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Consider any prime number $p$ and sequence $(n!+1)_{n=1}^{\infty}$

How many elements from this sequence are divisible by p? Let's denote this number as $f(p)$

I know that $f(p)$ is finite since for all $n\ge p$ ; $n!+1$ is not divisible by $p$.

I also know that from Wilson theorem we have $p|(p-1)!+1$ so $f(p)\ge1$.

What is upper bound for $f(p)$?

Regards

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  • $\begingroup$ as you indicated, $f(p)\lt p$, but I suppose you'd like a tighter bound $\endgroup$ Feb 10, 2020 at 15:00
  • $\begingroup$ Trying some examples: $f(2)=1$ ($n=1$); $f(3)=1$ ($n=2$); $f(5)=1$ ($n=4$); $f(7)=2$ ($n=3,6$); $f(11)=2$ ($n=5,10$); $f(13)=1$ ($n=12$); $f(17)=1$ ($n=16$); $f(19)=2$ ($n=9,18$); $\endgroup$
    – Paul
    Feb 10, 2020 at 15:10

2 Answers 2

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The residue classes modulo $p$ form a field and so, for $p$ odd, all the non-zero elements are in pairs of multiplicative inverses except for $\pm1$.

If we placed the numbers $1,2, ... ,p-1$ in the order :- $$1,p-1,a,a^{-1},b,b^{-1},c,c^{-1}, ...$$ then every other successive product of numbers from the left is $-1$.

No other ordering of the elements can do better than this and so an upper bound for $f(p)$ is given by $$\frac{p-1}{2}.$$

This bound is only achieved when $p-1=2$ i.e. $p=3$. For larger primes it takes several elements before $1\times2\times3\times ...$ reaches $p-1$ and then various pairs $a,a^{-1}$ have been 'lost' by one or other of the pair having been used. In practice therefore, the bound is typically much less than $\frac{p-1}{2}.$

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An upper bound is $f(p) \leq p - \sqrt{p-1}$. The number of nonzero residue classes attained by $n!$ is at least $\sqrt{p-1}$; see among $ 1!,2!,...,p!$ there are at least $ \sqrt{p}$ different residues in modulo $ p$

So at least $\sqrt{p-1} - 1$ of the first $p-1$ residues of $n!$ are not $-1$.

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