A384148 Numbers k such that (2^k-1)^k == 1 (mod (2^k+1)*k^2) and 2^(k-1) != 1 (mod k).

30457, 33865, 80185, 82621, 86785, 104845, 212401, 250705

Offset: 1

Author

Thomas Ordowski, May 20 2025

Keywords

• nonn
• hard
• more

Comments

If p > 3 is prime, then (2^p-1)^p == 1 (mod (2^p+1)*p^2).

Generally, if m is not divisible by 3 and 2^(m-1) == 1 (mod m), then (2^m-1)^m == 1 (mod (2^m+1)*m^2).

However, there are composite numbers satisfying this congruence that are not Fermat pseudoprimes to base 2. These exceptions constitute this sequence.

Crossrefs

Cf. A001567, A066488 (Fermat pseudoprimes to base 2 that are not divisible by 3).

Programs

PARI

isok(k) = if (!isprime(k) && (Mod(2, k)^(k-1) != 1), Mod((2^k-1),(2^k+1)*k^2)^k == 1); \\ Michel Marcus, May 20 2025

Extensions

a(3)-a(6) from Michel Marcus, May 21 2025

a(7)-a(8) from Michael S. Branicky, May 28 2025