<s>[INST] Prove ${_2F_1}\left({{\tfrac16,\tfrac23}\atop{\tfrac56}}\middle|\,\frac{80}{81}\right)=\frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$ I've found the following hypergeometric function value by numerical observation. The identity matches at least for $100$ digits.
$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$
Or using a Pfaff transformation in an equivalent form
$$81^{1/6} \cdot {_2F_1}\left(\begin{array}c\tfrac16,\tfrac16\\\tfrac56\end{array}\middle|\,-80\right) \stackrel{?}{=} \frac 35 \cdot 5^{1/6} \cdot 3^{2/3}$$
How could we prove it?

Other related problem: How could we prove that
$${_2F_1}\left(\begin{array}c\tfrac16,\tfrac23\\\tfrac56\end{array}\middle|\,\frac{80}{81}\right) \stackrel{?}{=} {_2F_1}\left(\begin{array}c\tfrac12,\tfrac56\\\tfrac12\end{array}\middle|\,\frac{4}{9}\right) = {_1F_0}\left(\begin{array}c\tfrac56\\\ - \,\end{array}\middle|\,\frac{4}{9}\right)$$
 [/INST] The Tito Piezas answer is related to this parameterization for
${}_2F_1(\frac{1}{6},\frac{2}{3};\frac{5}{6};z)$ in terms of modular forms:
$$
{}_2F_1\left(\frac{1}{6},\frac{2}{3};\frac{5}{6};
\frac{j_{3A}}{j_{3A}-108}\right)
= \eta(\tau)\; \eta(3\tau)\;(j_{3A}-108)^{1/6}\;
\Gamma\left(\frac{2}{3}\right)^3
\frac{(-6i\tau+\sqrt{3}-3i)}{2^{7/3}\pi}
\tag{1}$$
where $\eta$ is the Dedekind eta function and $j_{3A}$ is
$$
j_{3A} = \left(\frac{\eta(\tau)^6}{\eta(3\tau)^6}+
\frac{3^3 \eta(3\tau)^6}{\eta(\tau)^6}\right)^2
$$
(Choosing the proper branches of the sixth root and the ${}_2F_1$).
Take $\tau = \frac{1}{2} + i \frac{5}{6}\sqrt{3}$ to get the required result from special values of $\eta$...
$$
\eta\left(\frac{1}{2} + i \frac{5}{6}\sqrt{3}\right)
=2^{-3/2} 3^{3/8} 5^{-5/12} (1+\sqrt5)^{1/2} \pi^{-1}
\Gamma\left(\frac{1}{3}\right)^{3/2} (-1)^{1/24}
$$
and something similar for $\eta(3\tau)$, so
$$
j_{3A}(\tau) = -8640,\qquad
\frac{j_{3A}(\tau)}{j_{3A}(\tau)-108} = \frac{80}{81}
$$
In $(1)$, the product of $\Gamma(1/3)$ and $\Gamma(2/3)$ simplify to something algebraic together with a $\pi$, and the powers of $\pi$ cancel.
</s>