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
On
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 $-1$.
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 $\frac{p-1}{2}.$