Ramanujan's theory "$q_{3}$"
How to calculate singular moduli $t=\alpha_{3,n}$ explicitly of this?
$$\frac{\,_2F_1\big(\tfrac13,\tfrac23,1,1-t\big)}{\,_2F_1\big(\tfrac13,\tfrac23,1,t\big)} =\sqrt{n}$$
Where can I find how to do this? In the Berndt's book "Ramanujan's Notebook, part V" at page 179-180, there is a tentative to explain how
$$\big(t(1-t)\big)^{\frac 1 3}=\frac{2}{3} \frac{2-(G G')^{1/4}} {(G G')^{1/6}}=\frac 2 3 \frac{2+(g g')^{1/4}}{(g g')^{1/6}}$$
using theory "$q_{2}$".
We think to have decrypted the demand equation.
To explain Entry at the buttom of p.25 of the III Ramanujan's Notebook,
we think so:
$$\frac{1}{\big(t(1-t)\big)^{\frac 1 3}}=\frac{2}{3} \frac{2-(G G')^{1/4}} {(G G')^{1/6}}=\frac 2 3 \frac{2+(g g')^{1/4}}{(g g')^{1/6}}$$ where
$t=\beta=\alpha_{3n}$ , $\frac{1}{G}=G_{9n}^{24}$ and $\frac{1}{G'}=(G_{n})^{24}$
while
$\frac{1}{g}=(g_{9n})^{24}$ and $\frac{1}{g'}=(g_{n})^{24}$
$g_{n}$,$G_{n}$ are Ramanujan's functions (Class Invariants).
From first Entry of page n. 254 of the Chapter XXI of the II Notebook (page numbering of Ramanujan, that here is shown)
it is clear that $\beta=\frac{27}{4}\frac{(p+p^{2})^{2}}{(1+p+p^{2})^{3}}$ and the link with $q_{2}$ theory implies that $\alpha=p^{3}\frac{2+p}{1+2p}$ where $p=\frac{m-1}{2}$ with multiplicator $m$ of degree 3, equal to:
$$m^{2}=1+2 \sqrt{2} \frac{(G_{n})^{3}} {(G_{9n})^{9}}=\frac {9}{1+2\sqrt{2}\frac{(G_{9n})^{3}} {(G_{n})^{9}}}$$
Set $\rho=\frac{G_{n}}{\sqrt{2}(G_{9n})^{3}}$ and $\tau=\frac{G_{9n}}{\sqrt{2}(G_{n})^{3}}$ then the modular equation of third degree can be written thus: $$\rho^3=\frac{1-\tau^{3}}{1+8 \tau^{3}}$$
The multiplicator $m$ is: $m^{2}=1+8\rho^{3}$ and although $m=1+2p$, we can write $m^{2}=1+4p+4p^{2}=1+8\rho^{3}$
We have $p+p^2=2\rho^{3}$.
Then
$$\beta=\frac{27}{4}\frac{(2\rho^{3})^{2}}{(1+2\rho^{3})^3}=\frac{27\rho^{6}}{(1+2\rho^{3})^3}.$$
We need to calculate $(1-\beta)$:
$$(1-\beta)=1-\frac{27\rho^{6}} {(1+2\rho^{3})^3}=\frac{(\rho^{3}-1)^{2}(1+8\rho^{3})} {(1+2\rho^{3})^3}=\frac{27\tau^{6}}{(1+2\tau^{3})^3}$$
Now we observe, always by modular equation of the third degree, that
$$\frac{\tau}{\rho^{2}}+\frac{\rho}{\tau^{2}}=\frac{1}{\rho^{2}\tau^{2}}-8\rho\tau$$.
Then
$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac{1+2\rho^{3}} {3\rho^{2}}\frac{1+2\tau^{3}}{3\tau^{2}}$
$=\frac {2}{9}\big(\frac{1}{2\rho^{2}\tau^{2}}+2\rho\tau+\frac{\tau}{\rho^{2}}+\frac{\rho}{\tau^{2}}\big)$
$=\frac{2}{9}\big(\frac{1}{2\rho^{2}\tau^{2}}+2\rho\tau+\frac{1}{\rho^{2}\tau^{2}}-8\rho\tau\big)$ $=\frac {2}{9}\big(\frac{3}{2\rho^{2}\tau^{2}}-6\rho\tau\big)$ $$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac {2}{3}\big(\frac{1}{2\rho^{2}\tau^{2}}-2\rho\tau\big)$$
With the following substitutions
$\rho\tau=\frac{G_{n}}{\sqrt{2}(G_{9n})^{3}}\frac{G_{9n}}{\sqrt{2}(G_{n})^{3}}=\frac{1}{2(G_{9n})^{2}(G_{n})^{2}}$
$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac{2}{3}\big(2(G_{9n})^{4}(G_{n})^{4}-\frac{1}{(G_{9n})^{2}(G_{n})^{2}}\big)$.
Also we have
$$(G G')^{1/12}=\frac{2}{\rho\tau}=\frac{1}{(G_{9n})^{2}(G_{n})^{2}}$$
At the end we arrive to write
$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac{2}{3}\big(\frac{2}{(G G')^{1/6}}-(G G')^{1/12}\big)=$
$$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac{2}{3} \frac{2-(G G')^{1/4}} {(G G')^{1/6}}$$ Q.E.D.
2.We 'll show the second identity:
$$\frac{1}{\big(t(1-t)\big)^{\frac 1 3}}=\frac 2 3 \frac{2+(g g')^{1/4}}{(g g')^{1/6}}$$
We set $r=\frac{g_{n}}{\sqrt{2}(g_{9n})^{3}}$ and $s=\frac{g_{9n}}{\sqrt{2}(g_{n})^{3}}$ we use the modular equation of third degree for $g_{n}$ and $g_{9n}$:
$$1-8r^{3}=\frac{9}{1+8s^{3}}$$
Setting $m'=1+2p$ and $m'^{2}=1-8r^{3}$, the value of $p+p^{2}=-2r^{3}$,
moreover the value of singular modulus will be:
$\beta=\frac{27}{4}\frac{(-2r^{3})^{2}}{(1-2r^{3})^{3}}$
that is equal to: $\beta=\frac{27r^{6}}{(1-2r^{3})^{3}}$
It is necessary to calculate $(1-\beta)$:
$(1-\beta)=1-\frac{27r^{6}}{(1-2r^{3})^{3}}=\frac{(r^{3}+1)^{2}(1-8r^{3})}{(1-2r^{3})^{3}}=\frac{27s^{6}}{(1+2s^{3})^{3}}$
Now we observe by the modular equation of the third degree that
$$\frac{s}{r^{2}}-\frac{r}{s^{2}}=\frac{1}{r^{2}s^{2}}+8rs$$
Thus
$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac{(1-2r^{3})} {3 r^{2}}\frac{(1+2 s^{3})}{3 s^{2}}$
$=\frac {2}{9}\big(\frac{1}{2 r^{2} s^{2}}-2 r s+\frac{s}{r^{2}}-\frac{r}{s^{2}}\big)$
$=\frac{2}{9}\big(\frac{1}{2 r^{2}s^{2}}-2 r s+\frac{1}{r^{2}s^{2}}+8 r s\big)$
$=\frac {2}{9}\big(\frac{3}{2 r^{2}s^{2}}+6 r s\big)$
$$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac {2}{3}\big(\frac{1}{2 r^{2} s^{2}}+2 r s\big)$$
Also we have
$$r s=\frac{g_{n}}{\sqrt{2} (g_{9n})^{3}}\frac{g_9n}{(\sqrt{2}g_{n})^{3}}=\frac{1}{2 (g_{9n}g_{n})^{2}}$$
Moreover
$$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac{2}{3}\big(2(g_{9n}g_{n})^{4}+\frac{1}{(g_{9n}g_{n})^{2}}\big)$$
Also we have
$(g g')^{1/12}=\frac{1}{(g_{9n}g_{n})^{2}}$
making substitution in the last expression we arrive at
$$\frac{1}{\big(\beta(1-\beta)\big)^{1/3}}=\frac {2} {3} \big(\frac{2}{(g g')^{1/6}}+{(g g')^{1/12}}\big)$$
$$=\frac 2 3 \frac{2+(g g')^{1/4}}{(g g')^{1/6}}$$
Q.E.D.
Now the link between $q_{3}$ theory and $q_{2}$ theory:
$$\big(\frac{1} {\alpha_{3n}}\big)^{1/3}=\frac{2}{3} \frac{(G_{9n})^{6}}{(G_{n})^{2}}+\frac{\sqrt{2}}{3} \frac{G_{n}} {(G_{9n})^{3}}$$
$$\big(\frac{1} {\alpha_{3n}}\big)^{1/3}=\frac{2}{3}\frac{(g_{9n})^{6}} {(g_{n})^{2}}-\frac{\sqrt{2}}{3}\frac{g_{n}}{(g_{9n})^{3}}$$
$$\big(\frac{1}{\beta_{3n}}\big)^{1/3}=\frac{2}{3}\frac{(G_{n})^{6}}{(G_{9n})^{2}}+\frac{\sqrt{2}}{3}\frac{G_{9n}}{(G_{n})^{3}}$$
$$\big(\frac{1}{\beta_{3n}}\big)^{1/3}=\frac{2}{3}\frac{(g_{n})^{6}} {(g_{9n})^{2}}+\frac{\sqrt{2}}{3}\frac{g_{9n}}{(g_{n})^{3}}$$
Knowing the invariant values $G_{n}$, $G_{9n}$, $g_{n}$ and $g_{9n}$, we calculated the first ten singular moduli:
$\alpha_{1}=\frac{1}{2}$
$\alpha_{2}=\frac{1}{2}-\frac{\sqrt{2}}{4}$
$\alpha_{3}=\big(\frac{\sqrt{3}-1}{2}\big)^3$
$\alpha_{4}=\frac{9-5\sqrt{3}}{18}$
$\alpha_{5}=\frac{25-11\sqrt{5}}{50}$
$\alpha_{6}=\big(\frac{4-\sqrt{6}}{10}\big)^3$
$\alpha_{7}=\frac{18-13\sqrt{3}+\sqrt{21}}{36}$
$\alpha_{8}=\frac{2-17\sqrt{2}+9\sqrt{6}}{4}$
$\alpha_{9}=\big(\frac{2^{1/3}-1}{2^{1/3}+2}\big)^3$
$\alpha_{10}=\frac{54-35\sqrt{2}-2\sqrt{5}}{108}$
These solve:
$$\frac{\,_2F_1\big(\tfrac13,\tfrac23,1,\,1-\alpha_n\big)}{\,_2F_1\big(\tfrac13,\tfrac23,1,\,\alpha_n\big)} =\sqrt{n}$$
Other applications:
$$\frac{1}{(\alpha_{6})^{1/3}}=4+\sqrt{6}$$
$$(\frac{1}{\beta_{4}})^{1/3}=\frac{\sqrt{3}(\sqrt{3}-1)}{2^{1/3}}$$
$$\beta_{4}=\frac{1}{2}+\frac{5 \sqrt{3}}{18}.$$
and so on...