A117277 Number of partitions of n whose consecutive parts differ by 3.

1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 3, 2, 1, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 2, 1, 4, 2, 2, 2, 2, 3, 4, 1, 2, 3, 3, 1, 4, 2, 2, 3, 2, 2, 4, 1, 3, 3, 2, 1, 4, 4, 2, 2, 2, 2, 5, 1, 3, 3, 2, 2, 4, 2, 2, 3, 3, 2, 4, 1, 2, 4, 3, 2, 4, 2, 3, 2, 2, 3, 4, 3, 2, 3, 2, 1, 6

Offset: 1

Author

Emeric Deutsch, Mar 07 2006

Keywords

• nonn

Comments

Also number of partitions of n such that if k is the largest part, then each of the parts 1,2,...,k-1 occurs exactly 3 times. Example: a(15)=3 because we have [3,3,2,2,2,1,1,1],[2,2,2,2,2,2,1,1,1] and [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1].

Row sums of A330887. - Omar E. Pol, May 07 2020

Column 3 of A323345. - Omar E. Pol, Dec 03 2020

Examples

			a(15) = 3 because we have [15], [9,6] and [8,5,2].
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A001227, A038548, A323345, A330887, A338731.

Programs

Maple

g:=sum(x^((3*k^2-k)/2)/(1-x^k),k=1..10): gser:=series(g,x=0,140): seq(coeff(gser,x^n),n=1..135);

Mathematica

Table[Sum[If[n > 3*k*(k-1)/2 && IntegerQ[n/k - 3*(k-1)/2], 1, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)

PARI

seq(N,d)=my(x='x+O('x^N));Vec(sum(k=1,N,x^(k*(d*k-d+2)/2)/(1-x^k)));
seq(100,3) \\ Joerg Arndt, May 05 2020

Formula

G.f.: Sum_{k>=1} x^((3*k^2-k)/2)/(1-x^k). In general, the generating function for the number of partitions in which consecutive parts differ by d is Sum_{k>=1} x^(k*(d*k-d+2)/2)/(1-x^k). For d=0, 1 and 2 one obtains A000005, A001227 and A038548, respectively.