OFFSET
1,3
COMMENTS
a(n)=NPC(2n;S;P) is the count of all neighbor-property cycles for a specific set S of 2n elements and a specific pair-property P. For more details, see the link and A242519.
Conjecture: in this case it seems that NPC(n;S;P)=0 for all odd n, so only the even ones are listed. This is definitely not the case when the property P is replaced by its negation (see A242534).
The cycle must alternate odd/even, so that 2 is not a common factor of sum and difference. Therefore only even lengths are possible. This proves the above conjecture. - Martin Fuller, Mar 12 2026
LINKS
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
EXAMPLE
For n=4, the only cycle is {1,2,3,4}.
The two solutions for n=6 are: C_1={1,2,3,4,5,6} and C_2={1,4,3,2,5,6}.
MATHEMATICA
A242533[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, 2 n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
lpf[x_] := Length[Select[cpf[x], ! # &]];
cpf[x_] := Module[{i},
Table[CoprimeQ[x[[i]] - x[[i + 1]], x[[i]] + x[[i + 1]]], {i,
Length[x] - 1}]];
Join[{1}, Table[A242533[n], {n, 2, 5}]]
(* OR, a less simple, but more efficient implementation. *)
A242533[n_, perm_, remain_] := Module[{opt, lr, i, new},
If[remain == {},
If[CoprimeQ[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[! CoprimeQ[Last[perm] + new, Last[perm] - new], Continue[]];
A242533[n, Join[perm, {new}],
Complement[Range[2, 2 n], perm, {new}]];
];
Return[ct];
];
];
Join[{1}, Table[ct = 0; A242533[n, {1}, Range[2, 2 n]]/2, {n, 2, 6}] ](* Robert Price, Oct 25 2018 *)
PROG
(C++) See the link.
CROSSREFS
KEYWORD
nonn,hard,more
AUTHOR
Stanislav Sykora, May 30 2014
EXTENSIONS
a(10)-a(11) from Fausto A. C. Cariboni, May 31 2017, Jun 01 2017
a(12)-a(18) from Martin Fuller, Mar 12 2026
STATUS
approved
