OFFSET
1,5
COMMENTS
Conjecture: a(n) > 0 for all n > 0. In general, for any n consecutive primes p_k,...,p_{k+n-1}, there always exists a permutation q_k,...,q_{k+n-1} of p_k,...,p_{k+n-1} with q_{k+n-1} = p_{k+n-1} such that the n-1 numbers |q_k-q_{k+1}|, |q_{k+1}-q_{k+2}|,...,|q_{k+n-2}-q_{k+n-1}| are pairwise distinct. (In the case k = 2, this implies that a(n) > 0.)
Clearly there is no permutation a,b,c of 3,5,7 such that the three numbers |a-b|,|b-c|,|c-a| are pairwise distinct. Also, for {a,b} = {7,11}, the three numbers |5-a|,|a-b|,|b-13| cannot be pairwise distinct.
On Aug 31 2013, Zhi-Wei Sun proved the following extension of the general conjecture: Let a_1 < a_2 < ... < a_n be a sequence of n distinct real numbers in ascending order. Then there is a permutation b_1, ..., b_n of a_1, ..., a_n with b_n = a_n such that |b_1-b_2|, |b_2-b_3|, ..., |b_{n-1}-b_n| are pairwise distinct. In fact, when n = 2*k is even we may take (b_1,...,b_n) = (a_k,a_{k+1},a_{k-1},a_{k+2},...,a_2,a_{2k-1},a_1,a_{2k}); when n = 2*k-1 is odd we may take (b_1,...,b_n) = (a_k,a_{k-1},a_{k+1},a_{k-2},a_{k+2},..., a_2,a_{2k-2},a_1,a_{2k-1}).
On Sep 01 2013, Zhi-Wei Sun made the following conjecture: (i) For any n distinct real numbers a_1, a_2, ..., a_n (not necessarily in ascending or descending order), there is a permutation b_1, ..., b_n of a_1, ..., a_n with b_1 = a_1 such that the n-1 distances |b_1-b_2|, |b_2-b_3|, ..., |b_{n-1}-b_n| are pairwise distinct.
(ii) Let a_1, ..., a_n be n distinct elements of a finite additive abelian group G. Suppose that |G| is not divisible by n, or n is even and G is cyclic. Then there exists a permutation b_1, ..., b_n of a_1, ..., a_n with b_1 = a_1 such that the n-1 differences b_{i+1}-b_i (i = 1, ..., n-1) are pairwise distinct.
We believe that part (ii) of the new conjecture holds at least when G is cyclic, and it might also hold when the group G is not abelian.
Note that if g is a primitive root modulo an odd prime p, then for any j = 0,...,p-2 the permutation g^j, g^{j+1},...,g^{j+p-2} of the p-1 nonzero residues modulo p has adjacent differences g^{i+j+1}-g^{i+j} = g^{i+j}*(g-1) (i = 0, ..., p-3) which are pairwise distinct modulo p.
LINKS
Z.-W. Sun, Some new problems in additive combinatorics, arXiv preprint arXiv:1309.1679 [math.NT], 2013-2014.
EXAMPLE
a(4) = 1 since (q_1,q_2,q_3,q_4) = (2,5,3,7) is the only suitable permutation.
a(5) = 3 since there are exactly three suitable permutations(q_1,q_2,q_3,q_4,q_5): (2,3,7,5,11), (2,5,7,3,11) and (2,7,3,5,11).
a(6) = 5 since there are exactly five suitable permutations (q_1,q_2,q_3,q_4,q_5,q_6): (2,5,3,11,7,13), (2,5,7,11,3,13), (2,7,5,11,3,13), (2,7,11,5,3,13), (2,11,5,7,3,13).
a(7) = 10, and the ten suitable permutations (q_1,...,q_7) are as follows:
(2,3,13,5,7,11,17), (2,7,3,13,11,5,17), (2,7,5,11,3,13,17),
(2,7,11,5,13,3,17), (2,11,3,13,7,5,17), (2,11,7,5,13,3,17),
(2,11,7,13,3,5,17), (2,11,7,13,5,3,17), (2,13,3,11,7,5,17),
(2,13,7,11,3,5,17).
MATHEMATICA
A185645[n_] := Module[{p, c = 0, i = 1, j, q},
If[n == 2, Return[1],
p = Permutations[Table[Prime[j], {j, 2, n - 1}]];
While[i <= Length[p],
q = Join[{2}, p[[i]], {Prime[n]}]; i++;
If[Length[Union[Join[Table[Abs[q[[j]] - q[[j + 1]]], {j, 1, n - 1}], {Abs[q[[n]] - q[[1]]]}]]] == n, c++]]; c]];
Table[A185645[n], {n, 1, 11}] (* Robert Price, Apr 04 2019 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Zhi-Wei Sun, Aug 29 2013
EXTENSIONS
Name clarified by Robert Price, Apr 04 2019
a(12)-a(18) from Bert Dobbelaere, Sep 08 2019
STATUS
approved