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

$$_2F_1\big(s,1-s,1,z\big)$$

with $s = \frac12, \frac13, \frac14, \frac16$ for infinitely many algebraic numbers $z.$

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**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][1],

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

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, 

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

and the equalities

$$\vartheta_3(0,q) = \sum_{m=-\infty}^\infty q^{m^2} = \frac{\eta^5(\tau)}{\eta^2\big(\tfrac{\tau}2\big)\,\eta^2(2\tau)}$$ 

has a nice cubic version in the next section.

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**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 the cubic analogue,

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

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**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}

Again, 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][2]. ***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][3] 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][4]*, and where all $q$ are $q=e^{2\pi i\tau}.$

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**IV. Parameter $s=\frac16$**

Given the golden ratio $\phi$, then,

$$_2F_1\left(\frac16,\frac56,1,\frac{\phi^{-5}}{5\sqrt5}\right) = \left(\frac{3\sqrt3}{8}\right)^{1/3}\left(\frac{5\sqrt5}{8}\right)^{1/6}\,\frac{\Gamma\big(\tfrac13\big)^3}{2\pi^2}$$

(Note: The rest of the section has been moved to a [MSE post][5] to trim this post.)

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**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 $\eta(\tau)$.

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**VI. Question**

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


  [1]: https://en.wikipedia.org/wiki/Modular_lambda_function#Relations_to_other_functions
  [2]: https://https%20:%20//%20oeis.org/A335227
  [3]: https://oeis.org/A004011
  [4]: https://mathworld.wolfram.com/OddDivisorFunction.html
  [5]: https://math.stackexchange.com/questions/4970412/the-golden-ratio-phi-for-2f-1-big-frac16-frac16-frac23-27-phi9-big