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A023137 Number of cycles of function f(x) = 5x mod n. 1
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; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

Number of factors in the factorization of the polynomial x^n-1 over the integers mod 5. - T. D. Noe, Apr 16 2003

REFERENCES

R. Lidl and H. Niederreiter, Finite Fields, Addison-Wesley, 1983, p. 65.

LINKS

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

FORMULA

a(n) = Sum_{d|m} phi(d)/ord(5, d), where m is n with all factors of 5 removed. - T. D. Noe, Apr 19 2003

EXAMPLE

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.

MATHEMATICA

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}]

CROSSREFS

Cf. A000005, A000374, A023135-A023136, A023138-A023142.

Sequence in context: A264799 A248503 A214714 * A065273 A140819 A138558

Adjacent sequences:  A023134 A023135 A023136 * A023138 A023139 A023140

KEYWORD

nonn

AUTHOR

David W. Wilson

STATUS

approved

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Last modified February 22 05:58 EST 2018. Contains 299430 sequences. (Running on oeis4.)