A384438 Composite numbers k such that ((2^k+1)/3)^k == 1 (mod k^2).

341, 1105, 1387, 1729, 1771, 2047, 2465, 2485, 2701, 2821, 3277, 3445, 4033, 4369, 4681, 5185, 5461, 6601, 7957, 8321, 8911, 9361, 10261, 10585, 11305, 11713, 11891, 13741, 13747, 13981, 14491, 15709, 15841, 16105, 16705, 18145, 18721, 19951, 23377, 28441, 29341

Offset: 1

Author

Thomas Ordowski, May 29 2025

Keywords

• nonn

Comments

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

Fermat pseudoprimes to base 2 not divisible by 3 (A066488) are a proper subsequence.

The terms k that are not 2^(k-1) == 1 (mod k) are 1771, 2485, 3445, 5185, 9361, ...

Crossrefs

Cf. A001567, A066488 (subsequence), A384148.

Programs

PARI

isok(k) = (k>1) && (k%2) && !isprime(k) && (Mod((2^k+1)/3, k^2)^k == 1); \\ Michel Marcus, May 29 2025

Extensions

a(18)-a(41) from Jinyuan Wang, May 29 2025