A348164 Number of partitions of n such that 5*(greatest part) = (number of parts).

0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 3, 3, 3, 3, 5, 5, 7, 8, 10, 11, 15, 16, 20, 22, 26, 28, 35, 38, 46, 52, 62, 70, 85, 95, 112, 127, 148, 166, 195, 219, 254, 288, 332, 375, 435, 489, 562, 635, 726, 817, 936, 1051, 1198, 1348, 1531, 1721, 1957, 2196, 2489

Offset

1,17

Comments

Also, the number of partitions of n such that (greatest part) = 5*(number of parts).

Links

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

Formula

G.f.: Sum_{k>=1} x^(6*k-1) * Product_{j=1..k-1} (1-x^(5*k+j-1))/(1-x^j).

a(n) ~ 5 * Pi^5 * exp(Pi*sqrt(2*n/3)) / (9 * 2^(3/2) * n^(7/2)). - Vaclav Kotesovec, Oct 17 2024

Example

a(19) = 3 counts these partitions:
[3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[3, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1],
[2, 2, 2, 2, 2, 2, 2, 2, 2, 1].

Mathematica

nmax = 100; Rest[CoefficientList[Series[Sum[x^(6*k-1) * Product[(1 - x^(5*k+j-1)) / (1 - x^j), {j, 1, k-1}], {k, 1, nmax/6 + 1}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Oct 15 2024 *)
nmax = 100; p = x^4; s = x^4; Do[p = Normal[Series[p*x^6*(1 - x^(6*k - 1))*(1 - x^(6*k))*(1 - x^(6*k + 1))*(1 - x^(6*k + 2))*(1 - x^(6*k + 3))*(1 - x^(6*k + 4))/((1 - x^(5*k + 4))*(1 - x^(5*k + 3))*(1 - x^(5*k + 2))*(1 - x^(5*k + 1))*(1 - x^(5*k))*(1 - x^k)), {x, 0, nmax}]]; s += p; , {k, 1, nmax/6 + 1}]; Take[CoefficientList[s, x], nmax] (* Vaclav Kotesovec, Oct 16 2024 *)

Prog

(PARI) my(N=66, x='x+O('x^N)); concat([0, 0, 0, 0], Vec(sum(k=1, N, x^(6*k-1)*prod(j=1, k-1, (1-x^(5*k+j-1))/(1-x^j)))))

Crossrefs

Column 5 of A350879.

Cf. A350897, A350899.

Sequence in context: A055377 A157524 A239496 * A299962 A340010 A128586

Adjacent sequences: A348161 A348162 A348163 * A348165 A348166 A348167

Keyword

nonn

Author

Seiichi Manyama, Jan 25 2022

Status

approved