A103477 Positive integers k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^(2^k).

207, 157167, 213319, 382447, 2145351

Offset: 1

Author

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

Keywords

• nonn
• hard
• more

Comments

The smallest prime factor any Fermat number 2^(2^k)+1 can have is 3*2^(k+2)+1 (for k >= 3). - Jeppe Stig Nielsen, Dec 03 2018

On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 3 and with n exactly 2 greater than m, then that m belongs to this sequence. - Jeppe Stig Nielsen, Dec 04 2018

All terms in this sequence are odd. Proof: by a theorem of Fermat, each prime p that is 1 mod 3 is expressible uniquely in the form a^2 + 3b^2, where a and b are positive integers. Gauss proved that 2 is a cubic residue iff 3|b. However, for even k and p = 1 + 3*2^(k+2), we have b = 2^(k/2+1) which is not divisible by 3. - William Hu, Oct 14 2024

Examples

			a(1)=207 because 207 is the smallest positive integer k for which 1 + 3*2^(k+2) divides the Fermat number 1 + 2^(2^k).
		

Links

• Wilfrid Keller, Prime factors k*2^n + 1 of Fermat numbers F_m

Crossrefs

Cf. A103478, A103479.

Programs

Mathematica

aQ[n_] := PowerMod[2, 2^n, 1 + 3*2^(n+2)] == 3*2^(n+2); Select[Range[100000], aQ] (* Amiram Eldar, Dec 04 2018 *)

PARI

isOK(n) = Mod(2, 1+3*2^(n+2))^(2^n) + 1 == 0 \\ Jeppe Stig Nielsen, Dec 03 2018

Extensions

Sequence name trimmed by Jeppe Stig Nielsen, Dec 03 2018