A nice pattern for the regularized beta function $I(\alpha^2,\frac{1}{4},\frac{1}{2})=\frac{1}{2^n}\ $? In this post, the problem was given integer/rational $N$, to solve for algebraic number $z$ in the equation,
$$\begin{aligned}\frac{1}{N}
&=I\left(z^2;\ a,b\right)\\[1.5mm]
&= \frac{B\left(z^2;\ a,b\right)}{B\left(a,b\right)}
\end{aligned} $$
using the beta function $B(a,b)$, incomplete beta $B(z;a,b)$ and regularized beta $I(z;a,b)$. It seems for some $a,b,$ there can be a pattern for the solutions. 
Given the equation for $n>1$,

$$I\left(x_1^2;\ \tfrac14,\tfrac12\right)=\frac{1}{2^n}\tag1$$

First define $\color{blue}{\gamma = u+\sqrt{-1+u^2}},$ with fundamental unit $u=1+\sqrt{2}.\ $ Then,
For $n=2$:
$$x_1=x_2-\sqrt{1+x_2^2},\quad \color{blue}{x_2 =\gamma}$$
For $n=3$:
$$x_1=x_2-\sqrt{1+x_2^2},\quad \color{green}{x_2 = x_3+\sqrt{-1+x_3^2},\quad x_3=x_4+\sqrt{1+x_4^2},}\quad \color{blue}{x_4 = \gamma}$$
For $n=4$:
$$x_1=x_2-\sqrt{1+x_2^2},\quad x_2 = x_3+\sqrt{-1+x_3^2},\quad x_3=x_4+\sqrt{1+x_4^2},\\ \quad \color{green}{x_4 = x_5+\sqrt{-1+x_5^2},\quad x_5=x_6+\sqrt{1+x_6^2},}\quad \color{blue}{x_6 = \gamma}$$
and so on, where we add the same two nested layers (in green) each time and so end with even index $x_m$.

Questions:



*

*Does this pattern really hold for all $2^n$ and $n>1?$ Why the regularity?

*What is the integral associated with $(1)$ similar to the ones in the post cited above?


We have
$$
B\left(z;\ \tfrac14,\tfrac12\right)=4\sqrt[4]{z}\, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};z\right).
$$
By formula 2.1.15 from Erdelyi, "Higher transcendental functions", vol.I
$$
_2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};z\right)=\sqrt{\frac{1}{1-z}} \, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};-\frac{4 z}{(1-z)^2}\right).
$$
Since $z$ and $-\frac{4 z}{(1-z)^2}$ have different signs when $z$ is real we need to apply this formula one more time
$$
\, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};z\right)=\frac{1}{2}\sqrt[4]{\frac{16 (1-z)^2}{(z+1)^4}} \, _2F_1\left(\frac{1}{2},\frac{1}{4};\frac{5}{4};\frac{16 z (1-z)^2}{(z+1)^4}\right).
$$
In terms of incomplete Beta function one gets
$$
B\left(z;\ \tfrac14,\tfrac12\right)=\frac12 B\left(\tfrac{16 z (1-z)^2}{(z+1)^4};\ \tfrac14,\tfrac12\right).
$$
I think this formula answers the question 1. From this formula one can work out the recursion for the argument.
