A229543 Number of undirected circular permutations i_0, i_1, ..., i_n of 0, 1, ..., n such that all the n+1 adjacent distances |i_0-i_1|, |i_1-i_2|, ..., |i_{n-1}-i_n|, |i_n-i_0| are perfect squares.

1, 0, 0, 1, 0, 1, 1, 1, 9, 14, 32, 184, 123, 696, 935, 6554, 21105, 60756, 241780, 517970, 1835109, 5741024, 16091004, 63590090, 285113492, 1098219807

Offset: 1

Author

Zhi-Wei Sun, Sep 25 2013

Keywords

• nonn
• more
• hard

Comments

Theorem: For any integer n > 5, there is a circular permutation i_0, i_1, ..., i_n of 0, 1, ..., n with both 1 and 4 adjacent to 0 such that all the n+1 adjacent distances |i_0-i_1|, |i_1-i_2|, ..., |i_{n-1}-i_n|, |i_n-i_0| are perfect squares.

Proof: By examples, the result holds for n = 6, ..., 11. Below we assume n > 11 and exhibit a circular permutation meeting the requirement.

If n == 0 (mod 6), then we may take the circular permutation (n,n-4,n-5,n-6,n-10,n-11,n-12,...,8,7,6,2,3,4,0,1,5,9,10,11,...,n-9,n-8,n-7,n-3,n-2,n-1).

If n == 3 (mod 6), then we may take the circular permutation

(n,n-4,n-5,n-6,n-10,n-11,n-12,...,11,10,9,5,1,0,4,3,2,6,7,8,...,n-9,n-8,n-7,n-3,n-2,n-1).

If n == 1 (mod 6), then we may take the circular permutation

(n,n-4,n-5,n-6,n-10,n-11,n-12,...,9,8,7,3,2,1,0,4,5,6,10,11,12,...,n-9,n-8,n-7,n-3,n-2,n-1).

If n == 4 (mod 6), then we may take the circular permutation (n,n-4,n-5,n-6,n-10,n-11,n-12,...,12,11,10,6,5,4,0,1,2,3,7,8,9,...,n-9,n-8,n-7,n-3,n-2,n-1).

If n == 2 (mod 6), then we may take the circular permutation (n,n-4,n-5,n-6,n-10,n-11,n-12,...,10,9,8,4,0,1,5,6,2,3,7,11,12,13,...,n-9,n-8,n-7,n-3,n-2,n-1).

If n == 5 (mod 6), then we may take the circular permutation

(n,n-4,n-5,n-6,n-10,n-11,n-12,...,13,12,11,7,3,2,6,5,1,0,4,8,9,10,...,n-9,n-8,n-7,n-3,n-2,n-1).

Zhi-Wei Sun also used a similar method to show that for any positive integer n not equal to 2 or 4 there is a circular permutation i_0, i_1, ..., i_n of 0, 1, ..., n such that all the n+1 adjacent distances |i_0-i_1|, |i_1-i_2|, ..., |i_{n-1}-i_n|, |i_n-i_0| are triangular numbers.

Examples

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

Links

• Zhi-Wei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679 [math.NT], 2013-2014.

Crossrefs

Cf. A000290, A185645, A228626, A228728, A228956, A229005.

Programs

Mathematica

(* A program to compute required circular permutations for n = 8. 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. *)
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
f[i_,j_]:=f[i,j]=SQ[Abs[i-j]]
V[i_]:=V[i]=Part[Permutations[{1,2,3,4,5,6,7,8}],i]
m=0
Do[Do[If[f[If[j==0,0,Part[V[i],j]],If[j<8,Part[V[i],j+1],0]]==False,Goto[aa]],{j,0,8}];m=m+1;Print[m,":"," ",0," ",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!}]

Extensions

a(10)-a(24) from Alois P. Heinz, Sep 26 2013

a(25)-a(26) from Max Alekseyev, Jan 08 2015