Anton’s congruence


For every n (n!¯)p stands for the productPlanetmathPlanetmath of numbers between 1 and n which are not divisible by a given prime p. And we set (0!¯)p=1.

The corollary below generalizes a result first found by Anton, Stickelberger, and Hensel:

Let N0 be the least non-negative residue of n(modps) where p is a prime numberMathworldPlanetmath and n. Then

(n!¯)p(±1)n/ps(N0!¯)p(modps).
Proof.

We write each r in the product below as ips+j to get

(n!¯)p = 1rnps÷̸rr
= (0in/ps-11j<psps÷̸jips+j)(i=n/ps1jN0ps÷̸jips+j)
i=0n/ps-11j<psps÷̸jjj=1ps÷̸jN0j)
(ps!¯)pn/ps(N0!¯)p(modps).

From Wilson’s theorem for prime powers it follows that

(n!¯)p{(N0!¯)pifp=2,s3(-1)n/ps(N0!¯)potherwise.(modps).

Title Anton’s congruenceMathworldPlanetmathPlanetmathPlanetmath
Canonical name AntonsCongruence
Date of creation 2013-03-22 13:22:49
Last modified on 2013-03-22 13:22:49
Owner Thomas Heye (1234)
Last modified by Thomas Heye (1234)
Numerical id 10
Author Thomas Heye (1234)
Entry type Theorem
Classification msc 11A07
Related topic FactorialMathworldPlanetmath