OFFSET
0,2
COMMENTS
Partial sums of A003108.
Number of partitions of n into cubes if there are two kinds of 1's.
LINKS
FORMULA
G.f.: (1/(1 - x))*Sum_{j>=0} x^(j^3)/Product_{k=1..j} (1 - x^(k^3)).
From Vaclav Kotesovec, Apr 13 2018: (Start)
a(n) ~ sqrt(3) * exp(4*(Gamma(1/3)*Zeta(4/3))^(3/4) * n^(1/4) / 3^(3/2)) / (8 * Pi^2 * sqrt(n)).
a(n) ~ 3^(3/2) * n^(3/4) / (Gamma(1/3)*Zeta(4/3))^(3/4) * A003108(n). (End)
MAPLE
b:= proc(n, i) option remember; `if`(n=0 or i=1, n+1,
b(n, i-1)+ `if`(i^3>n, 0, b(n-i^3, i)))
end:
a:= n-> b(n, iroot(n, 3)):
seq(a(n), n=0..100); # Alois P. Heinz, Apr 13 2018
MATHEMATICA
nmax = 64; CoefficientList[Series[1/(1 - x) Product[1/(1 - x^k^3), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 64; CoefficientList[Series[1/(1 - x) Sum[x^j^3/Product[(1 - x^k^3), {k, 1, j}], {j, 0, nmax}], {x, 0, nmax}], x]
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Apr 13 2018
STATUS
approved