A291692 Expansion of Product_{k>=1} (1+x^(k^3))^(k^3).

1, 1, 0, 0, 0, 0, 0, 0, 8, 8, 0, 0, 0, 0, 0, 0, 28, 28, 0, 0, 0, 0, 0, 0, 56, 56, 0, 27, 27, 0, 0, 0, 70, 70, 0, 216, 216, 0, 0, 0, 56, 56, 0, 756, 756, 0, 0, 0, 28, 28, 0, 1512, 1512, 0, 351, 351, 8, 8, 0, 1890, 1890, 0, 2808, 2808, 65, 65, 0, 1512, 1512, 0

Offset: 0

Author

Vaclav Kotesovec, Aug 30 2017

Keywords

• nonn

Comments

In general, if m > 0 and g.f. = Product_{k>=1} (1+x^(k^m))^(k^m), then a(n, m) ~ c * exp((2*m+1) * ((2^(1+1/m)-1) * Gamma(1/m) * Zeta(2+1/m))^(m/(2*m+1)) * n^((m+1)/(2*m+1)) / ((2*m+2)^((m+1)/(2*m+1)) * m^(3*m/(2*m+1)))) * ((2^(1+1/m)-1) * (m+1) * Gamma(1/m) * Zeta(2+1/m))^(m/(4*m+2)) / (sqrt(2*m+1) * sqrt(Pi) * 2^((3*m+2)/(4*m+2)) * m^((m-1)/(4*m+2)) * n^((3*m+1)/(4*m+2))), where c = 2^(-1/12) for m = 1 and c = 1 for m > 1.

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A026007 (m=1), A291649 (m=2).

Cf. A033461, A279329.

Programs

Mathematica

nmax = 100; CoefficientList[Series[Product[(1 + x^(k^3))^(k^3), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 100; s = 1 + x; Do[s *= Sum[Binomial[k^3, j]*x^(j*k^3), {j, 0, Floor[nmax/k^3] + 1}]; s = Select[Expand[s], Exponent[#, x] <= nmax &];, {k, 2, nmax}]; Take[CoefficientList[s, x], nmax]

Formula

a(n) ~ exp(7*((2^(4/3)-1) * Gamma(1/3) * Zeta(7/3))^(3/7) * n^(4/7) / (2^(12/7) * 3^(9/7))) * ((2^(4/3)-1) * Gamma(1/3) * Zeta(7/3))^(3/14) / (2^(5/14) * 3^(1/7) * sqrt(7*Pi) * n^(5/7)).