A228917 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_{n-1}+i_n, i_n+i_0 are among those k with 6*k-1 and 6*k+1 twin primes.

1, 1, 1, 2, 2, 2, 5, 2, 12, 39, 98, 526, 2117, 6663, 15043, 68403, 791581, 4826577, 19592777, 102551299, 739788968, 4449585790, 36547266589, 324446266072, 2743681178070

Offset: 1

Author

Zhi-Wei Sun, Sep 08 2013

Keywords

• nonn
• more

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.

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.

Examples

			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).
		

Links

• Zhi-Wei Sun, Twin primes and circular permutations, a message to Number Theory List, Sept. 8, 2013.
• Z.-W. Sun, Some new problems in additive combinatorics, arXiv preprint arXiv:1309.1679 [math.NT], 2013-2014.

Crossrefs

Cf. A000040, A001359, A006512, A023200, A046132, A023201, A046117, A005384, A051252, A228766, A228860, A228886.

Programs

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[6n-1]&&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!}]

Extensions

a(10)-a(25) from Max Alekseyev, Sep 12 2013