OFFSET
1,5
COMMENTS
Note that when the first and last numbers of an arrangement sum to a prime, then there are n rotations that are treated as one arrangement. The case n=6 exhibits rotational solutions: {1,4,3,2,5,6}, which is actually a prime circle. See A051252 for more details about prime circles.
The Mathematica program uses a backtracking algorithm to count the arrangements. To print the unique arrangements, remove the comments from around the print statement. It seems that a greedy algorithm can be used to quickly find a solution for any n.
LINKS
Carlos Rivera, Conjecture 20. The first N natural numbers listed in an order such that the sum of each two adjacent of them is a prime number, and the Rivera's Algorithm, The Prime Puzzles & Problems Connection.
Carlos Rivera, Puzzle 189. Squares and primes in a row, The Prime Puzzles & Problems Connection.
EXAMPLE
a(5)=2 because there are two essential different arrangements: {1,4,3,2,5} and {3,4,1,2,5}.
MATHEMATICA
nMax=12; $RecursionLimit=500; try[lev_] := Module[{t, j, circular}, If[lev>n, circular=PrimeQ[soln[[1]]+soln[[n]]]; If[(!circular&&soln[[1]]<soln[[n]])||(circular&&soln[[1]]==1&&soln[[2]]<=soln[[n]]), (*Print[soln]; *)cnt++ ], (*else append another number to the soln list*) t=soln[[lev-1]]; For[j=1, j<=Length[s[[t]]], j++, If[ !MemberQ[soln, s[[t]][[j]]], soln[[lev]]=s[[t]][[j]]; try[lev+1]; soln[[lev]]=0]]]]; For[lst={1}; n=2, n<=nMax, n++, s=Table[{}, {n}]; For[i=1, i<=n, i++, For[j=1, j<=n, j++, If[i!=j&&PrimeQ[i+j], AppendTo[s[[i]], j]]]]; soln=Table[0, {n}]; For[cnt=0; i=1, i<=n, i++, soln[[1]]=i; try[2]]; AppendTo[lst, cnt]]; lst
CROSSREFS
KEYWORD
hard,more,nice,nonn
AUTHOR
T. D. Noe, Aug 02 2002
EXTENSIONS
a(21)-a(25) from Martin Ehrenstein, Jul 19 2023
STATUS
approved