OFFSET
0,10
COMMENTS
Convolution inverse of A033461.
The difference between the number of partitions of n into an even number of squares and the number of partitions of n into an odd number of squares.
LINKS
Vaclav Kotesovec, Table of n, a(n) for n = 0..20000
Martin Klazar, What is an answer? — remarks, results and problems on PIO formulas in combinatorial enumeration, part I, arXiv:1808.08449 [math.CO], 2018.
FORMULA
G.f.: Product_{k>=1} 1/(1 + x^(k^2)).
a(n) ~ (-1)^n * exp(3 * Pi^(1/3) * Zeta(3/2)^(2/3) * n^(1/3) / 2^(7/3)) * Zeta(3/2)^(1/3) / (2^(5/3) * sqrt(3) * Pi^(1/3) * n^(5/6)). - Vaclav Kotesovec, Sep 19 2017
a(n) = Sum_{k=0..n} (-1)^k * A243148(n,k). - Alois P. Heinz, Jul 25 2022
MATHEMATICA
nmax = 100; CoefficientList[Series[Product[1/(1 + x^(k^2)), {k, 1, Floor[Sqrt[nmax]] + 1}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 19 2017 *)
CROSSREFS
KEYWORD
sign
AUTHOR
Ilya Gutkovskiy, Sep 18 2017
STATUS
approved