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Search: id:A111041
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| A111041 |
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Numbers n such that (2*n^2) + 25 is prime. |
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+0
1
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| 3, 6, 12, 18, 21, 27, 33, 36, 39, 51, 54, 63, 66, 69, 81, 96, 114, 138, 159, 168, 177, 183, 204, 216, 219, 228, 231, 234, 237, 252, 258, 276, 279, 282
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OFFSET
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1,1
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COMMENT
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Necessarily n = 0 (mod 3) because: (a) if n = 3k+1 then 2*n^2 + 25 = 2*(3k+1)^2 + 25 = 2*(9*k^2 + 6*k + 1) + 25 = 18*k^2 + 12*k + 27 = 0 mod 3; (b) if n = 3k-1 then 2*n^2 + 25 = 2*(3k-1)^2 + 25 = 2*(9*k^2 - 6*k + 1) + 25 = 18*k^2 - 12*k + 27 = 0 mod 3; (c) while n = 3k then 2*n^2 + 25 = 2*(3k)^2 + 25 = 18*k^2 + 25 = 1 mod 3, which can be prime. - Jonathan Vos Post (jvospost3(AT)gmail.com), Oct 06 2005
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EXAMPLE
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If n=96 then (2*n^2) + 25 = 18457 (prime).
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CROSSREFS
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Sequence in context: A038046 A162845 A038588 this_sequence A079830 A160744 A160738
Adjacent sequences: A111038 A111039 A111040 this_sequence A111042 A111043 A111044
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KEYWORD
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nonn
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AUTHOR
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Parthasarathy Nambi (PachaNambi(AT)yahoo.com), Oct 05 2005
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EXTENSIONS
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More terms from Jonathan Vos Post (jvospost3(AT)gmail.com), Oct 06 2005
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