The post has been divided into sections to show some patterns, as well as possible evaluations of,

$$_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)\\[6pt] _2F_1\big(\tfrac13,\tfrac23,1,\beta\big)\\[6pt] _2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)\\[6pt] _2F_1\big(\tfrac16,\tfrac56,1,\delta\big)$$

for infinitely many algebraic numbers $\alpha, \beta, \gamma, \delta.$

I. Parameter $s=\frac12$

Given the nome $q = e^{\pi i\tau}$ and the Jacobi theta functions $\vartheta_n(0,q)$. Define the modular lambda function,

$$\alpha = \frac{16}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+16} = \left(\frac{\sqrt2\,\eta(\tau/2)\eta^2(2\tau)}{\eta^3(\tau)}\right)^8 = \left(\frac{\vartheta_2(0,q)}{\vartheta_3(0,q)}\right)^4$$

Then we propose for appropriate $\tau$ such as $\tau = \sqrt{-d}$ that the ratios below are algebraic numbers,

\begin{align} \left(\frac{\vartheta_2(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=\alpha\\ \left(\frac{\vartheta_4(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=1-\alpha\\ \left(\frac{\vartheta_3(0,q)}{\sqrt{_2F_1\big(\tfrac12,\tfrac12,1,\alpha\big)}}\right)^4 &\overset{\color{red}?}=1 \end{align}

Note that adding the first two implies the third. Hence,

$$\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_4(0,q)\big)^4 = \big(\vartheta_3(0,q)\big)^4$$

which is known to be true. As eta quotients in the same order above,

$$\left(\frac{2\eta^2(2\tau)}{\eta(\tau)}\right)^4+\left(\frac{\eta^2\big(\tfrac{\tau}2\big)}{\eta(\tau)}\right)^4 = \left(\frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}\right)^4$$

For $\tau = \sqrt{-d}$, then $\vartheta_3(0,q)$ also seems to have a nice alternative form to be consistent with the cubic version in the next section,

$$\vartheta_3(0,q) = \sum_{m=-\infty}^\infty q^{m^2} \,\overset{\color{red}?}=\, \frac{\eta^4\big(\tfrac{\tau}2\big)+4\eta^4(2\tau)}{2\eta^3(\tau)}$$

II. Parameter $s=\frac13$

Given the square of the nome, so $q = e^{2\pi i\tau}$ and the Borwein cubic theta functions $a(q),b(q),c(q)$. Define,

$$\beta = \left(\frac{3}{\left(\tfrac{\eta(\tau/3)}{\eta(3\tau)}\right)^3+3}\right)^3=\left(\frac{c(q)}{a(q)}\right)^3$$

Then we propose,

\begin{align} \left(\frac{c(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=\beta\\ \left(\frac{b(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1-\beta\\ \left(\frac{a(q)}{_2F_1\big(\tfrac13,\tfrac23,1,\beta\big)}\right)^3 &\overset{\color{red}?}=1 \end{align}

Adding the first two implies the third,

$$\big(c(q)\big)^3+\big(b(q)\big)^3=\big(a(q)\big)^3$$

which is also known to be true. As eta quotients,

$$\left(\frac{3\eta^3(3\tau)}{\eta(\tau)}\right)^3+\left(\frac{\eta^3(\tau)}{\eta(3\tau)}\right)^3 =\left(\frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}\right)^3$$

For the square of the nome, so $q = e^{2\pi i\tau}$, then we have cubic analogue,

$$a(q) = \sum_{m,n\,=-\infty}^\infty q^{m^2+mn+n^2} = \frac{\eta^3(\tau)+9\eta^3(9\tau)}{\eta(3\tau)}$$

III. Parameter $s=\frac14$

Given the square of the nome $q = e^{2\pi i\tau}$. Define, $$\gamma = \left(\frac{8}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{C(q)}{A(q)}\right)^2$$

Then,

\begin{align} \left(\frac{C(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=\gamma\\ \left(\frac{B(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1-\gamma\\ \left(\frac{A(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1 \end{align}

The first two implies the third,

$$\big(C(q)\big)^2+\big(B(q)\big)^2=\big(A(q)\big)^2$$

where $C(q), B(q), A(q)$ are defined by the eta quotients,

$$\left(\frac{8\eta^8(2\tau)}{\eta^4(\tau)}\right)^2+\left(\frac{\eta^8(\tau)}{\eta^4(2\tau)}\right)^2=\left(\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\right)^2$$

Unlike $a(q)$, I am not aware of a sum for $A(q)$,

$$A(q) = \sum_{m=-\infty}^\infty ??\, = \frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}$$

But note that,

$$\frac{24}{A(q)-1}= \frac1{q} -1 - 3q + 6q^2 + q^3 - 20q^4 + 24q^5 + 38q^6 - 132q^7 + \dots$$

which seems to be A335227. Update: Michael Somos pointed out that,

$$A(q) = 1 + 24q + 24q^2 + 96q^3 + 24q^4 + 144q^5 + 96q^6 + 192q^7 + \dots$$

which is A004011 and is the theta series of $D_4$ lattice. So we finally have a sum,

\begin{align}A(q) &= 1+24\sum_{n=1}^\infty\frac{n q^n}{1+q^n}\\[4pt] &=\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_3(0,q))^4\qquad \end{align}

related to the odd divisor function, and where all $q$ are $q=e^{2\pi i\tau}.$

IV. Parameter $s=\frac16$

This parameter is different since it is hard to find analogous identities of form $a^n+b^n=c^n$. But given the j-function $j=j(\tau)$. The proposed relation,

$$j(\tau) \overset{\color{red}?}= \left(\frac{\sqrt{_2F_1\big(\tfrac16,\tfrac56,1,\delta\big)}}{\eta(\tau)}\right)^{24}$$

seems true if,

Case 1. For $\tau = \sqrt{-d},\,$ integer $d>0,\,$ and $\delta = \dfrac{j\color{red}-\sqrt{j(j-1728)}}{2j}$.

Case 2. For $\tau = \frac{1+\sqrt{-d}}2,\,$ integer $d>3,\,$ and $\delta = \dfrac{j\color{red}+\sqrt{j(j-1728)}}{2j}$.

Example: Let $\tau = \sqrt{-7}$, so $j=j(\tau)=255^3$, and $\delta = \frac{7225-171\sqrt{1785}}{14450}$. Then we have the explicit evaluation,

$$_2F_1\left(\tfrac16,\tfrac56,1,\tfrac{7225-171\sqrt{1785}}{14450}\right) = \left(\frac{255}7\right)^{1/4}\frac{\Gamma\big(\frac17\big) \Gamma\big(\frac27\big) \Gamma\big(\frac47\big)}{8\pi^2}$$

And so on for other $\tau$.

V. Context

These observations arose from evaluations of the complete elliptic integral of the first kind, $K(k)$. For ex., given the tribonacci constant $T$, the real root of $T^3-T^2-T-1=0$, then,

$$K(k_{11}) = \frac{\pi\,(2T)^{2/3}}{2\;}\times \frac{\Gamma\big(\tfrac1{11}\big) \Gamma\big(\tfrac3{11}\big) \Gamma\big(\tfrac4{11}\big) \Gamma\big(\tfrac5{11}\big) \Gamma\big(\tfrac9{11}\big)}{11^{1/4}(2\pi)^3}$$

Manipulating the $s=\frac12$ relations above, we can have a much shorter version,

$$K(k_{11}) = \frac{\pi\,(2T)^{4/3}}{2\;}\times\eta^2\big(\sqrt{-11}\big)\quad\quad$$

Equating the two formulas, this also gives the explicit evaluation of $\eta(\tau)$.

VI. Question

Q: Are the proposed relations $M\overset{\color{red}?}=N$ with the red question marks in fact true?