The question of knowing whether there are infinitely many Fibonacci primes is an open question. As $F\_p$ is prime only if $p$ is prime, one has $\pi\_{FP}(x)\le \pi(\log\_{\phi} x+0.5\log 5)$, but numerical computations seem to show that this quantity is roughly equal to $\log\_{\phi}\log\_{\phi}x$, where $\pi\_{FP}(x)$ is the number of Fibonacci primes below $x$. Is there any heuristics suggesting this estimation should be true?  

Thanks in advance.



is there any heuristics suggesting that the number of Fibonacci primes below $x$ is equivalent to $\log_{\phi}\log_{\phi}x$?
We need to find a reasonable-sounding answer for the following question: for a fixed prime $p$; what is the probability that $F\_p$ is prime? 


A prime $q<F\_p$ divides $F\_p$ if and only if $z(q)$ divides $p$, where $z$ is the classical [Fibonacci entry point](http://oeis.org/A001177). Therefore, we must have $z(q)=p$ for our candidate prime factor.


First, we must limit the size of $q$-it can't be any number. Since $q$ is prime, $z(q)$ is a divisor of $q-\left ( \frac{q}{5} \right )$ (exercise), and in particular it is $\leq q+1 \approx q$. We conclude that $q\gtrsim p$.


On the other hand, from $F\_n \sim \varphi^n/\sqrt 5$ and $q<F\_p$ one has $q\lesssim\varphi ^p$ (and this is the best we can hope for as there is equality when $q=F\_p$ is prime).


What is the probability that $z(q)=p$? Since we know nothing about $q$ and $p$ except that they are prime, it is not too absurd to assume that $z(q)$ is a random integer between $\log q/\log \varphi$ and $q+1$, so its probability to be $p$ is about $\frac 1 q$ (I will return on this point later).


The probability that for a fixed $p$ a candidate prime factor $q$ of $F\_p$ does not exist is then about
$$ \prod\_{p<q<\varphi^p} \left ( 1-\frac 1 q \right ) \approx \frac{\text{e}^{-\gamma}}{\log \varphi ^p} \Big/ \frac{\text{e}^{-\gamma}}{\log p}=\frac 1 {\log \varphi} \frac{\log p}{p}$$
by Mertens.


The expected count of those primes up to $x$ is the sum of the individual probabilities up to $\log\_\varphi x$ (the quantity of Fibonacci numbers up to $x$), that is
$$ \frac 1 {\log \varphi}\sum\_{p <\log\_\varphi x} \frac{\log p}{p} ,$$
and Čebyšëv tells us this is about $\log\_\varphi \log\_\varphi x$.


Now, a comment:


The assumption I made about the probability of $z(q)=p$ is very rough, and has a high probability of being wildly inaccurate. The reason is that we know something more than only the primality of $p$ and $q$: $z(q)$ isn't just any random number between $\log\_\varphi q$ and $q+1$, it is a divisor either of $q+1$ or $q-1$! So, in order for $z(q)=p$ to be true, necessarily $q \equiv \pm 1 \pmod p$. For a more refined heuristic one would then have to take the product only over those $q$, and use some widely-believed statistic on the factorization of shifted primes-the calculations are not immediate and I haven't carried them out, but this is a thorny point and I am open to discussion about more sensible assumptions.


Another good starting point would be the [nice paper](http://arxiv.org/abs/1212.6221) by Cubre and Rouse.


