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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

  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 and this.

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  • $\begingroup$ Like $(1)$ which corresponding to Rogers-Ramanujan, do you have the continued fraction expansions for $B_{2}/B_{1}$ and $C_{2}/C_{1}$? If so please include that also in your post. Perhaps there are series expansions for each of these functions also. $\endgroup$ – Paramanand Singh Jul 7 '15 at 5:33
  • $\begingroup$ @ParamanandSingh: They were also studied by Ramanujan. In Duke's paper they 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 7:24
  • $\begingroup$ @ParamanandSingh: Just like $(1)$ has at least forty relatives, I wouldn't be surprised if $(2)$ and $(3)$ have relatives as well. $\endgroup$ – Tito Piezas III Jul 7 '15 at 7:32
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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 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.)

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  • $\begingroup$ I made a small tweak expressing $j_4$ as eta quotients. This is just fine, at least I now know the $C_i$'s hypergeometric formula. And thanks for giving the link to Chadwick Gugg's thesis, "Modular Identities for the Rogers'-Ramanujan Functions and Analogues". It's a great collection. $\endgroup$ – Tito Piezas III Jul 8 '15 at 5:00

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