

A103478


Positive integers k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.


2




OFFSET

1,1


COMMENTS

On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 5 and with n exactly 2 greater than m, then that m belongs to this sequence.  Jeppe Stig Nielsen, Dec 04 2018


LINKS

Table of n, a(n) for n=1..6.
Wilfrid Keller, Prime factors k*2^n + 1 of Fermat numbers F_m


EXAMPLE

a(1)=5 because 5 is the smallest positive integer k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.


MATHEMATICA

Select[Range[1, 2000], Mod[1 + PowerMod[2, 2^#, 1 + 5*2^(# + 2)], 1 + 5*2^(# + 2)] == 0 &] (* Julien Kluge, Jul 08 2016 *)


PROG

(PARI) isok(n) = Mod(2, 1+5*2^(n+2))^(2^n) + 1 == 0; \\ Michel Marcus, Apr 29 2016


CROSSREFS

Cf. A000215, A083575, A103477, A103479.
Sequence in context: A230497 A138905 A125955 * A327976 A121868 A111584
Adjacent sequences: A103475 A103476 A103477 * A103479 A103480 A103481


KEYWORD

nonn,more,hard


AUTHOR

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


STATUS

approved



