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!

This is close to the Babylonian method or Newton's method for computing $\sqrt n$  You use $a_{n-1}$ (which will be close to $\sqrt{n-1}$) as a starting guess and do one iteration.  If you want, you can show that the error is positive, decreasing, and (after a while) bounded by a small enough fraction that you know $44 \lt a_{2015} \lt 45$  
Added:  a little experimentation shows that $a_n$ is a little greater than $\sqrt n$.  For example, $a_{16} \approx 4.06653$.  Define $b_n=a_n-\sqrt n$.  If $\frac 1n \lt b_n \lt \frac 1{\sqrt n}$ (as is true for $n=16$), we can approximate $$b_{n+1}=\frac n{a_n}+\frac {a_n}n-\sqrt n\\
b_{n+1}=\frac n{b_n+\sqrt n}+\frac{b_n+\sqrt n}n-\sqrt n\\
\approx \sqrt n(1-\frac {b_n}{\sqrt n})+\frac 1{\sqrt n}+\frac {b_n}n-\sqrt n\\
=-b_n+\frac 1{\sqrt n}+\frac {b_n}n\\
\approx\frac 1{\sqrt{n+1}}+\frac 2{n^{3/2}}-b_n+\frac {b_n}n$$  We need to argue that $\frac 1{n+1} \le b_{n+1} \lt \frac 1{\sqrt {n+1}}$, but I can't quite get there.  It seems clear, and this gives us an induction that says $b_n$ is small and $a_n \approx \sqrt n$
