|
| |
|
|
A072676
|
|
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.
|
|
2
|
|
|
|
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, 29, 30, 31, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 45, 46, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 60, 61, 63, 64, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 78, 79, 81, 82, 83, 84
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
COMMENTS
|
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.
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
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.
|
|
|
MATHEMATICA
|
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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nice,nonn
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
