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A117517
Numbers k such that F(2*k + 1) is prime where F(m) is a Fibonacci number.
1
1, 2, 3, 5, 6, 8, 11, 14, 21, 23, 41, 65, 68, 179, 215, 216, 224, 254, 284, 285, 1485, 2361, 2693, 4655, 4838, 7215, 12780, 15378, 17999, 18755, 25416, 40919, 52455, 65010, 74045, 100553, 198689, 216890, 295020, 296844, 302355, 465758, 524948, 642803, 818003, 901529, 984360, 1452176
OFFSET
1,2
COMMENTS
For F(k) to be prime, with k > 4, it is necessary but not sufficient for k to be prime. Hence after F(4) = 3, every prime F(m) is of the form F(2*k+1) for some k. Every prime divides some Fibonacci number. See also comment to A093062. - Jonathan Vos Post, Apr 29 2006
LINKS
H. Dubner and W. Keller, New Fibonacci and Lucas Primes, Math. Comp. 68 (1999) 417-427.
FORMULA
a(n) = (A083668(n)-1)/2. - R. J. Mathar, Jul 08 2009
a(n) = (A001605(n+1)-1)/2, n > 1. - Vincenzo Librandi, May 24 2016
EXAMPLE
If k=68 then F(2*k + 1) = 19134702400093278081449423917, a prime, so 68 is a term.
MATHEMATICA
Select[Range[0, 5000], PrimeQ[Fibonacci[2 # + 1]] &] (* Vincenzo Librandi, May 24 2016 *)
PROG
(Magma) [n: n in [0..1000] | IsPrime(Fibonacci(2*n+1))]; // Vincenzo Librandi, May 24 2016
KEYWORD
nonn
AUTHOR
Parthasarathy Nambi, Apr 26 2006
EXTENSIONS
More terms from Vincenzo Librandi, May 24 2016
STATUS
approved