A228846 Largest m such that (2k+1)*2^m + 1 is a prime factor of the Fermat number 2^(2^n) + 1.

1, 2, 4, 8, 16, 7, 8, 9, 11, 16, 14, 14

Offset

0,2

Comments

a(n) >= n + 2 for n >= 2.

a(n) = A228845(n) if 2^(2^n) + 1 is prime or semiprime.

a(n) = max (A007814(p_i-1)), where p_i are the prime factors of 2^(2^n)+1. - Ralf Stephan, Sep 09 2013

For n >= 2, a(n) >= n + 3 if A046052(n) is an odd number. - Arkadiusz Wesolowski, Aug 10 2021

Links

Table of n, a(n) for n=0..11.

Wilfrid Keller, Fermat factoring status

Eric Weisstein's World of Mathematics, Fermat Number

Example

F(5) = 641*6700417 and max(A007814(640),A007814(6700416))=7, so a(5)=7.

Crossrefs

Cf. A000215, A019434, A023394, A046052, A228845.

Sequence in context: A016018 A333492 A228845 * A070336 A348433 A053026

Adjacent sequences: A228843 A228844 A228845 * A228847 A228848 A228849

Keyword

nonn,hard,more

Author

Arkadiusz Wesolowski, Sep 05 2013

Status

approved