Hypergeometric formulas for the Rogers-Ramanujan identities?

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

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

and define,

$$k = j-1728$$

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