A334464 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 4.

1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 1, 6, 1, 7, 9, 3, 1, 10, 1, 8, 4, 7, 1, 6, 6, 7, 4, 3, 1, 15, 1, 3, 4, 7, 6, 12, 1, 7, 4, 8, 1, 16, 1, 3, 9, 7, 1, 12, 1, 12, 4, 3, 1, 16, 6, 3, 4, 7, 1, 17, 8, 7, 4

Offset: 1

Author

Omar E. Pol, May 05 2020

Keywords

• nonn

Comments

The one-part partition n = n is included in the count.

For the relation to hexagonal numbers see also A334462.

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]. The number of parts of these partitions are 1, 2, 4 respectively. The total number of parts is 1 + 2 + 4 = 7, so a(28) = 7.
		

Links

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

Crossrefs

Row sums of A334462.

Column k=4 of A334466.

Cf. A000384.

Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), this sequence (k=4), A334732 (k=5), A334949 (k=6).

Cf. A334461.

Programs

Mathematica

nmax = 100;
CoefficientList[Sum[n x^(n(2n-1)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 2*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)

PARI

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(2*k-1))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

Formula

G.f.: Sum_{n>=1} n*x^(n*(2*n-1))/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)

Sum_{k=1..n} a(k) ~ sqrt(2) * n^(3/2) / 3. - Vaclav Kotesovec, Oct 23 2024