[seqfan] Puzzling equivalences

[Tomasz Ordowski]

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).