Entry 103

Level 12. Given \(q = e^{2\pi i \tau}\),  the q-Pochhammer symbol \((a;q)_n\), and Ramanujan's functions \(f(a,b)\) and \(f(-q)\) discussed in Entry 92. We have the sum-product identities $$\begin{align}A(q) &= \frac{f(-q,-q^{11})}{f(-q^4)} = \sum_{n=0}^\infty \frac {q^{4n^2+4n}\,(q;q^2)_{2n+1}} {(q^4;q^4)_{2n+1}}  = \prod_{n=1}^\infty(1-q^{12n-1})(1-q^{12n-11})\,\alpha_{12}\\ B(q) &=\frac{f(-q^5,-q^{7})}{f(-q^4)} \,=\, \sum_{n=0}^\infty \frac {q^{4n^2}\,(q;q^2)_{2n}} {(q^4;q^4)_{2n}} \quad = \; \prod_{n=1}^\infty(1-q^{12n-5})(1-q^{12n-7})\,\alpha_{12}\end{align}$$

where \(\alpha_{12} = \dfrac{(1-q^{12n})}{(1-q^{4n})}\). 

Let \(a = q^{7/8}A(q)\) and \(b = q^{-1/8}B(q)\) then \((a,b)\) are radicals for appropriate \(\tau\). Define their ratio \(a/b\) $$\text{K}_{12}(q) = q\,\prod_{n=1}^\infty\frac{(1-q^{12n-1})(1-q^{12n-11})}{(1-q^{12n-5})(1-q^{12n-7})}\quad\qquad$$ Surprisingly, this has a continued fraction (studied by Naika) based on a general form from Ramanujan for Level \(4m\), though its form is not as simple as the others. It is connected to the mod 6 version, $$\quad\text{K}_{6}(q) = q^{1/3}\prod_{n=1}^\infty\frac{(1-q^{6n-1})(1-q^{6n-5})}{(1-q^{6n-3})(1-q^{6n-3})} = \frac{\eta(\tau)\,\eta^3(6\tau)}{\eta(2\tau)\,\eta^3(3\tau)}$$ or the previously discussed cubic continued fraction \(\text{K}_{6}(q)\) by the quadratic relation 

$$j_{12I}=\frac1{\text{K}_{12}(q)}+\text{K}_{12}(q) = \frac1{\text{K}_6(q)\,\text{K}_6(q^2)} = \frac{\eta^3(3\tau)\,\eta(4\tau)}{\eta(\tau)\,\eta^3(12\tau)} = \left(\frac{\eta^2(4\tau)\,\eta(6\tau)}{\eta(2\tau)\,\eta^2(12\tau)}\right)^2+1$$

 where \(j_{12I}\) is the McKay-Thompson series of class 12I for the Monster (A187144).