3
$\begingroup$

$\mathbb{P}$ is the prime numbers set.

$p \in \mathbb{P}$

$a,b,c \in \mathbb{N}$

$n=a p^b+c$ where

$c= n\bmod p$

$b$ is the highest power of $p$ who divides $n-c$

How to find $\beta$ where $\beta$ is the highest power of $p$ who divides $n!$?

And how to find $\alpha$ where $\alpha$ is $\dfrac{n!}{p^\beta}\bmod p$?

$n!=\alpha p^\beta$

Probably the answer will appear $a$, $b$, $c$, $p$ and factorials.

Obs.:

$x = y\bmod z \iff x \equiv y \pmod{z}$ and $0 \leq x \leq z-1$

$x,y,z \in \mathbb{N}$

  • 0
    @GarouDan: OK. For the $a$ term, if it is $\ge p$, we will have to add $\lfloor a/p\rfloor +\lfloor a/p^2\rfloor +\cdots$ to my earlier closed form. This sum is less attractive. We can characterize it in terms of coefficients in the base $p$ expansion of $a$, but that really no longer quite qualifies as$a$closed form.2011-11-09

0 Answers 0