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Let $q = e^{2\pi i \tau}$. Given the *j-function*,

$$j = j(q) = 1/q + 744 + 196884q + 21493760q^2 + \dots$$

and define,

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$$k = j-1728$$

Let $\tau =\sqrt{-N}$, where $N > 1$. *Anybody knows how to prove the RHS of these conjectured relations?*:

$$\begin{align}q^{-1/60} G(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{19}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{1728}{j}\right)\\

&= k\,^{1/60}\,_2F_1\left(\tfrac{29}{60},\tfrac{-1}{60},\tfrac{4}{5},\tfrac{-1728}{k}\right)\\[2.5mm]

q^{11/60} H(q) = q^{11/60} \prod_{n=1}^\infty \frac{1}{(1-q^{5n-2})(1-q^{5n-3})} &= j\,^{-11/60}\,_2F_1\left(\tfrac{31}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{1728}{j}\right)\\

&= k\,^{-11/60}\,_2F_1\left(\tfrac{41}{60},\tfrac{11}{60},\tfrac{6}{5},\tfrac{-1728}{k}\right)\end{align}$$

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Let

$$\begin{align}

g(\tau) &= q^{-1/60} G(q)

\\ h(\tau) &= q^{11/60} H(q)

\end{align}$$

First of all, the equalities between the hypergeometric $j$ and $k$ expressions

follow from the eulerian transformation

$${}_2F_1(a,b;c;z) = (1-z)^{-b}\,{}_2F_1\left(c-a,b;c;\frac{z}{z-1}\right)$$

Therefore it suffices to prove the identities between $g$ resp. $h$

and the corresponding hypergeometric $j$ expressions.

I will translate those to more familiar identities.

We will use the Rogers-Ramanujan continued fraction

(RRCF),

$$\rho(\tau) = \frac{h(\tau)}{g(\tau)} = q^{1/5}\frac{H(q)}{G(q)}$$

Formula $(22)$ from the above MathWorld entry on RRCF essentially states that

$$\frac{1}{\rho^{5}} – 11 – \rho^5 = \frac{1}{g^6\,h^6}$$

(Use the product representation of $g$ and $h$ to identify the right-hand side).

From this, we can easily deduce

$$\begin{align}

g &= \frac{1}

{\left(\rho – 11\,\rho^6 – \rho^{11}\right)^{1/12}}

\\ h &= \frac{\rho}

{\left(\rho – 11\,\rho^6 – \rho^{11}\right)^{1/12}}

\end{align}$$

assuming that the arguments to the radicals are small positive reals,

which should follow from the restrictions you have placed on $\tau$.

Furthermore, formula $(46)$ from the above Mathworld entry on RRCF

gives the relation of $\rho$ with Klein’s $j$:

$$j = \frac

{\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^3}

{\left(\rho – 11\,\rho^6 – \rho^{11}\right)^5}$$

which allows us to write

$$\begin{align}

g\,j^{-1/60} &=

\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^{-1/20}

\\ h\,j^{11/60} &= \frac

{\left(1+228\,\rho^5+494\,\rho^{10}-228\,\rho^{15}+\rho^{20}\right)^{11/20}}

{1 – 11\,\rho^5 – \rho^{10}}

\end{align}$$

again assuming that the arguments to the radicals are small positive reals.

I need a *deus ex machina* now, and Raimundas Vidūnas arXiv:0807.4808v1 comes to the rescue.

His formulae $(59)$ and $(61)$

in section 6.3 (“icosahedral hypergeometric equations”, p. 20)

state precisely that

$$\begin{align}

{}_2F_1\left(\tfrac{19}{60},-\tfrac{1}{60};\tfrac{4}{5};\varphi_1(x)\right)

&= \left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^{-1/20}

\\{}_2F_1\left(\tfrac{31}{60},\tfrac{11}{60};\tfrac{6}{5};\varphi_1(x)\right)

&= \frac{\left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^{11/20}}

{1 + 11\,x – x^2}

\\\text{where}\qquad

\varphi_1(x) &= \frac{-1728\,x\,(1+11\,x-x^2)^5}

{\left(1-228\,x+494\,x^2+228\,x^3+x^4\right)^3}

\end{align}$$

And your conjecture follows from setting $x=-\rho^5$ which implies

$\varphi_1(x) = \frac{1728}{j}$.

Summarizing, there is an algebraic relation between $j$ and $g$ resp. $h$,

and the hypergeometric ${}_2F_1$ expressions are designed to solve those

algebraic relations for $g\,j^{-1/60}$ resp. $h\,j^{11/60}$, given $j$.

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