Does anyone know a closed expression for the following continued fraction? $$G(x) = \cfrac{1}{x+1+\cfrac{1}{x+3+\cfrac{4}{x+5+\cfrac{9}{x+7+\cfrac{16}{x+9+\cdots}}}}}$$ All I know is that $G(0) = \frac{\pi}{4}$ and $G(x)$ converges for $x \geq 0$, while also $G(x) \sim \frac{1}{x} \ (x \to \infty)$.
The equation $G(0) = \frac{\pi}{4}$ follows from a well-known continued fraction expansion for $\arctan$. However, the expansion above is different. It came up in my research, and I can't find it in any continued fraction tables.
The form of the coefficients of the proposed continued fraction is rather similar to that of the ratio of two hypergeometric functions tabulated here: \begin{equation} \frac{\mathbf{F}\left(a,b;c;z\right)}{\mathbf{F}\left(a+1,b+1;c+1;z\right)}={x _{0}+\cfrac{y_{1}}{x_{1}+\cfrac{y_{2}}{x_{2}+\cfrac{y_{3}}{x_{3}+\cdots}}}} \end{equation} where \begin{align} x_{n}&=c+n-(a+b+2n+1)z\\ y_{n}&=(a+n)(b+n)z(1-z) \end{align} and $\mathbf{F}$ are regularized hypergeometric functions.
In the following, the continued fraction is adapted to this representation by finding a parameter $\alpha$ to fit the above formula: \begin{align} G(x) &= \cfrac{1}{x+1+\cfrac{1}{x+3+\cfrac{4}{x+5+\cfrac{9}{x+7+\cfrac{16}{x+9+\cdots}}}}}\\ \frac{\alpha}{G(x)}&=\alpha(x+1)+\cfrac{1^2\alpha^2}{\alpha(x+3)+\cfrac{2^2\alpha^2}{\alpha(x+5)+\cfrac{3^2\alpha^2}{\alpha(x+7)+\cfrac{4^2\alpha^2}{\alpha(x+9)+\cdots}}}} \end{align} then we have \begin{align} x_n&=\alpha(x+2n+1)\\ y_n&=n^2\alpha^2 \end{align} From the $y_n$ definition, we must have $a=b=0$ and $z(1-z)=\alpha^2$. Then, from $x_n$, we find $c=1/2+\alpha x$ and $\alpha=1/2-z$. The compatibiliy condition for the parameter $z$ are \begin{align} \alpha^2&=z(1-z)\\ \alpha&=\frac12-z \end{align} and thus $z=(1-1/\sqrt{2})/2$ and $\alpha=2^{-3/2}$ which imposes $c=(1+x/\sqrt{2})/2$. (The root corresponding to $\left|z\right|<1$ was chosen to avoid possible divergence of the hypergeometric functions). With these parameters, \begin{align} \frac{2^{-3/2}}{G(x)}&=\frac{\mathbf{F}\left(0,0;(1+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}{\mathbf{F}\left(1,1;(3+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}\\ &=\frac{(1+x/\sqrt{2})/2}{{}_2F_1\left(1,1;(3+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)} \end{align} Finally, \begin{equation} G(x)=\frac{{}_2F_1\left(1,1;(3+x/\sqrt{2})/2;(1-1/\sqrt{2})/2\right)}{x+\sqrt{2}} \end{equation} From $\sin \pi/8=\sqrt{2-\sqrt{2}}/2$, it comes \begin{equation} G(x)=\frac{{}_2F_1\left(1,1;(3+x/\sqrt{2})/2;\sin^2\pi/8\right)}{x+\sqrt{2}} \end{equation} which seems to be numerically correct.
For $x=0$, we have from here \begin{align} {}_2F_1\left(1,1;3/2;s\right)=\frac{\sin^{-1}\sqrt{s}}{\sqrt{1-s}\sqrt{s}} \end{align} and thus $G(0)=\pi/4$ as expected. Other special cases can be deduced for $x=p\sqrt{2}$ or $x=p/\sqrt2$ where $p$ is an integer, as sevral explicit expressions for the hypergeometric function ${}_2F_1(1,1;c;s)$ are known. For example \begin{equation} G(\sqrt{2})=\frac{1+\sqrt{2}}{2}\ln\left( \frac{8}{3+2\sqrt{2}} \right) \end{equation}