\[F\left(a,\tfrac{1}{2}+a;\tfrac{3}{2}-2a;-\tfrac{1}{3}\right)=\left(\frac{8}{9}%
\right)^{-2a}\frac{\Gamma\left(\tfrac{4}{3}\right)\Gamma\left(\tfrac{3}{2}-2a%
\right)}{\Gamma\left(\tfrac{3}{2}\right)\Gamma\left(\tfrac{4}{3}-2a\right)}.\]