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A282779 Period of cubes mod n. 1

%I

%S 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,

%T 9,28,29,30,31,32,33,34,35,12,37,38,39,40,41,42,43,44,15,46,47,48,49,

%U 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

%N Period of cubes mod n.

%C 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_).

%C 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.

%C 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.

%H Ilya Gutkovskiy, <a href="/A282779/a282779.pdf">Extended graphical example</a>

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

%F 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

%e 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.

%t 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}]

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

%K nonn,easy

%O 1,2

%A _Ilya Gutkovskiy_, Feb 21 2017

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Last modified February 18 02:20 EST 2018. Contains 299297 sequences. (Running on oeis4.)