|
| |
|
|
A023135
|
|
Number of cycles of function f(x) = 3x mod n.
|
|
11
|
|
|
|
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
(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 GF(3). - T. D. Noe, Apr 16 2003
|
|
|
REFERENCES
|
R. Lidl and H. Niederreiter, Finite Fields, Addison-Wesley, 1983, p. 65.
|
|
|
LINKS
|
|
|
|
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))); }
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
