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Number of partitions of n^n into parts that are at most n with at least one part of each size.
5

%I #20 Nov 16 2018 02:33:53

%S 0,1,1,48,109809,32796849930,2555847904495965819,

%T 85962759806610904434664386174,

%U 1841132100297745277187328924904656111127054,34687813181057391872792859998288408847592250236051615502024

%N Number of partitions of n^n into parts that are at most n with at least one part of each size.

%H Alois P. Heinz, <a href="/A238001/b238001.txt">Table of n, a(n) for n = 0..27</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^(n^n-n*(n+1)/2)] Product_{j=1..n} 1/(1-x^j).

%F a(n) ~ n^(n*(n-1)) / (n!*(n-1)!) ~ exp(2*n) * n^(n*(n-3)) / (2*Pi). - _Vaclav Kotesovec_, Jun 05 2015

%e a(1) = 1: 1.

%e a(2) = 1: 211.

%e a(3) = 48: 3333333321, ..., 321111111111111111111111.

%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[n^n-n(n+1)/2 + 1] // Round];

%t Table[an = a[n]; Print[n, " ", an]; an, {n, 0, 9}] (* _Jean-François Alcover_, Nov 15 2018 *)

%Y Main diagonal of A238012.

%Y Cf. A236810, A237512, A237998, A237999, A238000.

%K nonn

%O 0,4

%A _Alois P. Heinz_, Feb 16 2014