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A023136
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Number of cycles of function f(x) = 4x mod n.
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7
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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; internal format)
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OFFSET
| 1,3
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LINKS
| T. D. Noe, Table of n, a(n) for n=1..10000
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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 (noe(AT)sspectra.com), 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 (qntmpkt(AT)yahoo.com), Sep 21 2007
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EXAMPLE
| a(9) = 5 because the function 4x mod 9 has the five cycles (0),(3),(6),(1,4,7),(2,8,5).
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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[4, n], {n, 100}]
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CROSSREFS
| Cf. A000374, A023135-A023142.
Cf. A133703, A133702.
Sequence in context: A072219 A173854 A059789 * A152774 A068074 A063195
Adjacent sequences: A023133 A023134 A023135 * A023137 A023138 A023139
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KEYWORD
| nonn
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AUTHOR
| David W. Wilson (davidwwilson(AT)comcast.net)
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