A242532 - OEIS

%I #20 Oct 25 2018 21:25:43

%S 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,

%T 44370,0,0,0,0,2032951,8840796,12791922,101519154,0,0

%N 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.

%C 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.

%C For this property P and sets {0,1,2,...,n-1} or {1,2,...,n} the problem does not appear to have any solution.

%C a(40)=a(41)=a(42)=a(43)=a(46)=a(47)=0. - _Fausto A. C. Cariboni_, May 17 2017

%H S. Sykora, <a href="http://dx.doi.org/10.3247/SL5Math14.002">On Neighbor-Property Cycles</a>, <a href="http://ebyte.it/library/Library.html#math">Stan's Library</a>, Volume V, 2014.

%e The shortest such cycle is of length n=9: {2,4,8,10,5,7,9,3,6}.

%e The next a(n)>0 occurs for n=14 and has 20 solutions.

%e The first and the last of these are:

%e C_1={2,4,8,10,5,7,14,12,15,13,11,9,3,6},

%e C_2={2,4,12,15,13,11,9,3,5,7,14,10,8,6}.

%t A242532[n_] := Count[Map[lpf, Map[j2f, Permutations[Range[3, n + 1]]]], 0]/2;

%t j2f[x_] := Join[{2}, x, {2}];

%t dvf[x_] := Module[{i},

%t    Table[Abs[x[[i]] - x[[i + 1]]] > 1 &&

%t      Divisible[x[[i]] + x[[i + 1]], x[[i]] - x[[i + 1]]], {i,

%t      Length[x] - 1}]];

%t lpf[x_] := Length[Select[dvf[x], ! # &]];

%t Table[A242532[n], {n, 1, 10}]

%t (* OR, a less simple, but more efficient implementation. *)

%t A242532[n_, perm_, remain_] := Module[{opt, lr, i, new},

%t    If[remain == {},

%t      If[Abs[First[perm] - Last[perm]] > 1 &&

%t        Divisible[First[perm] + Last[perm], First[perm] - Last[perm]],

%t       ct++];

%t      Return[ct],

%t      opt = remain; lr = Length[remain];

%t      For[i = 1, i <= lr, i++,

%t       new = First[opt]; opt = Rest[opt];

%t       If[Abs[Last[perm] - new] <= 1 || !

%t          Divisible[Last[perm] + new, Last[perm] - new], Continue[]];

%t       A242532[n, Join[perm, {new}],

%t        Complement[Range[3, n + 1], perm, {new}]];

%t       ];

%t      Return[ct];

%t      ];

%t    ];

%t Table[ct = 0; A242532[n, {2}, Range[3, n + 1]]/2, {n, 1, 15}] (* _Robert Price_, Oct 25 2018 *)

%o (C++) See the link.

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

%K nonn,hard,more

%O 1,14

%A _Stanislav Sykora_, May 30 2014

%E a(29)-a(37) from _Fausto A. C. Cariboni_, May 17 2017