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Where Fibonacci numbers surpass prime numbers

  1. Dec 13, 2008 #1
    The series of prime numbers pn=2, 3, 5, 7, 11, 13, 17, 19, 23, 27..., and Fibonacci numbers Fn=0, 1, 1, 2, 3, 5, 8, 13, 21, 34..., suggest that Fn might be considered to surpass pn exactly at an irrational value ns such that 9<ns<10 and can be determined most exactly from both series as n-->infinity.

    How would you determine ns?
  2. jcsd
  3. Dec 13, 2008 #2


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    There's a nice closed-form for the Fibonacci numbers, but there's nothing so nice for the primes that extends them continuously and 'nicely' to the noninteger reals. So I wouldn't know of a good way to do this.
  4. Dec 13, 2008 #3
    Thanks for your contribution, CRGreathouse. You seem to have addressed the heart of my problem.
  5. Dec 14, 2008 #4


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    It seems like you want to describe the primes and the fibonnaci numbers as some functions p(n) and f(n) and then extend those functions to the real numbers...how would you do this?
  6. Dec 14, 2008 #5
    I think your conjecture is wrong although to find a counterexample you need to go so so far away. The Fibonacci sequence satisfies the recurrence relation: $F_n=F_{n-1}+F_{n-2}$ with $F_1=1$ (or $0$ depends how you define it but it does not matter). Now, if you consider the recurrence relation: $L_n=L_{n-2}+L_{n-3}$ (looks like similar) with initial conditions $L_1=0$, $L_2=2$, $L_3=3$ it is 'simple' (you need some mathematic's background) to proof that if N is prime $L_N$ is also prime but the reverse is not true but to find a counterexample you need to go, as I said before, so far away, indeed it is possible to find a prime P such that $L_{P^2}$ is prime but this number is large but, of courseit , is possible to compute. By the way this last sequence I think is called Lucas sequence.
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