A350897 Number of partitions of n such that 5*(smallest part) = (number of parts).

0, 0, 0, 0, 1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 55, 65, 74, 87, 99, 115, 131, 151, 172, 199, 226, 260, 298, 343, 393, 454, 522, 603, 696, 804, 929, 1076, 1243, 1438, 1664, 1924, 2222, 2567, 2961, 3413, 3931, 4520, 5193, 5959, 6827, 7811, 8928, 10186, 11607, 13208, 15008, 17028, 19297

Offset: 1

Author

Seiichi Manyama, Jan 21 2022

Keywords

• nonn

Comments

In general, for m >= 1, if g.f.= Sum_{k>=1} x^(m*k^2)/Product_{j=1..m*k-1} (1-x^j), then a(n) ~ r^2 * (m*log(r)^2 + polylog(2, r^2))^(1/4) * exp(2*sqrt((m*log(r)^2 + polylog(2, r^2))*n)) / (2*sqrt(Pi*m*(m - (m-2)*r^2)) * n^(3/4)), where r is the positive real root of the equation r^2 = 1 - r^m. - Vaclav Kotesovec, Oct 14 2024

Links

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

Crossrefs

Column 5 of A350889.

Cf. A168657.

Programs

Mathematica

CoefficientList[Series[Sum[x^(5k^2)/Product[1-x^j,{j,5k-1}],{k,62}],{x,0,62}],x] (* Stefano Spezia, Jan 22 2022 *)

PARI

my(N=66, x='x+O('x^N)); concat([0, 0, 0, 0], Vec(sum(k=1, sqrtint(N\5), x^(5*k^2)/prod(j=1, 5*k-1, 1-x^j))))

Formula

G.f.: Sum_{k>=1} x^(5*k^2)/Product_{j=1..5*k-1} (1-x^j).

a(n) ~ c * exp(Pi*sqrt(r*n)) / n^(3/4), where r = 0.42067169741517... and c = 0.04778365700734... - Vaclav Kotesovec, Jan 26 2022

a(n) ~ r^2 * (5*log(r)^2 + polylog(2, r^2))^(1/4) * exp(2*sqrt((5*log(r)^2 + polylog(2, r^2))*n)) / (2*sqrt(5*Pi*(5 - 3*r^2)) * n^(3/4)), where r = 0.808730600479392... is the real root of the equation r^2 = 1 - r^5. - Vaclav Kotesovec, Oct 14 2024