Entry 102

Level 10. Given $q = e^{2\pi i \tau}$,  the q-Pochhammer symbol $(a;q)_n$, and Ramanujan's functions $f(a,b)$ and $\varphi(q)$ discussed in Entry 92. We have the sum-product identities

$$\begin{align}A(q) &= \frac{f(-q,-q^9)}{\varphi(-q)} \;=\; \sum_{n=0}^\infty \frac {q^{n(n+3)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}}  = \prod_{n=1}^\infty\frac{\alpha_{10}}{(1-q^{10n-3})(1-q^{10n-7})}\\ B(q) &= \frac{f(-q^3,-q^7)}{\varphi(-q)} = \sum_{n=0}^\infty \frac {q^{n(n+1)/2}\,(-q;q)_n} {(q;q)_n (q;q^2)_{n+1}}  = \prod_{n=1}^\infty\frac{\alpha_{10}}{(1-q^{10n-1})(1-q^{10n-9})}\end{align}$$

where $\alpha_{10} = \dfrac{(1-q^{10n})^2}{(1-q^{n})(1-q^{5n})}$. 

Let $a = q^{4/5}A(q)$ and $b = q^{1/5}B(q)$ then $(a,b)$ are radicals for appropriate $\tau$. Define their ratio $a/b$, $$\text{K}_{10}(q) = q^{3/5}\prod_{n=1}^\infty\frac{(1-q^{10n-1})(1-q^{10n-9})}{(1-q^{10n-3})(1-q^{10n-7})}$$ While no continued fraction is yet known for this, it is connected to the mod 5 version $$\text{K}_{5}(q) = q^{1/5}\prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}$$

simply as, $$\text{K}_{10}(q) = \text{K}_5(q)\,\text{K}_5(q^2)$$

where $\text{K}_5(q)=R(q)$ is just the Rogers-Ramanujan continued fraction.