|
%I #19 Jul 17 2021 04:30:27
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,19,20,21,22,24,25,26,27,28,
%T 29,30,31,33,34,35,36,37,39,40,41,42,43,45,46,48,49,50,51,52,53,54,55,
%U 56,57,58,60,61,63,64,66,67,68,69,70,72,73,74,75,76,78,79,81,82,83,84
%N Numbers k for which the prime circle problem has a solution composed of disjoint subsets: the arrangement of numbers 1 through 2k around a circle is such that the sum of each pair of adjacent numbers is prime, the odd numbers are in order and the even numbers are in groups of decreasing sequences.
%C This is a generalization of A072618. The integer k is in this sequence if either (a) 4k-1 and 2k+1 are prime, or (b) 2k+2i-1, 2k+2i+1 and 2i+1 are prime for some 0 < i < k. The Mathematica program computes a prime circle for such k. It is very easy to show that there are prime circles for large k, such as 10^10.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/PrimeCircle.html">Prime Circle</a>.
%e k=10 is a term because one solution is {1, 2, 3, 8, 5, 6, 7, 4, 9, 20, 11, 18, 13, 16, 15, 14, 17, 12, 19, 10} and the even numbers are in three decreasing sequences {2}, {8, 6, 4} and {20, 18, 16, 14, 12, 10}. Note that this solution contains {1, 2} and {1, 2, 3, 8, 5, 6, 7, 4}, which are solutions for k=1 and k=4.
%t n=10; lst={}; i=0; found=False; While[i<n&&!found, i++; If[i==n, found=PrimeQ[4n-1]&&PrimeQ[2n+1], found=PrimeQ[2n+2i-1]&&PrimeQ[2n+2i+1]&&PrimeQ[2i+1]]]; If[found, lst=Flatten[Table[{2j-1, 2n-2(j-i)}, {j, i, n}]], Print["no solution using this method"]]; If[found, While[n=i-1; n>0, i=0; found=False; While[i<n&&!found, i++; found=PrimeQ[2n+2i-1]&&PrimeQ[2n+2i+1]]; If[found, lst=Flatten[Append[Table[{2j-1, 2n-2(j-i)}, {j, i, n}], lst]]]]]; lst
%Y Cf. A051252, A072616, A072617, A072618, A072184.
%K nice,nonn
%O 1,2
%A _T. D. Noe_, Jul 01 2002
|
