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A228772
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Number of undirected circular permutations i_0,i_1,...,i_{n-1} of 0,1,...,n-1 such that i_0+i_1+i_2, i_1+i_2+i_3, ..., i_{n-3}+i_{n-2}+i_{n-1}, i_{n-2}+i_{n-1}+i_0, i_{n-1}+i_0+i_1 are pairwise distinct modulo n.
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2
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0, 3, 2, 24, 24, 392, 513, 4080, 8090, 96816, 238296, 2023896, 7325520, 63277376, 277838352, 2185076682, 12898278126
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OFFSET
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3,2
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COMMENTS
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Note that if n > 3 is not a multiple of 3 then a(n) > 0 since the natural circular permutation (0,1,2,...,n-1) meets the requirement.
Conjecture: Let G be an additive abelian group. If G is cyclic or G contains no involution, then for any finite subset A of G with |A| = n > 3, there is a numbering a_1,...,a_n of the elements of A such that the n sums a_1+a+2+a_3, a_2+a_3+a_4, ..., a_{n-2}+a_{n-1}+a_n, a_{n-1}+a_n+a_1, a_n+a_1+a_2 are pairwise distinct.
On Sep 13 2013, the author proved the conjecture for any torsion-free abelian group G.
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LINKS
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EXAMPLE
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a(4) = 3 due to the circular permutations (0,1,2,3), (0,1,3,2) and (0,2,1,3).
a(5) = 2 due to the circular permutations (0,1,2,3,4) and(0,2,4,1,3).
a(6) > 0 due to the circular permutation (0,1,2,4,5,3).
a(9) > 0 due to the circular permutation (0,1,2,3,8,5,6,7,4).
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MATHEMATICA
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(* A program to compute required circular permutations for n = 9. To get "undirected" circular permutations, we should identify a circular permutation with the one of the opposite direction; for example, (0, 4, 7, 6, 5, 8, 3, 2, 1) is identical to (0, 1, 2, 3, 8, 5, 6, 7, 4) if we ignore direction.*)
V[i_]:=Part[Permutations[{1, 2, 3, 4, 5, 6, 7, 8}], i]
m=0
Do[If[Length[Union[Table[Mod[If[j==0, 0, Part[V[i], j]]+If[j<8, Part[V[i], j+1], 0]+If[j<7, Part[V[i], j+2], If[j==7, 0, Part[V[i], 1]]], 9], {j, 0, 8}]]]<9, Goto[aa]];
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], " ", Part[V[i], 8]]; Label[aa]; Continue, {i, 1, 8!}]
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CROSSREFS
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KEYWORD
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nonn,more,hard
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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