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.

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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

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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

A007318 Pascal's triangle read by rows: C(n,k) = binomial(n,k) = n!/(k!*(n-k)!), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 5, 10, 10, 5, 1, 1, 6, 15, 20, 15, 6, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1
Offset: 0

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Author

N. J. A. Sloane and Mira Bernstein, Apr 28 1994

Keywords

Comments

A. W. F. Edwards writes: "It [the triangle] was first written down long before 1654, the year in which Blaise Pascal wrote his Traité du triangle arithmétique, but it was this work that brought together all the different aspects of the numbers for the first time. In it Pascal developed the properties of the number as a piece of pure mathematics ... and then, in a series of appendices, showed how these properties were relevant to the study of the figurate numbers, to the theory of combinations, to the expansion of binomial expressions, and to the solution of an important problem in the theory of probability." (A. W. F. Edwards, Pascal's Arithmetical Triangle, Johns Hopkins University Press (2002), p. xiii)
Edwards reports that the naming of the triangle after Pascal was done first by Montmort in 1708 as the "Table de M. Pascal pour les combinaisons" and then by De Moivre in 1730 as the "Triangulum Arithmeticum PASCALANIUM". (Edwards, p. xiv)
In China, Yang Hui in 1261 listed the coefficients of (a+b)^n up to n=6, crediting the expansion to Chia Hsein's Shih-so suan-shu circa 1100. Another prominent early use was in Chu Shih-Chieh's Precious Mirror of the Four Elements in 1303. (Edwards, p. 51)
In Persia, Al-Karaji discovered the binomial triangle "some time soon after 1007", and Al-Samawal published it in the Al-bahir some time before 1180. (Edwards, p. 52)
In India, Halayuda's commentary (circa 900) on Pingala's treatise on syllabic combinations (circa 200 B.C.E.) contains a clear description of the additive computation of the triangle. (Amulya Kumar Bag, Binomial Theorem in Ancient India, p. 72)
Also in India, the multiplicative formula for C(n,k) was known to Mahavira in 850 and restated by Bhaskara in 1150. (Edwards, p. 27)
In Italy, Tartaglia published the triangle in his General trattato (1556), and Cardano published it in his Opus novum (1570). (Edwards, p. 39, 44) - Russ Cox, Mar 29 2022
Also sometimes called Omar Khayyam's triangle.
Also sometimes called Yang Hui's triangle.
C(n,k) = number of k-element subsets of an n-element set.
Row n gives coefficients in expansion of (1+x)^n.
Binomial(n+k-1,n-1) is the number of ways of placing k indistinguishable balls into n boxes (the "bars and stars" argument - see Feller).
Binomial(n-1,k-1) is the number of compositions (ordered partitions) of n with k summands.
Binomial(n+k-1,k-1) is the number of weak compositions (ordered weak partitions) of n into exactly k summands. - Juergen Will, Jan 23 2016
Binomial(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0) and (1,1). - Joerg Arndt, Jul 01 2011
If thought of as an infinite lower triangular matrix, inverse begins:
+1
-1 +1
+1 -2 +1
-1 +3 -3 +1
+1 -4 +6 -4 +1
All 2^n palindromic binomial coefficients starting after the A006516(n)-th entry are odd. - Lekraj Beedassy, May 20 2003
Binomial(n+k-1,n-1) is the number of standard tableaux of shape (n,1^k). - Emeric Deutsch, May 13 2004
Can be viewed as an array, read by antidiagonals, where the entries in the first row and column are all 1's and A(i,j) = A(i-1,j) + A(i,j-1) for all other entries. The determinant of each of its n X n subarrays starting at (0,0) is 1. - Gerald McGarvey, Aug 17 2004
Also the lower triangular readout of the exponential of a matrix whose entry {j+1,j} equals j+1 (and all other entries are zero). - Joseph Biberstine (jrbibers(AT)indiana.edu), May 26 2006
Binomial(n-3,k-1) counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 132 and k descents. Binomial(n-3,k-1) also counts the permutations in S_n which have zero occurrences of the pattern 231 and one occurrence of the pattern 213 and k descents. - David Hoek (david.hok(AT)telia.com), Feb 28 2007
Inverse of A130595 (as an infinite lower triangular matrix). - Philippe Deléham, Aug 21 2007
Consider integer lists LL of lists L of the form LL = [m#L] = [m#[k#2]] (where '#' means 'times') like LL(m=3,k=3) = [[2,2,2],[2,2,2],[2,2,2]]. The number of the integer list partitions of LL(m,k) is equal to binomial(m+k,k) if multiple partitions like [[1,1],[2],[2]] and [[2],[2],[1,1]] and [[2],[1,1],[2]] are counted only once. For the example, we find 4*5*6/3! = 20 = binomial(6,3). - Thomas Wieder, Oct 03 2007
The infinitesimal generator for Pascal's triangle and its inverse is A132440. - Tom Copeland, Nov 15 2007
Row n>=2 gives the number of k-digit (k>0) base n numbers with strictly decreasing digits; e.g., row 10 for A009995. Similarly, row n-1>=2 gives the number of k-digit (k>1) base n numbers with strictly increasing digits; see A009993 and compare A118629. - Rick L. Shepherd, Nov 25 2007
From Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008: (Start)
Binomial(n+k-1, k) is the number of ways a sequence of length k can be partitioned into n subsequences (see the Naish link).
Binomial(n+k-1, k) is also the number of n- (or fewer) digit numbers written in radix at least k whose digits sum to k. For example, in decimal, there are binomial(3+3-1,3)=10 3-digit numbers whose digits sum to 3 (see A052217) and also binomial(4+2-1,2)=10 4-digit numbers whose digits sum to 2 (see A052216). This relationship can be used to generate the numbers of sequences A052216 to A052224 (and further sequences using radix greater than 10). (End)
From Milan Janjic, May 07 2008: (Start)
Denote by sigma_k(x_1,x_2,...,x_n) the elementary symmetric polynomials. Then:
Binomial(2n+1,2k+1) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2(i*Pi/(2n+1)), (i=1,2,...,n).
Binomial(2n,2k+1) = 2n*sigma_{n-1-k}(x_1,x_2,...,x_{n-1}), where x_i = tan^2(i*Pi/(2n)), (i=1,2,...,n-1).
Binomial(2n,2k) = sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n)), (i=1,2,...,n).
Binomial(2n+1,2k) = (2n+1)sigma_{n-k}(x_1,x_2,...,x_n), where x_i = tan^2((2i-1)Pi/(4n+2)), (i=1,2,...,n). (End)
Given matrices R and S with R(n,k) = binomial(n,k)*r(n-k) and S(n,k) = binomial(n,k)*s(n-k), then R*S = T where T(n,k) = binomial(n,k)*[r(.)+s(.)]^(n-k), umbrally. And, the e.g.f.s for the row polynomials of R, S and T are, respectively, exp(x*t)*exp[r(.)*x], exp(x*t)*exp[s(.)*x] and exp(x*t)*exp[r(.)*x]*exp[s(.)*x] = exp{[t+r(.)+s(.)]*x}. The row polynomials are essentially Appell polynomials. See A132382 for an example. - Tom Copeland, Aug 21 2008
As the rectangle R(m,n) = binomial(m+n-2,m-1), the weight array W (defined generally at A144112) of R is essentially R itself, in the sense that if row 1 and column 1 of W=A144225 are deleted, the remaining array is R. - Clark Kimberling, Sep 15 2008
If A007318 = M as an infinite lower triangular matrix, M^n gives A130595, A023531, A007318, A038207, A027465, A038231, A038243, A038255, A027466, A038279, A038291, A038303, A038315, A038327, A133371, A147716, A027467 for n=-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 respectively. - Philippe Deléham, Nov 11 2008
The coefficients of the polynomials with e.g.f. exp(x*t)*(cosh(t)+sinh(t)). - Peter Luschny, Jul 09 2009
The triangle or chess sums, see A180662 for their definitions, link Pascal's triangle with twenty different sequences, see the crossrefs. All sums come in pairs due to the symmetrical nature of this triangle. The knight sums Kn14 - Kn110 have been added. It is remarkable that all knight sums are related to the Fibonacci numbers, i.e., A000045, but none of the others. - Johannes W. Meijer, Sep 22 2010
Binomial(n,k) is also the number of ways to distribute n+1 balls into k+1 urns so that each urn gets at least one ball. See example in the example section below. - Dennis P. Walsh, Jan 29 2011
Binomial(n,k) is the number of increasing functions from {1,...,k} to {1,...,n} since there are binomial(n,k) ways to choose the k distinct, ordered elements of the range from the codomain {1,...,n}. See example in the example section below. - Dennis P. Walsh, Apr 07 2011
Central binomial coefficients: T(2*n,n) = A000984(n), T(n, floor(n/2)) = A001405(n). - Reinhard Zumkeller, Nov 09 2011
Binomial(n,k) is the number of subsets of {1,...,n+1} with k+1 as median element. To see this, note that Sum_{j=0..min(k,n-k)}binomial(k,j)*binomial(n-k,j) = binomial(n,k). See example in Example section below. - Dennis P. Walsh, Dec 15 2011
This is the coordinator triangle for the lattice Z^n, see Conway-Sloane, 1997. - N. J. A. Sloane, Jan 17 2012
One of three infinite families of integral factorial ratio sequences of height 1 (see Bober, Theorem 1.2). The other two are A046521 and A068555. For real r >= 0, C_r(n,k) := floor(r*n)!/(floor(r*k)!*floor(r*(n-k))!) is integral. See A211226 for the case r = 1/2. - Peter Bala, Apr 10 2012
Define a finite triangle T(m,k) with n rows such that T(m,0) = 1 is the left column, T(m,m) = binomial(n-1,m) is the right column, and the other entries are T(m,k) = T(m-1,k-1) + T(m-1,k) as in Pascal's triangle. The sum of all entries in T (there are A000217(n) elements) is 3^(n-1). - J. M. Bergot, Oct 01 2012
The lower triangular Pascal matrix serves as a representation of the operator exp(RLR) in a basis composed of a sequence of polynomials p_n(x) characterized by ladder operators defined by R p_n(x) = p_(n+1)(x) and L p_n(x) = n p_(n-1)(x). See A132440, A218272, A218234, A097805, and A038207. The transposed and padded Pascal matrices can be associated to the special linear group SL2. - Tom Copeland, Oct 25 2012
See A193242. - Alexander R. Povolotsky, Feb 05 2013
A permutation p_1...p_n of the set {1,...,n} has a descent at position i if p_i > p_(i+1). Let S(n) denote the subset of permutations p_1...p_n of {1,...,n} such that p_(i+1) - p_i <= 1 for i = 1,...,n-1. Then binomial(n,k) gives the number of permutations in S(n+1) with k descents. Alternatively, binomial(n,k) gives the number of permutations in S(n+1) with k+1 increasing runs. - Peter Bala, Mar 24 2013
Sum_{n=>0} binomial(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where binomial(n,k) = 0. Also Sum_{n>=0} binomial(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
The square n X n submatrix (first n rows and n columns) of the Pascal matrix P(x) defined in the formulas below when multiplying on the left the Vandermonde matrix V(x_1,...,x_n) (with ones in the first row) translates the matrix to V(x_1+x,...,x_n+x) while leaving the determinant invariant. - Tom Copeland, May 19 2014
For k>=2, n>=k, k/((k/(k-1) - Sum_{n=k..m} 1/binomial(n,k))) = m!/((m-k+1)!*(k-2)!). Note: k/(k-1) is the infinite sum. See A000217, A000292, A000332 for examples. - Richard R. Forberg, Aug 12 2014
Let G_(2n) be the subgroup of the symmetric group S_(2n) defined by G_(2n) = {p in S_(2n) | p(i) = i (mod n) for i = 1,2,...,2n}. G_(2n) has order 2^n. Binomial(n,k) gives the number of permutations in G_(2n) having n + k cycles. Cf. A130534 and A246117. - Peter Bala, Aug 15 2014
C(n,k) = the number of Dyck paths of semilength n+1, with k+1 "u"'s in odd numbered positions and k+1 returns to the x axis. Example: {U = u in odd position and = return to x axis} binomial(3,0)=1 (Uudududd); binomial(3,1)=3 [(Uududd_Ud_), (Ud_Uududd_), (Uudd_Uudd_)]; binomial(3,2)=3 [(Ud_Ud_Uudd_), (Uudd_Ud_Ud_), (Ud_Uudd_Ud_)]; binomial(3,3)=1 (Ud_Ud_Ud_Ud_). - Roger Ford, Nov 05 2014
From Daniel Forgues, Mar 12 2015: (Start)
The binomial coefficients binomial(n,k) give the number of individuals of the k-th generation after n population doublings. For each doubling of population, each individual's clone has its generation index incremented by 1, and thus goes to the next row. Just tally up each row from 0 to 2^n - 1 to get the binomial coefficients.
0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
This is a fractal process: to get the pattern from 0 to 2^n - 1, append a shifted down (by one row) copy of the pattern from 0 to 2^(n-1) - 1 to the right of the pattern from 0 to 2^(n-1) - 1. (Inspired by the "binomial heap" data structure.)
Sequence of generation indices: 1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n) (see A000120):
{0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, ...}
Binary expansion of 0 to 15:
0 1 10 11 100 101 110 111 1000 1001 1010 1011 1100 1101 1111
(End)
A258993(n,k) = T(n+k,n-k), n > 0. - Reinhard Zumkeller, Jun 22 2015
T(n,k) is the number of set partitions w of [n+1] that avoid 1/2/3 with rb(w)=k. The same holds for ls(w)=k, where avoidance is in the sense of Klazar and ls,rb defined by Wachs and White.
Satisfies Benford's law [Diaconis, 1977] - N. J. A. Sloane, Feb 09 2017
Let {A(n)} be a set with exactly n identical elements, with {A(0)} being the empty set E. Let {A(n,k)} be the k-th iteration of {A(n)}, with {A(n,0)} = {A(n)}. {A(n,1)} = The set of all the subsets of A{(n)}, including {A(n)} and E. {A(n,k)} = The set of all subsets of {A(n,k-1)}, including all of the elements of {A(n,k-1)}. Let A(n,k) be the number of elements in {A(n,k)}. Then A(n,k) = C(n+k,k), with each successive iteration replicating the members of the k-th diagonal of Pascal's Triangle. See examples. - Gregory L. Simay, Aug 06 2018
Binomial(n-1,k) is also the number of permutations avoiding both 213 and 312 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n-1,k) is also the number of permutations avoiding both 132 and 213 with k ascents. - Lara Pudwell, Dec 19 2018
Binomial(n,k) is the dimension of the k-th exterior power of a vector space of dimension n. - Stefano Spezia, Dec 22 2018
C(n,k-1) is the number of unoriented colorings of the facets (or vertices) of an n-dimensional simplex using exactly k colors. Each chiral pair is counted as one when enumerating unoriented arrangements. - Robert A. Russell, Oct 20 2020
From Dilcher and Stolarsky: "Two of the most ubiquitous objects in mathematics are the sequence of prime numbers and the binomial coefficients (and thus Pascal's triangle). A connection between the two is given by a well-known characterization of the prime numbers: Consider the entries in the k-th row of Pascal's triangle, without the initial and final entries. They are all divisible by k if and only if k is a prime." - Tom Copeland, May 17 2021
Named "Table de M. Pascal pour les combinaisons" by Pierre Remond de Montmort (1708) after the French mathematician, physicist and philosopher Blaise Pascal (1623-1662). - Amiram Eldar, Jun 11 2021
Consider the n-th diagonal of the triangle as a sequence b(n) with n starting at 0. From it form a new sequence by leaving the 0th term as is, and thereafter considering all compositions of n, taking the product of b(i) over the respective numbers i in each composition, adding terms corresponding to compositions with an even number of parts subtracting terms corresponding to compositions with an odd number of parts. Then the n-th row of the triangle is obtained, with every second term multiplied by -1, followed by infinitely many zeros. For sequences starting with 1, this operation is a special case of a self-inverse operation, and therefore the converse is true. - Thomas Anton, Jul 05 2021
C(n,k) is the number of vertices in an n-dimensional unit hypercube, at an L1 distance of k (or: with a shortest path of k 1d-edges) from a given vertex. - Eitan Y. Levine, May 01 2023
C(n+k-1,k-1) is the number of vertices at an L1 distance from a given vertex in an infinite-dimensional box, which has k sides of length 2^m for each m >= 0. Equivalently, given a set of tokens containing k distinguishable tokens with value 2^m for each m >= 0, C(n+k-1,k-1) is the number of subsets of tokens with a total value of n. - Eitan Y. Levine, Jun 11 2023
Numbers in the k-th column, i.e., numbers of the form C(n,k) for n >= k, are known as k-simplex numbers. - Pontus von Brömssen, Jun 26 2023
Let r(k) be the k-th row and c(k) the k-th column. Denote convolution by * and repeated convolution by ^. Then r(k)*r(m)=r(k+m) and c(k)*c(m)=c(k+m+1). This is because r(k) = r(1) ^ k and c(k) = c(0) ^ k+1. - Eitan Y. Levine, Jul 23 2023
Number of permutations of length n avoiding simultaneously the patterns 231 and 312(resp., 213 and 231; 213 and 312) with k descents (equivalently, with k ascents). An ascent (resp., descent) in a permutation a(1)a(2)...a(n) is position i such that a(i)a(i+1)). - Tian Han, Nov 25 2023
C(n,k) are generalized binomial coefficients of order m=0. Calculated by the formula C(n,k) = Sum_{i=0..n-k} binomial(n+1, n-k-i)*Stirling2(i+ m+ 1, i+1) *(-1)^i, where m = 0 for n>= 0, 0 <= k <= n. - Igor Victorovich Statsenko, Feb 26 2023
The Akiyama-Tanigawa algorithm applied to the diagonals, binomial(n+k,k), yields the powers of n. - Shel Kaphan, May 03 2024

Examples

			Triangle T(n,k) begins:
   n\k 0   1   2   3   4   5   6   7   8   9  10  11 ...
   0   1
   1   1   1
   2   1   2   1
   3   1   3   3   1
   4   1   4   6   4   1
   5   1   5  10  10   5   1
   6   1   6  15  20  15   6   1
   7   1   7  21  35  35  21   7   1
   8   1   8  28  56  70  56  28   8   1
   9   1   9  36  84 126 126  84  36   9   1
  10   1  10  45 120 210 252 210 120  45  10   1
  11   1  11  55 165 330 462 462 330 165  55  11   1
  ...
There are C(4,2)=6 ways to distribute 5 balls BBBBB, among 3 different urns, < > ( ) [ ], so that each urn gets at least one ball, namely, <BBB>(B)[B], <B>(BBB)[B], <B>(B)[BBB], <BB>(BB)[B], <BB>(B)[BB], and <B>(BB)[BB].
There are C(4,2)=6 increasing functions from {1,2} to {1,2,3,4}, namely, {(1,1),(2,2)},{(1,1),(2,3)}, {(1,1),(2,4)}, {(1,2),(2,3)}, {(1,2),(2,4)}, and {(1,3),(2,4)}. - _Dennis P. Walsh_, Apr 07 2011
There are C(4,2)=6 subsets of {1,2,3,4,5} with median element 3, namely, {3}, {1,3,4}, {1,3,5}, {2,3,4}, {2,3,5}, and {1,2,3,4,5}. - _Dennis P. Walsh_, Dec 15 2011
The successive k-iterations of {A(0)} = E are E;E;E;...; the corresponding number of elements are 1,1,1,... The successive k-iterations of {A(1)} = {a} are (omitting brackets) a;a,E; a,E,E;...; the corresponding number of elements are 1,2,3,... The successive k-iterations of {A(2)} = {a,a} are aa; aa,a,E; aa, a, E and a,E and E;...; the corresponding number of elements are 1,3,6,... - _Gregory L. Simay_, Aug 06 2018
Boas-Buck type recurrence for column k = 4: T(8, 4) = (5/4)*(1 + 5 + 15 + 35) = 70. See the Boas-Buck comment above. - _Wolfdieter Lang_, Nov 12 2018

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.
  • Amulya Kumar Bag, Binomial theorem in ancient India, Indian Journal of History of Science, vol. 1 (1966), pp. 68-74.
  • Arthur T. Benjamin and Jennifer Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 63ff.
  • Boris A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 306.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 68-74.
  • Paul Curtz, Intégration numérique des systèmes différentiels à conditions initiales, Centre de Calcul Scientifique de l'Armement, Arcueil, 1969.
  • A. W. F. Edwards, Pascal's Arithmetical Triangle, 2002.
  • William Feller, An Introduction to Probability Theory and Its Application, Vol. 1, 2nd ed. New York: Wiley, p. 36, 1968.
  • Ronald L. Graham, Donald E. Knuth, and Oren Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., 1994, p. 155.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §4.4 Powers and Roots, pp. 140-141.
  • David Hök, Parvisa mönster i permutationer [Swedish], 2007.
  • Donald E. Knuth, The Art of Computer Programming, Vol. 1, 2nd ed., p. 52.
  • Sergei K. Lando, Lecture on Generating Functions, Amer. Math. Soc., Providence, R.I., 2003, pp. 60-61.
  • Blaise Pascal, Traité du triangle arithmétique, avec quelques autres petits traitez sur la mesme matière, Desprez, Paris, 1665.
  • Clifford A. Pickover, A Passion for Mathematics, Wiley, 2005; see p. 71.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 271-275.
  • 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.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 6.
  • John Riordan, Combinatorial Identities, Wiley, 1968, p. 2.
  • Robert Sedgewick and Philippe Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996, p. 143.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, pages 43-52.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 13, 30-33.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers, Penguin Books, 1987, pp. 115-118.
  • Douglas B. West, Combinatorial Mathematics, Cambridge, 2021, p. 25.

Links

Crossrefs

Equals differences between consecutive terms of A102363. - David G. Williams (davidwilliams(AT)Paxway.com), Jan 23 2006
Row sums give A000079 (powers of 2).
Cf. A083093 (triangle read mod 3), A214292 (first differences of rows).
Partial sums of rows give triangle A008949.
The triangle of the antidiagonals is A011973.
Infinite matrix squared: A038207, cubed: A027465.
Cf. A101164. If rows are sorted we get A061554 or A107430.
Another version: A108044.
Cf. A008277, A132311, A132312, A052216, A052217, A052218, A052219, A052220, A052221, A052222, A052223, A144225, A202750, A211226, A047999, A026729, A052553, A051920, A193242.
Triangle sums (see the comments): A000079 (Row1); A000007 (Row2); A000045 (Kn11 & Kn21); A000071 (Kn12 & Kn22); A001924 (Kn13 & Kn23); A014162 (Kn14 & Kn24); A014166 (Kn15 & Kn25); A053739 (Kn16 & Kn26); A053295 (Kn17 & Kn27); A053296 (Kn18 & Kn28); A053308 (Kn19 & Kn29); A053309 (Kn110 & Kn210); A001519 (Kn3 & Kn4); A011782 (Fi1 & Fi2); A000930 (Ca1 & Ca2); A052544 (Ca3 & Ca4); A003269 (Gi1 & Gi2); A055988 (Gi3 & Gi4); A034943 (Ze1 & Ze2); A005251 (Ze3 & Ze4). - Johannes W. Meijer, Sep 22 2010
Fibonacci-Pascal triangles: A027926, A036355, A037027, A074829, A105809, A109906, A111006, A114197, A162741, A228074, A228196, A228576.
Cf. A137948, A245334.
Cf. A085478, A258993.
Cf. A115940 (pandigital binomial coefficients C(m,k) with k>1).
Cf. (simplex colorings) A325002 (oriented), [k==n+1] (chiral), A325003 (achiral), A325000 (k or fewer colors), A325009 (orthotope facets, orthoplex vertices), A325017 (orthoplex facets, orthotope vertices).
Triangles of generalized binomial coefficients (n,k)_m (or generalized Pascal triangles) for m = 2..12: A001263, A056939, A056940, A056941, A142465, A142467, A142468, A174109, A342889, A342890, A342891.

Programs

  • Axiom
    -- (start)
)set expose add constructor OutputForm
pascal(0,n) == 1
pascal(n,n) == 1
pascal(i,j | 0 < i and i < j) == pascal(i-1,j-1) + pascal(i,j-1)
pascalRow(n) == [pascal(i,n) for i in 0..n]
displayRow(n) == output center blankSeparate pascalRow(n)
for i in 0..20 repeat displayRow i -- (end)
  • GAP
    Flat(List([0..12],n->List([0..n],k->Binomial(n,k)))); # Stefano Spezia, Dec 22 2018
  • Haskell
    a007318 n k = a007318_tabl !! n !! k
a007318_row n = a007318_tabl !! n
a007318_list = concat a007318_tabl
a007318_tabl = iterate (\row -> zipWith (+) ([0] ++ row) (row ++ [0])) [1]
-- Cf. http://www.haskell.org/haskellwiki/Blow_your_mind#Mathematical_sequences
-- Reinhard Zumkeller, Nov 09 2011, Oct 22 2010
  • Magma
    /* As triangle: */ [[Binomial(n, k): k in [0..n]]: n in [0.. 10]]; // Vincenzo Librandi, Jul 29 2015
  • Maple
    A007318 := (n,k)->binomial(n,k);
  • Mathematica
    Flatten[Table[Binomial[n, k], {n, 0, 11}, {k, 0, n}]] (* Robert G. Wilson v, Jan 19 2004 *)
Flatten[CoefficientList[CoefficientList[Series[1/(1 - x - x*y), {x, 0, 12}], x], y]] (* Mats Granvik, Jul 08 2014 *)
  • Maxima
    create_list(binomial(n,k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
  • PARI
    C(n,k)=binomial(n,k) \\ Charles R Greathouse IV, Jun 08 2011
  • Python
    # See Hobson link. Further programs:
from math import prod,factorial
def C(n,k): return prod(range(n,n-k,-1))//factorial(k) # M. F. Hasler, Dec 13 2019, updated Apr 29 2022, Feb 17 2023
  • Python
    from math import comb, isqrt
def A007318(n): return comb(r:=(m:=isqrt(k:=n+1<<1))-(k<=m*(m+1)),n-comb(r+1,2)) # Chai Wah Wu, Nov 11 2024
  • Sage
    def C(n,k): return Subsets(range(n), k).cardinality() # Ralf Stephan, Jan 21 2014

Formula

a(n, k) = C(n,k) = binomial(n, k).
C(n, k) = C(n-1, k) + C(n-1, k-1).
The triangle is symmetric: C(n,k) = C(n,n-k).
a(n+1, m) = a(n, m) + a(n, m-1), a(n, -1) := 0, a(n, m) := 0, n
C(n, k) = n!/(k!(n-k)!) if 0<=k<=n, otherwise 0.
C(n, k) = ((n-k+1)/k) * C(n, k-1) with C(n, 0) = 1. - Michael B. Porter, Mar 23 2025
G.f.: 1/(1-y-x*y) = Sum_(C(n, k)*x^k*y^n, n, k>=0)
G.f.: 1/(1-x-y) = Sum_(C(n+k, k)*x^k*y^n, n, k>=0).
G.f. for row n: (1+x)^n = Sum_{k=0..n} C(n, k)*x^k.
G.f. for column k: x^k/(1-x)^(k+1); [corrected by Werner Schulte, Jun 15 2022].
E.g.f.: A(x, y) = exp(x+x*y).
E.g.f. for column n: x^n*exp(x)/n!.
In general the m-th power of A007318 is given by: T(0, 0) = 1, T(n, k) = T(n-1, k-1) + m*T(n-1, k), where n is the row-index and k is the column; also T(n, k) = m^(n-k)*C(n, k).
Triangle T(n, k) read by rows; given by A000007 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
Let P(n+1) = the number of integer partitions of (n+1); let p(i) = the number of parts of the i-th partition of (n+1); let d(i) = the number of different parts of the i-th partition of (n+1); let m(i, j) = multiplicity of the j-th part of the i-th partition of (n+1). Define the operator Sum_{i=1..P(n+1), p(i)=k+1} as the sum running from i=1 to i=P(n+1) but taking only partitions with p(i)=(k+1) parts into account. Define the operator Product_{j=1..d(i)} = product running from j=1 to j=d(i). Then C(n, k) = Sum_{p(i)=(k+1), i=1..P(n+1)} p(i)! / [Product_{j=1..d(i)} m(i, j)!]. E.g., C(5, 3) = 10 because n=6 has the following partitions with m=3 parts: (114), (123), (222). For their multiplicities one has: (114): 3!/(2!*1!) = 3; (123): 3!/(1!*1!*1!) = 6; (222): 3!/3! = 1. The sum is 3 + 6 + 1 = 10 = C(5, 3). - Thomas Wieder, Jun 03 2005
C(n, k) = Sum_{j=0..k} (-1)^j*C(n+1+j, k-j)*A000108(j). - Philippe Deléham, Oct 10 2005
G.f.: 1 + x*(1 + x) + x^3*(1 + x)^2 + x^6*(1 + x)^3 + ... . - Michael Somos, Sep 16 2006
Sum_{k=0..floor(n/2)} x^(n-k)*T(n-k,k) = A000007(n), A000045(n+1), A002605(n), A030195(n+1), A057087(n), A057088(n), A057089(n), A057090(n), A057091(n), A057092(n), A057093(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. Sum_{k=0..floor(n/2)} (-1)^k*x^(n-k)*T(n-k,k) = A000007(n), A010892(n), A009545(n+1), A057083(n), A001787(n+1), A030191(n), A030192(n), A030240(n), A057084(n), A057085(n+1), A057086(n), A084329(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, respectively. - Philippe Deléham, Sep 16 2006
C(n,k) <= A062758(n) for n > 1. - Reinhard Zumkeller, Mar 04 2008
C(t+p-1, t) = Sum_{i=0..t} C(i+p-2, i) = Sum_{i=1..p} C(i+t-2, t-1). A binomial number is the sum of its left parent and all its right ancestors, which equals the sum of its right parent and all its left ancestors. - Lee Naish (lee(AT)cs.mu.oz.au), Mar 07 2008
From Paul D. Hanna, Mar 24 2011: (Start)
Let A(x) = Sum_{n>=0} x^(n*(n+1)/2)*(1+x)^n be the g.f. of the flattened triangle:
A(x) = 1 + (x + x^2) + (x^3 + 2*x^4 + x^5) + (x^6 + 3*x^7 + 3*x^8 + x^9) + ...
then A(x) equals the series Sum_{n>=0} (1+x)^n*x^n*Product_{k=1..n} (1-(1+x)*x^(2*k-1))/(1-(1+x)*x^(2*k));
also, A(x) equals the continued fraction 1/(1- x*(1+x)/(1+ x*(1-x)*(1+x)/(1- x^3*(1+x)/(1+ x^2*(1-x^2)*(1+x)/(1- x^5*(1+x)/(1+ x^3*(1-x^3)*(1+x)/(1- x^7*(1+x)/(1+ x^4*(1-x^4)*(1+x)/(1- ...))))))))).
These formulas are due to (1) a q-series identity and (2) a partial elliptic theta function expression. (End)
For n > 0: T(n,k) = A029600(n,k) - A029635(n,k), 0 <= k <= n. - Reinhard Zumkeller, Apr 16 2012
Row n of the triangle is the result of applying the ConvOffs transform to the first n terms of the natural numbers (1, 2, 3, ..., n). See A001263 or A214281 for a definition of this transformation. - Gary W. Adamson, Jul 12 2012
From L. Edson Jeffery, Aug 02 2012: (Start)
Row n (n >= 0) of the triangle is given by the n-th antidiagonal of the infinite matrix P^n, where P = (p_{i,j}), i,j >= 0, is the production matrix
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,
... (End)
Row n of the triangle is also given by the n+1 coefficients of the polynomial P_n(x) defined by the recurrence P_0(x) = 1, P_1(x) = x + 1, P_n(x) = x*P_{n-1}(x) + P_{n-2}(x), n > 1. - L. Edson Jeffery, Aug 12 2013
For a closed-form formula for arbitrary left and right borders of Pascal-like triangles see A228196. - Boris Putievskiy, Aug 18 2013
For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013
(1+x)^n = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*Sum_{i=0..k} k^(n-i)*binomial(k,i)*x^(n-i)/(n-i)!. - Vladimir Kruchinin, Oct 21 2013
E.g.f.: A(x,y) = exp(x+x*y) = 1 + (x+y*x)/( E(0)-(x+y*x)), where E(k) = 1 + (x+y*x)/(1 + (k+1)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 08 2013
E.g.f.: E(0) -1, where E(k) = 2 + x*(1+y)/(2*k+1 - x*(1+y)/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
G.f.: 1 + x*(1+x)*(1+x^2*(1+x)/(W(0)-x^2-x^3)), where W(k) = 1 + (1+x)*x^(k+2) - (1+x)*x^(k+3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
Sum_{n>=0} C(n,k)/n! = e/k!, where e = exp(1), while allowing n < k where C(n,k) = 0. Also Sum_{n>=0} C(n+k-1,k)/n! = e * A000262(k)/k!, and for k>=1 equals e * A067764(k)/A067653(k). - Richard R. Forberg, Jan 01 2014
Sum_{n>=k} 1/C(n,k) = k/(k-1) for k>=1. - Richard R. Forberg, Feb 10 2014
From Tom Copeland, Apr 26 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result by A007318(x) = P(x). Then with :xD:^n = x^n*(d/dx)^n and B(n,x), the Bell polynomials (A008277),
A) P(x)= exp(x*dP) = exp[x*(e^M-I)] = exp[M*B(.,x)] = (I+dP)^B(.,x)
with dP = A132440, M = A238385-I, and I = identity matrix, and
B) P(:xD:) = exp(dP:xD:) = exp[(e^M-I):xD:] = exp[M*B(.,:xD:)] = exp[M*xD] = (I+dP)^(xD) with action P(:xD:)g(x) = exp(dP:xD:)g(x) = g[(I+dP)*x] (cf. also A238363).
C) P(x)^y = P(y*x). P(2x) = A038207(x) = exp[M*B(.,2x)], the face vectors of the n-dim hypercubes.
D) P(x) = [St2]*exp(x*M)*[St1] = [St2]*(I+dP)^x*[St1]
E) = [St1]^(-1)*(I+dP)^x*[St1] = [St2]*(I+dP)^x*[St2]^(-1)
where [St1]=padded A008275 just as [St2]=A048993=padded A008277 and exp(x*M) = (I+dP)^x = Sum_{k>=0} C(x,k) dP^k. (End)
T(n,k) = A245334(n,k) / A137948(n,k), 0 <= k <= n. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, Dec 21 2014: (Start)
Recurrence equation: T(n,k) = T(n-1,k)*(n + k)/(n - k) - T(n-1,k-1) for n >= 2 and 1 <= k < n, with boundary conditions T(n,0) = T(n,n) = 1. Note, changing the minus sign in the recurrence to a plus sign gives a recurrence for the square of the binomial coefficients - see A008459.
There is a relation between the e.g.f.'s of the rows and the diagonals of the triangle, namely, exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + 3*x + 3*x^2/2! + x^3/3!) = 1 + 4*x + 10*x^2/2! + 20*x^3/3! + 35*x^4/4! + .... This property holds more generally for the Riordan arrays of the form ( f(x), x/(1 - x) ), where f(x) is an o.g.f. of the form 1 + f_1*x + f_2*x^2 + .... See, for example, A055248 and A106516.
Let P denote the present triangle. For k = 0,1,2,... define P(k) to be the lower unit triangular block array
/I_k 0\
\ 0 P/ having the k X k identity matrix I_k as the upper left block; in particular, P(0) = P. The infinite product P(0)*P(1)*P(2)*..., which is clearly well-defined, is equal to the triangle of Stirling numbers of the second kind A008277. The infinite product in the reverse order, that is, ...*P(2)*P(1)*P(0), is equal to the triangle of Stirling cycle numbers A130534. (End)
C(a+b,c) = Sum_{k=0..a} C(a,k)*C(b,b-c+k). This is a generalization of equation 1 from section 4.2.5 of the Prudnikov et al. reference, for a=b=c=n: C(2*n,n) = Sum_{k=0..n} C(n,k)^2. See Links section for animation of new formula. - Hermann Stamm-Wilbrandt, Aug 26 2015
The row polynomials of the Pascal matrix P(n,x) = (1+x)^n are related to the Bernoulli polynomials Br(n,x) and their umbral compositional inverses Bv(n,x) by the umbral relation P(n,x) = (-Br(.,-Bv(.,x)))^n = (-1)^n Br(n,-Bv(.,x)), which translates into the matrix relation P = M * Br * M * Bv, where P is the Pascal matrix, M is the diagonal matrix diag(1,-1,1,-1,...), Br is the matrix for the coefficients of the Bernoulli polynomials, and Bv that for the umbral inverse polynomials defined umbrally by Br(n,Bv(.,x)) = x^n = Bv(n,Br(.,x)). Note M = M^(-1). - Tom Copeland, Sep 05 2015
1/(1-x)^k = (r(x) * r(x^2) * r(x^4) * ...) where r(x) = (1+x)^k. - Gary W. Adamson, Oct 17 2016
Boas-Buck type recurrence for column k for Riordan arrays (see the Aug 10 2017 remark in A046521, also for the reference) with the Boas-Buck sequence b(n) = {repeat(1)}. T(n, k) = ((k+1)/(n-k))*Sum_{j=k..n-1} T(j, k), for n >= 1, with T(n, n) = 1. This reduces, with T(n, k) = binomial(n, k), to a known binomial identity (e.g, Graham et al. p. 161). - Wolfdieter Lang, Nov 12 2018
C((p-1)/a, b) == (-1)^b * fact_a(a*b-a+1)/fact_a(a*b) (mod p), where fact_n denotes the n-th multifactorial, a divides p-1, and the denominator of the fraction on the right side of the equation represents the modular inverse. - Isaac Saffold, Jan 07 2019
C(n,k-1) = A325002(n,k) - [k==n+1] = (A325002(n,k) + A325003(n,k)) / 2 = [k==n+1] + A325003(n,k). - Robert A. Russell, Oct 20 2020
From Hermann Stamm-Wilbrandt, May 13 2021: (Start)
Binomial sums are Fibonacci numbers A000045:
Sum_{k=0..n} C(n + k, 2*k + 1) = F(2*n).
Sum_{k=0..n} C(n + k, 2*k) = F(2*n + 1). (End)
C(n,k) = Sum_{i=0..k} A000108(i) * C(n-2i-1, k-i), for 0 <= k <= floor(n/2)-1. - Tushar Bansal, May 17 2025

Extensions

Checked all links, deleted 8 that seemed lost forever and were probably not of great importance. - N. J. A. Sloane, May 08 2018

A000332 Binomial coefficient binomial(n,4) = n*(n-1)*(n-2)*(n-3)/24.

Original entry on oeis.org

0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of intersection points of diagonals of convex n-gon where no more than two diagonals intersect at any point in the interior.
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation. - Ignacio Larrosa Cañestro, Apr 09 2002. [See Les Reid link for proof. - N. J. A. Sloane, Apr 02 2016] [See Peter Kagey link for alternate proof. - Sameer Gauria, Jul 29 2025]
Start from cubane and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink on chemistry. - Robert G. Wilson v, Aug 02 2002
For n>0, a(n) = (-1/8)*(coefficient of x in Zagier's polynomial P_(2n,n)). (Zagier's polynomials are used by PARI/GP for acceleration of alternating or positive series.)
Figurate numbers based on the 4-dimensional regular convex polytope called the regular 4-simplex, pentachoron, 5-cell, pentatope or 4-hypertetrahedron with Schlaefli symbol {3,3,3}. a(n)=((n*(n-1)*(n-2)*(n-3))/4!). - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 07 2009
Maximal number of crossings that can be created by connecting n vertices with straight lines. - Cameron Redsell-Montgomerie (credsell(AT)uoguelph.ca), Jan 30 2007
If X is an n-set and Y a fixed (n-1)-subset of X then a(n) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Product of four consecutive numbers divided by 24. - Artur Jasinski, Dec 02 2007
The only prime in this sequence is 5. - Artur Jasinski, Dec 02 2007
For strings consisting entirely of 0's and 1's, the number of distinct arrangements of four 1's such that 1's are not adjacent. The shortest possible string is 7 characters, of which there is only one solution: 1010101, corresponding to a(5). An eight-character string has 5 solutions, nine has 15, ten has 35 and so on, congruent to A000332. - Gil Broussard, Mar 19 2008
For a(n)>0, a(n) is pentagonal if and only if 3 does not divide n. All terms belong to the generalized pentagonal sequence (A001318). Cf. A000326, A145919, A145920. - Matthew Vandermast, Oct 28 2008
Nonzero terms = row sums of triangle A158824. - Gary W. Adamson, Mar 28 2009
Except for the 4 initial 0's, is equivalent to the partial sums of the tetrahedral numbers A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
If the first 3 zeros are disregarded, that is, if one looks at binomial(n+3, 4) with n>=0, then it becomes a 'Matryoshka doll' sequence with alpha=0: seq(add(add(add(i,i=alpha..k),k=alpha..n),n=alpha..m),m=alpha..50). - Peter Luschny, Jul 14 2009
For n>=1, a(n) is the number of n-digit numbers the binary expansion of which contains two runs of 0's. - Vladimir Shevelev, Jul 30 2010
For n>0, a(n) is the number of crossing set partitions of {1,2,..,n} into n-2 blocks. - Peter Luschny, Apr 29 2011
The Kn3, Ca3 and Gi3 triangle sums of A139600 are related to the sequence given above, e.g., Gi3(n) = 2*A000332(n+3) - A000332(n+2) + 7*A000332(n+1). For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 29 2011
For n > 3, a(n) is the hyper-Wiener index of the path graph on n-2 vertices. - Emeric Deutsch, Feb 15 2012
Except for the four initial zeros, number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Aug 31 2012
a(n+3) is the number of different ways to color the faces (or the vertices) of a regular tetrahedron with n colors if we count mirror images as the same.
a(n) = fallfac(n,4)/4! is also the number of independent components of an antisymmetric tensor of rank 4 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Does not satisfy Benford's law [Ross, 2012] - N. J. A. Sloane, Feb 12 2017
Number of chiral pairs of colorings of the vertices (or faces) of a regular tetrahedron with n available colors. Chiral colorings come in pairs, each the reflection of the other. - Robert A. Russell, Jan 22 2020
From Mircea Dan Rus, Aug 26 2020: (Start)
a(n+3) is the number of lattice rectangles (squares included) in a staircase of order n; this is obtained by stacking n rows of consecutive unit lattice squares, aligned either to the left or to the right, which consist of 1, 2, 3, ..., n squares and which are stacked either in the increasing or in the decreasing order of their lengths. Below, there is a staircase or order 4 which contains a(7) = 35 rectangles. [See the Teofil Bogdan and Mircea Dan Rus link, problem 3, under A004320]
_
||
|||_
|||_|_
|||_|_|
(End)
a(n+4) is the number of strings of length n on an ordered alphabet of 5 letters where the characters in the word are in nondecreasing order. E.g., number of length-2 words is 15: aa,ab,ac,ad,ae,bb,bc,bd,be,cc,cd,ce,dd,de,ee. - Jim Nastos, Jan 18 2021
From Tom Copeland, Jun 07 2021: (Start)
Aside from the zeros, this is the fifth diagonal of the Pascal matrix A007318, the only nonvanishing diagonal (fifth) of the matrix representation IM = (A132440)^4/4! of the differential operator D^4/4!, when acting on the row vector of coefficients of an o.g.f., or power series.
M = e^{IM} is the matrix of coefficients of the Appell sequence p_n(x) = e^{D^4/4!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^4/4!}, which is that for A025036 aerated with triple zeros, the first column of M.
See A099174 and A000292 for analogous relationships for the third and fourth diagonals of the Pascal matrix. (End)
For integer m and positive integer r >= 3, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (3 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

Examples

			a(5) = 5 from the five independent components of an antisymmetric tensor A of rank 4 and dimension 5, namely A(1,2,3,4), A(1,2,3,5), A(1,2,4,5), A(1,3,4,5) and A(2,3,4,5). See the Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015

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.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 8.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 70.
  • 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. 7.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 294.
  • 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).
  • Charles W. Trigg, Mathematical Quickies, New York: Dover Publications, Inc., 1985, p. 53, #191.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 127.

Links

Crossrefs

binomial(n, k): A161680 (k = 2), A000389 (k = 5), A000579 (k = 6), A000580 (k = 7), A000581 (k = 8), A000582 (k = 9).
Cf. A053134, A053126, A075733, A006322, A006325.
Cf. also A000583, A014820, A092181, A092182, A092183.
Cf. A000217, A000292, A007318 (column k = 4).
Cf. A158824.
Cf. A006008 (Number of ways to color the faces (or vertices) of a regular tetrahedron with n colors when mirror images are counted as two).
Cf. A104712 (third column, k=4).
See A269747 for a 3-D analog.
Cf. A006008 (oriented), A006003 (achiral) tetrahedron colorings.
Row 3 of A325000, col. 4 of A007318.
Cf. A000292, A025036, A099174.

Programs

  • GAP
    A000332 := List([1..10^2], n -> Binomial(n, 4)); # Muniru A Asiru, Oct 16 2017
  • Magma
    [Binomial(n,4): n in [0..50]]; // Vincenzo Librandi, Nov 23 2014
  • Maple
    A000332 := n->binomial(n,4); [seq(binomial(n,4), n=0..100)];
  • Mathematica
    Table[ Binomial[n, 4], {n, 0, 45} ] (* corrected by Harvey P. Dale, Aug 22 2011 *)
Table[(n-4)(n-3)(n-2)(n-1)/24, {n, 100}] (* Artur Jasinski, Dec 02 2007 *)
LinearRecurrence[{5,-10,10,-5,1}, {0,0,0,0,1}, 45] (* Harvey P. Dale, Aug 22 2011 *)
CoefficientList[Series[x^4 / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 23 2014 *)
  • PARI
    a(n)=binomial(n,4);
  • Python
    # Starts at a(3), i.e. computes n*(n+1)*(n+2)*(n+3)/24
# which is more in line with A000217 and A000292.
def A000332():
    x, y, z, u = 1, 1, 1, 1
    yield 0
    while True:
        yield x
        x, y, z, u = x + y + z + u + 1, y + z + u + 1, z + u + 1, u + 1
a = A000332(); print([next(a) for i in range(41)]) # Peter Luschny, Aug 03 2019
  • Python
    print([n*(n-1)*(n-2)*(n-3)//24 for n in range(50)])
# Gennady Eremin, Feb 06 2022

Formula

a(n) = n*(n-1)*(n-2)*(n-3)/24.
G.f.: x^4/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = n*a(n-1)/(n-4). - Benoit Cloitre, Apr 26 2003, R. J. Mathar, Jul 07 2009
a(n) = Sum_{k=1..n-3} Sum_{i=1..k} i*(i+1)/2. - Benoit Cloitre, Jun 15 2003
Convolution of natural numbers {1, 2, 3, 4, ...} and A000217, the triangular numbers {1, 3, 6, 10, ...}. - Jon Perry, Jun 25 2003
a(n) = A110555(n+1,4). - Reinhard Zumkeller, Jul 27 2005
a(n+1) = ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24 - (n^5-(n-1)^5-1)/30; a(n) = A006322(n-2)-A006325(n-1). - Xavier Acloque, Oct 20 2003; R. J. Mathar, Jul 07 2009
a(4*n+2) = Pyr(n+4, 4*n+2) where the polygonal pyramidal numbers are defined for integers A>2 and B>=0 by Pyr(A, B) = B-th A-gonal pyramid number = ((A-2)*B^3 + 3*B^2 - (A-5)*B)/6; For all positive integers i and the pentagonal number function P(x) = x*(3*x-1)/2: a(3*i-2) = P(P(i)) and a(3*i-1) = P(P(i) + i); 1 + 24*a(n) = (n^2 + 3*n + 1)^2. - Jonathan Vos Post, Nov 15 2004
First differences of A000389(n). - Alexander Adamchuk, Dec 19 2004
For n > 3, the sum of the first n-2 tetrahedral numbers (A000292). - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005 [Corrected by Doug Bell, Jun 25 2017]
Starting (1, 5, 15, 35, ...), = binomial transform of [1, 4, 6, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
Sum_{n>=4} 1/a(n) = 4/3, from the Taylor expansion of (1-x)^3*log(1-x) in the limit x->1. - R. J. Mathar, Jan 27 2009
A034263(n) = (n+1)*a(n+4) - Sum_{i=0..n+3} a(i). Also A132458(n) = a(n)^2 - a(n-1)^2 for n>0. - Bruno Berselli, Dec 29 2010
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1. - Harvey P. Dale, Aug 22 2011
a(n) = (binomial(n-1,2)^2 - binomial(n-1,2))/6. - Gary Detlefs, Nov 20 2011
a(n) = Sum_{k=1..n-2} Sum_{i=1..k} i*(n-k-2). - Wesley Ivan Hurt, Sep 25 2013
a(n) = (A000217(A000217(n-2) - 1))/3 = ((((n-2)^2 + (n-2))/2)^2 - (((n-2)^2 + (n-2))/2))/(2*3). - Raphie Frank, Jan 16 2014
Sum_{n>=0} a(n)/n! = e/24. Sum_{n>=3} a(n)/(n-3)! = 73*e/24. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
Sum_{n>=4} (-1)^(n+1)/a(n) = 32*log(2) - 64/3 = A242023 = 0.847376444589... . - Richard R. Forberg, Aug 11 2014
4/(Sum_{n>=m} 1/a(n)) = A027480(m-3), for m>=4. - Richard R. Forberg, Aug 12 2014
E.g.f.: x^4*exp(x)/24. - Robert Israel, Nov 23 2014
a(n+3) = C(n,1) + 3*C(n,2) + 3*C(n,3) + C(n,4). Each term indicates the number of ways to use n colors to color a tetrahedron with exactly 1, 2, 3, or 4 colors.
a(n) = A080852(1,n-4). - R. J. Mathar, Jul 28 2016
From Gary W. Adamson, Feb 06 2017: (Start)
G.f.: Starting (1, 5, 14, ...), x/(1-x)^5 can be written
as (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1+x)^5;
as (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1+x+x^2)^5;
as (x * r(x) * r(x^4) * r(x^16) * r(x^64) * ...) where r(x) = (1+x+x^2+x^3)^5;
... (as a conjectured infinite set). (End)
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = A006008(n) - a(n+3) = (A006008(n) - A006003(n)) / 2 = a(n+3) - A006003(n).
a(n+3) = A006008(n) - a(n) = (A006008(n) + A006003(n)) / 2 = a(n) + A006003(n).
a(n) = A007318(n,4).
a(n+3) = A325000(3,n). (End)
Product_{n>=5} (1 - 1/a(n)) = cosh(sqrt(15)*Pi/2)/(100*Pi). - Amiram Eldar, Jan 21 2021

Extensions

Some formulas that referred to another offset corrected by R. J. Mathar, Jul 07 2009

A000262 Number of "sets of lists": number of partitions of {1,...,n} into any number of lists, where a list means an ordered subset.

Original entry on oeis.org

1, 1, 3, 13, 73, 501, 4051, 37633, 394353, 4596553, 58941091, 824073141, 12470162233, 202976401213, 3535017524403, 65573803186921, 1290434218669921, 26846616451246353, 588633468315403843, 13564373693588558173, 327697927886085654441, 8281153039765859726341
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Determinant of n X n matrix M=[m(i,j)] where m(i,i)=i, m(i,j)=1 if i > j, m(i,j)=i-j if j > i. - Vladeta Jovovic, Jan 19 2003
With p(n) = the number of integer partitions of n, d(i) = the number of different parts of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, Sum_{i=1..p(n)} = sum over i and Product_{j=1..d(i)} = product over j, one has: a(n) = Sum_{i=1..p(n)} n!/(Product_{j=1..d(i)} m(i,j)!). - Thomas Wieder, May 18 2005
Consider all n! permutations of the integer sequence [n] = 1,2,3,...,n. The i-th permutation, i=1,2,...,n!, consists of Z(i) permutation cycles. Such a cycle has the length lc(i,j), j=1,...,Z(i). For a given permutation we form the product of all its cycle lengths Product_{j=1..Z(i)} lc(i,j). Furthermore, we sum up all such products for all permutations of [n] which gives Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = A000262(n). For n=4 we have Sum_{i=1..n!} Product_{j=1..Z(i)} lc(i,j) = 1*1*1*1 + 2*1*1 + 3*1 + 2*1*1 + 3*1 + 2*1 + 3*1 + 4 + 3*1 + 4 + 2*2 + 2*1*1 + 3*1 + 4 + 3*1 + 2*1*1 + 2*2 + 4 + 2*2 + 4 + 3*1 + 2*1*1 + 3*1 + 4 = 73 = A000262(4). - Thomas Wieder, Oct 06 2006
For a finite set S of size n, a chain gang G of S is a partially ordered set (S,<=) consisting only of chains. The number of chain gangs of S is a(n). For example, with S={a, b}and n=2, there are a(2)=3 chain gangs of S, namely, {(a,a),(b,b)}, {(a,a),(a,b),(b,b)} and {(a,a),(b,a),(b,b)}. - Dennis P. Walsh, Feb 22 2007
(-1)*A000262 with the first term set to 1 and A084358 form a reciprocal pair under the list partition transform and associated operations described in A133314. Cf. A133289. - Tom Copeland, Oct 21 2007
Consider the distribution of n unlabeled elements "1" onto n levels where empty levels are allowed. In addition, the empty levels are labeled. Their names are 0_1, 0_2, 0_3, etc. This sequence gives the total number of such distributions. If the empty levels are unlabeled ("0"), then the answer is A001700. Let the colon ":" separate two levels. Then, for example, for n=3 we have a(3)=13 arrangements: 111:0_1:0_2, 0_1:111:0_2, 0_1:0_2:111, 111:0_2:0_1, 0_2:111:0_1, 0_2:0_1:111, 11:1:0, 11:0:1, 0:11:1, 1:11:0, 1:0:11, 0:1:11, 1:1:1. - Thomas Wieder, May 25 2008
Row sums of exponential Riordan array [1,x/(1-x)]. - Paul Barry, Jul 24 2008
a(n) is the number of partitions of [n] (A000110) into lists of noncrossing sets. For example, a(3)=3 counts 12, 1-2, 2-1 and a(4) = 73 counts the 75 partitions of [n] into lists of sets (A000670) except for 13-24, 24-13 which fail to be noncrossing. - David Callan, Jul 25 2008
a(i-j)/(i-j)! gives the value of the non-null element (i, j) of the lower triangular matrix exp(S)/exp(1), where S is the lower triangular matrix - of whatever dimension - having all its (non-null) elements equal to one. - Giuliano Cabrele, Aug 11 2008, Sep 07 2008
a(n) is also the number of nilpotent partial one-one bijections (of an n-element set). Equivalently, it is the number of nilpotents in the symmetric inverse semigroup (monoid). - Abdullahi Umar, Sep 14 2008
A000262 is the exp transform of the factorial numbers A000142. - Thomas Wieder, Sep 10 2008
If n is a positive integer then the infinite continued fraction (1+n)/(1+(2+n)/(2+(3+n)/(3+...))) converges to the rational number A052852(n)/A000262(n). - David Angell (angell(AT)maths.unsw.edu.au), Dec 18 2008
Vladeta Jovovic's formula dated Sep 20 2006 can be restated as follows: a(n) is the n-th ascending factorial moment of the Poisson distribution with parameter (mean) 1. - Shai Covo (green355(AT)netvision.net.il), Jan 25 2010
The umbral exponential generating function for a(n) is (1-x)^{-B}. In other words, write (1-x)^{-B} as a power series in x whose coefficients are polynomials in B, and then replace B^k with the Bell number B_k. We obtain a(0) + a(1)x + a(2)x^2/2! + ... . - Richard Stanley, Jun 07 2010
a(n) is the number of Dyck n-paths (A000108) with its peaks labeled 1,2,...,k in some order, where k is the number of peaks. For example a(2)=3 counts U(1)DU(2)D, U(2)DU(1)D, UU(1)DD where the label at each peak is in parentheses. This is easy to prove using generating functions. - David Callan, Aug 23 2011
a(n) = number of permutations of the multiset {1,1,2,2,...,n,n} such that for 1 <= i <= n, all entries between the two i's exceed i and if any such entries are present, they include n. There are (2n-1)!! permutations satisfying the first condition, and for example: a(3)=13 counts all 5!!=15 of them except 331221 and 122133 which fail the second condition. - David Callan, Aug 27 2014
a(n) is the number of acyclic, injective functions from subsets of [n] to [n]. Let subset D of [n] have size k. The number of acyclic, injective functions from D to [n] is (n-1)!/(n-k-1)! and hence a(n) = Sum_{k=0..n-1} binomial(n,k)*(n-1)!/(n-k-1)!. - Dennis P. Walsh, Nov 05 2015
a(n) is the number of acyclic, injective, labeled directed graphs on n vertices with each vertex having outdegree at most one. - Dennis P. Walsh, Nov 05 2015
For n > 0, a(n) is the number of labeled-rooted skinny-tree forests on n nodes. A skinny tree is a tree in which each vertex has at most one child. Let k denote the number of trees. There are binomial(n,k) ways to choose the roots, binomial(n-1,k-1) ways to choose the number of descendants for each root, and (n-k)! ways to permute those descendants. Summing over k, we obtain a(n) = Sum_{k=1..n} C(n,k)*C(n-1,k-1)*(n-k)!. - Dennis P. Walsh, Nov 10 2015
This is the Sheffer transform of the Bell numbers A000110 with the Sheffer matrix S = |Stirling1| = (1, -log(1-x)) = A132393. See the e.g.f. formula, a Feb 21 2017 comment on A048993, and R. Stanley's Jun 07 2010 comment above. - Wolfdieter Lang, Feb 21 2017
All terms = {1, 3} mod 10. - Muniru A Asiru, Oct 01 2017
We conjecture that for k = 2,3,4,..., the difference a(n+k) - a(n) is divisible by k: if true, then for each k the sequence a(n) taken modulo k is periodic with period dividing k. - Peter Bala, Nov 14 2017
The above conjecture is true - see the Bala link. - Peter Bala, Jan 20 2018
The terms of this sequence can be derived from the numerators of the fractions, produced by the recursion: b(0) = 1, b(n) = 1 + ((n-1) * b(n-1)) / (n-1 + b(n-1)) for n > 0. The denominators give A002720. - Dimitris Valianatos, Aug 01 2018
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 123. It is also the number of rooted labeled forests that avoid 312, 213, and 132, as well as the number of rooted labeled forests that avoid 132, 213, and 321. - Kassie Archer, Aug 30 2018
For n > 0, a(n) is the number of partitions of [2n-1] whose nontrivial blocks are of type {a,b}, with a <= n and b > n. In fact, for n > 0, a(n) = A056953(2n-1). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) is the number of ways to split n people into nonempty groups, have each group sit around a circular table, and select one person from each table (where two seating arrangements are considered identical if each person has the same left neighbors in both of them). - Enrique Navarrete, Oct 01 2023

Examples

			Illustration of first terms as sets of ordered lists of the first n integers:
  a(1) = 1  : (1)
  a(2) = 3  : (12), (21), (1)(2).
  a(3) = 13 : (123) (6 ways), (12)(3) (2*3 ways) (1)(2)(3) (1 way)
  a(4) = 73 : (1234) (24 ways), (123)(4) (6*4 ways), (12)(34) (2*2*3 ways), (12)(3)(4) (2*6 ways), (1)(2)(3)(4) (1 way).
G.f. = 1 + x + 3*x^2 + 13*x^3 + 73*x^4 + 501*x^4 + 4051*x^5 + 37633*x^6 + 394353*x^7 + ...

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 194.
  • 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

Row sums of A271703 and for n >= 1 of A008297. Unsigned row sums of A111596.
A002868 is maximal element of the n-th row of A271703 and for n >= 1 of A008297.
Main diagonal of A257740 and of A319501.
Cf. A001263, A001700, A002378, A005408, A066668, A056953, A002720.
Cf. A000110, A052852, A082579, A132393, A255807, A255819, A318976.

Programs

  • GAP
    a:=[1,1];; for n in [3..10^2] do a[n]:=(2*n-3)*a[n-1]-(n-2)*(n-3)*a[n-2]; od; A000262:=a;  # Muniru A Asiru, Oct 01 2017
  • Haskell
    a000262 n = a000262_list !! n
a000262_list = 1 : 1 : zipWith (-)
               (tail $ zipWith (*) a005408_list a000262_list)
                      (zipWith (*) a002378_list a000262_list)
-- Reinhard Zumkeller, Mar 06 2014
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (2*n-1)*Self(n-1) - (n-1)*(n-2)*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 14 2019
  • Magma
    [Factorial(n)*Evaluate(LaguerrePolynomial(n, -1), -1): n in [0..30]]; // G. C. Greubel, Feb 23 2021
  • Maple
    A000262 := proc(n) option remember: if n=0 then RETURN(1) fi: if n=1 then RETURN(1) fi: (2*n-1)*procname(n-1) - (n-1)*(n-2)*procname(n-2) end proc:
for n from 0 to 20 do printf(`%d,`,a(n)) od:
spec := [S, {S = Set(Prod(Z,Sequence(Z)))}, labeled]; [seq(combstruct[count](spec, size=n), n=0..40)];
with(combinat):seq(sum(abs(stirling1(n, k))*bell(k), k=0..n), n=0..18); # Zerinvary Lajos, Nov 26 2006
B:=[S,{S = Set(Sequence(Z,1 <= card),card <=13)},labelled]: seq(combstruct[count](B, size=n), n=0..19);# Zerinvary Lajos, Mar 21 2009
a := n -> `if`(n=0, 1, n!*hypergeom([1 - n], [2], -1)): seq(simplify(a(n)), n=0..19); # Peter Luschny, Jun 05 2014
  • Mathematica
    Range[0, 19]! CoefficientList[ Series[E^(x/(1-x)), {x, 0, 19}], x] (* Robert G. Wilson v, Apr 04 2005 *)
a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Exp[x/(1-x)], {x, 0, n}]]; (* Michael Somos, Jul 19 2005 *)
a[n_]:=If[n==0, 1, n! Sum[Binomial[n-1, j]/(j+1)!, {j, 0, n-1}]]; Table[a[n], {n, 0, 30}] (* Wilf *)
a[0] = 1; a[n_]:= n!*Hypergeometric1F1[n+1, 2, 1]/E; Table[a[n], {n, 0, 19}] (* Jean-François Alcover, Jun 18 2012, after Shai Covo *)
Table[Sum[BellY[n, k, Range[n]!], {k, 0, n}], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 09 2016 *)
a[ n_]:= If[ n<0, 0, n! SeriesCoefficient[ Product[ QPochhammer[x^k]^(-MoebiusMu[k]/k), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 02 2019 *)
Table[n!*LaguerreL[n, -1, -1], {n, 0, 30}] (* G. C. Greubel, Feb 23 2021 *)
RecurrenceTable[{a[n] == (2*n - 1)*a[n-1] - (n-1)*(n-2)*a[n-2], a[0] == 1, a[1] == 1}, a, {n, 0, 30}] (* Vaclav Kotesovec, Jul 21 2022 *)
  • Maxima
    makelist(sum(abs(stirling1(n,k))*belln(k),k,0,n),n,0,24); /* Emanuele Munarini, Jul 04 2011 */
  • Maxima
    makelist(hypergeometric([-n+1,-n],[],1),n,0,12); /* Emanuele Munarini, Sep 27 2016 */
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( x / (1 - x) + x * O(x^n)), n))}; /* Michael Somos, Feb 10 2005 */
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, eta(x^k + x * O(x^n))^( -moebius(k) / k)), n))}; /* Michael Somos, Feb 10 2005 */
  • PARI
    {a(n) = s = 1; for(k = 1, n-1, s = 1 + k * s / (k + s)); return( numerator(s))}; /* Dimitris Valianatos, Aug 03 2018 */
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( prod( k=1, n, (1 - x^k + x * O(x^n))^(-eulerphi(k) / k)), n))}; /* Michael Somos, Jun 02 2019 */
  • PARI
    a(n) = if (n==0, 1, (n-1)!*pollaguerre(n-1,1,-1)); \\ Michel Marcus, Feb 23 2021; Jul 13 2024
  • Python
    from sympy import hyper, hyperexpand
def A000262(n): return hyperexpand(hyper((-n+1, -n), [], 1)) # Chai Wah Wu, Jan 14 2022
  • Sage
    A000262 = lambda n: hypergeometric([-n+1, -n], [], 1)
[simplify(A000262(n)) for n in (0..19)] # Peter Luschny, Apr 08 2015

Formula

D-finite with recurrence: a(n) = (2*n-1)*a(n-1) - (n-1)*(n-2)*a(n-2).
E.g.f.: exp( x/(1-x) ).
a(n) = Sum_{k=0..n} |A008275(n,k)| * A000110(k). - Vladeta Jovovic, Feb 01 2003
a(n) = (n-1)!*LaguerreL(n-1,1,-1) for n >= 1. - Vladeta Jovovic, May 10 2003
Representation as n-th moment of a positive function on positive half-axis: a(n) = Integral_{x=0..oo} x^n*exp(-x-1)*BesselI(1, 2*x^(1/2))/x^(1/2) dx, n >= 1. - Karol A. Penson, Dec 04 2003
a(n) = Sum_{k=0..n} A001263(n, k)*k!. - Philippe Deléham, Dec 10 2003
a(n) = n! Sum_{j=0..n-1} binomial(n-1, j)/(j+1)!, for n > 0. - Herbert S. Wilf, Jun 14 2005
Asymptotic expansion for large n: a(n) -> (0.4289*n^(-1/4) + 0.3574*n^(-3/4) - 0.2531*n^(-5/4) + O(n^(-7/4)))*(n^n)*exp(-n + 2*sqrt(n)). - Karol A. Penson, Aug 28 2002
Minor part of this asymptotic expansion is wrong! Right is (in closed form): a(n) ~ n^(n-1/4)*exp(-1/2+2*sqrt(n)-n)/sqrt(2) * (1 - 5/(48*sqrt(n)) - 95/(4608*n)), numerically a(n) ~ (0.42888194248*n^(-1/4) - 0.0446752023417*n^(-3/4) - 0.00884196713*n^(-5/4) + O(n^(-7/4))) *(n^n)*exp(-n+2*sqrt(n)). - Vaclav Kotesovec, Jun 02 2013
a(n) = exp(-1)*Sum_{m>=0} [m]^n/m!, where [m]^n = m*(m+1)*...*(m+n-1) is the rising factorial. - Vladeta Jovovic, Sep 20 2006
Recurrence: D(n,k) = D(n-1,k-1) + (n-1+k) * D(n-1,k) n >= k >= 0; D(n,0)=0. From this, D(n,1) = n! and D(n,n)=1; a(n) = Sum_{i=0..n} D(n,i). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Proof: Notice two distinct subsets of the lists for [n]: 1) n is in its own list, then there are D(n-1,k-1); 2) n is in a list with other numbers. Denoting the separation of lists by |, it is not hard to see n has (n-1+k) possible positions, so (n-1+k) * D(n-1,k). - Stephen Dalton (StephenMDalton(AT)gmail.com), Jan 05 2007
Define f_1(x), f_2(x), ... such that f_1(x) = exp(x), f_{n+1}(x) = (d/dx)(x^2*f_n(x)), for n >= 2. Then a(n-1) = exp(-1)*f_n(1). - Milan Janjic, May 30 2008
a(n) = (n-1)! * Sum_{k=1..n} (a(n-k)*k!)/((n-k)!*(k-1)!), where a(0)=1. - Thomas Wieder, Sep 10 2008
a(n) = exp(-1)*n!*M(n+1,2,1), n >= 1, where M (=1F1) is the confluent hypergeometric function of the first kind. - Shai Covo (green355(AT)netvision.net.il), Jan 20 2010
a(n) = n!* A067764(n) / A067653(n). - Gary Detlefs, May 15 2010
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)^2*d/dx. Cf. A000110, A049118, A049119 and A049120. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, Nov 17 2011, Aug 02 2012, Dec 11 2012, Jan 27 2013, Jul 31 2013, Dec 25 2013: (Start)
Continued fractions:
E.g.f.: Q(0) where Q(k) = 1+x/((1-x)*(2k+1)-x*(1-x)*(2k+1)/(x+(1-x)*(2k+2)/Q(k+1))).
E.g.f.: 1 + x/(G(0)-x) where G(k) = (1-x)*k + 1 - x*(1-x)*(k+1)/G(k+1).
E.g.f.: exp(x/(1-x)) = 4/(2-(x/(1-x))*G(0))-1 where G(k) = 1 - x^2/(x^2 + 4*(1-x)^2*(2*k+1)*(2*k+3)/G(k+1) ).
E.g.f.: 1 + x*(E(0)-1)/(x+1) where E(k) = 1 + 1/(k+1)/(1-x)/(1-x/(x+1/E(k+1) )).
E.g.f.: E(0)/2, where E(k) = 1 + 1/(1 - x/(x + (k+1)*(1-x)/E(k+1) )).
E.g.f.: E(0)-1, where E(k) = 2 + x/( (2*k+1)*(1-x) - x/E(k+1) ).
(End)
E.g.f.: Product {n >= 1} ( (1 + x^n)/(1 - x^n) )^( phi(2*n)/(2*n) ), where phi(n) = A000010(n) is the Euler totient function. Cf. A088009. - Peter Bala, Jan 01 2014
a(n) = n!*hypergeom([1-n],[2],-1) for n >= 1. - Peter Luschny, Jun 05 2014
a(n) = (-1)^(n-1)*KummerU(1-n,2,-1). - Peter Luschny, Sep 17 2014
a(n) = hypergeom([-n+1, -n], [], 1) for n >= 0. - Peter Luschny, Apr 08 2015
E.g.f.: Product_{k>0} exp(x^k). - Franklin T. Adams-Watters, May 11 2016
0 = a(n)*(18*a(n+2) - 93*a(n+3) + 77*a(n+4) - 17*a(n+5) + a(n+6)) + a(n+1)*(9*a(n+2) - 80*a(n+3) + 51*a(n+4) - 6*a(n+5)) + a(n+2)*(3*a(n+2) - 34*a(n+3) + 15*a(n+4)) + a(n+3)*(-10*a(n+3)) if n >= 0. - Michael Somos, Feb 27 2017
G.f. G(x)=y satisfies a differential equation: (1-x)*y-2*(1-x)*x^2*y'+x^4*y''=1. - Bradley Klee, Aug 13 2018
a(n) = n! * LaguerreL(n, -1, -1) = c_{n}(n-1; -1) where c_{n}(x; a) are the Poisson - Charlier polynomials. - G. C. Greubel, Feb 23 2021
3 divides a(3*n-1); 9 divides a(9*n-1); 11 divides a(11*n-1). - Peter Bala, Mar 26 2022
For n > 0, a(n) = Sum_{k=0..n-1}*k!*C(n-1,k)*C(n,k). - Francesca Aicardi, Nov 03 2022
For n > 0, a(n) = (n-1)! * (Sum_{i=0..n-1} A002720(i) / i!). - Werner Schulte, Mar 29 2024
a(n+1) = numerator of (1 + n/(1 + n/(1 + (n-1)/(1 + (n-1)/(1 + ... + 1/(1 + 1/(1))))))). See A002720 for the denominators. - Peter Bala, Feb 11 2025

A000389 Binomial coefficients C(n,5).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 6, 21, 56, 126, 252, 462, 792, 1287, 2002, 3003, 4368, 6188, 8568, 11628, 15504, 20349, 26334, 33649, 42504, 53130, 65780, 80730, 98280, 118755, 142506, 169911, 201376, 237336, 278256, 324632, 376992, 435897, 501942, 575757, 658008, 749398
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n+4) is the number of inequivalent ways of coloring the vertices of a regular 4-dimensional simplex with n colors, under the full symmetric group S_5 of order 120, with cycle index (x1^5 + 10*x1^3*x2 + 20*x1^2*x3 + 15*x1*x2^2 + 30*x1*x4 + 20*x2*x3 + 24*x5)/120.
Figurate numbers based on 5-dimensional regular simplex. According to Hyun Kwang Kim, it appears that every nonnegative integer can be represented as the sum of g = 10 of these 5-simplex(n) numbers (compared with g=3 for triangular numbers, g=5 for tetrahedral numbers and g=8 for pentatope numbers). - Jonathan Vos Post, Nov 28 2004
The convolution of the nonnegative integers (A001477) with the tetrahedral numbers (A000292), which are the convolution of the nonnegative integers with themselves (making appropriate allowances for offsets of all sequences). - Graeme McRae, Jun 07 2006
a(n) is the number of terms in the expansion of (a_1 + a_2 + a_3 + a_4 + a_5 + a_6)^n. - Sergio Falcon, Feb 12 2007
Product of five consecutive numbers divided by 120. - Artur Jasinski, Dec 02 2007
Equals binomial transform of [1, 5, 10, 10, 5, 1, 0, 0, 0, ...]. - Gary W. Adamson, Feb 02 2009
Equals INVERTi transform of A099242 (1, 7, 34, 153, 686, 3088, ...). - Gary W. Adamson, Feb 02 2009
For a team with n basketball players (n>=5), this sequence is the number of possible starting lineups of 5 players, without regard to the positions (center, forward, guard) of the players. - Mohammad K. Azarian, Sep 10 2009
a(n) is the number of different patterns, regardless of order, when throwing (n-5) 6-sided dice. For example, one die can display the 6 numbers 1, 2, ..., 6; two dice can display the 21 digit-pairs 11, 12, ..., 56, 66. - Ian Duff, Nov 16 2009
Sum of the first n pentatope numbers (1, 5, 15, 35, 70, 126, 210, ...), see A000332. - Paul Muljadi, Dec 16 2009
Sum_{n>=0} a(n)/n! = e/120. Sum_{n>=4} a(n)/(n-4)! = 501*e/120. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
For a set of integers {1,2,...,n}, a(n) is the sum of the 2 smallest elements of each subset with 4 elements, which is 3*C(n+1,5) (for n>=4), hence a(n) = 3*C(n+1,5) = 3*A000389(n+1). - Serhat Bulut, Mar 11 2015
a(n) = fallfac(n,5)/5! is also the number of independent components of an antisymmetric tensor of rank 5 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Number of compositions (ordered partitions) of n+1 into exactly 6 parts. - Juergen Will, Jan 02 2016
Number of weak compositions (ordered weak partitions) of n-5 into exactly 6 parts. - Juergen Will, Jan 02 2016
a(n+3) could be the general number of all geodetic graphs of diameter n>=2 homeomorphic to the Petersen Graph. - Carlos Enrique Frasser, May 24 2018
From Robert A. Russell, Dec 24 2020: (Start)
a(n) is the number of chiral pairs of colorings of the 5 tetrahedral facets (or vertices) of the regular 4-D simplex (5-cell, pentachoron, Schläfli symbol {3,3,3}) using subsets of a set of n colors. Each member of a chiral pair is a reflection but not a rotation of the other.
a(n+4) is the number of unoriented colorings of the 5 tetrahedral facets of the regular 4-D simplex (5-cell, pentachoron) using subsets of a set of n colors. Each chiral pair is counted as one when enumerating unoriented arrangements. (End)
For integer m and positive integer r >= 4, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (4 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

Examples

			G.f. = x^5 + 6*x^6 + 21*x^7 + 56*x^8 + 126*x^9 + 252*x^10 + 462*x^11 + ...
For A={1,2,3,4}, the only subset with 4 elements is {1,2,3,4}; sum of 2 minimum elements of this subset: a(4) = 1+2 = 3 = 3*C(4+1,5).
For A={1,2,3,4,5}, the subsets with 4 elements are {1,2,3,4}, {1,2,3,5}, {1,2,4,5}, {1,3,4,5}, {2,3,4,5}; sum of 2 smallest elements of each subset: a(5) = (1+2)+(1+2)+(1+2)+(1+3)+(2+3) = 18 = 3*C(5+1,5). - _Serhat Bulut_, Mar 11 2015
a(6) = 6 from the six independent components of an antisymmetric tensor A of rank 5 and dimension 6: A(1,2,3,4,5), A(1,2,3,4,6), A(1,2,3,5,6), A(1,2,4,5,6), A(1,3,4,5,6), A(2,3,4,5,6). See the Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015

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.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • 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. 7.
  • Gupta, Hansraj; Partitions of j-partite numbers into twelve or a smaller number of parts. Collection of articles dedicated to Professor P. L. Bhatnagar on his sixtieth birthday. Math. Student 40 (1972), 401-441 (1974).
  • 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. A002299, A053127, A000332, A000579, A000580, A000581, A000582.
Cf. A000217, A005583, A051747, A000292.
Cf. A099242. - Gary W. Adamson, Feb 02 2009
Cf. A242023. A104712 (fourth column, k=5).
Cf. A001477, A049310, A052787, A067764, A110555, A277935.
5-cell colorings: A337895 (oriented), A132366(n-1) (achiral).
Unoriented colorings: A063843 (5-cell edges, faces), A128767 (8-cell vertices, 16-cell facets), A337957 (16-cell vertices, 8-cell facets), A338949 (24-cell), A338965 (600-cell vertices, 120-cell facets).
Chiral colorings: A331352 (5-cell edges, faces), A337954 (8-cell vertices, 16-cell facets), A234249 (16-cell vertices, 8-cell facets), A338950 (24-cell), A338966 (600-cell vertices, 120-cell facets).

Programs

  • Haskell
    a000389 n = a000389_list !! n
a000389_list = 0 : 0 : f [] a000217_list where
   f xs (t:ts) = (sum $ zipWith (*) xs a000217_list) : f (t:xs) ts
-- Reinhard Zumkeller, Mar 03 2015, Apr 13 2012
  • Magma
    [Binomial(n, 5): n in [0..40]]; // Vincenzo Librandi, Mar 12 2015
  • Maple
    f:=n->(1/120)*(n^5-10*n^4+35*n^3-50*n^2+24*n): seq(f(n), n=0..60);
ZL := [S, {S=Prod(B,B,B,B,B,B), B=Set(Z, 1 <= card)}, unlabeled]: seq(combstruct[count](ZL, size=n+1), n=0..42); # Zerinvary Lajos, Mar 13 2007
A000389:=1/(z-1)**6; # Simon Plouffe, 1992 dissertation
  • Mathematica
    Table[Binomial[n, 5], {n, 5, 50}] (* Stefan Steinerberger, Apr 02 2006 *)
CoefficientList[Series[x^5 / (1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Mar 12 2015 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{0,0,0,0,0,1},50] (* Harvey P. Dale, Jul 17 2016 *)
  • PARI
    (conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w);
(t(n)=n*(n+1)/2); u=vector(10,i,t(i)); conv(u,u)

Formula

G.f.: x^5/(1-x)^6.
a(n) = n*(n-1)*(n-2)*(n-3)*(n-4)/120.
a(n) = (n^5-10*n^4+35*n^3-50*n^2+24*n)/120. (Replace all x_i's in the cycle index with n.)
a(n+2) = Sum_{i+j+k=n} i*j*k. - Benoit Cloitre, Nov 01 2002
Convolution of triangular numbers (A000217) with themselves.
Partial sums of A000332. - Alexander Adamchuk, Dec 19 2004
a(n) = -A110555(n+1,5). - Reinhard Zumkeller, Jul 27 2005
a(n+3) = (1/2!)*(d^2/dx^2)S(n,x)|A049310.%20-%20_Wolfdieter%20Lang">{x=2}, n>=2, one half of second derivative of Chebyshev S-polynomials evaluated at x=2. See A049310. - _Wolfdieter Lang, Apr 04 2007
a(n) = A052787(n+5)/120. - Zerinvary Lajos, Apr 26 2007
Sum_{n>=5} 1/a(n) = 5/4. - R. J. Mathar, Jan 27 2009
For n>4, a(n) = 1/(Integral_{x=0..Pi/2} 10*(sin(x))^(2*n-9)*(cos(x))^9). - Francesco Daddi, Aug 02 2011
Sum_{n>=5} (-1)^(n + 1)/a(n) = 80*log(2) - 655/12 = 0.8684411114... - Richard R. Forberg, Aug 11 2014
a(n) = -a(4-n) for all n in Z. - Michael Somos, Oct 07 2014
0 = a(n)*(+a(n+1) + 4*a(n+2)) + a(n+1)*(-6*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Oct 07 2014
a(n) = 3*C(n+1, 5) = 3*A000389(n+1). - Serhat Bulut, Mar 11 2015
From Ilya Gutkovskiy, Jul 23 2016: (Start)
E.g.f.: x^5*exp(x)/120.
Inverse binomial transform of A054849. (End)
From Robert A. Russell, Dec 24 2020: (Start)
a(n) = A337895(n) - a(n+4) = (A337895(n) - A132366(n-1)) / 2 = a(n+4) - A132366(n-1).
a(n+4) = A337895(n) - a(n) = (A337895(n) + A132366(n-1)) / 2 = a(n) + A132366(n-1).
a(n+4) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4) + 1*C(n,5), where the coefficient of C(n,k) is the number of unoriented pentachoron colorings using exactly k colors. (End)

Extensions

Corrected formulas that had been based on other offsets. - R. J. Mathar, Jun 16 2009
I changed the offset to 0. This will require some further adjustments to the formulas. - N. J. A. Sloane, Aug 01 2010

A067653 Denominators of the coefficients in exp(x/(1-x)) power series.

Original entry on oeis.org

1, 1, 2, 6, 24, 40, 720, 5040, 4480, 362880, 3628800, 13305600, 479001600, 6227020800, 29059430400, 1307674368000, 1609445376000, 13173608448000, 6402373705728000, 121645100408832000, 810967336058880000, 4644631106519040000, 86461594444431360000
Offset: 0

Views

Author

Benoit Cloitre, Feb 03 2002

Keywords

Comments

Define c(n)=A067764(n)/A067653(n). For a given sequence s(n) consider P[s(n)](z):=e^(-z/(1-z))*Sum_{k>=0} s(k)c(k)z^k. Regarding complex valued abelian limitation the following holds true: if s(n) is convergent (to the limit s) then lim P[s(n)](z)=s as z tends to +1 in a certain subdomain D of the unit circle. There are two constraints: (1) D contains the line [0,1[. (2) There is a d>0 such that the intersection of {w|Re(w)>1-d} and D is a nonempty subset of a generalized Stolz set defined by {w||Im(w)|<=t*(1-Re(w))^(3/2)}, t<1. If z tends to +1 from outside such a domain that limit doesn't exist in general. - Hieronymus Fischer, Oct 20 2010
There is a significant overlap between the terms of this sequence and the terms of factorials (A000142): 44 of the first 100 terms of this sequence are also factorials. - Harvey P. Dale, Oct 27 2011
If there is no cancellation in the defining sum, a(n) = n!. In particular, a(p) = p! for prime p. In any case a(n) | n!. - Charles R Greathouse IV, Oct 27 2011

Examples

			exp(x/(1-x)) = 1 + x + (3/2)*x^2 + (13/6)*x^3 + (73/24)*x^4 + (167/40)*x^5 + (4051/720)*x^6 + (37633/5040)*x^7 + (43817/4480)*x^8 + (4596553/362880)*x^9 + ... .

References

  • H. Fischer, Eine Theorie komplexwertiger Abelscher Limitierungsmethoden (A theory of complex valued abelian limitation methods), Dissertation (1987), pp. 29-32.
  • K. Knopp, Theory and application of infinite series, Dover, p. 547.
  • O. Perron, Über das infinitäre Verhalten der Koeffizienten einer gewissen Potenzreihe, Archiv d. Math. u. Phys. (3), Vol. 22, pp. 329-340, 1914.
  • K. Zeller, W. Beekmann, Theorie der Limitierungsverfahren, Springer-Verlag, Berlin (1970).

Links

Crossrefs

Cf. A067764.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(x/(1-x)))); [Denominators(b[n]): n in [1..m]]; // G. C. Greubel, Dec 04 2018
  • Maple
    b:= proc(n) option remember; `if`(n=0, 1,
      add((n-k)*b(k), k=0..n-1)/n)
    end:
a:= n-> denom(b(n)):
seq(a(n), n=0..25);  # Alois P. Heinz, May 12 2016
  • Mathematica
    Denominator[Rest[CoefficientList[Series[Exp[x/(1-x)],{x,0,20}],x]]] (* Harvey P. Dale, Oct 26 2011 *)
r[n_] := If[n == 0, 1, Hypergeometric1F1[1 - n, 2, -1]]; Table[Denominator@  r[n], {n, 0, 22}] (* Peter Luschny, Feb 02 2019 *)
  • PARI
    apply(x->denominator(x),Vec(exp(x/(1-x)))) \\ Charles R Greathouse IV, Oct 27 2011
  • SageMath
    [denominator(sum(binomial(n-1, j-1)/factorial(j) for j in (1..n))) for n in range(30)] # G. C. Greubel, Dec 04 2018

Formula

a(n) is the denominator of Sum_{i=1..n} binomial(n-1, i-1)/i!.

Extensions

a(0)=1 prepended by Alois P. Heinz, May 12 2016

A321939 Numerators in the asymptotic expansion of the Maclaurin coefficients of exp(x/(1-x)).

Original entry on oeis.org

1, -5, -479, -15313, 710401, -3532731539, -1439747442109, -34886932972781, -171887027703456763, -6317295244143234168127, -2059266220658860906379923, -16155159358654324183625719723, -125609753430605939189919003924509
Offset: 0

Views

Author

Richard P. Brent, Dec 05 2018

Keywords

Comments

If r(n) = A067764(n)/A067653(n) then r(n)/(exp(2*sqrt(n))/(2*n^(3/4)*sqrt(Pi*e))) has an asymptotic expansion in ascending powers of 1/sqrt(n) whose coefficients are rational numbers 1, -5/48, etc. The sequence gives the numerators of these rational numbers.
Another expression for r(n), n > 0, is r(n) = M(n+1,2,1)/e, where M(a,b,z) = 1F1(a;b;z) is a confluent hypergeometric function (Kummer function).
The same rational numbers, except for signs, occur in the asymptotic expansion of the Maclaurin coefficients of exp(1/(1-x))*E1(1/(1-x)), where E1(x) is an exponential integral. See Lemmas 1-2 and Theorem 5 of the preprint by Brent et al. (2018).

Examples

			The asymptotic expansion is 1 - 5*h/48 - 479*h^2/4608 - 15313*h^3/3317760 + ..., where h = 1/sqrt(n).

References

  • L. J. Slater, Confluent Hypergeometric Functions, Cambridge University Press, 1960.

Links

Crossrefs

The denominators are A321940. The formula for A321939(n)/A321940(n) in Theorem 5 of Brent et al. (2018) uses A321937(n)/A321938(n). The sequence A321941 can be defined using A321939 and A321940.

Formula

A formula is given in Theorem 5, and a recurrence in Lemma 7, of Brent et al. (2018).

A321940 Denominators in the asymptotic expansion of the Maclaurin coefficients of exp(x/(1-x)).

Original entry on oeis.org

1, 48, 4608, 3317760, 127401984, 214035333120, 308210879692800, 2958824445050880, 5680942934497689600, 134979204123665104896000, 18141205034220590098022400, 56600559706768241105829888000
Offset: 0

Views

Author

Richard P. Brent, Dec 08 2018

Keywords

Comments

If r(n) = A067764(n)/A067653(n) then r(n)/(exp(2*sqrt(n))/(2*n^(3/4)*sqrt(Pi*e))) has an asymptotic expansion in ascending powers of 1/sqrt(n) whose coefficients are rational numbers 1, -5/48, etc. The sequence gives the denominators of these rational numbers.
Another expression for r(n), n > 0, is r(n) = M(n+1,2,1)/e, where M(a,b,z) = 1F1(a;b;z) is a confluent hypergeometric function (Kummer function).
The same rational numbers, except for signs, occur in the asymptotic expansion of the Maclaurin coefficients of exp(1/(1-x))*E1(1/(1-x)), where E1(x) is an exponential integral. See Lemmas 1-2 and Theorem 5 of the preprint by Brent et al. (2018).

Examples

			The asymptotic expansion is 1 - 5*h/48 - 479*h^2/4608 - 15313*h^3/3317760 + ..., where h = 1/sqrt(n).

References

  • L. J. Slater, Confluent Hypergeometric Functions, Cambridge University Press, 1960.

Links

Crossrefs

The numerators are A321939. The formula in Theorem 5 of Brent et al. (2018) uses A321937(n)/A321938(n).

Formula

A formula is given in Theorem 5, and a recurrence in Lemma 7, of Brent et al. (2018).

A380171 Numerators of coefficients in expansion of exp(-1 + 1 / Product_{k>=1} (1 - x^k)).

Original entry on oeis.org

1, 1, 5, 31, 265, 2621, 31621, 85319, 6574961, 22334789, 2092318021, 42552808871, 187499032037, 22150499622421, 22390616112461, 15039597200385451, 428293292251548001, 103005657594642373, 407547173842501629061, 2708181047424714819491, 36245898714951203790797
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 14 2025

Keywords

Examples

			1, 1, 5/2, 31/6, 265/24, 2621/120, 31621/720, 85319/1008, 6574961/40320, ...

Crossrefs

Cf. A000041, A017665, A058892, A066186, A067764, A098987, A380271 (denominators).

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[Exp[-1 + 1/Product[1 - x^k, {k, 1, nmax}]], {x, 0, nmax}], x] // Numerator
b[0] = 1; b[n_] := b[n] = (1/n) Sum[k PartitionsP[k] b[n - k], {k, 1, n}]; a[n_] := Numerator[b[n]]; Table[a[n], {n, 0, 20}]

Formula

b(0) = 1, b(n) = (1/n) * Sum_{k=1..n} k * A000041(k) * b(n-k), a(n) = numerator of b(n).
Showing 1-9 of 9 results.

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