A337596 - OEIS

%I #33 Sep 11 2020 09:59:26

%S 3,5,9,16,11,13,4,32,27,25,23,16,4,29,31,64,4,37,4,41,49,23,47,32,11,

%T 53,81,29,59,61,4,128,67,8,71,73,4,8,79,41,83,49,4,89,31,47,4,97,4,

%U 125,103,53,107,109,121,113,9,59,4,61,4,8,127,256,131,67,4,137

%N Largest m such that k^n (mod m) is always either 0, +1, or -1.

%C For a given n, for all k, k^n mod a(n) will always be either 0, 1 or a(n)-1. This will not be true for numbers larger than a(n).

%C It appears that a(m) = 4 for m in A045979. - _Michel Marcus_, Sep 04 2020

%e For n = 5 all fifth powers of natural numbers: 1,32,243,1024, etc. are either a multiple of 11, or 1 greater or 1 less than a multiple of 11. There is no greater number than 11 for which all fifth powers are at most 1 different from a multiple. So a(5) = 11.

%Y Cf. A010872, A070430, A167176, A000035, A070596, A070636, A109718.

%Y Cf. residues: A096008 (for n=2), A096087 (for n=3).

%K nonn

%O 1,1

%A _Elliott Line_, Sep 02 2020

%E More terms from _Michel Marcus_, Sep 04 2020