A307532 a(n) is the smallest k > 2^(2^n)+1 such that 2^(k-1) == 1 (mod (2^(2^n)-1)*k).

5, 7, 29, 281, 65617, 4294967681, 18446744073709552577, 340282366920938463463374607431768211841, 115792089237316195423570985008687907853269984665640564039457584007913129642241

Offset: 0

Author

Amiram Eldar and Thomas Ordowski, Apr 13 2019

Keywords

• nonn

Comments

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

a(n) is the least k = 2^(2^n) + m*2^n + 1 for m > 0 such that 2^(k-1) == 1 (mod k).

The values of m = (a(n)-2^(2^n)-1)/2^n are 2, 1, 3, 3, 5, 12, 15, 3, 9, 202, 56, 304, 635, 11095, 8948, ...; is m = A307535(n) for all n > 4?

Conjecture: a(n) is prime for all n >= 0.

Crossrefs

Cf. A001567, A065091, A307512, A307535.

Programs

Mathematica

a[n_] := Module[{k = 2^(2^n) + 2}, While[PowerMod[2, k - 1, (2^(2^n) - 1)*k] != 1, k++]; k]; Array[a, 10, 0]

PARI

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

Formula

a(n) == 1 (mod 2^n).

Extensions

a(8) from Chai Wah Wu, Apr 29 2019