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}
