# Hypergeometric function values and the Baxter constant

While I was working on this question by @Vladimir Reshetnikov, I've found the following relations between Gaussian hypergeometric function values and the Baxter constant:

\begin{align}{_2F_1}\left(\begin{array}c\tfrac13,\tfrac13\\1\end{array}\middle|\,-1\right) &\stackrel{?}{=} \frac{1}{2^{\small2/3}}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac23,\tfrac23\\1\end{array}\middle|\,-1\right) &\stackrel{?}{=} \frac{1}{2}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac13,\tfrac13\\1\end{array}\middle|\,\frac19\right) &\stackrel{?}{=} \frac{1}{\sqrt[3]{3}}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac23,\tfrac23\\1\end{array}\middle|\,\frac19\right) &\stackrel{?}{=} \frac{\sqrt[3]{3}}{2}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac13,\tfrac13\\1\end{array}\middle|\,9\right) &\stackrel{?}{=} \frac{3-i\sqrt3}{6}\,C^2_\text{B4CC},\\ {_2F_1}\left(\begin{array}c\tfrac23,\tfrac23\\1\end{array}\middle|\,9\right) &\stackrel{?}{=} -\frac{i}{2\sqrt3}\,C^2_\text{B4CC},\\ \end{align}

where ${_2F_1}$ is the Gaussian hypergeometric function, and

$$C^2_\text{B4CC} = \frac{3}{4\pi^2}\Gamma^3\left(\tfrac{1}{3}\right) \approx 1.460998486206318358158873117846059697\dots$$

The first two identity are known, but with the last four relations I've never met before.

How could we prove these identities?

In this paper, there is another connection between a hypergeometric value and Baxter constant.

• Your third formula is the case $a=\frac13$ of this – nospoon Aug 21 '15 at 19:16

1. To prove the third identity, it suffices to take the limit $z\to -\frac12$ of the Ramanujan's cubic transformation. It yields $$_2F_1\left(\frac13,\frac23;1;-\frac18\right)=\lim_{z\to-\frac12^+} \frac{_2F_1\left(\frac13,\frac23;1;1-\left(\frac{1-z}{1+2z}\right)^3\right)}{1+2z}=\frac23\,C_{\mathrm{B4CC}}^2,\tag{1}$$ where the limit is evaluated using the connection formulae expressing $_2F_1(\ldots;z)$ as a combination of two $_2F_1(\ldots;z^{-1})$. Combining (1) with the well-known transformation $$_2F_1\left(a,b;c;z\right)=\left(1-z\right)^{-a}{}_2F_1\left(a,c-b;c;\frac{z}{z-1}\right),$$ we get $$_2F_1\left(\frac13,\frac13;1;\frac19\right)=\left(1-\frac19\right)^{-\frac13}{}_2F_1\left(\frac13,\frac23;1;-\frac18\right)=\frac{C_{\mathrm{B4CC}}^2}{\sqrt[3]{3}}.\tag{2}$$
2. The fourth identity immediately follows from (2) and the transformation $$_2F_1\left(a,b;c;z\right)=\left(1-z\right)^{c-a-b}{}_2F_1\left(c-a,c-b;c;z\right).\tag{3}$$
3. The fifth and sixth identity are meaningless in the present form: you have to specify the on which side of the branch cut $[1;\infty)$ we evaluate $_2F_1$. In any case the corresponding hypergeometric functions are related by the same transformation (3) - it suffices to correctly take into account the branching of $(1-z)^{c-a-b}$. Therefore only one of these identities is independent.