While playing with something totally irrelevant I stumbled upon the recurrence:
$$a\_{n+1} = \frac{1}{a\_n} + a\_{n-1}$$


It turns out that given $a\_0 = 1, a\_1 = 1$,


$$lim \frac{a\_{2n}}{a\_{2n-1}} = \frac{\pi}{2}$$


I have a very crude idea (or rather a hint) on proving it (the iterations sort of unfold into a sort of Viete product, which is sort of expected), but my technique is rusty at best.


With different initial conditions, things start getting really scary, for example
$ a\_0 = 2, 3, 4, 5 $ yield $\frac{8}{\pi}, \frac{9\pi}{8}, \frac{128}{9\pi}, \frac{225\pi}{128}$ respectively.


So, the questions are:
Is it a known fact? If so, where can I read more on it? If not, may anybody help me to prove/disprove it? Does it mean anything?



"Harmonacci" recurrence and identities for $\pi$
The sequence $a\_n$ is closely related to the Wallis product
$$a'\_n = \prod\_{i = 1}^n \left(\frac{2i}{2i - 1} \frac{2i}{2i + 1}\right),$$
which converges to $\pi/2$ as $n$ goes to infinity. Namely, we have
$$a'\_n = a\_{n + 1} \cdot \frac{2n}{2n + 1}$$.
This could be proven by induction or maybe more easily by defining $b\_n = a\_n a\_{n - 1}$ and noticing that the recursion for $a\_n$ implies the (very simple) recursion
$$b\_{n + 1} = 1 + b\_n$$
for $b\_n$ and expressing $a\_n$ in terms of the $b\_n$.


For more general values of $a\_0$ one gets similar formulas for $a\_n$ as (up to a factor converging to 1) a Wallis product or inverse of a Wallis product where a few of the lower terms in the product are missing.


