A374841 Numbers k such that 2^(2^k-2) == 1 (mod k^2).

3, 7, 19, 43, 73, 127, 163, 337, 341, 379, 487, 601, 881, 883, 937, 1387, 1459, 1801, 2593, 2647, 2857, 3079, 3529, 3673, 3943, 4057, 4201, 4681, 5419, 5461, 5881, 6121, 6481, 6529, 6553, 6571, 6841, 7481, 7993, 8233, 8911, 9001, 9199, 9241, 9721, 10261, 10657, 11161, 11827, 12241

Offset: 1

Author

Thomas Ordowski, Jul 22 2024

Keywords

• nonn

Comments

If p is an odd prime and 2^(2^p-2) == 1 (mod p), then 2^(2^p-2) == 1 (mod p^2).

If 2^(k-1) == 1 (mod k) and 2^(2^k-2) == 1 (mod k), then 2^(2^k-2) == 1 (mod k^2).

Composite terms that are not Fermat pseudoprimes to base 2 are 66709, 951481, ...

Note that 66709 = 19*3511 and 951481 = 271*3511, where 3511 is a Wieferich prime.

Examples

			3 is a term, because 3^2 divides 2^(2^3-2) - 1 = 2^6 - 1 = 63.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..250 from Michel Marcus)

Crossrefs

Cf. A001220, A001567, A069051 (> 2 is a subsequence), A217468 (subsequence).

Programs

Mathematica

Select[Range[12500], PowerMod[2, 2^# - 2, #^2] == 1 &] (* Amiram Eldar, Jul 22 2024 *)

PARI

isok(k) = Mod(2, k^2)^(2^k-2) == 1; \\ Michel Marcus, Jan 05 2025

Extensions

More terms from Amiram Eldar Jul 22 2024