A334461 a(n) is the number of partitions of n into consecutive parts that differ by 4.

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

Offset: 1

Author

Omar E. Pol, May 01 2020

Keywords

• nonn

Examples

			For n = 28 there are three partitions of 28 into consecutive parts that differ by 4, including 28 as a valid partition. They are [28], [16, 12] and [13, 9, 5, 1]. So a(28) = 3.
		

Links

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

Crossrefs

Row sums of A334460.

Column k=4 of A323345.

Sequences of this family whose consecutive parts differ by k are A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), this sequence (k=4), A334541 (k=5), A334948 (k=6).

Programs

Mathematica

nmax = 105;
col[k_] := col[k] = CoefficientList[Sum[x^(n(k n - k + 2)/2 - 1)/(1 - x^n), {n, 1, nmax}] + O[x]^nmax, x];
a[n_] := col[4][[n]];
Array[a, nmax] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 2*k*(k-1), 1, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 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, 4) \\ Joerg Arndt, May 05 2020

Formula

The g.f. for "consecutive parts that differ by d" is Sum_{k>=1} x^(k*(d*k-d+2)/2) / (1-x^k); cf. A117277. - Joerg Arndt, Nov 30 2020