\[F\left(a,\tfrac{1}{2}+a;\tfrac{5}{6}+\tfrac{2}{3}a;\tfrac{1}{9}\right)=\sqrt{%
\pi}\left(\frac{3}{4}\right)^{a}\frac{\Gamma\left(\tfrac{5}{6}+\tfrac{2}{3}a%
\right)}{\Gamma\left(\tfrac{1}{2}+\tfrac{1}{3}a\right)\Gamma\left(\tfrac{5}{6}%
+\tfrac{1}{3}a\right)}.\]