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%I #29 Jan 07 2020 08:31:39
%S 1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,
%T 0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,
%U 0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1
%N Expansion of Product_{k>=2} (1 + x^(k^3)).
%C Number of partitions of n into distinct cubes > 1.
%H Alois P. Heinz, <a href="/A280130/b280130.txt">Table of n, a(n) for n = 0..100000</a> (first 10001 terms from Antti Karttunen)
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%H <a href="/index/Par#partN">Index entries for related partition-counting sequences</a>
%F G.f.: Product_{k>=2} (1 + x^(k^3)).
%F From _Vaclav Kotesovec_, Dec 26 2016: (Start)
%F a(n) = Sum_{k=0..n} (-1)^(n-k) * A279329(k).
%F a(n) + a(n-1) = A279329(n).
%F a(n) ~ A279329(n)/2.
%F (End)
%e a(35) = 1 because 35 = 27 + 8. This is the first nonzero value for a noncube index.
%e From _Antti Karttunen_, Aug 30 2017: (Start)
%e a(72) = 1 because there is just one solution: 72 = 4^3 + 2^3.
%e a(216) = 2 because there are two solutions: 216 = 6^3 = 5^3 + 4^3 + 3^3. This is the first index where a(n) > 1. (End)
%t nmax = 130; CoefficientList[Series[Product[1 + x^k^3, {k, 2, nmax}], {x, 0, nmax}], x]
%o (PARI) A280130(n,m=2) = { my(s=0); if(!n,1,for(c=m,n,if(ispower(c,3), s+=A280130(n-c,c+1))); (s)); }; \\ _Antti Karttunen_, Aug 30 2017
%o (PARI)
%o A280130(n,m=2)={if(n, sum(c=m, sqrtnint(n,3), A280130(n-c^3, c+1)), 1)} \\ At n ~ 2500 this is about 100 times faster than code from 2017, but for larger n (needed for A030272(n)=a(n^3)) better use (with parisize (or allocmem) >= 201*Nmax):
%o V280130=Vecsmall(prod(k=2, (Nmax=3*10^4)^(1/3), 1+x^k^3+O(x^Nmax))); A280130(n)=V280130[n+1] \\ _M. F. Hasler_, Jan 05 2020
%Y Cf. A003108, A078128, A279329.
%K nonn
%O 0
%A _Ilya Gutkovskiy_, Dec 26 2016
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