A remarkable Continued Fraction for $\pi$ How to analyze the Continued Fraction $3+\cfrac{\color{red}1^2}{5+\cfrac{\color{blue}4^2}{7+\cfrac{\color{red}3^2}{9+\cfrac{\color{blue}6^2}{11+\cfrac{\color{red}5^2}{13+\cfrac{\color{blue}8^2}{15+\cfrac{\color{red}7^2}{17+\ddots}}}}}}}$
Numerically, it seems to evaluate to $\pi$. But why?
This remarkable Continued Fraction was sent to me by Paul Levrie.
The pattern in the nominators and denominators is obvious. Alternating $\color{red}3,\color{red}5,\color{red}7,\color{red}9,\color{red}{11}$ etc and $\color{blue}4^2,\color{blue}6^2,\color{blue}8^2,\color{blue}{10}^2$ etc.
Because of the appearance of squares, Euler's Differential Method can not be used. A clever trick is required here! Any idea's?

The difference between the even and odd squared coefficients is similar to that of the continued fraction of the ratio of two hypergeometric functions (DLMF):
\begin{equation}
 \frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a,b+1;c+1;z\right)}=t_{0
}-\cfrac{u_{1}z}{t_{1}-\cfrac{u_{2}z}{t_{2}-\cfrac{u_{3}z}{t_{3}-\cdots}}}
\end{equation}
where
\begin{align}
 t_n&=c+n\\
 u_{2n+1}&=(a+n)(c-b+n)\\
 u_{2n}&=(b+n)(c-a+n)
\end{align}
where $\mathbf{F}$ are regularized hypergeometric functions.
The continued fraction is adapted to this representation
\begin{align}
 \mathcal J=&3+\cfrac{1^2}{5+\cfrac{4^2}{7+\cfrac{3^2}{9+\cfrac{6^2}{11+\cdots}}}}\\
 \frac12 \mathcal J&=3/2+\cfrac{1^2/4}{5/2+\cfrac{4^2/4}{7/2+\cfrac{3^2/4}{9/2+\cfrac{6^2/4}{11/2+\cdots}}}}
\end{align}
It can be checked that $a=1/2,b=1,c=3/2,z=-1$ satisfies the recurrence relations, then
\begin{align}
 \mathcal J&=\frac{2\,\mathbf{F}\left(1/2,1;3/2;-1\right)}{\mathbf{F}\left(1/2,2;5/2;-1\right)}\\
 &=\frac{2\Gamma(5/2)}{\Gamma(3/2)}\,\frac{{}_2F_1\left(1/2,1;3/2;-1\right)}{{}_2F_1
\left(1/2,2;5/2;-1\right)}\\
&=3\,\frac{{}_2F_1\left(1/2,1;3/2;-1\right)}{{}_2F_1
\left(1/2,2;5/2;-1\right)}
\end{align}
From here and here
\begin{align}
 {}_2F_1\left(1/2,1;3/2;-z\right)&=\frac{\tan^{-1}\sqrt{z}}{\sqrt{z}}\\
 {}_2F_1\left(1/2,2;5/2;-z\right)&=\frac34\frac{(z-1)\tan^{-1}\sqrt{z}+\sqrt{z}}{z^{3/2}}
\end{align}
then
\begin{equation}
 \mathcal J=\pi
\end{equation}
