A050259 and a new solution to 2^n = 3 (mod n)

[Max A.]

I answered Richard privately.
But if other people also want to refer to my full name, it is
Max Alekseyev

Regards,
Max

P.S. It is not a conspiracy but rather my way to beat spamming. So
far, spam emails never addressed me by my full name, and I can easily
recognize important emails that do contain my full name in the To:
field.

On 11/13/06, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> Congratulations!  May I know Max's full name,
> so that he can earn undying fame in UPINT as
> well as in OEIS ?   Thanks,   R.
>
> On Mon, 13 Nov 2006, Max A. wrote:
>
> > SeqFans,
> >
> > In course of exploring all solutions of the congruence 2^n = 3
> > (mod n)
> > below 10^16, I have found a new solution:
> >
> > n = 3468371109448915
> >
> > This came as a big surprise to me since my original goal was
> > to prove
> > that 8365386194032363 is the second term of A050259. But now
> > it
> > appears to be at least the third term.
> >
> > So far I've submitted the new solution as a comment to
> > A050259, but
> > hopefully it will take its true place in this sequence soon.
> >
> > Regards,
> > Max
> >
> > P.S. See also http://mathworld.wolfram.com/2.html
>
>