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

Consider the sequence:

$$a_0=x,~~~b_0=y$$

$$a_{n+1}=\frac{1}{2} \left(a_n+\sqrt{\frac{a_n^2+b_n^2}{2}} \right),~b_{n+1}=\frac{1}{2} \left(b_n+\sqrt{\frac{a_n^2+b_n^2}{2}}\right)$$

$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=l(x,y)$$

I can't pin down the pattern for the limit. Numerically, I've got the following result:

$$\frac{1}{l(x,y)}=\arctan (f(x,y))$$

What is the explicit expression for $f(x,y)$?

Numerical examples:

$$\begin{array}( x & y & \frac{1}{l(x,y)} \\ 1 & 2 & \arctan \left( \frac{3}{4} \right) \\ 1 & 3 & \arctan \left( \frac{1}{2} \right) \\ 2 & 3 & \arctan \left( \frac{5}{12} \right) \\ 3 & 5 & \arctan \left( \frac{1}{4} \right) \\ 1 & 5 & \arctan \left( \frac{\sqrt{13}-3}{2} \right) \\ 4 & 5 & \arctan \left( \frac{9}{40} \right) \\ 3 & 4 & \arctan \left( \frac{7}{24} \right) \\ 3 & 7 & \arctan \left( \frac{\sqrt{29}-5}{2} \right) \end{array}$$

Edit

To summarize @IvanNeretin's comment (and adding my own numerical results), so far we have:

$$l(x,0)=\frac{2x}{\pi}$$

$$f(x,x+1)=\frac{2x+1}{2x(x+1)}$$

$$f(x,x+2)=\frac{1}{x+1}$$

$f(x,x+3)$ - see my answer below

$$f(x,x+4)=\frac{\sqrt{x^2+4x+8}-(x+2)}{2}$$

Here $x \in \mathbb{R}^{+}$, $x \neq 0$

Also by symmetry we have of course:

$$f(x,y)=f(y,x)$$

At this point it's obvious that no ordinary function works here.

My hope is to connect $f(x,y)$ to some special function or an integral.

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 $$ l(1,2)=\frac{1}{2\tan^{-1}(1/3)}=\frac{1}{\tan^{-1}(3/4)}. $$

Geometrically:

$(a_{n+1}, b_{n+1})= \frac 12 (a_n,b_n) + \frac 12 \sqrt{a_n^2+b_n^2} (\cos \pi/4, \sin \pi/4)$

So we draw a line between $(a_n,b_n)$ and connect it to a point equidistant from the origin on the line $x = y$, and find the mid-point.

$r_n = \sqrt{a_n^2 + b_n^2}\\ \phi_n = \tan^{-1}(\frac{b_n}{a_n})$

$\phi_{n+1} = \frac{\frac{\pi}{4} + \phi_n}{2}\\ r_{n+1} = r_n\cos(\frac{\frac{\pi}{4} - \phi_n}{2})$