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 A267709 Number of partitions of pentagonal numbers. 1

%I

%S 1,1,7,77,1002,14883,239943,4087968,72533807,1327710076,24908858009,

%T 476715857290,9275102575355,182973889854026,3652430836071053,

%U 73653287861850339,1498478743590581081,30724985147095051099,634350763653787028583,13177726323474524612308

%N Number of partitions of pentagonal numbers.

%H Vaclav Kotesovec, <a href="/A267709/b267709.txt">Table of n, a(n) for n = 0..730</a> (terms 0..90 from Ilya Gutkovskiy)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Partition.html">Partition</a>, <a href="http://mathworld.wolfram.com/PartitionFunctionP.html">Partition Function P</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PentagonalNumber.html">Pentagonal Number</a>

%H <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>

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

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

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

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

%o (PARI) a(n)=numbpart(n*(3*n-1)/2) \\ _Charles R Greathouse IV_, Jul 26 2016

%o (Python)

%o from sympy.ntheory import npartitions

%o print [npartitions(n*(3*n - 1)/2) for n in range(51)] # _Indranil Ghosh_, Apr 11 2017

%Y Cf. A000041, A000326, A066655, A072213.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Apr 07 2016

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Last modified August 6 14:51 EDT 2020. Contains 336247 sequences. (Running on oeis4.)