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%I #17 Nov 07 2020 14:15:16
%S 1,1,0,0,0,0,1,0,2,6,14,23,51,108,228,511,1158,2500,5603,12304,26969,
%T 59222,130115,285370,624965,1368603,2987117,6517822,14187920,30823278,
%U 66834822,144671698,312551894,673913968,1450292087,3114720013,6676277754,14281662079
%N Number of partitions of n^3 into exactly n positive cubes.
%H Vaclav Kotesovec, <a href="/A307643/b307643.txt">Table of n, a(n) for n = 0..79</a>
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%F a(n) = A320841(n^3,n).
%e 9^3 =
%e 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 =
%e 1^3 + 1^3 + 2^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 =
%e 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3 =
%e 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 =
%e 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 =
%e 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 7^3,
%e so a(9) = 6.
%p b:= proc(n, i, t) option remember; `if`(n=0, `if`(t=0, 1, 0),
%p `if`(i<1 or t<1, 0, b(n, i-1, t)+
%p `if`(i^3>n, 0, b(n-i^3, i, t-1))))
%p end:
%p a:= n-> b(n^3, n$2):
%p seq(a(n), n=0..25); # _Alois P. Heinz_, Oct 12 2019
%t b[n_, i_, t_] := b[n, i, t] = If[n == 0, If[t == 0, 1, 0], If[i < 1 || t < 1, 0, b[n, i - 1, t] + If[i^3 > n, 0, b[n - i^3, i, t - 1]]]];
%t a[n_] := b[n^3, n, n];
%t a /@ Range[0, 25] (* _Jean-François Alcover_, Nov 07 2020, after _Alois P. Heinz_ *)
%Y Cf. A000578, A030272, A259792, A298671, A298672, A307644, A319435, A320841.
%K nonn
%O 0,9
%A _Ilya Gutkovskiy_, Apr 19 2019
%E More terms from _Vaclav Kotesovec_, Apr 20 2019
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