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A267709 Number of partitions of pentagonal numbers.

%I A267709 #54 Feb 16 2025 08:33:29
%S A267709 1,1,7,77,1002,14883,239943,4087968,72533807,1327710076,24908858009,
%T A267709 476715857290,9275102575355,182973889854026,3652430836071053,
%U A267709 73653287861850339,1498478743590581081,30724985147095051099,634350763653787028583,13177726323474524612308
%N A267709 Number of partitions of pentagonal numbers.
%H A267709 Vaclav Kotesovec, <a href="/A267709/b267709.txt">Table of n, a(n) for n = 0..730</a> (terms 0..90 from Ilya Gutkovskiy)
%H A267709 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/Partition.html">Partition</a>, <a href="https://mathworld.wolfram.com/PartitionFunctionP.html">Partition Function P</a>
%H A267709 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/PentagonalNumber.html">Pentagonal Number</a>
%H A267709 <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>
%F A267709 a(n) = A000041(A000326(n)).
%F A267709 a(n) ~ exp((Pi*sqrt(n*(3*n - 1)))/sqrt(3))/(2*sqrt(3)*n*(3*n - 1)).
%F A267709 a(n) = [x^(n*(3*n-1)/2)] Product_{k>=1} 1/(1 - x^k). - _Ilya Gutkovskiy_, Apr 11 2017
%e A267709 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.
%t A267709 Table[PartitionsP[n ((3 n - 1)/2)], {n, 0, 19}]
%o A267709 (PARI) a(n)=numbpart(n*(3*n-1)/2) \\ _Charles R Greathouse IV_, Jul 26 2016
%o A267709 (Python)
%o A267709 from sympy.ntheory import npartitions
%o A267709 print([npartitions(n*(3*n - 1)//2) for n in range(51)]) # _Indranil Ghosh_, Apr 11 2017
%Y A267709 Cf.  A000041, A000326, A066655, A072213.
%K A267709 nonn
%O A267709 0,3
%A A267709 _Ilya Gutkovskiy_, Apr 07 2016