
COMMENTS

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.
ZhiWei 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_{n1}+i_n, i_n+i_0 are integers of the form (p1)/2, where p is an odd prime. Also, we may replace the above (p1)/2 by (p+1)/4 or (p1)/6; when n > 4 we may substitute (p1)/4 for (p1)/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_{n1}+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_{n1}+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_{n1}+i_n, i_n+i_0 are among those integers k with 2*k3 and 2*k+3 both prime.


EXAMPLE

a(n) = 1 for n = 1,2,3 due to the permutation (0,...,n).
a(4) = 2 due to the permutations (0,1,4,3,2) and (0,2,1,4,3).
a(5) = 2 due to the permutations (0,1,4,3,2,5), (0,3,4,1,2,5).
a(6) = 2 due to the permutations
(0,1,6,4,3,2,5) and (0,3,4,6,1,2,5).
a(7) = 5 due to the permutations
(0,1,6,4,3,2,5,7), (0,1,6,4,3,7,5,2), (0,2,1,6,4,3,7,5),
(0,3,4,6,1,2,5,7), (0,5,2,1,6,4,3,7).
a(8) = 2 due to the permutations
(0,1,6,4,8,2,3,7,5) and (0,1,6,4,8,2,5,7,3).
a(9) = 12 due to the permutations
(0,1,6,4,3,9,8,2,5,7), (0,1,6,4,8,9,3,2,5,7),
(0,1,6,4,8,9,3,7,5,2), (0,2,1,6,4,8,9,3,7,5),
(0,2,8,9,1,6,4,3,7,5), (0,3,4,6,1,9,8,2,5,7),
(0,3,9,1,6,4,8,2,5,7), (0,3,9,8,4,6,1,2,5,7),
(0,5,2,1,6,4,8,9,3,7), (0,5,2,8,4,6,1,9,3,7),
(0,5,2,8,9,1,6,4,3,7), (0,5,7,3,9,1,6,4,8,2).
a(10) > 0 due to the permutation (0,5,2,3,9,1,6,4,8,10,7).
a(11) > 0 due to the permutation (0,10,8,9,3,7,11,6,4,1,2,5).
a(12) > 0 due to the permutation
(0, 5, 2, 1, 6, 4, 3, 9, 8, 10, 7, 11, 12).


MATHEMATICA

(* A program to compute required circular permutations for n = 7. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, (0, 7, 5, 2, 3, 4, 6, 1) is identical to (0, 1, 6, 4, 3, 2, 5, 7) if we ignore direction. Thus a(7) is half of the number of circular permutations yielded by this program. *)
tp[n_]:=tp[n]=PrimeQ[6n1]&&PrimeQ[6n+1]
V[i_]:=Part[Permutations[{1, 2, 3, 4, 5, 6, 7}], i]
m=0
Do[Do[If[tp[If[j==0, 0, Part[V[i], j]]+If[j<7, Part[V[i], j+1], 0]]==False, Goto[aa]], {j, 0, 7}];
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]]; Label[aa]; Continue, {i, 1, 7!}]
