Krdytkyu

For any prime $p$, let $n_p(m)$ denote the exponent of $p$ in the factorisation of $m!$, i.e. $m!=p^{n_p(m)}cdot k$ with $pnotmid k$.

I wonder if there is a general formula for $frac{m!}{p^{n_p(m)}}$ modulo $p$?

I could not prove but I believe that the frequencies of $1,2,...,p-1$ are equal.

Does not seem to be that simple as we let things run high; I did primes 5, 7, 11. Seems reasonable to suggest 1, p-1 are the same and highest, after that not clear, although 5, 7 seem symmetric. I ran 11 pretty high, I guess if a+b = 11 then their counts are similar.

1 count 1330   2 count 1169   3 count 1169   4 count 1332 



1 count 2570   2 count 2220   3 count 2230   4 count 2210   5 count 2220   6 count 2550



1 count 121894   2 count 115087   3 count 99710   4 count 100842   5 count 113429   

6 count 112909   7 count 99693   8 count 99050   9 count 114123   10 count 123263 















int main()

{

   int p = 11;

   int count[p] ;

   int fac = 1;

   for(int i = 1; i < p; ++i) count[i] = 0;

   for(int i = 1; i <= 100000 * p; ++i)

   {

     int n = i;

     while ( n % p == 0 ) n /= p;

     n %= p;

     fac *= n;

     fac %= p;



     count[fac] = 1 + count[fac];

         cout << i << "  "  << fac <<  "  count " << count[fac] << endl;



   }

   for(int i = 1; i < p; ++i) cout << i << " count " << count[i] << "   " ;

   cout << endl << endl; 









    return 0 ;

}

With $k$ as in the question, define $F(m)$ to be the residue of $k$ modulo $p$.

A formula

Let $m=a+bp$, where $0le ale p-1$. Then, modulo $p$, $F(m)$ is congruent to $a!(p-1)!^bF(b)$. By Wilson's Theorem this simplifies to $a!(-1)^bF(b)$.

So, in general, let $$m=a_0+a_1p+a_2p^2+...+a_{2n-1}p^{2n-1},0le a_ile p-1.$$
Then, for $A=a_1+a_3+...+a_{2n-1}$, $F(m)$ is congruent, modulo $p$, to
$$a_0!a_1!...a_{2n-1}!(-1)^A.$$

Frequencies

The formula shows that $F(m)$ is the product modulo $p$ $$F(a_0+a_1p) F(a_2+a_3p)...F(a_{2n-2}+a_{2n-1}p).$$
Therefore the frequencies for the $p^2$numbers $0!,1!,...,(p^2-1)!$ determine the frequencies for the $p^4$numbers up to $(p^4-1)!$ etc. etc.

For example, modulo 3 there are 4 1s and 5 2s from 0! to 8!

So up to 80! there are $4^2+5^2=41$ 1s and $2.4.5=40$ 2s.

Same about

$F(m,p) = F(m mod p , p) dot F(m div p , p) dot (-1)^{m div p} $

with the integer division.