A282779 Period of cubes mod n.

1, 2, 3, 4, 5, 6, 7, 8, 3, 10, 11, 12, 13, 14, 15, 16, 17, 6, 19, 20, 21, 22, 23, 24, 25, 26, 9, 28, 29, 30, 31, 32, 33, 34, 35, 12, 37, 38, 39, 40, 41, 42, 43, 44, 15, 46, 47, 48, 49, 50, 51, 52, 53, 18, 55, 56, 57, 58, 59, 60, 61, 62, 21, 64, 65, 66, 67, 68, 69, 70, 71, 24, 73, 74, 75, 76, 77, 78, 79, 80, 27

Offset: 1

Author

Ilya Gutkovskiy, Feb 21 2017

Keywords

• nonn
• easy

Comments

The length of the period of A000035 (n=2), A010872 (n=3), A109718 (n=4), A070471 (n=5), A010875 (n=6), A070472 (n=7), A109753 (n=8), A167176 (n=9), A008960 (n = 10), etc. (see also comment in A000578 from R. J. Mathar).

Conjecture: let a_p(n) be the length of the period of the sequence k^p mod n where p is a prime, then a_p(n) = n/p if n == 0 (mod p^2) else a_p(n) = n.

For example: sequence k^7 mod 98 gives 1, 30, 31, 18, 19, 48, 49, 50, 79, 80, 67, 68, 97, 0, 1, 30, 31, 18, 19, 48, 49, 50, 79, 80, 67, 68, 97, 0, ... (period 14), 7 is a prime, 98 == 0 (mod 7^2) and 98/7 = 14.

Examples

			a(9) = 3 because reading 1, 8, 27, 64, 125, 216, 343, 512, ... modulo 9 gives 1, 8, 0, 1, 8, 0, 1, 8, 0, ... with period length 3.
		

Links

• Ray Chandler, Table of n, a(n) for n = 1..10000
• Ilya Gutkovskiy, Extended graphical example
• Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, -1).

Crossrefs

Cf. A000035, A000578, A008960, A010872, A010875, A046530, A070471, A070472, A109718, A109753, A167176, A186646.

Programs

Mathematica

a[1] = 1; a[n_] := For[k = 1, True, k++, If[Mod[k^3, n] == 0 && Mod[(k + 1)^3 , n] == 1, Return[k]]]; Table[a[n], {n, 1, 81}]

Formula

Apparently: a(n) = 2*a(n-9) - a(n-18).

Empirical g.f.: x*(1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + 8*x^7 + 3*x^8 + 8*x^9 + 7*x^10 + 6*x^11 + 5*x^12 + 4*x^13 + 3*x^14 + 2*x^15 + x^16) / ((1 - x)^2*(1 + x + x^2)^2*(1 + x^3 + x^6)^2). - Colin Barker, Feb 21 2017