Limit of the sequence $a_{n+1}=\frac{1}{2} (a_n+\sqrt{\frac{a_n^2+b_n^2}{2}})$ - can't recognize the pattern

Let $z_i=a_i + ib_i$. Then your recursion is $$ z_{n+1}=a_{n+1}+ib_{n+1}=\frac{1}{2}(a_n+ib_n)+\frac{1}{2}(1+i)\sqrt{\frac{a_n^2+b_n^2}{2}}=\frac{1}{2}z_n+\frac{1+i}{2\sqrt{2}}|z_n|=\frac{1}{2}z_n+\frac{1}{2}e^{i\pi/4}|z_n|. $$ Now, let $z_n=e^{i(\theta_n+\pi/4)}y_n$, where the $y_n$ and $\theta_n$ are real. Then $$ y_{n+1}e^{i\theta_{n+1}}=\frac{1}{2}y_{n}e^{i\theta_{n}}+\frac{1}{2}y_n=\frac{1}{2}y_n(1+e^{i\theta_n}), $$ so $$ y_{n+1}=\frac{1}{2}y_n\sqrt{2+2\cos\theta_n}=y_n\cos\left(\theta_n/2\right) $$ and $$ \cos(\theta_{n+1})=\frac{y_n}{2y_{n+1}}(1+\cos\theta_n)=\frac{1+\cos\theta_n}{2\cos(\theta_n/2)}=\cos(\theta_n/2),$$ or $$ \theta_{n+1}=\frac{1}{2}\theta_n. $$ This just gives $\theta_{n}=2^{-n}\theta_{0}$, so $\theta_{\infty}=0$, and $$ y_{\infty}=y_0\prod_{i=0}^{\infty}\cos(2^{-i}\theta_0/2)=y_0\frac{\sin(\theta_0)}{\theta_0}. $$ The limit of the real and imaginary parts of $z_n$ is $$ l(x,y)=\frac{1}{\sqrt{2}}y_{\infty}=\frac{y_0}{\sqrt{2}}\frac{\sin\theta_0}{\theta_0}=\sqrt{\frac{x^2+y^2}{2}}\frac{\sin\theta_0}{\theta_0}, $$ where $$ \theta_0 = \tan^{-1}(y/x) - \frac{\pi}{4}=\tan^{-1}(y/x)-\tan^{-1}(1)=\tan^{-1}\left(\frac{y/x-1}{1+y/x}\right)=\tan^{-1}\left(\frac{y-x}{y+x}\right). $$ Then $$ l(x,y)=\frac{y-x}{2\tan^{-1}\left(\frac{y-x}{y+x}\right)}. $$ For $(x,y)=(1,2)$, for instance, we have $$ l(1,2)=\frac{1}{2\tan^{-1}(1/3)}=\frac{1}{\tan^{-1}(3/4)}. $$