The j-invaraint has a q-series. What is the easiest way to compute this series?
This is the way I currently know how to do it. Using Maxima:
a(k):=coeff(q*taylor(12^3/(1-(1-504*sum(n^5*q^n/(1-q^n),n,1,inf))^2/(1+240*sum(n^3*q^n/(1-q^n),n,1,inf))^3),q,0,k),q,k)
Which is from the formula:
$$j(q) = \frac{12^3}{1-\frac{\left(1-504 \sum\limits_{n=1}^\infty n^3\frac{ q^n}{1-q^n}\right)^2}{\left(1+240\sum\limits_{n=1}^\infty n^5\frac{ q^n}{1-q^n}\right)^3}} = \frac{1}{q}+744+198884q+... = \frac{1}{q}\sum\limits_{n=0}^\infty a(k) q^n$$
Which is basically just expanded. Where $a(k)$ gives the $k^{th}$ coefficient.
This formula gives me:
[1,744,196884,21493760,864299970,20245856256,333202640600,4252023300096,44656994071935,401490886656000,3176440229784420]
Surely there must be a simpler formula? (Just using elementary functions). I suppose one could just expand the above series and get a formula in terms of sums of binomial coefficients.
This one (slightly shorter) also works:
a(k):=coeff(taylor((1+240*sum(n^3*q^n/(1-q^n),n,1,inf))^3*prod( (1-q^n)^(-24),n,1,inf),q,0,k),q,k);
(I'd like to know how internally the software works these out!)