Ratio of Gamma Functions

Question

Is it possible to show that: \begin{align} \frac{\Gamma\left(\frac{1}{78}\right) \Gamma\left(\frac{29}{78}\right) \Gamma\left(\frac{35}{78}\right) \Gamma\left(\frac{53}{78}\right) \Gamma\left(\frac{55}{78}\right) \Gamma\left(\frac{61}{78}\right)}{\Gamma\left(\frac{2}{78}\right) \Gamma\left(\frac{28}{78}\right) \Gamma\left(\frac{32}{78}\right) \Gamma\left(\frac{44}{78}\right) \Gamma\left(\frac{58}{78}\right) \Gamma\left(\frac{70}{78}\right) } = \sqrt{3} \end{align}

There are other known ratios of Gamma functions, but as always there seems to be a product rule or trick to evaluate the ratios.

Answer 1

This has been sitting around quite a while. This seems to rely on $$\operatorname{\Gamma}\left(z\right)\operatorname{\Gamma}\left(z+\frac13\right)\operatorname{\Gamma}\left(z+\frac23\right)=2\pi\frac{\sqrt3}{3^{3z}}\operatorname{\Gamma}(3z)$$ So the ratio is $$\begin{align}r&=\frac{\operatorname{\Gamma}\left(\frac1{78}\right)\operatorname{\Gamma}\left(\frac{29}{78}\right)\operatorname{\Gamma}\left(\frac{35}{78}\right)\operatorname{\Gamma}\left(\frac{53}{78}\right)\operatorname{\Gamma}\left(\frac{55}{78}\right)\operatorname{\Gamma}\left(\frac{61}{78}\right)}{\operatorname{\Gamma}\left(\frac2{78}\right)\operatorname{\Gamma}\left(\frac{28}{78}\right)\operatorname{\Gamma}\left(\frac{32}{78}\right)\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{58}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)}\\ &=\frac{\color{red}{\operatorname{\Gamma}\left(\frac1{78}\right)}\operatorname{\Gamma}\left(\frac{29}{78}\right)\operatorname{\Gamma}\left(\frac{35}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{53}{78}\right)}\operatorname{\Gamma}\left(\frac{55}{78}\right)\operatorname{\Gamma}\left(\frac{61}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{27}{78}\right)}}{\operatorname{\Gamma}\left(\frac2{78}\right)\operatorname{\Gamma}\left(\frac{28}{78}\right)\operatorname{\Gamma}\left(\frac{32}{78}\right)\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{58}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)\operatorname{\Gamma}\left(\frac{27}{78}\right)}\\ &=\frac{2\pi\sqrt3\color{red}{\operatorname{\Gamma}\left(\frac{29}{78}\right)}\operatorname{\Gamma}\left(\frac{35}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{55}{78}\right)}\operatorname{\Gamma}\left(\frac{61}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{3}{78}\right)}}{3^{3/78}\operatorname{\Gamma}\left(\frac2{78}\right)\operatorname{\Gamma}\left(\frac{28}{78}\right)\operatorname{\Gamma}\left(\frac{32}{78}\right)\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{58}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)\operatorname{\Gamma}\left(\frac{27}{78}\right)}\\ &=\frac{12\pi^2\color{red}{\operatorname{\Gamma}\left(\frac{35}{78}\right)\operatorname{\Gamma}\left(\frac{61}{78}\right)\operatorname{\Gamma}\left(\frac9{78}\right)}}{3^{12/78}\operatorname{\Gamma}\left(\frac2{78}\right)\operatorname{\Gamma}\left(\frac{28}{78}\right)\operatorname{\Gamma}\left(\frac{32}{78}\right)\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{58}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{27}{78}\right)}}\\ &=\frac{24\pi^3\sqrt3}{3^{39/78}\operatorname{\Gamma}\left(\frac2{78}\right)\operatorname{\Gamma}\left(\frac{28}{78}\right)\operatorname{\Gamma}\left(\frac{32}{78}\right)\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{58}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)}\\ &=\frac{24\pi^3\sqrt3\operatorname{\Gamma}\left(\frac{54}{78}\right)}{3^{39/78}\color{red}{\operatorname{\Gamma}\left(\frac2{78}\right)\operatorname{\Gamma}\left(\frac{28}{78}\right)}\operatorname{\Gamma}\left(\frac{32}{78}\right)\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{58}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{54}{78}\right)}}\\ &=\frac{12\pi^2\operatorname{\Gamma}\left(\frac{54}{78}\right)}{3^{33/78}\color{red}{\operatorname{\Gamma}\left(\frac{32}{78}\right)}\operatorname{\Gamma}\left(\frac{44}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{58}{78}\right)}\operatorname{\Gamma}\left(\frac{70}{78}\right)\color{red}{\operatorname{\Gamma}\left(\frac{6}{78}\right)}}\\ &=\frac{2\pi\sqrt3\color{red}{\operatorname{\Gamma}\left(\frac{54}{78}\right)}}{3^{15/78}\color{red}{\operatorname{\Gamma}\left(\frac{44}{78}\right)\operatorname{\Gamma}\left(\frac{70}{78}\right)\operatorname{\Gamma}\left(\frac{18}{78}\right)}}\\ &=3^{39/78}=\sqrt3\end{align}$$ Beautiful to see those dominos fall like that!

Answer 2

One method to find such formulas is the duplication formula $$ \Gamma(z) \Gamma(z+1/2) = 2^{1-2z} \sqrt{\pi} \Gamma(2z), $$ and apply it to nominator and denominator, e.g., to obtain $$ {\Gamma(1/8) \Gamma(5/8) \Gamma(6/8) \over \Gamma(2/8) \Gamma(3/8) \Gamma(7/8)} = \sqrt{2}. $$ There seem to be more advanced ideas in Deligne, P. Valeurs de fonctions L et périodes d'intégrales, Amer. Math. Soc., Providence, R.I., 1979.