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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.

Conjecture: a(n) > 0 for all n > 0.

This implies the twin prime conjecture, and it is similar to the prime circle problem mentioned in A051252.

For each n = 2,3,... construct an undirected simple graph T(n) with vertices 0,1,...,n which has an edge connecting two distinct vertices i and j if and only if 6*(i+j)-1 and 6*(i+j)+1 are twin primes. Then a(n) is just the number of Hamiltonian cycles contained in T(n). Thus a(n) > 0 if and only if T(n) is a Hamilton graph.

Zhi-Wei Sun also made the following similar conjectures for odd primes, Sophie Germain primes, cousin primes and sexy primes:

(1) For any integer n > 0, there is a permutation i_0, i_1, ..., i_n of 0, 1, ..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are integers of the form (p-1)/2, where p is an odd prime. Also, we may replace the above (p-1)/2 by (p+1)/4 or (p-1)/6; when n > 4 we may substitute (p-1)/4 for (p-1)/2.

(2) For any integer n > 2, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are integers of the form (p+1)/6, where p is a Sophie Germain prime.

(3) For any integer n > 3, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are among those integers k with 6*k+1 and 6*k+5 both prime.

(4) For any integer n > 4, there is a permutation i_0, i_1, ..., i_n of 0, 1,..., n such that i_0+i_1, i_1+i_2, ..., i_{n-1}+i_n, i_n+i_0 are among those integers k with 2*k-3 and 2*k+3 both prime.

If i_1,...,i_{n-1} is a permutation of 1,...,n-1 with i_1-i_2, ..., i_{n-2}-i_{n-1}, i_{n-1}-i_1 pairwise distinct modulo n, then 0 = (i_1-i_2)+...+(i_{n-1}-i_1) == 1+2+...+(n-1) = n(n-1)/2 (mod n) and hence n is odd. So a(n) = 0 for every n = 4,6,8,...

If g is a primitive root modulo an odd prime p, then 1-g, g-g^2, g^2-g^3, ..., g^{p-3}-g^{p-2},g^{p-2}-1 are pairwise distinct modulo p. Therefore a(p) > 0 for any odd prime p.

Conjecture: a(n) > 0 for any odd number n > 1. In general, if G is an additive abelian group with |G| = n odd and greater than one, then there is a permutation a_1,...,a_{n-1} of all the nonzero elements of G such that a_1-a_2, a_2-a_3, ..., a_{n-2}-a_{n-1}, a_{n-1}-a_1 are pairwise distinct.

For any integer n > 1, the author has showed that there is a permutation i_1, ..., i_n of 1, ..., n such that i_1-i_2, i_2-i_3, ..., i_{n-1}-i_n are pairwise distinct if and only if n is even.

Conjecture: a(n) > 0 except for n = 3.

If n is a power of two, then a(n) > 0 since the identical permutation 1,2,3,...,n meets the requirement. For any prime p > 3, we have a(p) > 0 since the permutation 1,...,(p-1)/2, (p+3)/2,(p+1)/2,(p+5)/2,...,p meets our purpose.

Let G(n) be the undirected simple graph with vertices 1,...,n which has an edge connecting two distinct vertices i and j if and only if i + j is relatively prime to n. Then, for any n > 2, the number a(n) is just the number of those Hamiltonian cycles in G(n) on which the vertices 1 and n are adjacent.

Let m be any integer relatively prime to n, and let i_k be the smallest positive residue of k*m modulo n. Then i_1, i_2, ..., i_n is a permutation of 1, ..., n with the n adjacent differences i_1-i_2, i_2-i_3, ..., i_{n-1}-i_n, i_n-i_1 all coprime to n.

On Sep 06 2013, the author's two former PhD students Hui-Qin Cao (from Nanjing Audit Univ.) and Hao Pan (from Nanjing Univ.) proved the conjecture fully.

Note that if n > 3 is not a multiple of 3 then a(n) > 0 since the natural circular permutation (0,1,2,...,n-1) meets the requirement.

Conjecture: Let G be an additive abelian group. If G is cyclic or G contains no involution, then for any finite subset A of G with |A| = n > 3, there is a numbering a_1,...,a_n of the elements of A such that the n sums a_1+a+2+a_3, a_2+a_3+a_4, ..., a_{n-2}+a_{n-1}+a_n, a_{n-1}+a_n+a_1, a_n+a_1+a_2 are pairwise distinct.

On Sep 13 2013, the author proved the conjecture for any torsion-free abelian group G.

If n is not divisible by 3 then a(n) > 0 since the identical permutation of 1 ,..., n works for the purpose. If n is even, then a(n) > 0 since the permutation 1, 2, n-1, 4, n-3, 6, n-5, ..., n-2, 3, n meets the requirement. We guess that a(n) > 0 in the remaining case n = 6q+3 with q > 0. If n = 2k+1 == 3 (mod 6) with n > 3, then, for the permutation (i_1,...,i_n) = (1,2k,k,2k-1,k-1,2k-2,...,3,k+2,2,k+1,2k+1), all the n sums i_1+2*i_2, i_2+2*i_3, ..., i_{n-1}+2*i_n, i_n+2*i_1 are pairwise distinct (but they are not pairwise incongruent modulo n = 2k+1 when n > 9).

Zhi-Wei Sun also made the following general conjecture:

(i) (Weak version) Let a_1,...,a_n be n distinct elements of an additive abelian group G. Then, there is a permutation b_1,...,b_n of a_1,...,a_n such that a_1+2*b_1, a_2+2*b_2, ..., a_n+2*b_n are pairwise distinct. (The author has proved this for n up to 4 in any abelian group G.)

(ii) (Strong version) Let A be any subset of an additive abelian group G with |A| = n > 4. Then there is a numbering a_1, ..., a_n of all the elements of A such that a_1+2*a_2, a_2+2*a_3, ..., a_{n-1}+2*a_n, a_n+2*a_1 are pairwise distinct. (The author has proved this for any torsion-free abelian group G.)

Recall that a conjecture of Snevily proved by Arsovski states that for any two n-subsets A and B of an additive abelian group of odd order there is a numbering a_1,...,a_n of all the elements of A and a numbering b_1, ..., b_n of all the elements of B such that the n sums a_1+b_1, ..., a_n+b_n are pairwise distinct.