[Integers here refer to positive integers. $n$ is a positive integer.]

I had been thinking about this problem:

Find all integers $n$ which are divisible by all integers $m \leq \sqrt{n}$ .

After some work, I figured that this was possible only for a finitely many such numbers (One reason is that, $ \lfloor {\sqrt{n} }\rfloor \# $ grows faster than $n$ , where $ x \# $ is the primorial of $x$).

Again, we note that for all $n$ , all integers below $ \sqrt[n]{n} $ divide $n$. It is easily seen by noting that $\sqrt[n]{n} < 2$ for all $n \in \mathbb{N} $.

So, I thought about developing a function that, for all $n \in \mathbb{N} $ , tells me the smallest possible value $m$ such that all integers less than (or equal to) $ \sqrt[m]{n} $ divides $n$.

The problem can be restated as:

Find the smallest function (here I mean that the function takes on smallest possible values) $f:\mathbb{N} \to \mathbb{N}$ such that all integers below(or equal to) $ \sqrt[f(n)]{n} $ divide $n$.