A023140 Number of cycles of function f(x) = 8x mod n.
1, 1, 2, 1, 2, 2, 7, 1, 5, 2, 2, 2, 4, 7, 5, 1, 3, 5, 4, 2, 14, 2, 3, 2, 3, 4, 8, 7, 2, 5, 7, 1, 5, 3, 14, 5, 4, 4, 11, 2, 3, 14, 4, 2, 14, 3, 3, 2, 13, 3, 8, 4, 2, 8, 5, 7, 11, 2, 2, 5, 4, 7, 35, 1, 17, 5, 4, 3, 6, 14, 3, 5, 25, 4, 8, 4, 14, 11, 7, 2, 11, 3, 2, 14, 12, 4, 5, 2, 9, 14, 28, 3, 14, 3, 11, 2, 7
Offset: 1
Keywords
Examples
a(10) = 2 because the function 8x mod 10 has the two cycles (0),(2,6,8,4).
Links
- T. D. Noe, Table of n, a(n) for n = 1..10000
Programs
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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[8, n], {n, 100}]
Formula
a(n) = Sum_{d|m} phi(d)/ord(8, d), where m is n with all factors of 2 removed. - T. D. Noe, Apr 21 2003
a(n) = (1/ord(8,m))*Sum_{j = 0..ord(8,m)-1} gcd(8^j - 1, m), where m is n with all factors of 2 removed. - Nihar Prakash Gargava, Nov 14 2018