# 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).

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A simple note: let $a_n=\frac{p_n}{q_n}$. Then $\displaystyle a_n=\frac{a_{n-1}^2+a_{n-2}^2}{a_{n-1}+a_{n-2}} = \frac{\frac{p_{n-1}^2}{q_{n-1}^2}+\frac{p_{n-2}^2}{q_{n-2}^2}}{\frac{p_{n-1}}{q_‌​{n-1}}+\frac{p_{n-2}}{q_{n-2}}} = \frac{p_{n-1}^2q_{n-2}^2+p_{n-2}^2q_{n-1}^2}{p_{n-1}q_{n-1}q_{n-2}^2+p_{n-2}q_{n‌​-2}q_{n-1}^2}$, so the original recurrence can be expressed as a pair of recurrences $p_n = p_{n-1}^2q_{n-2}^2+p_{n-2}^2q_{n-1}^2$ and $q_n=p_{n-1}q_{n-1}q_{n-2}^2+p_{n-2}q_{n-2}q_{n-1}^2$ for the numerators and denominators of the sequence. – Steven Stadnicki Nov 3 '12 at 8:24
The limit $1.7919405510508976479$ is not identified by the ISC isc.carma.newcastle.edu.au – GEdgar Nov 3 '12 at 16:20
Limit varies with $a_2$ as follows: (1, 1)(2, 1.7919)(3, 2.6879)(4,3.6226)(5,4.5775)(100, 99.3487) – Mark Bennet Nov 3 '12 at 21:32

## 2 Answers

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.

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[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: $$\ell(x) = x \; \ell \left(\frac {1+x^2}{x (1+x)}\right).$$ 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 $$\ell(\epsilon^{-1}) \sim_{\epsilon \to 0} \epsilon^{-1}, \qquad \ell(-1+\epsilon) \sim_{\epsilon \to 0} 2 \epsilon^{-1}.$$

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.

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