Take two real numbers $x$ and $y$ greater than one. Set $x_0=x$ and $y_0=y$, and iterate the following: $$x_{n+1}=e^{W(y_n\text{ln}(x_n))}=\text{ssrt}(x_n^{y_n}),y_{n+1}=e^{W(x_n\text{ln}(y_n))}=\text{ssrt}(y_n^{x_n})$$ Where $W$ is the Lambert $W$ function and ssrt is the square super-root. Let $P(x,y) = \lim_{n\to\infty}x_n = \lim_{n\to\infty}y_n$. Is there a closed form for $P$? This seems quite similar to the arithmetic-geometric mean AGM$(x,y)$, which does have a integral-form expression, so I'd believe there's a chance for $P$ to as well. I've tried a little bit of numerical computation, but not much hints were found:
\begin{array}{|c|c|c|c|} \hline \text{ }\text{ }\text{ }_y \text{ }^x& 2& 3 & 4 & 5 & 6 & 7 \\ \hline 2 &2 &2.419365 &2.745368 &3.021021&3.263696&3.482585\\ \hline 3 & &3 &3.451185&3.832902&4.169207&4.472786\\ \hline 4 & & &4&4.464683&4.874384&5.244457\\ \hline 5 & & &&5&5.472265&5.899077\\ \hline 6 & & &&&6&6.477148\\ \hline 7 & & &&&&7\\ \hline \end{array}
It's clear that $P(x,x)=x$ and $P(x,y)=P(y,x)$, but I did find one more identity: $$P\Bigg(x,-\frac{xW_n\Big(-\frac{\text{ln}(x)}{x}\Big)}{\text{ln}(x)}\Bigg)=e^{W\Big(-xW_n\big(-\frac{\text{ln}(x)}{x}\big)\Big)}$$ This is since $-\frac{xW_n(-{\text{ln}(x)}/{x})}{\text{ln}(x)}$ satisfies $x^y=y^x$.