A228728 a(1)=1, a(2)=2 and for n > 2, a(n) is the least integer > a(n-1) such that there is a permutation b(1), ..., b(n) of a(1), ..., a(n) with b(1) = a(1) and b(n) = a(n), and with the n numbers |b(1)-b(2)|, |b(2)-b(3)|, ..., |b(n-1)-b(n)|, |b(n)-b(1)| pairwise distinct.

1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26

Offset: 1

Author

Zhi-Wei Sun, Aug 31 2013

Keywords

• nonn
• more

Comments

Conjecture: For any n distinct real numbers a_1 < a_2 < ... < a_n, if there is a permutation b_1,b_2,...,b_n of a_1,...,a_n with |b_1-b_2|, |b_2-b_3|, ..., |b_{n-1}-b_n|, |b_n-b_1| pairwise distinct, then there exists a permutation c_1,c_2,...,c_n of a_1,...,a_n with c_1 = a_1 and c_n = a_n such that the n numbers |c_1-c_2|, |c_2-c_3|, ..., |c_{n-1}-c_n|, |c_n-c_1| are pairwise distinct.

This conjecture is somewhat curious but we are unable to find a counterexample.

Examples

			a(3) = 4 since the permutation 1,2,3 does not meet the requirement (since 2-1 = 3-2) but the permutation 1,2,4 is okay as 2-1, 4-2, 4-1 are pairwise distinct.
a(4) = 6 since none of the permutations 1,2,4,5 and 1,4,2,5 meets the requirement (since 5-4 = 2-1 and 5-2 = 4-1), but the permutation 1,4,2,6 is okay as 4-1, 4-2, 6-2, 6-1 are pairwise distinct.
a(5) = 7 due to the permutation 1,6,2,4,7.
a(6) = 8 due to the permutation 1,4,6,2,7,8.
a(7) = 9 due to the permutation 1,4,8,6,7,2,9.
a(8) = 10 due to the permutation 1,7,4,9,8,6,2,10.
a(9) = 11 due to the permutation 1,6,7,9,2,10,4,8,11.
a(10) = 12 due to the permutation 1,6,8,9,2,11,7,10,4,12.
a(11) = 13 due to the permutation 1,12,2,11,4,10,6,9,7,8,13.
a(12) = 14 due to the permutation
     1, 13, 2, 12, 4, 11, 6, 10, 7, 9, 8, 14.
a(13) = 15 due to the permutation
     1, 11, 6, 8, 12, 9, 10, 2, 14, 7, 13, 4, 15.
a(14) = 16 due to the permutation
     1, 12, 9, 8, 10, 15, 2, 11, 7, 13, 6, 14, 4, 16.
a(15) = 17 due to the permutation
     1, 12, 9, 13, 4, 16, 6, 11, 10, 8, 14, 7, 15, 2, 17.
a(16) = 18 or 19 or 20 due to the permutation
     1, 17, 2, 16, 4, 15, 6, 14, 7, 13, 8, 12, 9, 11, 10, 20.
Permutations for n = 13, 14, 15 were produced by Qing-Hu Hou at Nankai Univ. on the author's request.
From _Charlie Neder_, Aug 23 2018: (Start)
a(16) = 18 due to the permutation
     1, 11, 10, 12, 9, 13, 8, 14, 7, 15, 6, 17, 4, 16, 2, 18.
a(17) = 19 due to the permutation
     1, 11, 12, 10, 13, 9, 14, 8, 15, 7, 16, 2, 18, 6, 17, 4, 19.
a(18) = 20 due to
     1, 12, 11, 13, 10, 14, 9, 15, 8, 16, 7, 17, 2, 19, 6, 18, 4, 20. (End)
From _Bert Dobbelaere_, Sep 09 2019: (Start)
a(19) = 22 due to the permutation
     1, 18, 2, 17, 8, 19, 7, 20, 6, 16, 9, 15, 10, 14, 11, 13, 12, 4, 22.
a(20) = 23 due to the permutation
     1, 18, 7, 20, 6, 22, 4, 19, 9, 16, 8, 17, 11, 13, 12, 15, 10, 14, 2, 23.
a(21) = 24 due to the permutation
     1, 19, 10, 20, 7, 18, 4, 16, 9, 17, 11, 15, 12, 14, 13, 8, 23, 6, 22, 2, 24.
a(22) = 25 due to the permutation
     1, 22, 4, 23, 7, 24, 9, 18, 8, 20, 6, 19, 11, 15, 12, 17, 10, 16, 14, 13, 2, 25.
a(23) = 26 due to the permutation
     1, 22, 7, 25, 6, 23, 10, 24, 4, 20, 8, 19, 11, 17, 13, 18, 9, 16, 14, 15, 12, 2, 26. (End)
		

Links

• Zhi-Wei Sun, A problem on circular permutations, a message to Number Theory List, Sep 01 2013.
• Z.-W. Sun, Some new problems in additive combinatorics, arXiv preprint arXiv:1309.1679 [math.NT], 2013-2014.

Crossrefs

Cf. A185645, A187815.

Programs

Mathematica

A program to find a(16) in terms of the values a(1),...,a(15):
V[i_]:=V[i]=Part[Permutations[{2,4,6,7,8,9,10,11,12,13,14,15,16,17}],i]
Do[Do[Do[If[Length[Union[{Abs[1-Part[V[i],1]]},Table[Abs[Part[V[i],j]-If[j<14,Part[V[i],j+1],n]],{j,1,14}]]]<15,Goto[aa]];
Print[n," "," ",V[i]];Goto[bb];Label[aa];Continue,{i,1,14!}];Continue,{n,18,20}];Label[bb];Break]
A228728[n_] := Module[{p, i, j, k, b, lim = 100},
  If[n <= 2, A228728[n] = n,
   j = A228728[n - 1] + 1;
    While[j < lim, A228728[n] = j;
    p = Permutations[Table[A228728[k], {k, 2, n - 1}]];
    i = 1; While[i <= Length[p],
     b = Join[{A228728[1]}, p[[i]], {A228728[n]}]; i++;
     If[Length[Union[Join[Table[Abs[b[[k]] - b[[k + 1]]], {k, 1, n - 1}], {Abs[b[[n]] - b[[1]]]}]]] == n, Return[j]]]; j++]]]
Table[A228728[n], {n, 1, 11}]  (* Robert Price, Apr 04 2019 *)

Extensions

a(16)-a(18) from Charlie Neder, Aug 23 2018

Name clarified by Robert Price, Apr 04 2019

a(19)-a(23) from Bert Dobbelaere, Sep 09 2019