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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.

%I A384184 #72 Jun 23 2025 22:11:41
%S A384184 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,
%T A384184 12,32,5,8,70,12,18,18,24,16,10,8,7,20,210,56,126,16,110,20,60,24,26,
%U A384184 18,120,16,9,28,29,24,30,24,60,64,6,10,33,16
%N 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.
%C A384184 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.
%H A384184 Mia Boudreau, <a href="/A384184/b384184.txt">Table of n, a(n) for n = 1..10000</a>
%H A384184 Mia Boudreau, <a href="https://github.com/miabea73/MiaOEIS/tree/main/A384184">Java program, GitHub</a>
%F A384184 a(2*n) = 2*a(n).
%F A384184 a(2^n) = 2^n.
%F A384184 Conjecture: a(2^n + 2^x) = 2^n * (x-n) if x > n.
%F A384184 a(2^n - 1) = A003418(n-1).
%F A384184 s(2^n + 1) = A000027(n).
%F A384184 a(2*n - 1) = A051732(n).
%F A384184 a(A004626(n)) % 2 = 1.
%F A384184 a(A065119(n)) = n/3.
%F A384184 a(A001122(n)) = (n-1) / 2.
%F A384184 a(A155072(n)) = (n-1) / 4.
%F A384184 a(A001133(n)) = (n-1) / 6.
%F A384184 a(A001134(n)) = (n-1) / 8.
%F A384184 a(A001135(n)) = (n-1) / 10.
%F A384184 a(A225759(n)) = (n-1) / 16.
%e A384184 For n = 11, the permutation is {0,3,4,7,8,1,2,9,10,5,6} and it has order a(11) = 5.
%o A384184 (Python)
%o A384184 from sympy.combinatorics import Permutation
%o A384184 def a(n):
%o A384184    L = list(range(n))
%o A384184    for i in range(n):
%o A384184        if (j:= (i << 1) % n) != i:
%o A384184            L[i],L[j] = L[j],L[i]
%o A384184    return Permutation(L).order() # _Darío Clavijo_, Jun 05 2025
%Y A384184 Cf. A003418, A000027, A051732, A004626, A065119, A001122, A155072, A001133, A001134, A001135, A225759.
%K A384184 nonn
%O A384184 1,2
%A A384184 _Mia Boudreau_, May 29 2025