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A242531
Number of cyclic arrangements of S={1,2,...,n} such that the difference of any two neighbors is a divisor of their sum.
16
0, 1, 1, 1, 1, 4, 3, 9, 26, 82, 46, 397, 283, 1675, 9938, 19503, 10247, 97978, 70478, 529383, 3171795, 7642285, 3824927, 48091810, 116017829, 448707198, 1709474581, 6445720883, 3009267707, 51831264296, 32231861027, 204352130129, 1264753275210, 3238009264466
OFFSET
1,6
COMMENTS
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.
LINKS
S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
EXAMPLE
The only such cycle of length n=5 is {1,2,4,5,3}.
For n=7 there are three solutions: C_1={1,2,4,5,7,6,3}, C_2={1,2,4,6,7,5,3}, C_3={1,2,6,7,5,4,3}.
MATHEMATICA
A242531[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2, n]]]], 0]/2;
j1f[x_] := Join[{1}, x, {1}];
dvf[x_] := Module[{i},
Table[Divisible[x[[i]] + x[[i + 1]], x[[i]] - x[[i + 1]]], {i,
Length[x] - 1}]];
lpf[x_] := Length[Select[dvf[x], ! # &]];
Join[{0, 1}, Table[A242531[n], {n, 3, 10}]]
(* OR, a less simple, but more efficient implementation. *)
A242531[n_, perm_, remain_] := Module[{opt, lr, i, new},
If[remain == {},
If[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[! Divisible[Last[perm] + new, Last[perm] - new], Continue[]];
A242531[n, Join[perm, {new}],
Complement[Range[2, n], perm, {new}]];
];
Return[ct];
];
];
Join[{0, 1}, Table[ct = 0; A242531[n, {1}, Range[2, n]]/2, {n, 3, 13}]] (* Robert Price, Oct 25 2018 *)
PROG
(C++) See the link.
KEYWORD
nonn,hard
AUTHOR
Stanislav Sykora, May 30 2014
EXTENSIONS
a(24)-a(28) from Fausto A. C. Cariboni, May 25 2017
a(29) from Fausto A. C. Cariboni, Jul 09 2020
a(30) from Fausto A. C. Cariboni, Jul 14 2020
More terms from Martin Fuller, Mar 08 2026
STATUS
approved