

A229878


Number of undirected circular permutations tau(1), ..., tau((p_n1)/2) of 1, ..., (p_n1)/2 such that the (p_n1)/2 numbers tau(1)^2 + tau(2)^2, tau(2)^2 + tau(3)^2, ..., tau((p_n3)/2)^2 + tau((p_n1)/2)^2, tau((p_n1)/2)^2 + tau(1)^2 give all the (p_n1)/2 quadratic residues modulo p_n, where p_n is the nth prime.


1




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_{n1}+a_n, a_n+a_1 give all the n quadratic residues modulo p.
ZhiWei 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_{n1}+a_n, a_n+a_1} = S (or T). Also, there exists a circular permutation b_1, ..., b_n of S with {b_1b_2, ..., b_{n1}b_n, b_nb_1} = S (or T).


LINKS

Table of n, a(n) for n=3..8.
ZhiWei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 20132014.


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

Cf. A229038.
Sequence in context: A137979 A160338 A216579 * A235145 A266342 A285936
Adjacent sequences: A229875 A229876 A229877 * A229879 A229880 A229881


KEYWORD

nonn,more,hard


AUTHOR

ZhiWei Sun, Oct 02 2013


STATUS

approved



