

A242532


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.


16



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

1,14


COMMENTS

a(n)=NPC(n;S;P) is the count of all neighborproperty cycles for a specific set S of n elements and a specific pairproperty P. For more details, see the link and A242519.
For this property P and sets {0,1,2,...,n1} or {1,2,...,n} the problem does not appear to have any solution.


LINKS



EXAMPLE

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}.


MATHEMATICA

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];
];
];


PROG

(C++) See the link.


CROSSREFS

Cf. A242519, A242520, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242533, A242534.


KEYWORD

nonn,hard,more


AUTHOR



EXTENSIONS



STATUS

approved



