A384184 Order of the permutation of {0,...,n-1} formed by successively swapping elements at i and 2*i mod n, for i = 0,...,n-1.

1, 2, 1, 4, 2, 2, 2, 8, 3, 4, 5, 4, 6, 4, 6, 16, 4, 6, 9, 8, 4, 10, 28, 8, 10, 12, 9, 8, 14, 12, 12, 32, 5, 8, 70, 12, 18, 18, 24, 16, 10, 8, 7, 20, 210, 56, 126, 16, 110, 20, 60, 24, 26, 18, 120, 16, 9, 28, 29, 24, 30, 24, 60, 64, 6, 10, 33, 16

Offset

1,2

Comments

a(2*n) = 2*a(n) since the cycle lengths of the permutation with size 2*n is effectively that of size n twice, doubled. Thus, the LCM/order is doubled.

Links

Mia Boudreau, Table of n, a(n) for n = 1..10000

Mia Boudreau, Java program, GitHub

Formula

a(2*n) = 2*a(n).

a(2^n) = 2^n.

Conjecture: a(2^n + 2^x) = 2^n * (x-n) if x > n.

a(2^n - 1) = A003418(n-1).

s(2^n + 1) = A000027(n).

a(2*n - 1) = A051732(n).

a(A004626(n)) % 2 = 1.

a(A065119(n)) = n/3.

a(A001122(n)) = (n-1) / 2.

a(A155072(n)) = (n-1) / 4.

a(A001133(n)) = (n-1) / 6.

a(A001134(n)) = (n-1) / 8.

a(A001135(n)) = (n-1) / 10.

a(A225759(n)) = (n-1) / 16.

Example

For n = 11, the permutation is {0,3,4,7,8,1,2,9,10,5,6} and it has order a(11) = 5.

Prog

(Python)
from sympy.combinatorics import Permutation
def a(n):
   L = list(range(n))
   for i in range(n):
       if (j:= (i << 1) % n) != i:
           L[i], L[j] = L[j], L[i]
   return Permutation(L).order() # Darío Clavijo, Jun 05 2025

Crossrefs

Cf. A003418, A000027, A051732, A004626, A065119, A001122, A155072, A001133, A001134, A001135, A225759.

Sequence in context: A395523 A136693 A348045 * A086685 A343998 A374015

Adjacent sequences: A384181 A384182 A384183 * A384185 A384186 A384187

Keyword

nonn

Author

Mia Boudreau, May 29 2025

Status

approved