[seqfan] Re: Numbers n > 1 such that (n-1)^(2^n-2) == 1 (mod (2^n-1)n^2)
Max Alekseyev
maxale at gmail.com
Mon Oct 2 18:26:57 CEST 2023
The congruence (n-1)^(2^n-2) == 1 (mod n^2) restricts n to primes and
Fermat pseudoprimes (including even ones).
Also, it implies that n is coprime to 2^n-1, and thus it's just a matter if
such n satisfies (n-1)^(2^n-2) == 1 (mod 2^n-1), or in other words if 2^n-1
is prime or (generalized) Fermat pseudoprime to base n-1.
This condition becomes quite hard to test as n grows.
Regards,
Max
On Sun, Oct 1, 2023 at 11:23 AM Tomasz Ordowski <tomaszordowski at gmail.com>
wrote:
> Dear readers!
>
> Conjecture:
> For n > 1, (2^n-1)n^2 divides (n-1)^(2^n-2) - 1
> if and only if 2^n - 1 is prime or n = 257 and 65537.
>
> Is n = 65537 the second and probably final exception?
> As the largest known Fermat primes F_4 = 2^2^4 + 1.
>
> See my draft A366029:
> https://oeis.org/draft/A366029
> https://oeis.org/history/view?seq=A366029&v=24
>
> Best regards,
>
> Thomas Ordowski
> ___________
> Amiram Eldar:
> Yes, 65537 is also a term.
> I tried to check 2^32+1 but it seems that is too high to check.
>
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