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%I #19 Aug 12 2017 09:30:40
%S 1,1,2,120,290250,107320441096,21715974961480054078,
%T 8487986089807555456140271121440,
%U 22615863021403796876556069287242400147213424924,1449638083412288206280215383952017948209203861522683138464747658192
%N Number of compositions (ordered partitions) of n^3 into cubes.
%H Alois P. Heinz, <a href="/A290247/b290247.txt">Table of n, a(n) for n = 0..22</a>
%H <a href="/index/Com#comp">Index entries for sequences related to compositions</a>
%F a(n) = [x^(n^3)] 1/(1 - Sum_{k>=1} x^(k^3)).
%F a(n) = A023358(A000578(n)).
%e a(2) = 2 because 2^3 = 8 and we have [8], [1, 1, 1, 1, 1, 1, 1, 1].
%p b:= proc(n) option remember; local i; if n=0 then 1
%p else 0; for i while i^3<=n do %+b(n-i^3) od fi
%p end:
%p a:= n-> b(n^3):
%p seq(a(n), n=0..10); # _Alois P. Heinz_, Aug 12 2017
%t Table[SeriesCoefficient[1/(1 - Sum[x^k^3, {k, 1, n}]), {x, 0, n^3}], {n, 0, 9}]
%Y Cf. A000578, A023358, A030272, A224366, A259792.
%K nonn
%O 0,3
%A _Ilya Gutkovskiy_, Jul 24 2017
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