OFFSET
3,6
COMMENTS
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.
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.
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.
LINKS
Zhi-Wei Sun, Some new problems in additive combinatorics, arXiv preprint arXiv:1309.1679 [math.NT], 2013-2014.
EXAMPLE
a(4) = 1 due to the circular permutation (1,2,3).
a(5) = 1 due to the circular permutation (1,2,4,3).
a(6) = 1 due to the circular permutation (1,3,5,2,4).
a(7) = 1 due to the circular permutation (1,3,2,6,4,5).
a(8) = 12 due to the circular permutations
(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).
a(9) > 0 due to the permutation (1,2,3,4,6,5,8,7).
a(10) > 0 due to the permutation (1,2,4,5,6,8,9,3,7).
a(11) > 0 due to the permutation (1,2,3,4,6,7,5,10,9,8).
MATHEMATICA
(* 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. *)
V[i_]:=Part[Permutations[{2, 3, 4, 5, 6, 7, 8}], i]
m=0
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]];
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!}]
PROG
(Sage)
import itertools
def a(n):
ans = 0
for p in itertools.permutations([i for i in range(1, n)]):
if len(set((p[i]+p[(i+1)%(n-1)])%n for i in range(n-1))) == n-1: ans += 1
return ans/(2*n-2) # Robin Visser, Sep 27 2023
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Zhi-Wei Sun, Sep 03 2013
EXTENSIONS
a(12)-a(19) from Bert Dobbelaere, Sep 08 2019
a(20)-a(21) from Robin Visser, Sep 27 2023
STATUS
approved