

A242519


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,...


17



0, 1, 1, 1, 4, 8, 14, 32, 142, 426, 1204, 3747, 9374, 26306, 77700, 219877, 1169656, 4736264, 17360564, 69631372, 242754286, 891384309, 3412857926, 12836957200, 42721475348, 152125749587, 549831594988
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OFFSET

1,5


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


LINKS



FORMULA

For any S and any P, and for n>=3, NPC(n;S;P)<=A001710(n1).


EXAMPLE

The four such cycles of length 5 are:
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}.
The first and the last of the 426 such cycles of length 10 are:
C_1={1,2,3,4,5,6,7,8,10,9}, C_426={1,5,7,8,6,4,3,2,10,9}.


MATHEMATICA

A242519[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]], ! MemberQ[t, #] &]];
t = Table[2^k, {k, 0, 10}];
Join[{0, 1}, Table[A242519[n], {n, 3, 10}]]
(* OR, a less simple, but more efficient implementation. *)
A242519[n_, perm_, remain_] := Module[{opt, lr, i, new},
If[remain == {},
If[MemberQ[t, Abs[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[! MemberQ[t, Abs[Last[perm]  new]], Continue[]];
Complement[Range[2, n], perm, {new}]];
];
Return[ct];
];
];
t = Table[2^k, {k, 0, 10}];
Join[{0, 1}, Table[ct = 0; A242519[n, {1}, Range[2, n]]/2, {n, 3, 12}]] (* Robert Price, Oct 22 2018 *)


PROG

(C++) See the link.


CROSSREFS

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


KEYWORD

nonn,hard


AUTHOR



EXTENSIONS



STATUS

approved



