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A072617
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Number of essentially different ways of arranging numbers 1 through 2n around a circle so that the sum of each pair of adjacent numbers is prime, with the odd and even numbers in order in opposite directions.
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5
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1, 1, 1, 1, 1, 2, 1, 1, 1, 0, 0, 2, 1, 1, 2, 0, 0, 1, 1, 1, 3, 0, 0, 1, 0, 0, 2, 1, 1, 2, 0, 0, 2, 1, 1, 1, 0, 0, 3, 0, 0, 1, 0, 0, 3, 0, 0, 3, 1, 1, 1, 1, 1, 3, 0, 0, 0, 0, 0, 5, 0, 0, 3, 0, 0, 4, 1, 1, 4, 0, 0, 2, 1, 1, 2, 0, 0, 2, 0, 0, 4, 0, 0, 5, 0, 0, 4, 1, 1, 5, 0, 0, 3, 1, 1, 2, 1, 1, 4, 0
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OFFSET
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1,6
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COMMENTS
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A very restricted form of the prime circle problem whose sequence is A051252. Finding these solutions is very fast because there are only n possible solutions to try. See A072616 for the case where only the odd numbers or only the even numbers are in order. Note that a(2)=1 because the two solutions are essentially the same. Solutions can be printed by removing comments from the Mathematica program.
There is a provable solution for n when either (a) 2n+1 and 2n+3 are prime, (b) 2k+1, 2k+3, 2k+2n+1 and 2k+2n+3 are prime for some 0 < k < n-1, or (c) 2n-1, 2n+1 and 4n-1 are primes. Part (a) is due to Mike Hennebry. Note that cases (a) and (b) involve 3 sets of twin primes. For n > 3, due to the form of twin primes, it can be shown that (a) implies not (b) and not (c).
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LINKS
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EXAMPLE
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a(6) = 2 because there are two ways: {1,10,3,8,5,6,7,4,9,2,11,12} and {1,4,3,2,5,12,7,10,9,8,11,6}.
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MATHEMATICA
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For[lst={}; n=1, n<=100, n++, oddTable=Append[Table[2i-1, {i, n}], 1]; evenTable=Table[2n+2-2i, {i, n}]; evenTable=Join[evenTable, evenTable]; For[cnt=0; i=1, i<=n, i++, j=0; allPrime=True; While[j<n&&allPrime, j++; allPrime= PrimeQ[oddTable[[j]]+evenTable[[i+j-1]]]&& PrimeQ[oddTable[[j+1]]+evenTable[[i+j-1]]]]; (*If[allPrime, For[soln={}; j=1, j<=n, j++, AppendTo[soln, oddTable[[j]]]; AppendTo[soln, evenTable[[i+j-1]]]]; Print[soln]]; *)If[allPrime, cnt++ ]]; AppendTo[lst, cnt]]; lst
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CROSSREFS
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KEYWORD
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nice,nonn
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AUTHOR
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STATUS
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approved
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