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A282779
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Period of cubes mod n.
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1
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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
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OFFSET
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1,2
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COMMENTS
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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.
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LINKS
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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).
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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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}]
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CROSSREFS
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Cf. A000035, A000578, A008960, A010872, A010875, A046530, A070471, A070472, A109718, A109753, A167176, A186646.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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