Entry 130

Given \(_2F_1(a,b;c;z)\) and j-function \(j = j(\tau)\) where \(\tau = \frac{1+n\sqrt{-3}}2\) for positive integer \(n\). Then for type \(a+b=c=\color{blue}{\tfrac23}\) $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac5{12};\tfrac23;(1-2\beta_1)^2\big),\qquad \color{red}{\beta_1 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac12;\tfrac23;(1-2\beta_2)^2\big),\qquad \frac{1}{\beta_2}-1=\sqrt{\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}}\\ &\,_2F_1\big(\tfrac18,\tfrac{13}{24};\tfrac23;(1-2\beta_3)^2\big),\qquad \color{red}{\beta_3 =\,?} \\ &\,_2F_1\big(\tfrac1{12},\tfrac7{12};\tfrac23;(1-2\beta_4)^2\big),\qquad \frac{1}{\beta_4}-1=\frac{-2j+1728-2\sqrt{j(j-1728)}}{1728}\\\end{align}$$ For this type, there are infinitely many hypergeometrics such that both \((z_1, z_2)\) in $$_2F_1(a,b;c;z_1) = z_2$$ are algebraic numbers when \(n\) is a positive integer. Examples: Let \(\tau = \frac{1+3\sqrt{-3}}2\), $$_2F_1\Big(\frac16,\frac12;\frac23;\frac{125}{128}\Big) =\frac43\times2^{1/6}$$ $$_2F_1\Big(\frac1{12},\frac7{12};\frac23;\frac{64000}{64009}\Big) =\frac23\times253^{1/6}$$ Let \(\tau = \frac{1+5\sqrt{-3}}2\), $$_2F_1\left(\frac16,\frac12;\frac23;\Big(\frac45\Big)^2\Big(\frac{15-\sqrt5}{11}\Big)^3\right) =\frac35(5+4\sqrt5)^{1/6}$$