

A229232


Number of undirected circular permutations pi(1), ..., pi(n) of 1, ..., n with the n numbers pi(1)*pi(2)1, pi(2)*pi(3)1, ..., pi(n1)*pi(n)1, pi(n)*pi(1)1 all prime.


0




OFFSET

1,6


COMMENTS

Conjecture: a(n) > 0 for all n > 5 with n not equal to 13.
ZhiWei Sun also made the following conjectures:
(1) For any integer n > 1, there is a permutation pi(1), ..., pi(n) of 1, ..., n such that the n numbers 2*pi(1)*pi(2)1, ..., 2*pi(n1)*pi(n)1, 2*pi(n)*pi(1)1 are all prime. Also, for any positive integer n not equal to 4, there is a permutation pi(1), ..., pi(n) of 1, ..., n such that the n numbers 2*pi(1)*pi(2)+1, ..., 2*pi(n1)*pi(n)+1, 2*pi(n)*pi(1)+1 are all prime.
(2) Let F be a finite field with q > 7 elements. Then, there is a circular permutation a_1,...,a_{q1} of the q1 nonzero elements of F such that all the q1 elements a_1*a_21, a_2*a_31, ..., a_{q2}*a_{q1}1, a_{q1}*a_11 are primitive elements of the field F (i.e., generators of the multiplicative group F\{0}). Also, there is a circular permutation b_1,...,b_{q1} of the q1 nonzero elements of F such that all the q1 elements b_1*b_2+1, b_2*b_3+1, ..., b_{q2}*b_{q1}+1, b_{q1}*b_1+1 are primitive elements of the field F.


LINKS



EXAMPLE

a(4) = 1 due to the circular permutation (1,3,2,4).
a(6) = 2 due to the circular permutations
(1,3,2,4,5,6) and (1,3,2,6,5,4).
a(7) = 1 due to the circular permutation (1,3,2,7,6,5,4).
a(8) = 2 due to the circular permutations
(1,3,2,7,6,5,4,8) and (1,4,5,6,7,2,3,8).
a(9) = 2 due to the circular permutations
(1,3,4,5,6,7,2,9,8) and (1,3,8,9,2,7,6,5,4).
a(10) = 8 due to the circular permutations
(1,3,4,5,6,7,2,9,10,8), (1,3,4,5,6,7,2,10,9,8),
(1,3,8,9,10,2,7,6,5,4), (1,3,8,10,9,2,7,6,5,4),
(1,3,10,8,9,2,7,6,5,4), (1,3,10,9,2,7,6,5,4,8),
(1,4,5,6,7,2,3,10,9,8), (1,4,5,6,7,2,9,10,3,8).
a(13) = 0 since 8 is the unique j among 1 ,..., 12 with 13*j1 prime.


MATHEMATICA

(* A program to compute required circular permutations for n = 8. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, (1, 8, 4, 5, 6, 7, 2, 3) is identical to (1, 3, 2, 7, 6, 5, 4, 8) if we ignore direction. Thus, a(8) is half of the number of circular permutations yielded by this program. *)
V[i_]:=V[i]=Part[Permutations[{2, 3, 4, 5, 6, 7, 8}], i]
f[i_, j_]:=f[i, j]=PrimeQ[i*j1]
m=0
Do[Do[If[f[If[j==0, 1, Part[V[i], j]], If[j<7, Part[V[i], j+1], 1]]==False, Goto[aa]], {j, 0, 7}];
m=m+1; Print[m, ":", " ", 1, " ", 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!}]


CROSSREFS



KEYWORD

nonn,more,hard


AUTHOR



STATUS

approved



