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A073452
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a(n) is the number of essentially different ways in which the integers 1,2,3,...,n can be arranged in a sequence such that all pairs of adjacent integers sum to a prime number. Rotations and reversals are counted only once.
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3
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1, 1, 1, 1, 2, 3, 12, 16, 70, 232, 1072, 3136, 11648, 18388, 95772, 452136, 2047488, 5565488, 22802028, 60841609, 337801784, 2116714332, 11425028900, 69023494710, 429917269469
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OFFSET
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1,5
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COMMENTS
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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.
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LINKS
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EXAMPLE
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a(5)=2 because there are two essential different arrangements: {1,4,3,2,5} and {3,4,1,2,5}.
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MATHEMATICA
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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
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CROSSREFS
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KEYWORD
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hard,more,nice,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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