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Expansion of Product_{k>=1} 1/(1 + x^(k^3)).
3

%I #12 Sep 20 2017 15:22:28

%S 1,-1,1,-1,1,-1,1,-1,0,0,0,0,0,0,0,0,1,-1,1,-1,1,-1,1,-1,0,0,0,-1,1,

%T -1,1,-1,2,-2,2,-1,1,-1,1,-1,0,0,0,-1,1,-1,1,-1,2,-2,2,-1,1,-1,2,-2,1,

%U -1,1,-2,2,-2,1,-1,1,-1,1,0,0,0,1,-1,1,-1,1,-2,2,-2,1,-1,1,-2,2,-1,1,-1,2,-2,2,-1,1

%N Expansion of Product_{k>=1} 1/(1 + x^(k^3)).

%C Convolution inverse of A279329.

%C The difference between the number of partitions of n into an even number of cubes and the number of partitions of n into an odd number of cubes.

%C In general, if m > 0 and g.f. = Product_{k>=1} 1/(1 + x^(k^m)), then a(n) ~ (-1)^n * exp((m+1) * (Gamma(1/m) * Zeta(1 + 1/m) / m^2)^(m/(m+1)) * n^(1/(m+1)) / 2) * (Gamma(1/m) * Zeta(1 + 1/m))^(m/(2*(m+1))) / (sqrt(Pi*(m+1)) * 2^((m+1)/2) * m^((m-1)/(2*(m+1))) * n^((2*m+1)/(2*(m+1)))). - _Vaclav Kotesovec_, Sep 19 2017

%H Vaclav Kotesovec, <a href="/A292560/b292560.txt">Table of n, a(n) for n = 0..20000</a>

%H <a href="/index/Su#ssq">Index entries for sequences related to sums of cubes</a>

%H <a href="/index/Par#part">Index entries for related partition-counting sequences</a>

%F G.f.: Product_{k>=1} 1/(1 + x^(k^3)).

%F a(n) ~ (-1)^n * exp(2 * (Gamma(1/3) * Zeta(4/3))^(3/4) * n^(1/4) / 3^(3/2)) * (Gamma(1/3) * Zeta(4/3))^(3/8) / (8 * 3^(1/4) * sqrt(Pi) * n^(7/8)). - _Vaclav Kotesovec_, Sep 19 2017

%t nmax = 100; CoefficientList[Series[Product[1/(1 + x^(k^3)), {k, 1, Floor[nmax^(1/3)] + 1}], {x, 0, nmax}], x]

%Y Cf. A081362 (m=1), A292520 (m=2).

%Y Cf. A003108, A279329, A279484.

%K sign

%O 0,33

%A _Ilya Gutkovskiy_, Sep 19 2017