A294407 Expansion of 1/(1 + Sum_{i>=1} q^(i^2)/Product_{j=1..i} (1 + q^j)^2).

1, -1, 3, -8, 20, -51, 132, -339, 868, -2228, 5720, -14676, 37659, -96644, 248004, -636413, 1633144, -4190920, 10754580, -27598012, 70821032, -181738372, 466370429, -1196782952, 3071141180, -7881051500, 20224069573, -51898276576, 133179482008, -341760374284, 877013123076, -2250559385788

Offset: 0

Author

Ilya Gutkovskiy, Oct 30 2017

Keywords

• sign

Comments

Convolution inverse of the 3rd order mock theta function f(q) (A000025).

Links

• Robert Israel, Table of n, a(n) for n = 0..2441
• Eric Weisstein's World of Mathematics, Mock Theta Function

Crossrefs

Cf. A000025, A000039, A000199, A010815.

Programs

Maple

N:= 50: # to get a(0)..a(N)
g:= 1/(1+add(q^(i^2)/mul(1+q^j,j=1..i)^2, i=1..floor(sqrt(N)))):
S:= series(g, q, N+1):
seq(coeff(S,q,n),n=0..N); # Robert Israel, Nov 06 2017

Mathematica

nmax = 31; CoefficientList[Series[1/(1 + Sum[q^(i^2)/Product[(1 + q^j)^2, {j, 1, i}], {i, 1, nmax}]), {q, 0, nmax}], q]

Formula

G.f.: 1/(1 + Sum_{i>=1} q^(i^2)/Product_{j=1..i} (1 + q^j)^2).