A307512 a(n) is the smallest k > 2^n such that 2^(k-1) == 1 (mod (2^n-1)*k).

3, 5, 13, 17, 41, 67, 197, 257, 523, 1031, 2069, 4129, 8243, 16451, 32911, 65537, 131479, 262153, 524591, 1048601, 2097229, 4194389, 8388791, 16777441, 33554501, 67108913, 134217757, 268435889, 536871259, 1073741971, 2147484949, 4294967297, 8589934651, 17179869827

Offset: 1

Author

Thomas Ordowski, Apr 12 2019

Keywords

• nonn

Comments

a(n) = smallest k > 2^n such that k == 1 (mod n) and 2^(k-1) == 1 (mod k), so a(n) is an odd prime or a pseudoprime (Fermat pseudoprime to base 2).

Conjecture: a(n) is composite if and only if n = 2^j and 2^(2^j) + 1 is composite (presumably for all j > 4).

Note that a(2^j) = 2^(2^j) + 1 = A000215(j), the Fermat numbers.

For n <> 2^j, a(n) is the smallest k = 2^n - (2^n mod n) + m*n + 1 for m > 0 such that 2^(k-1) == 1 (mod k).

The last definition, also without the condition n <> 2^j, probably gives only primes.

Examples

			a(32) = 2^(2^5) + 1 = 641*6700417 is the smallest composite term.
		

Crossrefs

Cf. A000215, A000225, A001567, A065091, A270427, A306826.

Programs

Mathematica

a[n_] := Module[{k = 2^n + 1}, While[PowerMod[2, k - 1, (2^n - 1)*k] != 1, k++]; k]; Array[a, 50] (* Amiram Eldar, Apr 12 2019 *)

PARI

a(n) = my(k=2^n+1); while( Mod(2, (2^n-1)*k)^(k-1) != 1, k++); k; \\ Michel Marcus, Apr 25 2019

Formula

a(n) == 1 (mod n).

a(2^j) = A000215(j).

Extensions

More terms from Amiram Eldar, Apr 12 2019