A237827 Number of partitions of n such that 5*(least part) = greatest part.

0, 0, 0, 0, 0, 1, 1, 2, 3, 5, 7, 11, 13, 19, 24, 32, 39, 52, 61, 77, 93, 114, 133, 164, 188, 226, 261, 309, 353, 417, 471, 549, 622, 717, 808, 933, 1042, 1191, 1334, 1516, 1690, 1921, 2131, 2407, 2674, 3006, 3330, 3744, 4135, 4628, 5116, 5708, 6294, 7020

Offset: 1

Author

Clark Kimberling, Feb 16 2014

Keywords

• nonn
• easy

Examples

			a(8) = 2 counts these partitions:  521, 5111.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Alois P. Heinz)

Crossrefs

Cf. A000041, A117086, A237757, A237828, A237829.

Cf. A118096, A237825, A237826.

Programs

Mathematica

z = 64; q[n_] := q[n] = IntegerPartitions[n];
Table[Count[q[n], p_ /; 3 Min[p] = = Max[p]], {n, z}]     (* A237825*)
Table[Count[q[n], p_ /; 4 Min[p] = = Max[p]], {n, z}]     (* A237826 *)
Table[Count[q[n], p_ /; 5 Min[p] = = Max[p]], {n, z}]     (* A237827 *)
Table[Count[q[n], p_ /; 2 Min[p] + 1 = = Max[p]], {n, z}] (* A237828 *)
Table[Count[q[n], p_ /; 2 Min[p] - 1 = = Max[p]], {n, z}] (* A237829 *)
(* Second program: *)
kmax = 54;
Sum[x^(6 k)/Product[1 - x^j, {j, k, 5 k}], {k, 1, kmax}]/x + O[x]^kmax // CoefficientList[#, x]& (* Jean-François Alcover, May 30 2024, after Seiichi Manyama *)

PARI

my(N=60, x='x+O('x^N)); concat([0, 0, 0, 0, 0], Vec(sum(k=1, N, x^(6*k)/prod(j=k, 5*k, 1-x^j)))) \\ Seiichi Manyama, May 14 2023

Formula

G.f.: Sum_{k>=1} x^(6*k)/Product_{j=k..5*k} (1-x^j). - Seiichi Manyama, May 14 2023

a(n) ~ c * d^sqrt(n) / sqrt(n), where d = 5.4930955... and c = 0.135509... - Vaclav Kotesovec, Jun 19 2025