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A023140
Number of cycles of function f(x) = 8x mod n.
7
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
1,3
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
EXAMPLE
a(10) = 2 because the function 8x mod 10 has the two cycles (0),(2,6,8,4).
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}]
KEYWORD
nonn
STATUS
approved