Alois P. Heinz and Vaclav Kotesovec, <a href="/A238608/b238608_1.txt">Table of n, a(n) for n = 0..122</a> (terms 0..70 from Alois P. Heinz)
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Alois P. Heinz, and Vaclav Kotesovec, <a href="/A238608/b238608_1.txt">Table of n, a(n) for n = 0..122</a> (terms 0..70</a> from Alois P. Heinz)
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_
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a[n_] := SeriesCoefficient[1/QPochhammer[q, q, n], {q, 0, n^3}]; Table[ a[n], {n, 0, 20}] (* Jean-François Alcover, Dec 03 2015 *)
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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
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