I'm looking for a good closed form for $\Gamma(n+\frac{2}{3})$ involving at most $\Gamma(\frac{1}{3})$.
I know that $\Gamma(n+\frac{1}{3})=\Gamma(\frac{1}{3})\frac{(3n-2)!!!}{3^n}$ and my question is, is there an analogous for $\Gamma(n+\frac{2}{3})$?
I tried using WolframAlpha and ended up finding out that:
$$ \Gamma\left(\frac{2}{3}\right)=\frac{2\pi}{\sqrt{3}\Gamma\left(\frac{1}{3}\right)} $$ $$ \Gamma\left(\frac{5}{3}\right)=\frac{4\pi}{3\sqrt{3}\Gamma\left(\frac{1}{3}\right)} $$ $$ \Gamma\left(\frac{8}{3}\right)=\frac{20\pi}{9\sqrt{3}\Gamma\left(\frac{1}{3}\right)} $$ $$...$$ and noticed the possible pattern: $$ \Gamma\left(n+\frac{2}{3}\right)=\frac{2t_n\pi}{3^n\sqrt{3}\Gamma\left(\frac{1}{3}\right)} $$ where $t_n=\prod_{k=0}^{n-1}\left(3k+2\right)$ is OEIS A008544.
Is this true? How can we prove it?