[seqfan] Numbers n > 1 such that (n-1)^(2^n-2) == 1 (mod (2^n-1)n^2)
Tomasz Ordowski
tomaszordowski at gmail.com
Sun Oct 1 17:21:50 CEST 2023
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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