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From: Robert Israel on 14 Dec 2009 14:04
Gerry <gerrymrt(a)gmail.com> writes:

> Hi All,
>
> something i didn't find a reference for and might be of interest.
>
> Since the denominators of the Bernoulli numbers are based on the
> primes 2 * f + 1 for every divisor f of 2^n one can write the
> following equation which holds for the Fermat primes :
>
> gcd(2^n+1 , denominator(bernoulli(2^(n+1))) = 2^n+1
>
> So is there a chance to compute the denominators of the Bernoulli
> numbers without the numerators?

<http://www.research.att.com/~njas/sequences/A002445>

From the Von Staudt-Clausen theorem, denominator(B_2n) = product of primes p
such that (p-1)|2n.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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