A229850 Number of prime factors congruent to 1 mod 3 that divide the Fermat number 2^(2^n) + 1.

0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 3, 2

Offset: 0

Author

Arkadiusz Wesolowski, Oct 01 2013

Keywords

• nonn
• hard
• more

Comments

a(n) < A046052(n) because all Fermat numbers greater than 3 are equal to 2 (mod 3).

a(n) = 1 if A046052(n) = 2.

If A046052(n) = 3, then a(n) = 0 or 2.

a(n) <= A228846(n) - n - 1 for n = 0 to 11.

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. 61-63, 65-66.

Links

• Wilfrid Keller, Fermat factoring status
• Eric Weisstein's World of Mathematics, Fermat Number

Crossrefs

Cf. A000215, A002476, A023394, A046052, A229854.