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A242532
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Number of cyclic arrangements of S={2,3,...,n+1} such that the difference of any two neighbors is greater than 1, and a divisor of their sum.
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16
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0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 20, 39, 0, 0, 0, 0, 319, 967, 0, 0, 1464, 6114, 16856, 44370, 0, 0, 0, 0, 2032951, 8840796, 12791922, 101519154, 0, 0
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OFFSET
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1,14
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COMMENTS
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a(n)=NPC(n;S;P) is the count of all neighbor-property cycles for a specific set S of n elements and a specific pair-property P. For more details, see the link and A242519.
For this property P and sets {0,1,2,...,n-1} or {1,2,...,n} the problem does not appear to have any solution.
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LINKS
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EXAMPLE
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The shortest such cycle is of length n=9: {2,4,8,10,5,7,9,3,6}.
The next a(n)>0 occurs for n=14 and has 20 solutions.
The first and the last of these are:
C_1={2,4,8,10,5,7,14,12,15,13,11,9,3,6},
C_2={2,4,12,15,13,11,9,3,5,7,14,10,8,6}.
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MATHEMATICA
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A242532[n_] := Count[Map[lpf, Map[j2f, Permutations[Range[3, n + 1]]]], 0]/2;
j2f[x_] := Join[{2}, x, {2}];
dvf[x_] := Module[{i},
Table[Abs[x[[i]] - x[[i + 1]]] > 1 &&
Divisible[x[[i]] + x[[i + 1]], x[[i]] - x[[i + 1]]], {i,
Length[x] - 1}]];
lpf[x_] := Length[Select[dvf[x], ! # &]];
(* OR, a less simple, but more efficient implementation. *)
A242532[n_, perm_, remain_] := Module[{opt, lr, i, new},
If[remain == {},
If[Abs[First[perm] - Last[perm]] > 1 &&
Divisible[First[perm] + Last[perm], First[perm] - Last[perm]],
ct++];
Return[ct],
opt = remain; lr = Length[remain];
For[i = 1, i <= lr, i++,
new = First[opt]; opt = Rest[opt];
If[Abs[Last[perm] - new] <= 1 || !
Divisible[Last[perm] + new, Last[perm] - new], Continue[]];
Complement[Range[3, n + 1], perm, {new}]];
];
Return[ct];
];
];
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PROG
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(C++) See the link.
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CROSSREFS
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Cf. A242519, A242520, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242533, A242534.
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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