2

For prime $p$, show whether $$\prod_{p \geq 1} p^{\lfloor \frac{x}{p-1} \rfloor} \sim x!$$ as $x$ approaches infinity, and explain.

I don’t know that it’s true, but I thought that it followed, if perhaps somewhat indirectly, from the fact that an arbitrarily large positive integer has, on average, 1 factor of 2, $1/2$ factors of 3, $1/4$ factors of 5, et cetera, so the product of the integers should approach the product of the approximations.

Lieutenant Zipp
  • 1,767
  • 1
    For clarification, do you mean for the product to be running through all primes less than $x$? – Vedvart1 Jun 26 '19 at 12:49
  • 1
    I think the claim is true, as $\lfloor\frac{x}{p-1}\rfloor$ measures the highest power of $p$ that divides $x$ – vidyarthi Jun 26 '19 at 12:50
  • 4
    @Vedvart1 I dont think less than $x$ is required, as when $p>x$, the term in the exponent is zero, thereby giving us $1$ to multiply with – vidyarthi Jun 26 '19 at 12:52
  • 3
    The formula for the highest power of a prime $p$ dividing $x!$ is well approximated by $\lfloor x/(p-1)\rfloor$. The sum is approximately geometric! – Jyrki Lahtonen Jun 26 '19 at 12:59
  • 2
    From calculating the first few terms of the sequence, you can find it at https://oeis.org/A091137, which references the relationship to the factorial and uses the same argument as vidyarthi. "This is always a relatively small multiple of n!, since the multiplicity with which a prime p divides n! is always <= n/(p-1); it is equal to floor(n/(p-1)) at least when n is a power of p." - Franklin T. Adams-Watters – Jam Jun 26 '19 at 12:59

0 Answers0