# Where Fibonacci numbers surpass prime numbers

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?

CRGreathouse
Homework Helper
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.

Thanks for your contribution, CRGreathouse. You seem to have addressed the heart of my problem.

Office_Shredder
Staff Emeritus
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.