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?
