User:Alanonala/sandbox

&= \sum_{n=0}^\infty \frac {q^{n^2}} {(1-q)(1-q^2)\cdots(1-q^n)} =\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty}\\ &= \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\ &=\sqrt[60]{qj}\,_2F_1\left(-\tfrac{1}{60},\tfrac{19}{60};\tfrac{4}{5};\tfrac{1728}{j}\right)\\ &=\sqrt[60]{q\left(j-1728\right)}\,_2F_1\left(-\tfrac{1}{60},\tfrac{29}{60};\tfrac{4}{5};-\tfrac{1728}{j-1728}\right)\\ &= 1+ q +q^2 +q^3 +2q^4+2q^5 +3q^6+\cdots \end{align}$$
 * $$\begin{align}G(q)


 * $$\begin{align}G(q)

&= \sum_{n=0}^\infty \frac {q^{n^2}} {(1-q)(1-q^2)\cdots(1-q^n)} =\sum_{n=0}^\infty \frac {q^{n^2}} {(q;q)_n} = \frac {1}{(q;q^5)_\infty (q^4; q^5)_\infty}\\ &= \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})}\\ &=\sqrt[60]{qj}\,_2F_1\left(-\tfrac{1}{60},\tfrac{19}{60};\tfrac{4}{5};\tfrac{1728}{j}\right)\\ &=\sqrt[60]{q\left(j-1728\right)}\,_2F_1\left(-\tfrac{1}{60},\tfrac{29}{60};\tfrac{4}{5};-\tfrac{1728}{j-1728}\right)\\ &= 1+ q +q^2 +q^3 +2q^4+2q^5 +3q^6+\cdots \end{align}$$