A question on A068480

[Max A.]

On 11/16/06, Max A. <maxale at gmail.com> wrote:

> 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.

Oh, my bad! Correct statement:

The special case is when n is of the form 4k+1 and p=2n+1 is prime
(implying that 2 and -1 are non-squares modulo p), and p also belongs
to A058302.
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.

The hypothesis about gcd(n!-1,2^n+1) implies that:
8k+3 belong to A058302 if and only if 4k+1 belong to A068480.

Max