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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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    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, $x_n = x_{n+2}-x_{n+1}$.2012-05-25

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