 27 2013: (Start)
Prime p divides a((p+1)/2) for p > 3. See A120303(n) = Largest prime factor of Catalan number.
Reciprocal Catalan Constant C = 1 + 4*sqrt(3)*Pi/27 = 1.80613.. = A121839.
Log(Phi) = (125*C - 55) / (24*sqrt(5)), where C = Sum_{k>=1} (-1)^(k+1)*1/a(k). See A002390 = Decimal expansion of natural logarithm of golden ratio.
3-d analog of the Catalan numbers: (3n)!/(n!(n+1)!(n+2)!) = A161581(n) = A006480(n) / ((n+1)^2*(n+2)), where A006480(n) = (3n)!/(n!)^3 De Bruijn's S(3,n). (End)
For a relation to the inviscid Burgers's, or Hopf, equation, see A001764. - Tom Copeland, Feb 15 2014
From Fung Lam, May 01 2014: (Start)
One class of generalized Catalan numbers can be defined by g.f. A(x) = (1-sqrt(1-q*4*x*(1-(q-1)*x)))/(2*q*x) with nonzero parameter q. Recurrence: (n+3)*a(n+2) -2*q*(2*n+3)*a(n+1) +4*q*(q-1)*n*a(n) = 0 with a(0)=1, a(1)=1.
Asymptotic approximation for q >= 1: a(n) ~ (2*q+2*sqrt(q))^n*sqrt(2*q*(1+sqrt(q))) /sqrt(4*q^2*Pi*n^3).
For q <= -1, the g.f. defines signed sequences with asymptotic approximation: a(n) ~ Re(sqrt(2*q*(1+sqrt(q)))*(2*q+2*sqrt(q))^n) / sqrt(q^2*Pi*n^3), where Re denotes the real part. Due to Stokes' phenomena, accuracy of the asymptotic approximation deteriorates at/near certain values of n.
Special cases are A000108 (q=1), A068764 to A068772 (q=2 to 10), A240880 (q=-3).
(End)
Number of sequences [s(0), s(1), ..., s(n)] with s(n)=0, Sum_{j=0..n} s(j) = n, and Sum_{j=0..k} s(j)-1 >= 0 for k < n-1 (and necessarily Sum_{j=0..n-1} s(j)-1 = 0). These are the branching sequences of the (ordered) trees with n non-root nodes, see example. - Joerg Arndt, Jun 30 2014
Number of stack-sortable permutations of [n], these are the 231-avoiding permutations; see the Bousquet-Mélou reference. - Joerg Arndt, Jul 01 2014
a(n) is the number of increasing strict binary trees with 2n-1 nodes that avoid 132. For more information about increasing strict binary trees with an associated permutation, see A245894. - Manda Riehl, Aug 07 2014
In a one-dimensional medium with elastic scattering (zig-zag walk), first recurrence after 2n+1 scattering events has the probability C(n)/2^(2n+1). - Joachim Wuttke, Sep 11 2014
The o.g.f. C(x) = (1 - sqrt(1-4x))/2, for the Catalan numbers, with comp. inverse Cinv(x) = x*(1-x) and the functions P(x) = x / (1 + t*x) and its inverse Pinv(x,t) = -P(-x,t) = x / (1 - t*x) form a group under composition that generates or interpolates among many classic arrays, such as the Motzkin (Riordan, A005043), Fibonacci (A000045), and Fine (A000957) numbers and polynomials (A030528), and enumerating arrays for Motzkin, Dyck, and Łukasiewicz lattice paths and different types of trees and non-crossing partitions (A091867, connected to sums of the refined Narayana numbers A134264). - Tom Copeland, Nov 04 2014
Conjecture: All the rational numbers Sum_{i=j..k} 1/a(i) with 0 < min{2,k} <= j <= k have pairwise distinct fractional parts. - Zhi-Wei Sun, Sep 24 2015
The Catalan number series A000108(n+3), offset n=0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset n=0 (empirical observation). - Tony Foster III, Sep 05 2016
Hankel transforms of the Catalan numbers with the first 2, 4, and 5 terms omitted give A001477, A006858, and A091962, respectively, without the first 2 terms in all cases. More generally, the Hankel transform of the Catalan numbers with the first k terms omitted is H_k(n) = Product_{j=1..k-1} Product_{i=1..j} (2*n+j+i)/(j+i) [see Cigler (2011), Eq. (1.14) and references therein]; together they form the array A078920/A123352/A368025. - Andrey Zabolotskiy, Oct 13 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear. See S. J. Miller, ed., 2015, p. 5. - N. J. A. Sloane, Feb 09 2017
Coefficients of the generating series associated to the Magmatic and Dendriform operadic algebras. Cf. p. 422 and 435 of the Loday et al. paper. - Tom Copeland, Jul 08 2018
Let M_n be the n X n matrix with M_n(i,j) = binomial(i+j-1,2j-2); then det(M_n) = a(n). - Tony Foster III, Aug 30 2018
Also the number of Catalan trees, or planted plane trees (Bona, 2015, p. 299, Theorem 4.6.3). - N. J. A. Sloane, Dec 25 2018
Number of coalescent histories for a caterpillar species tree and a matching caterpillar gene tree with n+1 leaves (Rosenberg 2007, Corollary 3.5). - Noah A Rosenberg, Jan 28 2019
Finding solutions of eps*x^2+x-1 = 0 for eps small, that is, writing x = Sum_{n>=0} x_{n}*eps^n and expanding, one finds x = 1 - eps + 2*eps^2 - 5*eps^3 + 14*eps^3 - 42*eps^4 + ... with x_{n} = (-1)^n*C(n). Further, letting x = 1/y and expanding y about 0 to find large roots, that is, y = Sum_{n>=1} y_{n}*eps^n, one finds y = 0 - eps + eps^2 - 2*eps^3 + 5*eps^3 - ... with y_{n} = (-1)^n*C(n-1). - Derek Orr, Mar 15 2019
Permutations of length n that produce a bipartite permutation graph of order n [see Knuth (1973), Busch (2006), Golumbic and Trenk (2004)]. - Elise Anderson, R. M. Argus, Caitlin Owens, Tessa Stevens, Jun 27 2019
For n > 0, a random selection of n + 1 objects (the minimum number ensuring one pair by the pigeonhole principle) from n distinct pairs of indistinguishable objects contains only one pair with probability 2^(n-1)/a(n) = b(n-1)/A098597(n), where b is the 0-offset sequence with the terms of A120777 repeated (1,1,4,4,8,8,64,64,128,128,...). E.g., randomly selecting 6 socks from 5 pairs that are black, blue, brown, green, and white, results in only one pair of the same color with probability 2^(5-1)/a(5) = 16/42 = 8/21 = b(4)/A098597(5). - Rick L. Shepherd, Sep 02 2019
See Haran & Tabachnikov link for a video discussing Conway-Coxeter friezes. The Conway-Coxeter friezes with n nontrivial rows are generated by the counts of triangles at each vertex in the triangulations of regular n-gons, of which there are a(n). - Charles R Greathouse IV, Sep 28 2019
For connections to knot theory and scattering amplitudes from Feynman diagrams, see Broadhurst and Kreimer, and Todorov. Eqn. 6.12 on p. 130 of Bessis et al. becomes, after scaling, -12g * r_0(-y/(12g)) = (1-sqrt(1-4y))/2, the o.g.f. (expressed as a Taylor series in Eqn. 7.22 in 12gx) given for the Catalan numbers in Copeland's (Sep 30 2011) formula below. (See also Mizera p. 34, Balduf pp. 79-80, Keitel and Bartosch.) - Tom Copeland, Nov 17 2019
Number of permutations in S_n whose principal order ideals in the weak order are modular lattices. - Bridget Tenner, Jan 16 2020
Number of permutations in S_n whose principal order ideals in the weak order are distributive lattices. - Bridget Tenner, Jan 16 2020
Legendre gives the following formula for computing the square root modulo 2^m:
sqrt(1 + 8*a) mod 2^m = (1 + 4*a*Sum_{i=0..m-4} C(i)*(-2*a)^i) mod 2^m
as cited by L. D. Dickson, History of the Theory of Numbers, Vol. 1, 207-208. - Peter Schorn, Feb 11 2020
a(n) is the number of length n permutations sorted to the identity by a consecutive-132-avoiding stack followed by a classical-21-avoiding stack. - Kai Zheng, Aug 28 2020
Number of non-crossing partitions of a 2*n-set with n blocks of size 2. Also number of non-crossing partitions of a 2*n-set with n+1 blocks of size at most 3, and without cyclical adjacencies. The two partitions can be mapped by rotated Kreweras bijection. - Yuchun Ji, Jan 18 2021
Named by Riordan (1968, and earlier in Mathematical Reviews, 1948 and 1964) after the French and Belgian mathematician Eugène Charles Catalan (1814-1894) (see Pak, 2014). - Amiram Eldar, Apr 15 2021
For n >= 1, a(n-1) is the number of interpretations of x^n is an algebra where power-associativity is not assumed. For example, for n = 4 there are a(3) = 5 interpretations: x(x(xx)), x((xx)x), (xx)(xx), (x(xx))x, ((xx)x)x. See the link "Non-associate powers and a functional equation" from I. M. H. Etherington and the page "Nonassociative Product" from Eric Weisstein's World of Mathematics for detailed information. See also A001190 for the case where multiplication is commutative. - Jianing Song, Apr 29 2022
Number of states in the transition diagram associated with the Laplacian system over the complete graph K_N, corresponding to ordered initial conditions x_1 < x_2 < ... < x_N. - Andrea Arlette España, Nov 06 2022
a(n) is the number of 132-avoiding stabilized-interval-free permutations of size n+1. - Juan B. Gil, Jun 22 2023
Number of rooted polyominoes composed of n triangular cells of the hyperbolic regular tiling with Schläfli symbol {3,oo}. A rooted polyomino has one external edge identified, and chiral pairs are counted as two. A stereographic projection of the {3,oo} tiling on the Poincaré disk can be obtained via the Christensson link. - Robert A. Russell, Jan 27 2024
a(n) is the number of extremely lucky Stirling permutations of order n; i.e., the number of Stirling permutations of order n that have exactly n lucky cars. (see Colmenarejo et al. reference) - Bridget Tenner, Apr 16 2024

Examples

			From _Joerg Arndt_ and Greg Stevenson, Jul 11 2011: (Start)
The following products of 3 transpositions lead to a 4-cycle in S_4:
(1,2)*(1,3)*(1,4);
(1,2)*(1,4)*(3,4);
(1,3)*(1,4)*(2,3);
(1,4)*(2,3)*(2,4);
(1,4)*(2,4)*(3,4). (End)
G.f. = 1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...
For n=3, a(3)=5 since there are exactly 5 binary sequences of length 7 in which the number of ones first exceed the number of zeros at entry 7, namely, 0001111, 0010111, 0011011, 0100111, and 0101011. - _Dennis P. Walsh_, Apr 11 2012
From _Joerg Arndt_, Jun 30 2014: (Start)
The a(4) = 14 branching sequences of the (ordered) trees with 4 non-root nodes are (dots denote zeros):
01:  [ 1 1 1 1 . ]
02:  [ 1 1 2 . . ]
03:  [ 1 2 . 1 . ]
04:  [ 1 2 1 . . ]
05:  [ 1 3 . . . ]
06:  [ 2 . 1 1 . ]
07:  [ 2 . 2 . . ]
08:  [ 2 1 . 1 . ]
09:  [ 2 1 1 . . ]
10:  [ 2 2 . . . ]
11:  [ 3 . . 1 . ]
12:  [ 3 . 1 . . ]
13:  [ 3 1 . . . ]
14:  [ 4 . . . . ]
(End)

References

  • The large number of references and links demonstrates the ubiquity of the Catalan numbers.
  • R. Alter, Some remarks and results on Catalan numbers, pp. 109-132 in Proceedings of the Louisiana Conference on Combinatorics, Graph Theory and Computer Science. Vol. 2, edited R. C. Mullin et al., 1971.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, many references.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 53.
  • J. H. Conway and R. K. Guy, The Book of Numbers, New York: Springer-Verlag, 1995, ch. 4, pp. 96-106.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see pp. 183, 196, etc.).
  • Michael Dairyko, Samantha Tyner, Lara Pudwell, and Casey Wynn, Non-contiguous pattern avoidance in binary trees. Electron. J. Combin. 19 (2012), no. 3, Paper 22, 21 pp. MR2967227.
  • E. Deutsch, Dyck path enumeration, Discrete Math., 204, 167-202, 1999.
  • E. Deutsch and L. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31, 31-38, 2001.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 1, 207-208.
  • Tomislav Doslic and Darko Veljan, Logarithmic behavior of some combinatorial sequences. Discrete Math. 308 (2008), no. 11, 2182-2212. MR2404544 (2009j:05019)
  • S. Dulucq and J.-G. Penaud, Cordes, arbres et permutations. Discrete Math. 117 (1993), no. 1-3, 89-105.
  • A. Errera, Analysis situs - Un problème d'énumération, Mémoires Acad. Bruxelles, Classe des sciences, Série 2, Vol. XI, Fasc. 6, No. 1421 (1931), 26 pp.
  • Ehrenfeucht, Andrzej; Haemer, Jeffrey; Haussler, David. Quasimonotonic sequences: theory, algorithms and applications. SIAM J. Algebraic Discrete Methods 8 (1987), no. 3, 410-429. MR0897739 (88h:06026)
  • I. M. H. Etherington, Non-associate powers and a functional equation. The Mathematical Gazette, 21 (1937): 36-39; addendum 21 (1937), 153.
  • I. M. H. Etherington, On non-associative combinations, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162.
  • I. M. H. Etherington, Some problems of non-associative combinations (I), Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
  • K. Fan, Structure of a Hecke algebra quotient, J. Amer. Math. Soc., 10 (1997), 139-167.
  • Susanna Fishel, Myrto Kallipoliti and Eleni Tzanaki, Facets of the Generalized Cluster Complex and Regions in the Extended Catalan Arrangement of Type A, The electronic Journal of Combinatorics 20(4) (2013), #P7.
  • D. Foata and D. Zeilberger, A classic proof of a recurrence for a very classical sequence, J. Comb Thy A 80 380-384 1997.
  • H. G. Forder, Some problems in combinatorics, Math. Gazette, vol. 45, 1961, 199-201.
  • Fürlinger, J.; Hofbauer, J., q-Catalan numbers. J. Combin. Theory Ser. A 40 (1985), no. 2, 248-264. MR0814413 (87e:05017)
  • M. Gardner, Time Travel and Other Mathematical Bewilderments, Chap. 20 pp. 253-266, W. H. Freeman NY 1988.
  • James Gleick, Faster, Vintage Books, NY, 2000 (see pp. 259-261).
  • M. C. Golumbic and A. N. Trenk, Tolerance graphs, Vol. 89, Cambridge University Press, 2004, pp. 32.
  • S Goodenough, C Lavault, Overview on Heisenberg—Weyl Algebra and Subsets of Riordan Subgroups, The Electronic Journal of Combinatorics, 22(4) (2015), #P4.16,
  • H. W. Gould, Research bibliography of two special number sequences, Mathematica Monongaliae, Vol. 12, 1971.
  • D. Gouyou-Beauchamps, Chemins sous-diagonaux et tableau de Young, pp. 112-125 of "Combinatoire Enumerative (Montreal 1985)", Lect. Notes Math. 1234, 1986.
  • M. Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 53-63 and 85-93.
  • J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 530.
  • N. S. S. Gu, N. Y. Li and T. Mansour, 2-Binary trees: bijections and related issues, Discr. Math., 308 (2008), 1209-1221.
  • R. K. Guy, Dissecting a polygon into triangles, Research Paper #9, Math. Dept., Univ. Calgary, 1967.
  • R. K. Guy and J. L. Selfridge, The nesting and roosting habits of the laddered parenthesis. Amer. Math. Monthly 80 (1973), 868-876.
  • Peter Hajnal and Gabor V. Nagy, A bijective proof of Shapiro's Catalan convolution, Elect. J. Combin., 21 (2014), #P2.42.
  • F. Harary and E. M. Palmer, Graphical Enumeration, Academic Press, NY, 1973, p. 67, (3.3.23).
  • F. Harary, G. Prins, and W. T. Tutte, The number of plane trees. Indag. Math. 26, 319-327, 1964.
  • J. Harris, Algebraic Geometry: A First Course (GTM 133), Springer-Verlag, 1992, pages 245-247.
  • S. Heubach, N. Y. Li and T. Mansour, Staircase tilings and k-Catalan structures, Discrete Math., 308 (2008), 5954-5964.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • Higgins, Peter M. Combinatorial results for semigroups of order-preserving mappings. Math. Proc. Camb. Phil. Soc. (1993), 113: 281-296.
  • B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 513, Eq. (7.282).
  • F. Hurtado, M. Noy, Ears of triangulations and Catalan numbers, Discrete Mathematics, Volume 149, Issues 1-3, Feb 22 1996, Pages 319-324.
  • M. Janjic, Determinants and Recurrence Sequences, Journal of Integer Sequences, 2012, Article 12.3.5.
  • R. H. Jeurissen, Raney and Catalan, Discrete Math., 308 (2008), 6298-6307.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 36.
  • Kim, Ki Hang; Rogers, Douglas G.; Roush, Fred W. Similarity relations and semiorders. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 577-594, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561081 (81i:05013)
  • Klarner, D. A. A Correspondence Between Sets of Trees. Indag. Math. 31, 292-296, 1969.
  • M. Klazar, On numbers of Davenport-Schinzel sequences, Discr. Math., 185 (1998), 77-87.
  • D. E. Knuth, The Art of Computer Programming, 2nd Edition, Vol. 1, Addison-Wesley, 1973, pp. 238.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.6 (p. 450).
  • Thomas Koshy and Mohammad Salmassi, "Parity and Primality of Catalan Numbers", College Mathematics Journal, Vol. 37, No. 1 (Jan 2006), pp. 52-53.
  • M. Kosters, A theory of hexaflexagons, Nieuw Archief Wisk., 17 (1999), 349-362.
  • E. Krasko, A. Omelchenko, Brown's Theorem and its Application for Enumeration of Dissections and Planar Trees, The Electronic Journal of Combinatorics, 22 (2015), #P1.17.
  • C. Krishnamachary and M. Bheemasena Rao, Determinants whose elements are Eulerian, prepared Bernoullian and other numbers, J. Indian Math. Soc., 14 (1922), 55-62, 122-138 and 143-146.
  • P. Lafar and C. T. Long, A combinatorial problem, Amer. Math. Mnthly, 69 (1962), 876-883.
  • Laradji, A. and Umar, A. On certain finite semigroups of order-decreasing transformations I, Semigroup Forum 69 (2004), 184-200.
  • P. J. Larcombe, On pre-Catalan Catalan numbers: Kotelnikow (1766), Mathematics Today, 35 (1999), p. 25.
  • P. J. Larcombe, On the history of the Catalan numbers: a first record in China, Mathematics Today, 35 (1999), p. 89.
  • P. J. Larcombe, The 18th century Chinese discovery of the Catalan numbers, Math. Spectrum, 32 (1999/2000), 5-7.
  • P. J. Larcombe and P. D. C. Wilson, On the trail of the Catalan sequence, Mathematics Today, 34 (1998), 114-117.
  • P. J. Larcombe and P. D. C. Wilson, On the generating function of the Catalan sequence: a historical perspective, Congress. Numer., 149 (2001), 97-108.
  • G. S. Lueker, Some techniques for solving recurrences, Computing Surveys, 12 (1980), 419-436.
  • J. J. Luo, Antu Ming, the first inventor of Catalan numbers in the world [in Chinese], Neimenggu Daxue Xuebao, 19 (1998), 239-245.
  • C. L. Mallows, R. J. Vanderbei, Which Young Tableaux Can Represent an Outer Sum?, Journal of Integer Sequences, Vol. 18, 2015, #15.9.1.
  • Toufik Mansour, Matthias Schork, and Mark Shattuck, Catalan numbers and pattern restricted set partitions. Discrete Math. 312(2012), no. 20, 2979-2991. MR2956089
  • Toufik Mansour and Simone Severini, Enumeration of (k,2)-noncrossing partitions, Discrete Math., 308 (2008), 4570-4577.
  • M. E. Mays and Jerzy Wojciechowski, A determinant property of Catalan numbers. Discrete Math. 211, No. 1-3, 125-133 (2000). Zbl 0945.05037
  • D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344.
  • A. Milicevic and N. Trinajstic, "Combinatorial Enumeration in Chemistry", Chem. Modell., Vol. 4, (2006), pp. 405-469.
  • Miller, Steven J., ed. Benford's Law: Theory and Applications. Princeton University Press, 2015.
  • David Molnar, "Wiggly Games and Burnside's Lemma", Chapter 8, The Mathematics of Various Entertaining Subjects: Volume 3 (2019), Jennifer Beineke & Jason Rosenhouse, eds. Princeton University Press, Princeton and Oxford, p. 102.
  • C. O. Oakley and R. J. Wisner, Flexagons, Amer. Math. Monthly, 64 (1957), 143-154.
  • A. Panholzer and H. Prodinger, Bijections for ternary trees and non-crossing trees, Discrete Math., 250 (2002), 181-195 (see Eq. 4).
  • Papoulis, Athanasios. "A new method of inversion of the Laplace transform."Quart. Appl. Math 14.405-414 (1957): 124.
  • S. G. Penrice, Stacks, bracketings and CG-arrangements, Math. Mag., 72 (1999), 321-324.
  • C. A. Pickover, Wonders of Numbers, Chap. 71, Oxford Univ. Press NY 2000.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • G. Pólya, On the number of certain lattice polygons. J. Combinatorial Theory 6 1969 102-105. MR0236031 (38 #4329)
  • C. Pomerance, Divisors of the middle binomial coefficient, Amer. Math. Monthly, 112 (2015), 636-644.
  • Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.
  • Ronald C. Read, "The Graph Theorists who Count -- and What They Count", in 'The Mathematical Gardner', in D. A. Klarner, Ed., pp. 331-334, Wadsworth CA 1989.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 101.
  • J. Riordan, The distribution of crossings of chords joining pairs of 2n points on a circle, Math. Comp., 29 (1975), 215-222.
  • T. Santiago Costa Oliveira, "Catalan traffic" and integrals on the Grassmannian of lines, Discr. Math., 308 (2007), 148-152.
  • A. Sapounakis, I. Tasoulas and P. Tsikouras, Counting strings in Dyck paths, Discrete Math., 307 (2007), 2909-2924.
  • E. Schröder, Vier combinatorische Probleme, Z. f. Math. Phys., 15 (1870), 361-376.
  • Shapiro, Louis W. Catalan numbers and "total information" numbers. Proceedings of the Sixth Southeastern Conference on Combinatorics, Graph Theory, and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1975), pp. 531-539. Congressus Numerantium, No. XIV, Utilitas Math., Winnipeg, Man., 1975. MR0398853 (53 #2704).
  • L. W. Shapiro, A short proof of an identity of Touchard's concerning Catalan numbers, J. Combin. Theory, A 20 (1976), 375-376.
  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.
  • L. W. Shapiro, W.-J. Woan and S. Getu, The Catalan numbers via the World Series, Math. Mag., 66 (1993), 20-22.
  • D. M. Silberger, Occurrences of the integer (2n-2)!/n!(n-1)!, Roczniki Polskiego Towarzystwa Math. 13 (1969): 91-96.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • S. Snover and S. Troyer, Multidimensional Catalan numbers, Abstracts 848-05-94 and 848-05-95, 848th Meeting, Amer. Math. Soc., Worcester Mass., March 15-16, 1989.
  • R. P. Stanley, Enumerative Combinatorics, Wadsworth, Vol. 1, 1986, Vol. 2, 1999; see especially Chapter 6.
  • R. P. Stanley, Recent Progress in Algebraic Combinatorics, Bull. Amer. Math. Soc., 40 (2003), 55-68.
  • Richard P. Stanley, "Catalan Numbers", Cambridge University Press, 2015.
  • J. J. Sylvester, On reducible cyclodes, Coll. Math. Papers, Vol. 2, see especially page 670, where Catalan numbers appear.
  • Thiel, Marko. "A new cyclic sieving phenomenon for Catalan objects." Discrete Mathematics 340.3 (2017): 426-429.
  • I. Vun and P. Belcher, Catalan numbers, Mathematical Spectrum, 30 (1997/1998), 3-5.
  • D. Wells, Penguin Dictionary of Curious and Interesting Numbers, Entry 42 p 121, Penguin Books, 1987.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 41.
  • J. Wuttke, The zig-zag walk with scattering and absorption on the real half line and in a lattice model, J. Phys. A 47 (2014), 215203, 1-9.

Links

Crossrefs

Cf. A000142, A000245, A000344, A000588, A000957, A000984, A001392, A001453, A001791, A002057, A002420, A003046, A003517, A003518, A003519, A006480, A008276, A008549, A014137, A014138, A014140, A022553 (inv. Eul. trans.), A024492, A032357, A032443, A039599, A048990, A059288, A068875, A069640, A086117, A088327 (Eul. trans.), A094216, A094638, A094639, A098597, A099731, A119822, A120304, A124926, A129763, A137697, A154559, A161581, A167892, A167893, A179277, A211611, A275431 (multisets).
A row of A060854.
See A001003, A001190, A001699, A000081 for other ways to count parentheses.
Enumerates objects encoded by A014486.
A diagonal of any of the essentially equivalent arrays A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Cf. A051168 (diagonal of the square array described).
Cf. A033552, A176137 (partitions into Catalan numbers).
Cf. A000753, A000736 (Boustrophedon transforms).
Cf. A120303 (largest prime factor of Catalan number).
Cf. A121839 (reciprocal Catalan constant), A268813.
Cf. A038003, A119861, A119908, A120274, A120275 (odd Catalan number).
Cf. A002390 (decimal expansion of natural logarithm of golden ratio).
Coefficients of square root of the g.f. are A001795/A046161.
For a(n) mod 6 see A259667.
For a(n) in base 2 see A264663.
Hankel transforms with first terms omitted: A001477, A006858, A091962, A078920, A123352, A368025.
Cf. A001147, A163960.
Cf. A332602 (conjectured production matrix).
Polyominoes: A001683(n+2) (oriented), A000207 (unoriented), A369314 (chiral), A208355(n-1) (achiral), A001764 {4,oo}.

Programs

  • GAP
    A000108:=List([0..30],n->Binomial(2*n,n)/(n+1)); # Muniru A Asiru, Feb 17 2018
  • Haskell
    import Data.List (genericIndex)
a000108 n = genericIndex a000108_list n
a000108_list = 1 : catalan [1] where
   catalan cs = c : catalan (c:cs) where
      c = sum $ zipWith (*) cs $ reverse cs
-- Reinhard Zumkeller, Nov 12 2011
a000108 = map last $ iterate (scanl1 (+) . (++ [0])) [1]
-- David Spies, Aug 23 2015
  • Magma
    C:= func< n | Binomial(2*n,n)/(n+1) >; [ C(n) : n in [0..60]];
  • Magma
    [Catalan(n): n in [0..40]]; // Vincenzo Librandi, Apr 02 2011
  • Maple
    A000108 := n->binomial(2*n,n)/(n+1);
G000108 := (1 - sqrt(1 - 4*x)) / (2*x);
spec := [ A, {A=Prod(Z,Sequence(A))}, unlabeled ]: [ seq(combstruct[count](spec, size=n+1), n=0..42) ];
with(combstruct): bin := {B=Union(Z,Prod(B,B))}: seq(count([B,bin,unlabeled],size=n+1), n=0..25); # Zerinvary Lajos, Dec 05 2007
gser := series(G000108, x=0, 42): seq(coeff(gser, x, n), n=0..41); # Zerinvary Lajos, May 21 2008
seq((2*n)!*coeff(series(hypergeom([],[2],x^2),x,2*n+2),x,2*n),n=0..30); # Peter Luschny, Jan 31 2015
A000108List := proc(m) local A, P, n; A := [1, 1]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), A[-1]]);
A := [op(A), P[-1]] od; A end: A000108List(31); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[(2 n)!/n!/(n + 1)!, {n, 0, 20}]
Table[4^n Gamma[n + 1/2]/(Sqrt[Pi] Gamma[n + 2]), {n, 0, 20}] (* Eric W. Weisstein, Oct 31 2024 *)
Table[Hypergeometric2F1[1 - n, -n, 2, 1], {n, 0, 20}] (* Richard L. Ollerton, Sep 13 2006 *)
Table[CatalanNumber @ n, {n, 0, 20}] (* Robert G. Wilson v, Feb 15 2011 *)
CatalanNumber[Range[0, 20]] (* Eric W. Weisstein, Oct 31 2024 *)
CoefficientList[InverseSeries[Series[x/Sum[x^n, {n, 0, 31}], {x, 0, 31}]]/x, x] (* Mats Granvik, Nov 24 2013 *)
CoefficientList[Series[(1 - Sqrt[1 - 4 x])/(2 x), {x, 0, 20}], x] (* Stefano Spezia, Aug 31 2018 *)
  • Maxima
    A000108(n):=binomial(2*n,n)/(n+1)$ makelist(A000108(n),n,0,30); /* Martin Ettl, Oct 24 2012 */
  • MuPAD
    combinat::dyckWords::count(n) $ n = 0..38 // Zerinvary Lajos, Apr 14 2007
  • PARI
    a(n)=binomial(2*n,n)/(n+1) \\ M. F. Hasler, Aug 25 2012
  • PARI
    a(n) = (2*n)! / n! / (n+1)!
  • PARI
    a(n) = my(A, m); if( n<0, 0, m=1; A = 1 + x + O(x^2); while(m<=n, m*=2; A = sqrt(subst(A, x, 4*x^2)); A += (A - 1) / (2*x*A)); polcoeff(A, n));
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( serreverse( x / (1 + x)^2 + x * O(x^n)), n))}; /* Michael Somos */
  • PARI
    (recur(a,b)=if(b<=2,(a==2)+(a==b)+(a!=b)*(1+a/2), (1+a/b)*recur(a,b-1))); a(n)=recur(n,n); \\ R. J. Cano, Nov 22 2012
  • PARI
    x='x+O('x^40); Vec((1-sqrt(1-4*x))/(2*x)) \\ Altug Alkan, Oct 13 2015
  • Python
    from gmpy2 import divexact
A000108 = [1, 1]
for n in range(1, 10**3):
    A000108.append(divexact(A000108[-1]*(4*n+2),(n+2))) # Chai Wah Wu, Aug 31 2014
  • Python
    # Works in Sage also.
A000108 = [1]
for n in range(1000):
    A000108.append(A000108[-1]*(4*n+2)//(n+2)) # Günter Rote, Nov 08 2023
  • Sage
    [catalan_number(i) for i in range(27)] # Zerinvary Lajos, Jun 26 2008
  • Sage
    # Generalized algorithm of L. Seidel
def A000108_list(n) :
    D = [0]*(n+1); D[1] = 1
    b = True; h = 1; R = []
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1; R.append(D[1])
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
    return R
A000108_list(31) # Peter Luschny, Jun 02 2012

Formula

a(n) = binomial(2*n, n)/(n+1) = (2*n)!/(n!*(n+1)!) = A000984(n)/(n+1).
Recurrence: a(n) = 2*(2*n-1)*a(n-1)/(n+1) with a(0) = 1.
Recurrence: a(n) = Sum_{k=0..n-1} a(k)a(n-1-k).
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x), and satisfies A(x) = 1 + x*A(x)^2.
a(n) = Product_{k=2..n} (1 + n/k).
a(n+1) = Sum_{i} binomial(n, 2*i)*2^(n-2*i)*a(i). - Touchard
It is known that a(n) is odd if and only if n=2^k-1, k=0, 1, 2, 3, ... - Emeric Deutsch, Aug 04 2002, corrected by M. F. Hasler, Nov 08 2015
Using the Stirling approximation in A000142 we get the asymptotic expansion a(n) ~ 4^n / (sqrt(Pi * n) * (n + 1)). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 13 2001
Integral representation: a(n) = (1/(2*Pi))*Integral_{x=0..4} x^n*sqrt((4-x)/x). - Karol A. Penson, Apr 12 2001
E.g.f.: exp(2*x)*(I_0(2*x)-I_1(2*x)), where I_n is Bessel function. - Karol A. Penson, Oct 07 2001
a(n) = polygorial(n, 6)/polygorial(n, 3). - Daniel Dockery (peritus(AT)gmail.com), Jun 24 2003
G.f. A(x) satisfies ((A(x) + A(-x)) / 2)^2 = A(4*x^2). - Michael Somos, Jun 27 2003
G.f. A(x) satisfies Sum_{k>=1} k(A(x)-1)^k = Sum_{n>=1} 4^{n-1}*x^n. - Shapiro, Woan, Getu
a(n+m) = Sum_{k} A039599(n, k)*A039599(m, k). - Philippe Deléham, Dec 22 2003
a(n+1) = (1/(n+1))*Sum_{k=0..n} a(n-k)*binomial(2k+1, k+1). - Philippe Deléham, Jan 24 2004
a(n) = Sum_{k>=0} A008313(n, k)^2. - Philippe Deléham, Feb 14 2004
a(m+n+1) = Sum_{k>=0} A039598(m, k)*A039598(n, k). - Philippe Deléham, Feb 15 2004
a(n) = Sum_{k=0..n} (-1)^k*2^(n-k)*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
Sum_{n>=0} 1/a(n) = 2 + 4*Pi/3^(5/2) = F(1,2;1/2;1/4) = A268813 = 2.806133050770763... (see L'Univers de Pi link). - Gerald McGarvey and Benoit Cloitre, Feb 13 2005
a(n) = Sum_{k=0..floor(n/2)} ((n-2*k+1)*binomial(n, n-k)/(n-k+1))^2, which is equivalent to: a(n) = Sum_{k=0..n} A053121(n, k)^2, for n >= 0. - Paul D. Hanna, Apr 23 2005
a((m+n)/2) = Sum_{k>=0} A053121(m, k)*A053121(n, k) if m+n is even. - Philippe Deléham, May 26 2005
E.g.f. Sum_{n>=0} a(n) * x^(2*n) / (2*n)! = BesselI(1, 2*x) / x. - Michael Somos, Jun 22 2005
Given g.f. A(x), then B(x) = x * A(x^3) satisfies 0 = f(x, B(X)) where f(u, v) = u - v + (u*v)^2 or B(x) = x + (x * B(x))^2 which implies B(-B(x)) = -x and also (1 + B^3) / B^2 = (1 - x^3) / x^2. - Michael Somos, Jun 27 2005
a(n) = a(n-1)*(4-6/(n+1)). a(n) = 2a(n-1)*(8a(n-2)+a(n-1))/(10a(n-2)-a(n-1)). - Franklin T. Adams-Watters, Feb 08 2006
Sum_{k>=1} a(k)/4^k = 1. - Franklin T. Adams-Watters, Jun 28 2006
a(n) = A047996(2*n+1, n). - Philippe Deléham, Jul 25 2006
Binomial transform of A005043. - Philippe Deléham, Oct 20 2006
a(n) = Sum_{k=0..n} (-1)^k*A116395(n,k). - Philippe Deléham, Nov 07 2006
a(n) = (1/(s-n))*Sum_{k=0..n} (-1)^k (k+s-n)*binomial(s-n,k) * binomial(s+n-k,s) with s a nonnegative free integer [H. W. Gould].
a(k) = Sum_{i=1..k} |A008276(i,k)| * (k-1)^(k-i) / k!. - André F. Labossière, May 29 2007
a(n) = Sum_{k=0..n} A129818(n,k) * A007852(k+1). - Philippe Deléham, Jun 20 2007
a(n) = Sum_{k=0..n} A109466(n,k) * A127632(k). - Philippe Deléham, Jun 20 2007
Row sums of triangle A124926. - Gary W. Adamson, Oct 22 2007
Limit_{n->oo} (1 + Sum_{k=0..n} a(k)/A004171(k)) = 4/Pi. - Reinhard Zumkeller, Aug 26 2008
a(n) = Sum_{k=0..n} A120730(n,k)^2 and a(k+1) = Sum_{n>=k} A120730(n,k). - Philippe Deléham, Oct 18 2008
Given an integer t >= 1 and initial values u = [a_0, a_1, ..., a_{t-1}], we may define an infinite sequence Phi(u) by setting a_n = a_{n-1} + a_0*a_{n-1} + a_1*a_{n-2} + ... + a_{n-2}*a_1 for n >= t. For example, the present sequence is Phi([1]) (also Phi([1,1])). - Gary W. Adamson, Oct 27 2008
a(n) = Sum_{l_1=0..n+1} Sum_{l_2=0..n}...Sum_{l_i=0..n-i}...Sum_{l_n=0..1} delta(l_1,l_2,...,l_i,...,l_n) where delta(l_1,l_2,...,l_i,...,l_n) = 0 if any l_i < l_(i+1) and l_(i+1) <> 0 for i=1..n-1 and delta(l_1,l_2,...,l_i,...,l_n) = 1 otherwise. - Thomas Wieder, Feb 25 2009
a(n) = A000680(n)/A006472(n+1). - Mark Dols, Jul 14 2010; corrected by M. F. Hasler, Nov 08 2015
Let A(x) be the g.f., then B(x)=x*A(x) satisfies the differential equation B'(x)-2*B'(x)*B(x)-1=0. - Vladimir Kruchinin, Jan 18 2011
Complement of A092459; A010058(a(n)) = 1. - Reinhard Zumkeller, Mar 29 2011
G.f.: 1/(1-x/(1-x/(1-x/(...)))) (continued fraction). - Joerg Arndt, Mar 18 2011
With F(x) = (1-2*x-sqrt(1-4*x))/(2*x) an o.g.f. in x for the Catalan series, G(x) = x/(1+x)^2 is the compositional inverse of F (nulling the n=0 term). - Tom Copeland, Sep 04 2011
With H(x) = 1/(dG(x)/dx) = (1+x)^3 / (1-x), the n-th Catalan number is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)), and H(x) is the o.g.f. for A115291. - Tom Copeland, Sep 04 2011
From Tom Copeland, Sep 30 2011: (Start)
With F(x) = (1-sqrt(1-4*x))/2 an o.g.f. in x for the Catalan series, G(x)= x*(1-x) is the compositional inverse and this relates the Catalan numbers to the row sums of A125181.
With H(x) = 1/(dG(x)/dx) = 1/(1-2x), the n-th Catalan number (offset 1) is given by (1/n!)*((H(x)*d/dx)^n)x evaluated at x=0, i.e., F(x) = exp(x*H(u)*d/du)u, evaluated at u = 0. Also, dF(x)/dx = H(F(x)). (End)
G.f.: (1-sqrt(1-4*x))/(2*x) = G(0) where G(k) = 1 + (4*k+1)*x/(k+1-2*x*(k+1)*(4*k+3)/(2*x*(4*k+3)+(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: exp(2*x)*(BesselI(0,2*x) - BesselI(1,2*x)) = G(0) where G(k) = 1 + (4*k+1)*x/((k+1)*(2*k+1)-x*(k+1)*(2*k+1)*(4*k+3)/(x*(4*k+3)+(k+1)*(2*k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Nov 30 2011
E.g.f.: Hypergeometric([1/2],[2],4*x) which coincides with the e.g.f. given just above, and also by Karol A. Penson further above. - Wolfdieter Lang, Jan 13 2012
A076050(a(n)) = n + 1 for n > 0. - Reinhard Zumkeller, Feb 17 2012
a(n) = A208355(2*n-1) = A208355(2*n) for n > 0. - Reinhard Zumkeller, Mar 04 2012
a(n+1) = A214292(2*n+1,n) = A214292(2*n+2,n). - Reinhard Zumkeller, Jul 12 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = k*(4*x+1) + 2*x + 2 - x*(2*k+3)*(2*k+4)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 20 2012
G.f.: hypergeom([1/2,1],[2],4*x). - Joerg Arndt, Apr 06 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,1,-1/2-n,-1)/(n+1). - Karol A. Penson, Jul 28 2013
For n > 0: a(n) = sum of row n in triangle A001263. - Reinhard Zumkeller, Oct 10 2013
a(n) = binomial(2n,n-1)/n and a(n) mod n = binomial(2n,n) mod n = A059288(n). - Jonathan Sondow, Dec 14 2013
a(n-1) = Sum_{t1+2*t2+...+n*tn=n} (-1)^(1+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*a(1)^t1*a(2)^t2*...*a(n)^tn. - Mircea Merca, Feb 27 2014
a(n) = Sum_{k=1..n} binomial(n+k-1,n)/n if n > 0. Alexander Adamchuk, Mar 25 2014
a(n) = -2^(2*n+1) * binomial(n-1/2, -3/2). - Peter Luschny, May 06 2014
a(n) = (4*A000984(n) - A000984(n+1))/2. - Stanislav Sykora, Aug 09 2014
a(n) = A246458(n) * A246466(n). - Tom Edgar, Sep 02 2014
a(n) = (2*n)!*[x^(2*n)]hypergeom([],[2],x^2). - Peter Luschny, Jan 31 2015
a(n) = 4^(n-1)*hypergeom([3/2, 1-n], [3], 1). - Peter Luschny, Feb 03 2015
a(2n) = 2*A000150(2n); a(2n+1) = 2*A000150(2n+1) + a(n). - John Bodeen, Jun 24 2015
a(n) = Sum_{t=1..n+1} n^(t-1)*abs(Stirling1(n+1, t)) / Sum_{t=1..n+1} abs(Stirling1(n+1, t)), for n > 0, see (10) in Cereceda link. - Michel Marcus, Oct 06 2015
a(n) ~ 4^(n-2)*(128 + 160/N^2 + 84/N^4 + 715/N^6 - 10180/N^8)/(N^(3/2)*Pi^(1/2)) where N = 4*n+3. - Peter Luschny, Oct 14 2015
a(n) = Sum_{k=1..floor((n+1)/2)} (-1)^(k-1)*binomial(n+1-k,k)*a(n-k) if n > 0; and a(0) = 1. - David Pasino, Jun 29 2016
Sum_{n>=0} (-1)^n/a(n) = 14/25 - 24*arccsch(2)/(25*sqrt(5)) = 14/25 - 24*A002390/(25*sqrt(5)) = 0.353403708337278061333... - Ilya Gutkovskiy, Jun 30 2016
C(n) = (1/n) * Sum_{i+j+k=n-1} C(i)*C(j)*C(k)*(k+1), n >= 1. - Yuchun Ji, Feb 21 2016
C(n) = 1 + Sum_{i+j+kYuchun Ji, Sep 01 2016
a(n) = A001700(n) - A162551(n) = binomial(2*n+1,n+1). - 2*binomial(2*n,n-1). - Taras Goy, Aug 09 2018
G.f.: A(x) = (1 - sqrt(1 - 4*x)) / (2*x) = 2F1(1/2,1;2;4*x). G.f. A(x) satisfies A = 1 + x*A^2. - R. J. Mathar, Nov 17 2018
C(n) = 1 + Sum_{i=0..n-1} A000245(i). - Yuchun Ji, Jan 10 2019
From A.H.M. Smeets, Apr 11 2020: (Start)
(1+sqrt(1+4*x))/2 = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4; and sqrt(x+sqrt(x+sqrt(x+...))) = 1-Sum_{i >= 0} a(i)*(-x)^(i+1), for any complex x with |x| < 1/4 and x <> 0. (End)
a(3n+1)*a(5n+4)*a(15n+10) = a(3n+2)*a(5n+2)*a(15n+11). The first case of Catalan product equation of a triple partition of 23n+15. - Yuchun Ji, Sep 27 2020
a(n) = 4^n * (-1)^(n+1) * 3F2[{n + 1,n + 1/2,n}, {3/2,1}, -1], n >= 1. - Sergii Voloshyn, Oct 22 2020
a(n) = 2^(1 + 2 n) * (-1)^(n)/(1 + n) * 3F2[{n, 1/2 + n, 1 + n}, {1/2, 1}, -1], n >= 1. - Sergii Voloshyn, Nov 08 2020
a(n) = (1/Pi)*4^(n+1)*Integral_{x=0..Pi/2} cos(x)^(2*n)*sin(x)^2 dx. - Greg Dresden, May 30 2021
From Peter Bala, Aug 17 2021: (Start)
G.f. A(x) satisfies A(x) = 1/sqrt(1 - 4*x) * A( -x/(1 - 4*x) ) and (A(x) + A(-x))/2 = 1/sqrt(1 - 4*x) * A( -2*x/(1 - 4*x) ); these are the cases k = 0 and k = -1 of the general formula 1/sqrt(1 - 4*x) * A( (k-1)*x/(1 - 4*x) ) = Sum_{n >= 0} ((k^(n+1) - 1)/(k - 1))*Catalan(n)*x^n.
2 - sqrt(1 - 4*x)/A( k*x/(1 - 4*x) ) = 1 + Sum_{n >= 1} (1 + (k + 1)^n) * Catalan(n-1)*x^n. (End)
Sum_{n>=0} a(n)*(-1/4)^n = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Mar 22 2022
0 = a(n)*(16*a(n+1) - 10*a(n+2)) + a(n+1)*(2*a(n+1) + a(n+2)) for all n>=0. - Michael Somos, Dec 12 2022
G.f.: (offset 1) 1/G(x), with G(x) = 1 - 2*x - x^2/G(x) (Jacobi continued fraction). - Nikolaos Pantelidis, Feb 01 2023
a(n) = K^(2n+1, n, 1) for all n >= 0, where K^(n, s, x) is the Krawtchouk polynomial defined to be Sum_{k=0..s} (-1)^k * binomial(n-x, s-k) * binomial(x, k). - Vladislav Shubin, Aug 17 2023
From Peter Bala, Feb 03 2024: (Start)
The g.f. A(x) satisfies the following functional equations:
A(x) = 1 + x/(1 - 4*x) * A(-x/(1 - 4*x))^2,
A(x^2) = 1/(1 - 2*x) * A(- x/(1 - 2*x))^2 and, for arbitrary k,
1/(1 - k*x) * A(x/(1 - k*x))^2 = 1/(1 - (k+4)*x) * A(-x/(1 - (k+4)*x))^2. (End)
a(n) = A363448(n) + A363449(n). - Julien Rouyer, Jun 28 2024

A000984 Central binomial coefficients: binomial(2*n,n) = (2*n)!/(n!)^2.

Original entry on oeis.org

1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, 2704156, 10400600, 40116600, 155117520, 601080390, 2333606220, 9075135300, 35345263800, 137846528820, 538257874440, 2104098963720, 8233430727600, 32247603683100, 126410606437752, 495918532948104, 1946939425648112
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Devadoss refers to these numbers as type B Catalan numbers (cf. A000108).
Equal to the binomial coefficient sum Sum_{k=0..n} binomial(n,k)^2.
Number of possible interleavings of a program with n atomic instructions when executed by two processes. - Manuel Carro (mcarro(AT)fi.upm.es), Sep 22 2001
Convolving a(n) with itself yields A000302, the powers of 4. - T. D. Noe, Jun 11 2002
Number of ordered trees with 2n+1 edges, having root of odd degree and nonroot nodes of outdegree 0 or 2. - Emeric Deutsch, Aug 02 2002
Also number of directed, convex polyominoes having semiperimeter n+2.
Also number of diagonally symmetric, directed, convex polyominoes having semiperimeter 2n+2. - Emeric Deutsch, Aug 03 2002
The second inverse binomial transform of this sequence is this sequence with interpolated zeros. Its g.f. is (1 - 4*x^2)^(-1/2), with n-th term C(n,n/2)(1+(-1)^n)/2. - Paul Barry, Jul 01 2003
Number of possible values of a 2n-bit binary number for which half the bits are on and half are off. - Gavin Scott (gavin(AT)allegro.com), Aug 09 2003
Ordered partitions of n with zeros to n+1, e.g., for n=4 we consider the ordered partitions of 11110 (5), 11200 (30), 13000 (20), 40000 (5) and 22000 (10), total 70 and a(4)=70. See A001700 (esp. Mambetov Bektur's comment). - Jon Perry, Aug 10 2003
Number of nondecreasing sequences of n integers from 0 to n: a(n) = Sum_{i_1=0..n} Sum_{i_2=i_1..n}...Sum_{i_n=i_{n-1}..n}(1). - J. N. Bearden (jnb(AT)eller.arizona.edu), Sep 16 2003
Number of peaks at odd level in all Dyck paths of semilength n+1. Example: a(2)=6 because we have U*DU*DU*D, U*DUUDD, UUDDU*D, UUDUDD, UUU*DDD, where U=(1,1), D=(1,-1) and * indicates a peak at odd level. Number of ascents of length 1 in all Dyck paths of semilength n+1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(2)=6 because we have uDuDuD, uDUUDD, UUDDuD, UUDuDD, UUUDDD, where an ascent of length 1 is indicated by a lower case letter. - Emeric Deutsch, Dec 05 2003
a(n-1) = number of subsets of 2n-1 distinct elements taken n at a time that contain a given element. E.g., n=4 -> a(3)=20 and if we consider the subsets of 7 taken 4 at a time with a 1 we get (1234, 1235, 1236, 1237, 1245, 1246, 1247, 1256, 1257, 1267, 1345, 1346, 1347, 1356, 1357, 1367, 1456, 1457, 1467, 1567) and there are 20 of them. - Jon Perry, Jan 20 2004
The dimension of a particular (necessarily existent) absolutely universal embedding of the unitary dual polar space DSU(2n,q^2) where q>2. - J. Taylor (jt_cpp(AT)yahoo.com), Apr 02 2004.
Number of standard tableaux of shape (n+1, 1^n). - Emeric Deutsch, May 13 2004
Erdős, Graham et al. conjectured that a(n) is never squarefree for sufficiently large n (cf. Graham, Knuth, Patashnik, Concrete Math., 2nd ed., Exercise 112). Sárközy showed that if s(n) is the square part of a(n), then s(n) is asymptotically (sqrt(2)-2) * (sqrt(n)) * zeta(1/2). Granville and Ramare proved that the only squarefree values are a(1)=2, a(2)=6 and a(4)=70. - Jonathan Vos Post, Dec 04 2004 [For more about this conjecture, see A261009. - N. J. A. Sloane, Oct 25 2015]
The MathOverflow link contains the following comment (slightly edited): The Erdős squarefree conjecture (that a(n) is never squarefree for n>4) was proved in 1980 by Sárközy, A. (On divisors of binomial coefficients. I. J. Number Theory 20 (1985), no. 1, 70-80.) who showed that the conjecture holds for all sufficiently large values of n, and by A. Granville and O. Ramaré (Explicit bounds on exponential sums and the scarcity of squarefree binomial coefficients. Mathematika 43 (1996), no. 1, 73-107) who showed that it holds for all n>4. - Fedor Petrov, Nov 13 2010. [From N. J. A. Sloane, Oct 29 2015]
p divides a((p-1)/2)-1=A030662(n) for prime p=5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, ... = A002144(n) Pythagorean primes: primes of form 4n+1. - Alexander Adamchuk, Jul 04 2006
The number of direct routes from my home to Granny's when Granny lives n blocks south and n blocks east of my home in Grid City. To obtain a direct route, from the 2n blocks, choose n blocks on which one travels south. For example, a(2)=6 because there are 6 direct routes: SSEE, SESE, SEES, EESS, ESES and ESSE. - Dennis P. Walsh, Oct 27 2006
Inverse: With q = -log(log(16)/(pi a(n)^2)), ceiling((q + log(q))/log(16)) = n. - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Feb 26 2007
Number of partitions with Ferrers diagrams that fit in an n X n box (including the empty partition of 0). Example: a(2) = 6 because we have: empty, 1, 2, 11, 21 and 22. - Emeric Deutsch, Oct 02 2007
So this is the 2-dimensional analog of A008793. - William Entriken, Aug 06 2013
The number of walks of length 2n on an infinite linear lattice that begins and ends at the origin. - Stefan Hollos (stefan(AT)exstrom.com), Dec 10 2007
The number of lattice paths from (0,0) to (n,n) using steps (1,0) and (0,1). - Joerg Arndt, Jul 01 2011
Integral representation: C(2n,n)=1/Pi Integral [(2x)^(2n)/sqrt(1 - x^2),{x,-1, 1}], i.e., C(2n,n)/4^n is the moment of order 2n of the arcsin distribution on the interval (-1,1). - N-E. Fahssi, Jan 02 2008
Also the Catalan transform of A000079. - R. J. Mathar, Nov 06 2008
Straub, Amdeberhan and Moll: "... it is conjectured that there are only finitely many indices n such that C_n is not divisible by any of 3, 5, 7 and 11." - Jonathan Vos Post, Nov 14 2008
Equals INVERT transform of A081696: (1, 1, 3, 9, 29, 97, 333, ...). - Gary W. Adamson, May 15 2009
Also, in sports, the number of ordered ways for a "Best of 2n-1 Series" to progress. For example, a(2) = 6 means there are six ordered ways for a "best of 3" series to progress. If we write A for a win by "team A" and B for a win by "team B" and if we list the played games chronologically from left to right then the six ways are AA, ABA, BAA, BB, BAB, and ABB. (Proof: To generate the a(n) ordered ways: Write down all a(n) ways to designate n of 2n games as won by team A. Remove the maximal suffix of identical letters from each of these.) - Lee A. Newberg, Jun 02 2009
Number of n X n binary arrays with rows, considered as binary numbers, in nondecreasing order, and columns, considered as binary numbers, in nonincreasing order. - R. H. Hardin, Jun 27 2009
Hankel transform is 2^n. - Paul Barry, Aug 05 2009
It appears that a(n) is also the number of quivers in the mutation class of twisted type BC_n for n>=2.
Central terms of Pascal's triangle: a(n) = A007318(2*n,n). - Reinhard Zumkeller, Nov 09 2011
Number of words on {a,b} of length 2n such that no prefix of the word contains more b's than a's. - Jonathan Nilsson, Apr 18 2012
From Pascal's triangle take row(n) with terms in order a1,a2,..a(n) and row(n+1) with terms b1,b2,..b(n), then 2*(a1*b1 + a2*b2 + ... + a(n)*b(n)) to get the terms in this sequence. - J. M. Bergot, Oct 07 2012. For example using rows 4 and 5: 2*(1*(1) + 4*(5) + 6*(10) + 4*(10) + 1*(5)) = 252, the sixth term in this sequence.
Take from Pascal's triangle row(n) with terms b1, b2, ..., b(n+1) and row(n+2) with terms c1, c2, ..., c(n+3) and find the sum b1*c2 + b2*c3 + ... + b(n+1)*c(n+2) to get A000984(n+1). Example using row(3) and row(5) gives sum 1*(5)+3*(10)+3*(10)+1*(5) = 70 = A000984(4). - J. M. Bergot, Oct 31 2012
a(n) == 2 mod n^3 iff n is a prime > 3. (See Mestrovic link, p. 4.) - Gary Detlefs, Feb 16 2013
Conjecture: For any positive integer n, the polynomial sum_{k=0}^n a(k)x^k is irreducible over the field of rational numbers. In general, for any integer m>1 and n>0, the polynomial f_{m,n}(x) = Sum_{k=0..n} (m*k)!/(k!)^m*x^k is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 23 2013
This comment generalizes the comment dated Oct 31 2012 and the second of the sequence's original comments. For j = 1 to n, a(n) = Sum_{k=0..j} C(j,k)* C(2n-j, n-k) = 2*Sum_{k=0..j-1} C(j-1,k)*C(2n-j, n-k). - Charlie Marion, Jun 07 2013
The differences between consecutive terms of the sequence of the quotients between consecutive terms of this sequence form a sequence containing the reciprocals of the triangular numbers. In other words, a(n+1)/a(n)-a(n)/a(n-1) = 2/(n*(n+1)). - Christian Schulz, Jun 08 2013
Number of distinct strings of length 2n using n letters A and n letters B. - Hans Havermann, May 07 2014
From Fung Lam, May 19 2014: (Start)
Expansion of G.f. A(x) = 1/(1+q*x*c(x)), where parameter q is positive or negative (except q=-1), and c(x) is the g.f. of A000108 for Catalan numbers. The case of q=-1 recovers the g.f. of A000108 as xA^2-A+1=0. The present sequence A000984 refers to q=-2. Recurrence: (1+q)*(n+2)*a(n+2) + ((q*q-4*q-4)*n + 2*(q*q-q-1))*a(n+1) - 2*q*q*(2*n+1)*a(n) = 0, a(0)=1, a(1)=-q. Asymptotics: a(n) ~ ((q+2)/(q+1))*(q^2/(-q-1))^n, q<=-3, a(n) ~ (-1)^n*((q+2)/(q+1))*(q^2/(q+1))^n, q>=5, and a(n) ~ -Kq*2^(2*n)/sqrt(Pi*n^3), where the multiplicative constant Kq is given by K1=1/9 (q=1), K2=1/8 (q=2), K3=3/25 (q=3), K4=1/9 (q=4). These formulas apply to existing sequences A126983 (q=1), A126984 (q=2), A126982 (q=3), A126986 (q=4), A126987 (q=5), A127017 (q=6), A127016 (q=7), A126985 (q=8), A127053 (q=9), and to A007854 (q=-3), A076035 (q=-4), A076036 (q=-5), A127628 (q=-6), A126694 (q=-7), A115970 (q=-8). (End)
a(n)*(2^n)^(j-2) equals S(n), where S(n) is the n-th number in the self-convolved sequence which yields the powers of 2^j for all integers j, n>=0. For example, when n=5 and j=4, a(5)=252; 252*(2^5)^(4-2) = 252*1024 = 258048. The self-convolved sequence which yields powers of 16 is {1, 8, 96, 1280, 17920, 258048, ...}; i.e., S(5) = 258048. Note that the convolved sequences will be composed of numbers decreasing from 1 to 0, when j<2 (exception being j=1, where the first two numbers in the sequence are 1 and all others decreasing). - Bob Selcoe, Jul 16 2014
The variance of the n-th difference of a sequence of pairwise uncorrelated random variables each with variance 1. - Liam Patrick Roche, Jun 04 2015
Number of ordered trees with n edges where vertices at level 1 can be of 2 colors. Indeed, the standard decomposition of ordered trees leading to the equation C = 1 + zC^2 (C is the Catalan function), yields this time G = 1 + 2zCG, from where G = 1/sqrt(1-4z). - Emeric Deutsch, Jun 17 2015
Number of monomials of degree at most n in n variables. - Ran Pan, Sep 26 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r, then V(n, 2^n) / Pi = V(n-1, 2^n) * a(n/2) for all even n. - Peter Luschny, Oct 12 2015
a(n) is the number of sets {i1,...,in} of length n such that n >= i1 >= i2 >= ... >= in >= 0. For instance, a(2) = 6 as there are only 6 such sets: (2,2) (2,1) (2,0) (1,1) (1,0) (0,0). - Anton Zakharov, Jul 04 2016
From Ralf Steiner, Apr 07 2017: (Start)
By analytic continuation to the entire complex plane there exist regularized values for divergent sums such as:
Sum_{k>=0} a(k)/(-2)^k = 1/sqrt(3).
Sum_{k>=0} a(k)/(-1)^k = 1/sqrt(5).
Sum_{k>=0} a(k)/(-1/2)^k = 1/3.
Sum_{k>=0} a(k)/(1/2)^k = -1/sqrt(7)i.
Sum_{k>=0} a(k)/(1)^k = -1/sqrt(3)i.
Sum_{k>=0} a(k)/2^k = -i. (End)
Number of sequences (e(1), ..., e(n+1)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) > e(j). [Martinez and Savage, 2.18] - Eric M. Schmidt, Jul 17 2017
The o.g.f. for the sequence equals the diagonal of any of the following the rational functions: 1/(1 - (x + y)), 1/(1 - (x + y*z)), 1/(1 - (x + x*y + y*z)) or 1/(1 - (x + y + y*z)). - Peter Bala, Jan 30 2018
From Colin Defant, Sep 16 2018: (Start)
Let s denote West's stack-sorting map. a(n) is the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 231, and 321. a(n) is also the number of permutations pi of [n+1] such that s(pi) avoids the patterns 132, 312, and 321.
a(n) is the number of permutations of [n+1] that avoid the patterns 1342, 3142, 3412, and 3421. (End)
All binary self-dual codes of length 4n, for n>0, must contain at least a(n) codewords of weight 2n. More to the point, there will always be at least one, perhaps unique, binary self-dual code of length 4n that will contain exactly a(n) codewords that have a hamming weight equal to half the length of the code (2n). This code can be constructed by direct summing the unique binary self-dual code of length 2 (up to permutation equivalence) to itself an even number of times. A permutation equivalent code can be constructed by augmenting two identity matrices of length 2n together. - Nathan J. Russell, Nov 25 2018
From Isaac Saffold, Dec 28 2018: (Start)
Let [b/p] denote the Legendre symbol and 1/b denote the inverse of b mod p. Then, for m and n, where n is not divisible by p,
[(m+n)/p] == [n/p]*Sum_{k=0..(p-1)/2} (-m/(4*n))^k * a(k) (mod p).
Evaluating this identity for m = -1 and n = 1 demonstrates that, for all odd primes p, Sum_{k=0..(p-1)/2} (1/4)^k * a(k) is divisible by p. (End)
Number of vertices of the subgraph of the (2n-1)-dimensional hypercube induced by all bitstrings with n-1 or n many 1s. The middle levels conjecture asserts that this graph has a Hamilton cycle. - Torsten Muetze, Feb 11 2019
a(n) is the number of walks of length 2n from the origin with steps (1,1) and (1,-1) that stay on or above the x-axis. Equivalently, a(n) is the number of walks of length 2n from the origin with steps (1,0) and (0,1) that stay in the first octant. - Alexander Burstein, Dec 24 2019
Number of permutations of length n>0 avoiding the partially ordered pattern (POP) {3>1, 1>2} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is larger than the second element but smaller than the third elements. - Sergey Kitaev, Dec 08 2020
From Gus Wiseman, Jul 21 2021: (Start)
Also the number of integer compositions of 2n+1 with alternating sum 1, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 1 through a(2) = 6 compositions are:
(1) (2,1) (3,2)
(1,1,1) (1,2,2)
(2,2,1)
(1,1,2,1)
(2,1,1,1)
(1,1,1,1,1)
The following relate to these compositions:
- The unordered version is A000070.
- The alternating sum -1 version is counted by A001791, ranked by A345910/A345912.
- The alternating sum 0 version is counted by A088218, ranked by A344619.
- Including even indices gives A126869.
- The complement is counted by A202736.
- Ranked by A345909 (reverse: A345911).
Equivalently, a(n) counts binary numbers with 2n+1 digits and one more 1 than 0's. For example, the a(2) = 6 binary numbers are: 10011, 10101, 10110, 11001, 11010, 11100.
(End)
From Michael Wallner, Jan 25 2022: (Start)
a(n) is the number of nx2 Young tableaux with a single horizontal wall between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021].
Example for a(2)=6:
3 4 2 4 3 4 3|4 4|3 2|4
1|2, 1|3, 2|1, 1 2, 1 2, 1 3
a(n) is also the number of nx2 Young tableaux with n "walls" between the first and second column.
Example for a(2)=6:
3|4 2|4 4|3 3|4 4|3 4|2
1|2, 1|3, 1|2, 2|1, 2|1, 3|1 (End)
From Shel Kaphan, Jan 12 2023: (Start)
a(n)/4^n is the probability that a fair coin tossed 2n times will come up heads exactly n times and tails exactly n times, or that a random walk with steps of +-1 will return to the starting point after 2n steps (not necessarily for the first time). As n becomes large, this number asymptotically approaches 1/sqrt(n*Pi), using Stirling's approximation for n!.
a(n)/(4^n*(2n-1)) is the probability that a random walk with steps of +-1 will return to the starting point for the first time after 2n steps. The absolute value of the n-th term of A144704 is denominator of this fraction.
Considering all possible random walks of exactly 2n steps with steps of +-1, a(n)/(2n-1) is the number of such walks that return to the starting point for the first time after 2n steps. See the absolute values of A002420 or A284016 for these numbers. For comparison, as mentioned by Stefan Hollos, Dec 10 2007, a(n) is the number of such walks that return to the starting point after 2n steps, but not necessarily for the first time. (End)
p divides a((p-1)/2) + 1 for primes p of the form 4*k+3 (A002145). - Jules Beauchamp, Feb 11 2023
Also the size of the shuffle product of two words of length n, such that the union of the two words consist of 2n distinct elements. - Robert C. Lyons, Mar 15 2023
a(n) is the number of vertices of the n-dimensional cyclohedron W_{n+1}. - Jose Bastidas, Mar 25 2025
Consider a stack of pancakes of height n, where the only allowed operation is reversing the top portion of the stack. First, perform a series of reversals of increasing sizes, followed by a series of reversals of decreasing sizes. The number of distinct permutations of the initial stack that can be reached through these operations is a(n). - Thomas Baruchel, May 12 2025

Examples

			G.f.: 1 + 2*x + 6*x^2 + 20*x^3 + 70*x^4 + 252*x^5 + 924*x^6 + ...
For n=2, a(2) = 4!/(2!)^2 = 24/4 = 6, and this is the middle coefficient of the binomial expansion (a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4. - _Michael B. Porter_, Jul 06 2016

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 828.
  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, id. 160.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 575, line -3, with a=b=n.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 101.
  • Emeric Deutsch and Louis W. Shapiro, Seventeen Catalan identities, Bulletin of the Institute of Combinatorics and its Applications, 31 (2001), 31-38.
  • Henry W. Gould, Combinatorial Identities, Morgantown, 1972, (3.66), page 30.
  • Ronald. L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, Second Ed., see Exercise 112.
  • Martin Griffiths, The Backbone of Pascal's Triangle, United Kingdom Mathematics Trust (2008), 3-124.
  • Leonard Lipshitz and A. van der Poorten, "Rational functions, diagonals, automata and arithmetic", in Number Theory, Richard A. Mollin, ed., Walter de Gruyter, Berlin (1990), 339-358.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Cf. A000108, A002420, A002457, A030662, A002144, A135091, A081696, A182400. Differs from A071976 at 10th term.
Bisection of A001405 and of A226302. See also A025565, the same ordered partitions but without all in which are two successive zeros: 11110 (5), 11200 (18), 13000 (2), 40000 (0) and 22000 (1), total 26 and A025565(4)=26.
Cf. A226078, A051924 (first differences).
Row sums of A059481, A008459, A152229, A158815, A205946.
Cf. A258290 (arithmetic derivative). Cf. A098616, A214377.
See A261009 for a conjecture about this sequence.
Cf. A046521 (first column).
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Cf. A000346, A001700, A001791, A008549, A032443, A073016, A097805, A126869.
Cf. A003418, A056040, A001316, A261130, A356637.

Programs

  • GAP
    List([1..1000], n -> Binomial(2*n,n)); # Muniru A Asiru, Jan 30 2018
  • Haskell
    a000984 n = a007318_row (2*n) !! n  -- Reinhard Zumkeller, Nov 09 2011
  • Magma
    a:= func< n | Binomial(2*n,n) >; [ a(n) : n in [0..10]];
  • Maple
    A000984 := n-> binomial(2*n,n); seq(A000984(n), n=0..30);
with(combstruct); [seq(count([S,{S=Prod(Set(Z,card=i),Set(Z,card=i))}, labeled], size=(2*i)), i=0..20)];
with(combstruct); [seq(count([S,{S=Sequence(Union(Arch,Arch)), Arch=Prod(Epsilon, Sequence(Arch),Z)},unlabeled],size=i), i=0..25)];
with(combstruct):bin := {B=Union(Z,Prod(B,B))}: seq (count([B,bin,unlabeled],size=n)*n, n=1..25); # Zerinvary Lajos, Dec 05 2007
A000984List := proc(m) local A, P, n; A := [1,2]; P := [1];
for n from 1 to m - 2 do P := ListTools:-PartialSums([op(P), 2*P[-1]]);
A := [op(A), 2*P[-1]] od; A end: A000984List(28); # Peter Luschny, Mar 24 2022
  • Mathematica
    Table[Binomial[2n, n], {n, 0, 24}] (* Alonso del Arte, Nov 10 2005 *)
CoefficientList[Series[1/Sqrt[1-4x],{x,0,25}],x]  (* Harvey P. Dale, Mar 14 2011 *)
  • Maxima
    A000984(n):=(2*n)!/(n!)^2$ makelist(A000984(n),n,0,30); /* Martin Ettl, Oct 22 2012 */
  • PARI
    A000984(n)=binomial(2*n,n) \\ much more efficient than (2n)!/n!^2. \\ M. F. Hasler, Feb 26 2014
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=prodeuler(p=2,2*n,p^(fv(2*n,p)-2*fv(n,p))) \\ Charles R Greathouse IV, Aug 21 2013
  • PARI
    fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=my(s=1);forprime(p=2,2*n,s*=p^(fv(2*n,p)-2*fv(n,p)));s \\ Charles R Greathouse IV, Aug 21 2013
  • Python
    from _future_ import division
A000984_list, b = [1], 1
for n in range(10**3):
    b = b*(4*n+2)//(n+1)
    A000984_list.append(b) # Chai Wah Wu, Mar 04 2016

Formula

a(n)/(n+1) = A000108(n), the Catalan numbers.
G.f.: A(x) = (1 - 4*x)^(-1/2) = 1F0(1/2;;4x).
a(n+1) = 2*A001700(n) = A030662(n) + 1. a(2*n) = A001448(n), a(2*n+1) = 2*A002458(n) =A099976.
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1)=0.
a(n) = 2^n/n! * Product_{k=0..n-1} (2*k+1).
a(n) = a(n-1)*(4-2/n) = Product_{k=1..n} (4-2/k) = 4*a(n-1) + A002420(n) = A000142(2*n)/(A000142(n)^2) = A001813(n)/A000142(n) = sqrt(A002894(n)) = A010050(n)/A001044(n) = (n+1)*A000108(n) = -A005408(n-1)*A002420(n). - Henry Bottomley, Nov 10 2000
Using Stirling's formula in A000142 it is easy to get the asymptotic expression a(n) ~ 4^n / sqrt(Pi * n). - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
Integral representation as n-th moment of a positive function on the interval [0, 4]: a(n) = Integral_{x=0..4}(x^n*((x*(4-x))^(-1/2))/Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Sep 17 2001
Sum_{n>=1} 1/a(n) = (2*Pi*sqrt(3) + 9)/27. [Lehmer 1985, eq. (15)] - Benoit Cloitre, May 01 2002 (= A073016. - Bernard Schott, Jul 20 2022)
a(n) = Max_{ (i+j)!/(i!j!) | 0<=i,j<=n }. - Benoit Cloitre, May 30 2002
a(n) = Sum_{k=0..n} binomial(n+k-1,k), row sums of A059481. - Vladeta Jovovic, Aug 28 2002
E.g.f.: exp(2*x)*I_0(2x), where I_0 is Bessel function. - Michael Somos, Sep 08 2002
E.g.f.: I_0(2*x) = Sum a(n)*x^(2*n)/(2*n)!, where I_0 is Bessel function. - Michael Somos, Sep 09 2002
a(n) = Sum_{k=0..n} binomial(n, k)^2. - Benoit Cloitre, Jan 31 2003
Determinant of n X n matrix M(i, j) = binomial(n+i, j). - Benoit Cloitre, Aug 28 2003
Given m = C(2*n, n), let f be the inverse function, so that f(m) = n. Letting q denote -log(log(16)/(m^2*Pi)), we have f(m) = ceiling( (q + log(q)) / log(16) ). - David W. Cantrell (DWCantrell(AT)sigmaxi.net), Oct 30 2003
a(n) = 2*Sum_{k=0..(n-1)} a(k)*a(n-k+1)/(k+1). - Philippe Deléham, Jan 01 2004
a(n+1) = Sum_{j=n..n*2+1} binomial(j, n). E.g., a(4) = C(7,3) + C(6,3) + C(5,3) + C(4,3) + C(3,3) = 35 + 20 + 10 + 4 + 1 = 70. - Jon Perry, Jan 20 2004
a(n) = (-1)^(n)*Sum_{j=0..(2*n)} (-1)^j*binomial(2*n, j)^2. - Helena Verrill (verrill(AT)math.lsu.edu), Jul 12 2004
a(n) = Sum_{k=0..n} binomial(2n+1, k)*sin((2n-2k+1)*Pi/2). - Paul Barry, Nov 02 2004
a(n-1) = (1/2)*(-1)^n*Sum_{0<=i, j<=n}(-1)^(i+j)*binomial(2n, i+j). - Benoit Cloitre, Jun 18 2005
a(n) = C(2n, n-1) + C(n) = A001791(n) + A000108(n). - Lekraj Beedassy, Aug 02 2005
G.f.: c(x)^2/(2*c(x)-c(x)^2) where c(x) is the g.f. of A000108. - Paul Barry, Feb 03 2006
a(n) = A006480(n) / A005809(n). - Zerinvary Lajos, Jun 28 2007
a(n) = Sum_{k=0..n} A106566(n,k)*2^k. - Philippe Deléham, Aug 25 2007
a(n) = Sum_{k>=0} A039599(n, k). a(n) = Sum_{k>=0} A050165(n, k). a(n) = Sum_{k>=0} A059365(n, k)*2^k, n>0. a(n+1) = Sum_{k>=0} A009766(n, k)*2^(n-k+1). - Philippe Deléham, Jan 01 2004
a(n) = 4^n*Sum_{k=0..n} C(n,k)(-4)^(-k)*A000108(n+k). - Paul Barry, Oct 18 2007
a(n) = Sum_{k=0..n} A039598(n,k)*A059841(k). - Philippe Deléham, Nov 12 2008
A007814(a(n)) = A000120(n). - Vladimir Shevelev, Jul 20 2009
From Paul Barry, Aug 05 2009: (Start)
G.f.: 1/(1-2x-2x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-... (continued fraction);
G.f.: 1/(1-2x/(1-x/(1-x/(1-x/(1-... (continued fraction). (End)
If n>=3 is prime, then a(n) == 2 (mod 2*n). - Vladimir Shevelev, Sep 05 2010
Let A(x) be the g.f. and B(x) = A(-x), then B(x) = sqrt(1-4*x*B(x)^2). - Vladimir Kruchinin, Jan 16 2011
a(n) = (-4)^n*sqrt(Pi)/(gamma((1/2-n))*gamma(1+n)). - Gerry Martens, May 03 2011
a(n) = upper left term in M^n, M = the infinite square production matrix:
2, 2, 0, 0, 0, 0, ...
1, 1, 1, 0, 0, 0, ...
1, 1, 1, 1, 0, 0, ...
1, 1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, 1, .... - Gary W. Adamson, Jul 14 2011
a(n) = Hypergeometric([-n,-n],[1],1). - Peter Luschny, Nov 01 2011
E.g.f.: hypergeometric([1/2],[1],4*x). - Wolfdieter Lang, Jan 13 2012
a(n) = 2*Sum_{k=0..n-1} a(k)*A000108(n-k-1). - Alzhekeyev Ascar M, Mar 09 2012
G.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 2*(2*k+1)*x + (k+1) - 2*(k+1)*(2*k+3)*x/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Jun 28 2012
a(n) = Sum_{k=0..n} binomial(n,k)^2*H(k)/(2*H(n)-H(2*n)), n>0, where H(n) is the n-th harmonic number. - Gary Detlefs, Mar 19 2013
G.f.: Q(0)*(1-4*x), where Q(k) = 1 + 4*(2*k+1)*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 2*x*(2*k+1)/(2*x*(2*k+1) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 24 2013
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - 2*x/(2*x + (k+1)^2/(2*k+1)/E(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
Special values of Jacobi polynomials, in Maple notation: a(n) = 4^n*JacobiP(n,0,-1/2-n,-1). - Karol A. Penson, Jul 27 2013
a(n) = 2^(4*n)/((2*n+1)*Sum_{k=0..n} (-1)^k*C(2*n+1,n-k)/(2*k+1)). - Mircea Merca, Nov 12 2013
a(n) = C(2*n-1,n-1)*C(4*n^2,2)/(3*n*C(2*n+1,3)), n>0. - Gary Detlefs, Jan 02 2014
Sum_{n>=0} a(n)/n! = A234846. - Richard R. Forberg, Feb 10 2014
0 = a(n)*(16*a(n+1) - 6*a(n+2)) + a(n+1)*(-2*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 17 2014
a(n+1) = 4*a(n) - 2*A000108(n). Also a(n) = 4^n*Product_{k=1..n}(1-1/(2*k)). - Stanislav Sykora, Aug 09 2014
G.f.: Sum_{n>=0} x^n/(1-x)^(2*n+1) * Sum_{k=0..n} C(n,k)^2 * x^k. - Paul D. Hanna, Nov 08 2014
a(n) = (-4)^n*binomial(-1/2,n). - Jean-François Alcover, Feb 10 2015
a(n) = 4^n*hypergeom([-n,1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(n-k,k)*2^(n-2*k). - Robert FERREOL, Aug 29 2015
a(n) ~ 4^n*(2-2/(8*n+2)^2+21/(8*n+2)^4-671/(8*n+2)^6+45081/(8*n+2)^8)/sqrt((4*n+1) *Pi). - Peter Luschny, Oct 14 2015
A(-x) = 1/x * series reversion( x*(2*x + sqrt(1 + 4*x^2)) ). Compare with the o.g.f. B(x) of A098616, which satisfies B(-x) = 1/x * series reversion( x*(2*x + sqrt(1 - 4*x^2)) ). See also A214377. - Peter Bala, Oct 19 2015
a(n) = GegenbauerC(n,-n,-1). - Peter Luschny, May 07 2016
a(n) = gamma(1+2*n)/gamma(1+n)^2. - Andres Cicuttin, May 30 2016
Sum_{n>=0} (-1)^n/a(n) = 4*(5 - sqrt(5)*log(phi))/25 = 0.6278364236143983844442267..., where phi is the golden ratio. - Ilya Gutkovskiy, Jul 04 2016
From Peter Bala, Jul 22 2016: (Start)
This sequence occurs as the closed-form expression for several binomial sums:
a(n) = Sum_{k = 0..2*n} (-1)^(n+k)*binomial(2*n,k)*binomial(2*n + 1,k).
a(n) = 2*Sum_{k = 0..2*n-1} (-1)^(n+k)*binomial(2*n - 1,k)*binomial(2*n,k) for n >= 1.
a(n) = 2*Sum_{k = 0..n-1} binomial(n - 1,k)*binomial(n,k) for n >= 1.
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y + k,n) = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x - k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... both Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y + k,n) and Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x - k,n)*binomial(y - k,n) appear to equal Kronecker's delta(n,0).
a(n) = (-1)^n*Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)*binomial(x + k,n)*binomial(y - k,n) for arbitrary x and y.
For m = 3,4,5,... Sum_{k = 0..m*n} (-1)^k*binomial(m*n,k)*binomial(x + k,n)*binomial(y - k,n) appears to equal Kronecker's delta(n,0).
a(n) = Sum_{k = 0..2n} (-1)^k*binomial(2*n,k)*binomial(3*n - k,n)^2 = Sum_{k = 0..2*n} (-1)^k*binomial(2*n,k)* binomial(n + k,n)^2. (Gould, Vol. 7, 5.23).
a(n) = Sum_{k = 0..n} (-1)^(n+k)*binomial(2*n,n + k)*binomial(n + k,n)^2. (End)
From Ralf Steiner, Apr 07 2017: (Start)
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for q in N, p in Z/{-4q< (some p) <-2}.
...
Sum_{k>=0} a(k)/(-4)^k = 1/sqrt(2).
Sum_{k>=0} a(k)/(17/4)^k = sqrt(17).
Sum_{k>=0} a(k)/(18/4)^k = 3.
Sum_{k>=0} a(k)/5^k = sqrt(5).
Sum_{k>=0} a(k)/6^k = sqrt(3).
Sum_{k>=0} a(k)/8^k = sqrt(2).
...
Sum_{k>=0} a(k)/(p/q)^k = sqrt(p/(p-4q)) for p>4q.(End)
Boas-Buck recurrence: a(n) = (2/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n, 0). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = Sum_{k = 0..n} (-1)^(n-k) * binomial(2*n+1, k) for n in N. - Rene Adad, Sep 30 2017
a(n) = A034870(n,n). - Franck Maminirina Ramaharo, Nov 26 2018
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*(sqrt(4-x) + sqrt(x)*arcsin(sqrt(x)/2)) / (4-x)^(3/2).
E.g.f. for {1/a(n)}: 1 + exp(x/4)*sqrt(Pi*x)*erf(sqrt(x)/2)/2.
Sum_{n>=0} (-1)^n/a(n) = 4*(1/5 - arcsinh(1/2)/(5*sqrt(5))). (End)
From Peter Luschny, Sep 08 2022: (Start)
a(n) = 2^(2*n)*Product_{k=1..2*n} k^((-1)^(k+1)) = A056040(2*n).
a(n) = A001316(n) * A356637(n) * A261130(n) for n >= 2. (End)
a(n) = 4^n*binomial(n-1/2,-1/2) = 4^n*GegenbauerC(n,1/4,1). - Gerry Martens, Oct 19 2022
Occurs on the right-hand side of the binomial sum identities Sum_{k = -n..n} (-1)^k * (n + x - k) * binomial(2*n, n+k)^2 = (x + n)*a(n) and Sum_{k = -n..n} (-1)^k * (n + x - k)^2 * binomial(2*n, n+k)^3 = x*(x + 2*n)*a(n) (x arbitrary). Compare with the identity: Sum_{k = -n..n} (-1)^k * binomial(2*n, n+k)^2 = a(n). - Peter Bala, Jul 31 2023
From Peter Bala, Mar 31 2024: (Start)
4^n*a(n) = Sum_{k = 0..2*n} (-1)^k*a(k)*a(2*n-k).
16^n = Sum_{k = 0..2*n} a(k)*a(2*n-k). (End)
From Gary Detlefs, May 28 2024: (Start)
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*binomial(2*k,k)*2^(n-2*k). (H. W. Gould) - Gary Detlefs, May 28 2024
a(n) = Sum_{k=0..2*n} (-1)^k*binomial(2n,k)*binomial(2*n+2*k,n+k)*3^(2*n-k). (H. W. Gould) (End)
a(n) = Product_{k>=n+1} k^2/(k^2 - n^2). - Antonio Graciá Llorente, Sep 08 2024
a(n) = Product_{k=1..n} A003418(floor(2*n/k))^((-1)^(k+1)) (Golomb, 2003). - Amiram Eldar, Aug 08 2025

A001045 Jacobsthal sequence (or Jacobsthal numbers): a(n) = a(n-1) + 2*a(n-2), with a(0) = 0, a(1) = 1; also a(n) = nearest integer to 2^n/3.

Original entry on oeis.org

0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, 1365, 2731, 5461, 10923, 21845, 43691, 87381, 174763, 349525, 699051, 1398101, 2796203, 5592405, 11184811, 22369621, 44739243, 89478485, 178956971, 357913941, 715827883, 1431655765, 2863311531, 5726623061, 11453246123
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Don Knuth points out (personal communication) that Jacobsthal may never have seen the actual values of this sequence. However, Horadam uses the name "Jacobsthal sequence", such an important sequence needs a name, and there is a law that says the name for something should never be that of its discoverer. - N. J. A. Sloane, Dec 26 2020
Number of ways to tile a 3 X (n-1) rectangle with 1 X 1 and 2 X 2 square tiles.
Also, number of ways to tile a 2 X (n-1) rectangle with 1 X 2 dominoes and 2 X 2 squares. - Toby Gottfried, Nov 02 2008
Also a(n) counts each of the following four things: n-ary quasigroups of order 3 with automorphism group of order 3, n-ary quasigroups of order 3 with automorphism group of order 6, (n-1)-ary quasigroups of order 3 with automorphism group of order 2 and (n-2)-ary quasigroups of order 3. See the McKay-Wanless (2008) paper. - Ian Wanless, Apr 28 2008
Also the number of ways to tie a necktie using n + 2 turns. So three turns make an "oriental", four make a "four in hand" and for 5 turns there are 3 methods: "Kelvin", "Nicky" and "Pratt". The formula also arises from a special random walk on a triangular grid with side conditions (see Fink and Mao, 1999). - arne.ring(AT)epost.de, Mar 18 2001
Also the number of compositions of n + 1 ending with an odd part (a(2) = 3 because 3, 21, 111 are the only compositions of 3 ending with an odd part). Also the number of compositions of n + 2 ending with an even part (a(2) = 3 because 4, 22, 112 are the only compositions of 4 ending with an even part). - Emeric Deutsch, May 08 2001
Arises in study of sorting by merge insertions and in analysis of a method for computing GCDs - see Knuth reference.
Number of perfect matchings of a 2 X n grid upon replacing unit squares with tetrahedra (C_4 to K_4):
o----o----o----o...
| \/ | \/ | \/ |
| /\ | /\ | /\ |
o----o----o----o... - Roberto E. Martinez II, Jan 07 2002
Also the numerators of the reduced fractions in the alternating sum 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 + ... - Joshua Zucker, Feb 07 2002
Also, if A(n), B(n), C(n) are the angles of the n-orthic triangle of ABC then A(1) = Pi - 2*A, A(n) = s(n)*Pi + (-2)^n*A where s(n) = (-1)^(n-1) * a(n) [1-orthic triangle = the orthic triangle of ABC, n-orthic triangle = the orthic triangle of the (n-1)-orthic triangle]. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Jun 05 2002
Also the number of words of length n+1 in the two letters s and t that reduce to the identity 1 by using the relations sss = 1, tt = 1 and stst = 1. The generators s and t and the three stated relations generate the group S3. - John W. Layman, Jun 14 2002
Sums of pairs of consecutive terms give all powers of 2 in increasing order. - Amarnath Murthy, Aug 15 2002
Excess clockwise moves (over counterclockwise) needed to move a tower of size n to the clockwise peg is -(-1)^n*(2^n - (-1)^n)/3; a(n) is its unsigned version. - Wouter Meeussen, Sep 01 2002
Also the absolute value of the number represented in base -2 by the string of n 1's, the negabinary repunit. The Mersenne numbers (A000225 and its subsequences) are the binary repunits. - Rick L. Shepherd, Sep 16 2002
Note that 3*a(n) + (-1)^n = 2^n is significant for Pascal's triangle A007318. It arises from a Jacobsthal decomposition of Pascal's triangle illustrated by 1 + 7 + 21 + 35 + 35 + 21 + 7 + 1 = (7 + 35 + 1) + (1 + 35 + 7) + (21 + 21) = 43 + 43 + 42 = 3a(7) - 1; 1 + 8 + 28 + 56 + 70 + 56 + 28 + 8 + 1 = (1 + 56 + 28) + (28 + 56 + 1) + (8 + 70 + 8) = 85 + 85 + 86 = 3a(8)+1. - Paul Barry, Feb 20 2003
Number of positive integers requiring exactly n signed bits in the nonadjacent form representation.
Equivalently, number of length-(n-1) words with letters {0, 1, 2} where no two consecutive letters are nonzero, see example and fxtbook link. - Joerg Arndt, Nov 10 2012
Counts walks between adjacent vertices of a triangle. - Paul Barry, Nov 17 2003
Every amphichiral rational knot written in Conway notation is a palindromic sequence of numbers, not beginning or ending with 1. For example, for 4 <= n <= 12, the amphichiral rational knots are: 2 2, 2 1 1 2, 4 4, 3 1 1 3, 2 2 2 2, 4 1 1 4, 3 1 1 1 1 3, 2 3 3 2, 2 1 2 2 1 2, 2 1 1 1 1 1 1 2, 6 6, 5 1 1 5, 4 2 2 4, 3 3 3 3, 2 4 4 2, 3 2 1 1 2 3, 3 1 2 2 1 3, 2 2 2 2 2 2, 2 2 1 1 1 1 2 2, 2 1 2 1 1 2 1 2, 2 1 1 1 1 1 1 1 1 2. For the number of amphichiral rational knots for n=2*k (k=1, 2, 3, ...), we obtain the sequence 0, 1, 1, 3, 5, 11, 21, 43, 85, 171, 341, 683, ... - Slavik Jablan, Dec 26 2003
a(n+2) counts the binary sequences of total length n made up of codewords from C = {0, 10, 11}. - Paul Barry, Jan 23 2004
Number of permutations with no fixed points avoiding 231 and 132.
The n-th entry (n > 1) of the sequence is equal to the 2,2-entry of the n-th power of the unnormalized 4 X 4 Haar matrix: [1 1 1 0 / 1 1 -1 0 / 1 1 0 1 / 1 1 0 -1]. - Simone Severini, Oct 27 2004
a(n) is the number of Motzkin (n+1)-sequences whose flatsteps all occur at level 1 and whose height is less than or equal to 2. For example, a(4) = 5 counts UDUFD, UFDUD, UFFFD, UFUDD, UUDFD. - David Callan, Dec 09 2004
a(n+1) gives row sums of A059260. - Paul Barry, Jan 26 2005
If (m + n) is odd, then 3*(a(m) + a(n)) is always of the form a^2 + 2*b^2, where a and b both equal powers of 2; consequently every factor of (a(m) + a(n)) is always of the form a^2 + 2*b^2. - Matthew Vandermast, Jul 12 2003
Number of "0,0" in f_{n+1}, where f_0 = "1" and f_{n+1} = a sequence formed by changing all "1"s in f_n to "1,0" and all "0"s in f_n to "0,1". - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Sep 22 2006
All prime Jacobsthal numbers A049883[n] = {3, 5, 11, 43, 683, 2731, 43691, ...} have prime indices except for a(4) = 5. All prime Jacobsthal numbers with prime indices (all but a(4) = 5) are of the form (2^p + 1)/3 - the Wagstaff primes A000979[n]. Indices of prime Jacobsthal numbers are listed in A107036[n] = {3, 4, 5, 7, 11, 13, 17, 19, 23, 31, 43, 61, ...}. For n>1 A107036[n] = A000978[n] Numbers n such that (2^n + 1)/3 is prime. - Alexander Adamchuk, Oct 03 2006
Correspondence: a(n) = b(n)*2^(n-1), where b(n) is the sequence of the arithmetic means of previous two terms defined by b(n) = 1/2*(b(n-1) + b(n-2)) with initial values b(0) = 0, b(1) = 1; the g.f. for b(n) is B(x) := x/(1-(x^1+x^2)/2), so the g.f. A(x) for a(n) satisfies A(x) = B(2*x)/2. Because b(n) converges to the limit lim (1-x)*B(x) = 1/3*(b(0) + 2*b(1)) = 2/3 (for x --> 1), it follows that a(n)/2^(n-1) also converges to 2/3 (see also A103770). - Hieronymus Fischer, Feb 04 2006
Inverse: floor(log_2(a(n))) = n - 2 for n >= 2. Also: log_2(a(n) + a(n-1)) = n - 1 for n >= 1 (see also A130249). Characterization: x is a Jacobsthal number if and only if there is a power of 4 (= c) such that x is a root of p(x) = 9*x*(x-c) + (c-1)*(2*c+1) (see also the indicator sequence A105348). - Hieronymus Fischer, May 17 2007
This sequence counts the odd coefficients in the expansion of (1 + x + x^2)^(2^n - 1), n >= 0. - Tewodros Amdeberhan (tewodros(AT)math.mit.edu), Oct 18 2007, Jan 08 2008
2^(n+1) = 2*A005578(n) + 2*a(n) + 2*A000975(n-1). Let A005578(n), a(n), A000975(n-1) = triangle (a, b, c). Then ((S-c), (S-b), (S-a)) = (A005578(n-1), a(n-1), A000975(n-2)). Example: (a, b, c) = (11, 11, 10) = (A005578(5), a(5), A000975(4)). Then ((S-c), (S-b), (S-a)) = (6, 5, 5) = (A005578(4), a(4), A000975(3)). - Gary W. Adamson, Dec 24 2007
Sequence is identical to the absolute values of its inverse binomial transform. A similar result holds for [0,A001045*2^n]. - Paul Curtz, Jan 17 2008
From a(2) on (i.e., 1, 3, 5, 11, 21, ...) also: least odd number such that the subsets of {a(2), ..., a(n)} sum to 2^(n-1) different values, cf. A138000 and A064934. It is interesting to note the pattern of numbers occurring (or not occurring) as such a sum (A003158). - M. F. Hasler, Apr 09 2008
a(n) is the term (5, 1) of n-th power of the 5 X 5 matrix shown in A121231. - Gary W. Adamson, Oct 03 2008
A147612(a(n)) = 1. - Reinhard Zumkeller, Nov 08 2008
a(n+1) = Sum(A153778(i): 2^n <= i < 2^(n+1)). - Reinhard Zumkeller, Jan 01 2009
It appears that a(n) is also the number of integers between 2^n and 2^(n+1) that are divisible by 3 with no remainder. - John Fossaceca (john(AT)fossaceca.net), Jan 31 2009
Number of pairs of consecutive odious (or evil) numbers between 2^(n+1) and 2^(n+2), inclusive. - T. D. Noe, Feb 05 2009
Equals eigensequence of triangle A156319. - Gary W. Adamson, Feb 07 2009
A three-dimensional interpretation of a(n+1) is that it gives the number of ways of filling a 2 X 2 X n hole with 1 X 2 X 2 bricks. - Martin Griffiths, Mar 28 2009
Starting with offset 1 = INVERTi transform of A002605: (1, 2, 6, 16, 44, ...). - Gary W. Adamson, May 12 2009
Convolved with (1, 2, 2, 2, ...) = A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 23 2009
The product of a pair of successive terms is always a triangular number. - Giuseppe Ottonello, Jun 14 2009
Let A be the Hessenberg matrix of order n, defined by: A[1, j] = 1, A[i, i] := -2, A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 1, a(n) = (-1)^(n-1)*det(A). - Milan Janjic, Jan 26 2010
Let R denote the irreducible representation of the symmetric group S_3 of dimension 2, and let s and t denote respectively the sign and trivial irreducible representations of dimension 1. The decomposition of R^n into irreducible representations consists of a(n) copies of R and a(n-1) copies of each of s and t. - Andrew Rupinski, Mar 12 2010
As a fraction: 1/88 = 0.0113636363... or 1/9898 = 0.00010103051121... - Mark Dols, May 18 2010
Starting with "1" = the INVERT transform of (1, 0, 2, 0, 4, 0, 8, ...); e.g., a(7) = 43 = (1, 1, 1, 3, 5, 11, 21) dot (8, 0, 4, 0, 2, 0, 1) = (8 + 4 + 10 + 21) = 43. - Gary W. Adamson, Oct 28 2010
Rule 28 elementary cellular automaton (A266508) generates this sequence. - Paul Muljadi, Jan 27 2011
This is a divisibility sequence. - Michael Somos, Feb 06 2011
From L. Edson Jeffery, Apr 04 2011: (Start)
Let U be the unit-primitive matrix (see [Jeffery])
U = U_(6,2) =
(0 0 1)
(0 2 0)
(2 0 1).
Then a(n+1) = (Trace(U^n))/3, a(n+1) = ((U^n){3, 3})/3, a(n) = ((U^n){1, 3})/3 and a(n) = ((U^(n+1))_{1, 1})/2. (End)
The sequence emerges in using iterated deletion of strictly dominated strategies to establish the best-response solution to the Cournot duopoly problem as a strictly dominant strategy. The best response of firm 1 to firm 2's chosen quantity is given by q*1 = 1/2*(a - c - q_2), where a is reservation price, c is marginal cost, and q_2 is firm two's chosen quantity. Given that q_2 is in [o, a - c], q*_1 must be in [o, 1/2*(a - c)]. Since costs are symmetric, we know q_2 is in [0, 1/2*(a - c)]. Then we know q*_1 is in [1/4*(a - c), 1/2*(a - c)]. Continuing in this way, the sequence of bounds we get (factoring out a - c) is {1/2, 1/4, 3/8, 5/16, ...}; the numerators are the Jacobsthal numbers. - _Michael Chirico, Sep 10 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n >= 2, 3*a(n-1) equals the number of 3-colored compositions of n with all parts greater than or equal to 2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
This sequence is connected with the Collatz problem. We consider the array T(i, j) where the i-th row gives the parity trajectory of i, for example for i = 6, the infinite trajectory is 6 -> 3 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1... and T(6, j) = [0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, ..., 1, 0, 0, 1, ...]. Now, we consider the sum of the digits "1" of each column. We obtain the sequence a(n) = Sum_{k = 1..2^n} T(k, n) = Sum {k = 1..2^n} digits "1" of the n-th column. Because a(n) + a(n+1) = 2^n, then a(n+1) = Number of digits "0" among the 2^n elements of the n-th column. - _Michel Lagneau, Jan 11 2012
3!*a(n-1) is apparently the trace of the n-th power of the adjacency matrix of the complete 3-graph, a 3 X 3 matrix with diagonal elements all zero and off-diagonal all ones. The off-diagonal elements for the n-th power are all equal to a(n) while each diagonal element seems to be a(n) + 1 for an even power and a(n) - 1 for an odd. These are related to the lengths of closed paths on the graph (see Delfino and Viti's paper). - Tom Copeland, Nov 06 2012
From Paul Curtz, Dec 11 2012: (Start)
2^n * a(-n) = (-1)^(n-1) * a(n), which extends the sequence to negative indices: ..., -5/16, 3/8, -1/4, 1/2, 0, 1, 1, 3, 5, ...
The "autosequence" property with respect to the binomial transform mentioned in my comment of Jan 17 2008 is still valid if the term a(-1) is added to the array of the sequence and its iterated higher-order differences in subsequent rows:
0 1/2 1/2 3/2 5/2 11/2 ...
1/2 0 1 1 3 5 ...
-1/2 1 0 2 2 6 ...
3/2 -1 2 0 4 4 ...
-5/2 3 -2 4 0 8 ...
11/2 -5 6 -4 8 0 ...
The main diagonal in this array contains 0's. (End)
Assign to a triangle T(n, 0) = 1 and T(n+1, 1) = n; T(r, c) = T(r-1, c-1) + T(r-1, c-2) + T(r-2, c-2). Then T(n+1, n) - T(n, n) = a(n). - J. M. Bergot, May 02 2013
a(n+1) counts clockwise walks on n points on a circle that take steps of length 1 and 2, return to the starting point after two full circuits, and do not duplicate any steps (USAMO 2013, problem 5). - Kiran S. Kedlaya, May 11 2013
Define an infinite square array m by m(n, 0) = m(0, n) = a(n) in top row and left column and m(i, j) = m(i, j-1) + m(i-1, j-1) otherwise, then m(n+1, n+1) = 3^(n-1). - J. M. Bergot, May 10 2013
a(n) is the number of compositions (ordered partitions) of n - 1 into one sort of 1's and two sorts of 2's. Example: the a(4) = 5 compositions of 3 are 1 + 1 + 1, 1 + 2, 1 + 2', 2 + 1 and 2' + 1. - Bob Selcoe, Jun 24 2013
Without 0, a(n)/2^n equals the probability that n will occur as a partial sum in a randomly-generated infinite sequence of 1's and 2's. The limiting ratio is 2/3. - Bob Selcoe, Jul 04 2013
Number of conjugacy classes of Z/2Z X Z/2Z in GL(2,2^(n+1)). - Jared Warner, Aug 18 2013
a(n) is the top left entry of the (n-1)-st power of the 3 X 3 matrix [1, 1, 1, 1, 0, 0, 1, 0, 0]. a(n) is the top left entry of the (n+1)-st power of any of the six 3 X 3 matrices [0, 1, 0; 1, 1, 1; 0, 1, 0], [0, 1, 1; 0, 1, 1; 1, 1, 0], [0, 0, 1; 1, 1, 1; 1, 1, 0], [0, 1, 1; 1, 0, 1; 0, 1, 1], [0, 0, 1; 0, 0, 1; 1, 1, 1] or [0, 1, 0; 1, 0, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014
This is the only integer sequence from the family of homogeneous linear recurrence of order 2 given by a(n) = k*a(n-1) + t*a(n-2) with positive integer coefficients k and t and initial values a(0) = 0 and a(1) = 1 whose ratio a(n+1)/a(n) converges to 2 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is the Lucas sequence U(1, -2). - Felix P. Muga II, Mar 21 2014
sqrt(a(n+1) * a(n-1)) -> a(n) + 3/4 if n is even, and -> a(n) - 3/4 if n is odd, for n >= 2. - Richard R. Forberg, Jun 24 2014
a(n+1) counts closed walks on the end vertices of P_3 containing one loop at the middle vertex. a(n-1) counts closed walks on the middle vertex of P_3 containing one loop at that vertex. - David Neil McGrath, Nov 07 2014
From César Eliud Lozada, Jan 21 2015: (Start)
Let P be a point in the plane of a triangle ABC (with sides a, b, c) and barycentric coordinates P = [x:y:z]. The Complement of P with respect to ABC is defined to be Complement(P) = [b*y + c*z : c*z + a*x : a*x + b*y].
Then, for n >= 1, Complement(Complement(...(Complement(P))..)) = (n times) =
[2*a(n-1)*a*x + (2*a(n-1) - (-1)^n)*(b*y + c*z):
2*a(n-1)*b*y + (2*a(n-1) - (-1)^n)*(c*z + a*x):
2*a(n-1)*c*z + (2*a(n-1) - (-1)^n)*(a*x + b*y)]. (End)
a(n) (n >= 2) is the number of induced hypercubes of the Fibonacci cube Gamma(n-2). See p. 513 of the Klavzar reference. Example: a(5) = 11. Indeed, the Fibonacci cube Gamma(3) is <>- (cycle C(4) with a pendant edge) and the hypercubes are: 5 vertices, 5 edges, and 1 square. - Emeric Deutsch, Apr 07 2016
If the sequence of points {P_i(x_i, y_i)} on the cubic y = a*x^3 + b*x^2 + c*x + d has the property that the segment P_i(x_i, y_i) P_i+1(x_i+1, y_i+1) is always tangent to the cubic at P_i+1(x_i+1, y_i+1) then a(n) = -2^n*a/b*(x_(n+1)-(-1/2)^n*x_1). - Michael Brozinsky, Aug 01 2016
With the quantum integers defined by [n+1]A000225%20are%20given%20by%20q%20=%20sqrt(2).%20Cf.%20A239473.%20-%20_Tom%20Copeland">q = (q^(n+1) - q^(-n-1)) / (q - q^(-1)), the Jacobsthal numbers are a(n+1) = (-1)^n*q^n [n+1]_q with q = i * sqrt(2) for i^2 = -1, whereas the signed Mersenne numbers A000225 are given by q = sqrt(2). Cf. A239473. - _Tom Copeland, Sep 05 2016
Every positive integer has a unique expression as a sum of Jacobsthal numbers in which the index of the smallest summand is odd, with a(1) and a(2) both allowed. See the L. Carlitz, R. Scoville, and V. E. Hoggatt, Jr. reference. - Ira M. Gessel, Dec 31 2016. See A280049 for these expansions. - N. J. A. Sloane, Dec 31 2016
For n > 0, a(n) equals the number of ternary words of length n-1 in which 0 and 1 avoid runs of odd lengths. - Milan Janjic, Jan 08 2017
For n > 0, a(n) equals the number of orbits of the finite group PSL(2,2^n) acting on subsets of size 4 for the 2^n+1 points of the projective line. - Paul M. Bradley, Jan 31 2017
For n > 1, number of words of length n-2 over alphabet {1,2,3} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 17 2017
Also, the decimal representation of the x-axis, from the origin to the right edge, of the n-th stage of growth of the two-dimensional cellular automaton defined by "Rule 678", based on the 5-celled von Neumann neighborhood, initialized with a single black (ON) cell at stage zero. See A283641. - Robert Price, Mar 12 2017
Also the number of independent vertex sets and vertex covers in the 2 X (n-2) king graph. - Eric W. Weisstein, Sep 21 2017
From César Eliud Lozada, Dec 14 2017: (Start)
Let T(0) be a triangle and let T(1) be the medial triangle of T(0), T(2) the medial triangle of T(1) and, in general, T(n) the medial triangle of T(n-1). The barycentric coordinates of the first vertex of T(n) are [2*a(n-1)/a(n), 1, 1], for n > 0.
Let S(0) be a triangle and let S(1) be the antimedial triangle of S(0), S(2) the antimedial triangle of S(1) and, in general, S(n) the antimedial triangle of S(n-1). The barycentric coordinates of the first vertex of S(n) are [-a(n+1)/a(n), 1, 1], for n > 0. (End)
a(n) is also the number of derangements in S_{n+1} with empty peak set. - Isabella Huang, Apr 01 2018
For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020
Number of 2-compositions of n+1 with 1 not allowed as a part; see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020
The number of Hamiltonian paths of the flower snark graph of even order 2n > 2 is 12*a(n-1). - Don Knuth, Dec 25 2020
When set S = {1, 2, ..., 2^n}, n>=0, then the largest subset T of S with the property that if x is in T, then 2*x is not in T, has a(n+1) elements. For example, for n = 4, #S = 16, a(5) = 11 with T = {1, 3, 4, 5, 7, 9, 11, 12, 13, 15, 16} (see Hassan Tarfaoui link, Concours Général 1991). - Bernard Schott, Feb 14 2022
a(n) is the number of words of length n over a binary alphabet whose position in the lexicographic order is one more than a multiple of three. a(3) = 3: aaa, abb, bba. - Alois P. Heinz, Apr 13 2022
Named by Horadam (1988) after the German mathematician Ernst Jacobsthal (1882-1965). - Amiram Eldar, Oct 02 2023
Define the sequence u(n) = (u(n-1) + u(n-2))/u(n-3) with u(0) = 0, u(1) = 1, u(2) = u(3) = -1. Then u(4*n) = -1 + (-1)^n/a(n+1), u(4*n+1) = 2 - (-1)^n/a(n+1), u(4*n+2) = u(4*n+3) = -1. For example, a(3) = 3 and u(8) = -2/3, u(9) = 5/3, u(10) = u(11) = -1. - Michael Somos, Oct 24 2023
From Miquel A. Fiol, May 25 2024: (Start)
Also, a(n) is the number of (3-color) states of a cycle (n+1)-pole C_{n+1} with n+1 terminals (or semiedges).
For instance, for n=3, the a(3)=3 states (3-coloring of the terminals) of C_4 are
a a a a a b
a a b b a b (End)
Also, with offset 1, the cogrowth sequence of the 6-element dihedral group D3. - Sean A. Irvine, Nov 04 2024

Examples

			a(2) = 3 because the tiling of the 3 X 2 rectangle has either only 1 X 1 tiles, or one 2 X 2 tile in one of two positions (together with two 1 X 1 tiles).
From _Joerg Arndt_, Nov 10 2012: (Start)
The a(6)=21 length-5 ternary words with no two consecutive letters nonzero are (dots for 0's)
[ 1]   [ . . . . ]
[ 2]   [ . . . 1 ]
[ 3]   [ . . . 2 ]
[ 4]   [ . . 1 . ]
[ 5]   [ . . 2 . ]
[ 6]   [ . 1 . . ]
[ 7]   [ . 1 . 1 ]
[ 8]   [ . 1 . 2 ]
[ 9]   [ . 2 . . ]
[10]   [ . 2 . 1 ]
[11]   [ . 2 . 2 ]
[12]   [ 1 . . . ]
[13]   [ 1 . . 1 ]
[14]   [ 1 . . 2 ]
[15]   [ 1 . 1 . ]
[16]   [ 1 . 2 . ]
[17]   [ 2 . . . ]
[18]   [ 2 . . 1 ]
[19]   [ 2 . . 2 ]
[20]   [ 2 . 1 . ]
[21]   [ 2 . 2 . ]
(End)
G.f. = x + x^2 + 3*x^3 + 5*x^4 + 11*x^5 + 21*x^6 + 43*x^7 + 85*x^8 + 171*x^9 + ...

References

  • Jathan Austin and Lisa Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
  • Thomas Fink and Yong Mao, The 85 ways to tie a tie, Fourth Estate, London, 1999; Die 85 Methoden eine Krawatte zu binden. Hoffmann und Kampe, Hamburg, 1999.
  • International Mathematical Olympiad 2001, Hong Kong Preliminary Selection Contest Problem #16.
  • Jablan S. and Sazdanovic R., LinKnot: Knot Theory by Computer, World Scientific Press, 2007. See p. 80.
  • Ernst Erich Jacobsthal, Fibonaccische Polynome und Kreisteilungsgleichungen, Sitzungsber. Berliner Math. Gesell. 17 (1919-1920), 43-57.
  • Tanya Khovanova, "Coins and Logic", Chapter 6, The Mathematics of Various Entertaining Subjects: Volume 3 (2019), Jennifer Beineke & Jason Rosenhouse, eds. Princeton University Press, Princeton and Oxford, p. 73. ISBN: 0691182582, 978-0691182582.
  • Donald E. Knuth, Art of Computer Programming, Vol. 3, Sect. 5.3.1, Eq. 13.
  • Thomas Koshy, Fibonacci and Lucas numbers with applications, Wiley, 2001, p. 98.
  • Steven Roman, Introduction to Coding and Information Theory, Springer Verlag, 1996, 41-42.
  • P. D. Seymour and D. J. A. Welsh, Combinatorial applications of an inequality form statistical mechanics, Math. Proc. Camb. Phil. Soc. 77 (1975), 485-495. [Although Daykin et al. (1979) claim that the present sequence is studied in this article, it does not seem to be explicitly mentioned. Note that definition of log-convex in (3.1) is wrong. - N. J. A. Sloane, Dec 26 2020]
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Robert M. Young, Excursions in Calculus, MAA, 1992, p. 239

Links

Crossrefs

Partial sums of this sequence give A000975, where there are additional comments from B. E. Williams and Bill Blewett on the tie problem.
A002487(a(n)) = A000045(n).
Row sums of A059260, A156667 and A134317. Equals A026644(n-2)+1 for n > 1.
a(n) = A073370(n-1, 0), n >= 1 (first column of triangle).
Cf. A266508 (binary), A081857 (base 4), A147612 (characteristic function).
Cf. A000978, A000979, A019322, A066845, A105348, A130249, A130250, A130253, A005578, A002083, A113405, A138000, A064934, A003158, A175286 (Pisano periods), A147613, A156319, A002605, A000225, A052129, A014551 (companion "autosequence"), A015266, A015287, A015305, A015323, A015338, A015356, A015371, A084175, A245489, A283641.
Cf. A049883 = primes in this sequence, A107036 = indices of primes, A129738.
Cf. A091084 (mod 10), A239473, A280049.
Bisections: A002450, A007583.
Cf. A077925 (signed version).

Programs

  • Haskell
    a001045 = (`div` 3) . (+ 1) . a000079
a001045_list = 0 : 1 :
   zipWith (+) (map (2 *) a001045_list) (tail a001045_list)
-- Reinhard Zumkeller, Mar 24 2013, Jan 05 2012, Feb 05 2011
  • Magma
    [n le 2 select n-1 else Self(n-1)+2*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Jun 27 2016
  • Maple
    A001045 := proc(n)
  (2^n-(-1)^n)/3 ;
end proc: # R. J. Mathar, Dec 18 2012
  • Mathematica
    Jacob0[n_] := (2^n - (-1)^n)/3; Table[Jacob0[n], {n, 0, 33}] (* Robert G. Wilson v, Dec 05 2005 *)
Array[(2^# - (-1)^#)/3 &, 33, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
LinearRecurrence[{1, 2}, {0, 1}, 40] (* Harvey P. Dale, Nov 30 2011 *)
CoefficientList[Series[x/(1 - x - 2 x^2), {x, 0, 34}], x] (* Robert G. Wilson v, Jul 21 2015 *)
Table[(2^n - (-1)^n)/3, {n, 0, 20}] (* Eric W. Weisstein, Sep 21 2017 *)
Table[Abs[QBinomial[n, 1, -2]], {n, 0, 35}] (* John Keith, Jan 29 2022 *)
  • Maxima
    a[0]:0$
a[n]:=2^(n-1)-a[n-1]$
A001045(n):=a[n]$
makelist(A001045(n),n,0,30); /* Martin Ettl, Nov 05 2012 */
  • PARI
    a(n) = (2^n - (-1)^n) / 3
  • PARI
    M=[1,1,0;1,0,1;0,1,1];for(i=0,34,print1((M^i)[2,1],",")) \\ Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
  • PARI
    a=0; for(n=0,34,print1(a,", "); a=2*(a-n%2)+1) \\ K. Spage, Aug 22 2014
  • Python
    # A001045.py
def A001045():
    a, b = 0, 1
    while True:
        yield a
        a, b = b, b+2*a
sequence = A001045()
[next(sequence) for i in range(20)] # David Radcliffe, Jun 26 2016
  • Python
    [(2**n-(-1)**n)//3 for n in range(40)] # Gennady Eremin, Mar 03 2022
  • Sage
    [lucas_number1(n, 1, -2) for n in range(34)]  # Zerinvary Lajos, Apr 22 2009
# Alternatively:
a = BinaryRecurrenceSequence(1,2)
[a(n) for n in (0..34)] # Peter Luschny, Aug 29 2016

Formula

a(n) = 2^(n-1) - a(n-1). a(n) = 2*a(n-1) - (-1)^n = (2^n - (-1)^n)/3.
G.f.: x/(1 - x - 2*x^2) = x/((x+1)*(1-2*x)). Simon Plouffe in his 1992 dissertation
E.g.f.: (exp(2*x) - exp(-x))/3.
a(2*n) = 2*a(2*n-1)-1 for n >= 1, a(2*n+1) = 2*a(2*n)+1 for n >= 0. - Lee Hae-hwang, Oct 11 2002; corrected by Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
Also a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x, y) = x*F(n-1)(x, y) + y*F(n-2)(x, y), with y=2*x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 04 2002
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*3^(k-1). - Paul Barry, Apr 02 2003
The ratios a(n)/2^(n-1) converge to 2/3 and every fraction after 1/2 is the arithmetic mean of the two preceding fractions. - Gary W. Adamson, Jul 05 2003
a(n) = U(n-1, i/(2*sqrt(2)))*(-i*sqrt(2))^(n-1) with i^2=-1. - Paul Barry, Nov 17 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 2^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(2*n) = A002450(n) = (4^n - 1)/3; a(2*n+1) = A007583(n) = (2^(2*n+1) + 1)/3. - Philippe Deléham, Mar 27 2004
a(n) = round(2^n/3) = (2^n + (-1)^(n-1))/3 so lim_{n->infinity} 2^n/a(n) = 3. - Gerald McGarvey, Jul 21 2004
a(n) = Sum_{k=0..n-1} (-1)^k*2^(n-k-1) = Sum_{k=0..n-1}, 2^k*(-1)^(n-k-1). - Paul Barry, Jul 30 2004
a(n+1) = Sum_{k=0..n} binomial(k, n-k)*2^(n-k). - Paul Barry, Oct 07 2004
a(n) = Sum_{k=0..n-1} W(n-k, k)*(-1)^(n-k)*binomial(2*k,k), W(n, k) as in A004070. - Paul Barry, Dec 17 2004
From Paul Barry, Jan 17 2005: (Start)
a(n) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*floor((2*k+1)/3).
a(n+1) = Sum_{k=0..n} k*binomial(n-1, (n-k)/2)*(1+(-1)^(n+k))*(A042965(k)+0^k). (End)
From Paul Barry, Jan 17 2005: (Start)
a(n+1) = ceiling(2^n/3) + floor(2^n/3) = (ceiling(2^n/3))^2 - (floor(2^n/3))^2.
a(n+1) = A005578(n) + A000975(n-1) = A005578(n)^2 - A000975(n-1)^2. (End)
a(n+1) = Sum_{k=0..n} Sum_{j=0..n} (-1)^(n-j)*binomial(j, k). - Paul Barry, Jan 26 2005
Let M = [1, 1, 0; 1, 0, 1; 0, 1, 1], then a(n) = (M^n)[2, 1], also matrix characteristic polynomial x^3 - 2*x^2 - x + 2 defines the three-step recursion a(0)=0, a(1)=1, a(2)=1, a(n) = 2*a(n-1) + a(n-2) - 2*a(n-3) for n > 2. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
a(n) = ceiling(2^(n+1)/3) - ceiling(2^n/3) = A005578(n+1) - A005578(n). - Paul Barry, Oct 08 2005
a(n) = floor(2^(n+1)/3) - floor(2^n/3) = A000975(n) - A000975(n-1). - Paul Barry, Oct 08 2005
From Paul Barry, Feb 20 2003: (Start)
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-1)+3*k);
a(n) = Sum_{k=0..floor(n/3)} binomial(n, f(n-2)+3*k), where f(n)=A080425(n). (End)
From Miklos Kristof, Mar 07 2007: (Start)
a(2*n) = (1/3)*Product_{d|n} cyclotomic(d,4).
a(2*n+1) = (1/3)*Product_{d|2*n+1} cyclotomic(2*d,2). (End)
From Hieronymus Fischer, Apr 23 2007: (Start)
The a(n) are closely related to nested square roots; this is 2*sin(2^(-n)*Pi/2*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0}.
Also 2*cos(2^(-n)*Pi*a(n)) = sqrt(2-sqrt(2-sqrt(2-sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1} as well as
2*sin(2^(-n)*3/2*Pi*a(n)) = sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n times, n >= 0} and
2*cos(2^(-n)*3*Pi*a(n)) = -sqrt(2+sqrt(2+sqrt(2+sqrt(...sqrt(2)))...) {with the '2' n-1 times, n >= 1}.
a(n) = 2^(n+1)/Pi*arcsin(b(n+1)/2) where b(n) is defined recursively by b(0)=2, b(n)=sqrt(2-b(n-1)).
There is a similar formula regarding the arccos function, this is a(n) = 2^n/Pi*arccos(b(n)/2).
With respect to the sequence c(n) defined recursively by c(0)=-2, c(n)=sqrt(2+c(n-1)), the following formulas hold true: a(n) = 2^n/3*(1-(-1)^n*(1-2/Pi*arcsin(c(n+1)/2))); a(n) = 2^n/3*(1-(-1)^n*(1-1/Pi*arccos(-c(n)/2))).
(End)
Sum_{k=0..n} A039599(n,k)*a(k) = A049027(n), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum_{k=0..n} A039599(n,k)*a(k+1) = A067336(n). - Philippe Deléham, Jun 10 2007
Let T = the 3 X 3 matrix [1,1,0; 1,0,1; 0,1,1]. Then T^n * [1,0,0,] = [A005578(n), a(n), A000975(n-1)]. - Gary W. Adamson, Dec 24 2007
a(n) + a(n+5) = 11*2^n. - Paul Curtz, Jan 17 2008
a(n) = Sum_{k=1..n} K(2, k)*a(n - k), where K(n,k) = k if 0 <= k <= n and K(n,k)=0 otherwise. (When using such a K-coefficient, several different arguments to K or several different definitions of K may lead to the same integer sequence. For example, the Fibonacci sequence can be generated in several ways using the K-coefficient.) - Thomas Wieder, Jan 13 2008
a(n) + a(n+2*k+1) = a(2*k+1)*2^n. - Paul Curtz, Feb 12 2008
a(n) = lower left term in the 2 X 2 matrix [0,2; 1,1]^n. - Gary W. Adamson, Mar 02 2008
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-2)^(n-k). -Philippe Deléham, Oct 26 2008
a(n) = sqrt(8*a(n-1)*a(n-2) + 1). E.g., sqrt(3*5*8+1) = 11, sqrt(5*11*8+1) = 21. - Giuseppe Ottonello, Jun 14 2009
Let p[i] = Fibonacci(i-1) and let A be the Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1], (i <= j), A[i,j] = -1, (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n-1) = det(A). - Milan Janjic, May 08 2010
a(p-1) = p*A007663(n)/3 if n > 1, and a(p-1) = p*A096060(n) if n > 2, with p=prime(n). - Jonathan Sondow, Jul 19 2010
Algebraically equivalent to replacing the 5's with 9's in the explicit (Binet) formula for the n-th term in the Fibonacci sequence: The formula for the n-th term in the Fibonacci sequence is F(n) = ((1+sqrt(5))^n - (1-sqrt(5))^n)/(2^n*sqrt(5)). Replacing the 5's with 9's gives ((1+sqrt(9))^n - (1-sqrt(9))^n)/(2^n*sqrt(9)) = (2^n+(-1)^(n+1))/3 = (2^n-(-1)^(n))/3 = a(n). - Jeffrey R. Goodwin, May 27 2011
For n > 1, a(n) = A000975(n-1) + (1 + (-1)^(n-1))/2. - Vladimir Shevelev, Feb 27 2012
From Sergei N. Gladkovskii, Jun 12 2012: (Start)
G.f.: x/(1-x-2*x^2) = G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)/G(k+1))); (continued fraction 3 kind, 3-step).
E.g.f.: G(0)/3; G(k) = 1 - ((-1)^k)/(2^k - 2*x*4^k/(2*x*2^k - ((-1)^k)*(k+1)/G(k+1))); (continued fraction 3rd kind, 3-step). (End)
a(n) = 2^k * a(n-k) + (-1)^(n+k)*a(k). - Paul Curtz, Jean-François Alcover, Dec 11 2012
a(n) = sqrt((A014551(n))^2 + (-1)^(n-1)*2^(n+2))/3. - Vladimir Shevelev, Mar 13 2013
G.f.: Q(0)/3, where Q(k) = 1 - 1/(4^k - 2*x*16^k/(2*x*4^k - 1/(1 + 1/(2*4^k - 8*x*16^k/(4*x*4^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 21 2013
G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(2*k+1 + 2*x)/( x*(2*k+2 + 2*x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 29 2013
G.f.: Q(0) -1, where Q(k) = 1 + 2*x^2 + (k+2)*x - x*(k+1 + 2*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n+2) = Sum_{k=0..n} A108561(n,k)*(-2)^k. - Philippe Deléham, Nov 17 2013
a(n) = (Sum_{k=1..n, k odd} C(n,k)*3^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 2^n for all n in Z. - Michael Somos, Mar 18 2014
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-3)^k = (2^n - (-1)^n)/3 = (-1)^(n-1)*Sum_{k=0..n-1} (-2)^k. Equals (-1)^(n-1)*Phi(n,-2), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
From Peter Bala, Apr 06 2015: (Start)
a(2*n)/a(n) = A014551(n) for n >= 1; a(3*n)/a(n) = 3*A245489(n) for n >= 1.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n.
exp( Sum_{n >= 1} a(3*n)/a(n)*x^n/n ) = Sum_{n >= 0} A084175(n+1)*x^n.
exp( Sum_{n >= 1} a(4*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015266(n+3)*(-x)^n.
exp( Sum_{n >= 1} a(5*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015287(n+4)*x^n.
exp( Sum_{n >= 1} a(6*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015305(n+5)*(-x)^n.
exp( Sum_{n >= 1} a(7*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015323(n+6)*x^n.
exp( Sum_{n >= 1} a(8*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015338(n+7)*(-x)^n.
exp( Sum_{n >= 1} a(9*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015356(n+8)*x^n.
exp( Sum_{n >= 1} a(10*n)/a(n)*x^n/n ) = Sum_{n >= 0} A015371(n+9)*(-x)^n. (End)
a(n) = (1-(-1)^n)/2 + floor((2^n)/3). - Reiner Moewald, Jun 05 2015
a(n+k)^2 - A014551(k)*a(n)*a(n+k) + (-2)^k*a(n)^2 = (-2)^n*a(k)^2, for n >= 0 and k >= 0. - Alexander Samokrutov, Jul 21 2015
Dirichlet g.f.: (PolyLog(s,2) + (1 - 2^(1-s))*zeta(s))/3. - Ilya Gutkovskiy, Jun 27 2016
From Yuchun Ji, Apr 08 2018: (Start)
a(m)*a(n) + a(m-1)*a(n-1) - 2*a(m-2)*a(n-2) = 2^(m+n-3).
a(m+n-1) = a(m)*a(n) + 2*a(m-1)*a(n-1); a(m+n) = a(m+1)*a(n+1) - 4*a(m-1)*a(n-1).
a(2*n-1) = a(n)^2 + 2*a(n-1)^2; a(2*n) = a(n+1)^2 - 4*a(n-1)^2. (End)
a(n+4) = a(n) + 5*2^n, a(0) = 0, a(1..4) = [1,1,3,5]. That is to say, for n > 0, the ones digits of Jacobsthal numbers follow the pattern 1,1,3,5,1,1,3,5,1,1,3,5,.... - Yuchun Ji, Apr 25 2019
a(n) mod 10 = A091084(n). - Alois P. Heinz, Apr 25 2019
The sequence starting with "1" is the second INVERT transform of (1, -1, 3, -5, 11, -21, 43, ...). - Gary W. Adamson, Jul 08 2019
From Kai Wang, Jan 14 2020: (Start)
a(n)^2 - a(n+1)*a(n-1) = (-2)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-2)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-2)^n*a(m-n).
a(m-n) = (-1)^n*(a(m)*A014551(n) - A014551(m)*a(n))/(2^(n+1)).
a(m+n) = (a(m)*A014551(n) + A014551(m)*a(n))/2.
A014551(n)^2 - A014551(n+r)*A014551(n-r) = 9*(-1)^(n-r-1)*2^(n-r)*a(r)^2 .
A014551(m)*A014551(n+1) - A014551(m+1)*A014551(n) = 9*(-1)^(n-1)*2^(n)*a(m-n).
A014551(m-n) = (-1)^(n)*(A014551(m)*A014551(n) - 9*a(m)*a(n))/2^(n+1).
A014551(m+n) = (A014551(m)*A014551(n) + 9*a(m)*a(n))/2.
a(n) = Sum_{i=0..n-1; j=0..n-1; i+2*j=n-1} 2^j*((i+j)!/(i!*j!)). (End)
For n > 0, 1/(2*a(n+1)) = Sum_{m>=n} a(m)/(a(m+1)*a(m+2)). - Kai Wang, Mar 03 2020
For 4 > h >= 0, k >= 0, a(4*k+h) mod 5 = a(h) mod 5. - Kai Wang, May 07 2020
From Kengbo Lu, Jul 27 2020: (Start)
a(n) = 1 + Sum_{k=0..n-1} a(k) if n odd; a(n) = Sum_{k=0..n-1} a(k) if n even.
a(n) = F(n) + Sum_{k=0..n-2} a(k)*F(n-k-1), where F denotes the Fibonacci numbers.
a(n) = b(n) + Sum_{k=0..n-1} a(k)*b(n-k), where b(n) is defined through b(0) = 0, b(1) = 1, b(n) = 2*b(n-2).
a(n) = 1 + 2*Sum_{k=0..n-2} a(k).
a(m+n) = a(m)*a(n+1) + 2*a(m-1)*a(n).
a(2*n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)*2^(i+j). (End)
G.f.: x/(1 - x - 2*x^2) = Sum_{n >= 0} x^(n+1) * Product_{k = 1..n} (k + 2*x)/(1 + k*x) (a telescoping series). - Peter Bala, May 08 2024
a(n) = Sum_{r>=0} F(n-2r, r), where F(n, 0) is the n-th Fibonacci number and F(n,r) = Sum_{j=1..n} F(n+1-j, r-1) F(j, r-1). - Gregory L. Simay, Aug 31 2024
From Peter Bala, Jun 27 2025: (Start)
The following are all examples of telescoping infinite products:
Product_{n >= 1} (1 + 2^n/a(2*n+2)) = 2, since 1 + 2^n/a(2*n+2) = b(n+1)/b(n), where b(n) = 2 - 3/(2^n + 1).
Product_{n >= 1} (1 - 2^n/a(2*n+2)) = 2/5, since 1 - 2^n/a(2*n+2) = c(n+1)/c(n), where c(n) = 2 + 3/(2^n - 1).
Product_{n >= 1} (1 + (-2)^n/a(2*n+2)) = 2/3, since 1 + (-2)^n/a(2*n+2) = d(n+1)/d(n), where d(n) = 2 - 1/(1 + (-2)^n).
Product_{n >= 1} (1 - (-2)^n/a(2*n+2)) = 6/5, since 1 - (-2)^n/a(2*n+2) = e(n+1)/e(n), where e(n) = 2 - 1/(1 - (-2)^n). (End)

Extensions

Thanks to Don Knuth, who pointed out several missing references, including Brocard (1880), which although it was mentioned in the 1973 Handbook of Integer Sequences, was omitted from the 1995 "Encyclopedia". - N. J. A. Sloane, Dec 26 2020

A088218 Total number of leaves in all rooted ordered trees with n edges.

Original entry on oeis.org

1, 1, 3, 10, 35, 126, 462, 1716, 6435, 24310, 92378, 352716, 1352078, 5200300, 20058300, 77558760, 300540195, 1166803110, 4537567650, 17672631900, 68923264410, 269128937220, 1052049481860, 4116715363800, 16123801841550, 63205303218876, 247959266474052
Offset: 0

Views

Author

Michael Somos, Sep 24 2003

Keywords

Comments

Essentially the same as A001700, which has more information.
Note that the unique rooted tree with no edges has no leaves, so a(0)=1 is by convention. - Michael Somos, Jul 30 2011
Number of ordered partitions of n into n parts, allowing zeros (cf. A097070) is binomial(2*n-1,n) = a(n) = essentially A001700. - Vladeta Jovovic, Sep 15 2004
Hankel transform is A000027; example: Det([1,1,3,10;1,3,10,35;3,10,35,126; 10,35,126,462]) = 4. - Philippe Deléham, Apr 13 2007
a(n) is the number of functions f:[n]->[n] such that for all x,y in [n] if xA045992(n). - Geoffrey Critzer, Apr 02 2009
Hankel transform of the aeration of this sequence is A000027 doubled: 1,1,2,2,3,3,... - Paul Barry, Sep 26 2009
The Fi1 and Fi2 triangle sums of A039599 are given by the terms of this sequence. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Alternating row sums of Riordan triangle A094527. See the Philippe Deléham formula. - Wolfdieter Lang, Nov 22 2012
(-2)*a(n) is the Z-sequence for the Riordan triangle A110162. For the notion of Z- and A-sequences for Riordan arrays see the W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 22 2012
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of integer compositions of 2n with alternating (or reverse-alternating) sum 0 (ranked by A344619). This is equivalent to Ran Pan's comment at A001700. For example, the a(0) = 1 through a(3) = 10 compositions are:
() (11) (22) (33)
(121) (132)
(1111) (231)
(1122)
(1221)
(2112)
(2211)
(11121)
(12111)
(111111)
For n > 0, a(n) is also the number of integer compositions of 2n with alternating sum 2.
(End)
Number of terms in the expansion of (x_1+x_2+...+x_n)^n. - César Eliud Lozada, Jan 08 2022

Examples

			G.f. = 1 + x + 3*x^2 + 10*x^3 + 35*x^4 + 126*x^5 + 462*x^6 + 1716*x^7 + ...
The five rooted ordered trees with 3 edges have 10 leaves.
..x........................
..o..x.x..x......x.........
..o...o...o.x..x.o..x.x.x..
..r...r....r....r.....r....

References

  • L. W. Shapiro and C. J. Wang, Generating identities via 2 X 2 matrices, Congressus Numerantium, 205 (2010), 33-46.

Links

Crossrefs

Same as A001700 modulo initial term and offset.
First differences are A024718.
Main diagonal of A071919 and of A305161.
A signed version is A110556.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A003242 counts anti-run compositions.
A025047 counts wiggly compositions (ascend: A025048, descend: A025049).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A106356 counts compositions by number of maximal anti-runs.
A124754 gives the alternating sum of standard compositions.
A345197 counts compositions by sum, length, and alternating sum.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218 (this sequence), ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218 (this sequence), ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.
Cf. A000027, A000070, A000097, A000108, A001622, A006232, A008965, A039599, A045992, A058696, A094527, A097070, A110162, A110555, A180662, A238279, A239830, A325534, A325535, A333213, A344607, A344611, A344617.

Programs

  • Magma
    [Binomial(2*n-1, n): n in [0..30]]; // Vincenzo Librandi, Aug 07 2014
  • Maple
    seq(binomial(2*n-1, n),n=0..24); # Peter Luschny, Sep 22 2014
  • Mathematica
    a[ n_] := SeriesCoefficient[(1 - x)^-n, {x, 0, n}];
c = (1 - (1 - 4 x)^(1/2))/(2 x);CoefficientList[Series[1/(1-(c-1)),{x,0,20}],x] (* Geoffrey Critzer, Dec 02 2010 *)
Table[Binomial[2 n - 1, n], {n, 0, 20}] (* Vincenzo Librandi, Aug 07 2014 *)
a[ n_] := If[ n < 0, 0, With[ {m = 2 n}, m! SeriesCoefficient[ (1 + BesselI[0, 2 x]) / 2, {x, 0, m}]]]; (* Michael Somos, Nov 22 2014 *)
  • PARI
    {a(n) = sum( i=0, n, binomial(n+i-2,i))};
  • PARI
    {a(n) = if( n<0, 0, polcoeff( (1 + 1 / sqrt(1 - 4*x + x * O(x^n))) / 2, n))};
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / (1 - x + x * O(x^n))^n, n))};
  • PARI
    {a(n) = if( n<0, 0, binomial( 2*n - 1, n))};
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( subst((1 - x) / (1 - 2*x), x, serreverse( x - x^2 + x * O(x^n))), n))};
  • Sage
    def A088218(n):
    return rising_factorial(n,n)/falling_factorial(n,n)
[A088218(n) for n in (0..24)]  # Peter Luschny, Nov 21 2012

Formula

G.f.: (1 + 1 / sqrt(1 - 4*x)) / 2.
a(n) = binomial(2*n - 1, n).
a(n) = (n+1)*A000108(n)/2, n>=1. - B. Dubalski (dubalski(AT)atr.bydgoszcz.pl), Feb 05 2002 (in A060150)
a(n) = (0^n + C(2n, n))/2. - Paul Barry, May 21 2004
a(n) is the coefficient of x^n in 1 / (1 - x)^n and also the sum of the first n coefficients of 1 / (1 - x)^n. Given B(x) with the property that the coefficient of x^n in B(x)^n equals the sum of the first n coefficients of B(x)^n, then B(x) = B(0) / (1 - x).
G.f.: 1 / (2 - C(x)) = (1 - x*C(x))/sqrt(1-4*x) where C(x) is g.f. for Catalan numbers A000108. Second equation added by Wolfdieter Lang, Nov 22 2012.
From Paul Barry, Nov 02 2004: (Start)
a(n) = Sum_{k=0..n} binomial(2*n, k)*cos((n-k)*Pi);
a(n) = Sum_{k=0..n} binomial(n, (n-k)/2)*(1+(-1)^(n-k))*cos(k*Pi/2)/2 (with interpolated zeros);
a(n) = Sum_{k=0..floor(n/2)} binomial(n, k)*cos((n-2*k)*Pi/2) (with interpolated zeros); (End)
a(n) = A110556(n)*(-1)^n, central terms in triangle A110555. - Reinhard Zumkeller, Jul 27 2005
a(n) = Sum_{0<=k<=n} A094527(n,k)*(-1)^k. - Philippe Deléham, Mar 14 2007
From Paul Barry, Mar 29 2010: (Start)
G.f.: 1/(1-x/(1-2x/(1-(1/2)x/(1-(3/2)x/(1-(2/3)x/(1-(4/3)x/(1-(3/4)x/(1-(5/4)x/(1-... (continued fraction);
E.g.f.: (of aerated sequence) (1 + Bessel_I(0, 2*x))/2. (End)
a(n + 1) = A001700(n). a(n) = A024718(n) - A024718(n - 1).
E.g.f.: E(x) = 1+x/(G(0)-2*x) ; G(k) = (k+1)^2+2*x*(2*k+1)-2*x*(2*k+3)*((k+1)^2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 21 2011
a(n) = Sum_{k=0..n}(-1)^k*binomial(2*n,n+k). - Mircea Merca, Jan 28 2012
a(n) = rf(n,n)/ff(n,n), where rf is the rising factorial and ff the falling factorial. - Peter Luschny, Nov 21 2012
D-finite with recurrence: n*a(n) +2*(-2*n+1)*a(n-1) = 0. - R. J. Mathar, Dec 04 2012
a(n) = hypergeom([1-n,-n],[1],1). - Peter Luschny, Sep 22 2014
G.f.: 1 + x/W(0), where W(k) = 4*k+1 - (4*k+3)*x/(1 - (4*k+1)*x/(4*k+3 - (4*k+5)*x/(1 - (4*k+3)*x/W(k+1) ))) ; (continued fraction). - Sergei N. Gladkovskii, Nov 13 2014
a(n) = A000984(n) + A001791(n). - Gus Wiseman, Jun 28 2021
E.g.f.: (1 + exp(2*x) * BesselI(0,2*x)) / 2. - Ilya Gutkovskiy, Nov 03 2021
From Amiram Eldar, Mar 12 2023: (Start)
Sum_{n>=0} 1/a(n) = 5/3 + 4*Pi/(9*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/5 - 8*log(phi)/(5*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024

A002457 a(n) = (2n+1)!/n!^2.

Original entry on oeis.org

1, 6, 30, 140, 630, 2772, 12012, 51480, 218790, 923780, 3879876, 16224936, 67603900, 280816200, 1163381400, 4808643120, 19835652870, 81676217700, 335780006100, 1378465288200, 5651707681620, 23145088600920, 94684453367400, 386971244197200, 1580132580471900
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Expected number of matches remaining in Banach's modified matchbox problem (counted when last match is drawn from one of the two boxes), multiplied by 4^(n-1). - Michael Steyer, Apr 13 2001
Hankel transform is (-1)^n*A014480(n). - Paul Barry, Apr 26 2009
Convolved with A000108: (1, 1, 1, 5, 14, 42, ...) = A000531: (1, 7, 38, 187, 874, ...). - Gary W. Adamson, May 14 2009
Convolution of A000302 and A000984. - Philippe Deléham, May 18 2009
1/a(n) is the integral of (x(1-x))^n on interval [0,1]. Apparently John Wallis computed these integrals for n=0,1,2,3,.... A004731, shifted left by one, gives numerators/denominators of related integrals (1-x^2)^n on interval [0,1]. - Marc van Leeuwen, Apr 14 2010
Extend the triangular peaks of Dyck paths of semilength n down to the baseline forming (possibly) larger and overlapping triangles. a(n) = sum of areas of these triangles. Also a(n) = triangular(n) * Catalan(n). - David Scambler, Nov 25 2010
Let H be the n X n Hilbert matrix H(i,j) = 1/(i+j-1) for 1 <= i,j <= n. Let B be the inverse matrix of H. The sum of the elements in row n of B equals a(n-1). - T. D. Noe, May 01 2011
Apparently the number of peaks in all symmetric Dyck paths with semilength 2n+1. - David Scambler, Apr 29 2013
Denominator of central elements of Leibniz's Harmonic Triangle A003506.
Central terms of triangle A116666. - Reinhard Zumkeller, Nov 02 2013
Number of distinct strings of length 2n+1 using n letters A, n letters B, and 1 letter C. - Hans Havermann, May 06 2014
Number of edges in the Hasse diagram of the poset of partitions in the n X n box ordered by containment (from Havermann's comment above, C represents the square added in the edge). - William J. Keith, Aug 18 2015
Let V(n, r) denote the volume of an n-dimensional sphere with radius r then V(n, 1/2^n) = V(n-1, 1/2^n) / a((n-1)/2) for all odd n. - Peter Luschny, Oct 12 2015
a(n) is the result of processing the n+1 row of Pascal's triangle A007318 with the method of A067056. Example: Let n=3. Given the 4th row of Pascal's triangle 1,4,6,4,1, we get 1*(4+6+4+1) + (1+4)*(6+4+1) + (1+4+6)*(4+1) + (1+4+6+4)*1 = 15+55+55+15 = 140 = a(3). - J. M. Bergot, May 26 2017
a(n) is the number of (n+1) X 2 Young tableaux with a two horizontal walls between the first and second column. If there is a wall between two cells, the entries may be decreasing; see [Banderier, Wallner 2021] and A000984 for one horizontal wall. - Michael Wallner, Jan 31 2022
a(n) is the number of facets of the symmetric edge polytope of the cycle graph on 2n+1 vertices. - Mariel Supina, May 12 2022
Diagonal of the rational function 1 / (1 - x - y)^2. - Ilya Gutkovskiy, Apr 23 2025

Examples

			G.f. = 1 + 6*x + 30*x^2 + 140*x^3 + 630*x^4 + 2772*x^5 + 12012*x^6 + 51480*x^7 + ...

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 159.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 83, Problem 25; p. 168, #30.
  • W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I.
  • C. Jordan, Calculus of Finite Differences. Röttig and Romwalter, Budapest, 1939; Chelsea, NY, 1965, p. 449.
  • M. Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 127-129.
  • C. Lanczos, Applied Analysis. Prentice-Hall, Englewood Cliffs, NJ, 1956, p. 514.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • J. Ser, Les Calculs Formels des Séries de Factorielles. Gauthier-Villars, Paris, 1933, p. 92.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • J. Wallis, Operum Mathematicorum, pars altera, Oxford, 1656, pp 31,34 [Marc van Leeuwen, Apr 14 2010]

Links

Crossrefs

Cf. A000531 (Banach's original match problem).
Cf. A033876, A000984, A001803, A132818, A046521 (second column).
Cf. A000108, A000217.
A diagonal of A331430.
The rightmost diagonal of the triangle A331431.

Programs

Formula

G.f.: (1-4x)^(-3/2) = 1F0(3/2;;4x).
a(n-1) = binomial(2*n, n)*n/2 = binomial(2*n-1, n)*n.
a(n-1) = 4^(n-1)*Sum_{i=0..n-1} binomial(n-1+i, i)*(n-i)/2^(n-1+i).
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n)*{1 + 3/8*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 21 2001
(2*n+2)!/(2*n!*(n+1)!) = (n+n+1)!/(n!*n!) = 1/beta(n+1, n+1) in A061928.
Sum_{i=0..n} i * binomial(n, i)^2 = n*binomial(2*n, n)/2. - Yong Kong (ykong(AT)curagen.com), Dec 26 2000
a(n) ~ 2*Pi^(-1/2)*n^(1/2)*2^(2*n). - Joe Keane (jgk(AT)jgk.org), Jun 07 2002
a(n) = 1/Integral_{x=0..1} x^n (1-x)^n dx. - Fred W. Helenius (fredh(AT)ix.netcom.com), Jun 10 2003
E.g.f.: exp(2*x)*((1+4*x)*BesselI(0, 2*x) + 4*x*BesselI(1, 2*x)). - Vladeta Jovovic, Sep 22 2003
a(n) = Sum_{i+j+k=n} binomial(2i, i)*binomial(2j, j)*binomial(2k, k). - Benoit Cloitre, Nov 09 2003
a(n) = (2*n+1)*A000984(n) = A005408(n)*A000984(n). - Zerinvary Lajos, Dec 12 2010
a(n-1) = Sum_{k=0..n} A039599(n,k)*A000217(k), for n >= 1. - Philippe Deléham, Jun 10 2007
Sum of (n+1)-th row terms of triangle A132818. - Gary W. Adamson, Sep 02 2007
Sum_{n>=0} 1/a(n) = 2*Pi/3^(3/2). - Jaume Oliver Lafont, Mar 07 2009
a(n) = Sum_{k=0..n} binomial(2k,k)*4^(n-k). - Paul Barry, Apr 26 2009
a(n) = A000217(n) * A000108(n). - David Scambler, Nov 25 2010
a(n) = f(n, n-3) where f is given in A034261.
a(n) = A005430(n+1)/2 = A002011(n)/4.
a(n) = binomial(2n+2, 2) * binomial(2n, n) / binomial(n+1, 1), a(n) = binomial(n+1, 1) * binomial(2n+2, n+1) / binomial(2, 1) = binomial(2n+2, n+1) * (n+1)/2. - Rui Duarte, Oct 08 2011
G.f.: (G(0) - 1)/(4*x) where G(k) = 1 + 2*x*((2*k + 3)*G(k+1) - 1)/(k + 1). - Sergei N. Gladkovskii, Dec 03 2011 [Edited by Michael Somos, Dec 06 2013]
G.f.: 1 - 6*x/(G(0)+6*x) where G(k) = 1 + (4*x+1)*k - 6*x - (k+1)*(4*k-2)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Aug 13 2012
G.f.: Q(0), where Q(k) = 1 + 4*(2*k + 1)*x*(2*k + 2 + Q(k+1))/(k+1). - Sergei N. Gladkovskii, May 10 2013 [Edited by Michael Somos, Dec 06 2013]
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 4*x*(2*k+3)/(4*x*(2*k+3) + 2*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
a(n) = 2^(4n)/Sum_{k=0..n} (-1)^k*C(2n+1,n-k)/(2k+1). - Mircea Merca, Nov 12 2013
a(n) = (2*n)!*[x^(2*n)] HeunC(0,0,-2,-1/4,7/4,4*x^2) where [x^n] f(x) is the coefficient of x^n in f(x) and HeunC is the Heun confluent function. - Peter Luschny, Nov 22 2013
0 = a(n) * (16*a(n+1) - 2*a(n+2)) + a(n+1) * (a(n+2) - 6*a(n+1)) for all n in Z. - Michael Somos, Dec 06 2013
a(n) = 4^n*binomial(n+1/2, 1/2). - Peter Luschny, Apr 24 2014
a(n) = 4^n*hypergeom([-2*n,-2*n-1,1/2],[-2*n-2,1],2)*(n+1)*(2*n+1). - Peter Luschny, Sep 22 2014
a(n) = 4^n*hypergeom([-n,-1/2],[1],1). - Peter Luschny, May 19 2015
a(n) = 2*4^n*Gamma(3/2+n)/(sqrt(Pi)*Gamma(1+n)). - Peter Luschny, Dec 14 2015
Sum_{n >= 0} 2^(n+1)/a(n) = Pi, related to Newton/Euler's Pi convergence transformation series. - Tony Foster III, Jul 28 2016. See the Weisstein Pi link, eq. (23). - Wolfdieter Lang, Aug 26 2016
Boas-Buck recurrence: a(n) = (6/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, and a(0) = 1. Proof from a(n) = A046521(n+1,1). See comment in A046521. - Wolfdieter Lang, Aug 10 2017
a(n) = (1/3)*Sum_{i = 0..n+1} C(n+1,i)*C(n+1,2*n+1-i)*C(3*n+2-i,n+1) = (1/3)*Sum_{i = 0..2*n+1} (-1)^(i+1)*C(2*n+1,i)*C(n+i+1,i)^2. - Peter Bala, Feb 07 2018
a(n) = (2*n+1)*binomial(2*n, n). - Kolosov Petro, Apr 16 2018
a(n) = (-4)^n*binomial(-3/2, n). - Peter Luschny, Oct 23 2018
a(n) = 1 / Sum_{s=0..n} (-1)^s * binomial(n, s) / (n+s+1). - Kolosov Petro, Jan 22 2019
a(n) = Sum_{k = 0..n} (2*k + 1)*binomial(2*n + 1, n - k). - Peter Bala, Feb 25 2019
4^n/a(n) = Integral_{x=0..1} (1 - x^2)^n. - Michael Somos, Jun 13 2019
D-finite with recurrence: 0 = a(n)*(6 + 4*n) - a(n+1)*(n + 1) for all n in Z. - Michael Somos, Jun 13 2019
Sum_{n>=0} (-1)^n/a(n) = 4*arcsinh(1/2)/sqrt(5). - Amiram Eldar, Sep 10 2020
From Jianing Song, Apr 10 2022: (Start)
G.f. for {1/a(n)}: 4*arcsin(sqrt(x)/2) / sqrt(x*(4-x)).
E.g.f. for {1/a(n)}: exp(x/4)*sqrt(Pi/x)*erf(sqrt(x)/2). (End)
G.f. for {1/a(n)}: 4*arctan(sqrt(x/(4-x))) / sqrt(x*(4-x)). - Michael Somos, Jun 17 2023
a(n) = Sum_{k = 0..n} (-1)^(n+k) * (n + 2*k + 1)*binomial(n+k, k). This is the particular case m = 1 of the identity Sum_{k = 0..m*n} (-1)^k * (n + 2*k + 1) * binomial(n+k, k) = (-1)^(m*n) * (m*n + 1) * binomial((m+1)*n+1, n). Cf. A090816 and A306290. - Peter Bala, Nov 02 2024
a(n) = (1/Pi)*(2*n + 1)*(2^(2*n + 1))*Integral_{x=0..oo} 1/(x^2 + 1)^(n + 1) dx. - Velin Yanev, Jan 28 2025

A053121 Catalan triangle (with 0's) read by rows.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 2, 0, 3, 0, 1, 0, 5, 0, 4, 0, 1, 5, 0, 9, 0, 5, 0, 1, 0, 14, 0, 14, 0, 6, 0, 1, 14, 0, 28, 0, 20, 0, 7, 0, 1, 0, 42, 0, 48, 0, 27, 0, 8, 0, 1, 42, 0, 90, 0, 75, 0, 35, 0, 9, 0, 1, 0, 132, 0, 165, 0, 110, 0, 44, 0, 10, 0, 1, 132, 0, 297, 0, 275, 0, 154, 0, 54, 0, 11, 0
Offset: 0

Views

Author

Wolfdieter Lang

Keywords

Comments

Inverse lower triangular matrix of A049310(n,m) (coefficients of Chebyshev's S polynomials).
Walks with a wall: triangle of number of n-step walks from (0,0) to (n,m) where each step goes from (a,b) to (a+1,b+1) or (a+1,b-1) and the path stays in the nonnegative quadrant.
T(n,m) is the number of left factors of Dyck paths of length n ending at height m. Example: T(4,2)=3 because we have UDUU, UUDU, and UUUD, where U=(1,1) and D=(1,-1). (This is basically a different formulation of the previous - walks with a wall - property.) - Emeric Deutsch, Jun 16 2011
"The Catalan triangle is formed in the same manner as Pascal's triangle, except that no number may appear on the left of the vertical bar." [Conway and Smith]
G.f. for row polynomials p(n,x) := Sum_{m=0..n} (a(n,m)*x^m): c(z^2)/(1-x*z*c(z^2)). Row sums (x=1): A001405 (central binomial).
In the language of the Shapiro et al. reference such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group. The g.f. Ginv(x) of the m=0 column of the inverse of a given Bell-matrix (here A049310) is obtained from its g.f. of the m=0 column (here G(x)=1/(1+x^2)) by Ginv(x)=(f^{(-1)}(x))/x, with f(x) := x*G(x) and f^{(-1)}is the compositional inverse function of f (here one finds, with Ginv(0)=1, c(x^2)). See the Shapiro et al. reference.
Number of involutions of {1,2,...,n} that avoid the patterns 132 and have exactly k fixed points. Example: T(4,2)=3 because we have 2134, 4231 and 3214. Number of involutions of {1,2,...,n} that avoid the patterns 321 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1324 and 2134. Number of involutions of {1,2,...,n} that avoid the patterns 213 and have exactly k fixed points. Example: T(4,2)=3 because we have 1243, 1432 and 4231. - Emeric Deutsch, Oct 12 2006
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1 . Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
Riordan array (c(x^2),xc(x^2)), where c(x) is the g.f. of Catalan numbers A000108. - Philippe Deléham, Nov 25 2007
A053121^2 = triangle A145973. Convolved with A001405 = triangle A153585. - Gary W. Adamson, Dec 28 2008
By columns without the zeros, n-th row = A000108 convolved with itself n times; equivalent to A = (1 + x + 2x^2 + 5x^3 + 14x^4 + ...), then n-th row = coefficients of A^(n+1). - Gary W. Adamson, May 13 2009
Triangle read by rows,product of A130595 and A064189 considered as infinite lower triangular arrays; A053121 = A130595*A064189 = B^(-1)*A097609*B where B = A007318. - Philippe Deléham, Dec 07 2009
From Mark Dols, Aug 17 2010: (Start)
As an upper right triangle, rows represent powers of 5-sqrt(24):
5 - sqrt(24)^1 = 0.101020514...
5 - sqrt(24)^2 = 0.010205144...
5 - sqrt(24)^3 = 0.001030928...
(Divided by sqrt(96) these powers give a decimal representation of the columns of A007318, with 1/sqrt(96) being the middle column.) (End)
T(n,k) is the number of dispersed Dyck paths of length n (i.e., Motzkin paths of length n with no (1,0) steps at positive heights) having k (1,0)-steps. Example: T(5,3)=4 because, denoting U=(1,1), D=(1,-1), H=(1,0), we have HHHUD, HHUDH, HUDHH, and UDHHH. - Emeric Deutsch, Jun 01 2011
Let S(N,x) denote the N-th Chebyshev S-polynomial in x (see A049310, cf. [W. Lang]). Then x^n = sum_{k=0..n} T(n,k)*S(k,x). - L. Edson Jeffery, Sep 06 2012
This triangle a(n,m) appears also in the (unreduced) formula for the powers rho(N)^n for the algebraic number over the rationals rho(N) = 2*cos(Pi/N) = R(N, 2), the smallest diagonal/side ratio R in the regular N-gon:
rho(N)^n = sum(a(n,m)*R(N,m+1),m=0..n), n>=0, identical in N >= 1. R(N,j) = S(j-1, x=rho(N)) (Chebyshev S (A049310)). See a comment on this under A039599 (even powers) and A039598 (odd powers). Proof: see the Sep 06 2012 comment by L. Edson Jeffery, which follows from T(n,k) (called here a(n,k)) being the inverse of the Riordan triangle A049310. - Wolfdieter Lang, Sep 21 2013
The so-called A-sequence for this Riordan triangle of the Bell type (c(x^2), x*c(x^2)) (see comments above) is A(x) = 1 + x^2. This proves the recurrence given in the formula section by Henry Bottomley for a(n, m) = a(n-1, m-1) + a(n-1, m+1) for n>=1 and m>=1, with inputs. The Z-sequence for this Riordan triangle is Z(x) = x which proves the recurrence a(n,0) = a(n-1,1), n>=1, a(0,0) = 1. For A- and Z-sequences for Riordan triangles see the W. Lang link under A006232. - Wolfdieter Lang, Sep 22 2013
Rows of the triangle describe decompositions of tensor powers of the standard (2-dimensional) representation of the Lie algebra sl(2) into irreducibles. Thus a(n,m) is the multiplicity of the m-th ((m+1)-dimensional) irreducible representation in the n-th tensor power of the standard one. - Mamuka Jibladze, May 26 2015
The Riordan row polynomials p(n, x) belong to the Boas-Buck class (see a comment and references in A046521), hence they satisfy the Boas-Buck identity: (E_x - n*1)*p(n, x) = (E_x + 1)*Sum_{j=0..n-1} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)*p(n-1-j, x), for n >= 0, where E_x = x*d/dx (Euler operator). For the triangle a(n, m) this entails a recurrence for the sequence of column m, given in the formula section. - Wolfdieter Lang, Aug 11 2017
From Roger Ford, Jan 22 2018: (Start)
For row n, the nonzero values represent the odd components (loops) formed by n+1 nonintersecting arches above and below the x-axis with the following constraints: The top has floor((n+3)/2) starting arches at position 1 and the next consecutive odd positions. All other starting top arches are in even positions. The bottom arches are a rainbow of arches. If the component=1 then the arch configuration is a semimeander solution.
Examples: For row 3 {0, 2, 0, 1} there are 3 arch configurations: 2 arch configurations have a component=1; 1 has a component=3. c=components, U=top arch starting in odd position, u=top arch starting in an even position, d=ending top arch:
.
top UuUdUddd c=3 top UdUuUddd c=1 top UdUdUudd c=1
/\ /\
//\\ / \
// \\ / /\ \ /\
// \\ / / \ \ / \
///\ /\\\ /\ / / /\ \ \ /\ /\ / /\ \
\\\ \/ /// \ \ \ \/ / / / \ \ \ \/ / / /
\\\ /// \ \ \ / / / \ \ \ / / /
\\\/// \ \ \/ / / \ \ \/ / /
\\// \ \ / / \ \ / /
\/ \ \/ / \ \/ /
\ / \ /
\/ \/
For row 4 {2, 0, 3, 0, 1} there are 6 arch configurations: 2 have a component=1; 3 have a component=3: 1 has a component=1. (End)

Examples

			Triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6  7  8  9 10 ...
  0:   1
  1:   0   1
  2:   1   0   1
  3:   0   2   0   1
  4:   2   0   3   0   1
  5:   0   5   0   4   0   1
  6:   5   0   9   0   5   0   1
  7:   0  14   0  14   0   6   0  1
  8:  14   0  28   0  20   0   7  0  1
  9:   0  42   0  48   0  27   0  8  0  1
  10: 42   0  90   0  75   0  35  0  9  0  1
  ... (Reformatted by _Wolfdieter Lang_, Sep 20 2013)
E.g., the fourth row corresponds to the polynomial p(3,x)= 2*x + x^3.
From _Paul Barry_, May 29 2009: (Start)
Production matrix is
  0, 1,
  1, 0, 1,
  0, 1, 0, 1,
  0, 0, 1, 0, 1,
  0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 1, 0, 1,
  0, 0, 0, 0, 0, 0, 0, 1, 0, 1 (End)
Boas-Buck recurrence for column k = 2, n = 6: a(6, 2) = (3/4)*(0 + 2*a(4 ,2) + 0 + 6*a(2, 2)) = (3/4)*(2*3 + 6) = 9. - _Wolfdieter Lang_, Aug 11 2017

References

  • J. H. Conway and D. A. Smith, On Quaternions and Octonions, A K Peters, Ltd., Natick, MA, 2003. See p. 60. MR1957212 (2004a:17002)
  • A. Nkwanta, Lattice paths and RNA secondary structures, in African Americans in Mathematics, ed. N. Dean, Amer. Math. Soc., 1997, pp. 137-147.

Links

Crossrefs

Cf. A008315, A049310, A000108, A001405 (row sums), A145973, A153585, A108786, A037952. Another version: A008313. A039598 and A039599 without zeros, and odd and even numbered rows.
Variant without zero-diagonals: A033184 and with rows reversed: A009766.

Programs

  • Haskell
    a053121 n k = a053121_tabl !! n !! k
a053121_row n = a053121_tabl !! n
a053121_tabl = iterate
   (\row -> zipWith (+) ([0] ++ row) (tail row ++ [0,0])) [1]
-- Reinhard Zumkeller, Feb 24 2012
  • Maple
    T:=proc(n,k): if n+k mod 2 = 0 then (k+1)*binomial(n+1,(n-k)/2)/(n+1) else 0 fi end: for n from 0 to 13 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form; Emeric Deutsch, Oct 12 2006
F:=proc(l,p) if ((l-p) mod 2) = 1 then 0 else (p+1)*l!/( ( (l-p)/2 )! * ( (l+p)/2 +1)! ); fi; end;
r:=n->[seq( F(n,p),p=0..n)]; [seq(r(n),n=0..15)]; # N. J. A. Sloane, Jan 29 2011
A053121 := proc(n,k) option remember; `if`(k>n or k<0,0,`if`(n=k,1,
procname(n-1,k-1)+procname(n-1,k+1))) end proc:
seq(print(seq(A053121(n,k), k=0..n)),n=0..12); # Peter Luschny, May 01 2011
  • Mathematica
    a[n_, m_] /; n < m || OddQ[n-m] = 0; a[n_, m_] = (m+1) Binomial[n+1, (n-m)/2]/(n+1); Flatten[Table[a[n, m], {n, 0, 12}, {m, 0, n}]] [[1 ;; 90]] (* Jean-François Alcover, May 18 2011 *)
T[0, 0] := 1; T[n_, k_]/;0<=k<=n := T[n, k] = T[n-1, k-1]+T[n-1, k+1]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 0, 12}, {k, 0, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    T(n, m)=if(nCharles R Greathouse IV, Mar 09 2016
  • Sage
    def A053121_triangle(dim):
    M = matrix(ZZ,dim,dim)
    for n in (0..dim-1): M[n,n] = 1
    for n in (1..dim-1):
        for k in (0..n-1):
            M[n,k] = M[n-1,k-1] + M[n-1,k+1]
    return M
A053121_triangle(13) # Peter Luschny, Sep 19 2012

Formula

a(n, m) := 0 if n
a(n, m) = (4*(n-1)*a(n-2, m) + 2*(m+1)*a(n-1, m-1))/(n+m+2), a(n, m)=0 if n
G.f. for m-th column: c(x^2)*(x*c(x^2))^m, where c(x) = g.f. for Catalan numbers A000108.
G.f.: G(t,z) = c(z^2)/(1 - t*z*c(z^2)), where c(z) = (1 - sqrt(1-4*z))/(2*z) is the g.f. for the Catalan numbers (A000108). - Emeric Deutsch, Jun 16 2011
a(n, m) = a(n-1, m-1) + a(n-1, m+1) if n > 0 and m >= 0, a(0, 0)=1, a(0, m)=0 if m > 0, a(n, m)=0 if m < 0. - Henry Bottomley, Jan 25 2001
Sum_{k>=0} T(m,k)^2 = A000108(m). - Paul D. Hanna, Apr 23 2005
Sum_{k>=0} T(m, k)*T(n, k) = 0 if m+n is odd; Sum_{k>=0} T(m, k)*T(n, k) = A000108((m+n)/2) if m+n is even. - Philippe Deléham, May 26 2005
T(n,k)=sum{i=0..n, (-1)^(n-i)*C(n,i)*sum{j=0..i, C(i,j)*(C(i-j,j+k)-C(i-j,j+k+2))}}; Column k has e.g.f. BesselI(k,2x)-BesselI(k+2,2x). - Paul Barry, Feb 16 2006
Sum_{k=0..n} T(n,k)*(k+1) = 2^n. - Philippe Deléham, Mar 22 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A054336(n,k). - Philippe Deléham, Mar 30 2007
T(2*n+1,2*k+1) = A039598(n,k), T(2*n,2*k) = A039599(n,k). - Philippe Deléham, Apr 16 2007
Sum_{k=0..n} T(n,k)^x = A000027(n+1), A001405(n), A000108(n), A003161(n), A129123(n) for x = 0,1,2,3,4 respectively. - Philippe Deléham, Nov 22 2009
Sum_{k=0..n} T(n,k)*x^k = A126930(n), A126120(n), A001405(n), A054341(n), A126931(n) for x = -1, 0, 1, 2, 3 respectively. - Philippe Deléham, Nov 28 2009
Sum_{k=0..n} T(n,k)*A000045(k+1) = A098615(n). - Philippe Deléham, Feb 03 2012
Recurrence for row polynomials C(n, x) := Sum_{m=0..n} a(n, m)*x^m = x*Sum_{k=0..n} Chat(k)*C(n-1-k, x), n >= 0, with C(-1, 1/x) = 1/x and Chat(k) = A000108(k/2) if n is even and 0 otherwise. From the o.g.f. of the row polynomials: G(z; x) := Sum_{n >= 0} C(n, x)*z^n = c(z^2)*(1 + x*z*G(z, x)), with the o.g.f. c of A000108. - Ahmet Zahid KÜÇÜK and Wolfdieter Lang, Aug 23 2015
The Boas-Buck recurrence (see a comment above) for the sequence of column m is: a(n, m) = ((m+1)/(n-m))*Sum_{j=0..n-1-m} (1/2)*(1 - (-1)^j)*binomial(j+1, (j+1)/2)* a(n-1-j, k), for n > m >= 0 and input a(m, m) = 1. - Wolfdieter Lang, Aug 11 2017
Sum_{m=1..n} a(n,m) = A037952(n). - R. J. Mathar, Sep 23 2021

Extensions

Edited by N. J. A. Sloane, Jan 29 2011

A033184 Catalan triangle A009766 transposed.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 5, 5, 3, 1, 14, 14, 9, 4, 1, 42, 42, 28, 14, 5, 1, 132, 132, 90, 48, 20, 6, 1, 429, 429, 297, 165, 75, 27, 7, 1, 1430, 1430, 1001, 572, 275, 110, 35, 8, 1, 4862, 4862, 3432, 2002, 1001, 429, 154, 44, 9, 1
Offset: 1

Views

Author

Christian G. Bower

Keywords

Comments

Triangle read by rows: T(n,k) = number of Dyck n-paths (A000108) containing k returns to ground level. E.g., the paths UDUUDD, UUDDUD each have 2 returns; so T(3,2)=2. Row sums over even-indexed columns are the Fine numbers A000957. - David Callan, Jul 25 2005
Triangular array of numbers a(n,k) = number of linear forests of k planted planar trees and n non-root nodes.
Catalan convolution triangle; with offset [0,0]: a(n,m)=(m+1)*binomial(2*n-m,n-m)/(n+1), n >= m >= 0, else 0. G.f. for column m: c(x)*(x*c(x))^m with c(x) g.f. for A000108 (Catalan). - Wolfdieter Lang, Sep 12 2001
a(n+1,m+1), n >= m >= 0, a(n,m) := 0, nA030528(n,m)*(-1)^(n-m).
a(n,k)=number of Dyck paths of semilength n and having k returns to the axis. Also number of Dyck paths of semilength n and having first peak at height k. Also number of ordered trees with n edges and root degree k. Also number of ordered trees with n edges and having the leftmost leaf at level k. Also number of parallelogram polyominoes of semiperimeter n+1 and having k cells in the leftmost column. - Emeric Deutsch, Mar 01 2004
Triangle T(n,k) with 1<=k<=n given by [0, 1, 1, 1, 1, 1, 1, 1, ...] DELTA [1, 0, 0, 0, 0, 0, 0, 0, ...] = 1; 0, 1; 0, 1, 1; 0, 2, 2, 1; 0, 5, 5, 3, 1; 0, 14, 14, 9, 4, 1; ... where DELTA is the operator defined in A084938; essentially the same triangle as A059365. - Philippe Deléham, Jun 14 2004
Number of Dyck paths of semilength and having k-1 peaks at height 2. - Emeric Deutsch, Aug 31 2004
Riordan array (c(x),x*c(x)), c(x) the g.f. of A000108. Inverse of Riordan array (1-x,x*(1-x)). - Paul Barry, Jun 22 2005
Subtriangle of triangle A106566. - Philippe Deléham, Jan 07 2007
T(n, k) is also the number of order-preserving and order-decreasing full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
Triangle read by rows, product of A065600 and A007318 considered as infinite lower triangular arrays; A033184 = A065600*A007318. - Philippe Deléham, Dec 07 2009
The formula stating "Column k is the k-fold convolution of column 1" is equivalent to repeatedly applying M to [1,0,0,0,...], where M is an upper triangular matrix of all 1's with an additional single subdiagonal of 1's. - Gary W. Adamson, Jun 06 2011
4^(n-1) = (n-th row terms) dot (first n terms in A001792), where A001792 = binomial transform of the natural numbers: (1, 3, 8, 20, 48, 112, ...). Example: 4^4 = 256 = (14, 14, 9, 4, 1) dot (1, 3, 8, 20, 48) = (42 + 42 + 28 + 14 + 5 + 1) = 256. - Gary W. Adamson, Jun 17 2011
The e.g.f. for the n-th subdiagonal of the triangle has the form exp(x)*P(n,x), where P(n,x) is the e.g.f. for row n of triangle A039599. For example, the third row of A039599 is [5, 9, 5, 1] and so the third subdiagonal sequence of this triangle [5, 14, 28, 48, 75, ...] has the e.g.f. exp(x)*(5 + 9*x + 5*x^2/2! + x^3/3!). - Peter Bala, Oct 15 2019
Antidiagonals of convolution matrix of Table 1.3, p. 397, of Hoggatt and Bicknell. - Tom Copeland, Dec 25 2019
Also the convolution triangle of A120588(n) = A000108(n-1) for n > 0. - Peter Luschny, Oct 07 2022

Examples

			Triangle begins:
  ---+-----------------------------------
  n\k|   1    2    3    4    5    6    7
  ---+-----------------------------------
   1 |   1
   2 |   1    1
   3 |   2    2    1
   4 |   5    5    3    1
   5 |  14   14    9    4    1
   6 |  42   42   28   14    5    1
   7 | 132  132   90   48   20    6    1
From _Peter Bala_, Feb 17 2025: (Start)
The array factorizes as an infinite product (read from right to left) of triangular arrays:
  / 1               \        / 1              \ / 1              \ / 1             \
  | 1    1           |       | 0   1          | | 0  1           | | 1  1          |
  | 2    2   1       | = ... | 0   0   1      | | 0  1   1       | | 1  1  1       |
  | 5    5   3   1   |       | 0   0   1  1   | | 0  1   1  1    | | 1  1  1  1    |
  |14   14   9   4  1|       | 0   0   1  1  1| | 0  1   1  1  1 | | 1  1  1  1  1 |
  |...               |       |...             | |...             | |...            |
See Bala, Example 2.1. (End)

Links

Crossrefs

Rows of Catalan triangle A009766 read backwards.
a(n, 1) = A000108(n-1). Row sums = A000108(n) (Catalan).
The following are all versions of (essentially) the same Catalan triangle: A009766, A030237, A033184, A059365, A099039, A106566, A130020, A047072.
Diagonals give A000108, A000245, A002057, A000344, A003517, A000588, A003518, A003519, A001392.
Cf. A116364 (row squared sums), A120588.

Programs

  • Haskell
    a033184 n k = a033184_tabl !! (n-1) !! (k-1)
a033184_row n = a033184_tabl !! (n-1)
a033184_tabl = map reverse a009766_tabl
-- Reinhard Zumkeller, Feb 19 2014
  • Magma
    /* As triangle: */ [[Binomial(2*n-k,n)*k/(2*n-k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 12 2015
  • Maple
    a := proc(n,k) if k<=n then k*binomial(2*n-k,n)/(2*n-k) else 0 fi end: seq(seq(a(n,k),k=1..n),n=1..10);
# Uses function PMatrix from A357368. Adds row and column for n, k = 0.
PMatrix(10, n -> binomial(2*(n-1), n-1) / n); # Peter Luschny, Oct 07 2022
  • Mathematica
    nn = 10; c = (1 - (1 - 4 x)^(1/2))/(2 x); f[list_] := Select[list, # > 0 &]; Map[f, Drop[CoefficientList[Series[y x c/(1 - y x c), {x, 0, nn}], {x, y}],1]] //Flatten (* Geoffrey Critzer, Jan 31 2012 *)
Flatten[Reverse /@ NestList[Append[Accumulate[#], Last[Accumulate[#]]] &, {1}, 9]] (* Birkas Gyorgy, May 19 2012 *)
T[1, 1] := 1; T[n_, k_]/;1<=k<=n := T[n, k] = T[n-1, k-1]+T[n, k+1]; T[n_, k_] := 0; Flatten@Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* Oliver Seipel, Dec 31 2024 *)
  • PARI
    T(n,k)=binomial(2*(n-k)+k,n-k)*(k+1)/(n+1) \\ Paul D. Hanna, Aug 11 2008
  • Sage
    # The simplest way to construct the triangle.
def A033184_triangle(n) :
    T = [0 for i in (0..n)]
    for k in (1..n) :
        T[k] = 1
        for i in range(k-1,0,-1) :
            T[i] = T[i-1] + T[i+1]
        print([T[i] for i in (1..k)])
A033184_triangle(10) # Peter Luschny, Jan 27 2012

Formula

Column k is the k-fold convolution of column 1. The triangle is also defined recursively by (i) entries outside triangle are 0, (ii) top left entry is 1, (iii) every other entry is sum of its east and northwest neighbor. - David Callan, Jul 25 2005
G.f.: t*x*c/(1-t*x*c), where c=(1-sqrt(1-4*x))/(2*x) is the g.f. of the Catalan numbers (A000108). - Emeric Deutsch, Mar 01 2004
T(n+1,k+1) = C(2*n-k, n-k)*(k+1)/(n+1). - Paul D. Hanna, Aug 11 2008
T((m+1)*n+r-1,m*n+r-1)*r/(m*n+r) = Sum_{k=1..n} (k/n)*T((m+1)*n-k-1,m*n-1)*T(r+k,r), n >= m > 1. - Vladimir Kruchinin, Mar 17 2011
T(n-1,m-1) = (m/n)*Sum_{k=1..n-m+1} (k*A000108(k-1)*T(n-k-1,m-2)), n >= m > 1. - Vladimir Kruchinin, Mar 17 2011
T(n,k) = C(2*n-k-1,n-k) - C(2*n-k-1,n-k-1). - Dennis P. Walsh, Mar 19 2012
T(n,k) = C(2*n-k,n)*k/(2*n-k). - Dennis P. Walsh, Mar 19 2012
T(n,k) = T(n,k-1) - T(n-1,k-2). - Dennis P. Walsh, Mar 19 2012
G.f.: 2*x*y / (1 + sqrt(1 - 4*x) - 2*x*y) = Sum_{n >= k > 0} T(n, k) * x^n * y^k. - Michael Somos, Jun 06 2016

A039598 Triangle formed from odd-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x). Sometimes called Catalan's triangle.

Original entry on oeis.org

1, 2, 1, 5, 4, 1, 14, 14, 6, 1, 42, 48, 27, 8, 1, 132, 165, 110, 44, 10, 1, 429, 572, 429, 208, 65, 12, 1, 1430, 2002, 1638, 910, 350, 90, 14, 1, 4862, 7072, 6188, 3808, 1700, 544, 119, 16, 1, 16796, 25194, 23256, 15504, 7752, 2907, 798, 152, 18, 1
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

T(n,k) is the number of leaves at level k+1 in all ordered trees with n+1 edges. - Emeric Deutsch, Jan 15 2005
Riordan array ((1-2x-sqrt(1-4x))/(2x^2),(1-2x-sqrt(1-4x))/(2x)). Inverse array is A053122. - Paul Barry, Mar 17 2005
T(n,k) is the number of walks of n steps, each in direction N, S, W, or E, starting at the origin, remaining in the upper half-plane and ending at height k (see the R. K. Guy reference, p. 5). Example: T(3,2)=6 because we have ENN, WNN, NEN, NWN, NNE and NNW. - Emeric Deutsch, Apr 15 2005
Triangle T(n,k), 0<=k<=n, read by rows given by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = 2*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + 2*T(n-1,k) + T(n-1,k+1) for k>=1. - Philippe Deléham, Mar 30 2007
Number of (2n+1)-step walks from (0,0) to (2n+1,2k+1) and consisting of steps u=(1,1) and d=(1,-1) in which the path stays in the nonnegative quadrant. Examples: T(2,0)=5 because we have uuudd, uudud, uuddu, uduud, ududu; T(2,1)=4 because we have uuuud, uuudu, uuduu, uduuu; T(2,2)=1 because we have uuuuu. - Philippe Deléham, Apr 16 2007, Apr 18 2007
Triangle read by rows: T(n,k)=number of lattice paths from (0,0) to (n,k) that do not go below the line y=0 and consist of steps U=(1,1), D=(1,-1) and two types of steps H=(1,0); example: T(3,1)=14 because we have UDU, UUD, 4 HHU paths, 4 HUH paths and 4 UHH paths. - Philippe Deléham, Sep 25 2007
This triangle belongs to the family of triangles defined by T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0) = x*T(n-1,0) + T(n-1,1), T(n,k) = T(n-1,k-1) + y*T(n-1,k) + T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y): (0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970; (1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877; (1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598; (2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954; (3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791; (4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. - Philippe Deléham, Sep 25 2007
With offset [1,1] this is the (ordinary) convolution triangle a(n,m) with o.g.f. of column m given by (c(x)-1)^m, where c(x) is the o.g.f. for Catalan numbers A000108. See the Riordan comment by Paul Barry.
T(n, k) is also the number of order-preserving full transformations (of an n-chain) with exactly k fixed points. - Abdullahi Umar, Oct 02 2008
T(n,k)/2^(2n+1) = coefficients of the maximally flat lowpass digital differentiator of the order N=2n+3. - Pavel Holoborodko (pavel(AT)holoborodko.com), Dec 19 2008
The signed triangle S(n,k) := (-1)^(n-k)*T(n,k) provides the transformation matrix between f(n,l) := L(2*l)*5^n*F(2*l)^(2*n+1) (F=Fibonacci numbers A000045, L=Lucas numbers A000032) and F(4*l*(k+1)), k = 0, ..., n, for each l>=0: f(n,l) = Sum_{k=0..n} S(n,k)*F(4*l*(k+1)), n>=0, l>=0. Proof: the o.g.f. of the l.h.s., G(l;x) := Sum_{n>=0} f(n,l)*x^n = F(4*l)/(1 - 5*F(2*l)^2*x) is shown to match the o.g.f. of the r.h.s.: after an interchange of the n- and k-summation, the Riordan property of S = (C(x)/x,C(x)) (compare with the above comments by Paul Barry), with C(x) := 1 - c(-x), with the o.g.f. c(x) of A000108 (Catalan numbers), is used, to obtain, after an index shift, first Sum_{k>=0} F(4*l*(k))*GS(k;x), with the o.g.f of column k of triangle S which is GS(k;x) := Sum_{n>=k} S(n,k)*x^n = C(x)^(k+1)/x. The result is GF(l;C(x))/x with the o.g.f. GF(l,x) := Sum_{k>=0} F(4*l*k)*x^k = x*F(4*l)/(1-L(4*l)*x+x^2) (see a comment on A049670, and A028412). If one uses then the identity L(4*n) - 5*F(2*n)^2 = 2 (in Koshy's book [reference under A065563] this is No. 15, p. 88, attributed to Lucas, 1876), the proof that one recovers the o.g.f. of the l.h.s. from above boils down to a trivial identity on the Catalan o.g.f., namely 1/c^2(-x) = 1 + 2*x - (x*c(-x))^2. - Wolfdieter Lang, Aug 27 2012
O.g.f. for row polynomials R(x) := Sum_{k=0..n} a(n,k)*x^k:
((1+x) - C(z))/(x - (1+x)^2*z) with C the o.g.f. of A000108 (Catalan numbers). From Riordan ((C(x)-1)/x,C(x)-1), compare with a Paul Barry comment above. This coincides with the o.g.f. given by Emeric Deutsch in the formula section. - Wolfdieter Lang, Nov 13 2012
The A-sequence for this Riordan triangle is [1,2,1] and the Z-sequence is [2,1]. See a W. Lang link under A006232 with details and references. - Wolfdieter Lang, Nov 13 2012
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n+1, 2*k+1). T(n, k) appears in the formula for the (2*n+1)-th power of the algebraic number rho(N) := 2*cos(Pi/N) = R(N, 2) in terms of the even-indexed diagonal/side length ratios R(N, 2*(k+1)) = S(2*k+1, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310): rho(N)^(2*n+1) = Sum_{k=0..n} T(n, k)*R(N, 2*(k+1)), n >= 0, identical in N >= 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears. For the even powers of rho(n) see A039599. (End)
The tridiagonal Toeplitz production matrix P in the Example section corresponds to the unsigned Cartan matrix for the simple Lie algebra A_n as n tends to infinity (cf. Damianou ref. in A053122). - Tom Copeland, Dec 11 2015 (revised Dec 28 2015)
T(n,k) is the number of pairs of non-intersecting walks of n steps, each in direction N or E, starting at the origin, and such that the end points of the two paths are separated by a horizontal distance of k. See Shapiro 1976. - Peter Bala, Apr 12 2017
Also the convolution triangle of the Catalan numbers A000108. - Peter Luschny, Oct 07 2022

Examples

			Triangle T(n,k) starts:
n\k     0      1      2      3      4     5    6    7   8  9 10
0:      1
1:      2      1
2:      5      4      1
3:     14     14      6      1
4:     42     48     27      8      1
5:    132    165    110     44     10     1
6:    429    572    429    208     65    12    1
7:   1430   2002   1638    910    350    90   14    1
8:   4862   7072   6188   3808   1700   544  119   16   1
9:  16796  25194  23256  15504   7752  2907  798  152  18  1
10: 58786  90440  87210  62016  33915 14364 4655 1120 189 20  1
... Reformatted and extended by _Wolfdieter Lang_, Nov 13 2012.
Production matrix begins:
2, 1
1, 2, 1
0, 1, 2, 1
0, 0, 1, 2, 1
0, 0, 0, 1, 2, 1
0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 1, 2, 1
0, 0, 0, 0, 0, 0, 1, 2, 1
- _Philippe Deléham_, Nov 07 2011
From _Wolfdieter Lang_, Nov 13 2012: (Start)
Recurrence: T(5,1) = 165 = 1*42 + 2*48 +1*27. The Riordan A-sequence is [1,2,1].
Recurrence from Riordan Z-sequence [2,1]: T(5,0) = 132 = 2*42 + 1*48. (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
  Example for rho(N) = 2*cos(Pi/N) powers:
  n=2: rho(N)^5 = 5*R(N, 2) + 4*R(N, 4) + 1*R(N, 6) = 5*S(1, rho(N)) + 4*S(3, rho(N)) + 1*S(5, rho(N)), identical in N >= 1. For N=5 (the pentagon with only one distinct diagonal) the degree delta(5) = 2, hence R(5, 4) and R(5, 6) can be reduced, namely to R(5, 1) = 1 and R(5, 6) = -R(5,1) = -1, respectively. Thus rho(5)^5 = 5*R(N, 2) + 4*1  + 1*(-1) = 3 + 5*R(N, 2) = 3 + 5*rho(5), with the golden section rho(5). (End)

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.

Links

Crossrefs

Mirror image of A050166. Row sums are A001700.
Cf. A000108, A008313, A039599, A183134, A094527, A033184, A035324, A053122, A172417.

Programs

  • Magma
    /* As triangle: */ [[Binomial(2*n,n-k) - Binomial(2*n,n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 22 2015
  • Maple
    T:=(n,k)->binomial(2*n, n-k) - binomial(2*n, n-k-2); # N. J. A. Sloane, Aug 26 2013
# Uses function PMatrix from A357368. Adds row and column above and to the left.
PMatrix(10, n -> binomial(2*n, n) / (n + 1)); # Peter Luschny, Oct 07 2022
  • Mathematica
    Flatten[Table[Binomial[2n, n-k] - Binomial[2n, n-k-2], {n,0,9}, {k,0,n}]] (* Jean-François Alcover, May 03 2011 *)
  • PARI
    T(n,k)=binomial(2*n,n-k) - binomial(2*n,n-k-2) \\ Charles R Greathouse IV, Nov 07 2016
  • Sage
    # Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle.
def A039598_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True; h = 1
    for i in range(2*n) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        b = not b
        if b : print([D[z] for z in (1..h-1) ])
A039598_triangle(10)  # Peter Luschny, May 01 2012

Formula

Row n: C(2n, n-k) - C(2n, n-k-2).
a(n, k) = C(2n+1, n-k)*2*(k+1)/(n+k+2) = A050166(n, n-k) = a(n-1, k-1) + 2*a(n-1, k)+ a (n-1, k+1) [with a(0, 0) = 1 and a(n, k) = 0 if n<0 or nHenry Bottomley, Sep 24 2001
From Philippe Deléham, Feb 14 2004: (Start)
T(n, 0) = A000108(n+1), T(n, k) = 0 if n0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+2) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
Sum_{k>=0} T(m, k)*T(n, k) = A000108(m+n+1). (End)
T(n, k) = A009766(n+k+1, n-k) = A033184(n+k+2, 2k+2). - Philippe Deléham, Feb 14 2004
Sum_{j>=0} T(k, j)*A039599(n-k, j) = A028364(n, k). - Philippe Deléham, Mar 04 2004
Antidiagonal Sum_{k=0..n} T(n-k, k) = A000957(n+3). - Gerald McGarvey, Jun 05 2005
The triangle may also be generated from M^n * [1,0,0,0,...], where M = an infinite tridiagonal matrix with 1's in the super- and subdiagonals and [2,2,2,...] in the main diagonal. - Gary W. Adamson, Dec 17 2006
G.f.: G(t,x) = C^2/(1-txC^2), where C = (1-sqrt(1-4x))/(2x) is the Catalan function. From here G(-1,x)=C, i.e., the alternating row sums are the Catalan numbers (A000108). - Emeric Deutsch, Jan 20 2007
Sum_{k=0..n} T(n,k)*x^k = A000957(n+1), A000108(n), A000108(n+1), A001700(n), A049027(n+1), A076025(n+1), A076026(n+1) for x=-2,-1,0,1,2,3,4 respectively (see square array in A067345). - Philippe Deléham, Mar 21 2007, Nov 04 2011
Sum_{k=0..n} T(n,k)*(k+1) = 4^n. - Philippe Deléham, Mar 30 2007
Sum_{j>=0} T(n,j)*binomial(j,k) = A035324(n,k), A035324 with offset 0 (0 <= k <= n). - Philippe Deléham, Mar 30 2007
T(n,k) = A053121(2*n+1,2*k+1). - Philippe Deléham, Apr 16 2007, Apr 18 2007
T(n,k) = A039599(n,k) + A039599(n,k+1). - Philippe Deléham, Sep 11 2007
Sum_{k=0..n+1} T(n+1,k)*k^2 = A029760(n). - Philippe Deléham, Dec 16 2007
Sum_{k=0..n} T(n,k)*A059841(k) = A000984(n). - Philippe Deléham, Nov 12 2008
G.f.: 1/(1-xy-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-2x-x^2/(1-.... (continued fraction).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A001700(n), A194723(n+1), A194724(n+1), A194725(n+1), A194726(n+1), A194727(n+1), A194728(n+1), A194729(n+1), A194730(n+1) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Nov 03 2011
From Peter Bala, Dec 21 2014: (Start)
This triangle factorizes in the Riordan group as ( C(x), x*C(x) ) * ( 1/(1 - x), x/(1 - x) ) = A033184 * A007318, where C(x) = (1 - sqrt(1 - 4*x))/(2*x) is the o.g.f. for the Catalan numbers A000108.
Let U denote the lower unit triangular array with 1's on or below the main diagonal and zeros elsewhere. For k = 0,1,2,... define U(k) to be the lower unit triangular block array
/I_k 0\
\ 0 U/ having the k X k identity matrix I_k as the upper left block; in particular, U(0) = U. Then this array equals the bi-infinite product (...*U(2)*U(1)*U(0))*(U(0)*U(1)*U(2)*...). (End)
From Peter Bala, Jul 21 2015: (Start)
O.g.f. G(x,t) = (1/x) * series reversion of ( x/f(x,t) ), where f(x,t) = ( 1 + (1 + t)*x )^2/( 1 + t*x ).
1 + x*d/dx(G(x,t))/G(x,t) = 1 + (2 + t)*x + (6 + 4*t + t^2)*x^2 + ... is the o.g.f for A094527. (End)
Conjecture: Sum_{k=0..n} T(n,k)/(k+1)^2 = H(n+1)*A000108(n)*(2*n+1)/(n+1), where H(n+1) = Sum_{k=0..n} 1/(k+1). - Werner Schulte, Jul 23 2015
From Werner Schulte, Jul 25 2015: (Start)
Sum_{k=0..n} T(n,k)*(k+1)^2 = (2*n+1)*binomial(2*n,n). (A002457)
Sum_{k=0..n} T(n,k)*(k+1)^3 = 4^n*(3*n+2)/2.
Sum_{k=0..n} T(n,k)*(k+1)^4 = (2*n+1)^2*binomial(2*n,n).
Sum_{k=0..n} T(n,k)*(k+1)^5 = 4^n*(15*n^2+15*n+4)/4. (End)
The o.g.f. G(x,t) is such that G(x,t+1) is the o.g.f. for A035324, but with an offset of 0, and G(x,t-1) is the o.g.f. for A033184, again with an offset of 0. - Peter Bala, Sep 20 2015
Denote this lower triangular array by L; then L * transpose(L) is the Cholesky factorization of the Hankel matrix ( 1/(i+j)*binomial(2*i+2*j-2, i+j-1) )A172417%20read%20as%20a%20square%20array.%20See%20Chamberland,%20p.%201669.%20-%20_Peter%20Bala">i,j >= 1 = A172417 read as a square array. See Chamberland, p. 1669. - _Peter Bala, Oct 15 2023

Extensions

Typo in one entry corrected by Philippe Deléham, Dec 16 2007
Previous Showing 11-20 of 134 results. Next

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.