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

1,6

Links

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

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

Example

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.

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 *)

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, 4) \\ Joerg Arndt, May 05 2020

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).

Sequence in context: A086435 A266226 A099305 * A338652 A033109 A321469

Adjacent sequences: A334458 A334459 A334460 * A334462 A334463 A334464

Keyword

nonn

Author

Omar E. Pol, May 01 2020

Status

approved