How do we reach the answer to the following recursive problem? $\large{a_{n+1} = \frac{a_n}{n} + \frac{n}{a_n}, a_1 = 1}$
Let the sequence $\large{\left< a_n \right>}$ be defined as above for all positive integers n. Evaluate $\large{\left \lfloor a_{2015} \right \rfloor}$.
I wrote a C++ program to solve the problem but I've no idea how to mathematically approach it.
Thanks in advance!

It is probably easier to prove directly by induction that $\sqrt{n-1}\le a_n\le\sqrt{n+2}$ from some $n$ onwards. 
We must prove at first that $a_{n+1}\le\sqrt{n+3}$, which is equivalent to $a_n/n+n/a_n\le\sqrt{n+3}$. But by the inductive hypothesis $a_n/n+n/a_n\le\sqrt{n+2}/n+n/\sqrt{n}$, so that we are left to prove that $\sqrt{n+2}/n+\sqrt{n}\le\sqrt{n+3}$. This last inequality boils down to $\sqrt{1+2/n}(\sqrt{1+3/n}+1)\le3$, which is true for $n\ge4$.
In a similar way, to prove that $a_{n+1}\ge\sqrt{n}$ one must prove $\sqrt{n-1}/n+n/\sqrt{n+2}\le\sqrt{n}$. After some tinkering this is equivalent to $\sqrt{1+(3n-2)/n^2}(1+\sqrt{1+2/n})\ge2$, which is true for $n\ge1$.
In the end we get then $\sqrt{2014}\le a_{2015}\le\sqrt{2017}$, that is $44.878\le a_{2015}\le44.911$.
