Given a complex number $\begin{aligned}\frac{z}{n}=x+iy\end{aligned}$ and a gamma function $\Gamma(z)$ with $x\gt0$, it is conjectured that the following continued fraction for $\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)$ is true
$$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=\frac{\displaystyle\Gamma\left(\frac{z}{4z+2n}\right)\Gamma\left(\frac{3z+2n}{4z+2n}\right)}{\displaystyle\Gamma\left(\frac{z+n}{4z+2n}\right)\Gamma\left(\frac{3z+n}{4z+2n}\right)}=\cfrac{2z+2n}{2z+n+\cfrac{(0z-n)(4z+3n)} {3(2z+n)+\cfrac{(2z+0n)(6z+4n)}{5(2z+n)+\cfrac{(4z+n)(8z+5n)}{7(2z+n)+\cfrac{(6z+2n)(10z+6n)}{9(2z+n)+\ddots}}}}}\end{aligned}$$ Or in gauss's notation $$\begin{aligned}\displaystyle\cot\left(\frac{z\pi}{4z+2n}\right)=-\frac{1}{2z+2n}\underset{m=0}{\overset{\infty}{\mathbf K}}\frac{((2m-2)z+(m-2)n)((2m+2)z+(m+2)n)}{((2m+1)(2z+n)}\end{aligned}$$
Corollaries:
1):let $z=1$ and $n=2$,then we obtain a beautiful continued fraction for square root 2
$$\begin{aligned}{-1+\cfrac{3}{2+\cfrac{\frac{(-1)(5)}{(1)(3)}} {2+\cfrac{\frac{(1)(7)}{(3)(5)}}{2+\cfrac{\frac{(3)(9)}{(5)(7)}}{2+\cfrac{\frac{(5)(11)}{(7)(9)}}{2+\ddots}}}}}}=\sqrt{2}\end{aligned}$$
2):However the most interesting case(for me at least),occurs when we take the limit to zero
$$\begin{aligned}\lim_{z\to0} \cfrac{z(z+1)}{2z+1+\cfrac{(0z-1)(4z+3)} {3(2z+1)+\cfrac{(2z+0)(6z+4)}{5(2z+1)+\cfrac{(4z+1)(8z+5)}{7(2z+1)+\cfrac{(6z+2)(10z+6)}{9(2z+1)+\ddots}}}}}=\frac{1}{\pi}\end{aligned}$$ yielding a new limit for $\pi$ from which one obtains the first few convergents $\begin{aligned}0,\frac{3}{8},\underline{\frac{5}{16},\frac{15}{47},\frac{7}{22}},\frac{1365}{4288},\frac{3015}{9472},\frac{1575}{4948},\ddots\end{aligned}$.
Where the underlined convergents appear in the stern-brocot tree for $\pi $ associated to its simple continued fraction.
Q: Is the conjectured continued fraction true (for all complex numbers $z$ with $x\gt0$)?
Update:I initially defined the continued fraction $\displaystyle\cot\left(\frac{z\pi}{4z+2}\right)$ for only natural numbers,but as a matter of fact it holds for all complex numbers $z$ with real part greater than zero.This continued fraction is one special case of the general continued fraction found here.