Does this deceptively simple nonlinear recurrence relation have a closed form solution? Given the base case $a_0 = 1$, does $a_n = a_{n-1} + \frac{1}{\left\lfloor{a_{n-1}}\right \rfloor}$ have a closed form solution? The sequence itself is divergent and simply goes {$1, 2, 2+\frac{1}{2}, 3, 3+\frac{1}{3}, 3+\frac{2}{3}, 4, 4+\frac{1}{4}, 4+\frac{2}{4}, 4+\frac{3}{4}, . . .$} and so forth. It seems like it should be easy but I can't seem to find a solution. Any suggestions?

The sequence $a_n$ for $n\geq 1$ has the following formula:
$$a_n=\left\lfloor \sqrt{2n}+\tfrac{1}{2}\right\rfloor +\frac{\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor-\left\lfloor \frac{1}{2} \left(\sqrt{8 n-7}+1\right)\right\rfloor ^2 +2 n}{2 \left\lfloor \sqrt{2n}+\frac{1}{2}\right\rfloor }.$$
Here is the Wolfram Alpha link to check it.
It is related to OEIS A002024 and OEIS A002262.
