

A103477


Positive integers k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^2^k.


2




OFFSET

1,1


COMMENTS

The smallest prime factor any Fermat number 2^(2^k)+1 can have is 3*2^(k+2)+1.  Jeppe Stig Nielsen, Dec 03 2018
On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 3 and with n exactly 2 greater than m, then that m belongs to this sequence.  Jeppe Stig Nielsen, Dec 04 2018


LINKS



EXAMPLE

a(1)=207 because 207 is the smallest positive integer k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^2^k.


MATHEMATICA

aQ[n_] := PowerMod[2, 2^n, 1 + 3*2^(n+2)] == 3*2^(n+2); Select[Range[100000], aQ] (* Amiram Eldar, Dec 04 2018 *)


PROG



CROSSREFS



KEYWORD

nonn,hard,more


AUTHOR

Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005


EXTENSIONS



STATUS

approved



