How many numbers satisfy this floor condition

Question

How many numbers n between 1 and 600 do exist such that the floor of $\sqrt[4]{n}$ divides n. please solved the problem in a general case.

Answer 1

Note that $\lfloor \sqrt[4]n \rfloor=2$ for $16 \le n \le 80$ so how many $n$'s in this range qualify?

If we let $k=\lfloor \sqrt[4] n \rfloor$ we have $\sum_{i=1}^{k-1}\left(\lfloor \frac {(i+1)^4-(i^4+2)}i\rfloor+1\right)+\lfloor \frac {n-(k^4+1)}k\rfloor+1$ for the general case. The idea is as the example above, to count the number of multiples of $i$ from $i^4$ through $(i+1)^4-1$. Note that we start and end with a multiple of $i$.