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%I #42 Aug 26 2023 11:53:51
%S 1,0,1,0,7,0,39,0,419,0,7208,0,226512,0,7885970,0,345718580,0,
%T 18478915794,0
%N Number of undirected circular permutations i_1,...,i_{n-1} of 1,...,n-1 with i_1-i_2, ..., i_{n-2}-i_{n-1}, i_{n-1}-i_1 pairwise distinct modulo n.
%C If i_1,...,i_{n-1} is a permutation of 1,...,n-1 with i_1-i_2, ..., i_{n-2}-i_{n-1}, i_{n-1}-i_1 pairwise distinct modulo n, then 0 = (i_1-i_2)+...+(i_{n-1}-i_1) == 1+2+...+(n-1) = n(n-1)/2 (mod n) and hence n is odd. So a(n) = 0 for every n = 4,6,8,...
%C If g is a primitive root modulo an odd prime p, then 1-g, g-g^2, g^2-g^3, ..., g^{p-3}-g^{p-2},g^{p-2}-1 are pairwise distinct modulo p. Therefore a(p) > 0 for any odd prime p.
%C Conjecture: a(n) > 0 for any odd number n > 1. In general, if G is an additive abelian group with |G| = n odd and greater than one, then there is a permutation a_1,...,a_{n-1} of all the nonzero elements of G such that a_1-a_2, a_2-a_3, ..., a_{n-2}-a_{n-1}, a_{n-1}-a_1 are pairwise distinct.
%C For any integer n > 1, the author has showed that there is a permutation i_1, ..., i_n of 1, ..., n such that i_1-i_2, i_2-i_3, ..., i_{n-1}-i_n are pairwise distinct if and only if n is even.
%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(3) = 1 since the circular permutation (1,2) of 1,2 meets the requirement.
%e a(5) = 1 due to the circular permutation (1,2,4,3).
%e a(7) = 7 due to the following circular permutations:
%e (1,2,5,4,6,3), (1,2,6,4,3,5), (1,3,2,5,6,4), (1,3,2,6,4,5),
%e (1,3,4,2,6,5), (1,4,5,3,2,6), (1,5,4,2,3,6).
%e a(9) > 0 due to the circular permutation (1,2,5,3,7,6,8,4).
%e a(15) > 0 due to the circular permutation
%e (1,3,14,7,4,11,5,10,9,12,13,2,8,6).
%e a(21) > 0 due to the circular permutation
%e (1,2,11,8,19,15,9,4,10,3,6,13,18,7,5,17,16,20,12,14).
%e Permutations for n = 15, 21 were produced by Qing-Hu Hou at Nankai Univ. after the author told him the conjecture.
%p A228762 := proc(n)
%p local a, pL,p,mset,per,i ;
%p a := 0 ;
%p pL := combinat[permute](n-2) ;
%p for p in pL do
%p mset := {} ;
%p per := [1,seq(op(i,p)+1,i=1..nops(p))] ;
%p # only directed
%p if op(2,per) <= op(-1,per) then
%p for i from 1 to nops(per) do
%p if i = nops(per) then
%p mset := mset union { modp(n+op(i,per)-op(1,per),n) } ;
%p else
%p mset := mset union { modp(n+op(i,per)-op(i+1,per),n) } ;
%p end if;
%p end do:
%p if nops(mset) = n-1 then
%p a := a+1 ;
%p end if;
%p end if;
%p end do:
%p return a;
%p end proc:
%p for n from 3 do
%p A228762(n) ;
%p end do; # _R. J. Mathar_, Sep 03 2013
%t A program to compute required circular permutations of 1,...,7 (beginning with 1). To get undirected circular permutations, we should identify one such a permutation with the one of the opposite direction; for example, (1,3,6,4,5,2) is identical to (1,2,5,4,6,3).
%t V[i_]:=Part[Permutations[{2,3,4,5,6}],i]
%t m=0
%t Do[If[Length[Union[{Mod[1-Part[V[i],1],7]},Table[Mod[Part[V[i],j]-If[j<5,Part[V[i],j+1],1],7],{j,1,5}]]]<6,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]];Label[aa];Continue,{i,1,5!}]
%Y Cf. A185645, A228728, A228766.
%K nonn,more
%O 3,5
%A _Zhi-Wei Sun_, Sep 03 2013
%E a(9) and a(11) added by _R. J. Mathar_, Sep 03 2013
%E a(12)-a(22) from _Robin Visser_, Aug 26 2023
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