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Search: id:A072184
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| A072184 |
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n for which the prime circle problem has a slightly more complex solution than given in A072676. Complement of A072676. |
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3
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| 17, 23, 32, 38, 44, 47, 59, 62, 65, 71, 77, 80, 86, 92, 101, 104, 107, 110, 122, 128, 137, 143, 146, 149, 152, 161, 164, 167, 170, 176, 179, 182, 185, 188, 191, 194, 197, 200, 203, 206, 212, 218, 224, 227, 233, 236, 239, 242, 245, 248, 251, 254, 257, 263, 266, 269, 272, 275, 278, 287, 290
(list; graph; listen)
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OFFSET
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0,1
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COMMENT
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n is in this sequence if (a) 2n+2i-1 and 2n+2i+1 are prime for some 0 < i < n and (b) there is an integer j < i such that the following four numbers are prime: 2n-1+2j, 2j+1, 2i+2j-1, and 2i+2j+1. The Mathematica program computes a prime circle for such n.
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LINKS
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Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
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MATHEMATICA
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n=17; lst={}; i=0; found=False; While[ !found&&i<n, i++; If[PrimeQ[2n+2i-1]&&PrimeQ[2n+2i+1], j=0; While[ !found&&j<i, j++; found=PrimeQ[2n-1+2j]&&PrimeQ[2j+1]&&PrimeQ[2i+2j-1]&&PrimeQ[2i+2j+1]]]]; If[found, lst=Flatten[Join[Table[{2k-1, 2i-2(k-j)}, {k, j, i-1}], Table[{2k-1, 2n-2(k-i)}, {k, i, n-1}], {2n-1, 2j}]], Print["no solution using this method"]]; If[found, i=j; 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
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CROSSREFS
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Cf. A051252.
Adjacent sequences: A072181 A072182 A072183 this_sequence A072185 A072186 A072187
Sequence in context: A050712 A068581 A080830 this_sequence A107644 A102874 A086532
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KEYWORD
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nonn
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AUTHOR
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T. D. Noe (noe(AT)sspectra.com), Jul 01 2002
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