A393840 Number of prime factors of the form k*2^A228845(n) + 1 (with k odd) that divide the Fermat number 2^(2^n) + 1.

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 4

Offset

0,6

Comments

For any n, a(n) = 1 or is a multiple of 2.

a(12) >= 4.

Conjecture 1: the sequence is not monotonic.

Conjecture 2: if A046052(n) = 2*m for some integer m > 1, then a(n) + 1 < A046052(n). According to this, if a Fermat number has an even number of prime factors and is not semiprime, at least two of them are in A351865.

References

M. Křížek, 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. 66-67.

Links

Table of n, a(n) for n=0..11.

Wilfrid Keller, Fermat factoring status.

Michal Křížek and Jan Chleboun, A note on factorization of the Fermat numbers and their factors of the form 3*h*2^n + 1, Mathematica Bohemica, Vol. 119 (1994), No. 4, pp. 442-444.

Eric Weisstein's World of Mathematics, Fermat Number.

Crossrefs

Cf. A046052, A228845.

Sequence in context: A307833 A331566 A358473 * A369978 A390107 A083261

Adjacent sequences: A393837 A393838 A393839 * A393841 A393843 A393844

Keyword

nonn,hard,more

Author

Arkadiusz Wesolowski, Apr 02 2026

Status

approved