A023135 Number of cycles of function f(x) = 3x mod n.

1, 2, 1, 3, 2, 2, 2, 5, 1, 4, 3, 3, 5, 4, 2, 7, 2, 2, 2, 7, 2, 6, 3, 5, 3, 10, 1, 7, 2, 4, 2, 9, 3, 4, 5, 3, 3, 4, 5, 13, 6, 4, 2, 9, 2, 6, 3, 7, 3, 6, 2, 15, 2, 2, 6, 13, 2, 4, 3, 7, 7, 4, 2, 11, 10, 6, 4, 7, 3, 10, 3, 5, 7, 6, 3, 7, 6, 10, 2, 23, 1, 12, 3, 7, 7, 4, 2, 15, 2, 4, 18, 9, 2, 6, 5, 9, 3, 6, 3, 11

Offset

1,2

Comments

Number of factors in the factorization of the polynomial x^n-1 over GF(3). - 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(3, d), where m is n with all factors of 3 removed. - T. D. Noe, Apr 19 2003

a(n) = (1/ord(3,m))*Sum_{j = 0..ord(3,m)-1} gcd(3^j - 1, m), where m is n with all factors of 3 removed. - Nihar Prakash Gargava, Nov 14 2018

Example

a(15) = 2 because (1) the function 3x mod 15 has the two cycles (0),(3,9,12,6) and (2) the factorization of x^15-1 over integers mod 3 is (2+x)^3 (1+x+x^2+x^3+x^4)^3, 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 -> 3]] - 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[3, n], {n, 100}]

Prog

(PARI) a(n)={sumdiv(n/3^valuation(n, 3), d, eulerphi(d)/znorder(Mod(3, d))); }
vector(100, n, a(n)) \\ Joerg Arndt, Jan 22 2024

Crossrefs

Cf. A000005, A000374.

Cf. A023136, A023137, A023138, A023139, A023140, A023141, A023142.

Sequence in context: A276166 A177062 A133924 * A345058 A191654 A327983

Adjacent sequences: A023132 A023133 A023134 * A023136 A023137 A023138

Keyword

nonn

Author

David W. Wilson

Status

approved