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A242525
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Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is at most 3.
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16
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1, 1, 1, 3, 6, 10, 17, 31, 57, 104, 188, 340, 616, 1117, 2025, 3670, 6651, 12054, 21847, 39596, 71764, 130065, 235730, 427238, 774328, 1403395, 2543518, 4609881, 8354965, 15142569, 27444447, 49740415, 90149708, 163387657, 296124381, 536696900
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OFFSET
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1,4
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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.
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LINKS
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FORMULA
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Empirical: a(n) = a(n-1)+a(n-2)+a(n-4)+a(n-5) for n>7. - Andrew Howroyd, Apr 08 2016
Empirical G.f.: x^2 + ((1-x)^2*(1+x)^2)/(1-x-x^2-x^4-x^5). - Andrew Howroyd, Apr 08 2016
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EXAMPLE
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For n=4, The three cycles are: C_1={1,2,3,4}, C_2={1,2,4,3}, C_3={1,3,2,4}.
The first and the last of the 104 such cycles of length n=10 are: C_1={1,2,3,5,6,8,9,10,7,4}, C_104={1,3,6,9,10,8,7,5,2,4}.
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MATHEMATICA
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A242525[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[Abs[Differences[x]], # > 3 &]];
Join[{1, 1}, Table[A242525[n], {n, 3, 10}]]
(* OR, a less simple, but more efficient implementation. *)
A242525[n_, perm_, remain_] := Module[{opt, lr, i, new},
If[remain == {},
If[Abs[First[perm] - Last[perm]] <= 3, 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] > 3, Continue[]];
Complement[Range[2, n], perm, {new}]];
];
Return[ct];
];
];
Join[{1, 1},
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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, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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