A358684 a(n) is the minimum integer k such that the smallest prime factor of the n-th Fermat number exceeds 2^(2^n - k).

0, 0, 0, 0, 0, 23, 46, 73, 206, 491, 999, 2030, 4080, 8151, 16208, 32738, 65507, 131028, 262121, 524252

Offset

0,6

Comments

a(14) is likely correct (see A093179); a(20) is unknown; a(21) to a(23) are 2097110, 4194189, 8388581; and a(24) is unknown.

2^(2^n - a(n)) < A093179(n).

Conjecture I: the dyadic valuation of A093179(n) - 1 does not exceed 2^n - a(n).

Conjecture II: a(n) ~ 2^n as n -> oo.

From Lorenzo Sauras Altuzarra, Jan 11 2026: (Start)

Moritz Firsching (personal communication) reports that AlphaProof solved Conjecture I by applying the formula below and the fact that A007814(k-1) <= floor(log_2(k)) for every integer k >= 2. See links for formal proof by AlphaProof.

Conjecture II is thus the question whether floor(log_2(A093179(n)))/2^n tends to zero or not. If there are infinitely many Fermat primes (which is currently unknown), then it cannot tend to zero. (End)

Links

Table of n, a(n) for n=0..19.

Lorenzo Sauras-Altuzarra, Some properties of the factors of Fermat numbers, Art Discrete Appl. Math. (2022).

Google DeepMind, Lean proof of Firsching's formula below.

Google DeepMind, Lean proof of Conjecture I.

Formula

a(n) = 2^n-floor(log_2(A093179(n))). - Lorenzo Sauras Altuzarra, Jan 11 2026

Example

For n=5, the smallest prime factor of F(5) = 2^(2^5) + 1 is 641 and it falls between 2^(2^5 - 23) = 512 < 641 < 1024 = 2^(2^5 - 22) so that a(5) = 23.

Crossrefs

Cf. A000215, A093179.

Sequence in context: A048845 A008605 A038152 * A344133 A103629 A253177

Adjacent sequences: A358681 A358682 A358683 * A358685 A358686 A358687

Keyword

nonn,hard,more

Author

Lorenzo Sauras Altuzarra, Nov 26 2022

Status

approved