# Functions about integers which are divisible by all numbers less than or equal to a fractional power of the integer

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

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 $n$.

• There's an obvious (and unique) answer to your question : $f(n)$ is the smallest $m$ such that all integers $\leq \sqrt[m]{n}$ divide $n$. – user25235 Jun 16 '13 at 15:07
• What do you want to do with this function, aside from defining it? – S. Carnahan Jun 16 '13 at 15:13
• Let g(n) be 1 + ceiling (log_2 n). Then n^1/g < 2, and so your divisibility relation holds. If you wish it to hold for odd numbers, you won't do much better than g(n). Gerhard "Close Enough Sometimes Good Enough" Paseman, 2013.06.16 – Gerhard Paseman Jun 16 '13 at 20:18
• For numbers k which are prime powers and n that are divisible by all integers up to but not including k, one can use ceil(log_k(n)) instead of ceil(log_2(n)). Gerhard "Ask Me About System Design" Paseman, 2013.06.16 – Gerhard Paseman Jun 16 '13 at 21:12