A227050 Number of essentially different ways of arranging numbers 1 through 2n around a circle so that the sum and absolute difference of each pair of adjacent numbers are prime.
0, 0, 0, 0, 0, 2, 1, 4, 88, 0, 976, 22277, 22365, 376002, 3172018, 5821944, 10222624, 424452210, 6129894510, 38164752224
Offset: 1
Examples
For n = 6 the a(6) = 2 solutions are (1, 4, 9, 2, 5, 12, 7, 10, 3, 8, 11, 6) and (1, 6, 11, 8, 3, 10, 7, 4, 9, 2, 5, 12) because abs(1 - 4) = 3 and 1 + 4 = 5 are prime, etc.
Links
- Gary Antonick, Numberplay: Bernardo Recamán’s Primes in a Circle Puzzle, Jun 17 2013.
- Stanislav Sykora, On Neighbor-Property Cycles, Stan's Library, Volume V, 2014; Table III.
Crossrefs
Programs
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Mathematica
A227050[n_] := Count[Map[lpf, Map[j1f, Permutations[Range[2,2 n]]]], 0]/2; j1f[x_] := Join[{1}, x, {1}]; lpf[x_] := Length[ Join[Select[asf[x], ! PrimeQ[#] &], Select[Differences[x], ! PrimeQ[#] &]]]; asf[x_] := Module[{i}, Table[x[[i]] + x[[i + 1]], {i, Length[x] - 1}]]; Table[A227050[n], {n, 1, 6}] (* OR, a less simple, but more efficient implementation. *) A227050[n_, perm_, remain_] := Module[{opt, lr, i, new}, If[remain == {}, If[PrimeQ[First[perm] - Last[perm]] && 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] && PrimeQ[Last[perm] + new]), Continue[]]; A227050[n, Join[perm, {new}], Complement[Range[2 n], perm, {new}]]; ]; Return[ct]; ]; ]; Table[ct = 0; A227050[n, {1}, Range[2, 2 n]]/2, {n, 1, 10}] (* Robert Price, Oct 22 2018 *)
Extensions
a(15)-a(18) added by Tim Cieplowski, Jan 04 2015
a(19) from Fausto A. C. Cariboni, Jun 06 2017
a(20) from Bert Dobbelaere, Feb 15 2020
Comments