A267709 Number of partitions of pentagonal numbers.

1, 1, 7, 77, 1002, 14883, 239943, 4087968, 72533807, 1327710076, 24908858009, 476715857290, 9275102575355, 182973889854026, 3652430836071053, 73653287861850339, 1498478743590581081, 30724985147095051099, 634350763653787028583, 13177726323474524612308

Offset: 0

Author

Ilya Gutkovskiy, Apr 07 2016

Keywords

• nonn

Examples

			a(2) = 7, because second pentagonal number is a 5 and 5 can be partitioned in 7 distinct ways: 5, 4 + 1, 3 + 2, 3 + 1 + 1, 3 + 2 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..730 (terms 0..90 from Ilya Gutkovskiy)
• Eric Weisstein's World of Mathematics, Partition, Partition Function P
• Eric Weisstein's World of Mathematics, Pentagonal Number
• Index entries for related partition-counting sequences

Crossrefs

Cf. A000041, A000326, A066655, A072213.

Programs

Mathematica

Table[PartitionsP[n ((3 n - 1)/2)], {n, 0, 19}]

PARI

a(n)=numbpart(n*(3*n-1)/2) \\ Charles R Greathouse IV, Jul 26 2016

Python

from sympy.ntheory import npartitions
print([npartitions(n*(3*n - 1)//2) for n in range(51)]) # Indranil Ghosh, Apr 11 2017

Formula

a(n) = A000041(A000326(n)).

a(n) ~ exp((Pi*sqrt(n*(3*n - 1)))/sqrt(3))/(2*sqrt(3)*n*(3*n - 1)).

a(n) = [x^(n*(3*n-1)/2)] Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Apr 11 2017