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%I #15 Mar 01 2019 15:48:24
%S 0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,
%T 1,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,1,0,2,0,0,1,0,
%U 0,1,0,2,0,0,1,0,0,1,0,2,1,0,1,0,0,1,0,2,1,0,2,0,0,1,0,2,1,0,2,0,0,1,0,2,1
%N Number of ways to write n as sum of cubes>1.
%C a(A078129(n))=0; a(A078130(n))=1; a(A078131(n))>0;
%C Conjecture (lower bound): for all k exists b(k) such that a(n)>k for n>b(k); see b(0)=A078129(83)=154 and b(1)=A078130(63)=218.
%H Vaclav Kotesovec, <a href="/A078128/b078128.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/CubicNumber.html">Cubic Number</a>.
%F a(n) = 1/n*Sum_{k=1..n} (b(k)-1)*a(n-k), a(0) = 1, where b(k) is sum of cube divisors of k. - _Vladeta Jovovic_, Nov 20 2002
%F From _Vaclav Kotesovec_, Jan 05 2017: (Start)
%F a(n) = A003108(n) - A003108(n-1).
%F a(n) ~ exp(4*(Gamma(1/3) * Zeta(4/3))^(3/4) * n^(1/4) / 3^(3/2)) * (Gamma(1/3) * Zeta(4/3))^(3/2) / (8 * 3^(5/2) * Pi^2 * n^2).
%F (End)
%e a(160)=4: 160 = 20*2^3 = 4^3+12*2^3 = 2*4^3+4*2^3 = 5^3+3^3+2^3.
%t nmax = 105; CoefficientList[Series[Product[1/(1 - x^(k^3)), {k, 1, nmax}], {x, 0, nmax}], x] // Differences (* _Jean-François Alcover_, Mar 01 2019, after _Vaclav Kotesovec_ *)
%Y Cf. A000578, A003108, A001235, A078132, A078133, A078134.
%K nonn
%O 1,64
%A _Reinhard Zumkeller_, Nov 19 2002
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