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Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is 2^k for some k=0,1,2,...
18

%I #29 Oct 23 2018 03:03:24

%S 0,1,1,1,4,8,14,32,142,426,1204,3747,9374,26306,77700,219877,1169656,

%T 4736264,17360564,69631372,242754286,891384309,3412857926,12836957200,

%U 42721475348,152125749587,549831594988

%N Number of cyclic arrangements of S={1,2,...,n} such that the difference between any two neighbors is 2^k for some k=0,1,2,...

%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. Evaluating this sequence for n>=3 is equivalent to counting Hamiltonian cycles in a pair-property graph with n vertices and is often quite hard. For more details, see the link.

%H Hiroaki Yamanouchi, <a href="/A242519/b242519.txt">Table of n, a(n) for n = 1..27</a> (first 21 terms from Stanislav Sykora)

%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.

%F For any S and any P, and for n>=3, NPC(n;S;P)<=A001710(n-1).

%e The four such cycles of length 5 are:

%e C_1={1,2,3,4,5}, C_2={1,2,4,3,5}, C_3={1,2,4,5,3}, C_4={1,3,2,4,5}.

%e The first and the last of the 426 such cycles of length 10 are:

%e C_1={1,2,3,4,5,6,7,8,10,9}, C_426={1,5,7,8,6,4,3,2,10,9}.

%t A242519[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;

%t j1f[x_] := Join[{1}, x, {1}];

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

%t t = Table[2^k, {k, 0, 10}];

%t Join[{0, 1}, Table[A242519[n], {n, 3, 10}]]

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

%t A242519[n_, perm_, remain_] := Module[{opt, lr, i, new},

%t If[remain == {},

%t If[MemberQ[t, Abs[First[perm] - Last[perm]]], 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[! MemberQ[t, Abs[Last[perm] - new]], Continue[]];

%t A242519[n, Join[perm, {new}],

%t Complement[Range[2, n], perm, {new}]];

%t ];

%t Return[ct];

%t ];

%t ];

%t t = Table[2^k, {k, 0, 10}];

%t Join[{0, 1}, Table[ct = 0; A242519[n, {1}, Range[2, n]]/2, {n, 3, 12}]] (* _Robert Price_, Oct 22 2018 *)

%o (C++) See the link.

%Y Cf. A001710, A236602, A242520, A242521, A242522, A242523, A242524, A242525, A242526, A242527, A242528, A242529, A242530, A242531, A242532, A242533, A242534.

%K nonn,hard

%O 1,5

%A _Stanislav Sykora_, May 27 2014

%E a(22)-a(27) from _Hiroaki Yamanouchi_, Aug 29 2014