Entry 93

Let $q = e^{2\pi i \tau}$.  Ramanujan's theta function $f(a,b)$ is given by $$f(a,b) = \sum_{n=-\infty}^\infty a^{n(n+1)/2}\; b^{n(n-1)/2}$$ Then the following have nice $q$-continued fractions $$\begin{align}\text{K}_4(q) &= q^{1/8}\frac{f(-q,-q^3)}{f(-q^2,-q^2)} = q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})}\\ \text{K}_5(q) &= q^{1/5}\frac{f(-q,-q^4)}{f(-q^2,-q^3)} = q^{1/5} \prod_{n=1}^\infty\frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})}\\ \text{K}_6(q) &= q^{1/3}\frac{f(-q,-q^5)}{f(-q^3,-q^3)} = q^{1/3} \prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})}\\ \text{K}_8(q) &= q^{1/2}\frac{f(-q,-q^7)}{f(-q^3,-q^5)} = q^{1/2} \prod_{n=1}^\infty\frac{(1-q^{8n-1})(1-q^{8n-7})}{(1-q^{8n-3})(1-q^{8n-5})}\\ \text{K}_{12}(q) &= q^{1/1} \frac{f(-q,-q^{11})}{f(-q^5,-q^7)} = q^{1/1} \prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\end{align}$$ with the most famous being $\text{K}_5(q)=R(q)$ since this is the Rogers-Ramanujan continued fraction. Each will be discussed in subsequent entries.