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A023395
Only Fermat number divisible by A023394(n) is 2^2^a(n) + 1.
1
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, 38, 32, 25, 17, 39, 6, 26, 27, 30, 30, 8, 12, 15, 29, 38, 7, 25, 27, 36, 42, 25, 13, 13, 55
OFFSET
1,3
COMMENTS
From Jianing Song, Mar 02 2021: (Start)
2^(a(n)+1) is the multiplicative order of 2 modulo A023394(n).
Each k occurs A046052(k) times in this sequence provided that F(k) = 2^2^k + 1 is squarefree (no counterexamples are known). (End)
Alternatively, a(n) is the only k such that A023394(n) divides A000215(k). - Lorenzo Sauras Altuzarra, Feb 01 2023
PROG
(PARI) forprime(p=3, , r=znorder(Mod(2, p)); hammingweight(r)==1&&print1(logint(r, 2)-1, ", ")) \\ Jeppe Stig Nielsen, Mar 04 2018
CROSSREFS
KEYWORD
nonn,more
EXTENSIONS
a(25)-a(41) computed using data from Wilfrid Keller by T. D. Noe, Feb 01 2009
Three more terms by T. D. Noe, Feb 03 2009
Six more terms from Wilfrid Keller by T. D. Noe, Jan 14 2013
STATUS
approved