OFFSET
1,5
COMMENTS
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.
EXAMPLE
a(4) = 1 due to the permutation 1,2,3,4.
a(5) = 2 due to the permutations 1,2,4,3,5 and 1,3,4,2,5.
a(6) = 1 due to the permutation 1,4,3,2,5,6.
a(7) > 0 due to the permutation 1,2,3,5,4,6,7.
a(8) > 0 due to the permutation 1,2,3,4,5,6,7,8.
a(9) > 0 due to the permutation 1,3,2,5,8,6,4,7,9.
a(10) > 0 due to the permutation 1,2,5,4,7,6,3,8,9,10.
a(11) > 0 due to the permutation 1,2,3,4,5,7,6,8,9,10,11.
a(12) > 0 due to the permutation 1,4,9,2,5,8,3,10,7,6,11,12.
MATHEMATICA
(*A program to compute the required permutations for n = 9.*)
V[i_]:=Part[Permutations[{2, 3, 4, 5, 6, 7, 8}], i]
m=0
Do[Do[If[GCD[If[j==0, 1, Part[V[i], j]]+If[j<7, Part[V[i], j+1], 9], 9]>1, 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], " ", 9]; Label[aa]; Continue, {i, 1, 7!}]
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Zhi-Wei Sun, Sep 05 2013
EXTENSIONS
a(12)-a(27) from Max Alekseyev, Sep 13 2013
STATUS
approved