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A242533
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Number of cyclic arrangements of S={1,2,...,2n} such that the difference of any two neighbors is coprime to their sum.
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16
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1, 1, 2, 36, 288, 3888, 200448, 4257792, 139511808, 11813990400, 532754620416
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OFFSET
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1,3
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COMMENTS
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a(n)=NPC(2n;S;P) is the count of all neighbor-property cycles for a specific set S of 2n elements and a specific pair-property P. For more details, see the link and A242519.
Conjecture: in this case it seems that NPC(n;S;P)=0 for all odd n, so only the even ones are listed. This is definitely not the case when the property P is replaced by its negation (see A242534).
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LINKS
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EXAMPLE
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For n=4, the only cycle is {1,2,3,4}.
The two solutions for n=6 are: C_1={1,2,3,4,5,6} and C_2={1,4,3,2,5,6}.
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MATHEMATICA
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A242533[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, 2 n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[cpf[x], ! # &]];
cpf[x_] := Module[{i},
Table[CoprimeQ[x[[i]] - x[[i + 1]], x[[i]] + x[[i + 1]]], {i,
Length[x] - 1}]];
Join[{1}, Table[A242533[n], {n, 2, 5}]]
(* OR, a less simple, but more efficient implementation. *)
A242533[n_, perm_, remain_] := Module[{opt, lr, i, new},
If[remain == {},
If[CoprimeQ[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[! CoprimeQ[Last[perm] + new, Last[perm] - new], Continue[]];
Complement[Range[2, 2 n], perm, {new}]];
];
Return[ct];
];
];
Join[{1}, Table[ct = 0; A242533[n, {1}, Range[2, 2 n]]/2, {n, 2, 6}] ](* Robert Price, Oct 25 2018 *)
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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, A242532, 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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