OFFSET
3,6
COMMENTS
Conjecture: a(n) > 0 for all n > 6. In other words, for any prime p = 2*n+1 > 13, there is a circular permutation a_1, ..., a_n of the n quadratic residues modulo p such that a_1+a_2, a_2+a_3, ..., a_{n-1}+a_n, a_n+a_1 give all the n quadratic residues modulo p.
Zhi-Wei Sun also made the following general conjecture:
Let F be a finite field with |F| = q = 2*n+1 > 13. Let S = {a^2: a is a nonzero element of F} and T = (F\{0})\S. Then there is a circular permutation a_1, ..., a_n of S such that {a_1+a_2, ..., a_{n-1}+a_n, a_n+a_1} = S (or T). Also, there exists a circular permutation b_1, ..., b_n of S with {b_1-b_2, ..., b_{n-1}-b_n, b_n-b_1} = S (or T).
LINKS
Zhi-Wei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 2013-2014.
EXAMPLE
a(5) = 1 due to the circular permutation (1,2,4,3,5). Note that 1^2+2^2, 2^2+4^2, 4^2+3^2, 3^2+5^2, 5^2+1^2 give the 5 quadratic residues modulo p_5 = 11.
a(7) = 1 due to the circular permutation (1,5,8,6,4,3,2,7).
a(8) = 2 due to the circular permutations
(1,2,4,8,3,6,7,5,9) and (1,4,3,7,9,2,8,6,5).
MATHEMATICA
(* A program to compute required circular permutations for n = 8. Note that p_8 = 19. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction. Thus a(8) is half of the number of circular permutations yielded by this program. *)
V[i_]:=Part[Permutations[{2, 3, 4, 5, 6, 7, 8, 9}], i]
m=0
Do[If[Union[Table[Mod[If[j==0, 1, Part[V[i], j]]^2+If[j<8, Part[V[i], j+1], 1]^2, 19], {j, 0, 8}]]!=Union[Table[Mod[k^2, 19], {k, 1, 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], " ", Part[V[i], 8]]; Label[aa]; Continue, {i, 1, 8!}]
CROSSREFS
KEYWORD
nonn,more,hard
AUTHOR
Zhi-Wei Sun, Oct 02 2013
STATUS
approved