\[\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{\left(q^{2};q^{2}\right)_{n}\left(q;q^{2}%
\right)_{n}}=\mbox{ coeff. of }z^{0}\mbox{ in }\frac{\left(-zq;q^{2}\right)_{%
\infty}\left(-z^{-1}q;q^{2}\right)_{\infty}\left(q^{2};q^{2}\right)_{\infty}}{%
\left(-z^{-1};q^{2}\right)_{\infty}\left(q;q^{2}\right)_{\infty}\left(z^{-1};q%
^{2}\right)_{\infty}}=\frac{1}{\left(q;q^{2}\right)_{\infty}}\mbox{ coeff. of %
}z^{0}\mbox{ in }\frac{\left(-zq;q^{2}\right)_{\infty}\left(-z^{-1}q;q^{2}%
\right)_{\infty}\left(q^{2};q^{2}\right)_{\infty}}{\left(z^{-2};q^{4}\right)_{%
\infty}}=\frac{G(q^{4})}{\left(q;q^{2}\right)_{\infty}},\]