A334949 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 6.

1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 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, 4, 3, 1, 10, 1, 3, 4, 7, 6, 6, 1, 7, 4, 8, 1, 10, 1, 3, 9, 7, 1, 6, 1, 12, 4, 3, 1, 10, 6, 3, 4, 7, 1, 11, 1, 7, 4

Offset

1,8

Comments

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

For the relation to the octagonal numbers see also A334947.

Links

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

Formula

G.f.: Sum_{n>=1} n*x^(n*(3*n-2))/(1-x^n). (For proof, see A330889. - Omar E. Pol, Nov 22 2020)

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

Example

For n = 24 there are three partitions of 24 into consecutive parts that differ by 6, including 24 as a valid partition. They are [24], [15, 9] and [14, 8, 2]. There are 1, 2 and 3 parts respectively, hence the total number of parts is 1 + 2 + 3 = 6, so a(24) = 6.

Mathematica

nmax = 100;
CoefficientList[Sum[n x^(n(3n-2)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 3*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*(3*k-2))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

Crossrefs

Row sums of A334947.

Column k=6 of A334466.

Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), A334732 (k=5), this sequence (k=6).

Cf. A000567, A334946, A334947, A334948, A334953.

Sequence in context: A125768 A377301 A377300 * A334732 A266875 A307193

Adjacent sequences: A334946 A334947 A334948 * A334950 A334951 A334952

Keyword

nonn

Author

Omar E. Pol, May 27 2020

Status

approved