Euler transform of period 24 sequence [ 1, 1, 1, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 1, 1, 1, 0, ...].
Given g.f. A(x), then B(x)=A(x)^2-A(x) satisfies 0=f(B(x), B(x^2)) where f(u, v)=(1+6*u)*v*(1+2*v)-u^2.
G.f.: {Sum_{k} q^(6k^2-k) }/{Sum_{k} (-1)^k q^((3k^2-k)/2) }.
G.f.: Product_{k>0} (1-q^(12k))(1+q^(12k-5))(1+q^(12k-7))/(1-q^k).
G.f.: 1+Sum_{k>0} Prod[i=1..k, (1+q^i)^2]*(1+q^k)*q^(k^2) /{(1-q)(1-q^2)...(1-q^(2k))}.
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(9/4) * 3^(3/4) * n^(3/4)). -
Vaclav Kotesovec, Aug 31 2015