A023137 - OEIS

A023137

Number of cycles of function f(x) = 5x mod n.

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

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OFFSET

1,2

COMMENTS

Number of factors in the factorization of the polynomial x^n-1 over GF(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

a(n) = (1/ord(5,m))*Sum_{j = 0..ord(5,m)-1} gcd(5^j - 1, m), where m is n with all factors of 5 removed. - Nihar Prakash Gargava, Nov 14 2018

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}]

PROG

(PARI) a(n)={sumdiv(n/5^valuation(n, 5), d, eulerphi(d)/znorder(Mod(5, d))); }

vector(100, n, a(n)) \\ Joerg Arndt, Jan 22 2024

(Python)

from sympy import totient, n_order, divisors

def A023137(n):

a, b = divmod(n, 5)

while not b:

n = a

a, b = divmod(n, 5)

return sum(totient(d)//n_order(5, d) for d in divisors(n, generator=True) if d>1)+1 # Chai Wah Wu, Apr 09 2024

CROSSREFS