A229827 Number of circular permutations i_1,...,i_n of 1,...,n such that the n numbers i_1^2+i_2, i_2^2+i_3, ..., i_{n-1}^2+i_n, i_n^2+i_1 form a complete system of residues modulo n.

0, 0, 0, 4, 0, 24, 0, 0, 0, 5308, 0, 123884, 0, 0, 0, 147288372, 0, 7238567052, 0, 0, 0

Offset: 2

Author

Zhi-Wei Sun, Sep 30 2013

Keywords

• nonn
• more
• hard

Comments

Conjecture: (i) a(p) > 0 for any prime p > 3. Moreover, for any finite field F(q) with q elements and a polynomial P(x) over F(q) of degree smaller than q-1, if P(x) is not of the form c-x, then there is a circular permutation a_1, ..., a_q of all the elements of F(q) with

{P(a_1)+a_2, P(a_2)+a_3, ..., P(a_{q-1})+a_q, P(a_q)+a_1}

equal to F(q).

(ii) Let F be any field with |F| > 7, and let A be a finite subset of F with |A| = n > 2. Let P(x) be a polynomial over F whose degree is smaller than p-1 if F is of prime characteristic p. If P(x) is not of the form c-x, then there is a circular permutation a_1, ..., a_n of all the elements of A such that the n sums P(a_1)+a_2, P(a_2)+a_3, ..., P(a_{n-1})+a_n, P(a_n)+a_1 are pairwise distinct.

Clearly, if a(n) > 0, then we must have 1^2+2^2+...+n^2 == 0 (mod n), i.e., (n+1)*(2n+1) == 0 (mod 6). Thus, when n is divisible by 2 or 3, we must have a(n) = 0.

It is well known that for any finite field F(q) with q elements we have sum_{x in F(q)} x^k = 0 for every k = 0, ..., q-2.

Verified for n < 256: a(n) > 0 iff n is not divisible by 2 or 3. - Bert Dobbelaere, Apr 23 2021

Examples

			a(5) = 4 due to the circular permutations
    (1,3,4,5,2), (1,4,2,3,5), (1,5,2,4,3), (1,5,4,2,3).
a(7) > 0 due to the circular permutation (1,2,3,7,4,6,5).
a(11) > 0 due to the circular permutation
    (1,2,3,4,6,5,9,11,10,8,7).
		

Links

• Zhi-Wei Sun, A combinatorial conjecture on finite fields, a message to Number Theory List, Sept. 30, 2013.
• Zhi-Wei Sun, Some new problems in additive combinatorics , preprint, arXiv:1309.1679 [math.NT], 2013-2014.

Crossrefs

Cf. A227456, A229082, A229141.

Programs

Mathematica

(* A program to compute desired circular permutations for n = 7. *)
  V[i_]:=Part[Permutations[{2,3,4,5,6,7}],i]
m=0
Do[If[Length[Union[Table[Mod[If[j==0,1,Part[V[i],j]]^2+If[j<6,Part[V[i],j+1],1],7],{j,0,6}]]]<7,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]];Label[aa];Continue,{i,1,6!}]

Extensions

a(11)-a(22) from Bert Dobbelaere, Apr 23 2021