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

1,6

Comments

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

For the relation to hexagonal numbers see also A334462.

Links

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

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

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]. 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.

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 *)

Prog

(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

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.

Sequence in context: A307193 A111742 A178220 * A316780 A338730 A104740

Adjacent sequences: A334461 A334462 A334463 * A334465 A334466 A334467

Keyword

nonn

Author

Omar E. Pol, May 05 2020

Status

approved