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

[Tomasz Ordowski]

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