From: Robin Chapman
Subject: Re: Bernoulli numbers: is this known?
Date: Thu, 22 Jul 1999 07:43:07 GMT
Newsgroups: sci.math
Keywords: Clausen and von Staudt's theorem (denominators)
In article <7n6c2k$inv$1@nnrp1.deja.com>,
feldmann4350@my-deja.com wrote:
> I already posted this, but with a wrong header; sorry for the repetition
>
> Trying to solve the x^3=x ring problem, i discovered that, if we put
> f(k)=gcd (2^k-2,3^k-3,4^k-4,5^k-5,...,n^k-n,...), we get f(k)=2 if k is
> even, and f(k)= denominator of k-th Bernouilli number if k is odd (i.e.
> f(3)=6, f(5)=30, f(7)=42, f(9)=30, f(11)=66, f(13)=2730, etc.)
You mean the denominator of the (k-1)-th Bernoulli number.
> (those above are (quite easily) proved; i checked the general result
> for k<500 with Maple)
>
> Is this well known? Conjectured? Any hint?
It's a well-known theorem of Clausen and von Staudt that for even m
the denominator of B_m is the product of the primes p with (p-1) | m.
Let's consider f(k) for k>=2. It must be square-free, for if
p is prime then p^2 doesn't divide p^k - p. Let's characterize the
primes p for which p | f(k). These are the primes p for which
a^k = a (mod p) for all a. As this congruence is true for a = 0 (mod p)
then it's true for all a iff a^{k-1} = 1 (mod p) for all a not
divisible by p. This occurs iff (p-1)|(k-1) since the multiplicative
group modulo p is cyclic of order p-1. Hence f(k) is the product
of all primes p with (p-1)|(k-1). If k is even then f(k) is 2 since
p = 2 is the only prime with p-1 dividing the odd number k-1. If
k is odd then f(k) is the denominator of B_{k-1} by Clausen-von Staudt.
--
Robin Chapman
http://www.maths.ex.ac.uk/~rjc/rjc.html
"They did not have proper palms at home in Exeter."
Peter Carey, _Oscar and Lucinda_
Sent via Deja.com http://www.deja.com/
Share what you know. Learn what you don't.