

A228886


Number of permutations i_0,i_1,...,i_n of 0,1,...,n with i_0 = 0 and i_n = n, and with i_0+i_1, i_1+i_2, ..., i_{n1}+i_n, i_n+i_0 all coprime to both n1 and n+1.


5



1, 0, 1, 0, 1, 8, 3, 24, 78, 1164, 34, 4021156, 400, 87180, 2499480, 7509358, 1700352, 39982182134232, 1427688, 212987263960, 9533487948, 36638961135462, 29317847040, 30258969747586970112, 1655088666624
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OFFSET

1,6


COMMENTS

Conjecture: a(n) > 0 except for n = 2, 4.
Note that this conjecture is stronger than the one stated in the comments in A228860. Also, there is no permutation i_0, ..., i_7 of 0, ..., 7 with i_0+i_1, i_1+i_2, ..., i_6+i_7, i_7+i_0 all coprime to 7*13  1 = 90.
For n > 1 let S(n) be an undirected simple graph with vertices 0, 1, ..., n which has an edge connecting two distinct vertices i and j if and only if i + j is relatively prime to both n1 and n+1. Then a(n) is just the number of those Hamiltonian cycles in S(n) on which the vertices 0 and n are adjacent.
We have some other similar conjectures. For example, for any positive integer n there is a permutation i_0,i_1,...,i_n of 0,1,...,n with i_0 = 0 and i_n = n such that i_0 + i_1, i_1 + i_2, ..., i_{n1} + i_n, i_n + i_0 are all relatively prime to both 2*n1 and 2*n+1.
On Sep 10 2013, ZhiWei Sun proved the conjecture for any positive odd integer n. In fact, if n == 1 or 3 (mod 6) then we may take the permutation 0,n2,2,n4,4,...,1,n1,n which meets the requirement; if n == 3 or 5 (mod 6) then the permutation 0,1,n1,3,n3,...,n2,2,n suffices for our purpose.


LINKS

Table of n, a(n) for n=1..25.
ZhiWei Sun, List of required permutations for n = 1..10
ZhiWei Sun, Some new problems in additive combinatorics, preprint, arXiv:1309.1679.


EXAMPLE

a(2) = 0 since 1+2 = 2^21.
a(3) = 1 due to the identical permutation (0,1,2,3).
a(4) = 0 since 1+2 divides 4^21.
a(5) = 1 due to the permutation (0,1,4,3,2,5).
a(6) = 8 due to the permutations
(0,1,2,4,5,3,6), (0,1,3,5,4,2,6),
(0,2,4,5,1,3,6), (0,3,1,2,4,5,6),
(0,3,1,5,4,2,6), (0,4,2,1,3,5,6),
(0,4,2,1,5,3,6), (0,4,5,3,1,2,6).
a(7) = 3 due to the permutations(0,1,4,3,2,5,6,7), (0,1,6,5,2,3,4,7), (0,5,2,3,4,1,6,7).
a(8) > 0 due to the permutation (0,1,4,7,3,2,6,5,8).
a(9) > 0 due to the permutation (0,1,2,5,4,7,6,3,8,9).
a(10) > 0 due to the permutation (0,1,3,2,5,8,6,4,9,7,10).


MATHEMATICA

(*A program to compute the required permutations for n = 8.*)
V[i_]:=Part[Permutations[{1, 2, 3, 4, 5, 6, 7}], i]
m=0
Do[Do[If[GCD[If[j==0, 0, Part[V[i], j]]+If[j<7, Part[V[i], j+1], 8], 8^21]>1, 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], " ", 8]; Label[aa]; Continue, {i, 1, 7!}]


CROSSREFS

Cf. A228860, A228917, A228956.
Sequence in context: A037206 A065530 A137481 * A182160 A213838 A248294
Adjacent sequences: A228883 A228884 A228885 * A228887 A228888 A228889


KEYWORD

nonn,more,hard


AUTHOR

ZhiWei Sun, Sep 07 2013


EXTENSIONS

a(11)a(25) from Max Alekseyev, Sep 13 2013


STATUS

approved



