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%I #7 Jan 29 2018 21:12:15
%S 1,1,0,6,6,20,120,7,1689,6636,36540,64020,963996,2894892,19555965,
%T 176079995,955611188,6684303780,42462792168,292378003753,
%U 1886275214112,13384059605364,87399249887334,624073002367892,5080120229014734,37587589611771480
%N Number of ordered ways of writing n^2 as a sum of n nonnegative cubes.
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%F a(n) = [x^(n^2)] (Sum_{k>=0} x^(k^3))^n.
%e a(3) = 6 because we have [8, 1, 0], [8, 0, 1], [1, 8, 0], [1, 0, 8], [0, 8, 1] and [0, 1, 8].
%p f:= n -> coeff(add(x^(k^3),k=0..floor(n^(2/3)))^n,x,n^2):
%p map(f, [$0..30]); # _Robert Israel_, Jan 29 2018
%t Table[SeriesCoefficient[Sum[x^k^3, {k, 0, Floor[n^(2/3) + 1]}]^n, {x, 0, n^2}], {n, 0, 25}]
%Y Cf. A000290, A000578, A023358, A030272, A218495, A259792, A291700, A298671, A298672, A298934, A298937.
%K nonn
%O 0,4
%A _Ilya Gutkovskiy_, Jan 29 2018
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