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A228846 Largest m such that (2k+1)*2^m + 1 is a prime factor of the Fermat number 2^(2^n) + 1. 4
1, 2, 4, 8, 16, 7, 8, 9, 11, 16, 14, 14 (list; graph; refs; listen; history; text; internal format)
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
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
Sequence in context: A016018 A333492 A228845 * A070336 A348433 A053026
KEYWORD
nonn,hard,more
AUTHOR
STATUS
approved

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Last modified March 29 09:59 EDT 2024. Contains 371268 sequences. (Running on oeis4.)