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For any positive integer $n$, write $n!=a_nb_n^2$, where $a_n$ is squarefree. Does there exist a constant $c$ such that for any $\epsilon>0$, there exists $N$ such that $c^{(1-\epsilon)n}<a_n<c^{(1+\epsilon)n}$ for all $n>N$?

From Stirling's approximation we know that $n!$ is on the order of $\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$. The product of all primes less than $n$ is roughly $e^n$, as shown by Erdos. The product we are considering can only be less than that.

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  • $\begingroup$ It may be possible to estimate this, probably a bit roughly. Why do you think $2^n$ is a reasonable attempt? $\endgroup$
    – Will Jagy
    Feb 6, 2017 at 0:41
  • $\begingroup$ I think it should be on the order of $c^n$ for some $c$, since the product of primes less than $n$ is roughly $e^n$. $\endgroup$
    – pi66
    Feb 6, 2017 at 0:47

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Note: the question first indicated $2^n$ as an estimate, which is correct. Edited since.

ORIGINAL: Not a bad guess. I get, with Chebyshev's first function $\theta(x),$ the log of your number as $$ \theta(n) - \theta \left( \frac{n}{2} \right) +\theta \left( \frac{n}{3} \right) -\theta \left( \frac{n}{4} \right) +\theta \left( \frac{n}{5} \right) -\theta \left( \frac{n}{6} \right) \cdots $$ where the alternating series $1 - \frac{1}{2} +$ has limit $\log 2.$

The main ingredient is Legendre's theorem on the power of a prime dividing a factorial. Much more work to establish explicit bounds, if true.

Note that oeis agrees https://oeis.org/A055204 but with no proof or error estimates.

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  • $\begingroup$ Hi @Will Jagy, correct me if i'm wrong, I think we should use the Chebyshev's Second Function instead. For example in 58! , the prime 7 has 8 multiples (7, 14, ... , 56, ) , but since $49=7^2$ contributes two 7-factors , it ended up with odd many 7's. So in order to have the upper-bound, we just need to make sure that any primes power $p^k$ appears odd many times on each $k$. $\endgroup$ Jun 3, 2017 at 17:34

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