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%I #37 Feb 17 2023 15:14:41
%S 0,1,2,3,5,4,12,6,11,11,9,5,18,12,10,12,23,16,15,10,19,12,19,13,36,21,
%T 38,32,25,17,39,6,26,27,30,30,8,12,15,29,38,7,25,27,36,42,25,13,13,55
%N Only Fermat number divisible by A023394(n) is 2^2^a(n) + 1.
%C From _Jianing Song_, Mar 02 2021: (Start)
%C 2^(a(n)+1) is the multiplicative order of 2 modulo A023394(n).
%C Each k occurs A046052(k) times in this sequence provided that F(k) = 2^2^k + 1 is squarefree (no counterexamples are known). (End)
%C Alternatively, a(n) is the only k such that A023394(n) divides A000215(k). - _Lorenzo Sauras Altuzarra_, Feb 01 2023
%H Wilfrid Keller, <a href="http://www.prothsearch.com/fermat.html">Prime factors k.2^n + 1 of Fermat numbers F_m</a>
%H Lorenzo Sauras-Altuzarra, <a href="https://doi.org/10.26493/2590-9770.1473.ec5">Some properties of the factors of Fermat numbers</a>, Art Discrete Appl. Math. (2022).
%H <a href="/index/Pri">Index entries for sequences that are related to primes dividing Fermat numbers</a>
%o (PARI) forprime(p=3,,r=znorder(Mod(2,p));hammingweight(r)==1&&print1(logint(r,2)-1,", ")) \\ _Jeppe Stig Nielsen_, Mar 04 2018
%Y Cf. A000215, A023394, A046052.
%K nonn,more
%O 1,3
%A _David W. Wilson_
%E a(25)-a(41) computed using data from Wilfrid Keller by _T. D. Noe_, Feb 01 2009
%E Three more terms by _T. D. Noe_, Feb 03 2009
%E Six more terms from Wilfrid Keller by _T. D. Noe_, Jan 14 2013
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