[seqfan] Re: Puzzling equivalences

[israel at math.ubc.ca]

Yes, next term is 2001907169.

Cheers,
Robert

On Jul 4 2024, Tomasz Ordowski wrote:

>Hello everyone!
>
>I have something puzzling,
>namely...
>
>   Numbers n such that
>((2^n-2)/n)^n == 2 (mod n).
>We have: 3, 29, 37, 3373, ...
>Is 2001907169 the next term?
>If so, they are probably A125854.
>See https://oeis.org/A125854
>How to explain it then?
>
>Note that numbers n such that
>((2^n-2)/n)^n == 2^n (mod n^2) *
>are also 3, 29, 37, 3373, ...
>Next number please.
>
>Best,
>
>Tom Ordo
>______________
>(*) The equivalent condition is
>((2^(n-1)-1)/n)^n == 1 (mod n^2).
>
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