Which starting conditions for the Fibonacci sequence, gives most primes

I found the following question (at http://aperiodical.com/2012/05/matt-parkers-twitter-puzzle-25-may/):

If you start the Fibonacci sequence 2,1 instead of 1,1 do you get more or fewer primes?

Looking at the first 10 numbers, we get more primes with the modified starting conditions:

1, 1, 2, 3, 5, 8, 13, 21, 34, 55

2, 1, 3, 4, 7, 11, 18, 29, 47, 76

The closed form for the normal Fibonacci sequence is $$F_n(1,1) = \frac{1}{\sqrt{5}}\left(\frac{1+ \sqrt{5}}{2}\right)^n - \frac{1}{\sqrt{5}}\left(\frac{1- \sqrt{5}}{2}\right)^n$$

and I worked out the closed form for the modified version

$$F_n(2,1) = \left(-\frac12 + \frac12 \sqrt{5}\right)\left(\frac12 + \frac12 \sqrt{5}\right)^n + \left(-\frac12 - \frac12 \sqrt{5}\right)\left(\frac12 - \frac12 \sqrt{5}\right)^n$$

I have not been able to find an integer $n$, where the two functions intersect.

Do you have a solution to the problem, or maybe just an idea to follow?

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It is wide open whether the number of prime Fibonacci numbers is finite (en.wikipedia.org/wiki/Fibonacci_prime). If it's infinite, then one can't talk about number of primes and must instead talk about density. For reference, the second sequence is called the Lucas numbers (en.wikipedia.org/wiki/Lucas_number), and I don't understand what the intersection of the sequences has to do with the question about primes. – Qiaochu Yuan May 25 '12 at 13:18
Thanks for the references. I did not know about the Lucas Numbers. My idea, was to see if the sequences became identical at some point. – utdiscant May 25 '12 at 13:22
They never become identical. – Qiaochu Yuan May 25 '12 at 13:24
And the reason why they never become identical is that you can run the defining equation backwards and continue the sequence as far as you want to the left in a unique way. Namely, $x_n = x_{n+2}-x_{n+1}$. – Henning Makholm May 25 '12 at 15:20