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!

If we let $x_n=a_n/\sqrt n$ then we have the recursive rule for $x_n$:
$$
x_{n+1}={x_n\over\sqrt{n(n+1)}}+{1\over x_n}\sqrt{n\over n+1}.
$$
It is easy to prove that $\lim_{n\to\infty}x_n=1$, so that $a_n/\sqrt{2015}$ should be very close to $1$. In other words we may conclude that
$$
\lfloor a_{2015} \rfloor = \lfloor \sqrt{2015} \rfloor=44.
$$
