A242527 Number of cyclic arrangements (up to direction) of {0,1,...,n-1} such that the sum of any two neighbors is a prime.
0, 0, 0, 0, 1, 1, 2, 6, 6, 22, 80, 504, 840, 6048, 3888, 37524, 72976, 961776, 661016, 11533030, 7544366, 133552142, 208815294, 5469236592, 6429567323, 153819905698, 182409170334, 4874589558919, 7508950009102, 209534365631599
Offset: 1
Examples
The first such cycle is of length n=5: {0,2,1,4,3}. The first case with 2 solutions is for cycle length n=7: C_1={0,2,3,4,1,6,5}, C_2={0,2,5,6,1,4,3}. The first and the last of the 22 such cycles of length n=10 are: C_1={0,3,2,1,4,9,8,5,6,7}, C_22={0,5,8,9,4,3,2,1,6,7}.
Links
- S. Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014.
Crossrefs
Programs
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Mathematica
A242527[n_] := Count[Map[lpf, Map[j0f, Permutations[Range[n - 1]]]], 0]/2; j0f[x_] := Join[{0}, x, {0}]; lpf[x_] := Length[Select[asf[x], ! PrimeQ[#] &]]; asf[x_] := Module[{i}, Table[x[[i]] + x[[i + 1]], {i, Length[x] - 1}]]; Table[A242527[n], {n, 1, 10}] (* OR, a less simple, but more efficient implementation. *) A242527[n_, perm_, remain_] := Module[{opt, lr, i, new}, If[remain == {}, If[PrimeQ[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[! PrimeQ[Last[perm] + new], Continue[]]; A242527[n, Join[perm, {new}], Complement[Range[n - 1], perm, {new}]]; ]; Return[ct]; ]; ]; Table[ct = 0; A242527[n, {0}, Range[n - 1]]/2, {n, 1, 15}] (* Robert Price, Oct 18 2018 *)
Extensions
a(23)-a(30) from Max Alekseyev, Jul 09 2014
Comments