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

A001840 Expansion of g.f. x/((1 - x)^2*(1 - x^3)).

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 9, 12, 15, 18, 22, 26, 30, 35, 40, 45, 51, 57, 63, 70, 77, 84, 92, 100, 108, 117, 126, 135, 145, 155, 165, 176, 187, 198, 210, 222, 234, 247, 260, 273, 287, 301, 315, 330, 345, 360, 376, 392, 408, 425, 442, 459, 477, 495, 513, 532, 551, 570, 590
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n-3) is the number of aperiodic necklaces (Lyndon words) with 3 black beads and n-3 white beads.
Number of triangular partitions (see Almkvist).
Consists of arithmetic progression quadruples of common difference n+1 starting at A045943(n). Refers to the least number of coins needed to be rearranged in order to invert the pattern of a (n+1)-rowed triangular array. For instance, a 5-rowed triangular array requires a minimum of a(4)=5 rearrangements (shown bracketed here) for it to be turned upside down.
.....{*}..................{*}*.*{*}{*}
.....*.*....................*.*.*.{*}
....*.*.*....---------\......*.*.*
..{*}*.*.*...---------/.......*.*
{*}{*}*.*{*}..................{*}
- Lekraj Beedassy, Oct 13 2003
Partial sums of 1,1,1,2,2,2,3,3,3,4,4,4,... - Jon Perry, Mar 01 2004
Sum of three successive terms is a triangular number in natural order starting with 3: a(n)+a(n+1)+a(n+2) = T(n+2) = (n+2)*(n+3)/2. - Amarnath Murthy, Apr 25 2004
Apply Riordan array (1/(1-x^3),x) to n. - Paul Barry, Apr 16 2005
Absolute values of numbers that appear in A145919. - Matthew Vandermast, Oct 28 2008
In the Moree definition, (-1)^n*a(n) is the 3rd Witt transform of A033999 and (-1)^n*A004524(n) with 2 leading zeros dropped is the 2nd Witt transform of A033999. - R. J. Mathar, Nov 08 2008
Column sums of:
1 2 3 4 5 6 7 8 9.....
1 2 3 4 5 6.....
1 2 3.....
........................
----------------------
1 2 3 5 7 9 12 15 18 - Jon Perry, Nov 16 2010
a(n) is the sum of the positive integers <= n that have the same residue modulo 3 as n. They are the additive counterpart of the triple factorial numbers. - Peter Luschny, Jul 06 2011
a(n+1) is the number of 3-tuples (w,x,y) with all terms in {0,...,n} and w=3*x+y. - Clark Kimberling, Jun 04 2012
a(n+1) is the number of pairs (x,y) with x and y in {0,...,n}, x-y = (1 mod 3), and x+y < n. - Clark Kimberling, Jul 02 2012
a(n+1) is the number of partitions of n into two sorts of part(s) 1 and one sort of (part) 3. - Joerg Arndt, Jun 10 2013
Arrange A004523 in rows successively shifted to the right two spaces and sum the columns:
1 2 2 3 4 4 5 6 6...
1 2 2 3 4 4 5...
1 2 2 3 4...
1 2 2...
1...
------------------------------
1 2 3 5 7 9 12 15 18... - L. Edson Jeffery, Jul 30 2014
a(n) = A258708(n+1,1) for n > 0. - Reinhard Zumkeller, Jun 23 2015
Also the number of triples of positive integers summing to n + 4, the first less than each of the other two. Also the number of triples of positive integers summing to n + 2, the first less than or equal to each of the other two. - Gus Wiseman, Oct 11 2020
Also the lower matching number of the (n+1)-triangular honeycomb king graph = n-triangular grid graph (West convention). - Eric W. Weisstein, Dec 14 2024

Examples

			G.f. = x + 2*x^2 + 3*x^3 + 5*x^4 + 7*x^5 + 9*x^6 + 12*x^7 + 15*x^8 + 18*x^9 + ...
1+2+3=6=t(3), 2+3+5=t(4), 5+7+9=t(5).
[n] a(n)
--------
[1] 1
[2] 2
[3] 3
[4] 1 + 4
[5] 2 + 5
[6] 3 + 6
[7] 1 + 4 + 7
[8] 2 + 5 + 8
[9] 3 + 6 + 9
a(7) = floor(2/3) +floor(3/3) +floor(4/3) +floor(5/3) +floor(6/3) +floor(7/3) +floor(8/3) +floor(9/3) = 12. - _Bruno Berselli_, Aug 29 2013

References

  • Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 73, problem 25.
  • Ulrich Faigle, Review of Gerhard Post and G.J. Woeginger, Sports tournaments, home-away assignments and the break minimization problem, MR2224983(2007b:90134), 2007.
  • Hansraj Gupta, 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).
  • Richard K. Guy, A problem of Zarankiewicz, in P. Erdős and G. Katona, editors, Theory of Graphs (Proceedings of the Colloquium, Tihany, Hungary), Academic Press, NY, 1968, pp. 119-150, (p. 126, divided by 2).
  • 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. A000031, A001037, A008748, A051168.
Ordered union of triangular matchstick numbers A045943 and generalized pentagonal numbers A001318.
Cf. A058937.
A column of triangle A011847.
Cf. A000217, A130205.
Cf. A258708.
A001399 counts 3-part partitions, ranked by A014612.
A337483 counts either weakly increasing or weakly decreasing triples.
A337484 counts neither strictly increasing nor strictly decreasing triples.
A014311 ranks 3-part compositions, with strict case A337453.
Cf. A007052, A007304, A011782, A069905, A156040, A218004.

Programs

  • Haskell
    a001840 n = a001840_list !! n
a001840_list = scanl (+) 0 a008620_list
-- Reinhard Zumkeller, Apr 16 2012
  • Magma
    [ n le 2 select n else n*(n+1)/2-Self(n-1)-Self(n-2): n in [1..58] ];  // Klaus Brockhaus, Oct 01 2009
  • Maple
    A001840 := n->floor((n+1)*(n+2)/6);
A001840:=-1/((z**2+z+1)*(z-1)**3); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
seq(floor(binomial(n-1,2)/3), n=3..61); # Zerinvary Lajos, Jan 12 2009
A001840 :=  n -> add(k, k = select(k -> k mod 3 = n mod 3, [$1 .. n])): seq(A001840(n), n = 0 .. 58); # Peter Luschny, Jul 06 2011
  • Mathematica
    a[0]=0; a[1]=1; a[n_]:= a[n]= n(n+1)/2 -a[n-1] -a[n-2]; Table[a[n], {n,0,100}]
f[n_] := Floor[(n + 1)(n + 2)/6]; Array[f, 59, 0] (* Or *)
CoefficientList[ Series[ x/((1 + x + x^2)*(1 - x)^3), {x, 0, 58}], x] (* Robert G. Wilson v *)
a[ n_] := With[{m = If[ n < 0, -3 - n, n]}, SeriesCoefficient[ x /((1 - x^3) (1 - x)^2), {x, 0, m}]]; (* Michael Somos, Jul 11 2011 *)
LinearRecurrence[{2,-1,1,-2,1},{0,1,2,3,5},60] (* Harvey P. Dale, Jul 25 2011 *)
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n+4,{3}],#[[1]]<#[[2]]&&#[[1]]<#[[3]]&]],{n,0,15}] (* Gus Wiseman, Oct 05 2020 *)
  • PARI
    {a(n) = (n+1) * (n+2) \ 6}; /* Michael Somos, Feb 11 2004 */
  • Sage
    [binomial(n, 2) // 3 for n in range(2, 61)] # Zerinvary Lajos, Dec 01 2009

Formula

a(n) = (A000217(n+1) - A022003(n-1))/3;
a(n) = (A016754(n+1) - A010881(A016754(n+1)))/24;
a(n) = (A033996(n+1) - A010881(A033996(n+1)))/24.
Euler transform of length 3 sequence [2, 0, 1].
a(3*k-1) = k*(3*k + 1)/2;
a(3*k) = 3*k*(k + 1)/2;
a(3*k+1) = (k + 1)*(3*k + 2)/2.
a(n) = floor( (n+1)*(n+2)/6 ) = floor( A000217(n+1)/3 ).
a(n+1) = a(n) + A008620(n) = A002264(n+3). - Reinhard Zumkeller, Aug 01 2002
From Michael Somos, Feb 11 2004: (Start)
G.f.: x / ((1-x)^2 * (1-x^3)).
a(n) = 1 + a(n-1) + a(n-3) - a(n-4).
a(-3-n) = a(n). (End)
a(n) = a(n-3) + n for n > 2; a(0)=0, a(1)=1, a(2)=2. - Paul Barry, Jul 14 2004
a(n) = binomial(n+3, 3)/(n+3) + cos(2*Pi*(n-1)/3)/9 + sqrt(3)sin(2*Pi*(n-1)/3)/9 - 1/9. - Paul Barry, Jan 01 2005
From Paul Barry, Apr 16 2005: (Start)
a(n) = Sum_{k=0..n} k*(cos(2*Pi*(n-k)/3 + Pi/3)/3 + sqrt(3)*sin(2*Pi*(n-k)/3 + Pi/3)/3 + 1/3).
a(n) = Sum_{k=0..floor(n/3)} n-3*k. (End)
For n > 1, a(n) = A000217(n) - a(n-1) - a(n-2); a(0)=0, a(1)=1.
G.f.: x/(1 + x + x^2)/(1 - x)^3. - Maksym Voznyy (voznyy(AT)mail.ru), Jul 27 2009
a(n) = (4 + 3*n^2 + 9*n)/18 + ((n mod 3) - ((n-1) mod 3))/9. - Klaus Brockhaus, Oct 01 2009
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5), with n>4, a(0)=0, a(1)=1, a(2)=2, a(3)=3, a(4)=5. - Harvey P. Dale, Jul 25 2011
a(n) = A214734(n + 2, 1, 3). - Renzo Benedetti, Aug 27 2012
G.f.: x*G(0), where G(k) = 1 + x*(3*k+4)/(3*k + 2 - 3*x*(k+2)*(3*k+2)/(3*(1+x)*k + 6*x + 4 - x*(3*k+4)*(3*k+5)/(x*(3*k+5) + 3*(k+1)/G(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Jun 10 2013
Empirical: a(n) = floor((n+3)/(e^(6/(n+3))-1)). - Richard R. Forberg, Jul 24 2013
a(n) = Sum_{i=0..n} floor((i+2)/3). - Bruno Berselli, Aug 29 2013
0 = a(n)*(a(n+2) + a(n+3)) + a(n+1)*(-2*a(n+2) - a(n+3) + a(n+4)) + a(n+2)*(a(n+2) - 2*a(n+3) + a(n+4)) for all n in Z. - Michael Somos, Jan 22 2014
a(n) = n/2 + floor(n^2/3 + 2/3)/2. - Bruno Berselli, Jan 23 2017
a(n) + a(n+1) = A000212(n+2). - R. J. Mathar, Jan 14 2021
Sum_{n>=1} 1/a(n) = 20/3 - 2*Pi/sqrt(3). - Amiram Eldar, Sep 27 2022
E.g.f.: (exp(x)*(4 + 12*x + 3*x^2) - 4*exp(-x/2)*cos(sqrt(3)*x/2))/18. - Stefano Spezia, Apr 05 2023

A007997 a(n) = ceiling((n-3)(n-4)/6).

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 5, 7, 10, 12, 15, 19, 22, 26, 31, 35, 40, 46, 51, 57, 64, 70, 77, 85, 92, 100, 109, 117, 126, 136, 145, 155, 166, 176, 187, 199, 210, 222, 235, 247, 260, 274, 287, 301, 316, 330, 345, 361, 376, 392, 409, 425, 442, 460, 477, 495, 514, 532, 551, 571, 590, 610
Offset: 3

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of solutions to x+y+z=0 (mod m) with 0<=x<=y<=z
Nonorientable genus of complete graph on n nodes.
Also (with different offset) Molien series for alternating group A_3.
(1+x^3 ) / ((1-x)*(1-x^2)*(1-x^3)) is the Poincaré series [or Poincare series] (or Molien series) for H^*(S_6, F_2).
a(n+5) is the number of necklaces with 3 black beads and n white beads.
The g.f./x^5 is Z(C_3,x), the 3-variate cycle index polynomial for the cyclic group C_3, with substitution x[i]->1/(1-x^i), i=1,2,3. Therefore by Polya enumeration a(n+5) is the number of cyclically inequivalent 3-necklaces whose 3 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... . See A102190 for Z(C_3,x). - Wolfdieter Lang, Feb 15 2005
a(n+1) is the number of pairs (x,y) with x and y in {0,...,n}, x = (y mod 3), and x+y < n. - Clark Kimberling, Jul 02 2012
From Gus Wiseman, Oct 17 2020: (Start)
Also the number of 3-part integer compositions of n - 2 that are either weakly increasing or strictly decreasing. For example, the a(5) = 1 through a(13) = 15 compositions are:
(111) (112) (113) (114) (115) (116) (117) (118) (119)
(122) (123) (124) (125) (126) (127) (128)
(222) (133) (134) (135) (136) (137)
(321) (223) (224) (144) (145) (146)
(421) (233) (225) (226) (155)
(431) (234) (235) (227)
(521) (333) (244) (236)
(432) (334) (245)
(531) (532) (335)
(621) (541) (344)
(631) (542)
(721) (632)
(641)
(731)
(821)
(End)

Examples

			For m=7 (n=12), the 12 solutions are xyz = 000 610 520 511 430 421 331 322 662 653 644 554.

References

  • A. Adem and R. J. Milgram, Cohomology of Finite Groups, Springer-Verlag, 2nd. ed., 2004, p. 204.
  • D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 105.
  • J. L. Gross and T. W. Tucker, Topological Graph Theory, Wiley, 1987; see \bar{I}(n) p. 221.
  • J. L. Gross and J. Yellen, eds., Handbook of Graph Theory, CRC Press, 2004; p. 740.
  • E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, 2004.

Links

Crossrefs

Cf. A000031, A007998, A003050, A047996, A048259, A053618.
Apart from initial term, same as A058212.
Cf. A000212, A000217, A001840, A128422, A156040.
A001399(n-6)*2 = A069905(n-3)*2 = A211540(n-1)*2 counts the strict case.
A014311 intersected with A225620 U A333256 ranks these compositions.
A218004 counts these compositions of any length.
A000009 counts strictly decreasing compositions.
A000041 counts weakly increasing compositions.
A001523 counts unimodal compositions, with complement counted by A115981.
A007318 and A097805 count compositions by length.
A032020 counts strict compositions, ranked by A233564.
A333149 counts neither increasing nor decreasing strict compositions.
Cf. A014311, A332834, A333147, A337481, A337482-A337484.

Programs

  • Haskell
    a007997 n = ceiling $ (fromIntegral $ (n - 3) * (n - 4)) / 6
a007997_list = 0 : 0 : 1 : zipWith (+) a007997_list [1..]
-- Reinhard Zumkeller, Dec 18 2013
  • Maple
    x^5*(1+x^3)/((1-x)*(1-x^2)*(1-x^3));
seq(ceil(binomial(n,2)/3), n=0..63); # Zerinvary Lajos, Jan 12 2009
a := n -> (n*(n-7)-2*([1,1,-1][n mod 3 +1]-7))/6;
seq(a(n), n=3..64); # Peter Luschny, Jan 13 2015
  • Mathematica
    k = 3; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)
Table[Ceiling[((n-3)(n-4))/6],{n,3,100}] (* or *) LinearRecurrence[ {2,-1,1,-2,1},{0,0,1,1,2},100] (* Harvey P. Dale, Jan 21 2014 *)
  • PARI
    a(n)=(n^2-7*n+16)\6 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = a(n-3) + n - 2, a(0)=0, a(1)=0, a(2)=1 [Offset 0]. - Paul Barry, Jul 14 2004
G.f.: x^5*(1+x^3)/((1-x)*(1-x^2)*(1-x^3)) = x^5*(1-x+x^2)/((1-x)^2*(1-x^3)).
a(n+5) = Sum_{k=0..floor(n/2)} C(n-k,L(k/3)), where L(j/p) is the Legendre symbol of j and p. - Paul Barry, Mar 16 2006
a(3)=0, a(4)=0, a(5)=1, a(6)=1, a(7)=2, a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5). - Harvey P. Dale, Jan 21 2014
a(n) = (n^2 - 7*n + 14 - 2*(-1)^(2^(n + 1 - 3*floor((n+1)/3))))/6. - Luce ETIENNE, Dec 27 2014
a(n) = A001399(n-3) + A001399(n-6). Compare to A140106(n) = A001399(n-3) - A001399(n-6). - Gus Wiseman, Oct 17 2020
a(n) = (40 + 3*(n - 7)*n - 4*cos(2*n*Pi/3) - 4*sqrt(3)*sin(2*n*Pi/3))/18. - Stefano Spezia, Dec 14 2021
Sum_{n>=5} 1/a(n) = 6 - 2*Pi/sqrt(3) + 2*Pi*tanh(sqrt(5/3)*Pi/2)/sqrt(15). - Amiram Eldar, Oct 01 2022

A211540 Number of ordered triples (w,x,y) with all terms in {1..n} and 2w = 3x + 4y.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 2, 3, 4, 5, 7, 8, 10, 12, 14, 16, 19, 21, 24, 27, 30, 33, 37, 40, 44, 48, 52, 56, 61, 65, 70, 75, 80, 85, 91, 96, 102, 108, 114, 120, 127, 133, 140, 147, 154, 161, 169, 176, 184, 192, 200, 208, 217, 225, 234, 243, 252, 261, 271, 280, 290
Offset: 0

Views

Author

Clark Kimberling, Apr 15 2012

Keywords

Comments

For a guide to related sequences, see A211422.
Also the number of partitions of n+1 into three parts, where each part > 1. - Peter Woodward, May 25 2015
a(n) is also equal to the number of partitions of n+4 into three distinct parts, where each part > 1. - Giovanni Resta, May 26 2015
Number of different distributions of n+1 identical balls in 3 boxes as x,y,z where 0 < x < y < z. - Ece Uslu and Esin Becenen, Dec 31 2015
After the first three terms, partial sums of A008615. - Robert Israel, Dec 31 2015
For n >= 2, also the number of partitions of n - 2 into 3 parts. The Heinz numbers of these partitions are given by A014612. - Gus Wiseman, Oct 11 2020

Examples

			a(5) = a(6) = 1 with only one ordered triple (5,2,1). - _Michael Somos_, Feb 02 2015
a(11) = 5 Number of different distributions of 11 identical balls in 3 boxes as x,y and z where 0 < x < y < z. - _Ece Uslu_, Esin Becenen, Dec 31 2015
a(1) = a(2) = a(3) = a(4) = a(5) = 0, since with fewer than 6 identical balls there is no such distribution with 3 boxes that holds for 0 < x < y < z. - _Ece Uslu_, Esin Becenen, Dec 31 2015
G.f.: x^5 + x^6 + 2*x^7 + 3*x^8 + 4*x^9 + 5*x^10 + 7*x^11 + 8*x^12 + ...
From _Gus Wiseman_, Oct 11 2020: (Start)
The a(5) = 1 through a(15) = 14 partitions of n + 1 into three parts > 1 [Woodward] are the following (A = 10, B = 11, C = 12). The ordered version is A000217(n - 4) and the Heinz numbers are A046316.
  222  322  332  333  433  443  444  544  554  555  655
            422  432  442  533  543  553  644  654  664
                 522  532  542  552  643  653  663  754
                      622  632  633  652  662  744  763
                           722  642  733  743  753  772
                                732  742  752  762  844
                                822  832  833  843  853
                                     922  842  852  862
                                          932  933  943
                                          A22  942  952
                                               A32  A33
                                               B22  A42
                                                    B32
                                                    C22
The a(5) = 1 through a(15) = 14 partitions of n + 4 into three distinct parts > 1 [Resta] are the following (A = 10, B = 11, C = 12, D = 13, E = 14). The ordered version is A211540*6 and the Heinz numbers are A046389.
  432  532  542  543  643  653  654  754  764  765  865
            632  642  652  743  753  763  854  864  874
                 732  742  752  762  853  863  873  964
                      832  842  843  862  872  954  973
                           932  852  943  953  963  982
                                942  952  962  972  A54
                                A32  A42  A43  A53  A63
                                     B32  A52  A62  A72
                                          B42  B43  B53
                                          C32  B52  B62
                                               C42  C43
                                               D32  C52
                                                    D42
                                                    E32
The a(5) = 1 through a(15) = 14 partitions of n + 1 into three distinct parts [Uslu and Becenen] are the following (A = 10, B = 11, C = 12, D = 13). The ordered version is A211540(n)*6 and the Heinz numbers are A007304.
  321  421  431  432  532  542  543  643  653  654  754
            521  531  541  632  642  652  743  753  763
                 621  631  641  651  742  752  762  853
                      721  731  732  751  761  843  862
                           821  741  832  842  852  871
                                831  841  851  861  943
                                921  931  932  942  952
                                     A21  941  951  961
                                          A31  A32  A42
                                          B21  A41  A51
                                               B31  B32
                                               C21  B41
                                                    C31
                                                    D21
(End)

Links

Crossrefs

Cf. A001399, A069905, A211422.
All of the following pertain to 3-part strict partitions.
- A000009 counts these partitions of any length, with non-strict version A000041.
- A007304 gives the Heinz numbers, with non-strict version A014612.
- A101271 counts the relatively prime case, with non-strict version A023023.
- A220377 counts the pairwise coprime case, with non-strict version A307719.
- A337605 counts the pairwise non-coprime case, with non-strict version A337599.
Cf. A000217, A001840, A156040, A284825, A337453, A337483, A337484, A337563.

Programs

  • Magma
    I:=[0,0,0,0,0,1]; [n le 6 select I[n] else Self(n-1)+Self(n-2)-Self(n-4)-Self(n-5)+Self(n-6): n in [1..70]]; // Vincenzo Librandi, Dec 31 2015
  • Maple
    f:= gfun:-rectoproc({a(n) = a(n-1)+a(n-2)-a(n-4)-a(n-5)+a(n-6),seq(a(i)=0,i=0..4),a(5)=1},a(n),remember):
seq(f(i),i=0..100); # Robert Israel, Dec 31 2015
  • Mathematica
    t[n_] := t[n] = Flatten[Table[-2 w + 3 x + 4 y, {w, n}, {x, n}, {y, n}]]
c[n_] := Count[t[n], 0]
t = Table[c[n], {n, 0, 80}]  (* A211540 *)
FindLinearRecurrence[t]
LinearRecurrence[{1, 1, 0, -1, -1, 1}, {0, 0, 0, 0, 0, 1}, 70] (* Vincenzo Librandi, Dec 31 2015 *)
Table[Length[Select[IntegerPartitions[n+1,{3}],UnsameQ@@#&]],{n,0,30}] (* Gus Wiseman, Oct 05 2020 *)
  • PARI
    {a(n) = round( (n-2)^2 / 12 )}; / * Michael Somos, Feb 02 2015 */
  • PARI
    concat(vector(5), Vec(x^5/(1-x-x^2+x^4+x^5-x^6) + O(x^100))) \\ Altug Alkan, Jan 10 2016

Formula

a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6).
a(n) = A069905(n-2) = A001399(n-5) for n >= 5. - Alois P. Heinz, Nov 03 2012
a(n) = 3*k^2-6*k+3 (for n = 6*k-3), 3*k^2-5*k+2 (for n = 6*k-2), 3*k^2-4*k+1 (for n = 6*k-1), 3*k^2-3*k+1 (for n = 6*k), 3*k^2-2*k (for n = 6*k+1), 3*k^2-k (for n = 6*k+2). - Ece Uslu, Esin Becenen, Dec 31 2015
a(n) = A004526(n-2) + a(n-2) for n > 2. - Ece Uslu, Esin Becenen, Dec 31 2015
G.f.: x^5/(1 - x - x^2 + x^4 + x^5 - x^6). - Robert Israel, Dec 31 2015
a(n) = Sum_{k=1..floor(n/3)} floor((n-k)/2)-k. - Wesley Ivan Hurt, Apr 27 2019
From Gus Wiseman, Oct 11 2020: (Start)
a(n+2) = A069905(n) = A001399(n-3) counts 3-part partitions.
a(n-1) = A069905(n-3) = A001399(n-6) counts 3-part strict partitions.
a(n-1) = A069905(n-3) = A001399(n-6) counts 3-part partitions with no 1's.
a(n-4) = A069905(n-6) = A001399(n-9) counts 3-part strict partitions with no 1's.
A000217(n-2) counts 3-part compositions.
a(n-1)*6 = A069905(n-3)*6 = A001399(n-6)*6 counts 3-part strict compositions.
A000217(n-5) counts 3-part compositions with no 1's.
a(n-4)*6 = A069905(n-6)*6 = A001399(n-9)*6 counts 3-part strict compositions with no 1's.
(End)

A071619 a(n) = ceiling(2*n^2 / 3).

Original entry on oeis.org

0, 1, 3, 6, 11, 17, 24, 33, 43, 54, 67, 81, 96, 113, 131, 150, 171, 193, 216, 241, 267, 294, 323, 353, 384, 417, 451, 486, 523, 561, 600, 641, 683, 726, 771, 817, 864, 913, 963, 1014, 1067, 1121, 1176, 1233, 1291, 1350, 1411, 1473, 1536, 1601, 1667, 1734, 1803, 1873
Offset: 0

Views

Author

N. J. A. Sloane, Jun 21 2002

Keywords

Comments

Old name was: If n mod 3 = 0 then 6*(n/3)^2 elif n mod 3 = 1 then 6*((n-2)/3)^2+8*(n-2)/3 + 3 else 6*((n-1)/3)^2+4*(n-1)/3+1.
From Ray G. Opao, Oct 20 2022: (Start)
For n >= 3, a(n) is the maximum number of objects that can be placed on an n X n grid such that no 3 adjacent grid points on the same row or column are occupied.
The first 5 terms of this description are illustrated in the Opao link. (End)

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.

Links

Crossrefs

Partial sums of A042968.
Essentially a bisection of A156040.

Programs

  • Magma
    [Ceiling(2*n^2/3): n in [0..100]]; // Wesley Ivan Hurt, Mar 12 2015
  • Maple
    A071619 := proc(n) if n mod 3 = 0 then 6*(n/3)^2 elif n mod 3 = 1 then 6*((n-1)/3)^2+4*(n-1)/3+1 else 6*((n-2)/3)^2+8*(n-2)/3 +3; fi; end;
  • Mathematica
    f[n_] := Which[Divisible[n, 3], 6(n/3)^2, Mod[n, 3] == 1, 6(((n - 1)/3)^2) + 4 (n - 1)/3 + 1, True, 6((n - 2)/3)^2 + 8((n - 2)/3) + 3]; Array[f, 60, 0] (* or *) LinearRecurrence[{2, -1, 1, -2, 1}, {0, 1, 3, 6, 11}, 60] (* Harvey P. Dale, Feb 28 2012 *)
CoefficientList[Series[((x * (1 + x) * (1 + x^2))/((1 + x + x^2) * (1 - x)^3)), {x, 0, 53}], x] (* L. Edson Jeffery, Jul 30 2014 *)
Ceiling[2Range[0, 49]^2/3] (* Alonso del Arte, Mar 13 2015 *)
Table[n^2 - Floor[n^2/3], {n, 0, 60}] (* Bruno Berselli, Jan 18 2017 *)
  • PARI
    f=(x,y)->6*((x-y)/3)^2+4*y*(x-y)/3+y*(y+1)/2;
a(n)=f(n,n%3); \\ R. J. Cano, Jul 20 2014

Formula

From Vladeta Jovovic, Jun 23 2002: (Start)
a(n) = (2/3)*n^2 if n mod 3 = 0, otherwise (2/3)*n^2 + 1/3.
Recurrence: a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
G.f.: x*(1 + x)*(1 + x^2)/(1 + x + x^2)/(1 - x)^3. (End)
a(n) = ceiling(2*n^2 / 3). - Wesley Ivan Hurt, Jun 20 2013
a(n) + a(n+1) + a(n+2) = A005893(n+1). - R. J. Mathar, Mar 01 2014
a(n+1) = A156040(2*n). - L. Edson Jeffery, Jul 30 2014
Let F(x,y) = 6*((x-y)/3)^2 + 4*y*(x-y)/3 + y*(y+1)/2; then a(n) = F(n,(n mod 3)). - R. J. Cano, Jul 30 2014
a(n) = Sum_{j=1..n} Sum_{i=1..n} ceiling((i+j-n)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = n^2 - floor(n^2/3) = (2/9)*(3*n^2 + 1 - cos(2*Pi*n/3)). - Bruno Berselli, Jan 18 2017
E.g.f.: (2*exp(x)*(1 + 3*x*(1 + x)) - 2*exp(-x/2)*cos(sqrt(3)*x/2))/9. - Stefano Spezia, Oct 17 2022
Sum_{n>=1} 1/a(n) = Pi^2/36 + 3*c*sinh(c)/(1+2*cosh(c)), where c = Pi*sqrt(2)/3. - Amiram Eldar, Jan 08 2023

Extensions

Corrected definition (Old Name) from Harvey P. Dale, Feb 28 2012
New name from Wesley Ivan Hurt, Mar 13 2015

A337484 Number of ordered triples of positive integers summing to n that are neither strictly increasing nor strictly decreasing.

Original entry on oeis.org

0, 0, 0, 1, 3, 6, 8, 13, 17, 22, 28, 35, 41, 50, 58, 67, 77, 88, 98, 111, 123, 136, 150, 165, 179, 196, 212, 229, 247, 266, 284, 305, 325, 346, 368, 391, 413, 438, 462, 487, 513, 540, 566, 595, 623, 652, 682, 713, 743, 776, 808, 841, 875, 910, 944, 981, 1017
Offset: 0

Views

Author

Gus Wiseman, Sep 11 2020

Keywords

Examples

			The a(3) = 1 through a(7) = 13 triples:
  (1,1,1)  (1,1,2)  (1,1,3)  (1,1,4)  (1,1,5)
           (1,2,1)  (1,2,2)  (1,3,2)  (1,3,3)
           (2,1,1)  (1,3,1)  (1,4,1)  (1,4,2)
                    (2,1,2)  (2,1,3)  (1,5,1)
                    (2,2,1)  (2,2,2)  (2,1,4)
                    (3,1,1)  (2,3,1)  (2,2,3)
                             (3,1,2)  (2,3,2)
                             (4,1,1)  (2,4,1)
                                      (3,1,3)
                                      (3,2,2)
                                      (3,3,1)
                                      (4,1,2)
                                      (5,1,1)

Crossrefs

A140106 is the unordered case.
A242771 allows strictly increasing but not strictly decreasing triples.
A337481 counts these compositions of any length.
A001399(n - 6) counts unordered strict triples.
A001523 counts unimodal compositions, with complement A115981.
A007318 and A097805 count compositions by length.
A069905 counts unordered triples.
A218004 counts strictly increasing or weakly decreasing compositions.
A332745 counts partitions with weakly increasing or weakly decreasing run-lengths.
A332835 counts compositions with weakly increasing or weakly decreasing run-lengths.
A337483 counts triples either weakly increasing or weakly decreasing.
Cf. A000212, A000217, A001840, A014311, A046691, A128422, A156040, A332834, A337461, A337482, A337561, A337603, A337604.

Programs

  • Mathematica
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{3}],!Less@@#&&!Greater@@#&]],{n,0,15}]

Formula

a(n) = 2*A242771(n - 1) - A000217(n - 1), n > 0.
2*A001399(n - 6) = 2*A069905(n - 3) = 2*A211540(n - 1) is the complement.
4*A001399(n - 6) = 4*A069905(n - 3) = 4*A211540(n - 1) is the strict case.
Conjectures from Colin Barker, Sep 13 2020: (Start)
G.f.: x^3*(1 + 2*x + 2*x^2 - x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n>6.
(End)

A337483 Number of ordered triples of positive integers summing to n that are either weakly increasing or weakly decreasing.

Original entry on oeis.org

0, 0, 0, 1, 2, 4, 5, 8, 10, 13, 16, 20, 23, 28, 32, 37, 42, 48, 53, 60, 66, 73, 80, 88, 95, 104, 112, 121, 130, 140, 149, 160, 170, 181, 192, 204, 215, 228, 240, 253, 266, 280, 293, 308, 322, 337, 352, 368, 383, 400, 416, 433, 450, 468, 485, 504, 522, 541, 560
Offset: 0

Views

Author

Gus Wiseman, Sep 07 2020

Keywords

Examples

			The a(3) = 1 through a(8) = 10 triples:
  (1,1,1)  (1,1,2)  (1,1,3)  (1,1,4)  (1,1,5)  (1,1,6)
           (2,1,1)  (1,2,2)  (1,2,3)  (1,2,4)  (1,2,5)
                    (2,2,1)  (2,2,2)  (1,3,3)  (1,3,4)
                    (3,1,1)  (3,2,1)  (2,2,3)  (2,2,4)
                             (4,1,1)  (3,2,2)  (2,3,3)
                                      (3,3,1)  (3,3,2)
                                      (4,2,1)  (4,2,2)
                                      (5,1,1)  (4,3,1)
                                               (5,2,1)
                                               (6,1,1)

Links

Crossrefs

A001399(n - 3) = A069905(n) = A211540(n + 2) counts the unordered case.
2*A001399(n - 6) = 2*A069905(n - 3) = 2*A211540(n - 1) counts the strict case.
A001399(n - 6) = A069905(n - 3) = A211540(n - 1) counts the strict unordered case.
A329398 counts these compositions of any length.
A218004 counts strictly increasing or weakly decreasing compositions.
A337484 counts neither strictly increasing nor strictly decreasing compositions.
Cf. A000212, A000217, A001840, A014311, A156040, A337461, A337603, A337604.

Programs

  • Mathematica
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{3}],LessEqual@@#||GreaterEqual@@#&]],{n,0,30}]

Formula

a(n > 0) = 2*A001399(n - 3) - A079978(n).
From Colin Barker, Sep 08 2020: (Start)
G.f.: x^3*(1 + x + x^2 - x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n>6. (End)
E.g.f.: (36 - 9*exp(-x) + exp(x)*(6*x^2 + 6*x - 19) - 8*exp(-x/2)*cos(sqrt(3)*x/2))/36. - Stefano Spezia, Apr 05 2023

A156041 Array A(n,k) (n>=1, k>=1) read by antidiagonals, where A(n,k) is the number of compositions (ordered partitions) of n into exactly k parts, some of which may be zero, with the first part greater than or equal to all the rest.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 3, 4, 4, 1, 1, 3, 6, 7, 5, 1, 1, 4, 8, 11, 11, 6, 1, 1, 4, 11, 17, 19, 16, 7, 1, 1, 5, 13, 26, 32, 31, 22, 8, 1, 1, 5, 17, 35, 54, 56, 48, 29, 9, 1, 1, 6, 20, 48, 82, 102, 93, 71, 37, 10, 1, 1, 6, 24, 63, 120, 172, 180, 148, 101, 46, 11, 1, 1, 7, 28, 81, 170
Offset: 1

Views

Author

Jack W Grahl, Feb 02 2009, Feb 11 2009

Keywords

Comments

A(n,k) is of course smaller than the number of ordered partitions of n into k parts and at least the number of partitions into k parts in descending order.
The sums of the antidiagonals give A079500 - 1. - N. J. A. Sloane, Feb 26 2011
For an alternative definition of essentially the same sequence, as a triangle, and which avoids the use of parts of size zero, see A184957. - N. J. A. Sloane, Feb 27 2011

Examples

			The array A(n,k) begins:
  1  1  1  1  1  1  1  1  1 ...
  1  2  3  4  5  6  7  8  9 ...
  1  2  4  7 11 16 22 29 ...
  1  3  6 11 19 31 48 ...
  1  3  8 17 32 56 ...
  1  4 11 26 54 ...
  1  4 13 35 ...
  ...
The antidiagonals are:
  1,
  1, 1,
  1, 2, 1,
  1, 2, 3, 1,
  1, 3, 4, 4, 1,
  1, 3, 6, 7, 5, 1,
  1, 4, 8, 11, 11, 6, 1,
  1, 4, 11, 17, 19, 16, 7, 1,
  1, 5, 13, 26, 32, 31, 22, 8, 1,
  ...
A(3,5) = 11 and the 11 partition of 3 into 5 parts of this type are: (3,0,0,0,0), (2,1,0,0,0), (2,0,1,0,0), (2,0,0,1,0), (2,0,0,0,1), (1,1,1,0,0), (1,1,0,1,0), (1,1,0,0,1), (1,0,1,1,0), (1,0,1,0,1), (1,0,0,1,1).

Links

Crossrefs

A156039 gives A(n,4) and A156040 gives A(n,3). A156042 is the part on or below the main diagonal. A(n,2) is A008619. A(2,n) is A000027. A(3,n) is A000124.
Cf. A079500.

Programs

  • Maple
    b:= proc(n, i, m) option remember;
      if n<0 then 0
    elif n=0 then 1
    elif i=1 then `if`(n<=m, 1, 0)
    else add(b(n-k, i-1, m), k=0..m)
      fi
    end:
A:= (n, k)-> add(b(n-m, k-1, m), m=ceil(n/k)..n):
seq(seq(A(d-k, k), k=1..d-1), d=1..14); # Alois P. Heinz, Jun 14 2009
  • Mathematica
    (* Returns rectangular array *) nn=10;Table[Table[Coefficient[Series[Sum[x^i((1-x^(i+1))/(1-x))^(k-1),{i,0,n}],{x,0,nn}],x^n],{k,1,nn}],{n,1,nn}]//Grid (* Geoffrey Critzer, Jul 15 2013 *)

Formula

A(n,k) = [x^n] Sum_{i=0..n} x^i*((1 - x^(i+1))/(1-x))^(k-1). - Geoffrey Critzer, Jul 15 2013

Extensions

More terms from Alois P. Heinz, Jun 14 2009
Edited by N. J. A. Sloane, Feb 26 2011

A014126 Number of partitions of 2*n into at most 4 parts.

Original entry on oeis.org

1, 2, 5, 9, 15, 23, 34, 47, 64, 84, 108, 136, 169, 206, 249, 297, 351, 411, 478, 551, 632, 720, 816, 920, 1033, 1154, 1285, 1425, 1575, 1735, 1906, 2087, 2280, 2484, 2700, 2928, 3169, 3422, 3689, 3969, 4263, 4571, 4894, 5231, 5584, 5952, 6336, 6736, 7153, 7586
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Bisection of A001400.
Molien series for 4-dimensional group of structure S_4 X C_2 and order 48, arising from complete weight enumerators of even trace-Hermitian self-dual additive codes over GF(4) containing the all-ones vector.
Partial sums of A156040. - Bob Selcoe, Feb 08 2014

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

Links

Crossrefs

Cf. A000631, A001400, A014125, A049347, A092498, A156040.

Programs

  • Maple
    with(combstruct): seq(count(Partition((2*n+4)), size=4), n=0..50); # Zerinvary Lajos, Mar 28 2008
  • Mathematica
    CoefficientList[Series[(1 + x^2) / ((1 - x)^2 (1 - x^2) (1 - x^3)), {x, 0, 100}], x] (* Vincenzo Librandi, Aug 15 2013 *)
LinearRecurrence[{2,0,-1,-1,0,2,-1},{1,2,5,9,15,23,34},50] (* Harvey P. Dale, Aug 31 2015 *)
  • PARI
    a(n)=(4*n^3+30*n^2+72*n+55+8*[1,-1,0][(n%3)+1]+9*(-1)^n)/72

Formula

G.f.: (1+x^2)/((1-x)^2*(1-x^2)*(1-x^3)). - James Sellers
a(n) = (1/72) * (4*n^3 + 30*n^2 + 72*n + 55 + 8*A049347(n) + 9*(-1)^n ). - Ralf Stephan, Aug 15 2013
E.g.f.: exp(-x)*(27 + 3*exp(2*x)*(55 + 106*x + 42*x^2 + 4*x^3) + 8*exp(x/2)*(3*cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2)))/216. - Stefano Spezia, Apr 05 2023

A156039 Number of compositions (ordered partitions) of n into 4 parts, where the first is at least as great as each of the others.

Original entry on oeis.org

1, 1, 4, 7, 11, 17, 26, 35, 48, 63, 81, 102, 127, 154, 187, 223, 263, 308, 359, 413, 474, 540, 612, 690, 775, 865, 964, 1069, 1181, 1301, 1430, 1565, 1710, 1863, 2025, 2196, 2377, 2566, 2767, 2977, 3197, 3428, 3671, 3923, 4188, 4464, 4752, 5052, 5365, 5689
Offset: 0

Views

Author

Jack W Grahl, Feb 02 2009, Feb 11 2009

Keywords

Comments

For n=1,2 these are just the tetrahedral numbers. a(n) is always at least 1/4 of the corresponding tetrahedral number, since each partition of this type gives up to four ordered partitions with the same cyclical order.
Diagonal sums of the irregular triangle A109439, for example a(0)=1, a(1)=1, a(2)=1+3, a(3)=1+3+3, a(4)=1+3+6+1. - Bob Selcoe, Feb 09 2014

Examples

			For n = 3 the a(3) = 7 compositions are: (3 0 0 0) (2 1 0 0) (2 0 1 0) (2 0 0 1) (1 1 1 0) (1 1 0 1) (1 0 1 1).

Links

Crossrefs

For partitions into 3 summands see A156040; also see A156041 and A156042.

Programs

  • Maple
    a:= proc(n) local m, r; m:= iquo(n, 12, 'r'); r:= r+1; (9 +(27 +72*m +18*r)*m +((9 +3*r) *r-12) /2)*m +[1, 1, 4, 7, 11, 17, 26, 35, 48, 63, 81, 102][r] end: seq(a(n), n=0..60); # Alois P. Heinz, Jun 14 2009
  • Mathematica
    LinearRecurrence[{2, -1, 1, -1, -1, 1, -1, 2, -1}, {1, 1, 4, 7, 11, 17, 26, 35, 48}, 60] (* Jean-François Alcover, May 17 2018 *)

Formula

G.f.: ( 1-x+3*x^2-x^3+x^4 ) / ( (1+x)*(1+x^2)*(1+x+x^2)*(x-1)^4 ). - Alois P. Heinz, Jun 14 2009

Extensions

More terms from Alois P. Heinz, Jun 14 2009
Showing 1-10 of 18 results. Next

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