A337596 Largest m such that k^n (mod m) is always either 0, +1, or -1.

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, 53, 81, 29, 59, 61, 4, 128, 67, 8, 71, 73, 4, 8, 79, 41, 83, 49, 4, 89, 31, 47, 4, 97, 4, 125, 103, 53, 107, 109, 121, 113, 9, 59, 4, 61, 4, 8, 127, 256, 131, 67, 4, 137

Offset: 1

Author

Elliott Line, Sep 02 2020

Keywords

• nonn

Comments

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).

It appears that a(m) = 4 for m in A045979. - Michel Marcus, Sep 04 2020

Examples

			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.
		

Crossrefs

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

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

Extensions

More terms from Michel Marcus, Sep 04 2020