OFFSET
1,1
COMMENTS
The prime k*2^(m+2) + 1 is a lucky Fermat factor if it divides 2^(2^m) + 1 and k = 3, 5, 6, 7, or 9 is the smallest value we can choose that is not excluded by congruence constraints modulo 12, which lead to divisibility of k*2^(m+2) + 1 by 3, 5, 7, or 13 (Krizek, Luca and Somer).
The m for which 2^(2^m) + 1 has a lucky factor are m = 5, 12, 23, 38, 73, 125, 207, 1945, 23471, 95328, 157167, 213319, 382447, 2145351, 2478782, ... From this it is trivial to write out a(1),...,a(15), but the numbers become too wide for a b-file. - Jeppe Stig Nielsen, Mar 13 2022
REFERENCES
M. Krizek, F. Luca, L. Somer, 17 Lectures on Fermat Numbers: From Number Theory to Geometry, CMS Books in Mathematics, vol. 9, Springer-Verlag, New York, 2001, pp. 77-79.
LINKS
Wilfrid Keller, Fermat factoring status
Eric Weisstein's World of Mathematics, Fermat Number
PROG
(PARI) r=vector(12, m, select(k->p=k*2^(m+2)+1; p%3!=0&&p%5!=0&&p%7!=0&&p%13!=0, [3, 5, 6, 7])[1]); for(m=0, +oo, k=r[(m+11)%12+1]; p=k*2^(m+2)+1; Mod(2, p)^(2^m)+1==0&&print1(p, ", ")) \\ Jeppe Stig Nielsen, Mar 13 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Arkadiusz Wesolowski, Oct 01 2013
STATUS
approved