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 A282779 Period of cubes mod n. 1
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 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. LINKS Ilya Gutkovskiy, Extended graphical example 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 EXAMPLE 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. 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}] CROSSREFS Cf. A000035, A000578, A008960, A010872, A010875, A046530, A070471, A070472, A109718, A109753, A167176, A186646. Sequence in context: A279649 A278060 A160597 * A305902 A245350 A245354 Adjacent sequences:  A282776 A282777 A282778 * A282780 A282781 A282782 KEYWORD nonn,easy AUTHOR Ilya Gutkovskiy, Feb 21 2017 STATUS approved

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Last modified October 15 22:46 EDT 2018. Contains 316252 sequences. (Running on oeis4.)