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Search: id:A023137
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| A023137 |
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Number of cycles of function f(x) = 5x mod n. |
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1
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| 1, 2, 2, 4, 1, 4, 2, 6, 3, 2, 3, 8, 4, 4, 2, 8, 2, 6, 3, 4, 5, 6, 2, 14, 1, 8, 4, 8, 3, 4, 11, 10, 6, 4, 2, 12, 2, 6, 11, 6, 3, 10, 2, 12, 3, 4, 2, 20, 3, 2, 5, 16, 2, 8, 3, 14, 6, 6, 3, 8, 3, 22, 12, 12, 4, 12, 4, 8, 5, 4, 15, 22, 2, 4, 2, 12, 6, 22, 3, 8, 5, 6, 2, 20, 2, 4, 8, 18, 3, 6, 11, 8, 22, 4, 3, 26
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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Number of factors in the factorization of the polynomial x^n-1 over the integers mod 5. - T. D. Noe (noe(AT)sspectra.com), Apr 16 2003
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REFERENCES
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R. Lidl and H. Niederreiter, Finite Fields, Addison-Wesley, 1983, p. 65.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..10000
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FORMULA
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a(n) = Sum_{d|m} phi(d)/ord(5, d), where m is n with all factors of 5 removed. - T. D. Noe (noe(AT)sspectra.com), Apr 19 2003
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EXAMPLE
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a(15) = 2 because (1) the function 5x mod 15 has the two cycles (0),(5,10) and (2) the factorization of x^15-1 over integers mod 5 is (4+x)^5 (1+x+x^2)^5, which has two unique factors. Note that the length of the cycles is the same as the degree of the factors.
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MATHEMATICA
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Table[Length[FactorList[x^n - 1, Modulus -> 5]] - 1, {n, 100}]
CountFactors[p_, n_] := Module[{sum=0, m=n, d, f, i}, While[Mod[m, p]==0, m/=p]; d=Divisors[m]; Do[f=d[[i]]; sum+=EulerPhi[f]/MultiplicativeOrder[p, f], {i, Length[d]}]; sum]; Table[CountFactors[5, n], {n, 100}]
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CROSSREFS
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Cf. A000005, A000374, A023135-A023136, A023138-A023142.
Sequence in context: A099320 A034951 A064848 this_sequence A065273 A111580 A138558
Adjacent sequences: A023134 A023135 A023136 this_sequence A023138 A023139 A023140
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KEYWORD
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nonn
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AUTHOR
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David W. Wilson (davidwwilson(AT)comcast.net)
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