OFFSET
1,1
COMMENTS
For such primes p, 2p-1 divides Fibonacci(p). Actually it is also true that (2m-1) divides Fibonacci(m) for *all* m > 7, m = 2 or 4 (mod 5) for which 2m-1 is prime.
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Vladimir Drobot, On primes in the Fibonacci sequence, Fib. Quart. 38 (1) (2000) 71.
EXAMPLE
Note that 19 is prime and so is 2*19-1 or 37.
MATHEMATICA
okQ[n_]:=Module[{x=Mod[n, 5]}, PrimeQ[2n-1]&&MemberQ[{2, 4}, x]]; Select[Prime[Range[5, 500]], okQ] (* Harvey P. Dale, Jan 14 2011 *)
PROG
(Haskell)
a053685 n = a053685_list !! (n-1)
a053685_list = dropWhile (<= 7) $ i a047211_list a005382_list where
i xs'@(x:xs) ys'@(y:ys) | x < y = i xs ys'
| x > y = i xs' ys
| otherwise = x : i xs ys
-- Reinhard Zumkeller, Oct 03 2012
CROSSREFS
KEYWORD
easy,nice,nonn
AUTHOR
James A. Sellers, Feb 15 2000
STATUS
approved