|
| |
|
|
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; 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 (noe(AT)sspectra.com), 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 (noe(AT)sspectra.com), 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: A205843 A171727 A171942 * A065273 A140819 A111580
Adjacent sequences: A023134 A023135 A023136 * A023138 A023139 A023140
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| David W. Wilson (davidwwilson(AT)comcast.net)
|
| |
|
|
