0
$\begingroup$

It is well known that the abc conjecture implies that the there are only finitely many solutions to Brocard problem, as shown by Overholt in Overholt, Marius (1993), "The diophantine equation $n! + 1 = m^2$", Bull. London Math. Soc., 25 (2): 104. While trying to extend this result and give some bound on the maximum number of solutions under the abc conjecture, I came out with an inequality of this form:

$$ p_{\pi(n)} \#> C (n!)^{2/3 - \epsilon} $$

for some small $\epsilon >0$ and some constant $C>0$. I would like to prove that this inequality does not hold for $n$ enough large, so I considered the following limit:

$$ \lim_{n \rightarrow \infty} \frac{p_{\pi(n)} \#}{(n!)^{2/3 - \epsilon}} $$

Wolfram alpha does not give me a direct answer, but I think that its value is $0$. Supporting this conjecture, we have for example this, this and this, where I used $\epsilon=0.0001$; these show that the function seems to be decreasing really rapidly to $0$.

Does anybody know how to prove/disprove this, or has any reference for this result?

$\endgroup$
8
  • 1
    $\begingroup$ Did you check the bounds for primorial and factorial in wikipedia? The explicit bounds appear to answer the question. $\endgroup$
    – joro
    Commented Nov 28, 2020 at 12:28
  • 4
    $\begingroup$ Isn't $p_{\pi(n)}\#$ simply the same as $n\#$? $n\#$ grows roughly like $e^n$ (by PNT), while $n!$ grows roughly like $n^n/e^n$ (by Stirling), hence the failure of your inequality. $\endgroup$
    – Wojowu
    Commented Nov 28, 2020 at 12:28
  • 3
    $\begingroup$ @joro: what part of the wikipedia article are you looking at? The growth rate Wojowu describes is immediate from the version of PNT which states that $\sum_{p\leq n} \log p \sim n$. $\endgroup$ Commented Nov 28, 2020 at 13:46
  • 1
    $\begingroup$ @joro Wikipedia states that $p_n\#\approx e^{n\log n}$, while $n\#\approx e^n$. $\endgroup$
    – Wojowu
    Commented Nov 28, 2020 at 14:05
  • 2
    $\begingroup$ I was wrong, sorry. deleted the comment. $\endgroup$
    – joro
    Commented Nov 28, 2020 at 14:37

0

You must log in to answer this question.

Browse other questions tagged .