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A227399
Number of permutations i_1, ..., i_n of 1, ..., n with i_1 = 1 and i_n = n such that i_1+2*i_2, i_2+2*i_3, ..., i_{n-1}+2*i_n, i_n+2*i_1 are pairwise distinct modulo n.
0
1, 1, 0, 1, 1, 2, 8, 20, 18, 166, 397, 2788, 5448, 78102, 149562, 2576896, 6003432, 91012592, 257246112, 5272355344, 12450552690
OFFSET
1,6
COMMENTS
If n is not divisible by 3 then a(n) > 0 since the identical permutation of 1 ,..., n works for the purpose. If n is even, then a(n) > 0 since the permutation 1, 2, n-1, 4, n-3, 6, n-5, ..., n-2, 3, n meets the requirement. We guess that a(n) > 0 in the remaining case n = 6q+3 with q > 0. If n = 2k+1 == 3 (mod 6) with n > 3, then, for the permutation (i_1,...,i_n) = (1,2k,k,2k-1,k-1,2k-2,...,3,k+2,2,k+1,2k+1), all the n sums i_1+2*i_2, i_2+2*i_3, ..., i_{n-1}+2*i_n, i_n+2*i_1 are pairwise distinct (but they are not pairwise incongruent modulo n = 2k+1 when n > 9).
Zhi-Wei Sun also made the following general conjecture:
(i) (Weak version) Let a_1,...,a_n be n distinct elements of an additive abelian group G. Then, there is a permutation b_1,...,b_n of a_1,...,a_n such that a_1+2*b_1, a_2+2*b_2, ..., a_n+2*b_n are pairwise distinct. (The author has proved this for n up to 4 in any abelian group G.)
(ii) (Strong version) Let A be any subset of an additive abelian group G with |A| = n > 4. Then there is a numbering a_1, ..., a_n of all the elements of A such that a_1+2*a_2, a_2+2*a_3, ..., a_{n-1}+2*a_n, a_n+2*a_1 are pairwise distinct. (The author has proved this for any torsion-free abelian group G.)
Recall that a conjecture of Snevily proved by Arsovski states that for any two n-subsets A and B of an additive abelian group of odd order there is a numbering a_1,...,a_n of all the elements of A and a numbering b_1, ..., b_n of all the elements of B such that the n sums a_1+b_1, ..., a_n+b_n are pairwise distinct.
LINKS
B. Arsovski, A proof of Snevily's conjecture, Israel J. Math. 182(2011), 505-508.
Zhi-Wei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 2013-2014.
EXAMPLE
a(6) = 2 due to the permutations 1,2,5,4,3,6 and 1,4,3,2,5,6.
a(9) > 0 due to the permutation 1,2,3,5,8,4,7,6,9.
a(12) > 0 due to the permutation 1,2,3,4,6,8,5,11,10,7,9,12.
MATHEMATICA
(* A program to compute desired permutations for n = 9. *)
V[i_]:=Part[Permutations[{2, 3, 4, 5, 6, 7, 8}], i]
m=0
Do[If[Length[Union[{2}, Table[Mod[If[j==0, 1, Part[V[i], j]]+2*If[j<7, Part[V[i], j+1], 9], 9], {j, 0, 7}]]]<9, Goto[aa]];
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], " ", 9]; Label[aa]; Continue, {i, 1, 7!}]
CROSSREFS
Sequence in context: A082821 A188893 A227127 * A327098 A030097 A136904
KEYWORD
nonn,more,hard
AUTHOR
Zhi-Wei Sun, Sep 20 2013
EXTENSIONS
a(12)-a(21) from Robin Visser, Aug 21 2023
STATUS
approved