%I
%S 1,1,1,2,2,2,5,2,12,39,98,526,2117,6663,15043,68403,791581,4826577,
%T 19592777,102551299,739788968,4449585790,36547266589,324446266072,
%U 2743681178070
%N Number of undirected circular permutations 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 k with 6*k1 and 6*k+1 twin primes.
%C Conjecture: a(n) > 0 for all n > 0.
%C This implies the twin prime conjecture, and it is similar to the prime circle problem mentioned in A051252.
%C 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.
%C ZhiWei Sun also made the following similar conjectures for odd primes, Sophie Germain primes, cousin primes and sexy primes:
%C (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.
%C (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.
%C (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.
%C (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.
%H ZhiWei Sun, <a href="http://listserv.nodak.edu/cgibin/wa.exe?A2=NMBRTHRY;1fd8fbc7.1309">Twin primes and circular permutations</a>, a message to Number Theory List, Sept. 8, 2013.
%H Z.W. Sun, <a href="http://arxiv.org/abs/1309.1679">Some new problems in additive combinatorics</a>, arXiv preprint arXiv:1309.1679, 2013
%e a(n) = 1 for n = 1,2,3 due to the permutation (0,...,n).
%e a(4) = 2 due to the permutations (0,1,4,3,2) and (0,2,1,4,3).
%e a(5) = 2 due to the permutations (0,1,4,3,2,5), (0,3,4,1,2,5).
%e a(6) = 2 due to the permutations
%e (0,1,6,4,3,2,5) and (0,3,4,6,1,2,5).
%e a(7) = 5 due to the permutations
%e (0,1,6,4,3,2,5,7), (0,1,6,4,3,7,5,2), (0,2,1,6,4,3,7,5),
%e (0,3,4,6,1,2,5,7), (0,5,2,1,6,4,3,7).
%e a(8) = 2 due to the permutations
%e (0,1,6,4,8,2,3,7,5) and (0,1,6,4,8,2,5,7,3).
%e a(9) = 12 due to the permutations
%e (0,1,6,4,3,9,8,2,5,7), (0,1,6,4,8,9,3,2,5,7),
%e (0,1,6,4,8,9,3,7,5,2), (0,2,1,6,4,8,9,3,7,5),
%e (0,2,8,9,1,6,4,3,7,5), (0,3,4,6,1,9,8,2,5,7),
%e (0,3,9,1,6,4,8,2,5,7), (0,3,9,8,4,6,1,2,5,7),
%e (0,5,2,1,6,4,8,9,3,7), (0,5,2,8,4,6,1,9,3,7),
%e (0,5,2,8,9,1,6,4,3,7), (0,5,7,3,9,1,6,4,8,2).
%e a(10) > 0 due to the permutation (0,5,2,3,9,1,6,4,8,10,7).
%e a(11) > 0 due to the permutation (0,10,8,9,3,7,11,6,4,1,2,5).
%e a(12) > 0 due to the permutation
%e (0, 5, 2, 1, 6, 4, 3, 9, 8, 10, 7, 11, 12).
%t (* 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. *)
%t tp[n_]:=tp[n]=PrimeQ[6n1]&&PrimeQ[6n+1]
%t V[i_]:=Part[Permutations[{1,2,3,4,5,6,7}],i]
%t m=0
%t 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}];
%t 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!}]
%Y Cf. A000040, A001359, A006512, A023200, A046132, A023201, A046117, A005384, A051252, A228766, A228860, A228886.
%K nonn,more
%O 1,4
%A _ZhiWei Sun_, Sep 08 2013
%E a(10)a(25) from _Max Alekseyev_, Sep 12 2013
