Entry 132

Given \(_2F_1(a,b;c;z)\) and Dedekind eta function \(\eta(\tau)\) where \(\tau = \frac{1+n\sqrt{-3}}2\) for positive integer \(n\). Then for type \(a+b=c=\color{blue}{\tfrac56}\) $$\begin{align}&\,_2F_1\big(\tfrac12,\tfrac13;\tfrac56;(1-2\delta_1)^2\big),\quad\;\delta_1 = \delta_2\\ &\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;(1-2\delta_2)^2\big),\quad\;\frac1{\delta_2}-1=\sqrt{\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}}\\ &\,_2F_1\big(\tfrac14,\tfrac7{12};\tfrac56;(1-2\delta_3)^2\big),\quad\color{red}{\delta_3 =\,?} \\ &\,_2F_1\big(\tfrac16,\tfrac23;\tfrac56;(1-2\delta_4)^2\big),\quad\;\frac1{\delta_4}-1\,=\,\frac1{27}\left(\tfrac{\eta\big(\frac{\tau+1}{3}\big)}{\eta(\tau)}\right)^{12}\end{align}$$ 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. Note that \( _2F_1\big(\tfrac12,\tfrac13;\tfrac56;z\big) =\,_2F_1\big(\tfrac13,\tfrac12;\tfrac56;z\big)\) so the first form is superfluous. Examples: Let \(\tau = \frac{1+5\sqrt{-3}}2\), $$_2F_1\Big(\frac13,\frac12;\frac56;\frac45\Big)=\;\frac35\sqrt5$$ $$\quad _2F_1\Big(\frac16,\frac23;\frac56;\frac{80}{81}\Big)=\frac35\,(9\sqrt5)^{1/3}$$

Entry 131

Given \(_2F_1(a,b;c;z)\) and Dedekind eta function \(\eta(\tau)\) where \(\tau = \frac{1+n\sqrt{-1}}2\) for positive integer \(n\). Then for type \(a+b=c=\color{blue}{\tfrac34}\) $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac12;\tfrac34;(1-2\gamma_1)^2\big), \quad\frac1{\gamma_1}-1=\sqrt{-\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}}\\ &\,_2F_1\big(\tfrac16,\tfrac7{12};\tfrac34;(1-2\gamma_2)^2\big),\quad\color{red}{\gamma_2 =\,?} \\ &\,_2F_1\big(\tfrac18,\tfrac58;\tfrac34;(1-2\gamma_3)^2\big),\quad\frac1{\gamma_3}-1\,=\, -\frac1{64}\Big(\tfrac{\sqrt2\,\eta(2\tau)}{\eta(\tau)}\Big)^{24}\\ &\,_2F_1\big(\tfrac1{12},\tfrac23;\tfrac34;(1-2\gamma_4)^2\big),\quad\color{red}{\gamma_4 =\,?}\end{align}$$ 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 \(\tau = \frac{1+5\sqrt{-1}}2\), 

$$_2F_1\Big(\frac14,\frac12;\frac34;\frac{80}{81}\Big)=\frac95$$

$$\quad _2F_1\Big(\frac18,\frac58;\frac34;\frac{25920}{25921}\Big)=\frac35\,161^{1/4}$$

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 \(\tau = \frac{1+5\sqrt{-3}}2\), $$_2F_1\left(\frac16,\frac12;\frac23;\Big(\frac45\Big)^2\Big(\frac{15-\sqrt5}{11}\Big)^3\right) =\frac35(5+4\sqrt5)^{1/6}$$

Entry 129

Given \(_2F_1(a,b;c;z)\) where \(a+b=c\). The previous four (Entries 125-128) discuss closed-forms and can be neatly summarized as type \(a+b=c=\color{blue}{\tfrac12}\)  $$\begin{align}&\,_2F_1\big(\tfrac14,\tfrac14;\tfrac12;(1-2\alpha_1)^2\big),\qquad \frac1{\alpha_1}-1=\frac1{16}\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8,\qquad \tau_1=n\sqrt{-4}\\ &\,_2F_1\big(\tfrac16,\tfrac13;\tfrac12;(1-2\alpha_2)^2\big),\qquad \frac1{\alpha_2}-1=\frac1{27}\Big(\tfrac{\eta(\tau/3)}{\eta(\tau)}\Big)^{12},\qquad \tau_2=n\sqrt{-3}\\ &\,_2F_1\big(\tfrac18,\tfrac38;\tfrac12;(1-2\alpha_3)^2\big),\qquad \frac1{\alpha_3}-1=\frac1{64}\Big(\tfrac{\eta(\tau/2)}{\eta(\tau)}\Big)^{24},\qquad \tau_3=n\sqrt{-2}\\ &\,_2F_1\big(\tfrac1{12},\tfrac5{12};\tfrac12;(1-2\alpha_4)^2\big),\quad\; \frac{1}{\alpha_4(1-\alpha_4)}=\frac1{432}\,j(\tau),\qquad\tau_4=n\sqrt{-1}\end{align}$$ 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. However, for the parameters \(c=\frac12, \frac23,\frac34. \frac56\), it seems only the first is easy. For type \(a+b=c=\color{blue}{\tfrac23}\), it's more challenging and will be discussed in the next entry.

Entry 128

Assume \(\tau=n\sqrt{-\color{blue}{4}}\) for some positive integer \(n\). Given \(_2F_1(a,b;c;z)\) where \(a+b=c=\frac12\) for the case \(a=\tfrac1{4}\). Let \(z_1 = (1-2w)^2\) where \(w\) is $$w=\frac{16}{16+\Big(\tfrac{\eta(\tau/4)}{\eta(\tau)}\Big)^8}$$ Then \((z_1, z_2)\) are algebraic numbers in

$$_2F_1\left(\tfrac14,\tfrac14;\tfrac12;z_1\right) = z_2$$

Examples: 

If \(n=2\) so \(\tau=2\sqrt{-4}\), then,

$$_2F_1\left(\frac14,\frac14;\frac12;\,\frac{9\,(3+\sqrt2)^2}{(1+\sqrt2)^6}\right)=\frac38\big(2+\sqrt2\big)$$ 

If \(n=3\) so \(\tau=3\sqrt{-4}\), then,

$$\quad _2F_1\left(\frac14,\frac14;\frac12;\,\frac{8\sqrt3}{(2+\sqrt3)^2}\right)=\frac23\big(3+2\sqrt3\big)^{1/2}$$