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.