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Number of partitions of n^3 into parts that are at most n.
13

%I #21 Sep 07 2016 08:04:17

%S 1,1,5,75,2280,106852,6889527,569704489,57733506640,6944433285769,

%T 968356321790171,153738253618009045,27396489338187214000,

%U 5417302365503826145732,1177436831956414016252071,279074576444362385794783853,71649589941044468875380333533

%N Number of partitions of n^3 into parts that are at most n.

%C In general, "number of partitions of j*n^3 into parts that are at most n" is (for j>0) asymptotic to exp(2*n + 1/(4*j)) * n^(n-3) * j^(n-1) / (2*Pi). - _Vaclav Kotesovec_, May 25 2015

%H Alois P. Heinz and Vaclav Kotesovec, <a href="/A238608/b238608.txt">Table of n, a(n) for n = 0..122</a> (terms 0..70 from Alois P. Heinz)

%F a(n) = [x^(n^3)] Product_{j=1..n} 1/(1-x^j).

%F a(n) ~ exp(2*n + 1/4) * n^(n-3) / (2*Pi). - _Vaclav Kotesovec_, May 25 2015

%p T:=proc(n,k) option remember; `if`(n=0 or k=1, 1, T(n,k-1) + `if`(n<k, 0, T(n-k,k))) end proc: seq(T(n^3, n), n=0..20); # _Vaclav Kotesovec_, May 25 2015 after _Alois P. Heinz_

%t a[n_] := SeriesCoefficient[1/QPochhammer[q, q, n], {q, 0, n^3}]; Table[ a[n], {n, 0, 20}] (* _Jean-François Alcover_, Dec 03 2015 *)

%Y Column k=3 of A238016.

%Y Cf. A258302 (j=2), A258303 (j=3), A258304 (j=4), A258305 (j=5).

%K nonn

%O 0,3

%A _Alois P. Heinz_, Mar 01 2014