Given the ** j-function** $j:=j(\tau)$, and $q=e^{2\pi i\tau} = \exp(2\pi i\tau)$ where, for convenience, we set $\tau=\sqrt{-n}$.

I.$\frac{A_2(q)}{A_1(q)} = \text{q-cfrac}:\;$ Icosahedral group

$$\begin{aligned} A_1(q) &= q^{-1/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})} = j\,^{1/60}\,_2F_1\left(\tfrac{-1}{60},\tfrac{19}{60},\tfrac{4}{5},\tfrac{1728}{j}\right)\\ A_2(q) &= q^{\color{blue}{11}/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})} = j\,^{-11/60}\,_2F_1\left(\tfrac{11}{60},\tfrac{31}{60},\tfrac{6}{5},\tfrac{1728}{j}\right) \end{aligned}$$

Then,

$$A_1(q^{\color{blue}{11}})A_2(q)-A_1(q)A_2(q^{\color{blue}{11}})=1\tag1$$

II.$\frac{B_2(q)}{B_1(q)} = \text{q-cfrac}:\;$ Octahedral group

$$\begin{aligned} B_1(q) &= q^{-1/24} \prod_{n=1}^\infty \frac{ (1-q^{4n-2})^2 }{ (1-q^{2n-1})^3}\quad =\quad j\,^{1/24}\,_2F_1\left(\tfrac{-1}{24},\tfrac{7}{24},\tfrac{3}{4},\tfrac{1728}{j}\right)\\ B_2(q) &= q^{\color{blue}{5}/24} \prod_{n=1}^\infty \frac{1}{(1-q^{2n-1})(1-q^{4n-2})^2} = j\,^{-5/24}\,_2F_1\left(\tfrac{5}{24},\tfrac{13}{24},\tfrac{5}{4},\tfrac{1728}{j}\right) \end{aligned}$$

Then,

$$B_1(q^{\color{blue}{5}})B_2(q)-B_1(q)B_2(q^{\color{blue}{5}})=1\tag2$$

III.$\frac{C_2(q)}{C_1(q)} = \text{q-cfrac}:\;$ Octahedral group?

$$\begin{aligned} C_1(q) &= q^{-1/16} \prod_{n=1}^\infty \frac{ 1 }{ (1-q^{8n-1})(1-q^{8n-4})(1-q^{8n-7})} =\,_2F_1\,\color{brown}{??}\\ C_2(q) &= q^{\color{blue}{7}/16} \prod_{n=1}^\infty \frac{1}{ (1-q^{8n-3})(1-q^{8n-4})(1-q^{8n-5})} =\,_2F_1\,\color{brown}{??} \end{aligned}$$

Then,

$$C_1(q^{\color{blue}{7}})C_2(q)-C_1(q)C_2(q^{\color{blue}{7}})=1\tag3$$

IV.$\frac{D_2(q)}{D_1(q)} = \text{not q-cfrac}:\;$ Tetrahedral group

$$\begin{aligned} D_1(q) &= q^{-1/12} \prod_{n=1}^\infty \frac{ (1-q^{n/3})^3 }{ (1-q^{n})^3}+3 D_2(q)= j\,^{1/12}\,_2F_1\left(\tfrac{-1}{12},\tfrac{1}{4},\tfrac{2}{3},\tfrac{1728}{j}\right)\\ D_2(q) &= q^{1/4} \prod_{n=1}^\infty \frac{ (1-q^{3n})^3 }{ (1-q^{n})^3}\quad = \quad j\,^{-1/4}\,_2F_1\left(\tfrac{1}{4},\tfrac{7}{12},\tfrac{4}{3},\tfrac{1728}{j}\right) \end{aligned}$$

Then,

$$D_1(q^{n})D_2(q)-D_1(q)D_2(q^{n})\overset{\color{brown}?}{=} 1\tag4$$

Note, however, that $\frac{D_1(q)}{D_2(q)} =4c^2+c^{-1}$ where $c$ is Ramanujan's *cubic continued fraction*.

V. Questions

- What is the hypergeometric formula for the two $C_i(q)$?
- It seems impossible to find integer $n$ for relation $(4)$. Why?
- Are there other infinite products with relations similar to $(1),(2),(3)$?

Duke's paperthey are given as eq. $(9.1)$ and $(9.2)$, $$u(\tau) = \frac{B_2(q)}{B_1(q)}= \cfrac{\sqrt{2}\,q^{1/8}}{1 + \cfrac{q} {1 +q+ \cfrac{q^2} {1+q^2+\ddots}}}$$ $$v(\tau) = \frac{C_2(q)}{C_1(q)}= \cfrac{q^{1/2}}{1 + q+\cfrac{q^2} {1 +q^3+ \cfrac{q^4}{1+q^5+\ddots}}}$$ $\endgroup$ – Tito Piezas III Jul 7 '15 at 7:24forty relatives, I wouldn't be surprised if $(2)$ and $(3)$ have relatives as well. $\endgroup$ – Tito Piezas III Jul 7 '15 at 7:32