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

1,5

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

Links

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

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.

Example

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

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);

Prog

(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

Crossrefs

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

Sequence in context: A087976 A227903 A353567 * A033831 A338650 A033105

Adjacent sequences: A117274 A117275 A117276 * A117278 A117279 A117280

Keyword

nonn

Author

Emeric Deutsch, Mar 07 2006

Status

approved