A question on A068480

[Max A.]

It is unlikely for n!-1 and 2^n+1 to have a common prime factor
(unless in a special case described below), especially since
1) all prime factors of n!-1 are greater than n;
2) all prime factors of 2^n+1 are of the form p=2kq+1 where q is a
divisor of n, and the multiplicative of 2 modulo p divides 2q.

The special case is when n is even and p=2n+1 is prime. In this case,
p divides both n!-1 and 2^n+1, and thus gcd(n!-1,2^n+1). And it is
very likely that gcd(n!-1,2^n+1) = p.

Overall, the statement "if gcd(n!-1,2^n+1)>1 then
gcd(n!-1,2^n+1)=2n+1" sounds plausible and I verified it for n up to
45,000. But it may be very hard to give a proof.

Max

On 11/15/06, Giovanni Resta <g.resta at iit.cnr.it> wrote:
> A068480 is the sequence on numbers n such that gcd(n!-1,2^n+1)>1.
>
> I noticed that for all the numbers listed gcd(n!-1,2^n+1) = 2n+1.
>
> Is this a rule (proof ?) or a coincidence (counterexample ?) .
>
> thanks,
> giovanni.
>
>
>