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%I #12 Nov 15 2018 15:53:05
%S 0,1,17,3781,14942231,1264608203048,2555847904495965819,
%T 132574244496779071303074376,185560862224740635595130202984468935,
%U 7271076505438083132065943012753686950455454055,8205115354631567886718289443554629632451344416164686337673
%N Number of partitions of 6^n into parts that are at most n with at least one part of each size.
%H Alois P. Heinz, <a href="/A239165/b239165.txt">Table of n, a(n) for n = 0..37</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])
%F a(n) = [x^(6^n-n*(n+1)/2)] Product_{j=1..n} 1/(1-x^j).
%F a(n) ~ 6^(n*(n-1)) / (n!*(n-1)!). - _Vaclav Kotesovec_, Jun 05 2015
%t maxExponent = 50; 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[6^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=6 of A238012.
%K nonn
%O 0,3
%A _Alois P. Heinz_, Mar 11 2014
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