A363066 Number of partitions p of n such that (1/3)*max(p) is a part of p.

1, 0, 0, 0, 1, 1, 2, 3, 5, 6, 9, 11, 16, 20, 27, 33, 45, 55, 72, 89, 116, 142, 181, 222, 281, 343, 429, 522, 649, 786, 967, 1168, 1429, 1719, 2088, 2504, 3026, 3615, 4345, 5174, 6192, 7349, 8755, 10360, 12297, 14507, 17154, 20182, 23788, 27910, 32790, 38374, 44955, 52480, 61307, 71402

Offset: 0

Author

Seiichi Manyama, May 16 2023

Keywords

• nonn

Examples

			a(7) = 3 counts these partitions:  331, 3211, 31111.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Seiichi Manyama)

Crossrefs

Cf. A002865, A238479, A363067, A363068.

Cf. A008484, A237825, A363045.

Cf. A035295.

Programs

Mathematica

nmax = 60; CoefficientList[Series[Sum[x^(4*k)/Product[1 - x^j, {j, 1, 3*k}], {k, 0, nmax}], {x, 0, nmax}], x]  (* Vaclav Kotesovec, Jun 18 2025 *)
nmax = 60; p=1; s=1; Do[p=Expand[p*(1-x^(3*k))*(1-x^(3*k-1))*(1-x^(3*k-2))]; p=Take[p, Min[nmax+1, Exponent[p, x]+1, Length[p]]]; s+=x^(4*k)/p; , {k, 1, nmax}]; CoefficientList[Series[s, {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 18 2025 *)
Join[{1},Table[Count[IntegerPartitions[n],?(MemberQ[#,#[[1]]/3]&)],{n,60}]] (* _Harvey P. Dale, Jun 29 2025 *)

PARI

a(n) = sum(k=0, n\4, #partitions(n-4*k, 3*k));

Formula

G.f.: Sum_{k>=0} x^(4*k)/Product_{j=1..3*k} (1-x^j).

a(n) ~ Gamma(1/3) * Pi^(1/3) * exp(Pi*sqrt(2*n/3)) / (2^(13/6) * 3^(8/3) * n^(7/6)). - Vaclav Kotesovec, Jun 19 2025