[seqfan] Re: Numbers n such that 2^(2^n-2) == 1 (mod n^2)

[Tomasz Ordowski]

Emmanuel, thanks for this next exception.

Note that A069051 > 2 is a subsequence.
https://oeis.org/history/view?seq=A069051&v=116
See my new comment.

Also A217468 is a subsequence.
https://oeis.org/A217468

Tom Ordo


pt., 19 lip 2024 o 11:57 Emmanuel Vantieghem <emmanuelvantieghem at gmail.com>
napisał(a):

> 951481 = 271*3511  is the next exception.
>
> Op do 18 jul 2024 om 23:57 schreef Tomasz Ordowski <
> tomaszordowski at gmail.com
> >:
>
> > 1, 3, 7, 19, 43, 73, 127, 163, 337, 341,
> > 379, 487, 601, 881, 883, 937, 1387, ...
> > These numbers are not in the OEIS.
> > Such composites 341, 1387, 4681,
> > 5461, 8911, 10261, 14491, 15841, ...
> > are Fermat pseudoprimes to base 2,
> > except the number 66709 = 19*3511,
> > that Amiram Eldar found. Note that
> > the factor 3511 is a Wieferich prime.
> >   Are there any other exceptions?
> >
> > If p is prime and 2^(2^p-2) == 1 (mod p),
> > then 2^(2^p-2) == 1 (mod p^2).
> >
> >    Generally,
> > if 2^(2^n-2) == 1 (mod n),
> > then 2^(2^n-2) == 1 (mod n^2)
> > if and only if 2^(n-1) == 1 (mod n),
> > with the exception n = 66709 = 19*3511.
> >    Find more counterexamples.
> >
> > Best,
> >
> > Tom Ordo
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
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