A question on A068480
Max Alekseyev
maxale at gmail.com
Fri Oct 26 19:53:41 CEST 2007
Rustem Aidagulov has found the following counterexamples:
Prime 521881 divides gcd(n!-1,2^n+1) for n=221799 and n=300081.
Prime 774593 divides gcd(n!-1,2^n+1) for n=338884.
Therefore, for these n, gcd(n!-1,2^n+1)>1 but not equal to 2n+1.
Regards,
Max
On 11/16/06, Max A. <maxale at gmail.com> wrote:
> 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
>
>
> 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.
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