|
%I #14 Nov 15 2018 11:10:41
%S 0,1,12,1240,1655004,32796849930,10743023668660275,
%T 62590747974586286694030,6826987264035710020018176749475,
%U 14471606032117455546329821353159274382372,613427607589897771307393494301176209875530879140211
%N Number of partitions of 5^n into parts that are at most n with at least one part of each size.
%H Alois P. Heinz, <a href="/A239164/b239164.txt">Table of n, a(n) for n = 0..40</a>
%H A. V. Sills and D. Zeilberger, <a href="https://arxiv.org/abs/1108.4391">Formulae for the number of partitions of n into at most m parts (using the quasi-polynomial ansatz)</a>, arXiv:1108.4391 [math.CO], 2011.
%F a(n) = [x^(5^n-n*(n+1)/2)] Product_{j=1..n} 1/(1-x^j).
%F a(n) ~ 5^(n*(n-1)) / (n!*(n-1)!). - _Vaclav Kotesovec_, Jun 05 2015
%e a(2) = 12: 2222222222221, 22222222222111, 222222222211111, 2222222221111111, 22222222111111111, 222222211111111111, 2222221111111111111, 22222111111111111111, 222211111111111111111, 2221111111111111111111, 22111111111111111111111, 211111111111111111111111.
%t maxExponent = 45; a[0] = 0; a[1] = 1;
%t a[n_] := Module[{}, aparts = List @@ (Product[1/(1 - x^j), {j, 1, n}] // Apart); cc = aparts + O[x]^maxExponent // CoefficientList[#, x]&; f[k_] = Total[FindSequenceFunction[#, k]& /@ cc];f[5^n - n(n+1)/2 + 1] // Round];
%t Table[an = a[n]; Print[n, " ", an]; an, {n, 0, 10}] (* _Jean-François Alcover_, Nov 15 2018 *)
%Y Column k=5 of A238012.
%K nonn
%O 0,3
%A _Alois P. Heinz_, Mar 11 2014
|
