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

Author

Seiichi Manyama, Jan 25 2022

Keywords

• nonn

Comments

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

Examples

			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].
		

Links

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

Crossrefs

Column 5 of A350879.

Cf. A350897, A350899.

Programs

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 *)

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)))))

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