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A023136 Number of cycles of function f(x) = 4x mod n. 8
1, 1, 3, 1, 3, 3, 3, 1, 5, 3, 3, 3, 3, 3, 9, 1, 5, 5, 3, 3, 9, 3, 3, 3, 5, 3, 7, 3, 3, 9, 7, 1, 9, 5, 9, 5, 3, 3, 9, 3, 5, 9, 7, 3, 15, 3, 3, 3, 5, 5, 15, 3, 3, 7, 9, 3, 9, 3, 3, 9, 3, 7, 23, 1, 13, 9, 3, 5, 9, 9, 3, 5, 9, 3, 15, 3, 9, 9, 3, 3, 9, 5, 3, 9, 23, 7, 9, 3, 9, 15, 17, 3, 21, 3, 9, 3, 5, 5, 15, 5, 3, 15 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,3

LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = Sum_{d|m} phi(d)/ord(4, d), where m is n with all factors of 2 removed. The formula was developed by extending the ideas in A000374 to composite multipliers. - T. D. Noe, Apr 21 2003

Mobius transform of A133702: (1, 2, 4, 3, 4, 8, 4, 4, 9, 8,...). = Row sums of triangle A133703. - Gary W. Adamson, Sep 21 2007

EXAMPLE

a(9) = 5 because the function 4x mod 9 has the five cycles (0),(3),(6),(1,4,7),(2,8,5).

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[4, n], {n, 100}]

PROG

(PARI) a(n)=sumdiv(n>>valuation(n, 2), d, eulerphi(d)/znorder(Mod(4, d))) \\ Charles R Greathouse IV, Aug 05 2016

CROSSREFS

Cf. A000374, A023135-A023142, A133703, A133702.

Sequence in context: A173854 A059789 A275367 * A152774 A275820 A210146

Adjacent sequences:  A023133 A023134 A023135 * A023137 A023138 A023139

KEYWORD

nonn

AUTHOR

David W. Wilson

STATUS

approved

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Last modified October 18 18:10 EDT 2018. Contains 316323 sequences. (Running on oeis4.)