Recursion equation $a_{n+2}= a_{n+1} + \frac{1}{a_n}$

Question

Today I was asked by some of my students, whether the following recursion equation admits a closed form solution

$$ a_{n+2} = a_{n+1} + \frac{1}{a_n}, $$

where $a_1, a_2 >0$. Their motivation is the following, they know the Fibonacci sequence and wondered, whether we can replace $a_n$ by $\frac{1}{a_n}$ and still solve it.

By considering the asymptotics, we can exclude that it is a rational function or an exponential function. Something similar to the square root would yield the correct asymptotics, but I am not familiar with the techniques to solve this kind of equation. Any pointer in the right direction would be appreciated (also on how to show that no closed form solution exists).

Added: As pointed out in the comments there is a very relevant post that I missed Any insight on the half reciprocal Fibonacci sequence?

They do not give a closed form, but a pretty good asymptotics. There was also this interesting link http://oeis.org/A296397

If only they had inverted that $a_{n+1}$ term instead, then this would be easy! — Theo Bendit, Dec 01 '22 at 01:20
@TheoBendit Can you elaborate why it would be easier? I am not familiar with these kind of questions. — Severin Schraven, Dec 01 '22 at 01:24
I'm not that familiar either; it's an ad-hoc trick. If you multiply both sides by $a_{n+1}$, you get $a_{n+2}a_{n+1} = 1 + a_{n+1}a_n$, which simplifies through a simple substitution to $a_{n+1}a_n = n + C$. From here, you can write out a solution with products or descending products, or something like that. Inverting $a_n$ instead gives me no such option. — Theo Bendit, Dec 01 '22 at 01:30
@SeverinSchraven FYI, using an Approach0 search, the AoPS threads Do we know any good bounds? and $a_{180}>19$ use your recursion equation, but deal with other properties (e.g., boundedness & a lower bound on a specific value, respectively). Also, there are several posts here which deal ... — John Omielan, Dec 01 '22 at 01:47
(cont.) with a variation of $a_{n+2}=a_{n+1}+\frac{1}{a_{n+1}}$, e.g., prove that the following sequence converges to infinity, Limit of the Sequence: $a_{n+1}=a_n+\frac{1}{a_n}$ with $a_1=1$, Assume that $x_1=1$ and $x_{n+1}=x_n+\dfrac{1}{x_n}.$ Prove $\lim\limits_{n \to \infty}x_n=+\infty.$, etc. — John Omielan, Dec 01 '22 at 01:50
@JohnOmielan Thanks! I am aware that it goes to infinity (monotonicity plus the recursion equation give an easy contradiction). The question is really about closed form solution (or lack thereof). — Severin Schraven, Dec 01 '22 at 01:58
@SeverinSchraven You're welcome. I only mentioned those threads because, since the solutions don't try to determine or use any closed form, this gives some (although admittedly only rather weak) indication that there's either no closed form solution, or it exists but it's difficult to get, complicated to state, etc. — John Omielan, Dec 01 '22 at 02:02
@JohnOmielan I was thinking the same. But for these kind of things I have no sense of how difficult they are. Maybe there is some smart trick. On the other hand, I would guess that proving absence of a closed form solution would be rather difficult. In any case, there are tons of smart people on this page, somebody will know the answer :) — Severin Schraven, Dec 01 '22 at 02:08
If you start with $a_1 = a_2 = 1$ then the numerators appear to be given by http://oeis.org/A296397. — JBL, Dec 01 '22 at 03:17
@JBL Thanks! It seems that this recursion is much more complicated than what I thought. — Severin Schraven, Dec 01 '22 at 04:49
@JBL's OEIS link also reveals another very relevant MSE post: https://math.stackexchange.com/questions/2561924/any-insight-on-the-half-reciprocal-fibonacci-sequence — Theo Bendit, Dec 01 '22 at 05:58
@TheoBendit Thanks, that confirms my guess about the asymptotics and also leads me to believe that a closed form might be difficult to obtain. — Severin Schraven, Dec 01 '22 at 07:04

Recursion equation $a_{n+2}= a_{n+1} + \frac{1}{a_n}$

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