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?



Does this deceptively simple nonlinear recurrence relation have a closed form solution?
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](https://www.wolframalpha.com/input/?i=Table%5BFloor%5B1%2F2+%2B+Sqrt%5B2%5D+Sqrt%5Bn%5D%5D+%2B+(++2+n+%2B+Floor%5B1%2F2+(1+%2B+Sqrt%5B-7+%2B+8+n%5D)%5D+-++++Floor%5B1%2F2+(1+%2B+Sqrt%5B-7+%2B+8+n%5D)%5D%5E2)%2F(2+Floor%5B1%2F2+%2B+Sqrt%5B2%5D+Sqrt%5Bn%5D%5D),%7Bn,1,15%7D%5D) link to check it.


It is related to [OEIS A002024](https://oeis.org/A002024) and [OEIS A002262.](https://oeis.org/A002262)


