A228766 - OEIS

%I #51 Sep 27 2023 13:44:00

%S 0,1,1,1,1,12,21,74,309,1376,5016,27198,138592,928544,4735266,

%T 31263708,206761952,1677199872,11111483094

%N Number of undirected circular permutations i_1,...,i_{n-1} of 1,...,n-1 with i_1 + i_2, i_2 + i_3, ..., i_{n-2} + i_{n-1}, i_{n-1} + i_1 pairwise distinct modulo n.

%C Conjecture: a(n) > 0 for all n > 3. In general, if a_1,...,a_n are n > 2 distinct elements of a finite additive abelian group G with n odd or |G| not divisible by n, then there exists a circular permutation b_1,...,b_n of a_1,...,a_n such that b_1+b_2, b_2+b_3, ..., b_{n-1}+b_n, b_n+b_1 are pairwise distinct.

%C Note that if g is a primitive root modulo a prime p > 3 then 1+g, g+g^2, ..., g^{p-3}+g^{p-2}, g^{p-2}+1 are pairwise distinct modulo p. So a(p) > 0 for any prime p > 3.

%C If n > 2 is odd, then  0+1, 1+2, ..., (n-2)+(n-1), (n-1)+0 are pairwise distinct modulo n, and hence the conjecture holds in the case {a_1,...,a_n} = G = Z/nZ.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1309.1679">Some new problems in additive combinatorics</a>, arXiv preprint arXiv:1309.1679 [math.NT], 2013-2014.

%e a(4) = 1 due to the circular permutation (1,2,3).

%e a(5) = 1 due to the circular permutation (1,2,4,3).

%e a(6) = 1 due to the circular permutation (1,3,5,2,4).

%e a(7) = 1 due to the circular permutation (1,3,2,6,4,5).

%e a(8) = 12 due to the circular permutations

%e   (1,2,4,5,3,7,6), (1,2,6,7,3,4,5), (1,2,7,6,4,3,5), (1,4,2,5,6,3,7), (1,4,2,7,3,5,6), (1,4,3,7,2,6,5), (1,4,7,3,6,2,5), (1,5,2,3,6,4,7), (1,5,3,2,7,4,6), (1,5,4,7,3,2,6), (1,5,6,4,3,2,7), (1,6,5,4,2,3,7).

%e a(9) > 0 due to the permutation (1,2,3,4,6,5,8,7).

%e a(10) > 0 due to the permutation (1,2,4,5,6,8,9,3,7).

%e a(11) > 0 due to the permutation (1,2,3,4,6,7,5,10,9,8).

%t (* A program to compute required circular permutations for n = 9. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, (1,7,8,5,6,4,3,2) is identitical to (1,2,3,4,6,5,8,7) if we ignore direction. *)

%t V[i_]:=Part[Permutations[{2,3,4,5,6,7,8}],i]

%t m=0

%t Do[If[Length[Union[{Mod[1+Part[V[i],1],9]},Table[Mod[Part[V[i],j]+If[j<7,Part[V[i],j+1],1],9],{j,1,7}]]]<8,Goto[aa]];

%t m=m+1;Print[m,":"," ",1," ",Part[V[i],1]," ",Part[V[i],2]," ",Part[V[i],3]," ",Part[V[i],4]," ",Part[V[i],5]," ",Part[V[i],6]," ",Part[V[i],7]];Label[aa];Continue,{i,1,7!}]

%o (Sage)

%o import itertools

%o def a(n):

%o     ans = 0

%o     for p in itertools.permutations([i for i in range(1, n)]):

%o         if len(set((p[i]+p[(i+1)%(n-1)])%n for i in range(n-1))) == n-1: ans += 1

%o     return ans/(2*n-2)  # _Robin Visser_, Sep 27 2023

%Y Cf. A228762, A228772, A185645, A228728.

%K nonn,more

%O 3,6

%A _Zhi-Wei Sun_, Sep 03 2013

%E a(12)-a(19) from _Bert Dobbelaere_, Sep 08 2019

%E a(20)-a(21) from _Robin Visser_, Sep 27 2023