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Search: id:A117517
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| A117517 |
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Numbers n such that F(2*n + 1) is prime where F(m) is the Fibonacci number. |
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+0
1
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| 1, 2, 3, 5, 6, 8, 11, 14, 21, 23, 41, 65, 68
(list; graph; listen)
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OFFSET
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1,2
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COMMENT
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For F(n) to be prime, with n>4, it is necessary but not sufficient for n to be prime. Hence after F(4) = 3, every prime F(m) is of the form F(2*n+1) for some n. Every prime divides some Fibonacci number. See also comment to A093062. - Jonathan Vos Post (jvospost2(AT)yahoo.com), Apr 29 2006
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REFERENCES
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H. Dubner and W. Keller, New Fibonacci and Lucas Primes, Math. Comp. 68 (1999) 417-427.
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EXAMPLE
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If n=68 then F(2*n + 1) is a prime with twenty nine digits.
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CROSSREFS
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Cf. A000045, A001605, A117595.
Cf. A001602, A022307, A030427, A051694, A075737, A083668, A099000.
Sequence in context: A095172 A127312 A081830 this_sequence A098491 A107947 A120768
Adjacent sequences: A117514 A117515 A117516 this_sequence A117518 A117519 A117520
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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), Apr 26 2006
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