Solving the recurrence relation $a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}}$ I would like to know if there is a way to get the recurrence relation
$$a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}},\qquad (a_1=1,a_2=2)$$
in closed form, or if there is no such way, how one could proceed to find the limit of $(a_n)$ to by some sort of estimation (or some other method I don't know about).
 A: The sequence of ratios $r_n=a_{n+1}/a_n$ is positive and such that $r_{n+1}-1=-\frac{r_n-1}{r_n(1+r_n)}$, hence $r_n-1$ is alternatively positive and negative. Since $r_1=a_2/a_1=2\gt1$, one gets $a_{2n}\gt a_{2n+1}$ and $a_{2n-1}\lt a_{2n}$ for every $n\geqslant1$. Likewise, one can show that $(a_{2n})_{n\geqslant1}$ is decreasing while $(a_{2n-1})_{n\geqslant1}$ is increasing. Finally, $r_n\to1$ hence $(a_{2n})_{n\geqslant1}$ and $(a_{2n-1})_{n\geqslant1}$ converge to the same limit $\ell$ and this limit is such that $a_{2n-1}\lt\ell\lt a_{2n}$ for every $n\geqslant1$ (this holds for every initial conditions $0\lt a_1\lt a_2$, otherwise one should exchange the odd numbered and the even numbered terms).
Until somebody finds a way to compute analytically the limit $\ell$ of $(a_n)_{n\geqslant1}$, one should rely on numerical approximations, based on the inequalities  $a_{2n-1}\lt\ell\lt a_{2n}$ mentioned above. These show in particular that  $\ell\gt a_3=\frac53$, hence $\ell$ is neither $\frac32$ the arithmetic mean of $a_1=1$ and $a_2=2$, nor $\frac43$ its harmonic mean, nor $\sqrt2$ its geometric mean.
A: [Not a complete answer yet.  I will come back and edit later if I find the complete solution.]
This problem is similar to the arithmetic-geometric mean.  Denote by $\ell(a_0,a_1)$ the limit of the sequence $a_0, a_1, \dotsc$.  Then, $\ell(a_0,a_1) = \ell(a_1,a_2)$.  We also have that $\ell(r a_0, r a_1) = r \ell(a_0,a_1)$.  Then if we define $\ell(x) = \ell(1,x)$ we obtain the following functional equation:
\begin{equation}
  \ell(x) = x \; \ell \left(\frac {1+x^2}{x (1+x)}\right).
\end{equation}  This only works if $a_0 \neq 0$, but the case $a_0=0$ is very simple.  We define $g(x) = \tfrac {1+x^2}{x(1+x)} = 1 + \tfrac 1 x - \tfrac 2 {1+x}$.  Then the functional equation becomes $\ell(x) = x \ell(g(x))$
Now we can study the singularities of $\ell$.  First of all, $\ell(0) = 1$ because in that case we can take the sequence $a_0, a_1=0$ and then the rest of the terms are equal to $a_0$.  Similarly we have $\ell(1)=1$.  Setting $x = \epsilon$ and $x = -1+\epsilon$ we learn that
\begin{equation}
  \ell(\epsilon^{-1}) \sim_{\epsilon \to 0} \epsilon^{-1}, \qquad
  \ell(-1+\epsilon) \sim_{\epsilon \to 0} 2 \epsilon^{-1}.
\end{equation}
Usually such problems are solved by finding a differential equation and solving it, but I haven't managed to find such a differential equation yet.
