If n is a positive integer and p is a prime factor of 2n,
then p divides 2(2^(2n)-1) ?? Presumably true, but is it easy?

The primes p in question make up the denominator of Bernoulli number
B_{2n}, so an alternate formulation is: 2(2^n-1)B_n is an integer,
where B_n is a Bernoulli number.

On Mon, 21 Jul 2008 16:00:02 +0000 (UTC), G. A. Edgar wrote:

n = 15
p = 5, is prime, divides 30
2(2^(30)-1) = 2147483646 is not divisible by 5. Does this
disprove the "presumably true" assertion?

"G. A. Edgar" math.ohio-state.edu.invalid> writes:

Huh? Try n=p=5. In fact it's never true for n=p>3.

If p is an odd prime and n = k p, then 2^p == 2 mod p
so 2(2^(2n)-1) == 2(2^(2k)-1) mod p. Your statement would be true
if p(p-1) divides 2n.

The primes dividing the denominator of B_{2n} don't necessarily divide 2n.
For example, the denominator of B_200 is 2*3*5*11*41*101.

In article news.acm.uiuc.edu>, Peter Pearson
wrote:

I checked my conjecture for n up to 1000 or so. In fact I stated
it wrong here. We want a p-1 in there in one spot.
New formulation:

If n is a positive integer and p-1 is a prime factor of 2n,
then p divides 2(2^(2n)-1) ?? Presumably true, but is it easy?

"G. A. Edgar" math.ohio-state.edu.invalid> writes: