A334541 a(n) is the number of partitions of n into consecutive parts that differ by 5.

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

Offset: 1

Author

Omar E. Pol, May 05 2020

Keywords

• nonn
• easy

Comments

Note that all sequences of this family as A000005, A001227, A038548, A117277, A334461, etc. could be prepended with a(0) = 1 when they are interpreted as sequences of number of partitions, since A000041(0) = 1. However here a(0) is omitted in accordance with the mentioned members of the same family.

For the relation to heptagonal numbers see also A334465.

Examples

			For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4], so a(27) = 3.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Row sums of A334465.

Column k=5 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), A334461 (k=4), this sequence (k=5).

Cf. A000041, A000566, A303300.

Programs

Mathematica

first[n_] := Module[{res = Array[1&, n]}, For[i = 2, True, i++, start = i + 5 Binomial[i, 2]; If[start > n, Return[res]]; For[j = start, j <= n, j += i, res[[j]]++]]];
first[105] (* Jean-François Alcover, Nov 30 2020, after David A. Corneth *)
Table[Sum[If[n > 5*k*(k-1)/2 && IntegerQ[n/k - 5*(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, 5) \\ Joerg Arndt, May 06 2020

PARI

first(n) = { my(res = vector(n, i, 1)); for(i = 2, oo, start = i + 5 * binomial(i, 2); if(start > n, return(res)); forstep(j = start, n, i, res[j]++ ) ); } \\ David A. Corneth, May 17 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