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%I #31 Jan 26 2022 08:13:17
%S 1,1,2,12,20,55,294,497,1224,2520,14410,21912,54300,104286,220710,
%T 1105215,1697552,3839382,7356762,14873580,26275620,132112596,
%U 188666126,423247104,772560600,1535398150,2632049290,4975242048,21273166572,30649985160,64824339630,116604788800,223181224992
%N a(n) = [x^n] Product_{k>=1} (1 + n*x^k).
%C The number of partitions of n into distinct parts where each part can be colored in n different ways. For example, there are 4 partitions of 6 into distinct parts, namely 6, 5 + 1, 4 + 2 and 3 + 2 + 1; allowing for the colorings gives a(6) = 6 + 6*6 + 6*6 + 6*6*6 = 294. - _Peter Bala_, Aug 31 2017
%H Vaclav Kotesovec, <a href="/A291698/b291698.txt">Table of n, a(n) for n = 0..10000</a> (terms 0..2000 from Robert Israel)
%F a(n) = A286957(n,n).
%F a(n) == 0 (mod n); a(n) == n (mod n^2). - _Peter Bala_, Aug 31 2017
%F Conjecture: a(n) ~ exp(sqrt(2*(log(n)^2 + Pi^2/3)*n)) * (log(n)^2 + Pi^2/3)^(1/4) / (sqrt(Pi) * (2*n)^(5/4)). - _Vaclav Kotesovec_, Sep 15 2017
%p seq(coeff(mul(1+n*x^k,k=1..n),x,n),n=0..50); # _Robert Israel_, Aug 30 2017
%t Table[SeriesCoefficient[Product[1 + n x^k, {k, 1, n}], {x, 0, n}], {n, 0, 32}]
%t Table[SeriesCoefficient[QPochhammer[-n, x]/(1 + n), {x, 0, n}], {n, 0, 32}]
%Y Main diagonal of A286957.
%Y Cf. A022629, A124577, A292190, A292304.
%K nonn,nice
%O 0,3
%A _Ilya Gutkovskiy_, Aug 30 2017
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