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A023138
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Number of cycles of function f(x) = 6x mod n.
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7
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1, 1, 1, 1, 5, 1, 4, 1, 1, 5, 2, 1, 2, 4, 5, 1, 2, 1, 3, 5, 4, 2, 3, 1, 9, 2, 1, 4, 3, 5, 6, 1, 2, 2, 20, 1, 10, 3, 2, 5, 2, 4, 15, 2, 5, 3, 3, 1, 7, 9, 2, 2, 3, 1, 10, 4, 3, 3, 2, 5, 2, 6, 4, 1, 10, 2, 3, 2, 3, 20, 3, 1, 3, 10, 9, 3, 11, 2, 2, 5, 1, 2, 2, 4, 10, 15, 3, 2, 2, 5, 11, 3, 6, 3, 15, 1, 9, 7, 2, 9, 11
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OFFSET
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1,5
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LINKS
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FORMULA
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a(n) = Sum_{d|m} phi(d)/ord(6, d), where m is n with all factors of 2 and 3 removed. - T. D. Noe, Apr 21 2003
a(n) = (1/ord(6,m))*Sum_{j = 0..ord(6,m)-1} gcd(6^j - 1, m), where m is n with all factors of 2 and 3 removed. - Nihar Prakash Gargava, Nov 14 2018
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EXAMPLE
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a(11) = 2 because the function 6x mod 11 has the two cycles (0),(1,6,3,7,9,10,5,8,4,2).
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MATHEMATICA
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CountFactors[p_, n_] := Module[{sum=0, m=n, d, f, i, ps, j}, ps=Transpose[FactorInteger[p]][[1]]; Do[While[Mod[m, ps[[j]]]==0, m/=ps[[j]]], {j, Length[ps]}]; d=Divisors[m]; Do[f=d[[i]]; sum+=EulerPhi[f]/MultiplicativeOrder[p, f], {i, Length[d]}]; sum]; Table[CountFactors[6, n], {n, 100}]
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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