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In this post, I was curious if the divisor bound could be improved for the product of two consecutive even numbers. It seems most likely it cannot by much. How could the upper bound of $\sigma_0(x^2 - 1)$ be improved if we impose the condition that $$n! = x^2 - 1$$ for a positive integer $n$? This restricts the primes that divide $x^2 - 1$ to be below $n$. The bound found here would consider all the primes under a given number. Am I right in thinking that the bound could be reduced because of the reduction in potential prime factors?

Further, I noticed that all the primes between $\frac{n}{2}$ and $n$ would divide $x^2 - 1$ exactly once. This forces lower primes to have higher powers. Could that improve any potential bound even more? From stronger forms of Bertrand's Postulate, it should be possible to estimate the number of primes between those two numbers.

PiMaster
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    Am I right in thinking that this question is simply about the size of $\sigma_0(n!)$ and, by this stage, has nothing to do with $x^2-1$? – Greg Martin Jun 27 '24 at 19:54
  • There are only $3$ known Brocard's numbers $4,5,7$. So it's more interesting to consider the general case of $\sigma_0(n!)$ as mentioned by Greg Martin. – user25406 Jun 27 '24 at 20:11
  • I guess my question is something like "If $x^2 - 1$ is close to $n!$ and $n$ is large, does $n!$ always have more factors than $x^2 - 1$?" But, I think there are infinitely many counterexamples. – PiMaster Jun 27 '24 at 20:35
  • It should be possible to find upper and lower bounds on $\sigma_0(n!)$ by using Legendre's formula with bounds on $p_n$ and $\pi(n)$. There are probably better ways, though. – PiMaster Jun 27 '24 at 21:01

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