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
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> 


$$\begin{aligned}
A\_1(q) &= q^{-1/60} \prod\_{n=1}^\infty \frac{1}{(1-q^{5n-1})(1-q^{5n-4})} = j\,^{1/60}\,\_2F\_1\left(\tfrac{-1}{60},\tfrac{19}{60},\tfrac{4}{5},\tfrac{1728}{j}\right)\\
A\_2(q) &= q^{\color{blue}{11}/60} \prod\_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})} = j\,^{-11/60}\,\_2F\_1\left(\tfrac{11}{60},\tfrac{31}{60},\tfrac{6}{5},\tfrac{1728}{j}\right)
\end{aligned}$$


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
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> 


$$\begin{aligned}
B\_1(q) &= q^{-1/24} \prod\_{n=1}^\infty \frac{ (1-q^{4n-2})^2 }{ (1-q^{2n-1})^3}\quad =\quad j\,^{1/24}\,\_2F\_1\left(\tfrac{-1}{24},\tfrac{7}{24},\tfrac{3}{4},\tfrac{1728}{j}\right)\\
B\_2(q) &= q^{\color{blue}{5}/24} \prod\_{n=1}^\infty \frac{1}{(1-q^{2n-1})(1-q^{4n-2})^2} = j\,^{-5/24}\,\_2F\_1\left(\tfrac{5}{24},\tfrac{13}{24},\tfrac{5}{4},\tfrac{1728}{j}\right)
\end{aligned}$$


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?
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> 
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$$\begin{aligned}
C\_1(q) &= q^{-1/16} \prod\_{n=1}^\infty \frac{ 1 }{ (1-q^{8n-1})(1-q^{8n-4})(1-q^{8n-7})} =\,\_2F\_1\,\color{brown}{??}\\
C\_2(q) &= q^{\color{blue}{7}/16} \prod\_{n=1}^\infty \frac{1}{ (1-q^{8n-3})(1-q^{8n-4})(1-q^{8n-5})} =\,\_2F\_1\,\color{brown}{??}
\end{aligned}$$


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
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$$\begin{aligned}
D\_1(q) &= q^{-1/12} \prod\_{n=1}^\infty \frac{ (1-q^{n/3})^3 }{ (1-q^{n})^3}+3 D\_2(q)= j\,^{1/12}\,\_2F\_1\left(\tfrac{-1}{12},\tfrac{1}{4},\tfrac{2}{3},\tfrac{1728}{j}\right)\\ 
D\_2(q) &= q^{1/4} \prod\_{n=1}^\infty \frac{ (1-q^{3n})^3 }{ (1-q^{n})^3}\quad = \quad j\,^{-1/4}\,\_2F\_1\left(\tfrac{1}{4},\tfrac{7}{12},\tfrac{4}{3},\tfrac{1728}{j}\right)
\end{aligned}$$


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


See also [this](https://mathoverflow.net/questions/181861/) and [this](https://mathoverflow.net/questions/180438/).



Relations between modular functions of certain $q$-continued fractions
Still thinking about (2).
As for (3), there are many such identities. For example, try the three level seven functions and the exponent 3. I can't at the moment find the reference that I was thinking of, but [this](https://www.ideals.illinois.edu/bitstream/handle/2142/18485/Gugg_Chadwick.pdf?sequence=1) seems to have some.
As for (1), I think you have to look at a group for which $\Gamma\_1(8)$ is a normal subgroup. Set $j\_4=\frac{(q;q)\_{\infty }^8}{q(q^4;q^4)\_{\infty}^8} = \frac{\eta^8(\tau)}{\eta^8(4\tau)}$, where $\eta(\tau)$ is the *Dedekind eta function*. Then,
\begin{equation\*}
C\_1 = j\_4^{1/16} \,
 \_2F\_1\left(-\frac{1}{8},\frac{3}{8};\frac{1}{2};\frac{-16}
 {j\_4}\right)\\
C\_2 = j\_4^{-7/16} \,
 \_2F\_1\left(\frac{7}{8},\frac{3}{8};\frac{3}{2};\frac{-16}{
 j\_4}\right)
\end{equation\*}
However, you will probably be disappointed because these 2F1's are just radicals. (And the symmetry is dihedral not octahedral.)


