A229851 Lucky Fermat factors.

641, 114689, 167772161, 6597069766657, 188894659314785808547841, 850705917302346158658436518579420528641, 2468256835981809063232453773836025757474103798450369795022913537

Offset: 1

Author

Arkadiusz Wesolowski, Oct 01 2013

Keywords

• nonn

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

Crossrefs

Cf. A000215. Subsequence of A023394.

Programs

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