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 A023138 Number of cycles of function f(x) = 6x mod n. 7
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,5 LINKS T. D. Noe, Table of n, a(n) for n = 1..10000 FORMULA 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 EXAMPLE a(11) = 2 because the function 6x mod 11 has the two cycles (0),(1,6,3,7,9,10,5,8,4,2). MATHEMATICA 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}] CROSSREFS Cf. A000374. Cf. A023135, A023136, A023137, A023139, A023140, A023141, A023142. Sequence in context: A295882 A127551 A256547 * A194704 A108170 A086988 Adjacent sequences:  A023135 A023136 A023137 * A023139 A023140 A023141 KEYWORD nonn AUTHOR STATUS approved

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Last modified January 23 13:09 EST 2019. Contains 319394 sequences. (Running on oeis4.)