Eyebrow-raising pattern of number of primes between terms of the Fibonacci number sequence? So, $$1,1,2,3,5,8,13,21...$$ Any connection to primes?...it appears not. However, in between the Fibonacci numbers are how much primes? Let's see:

*

*$1$ and $1$: $0$

*$1$ and $2$: $0$

*$2$ and $3$: $0$

*$2$ and $3$: $0$

*$5$ and $8$: $1$

*$8$ and $13$: $1$

*$13$ and $21$: $2$

*$21$ and $34$: $3$

*$34$ and $55$: $5$

*$55$ and $89$: $8$

*$89$ and $144$: $13$
Huh. What could this imply? Let me just close with the same annoying (but wonderful) pattern. $$1,2,3,5,8,13,21...$$
 A: Eyebrow raising indeed, though the pattern does not continue as you suggest. I get
$$
0, 1, 1, 2, 3, 5, 7, 10, 16, 23, 37, 55, 84, 125, 198
$$
Remember that the the number of primes has a well known growth rate (https://en.wikipedia.org/wiki/Prime_number_theorem). Since the Fibonacci numbers are relatively spread out, using $n/\log n$ to approximate the number of primes less than $n$ will cause the number of primes between them to behave like the growth rate of the primes.
A: You can make sense of this pattern (which as yoann points out does not go on very far) as saying that the density of primes is a constant $\phi^{-3}\approx0.236$, since the gap length is $F_{n+1}-F_n=F_{n-1}$ and the conjectured prime count in this gap is $F_{n-4}\approx F_{n-1}\phi^{-3}$.
Unfortunately, this goes against the prime number theorem or various weak versions of it that say that the density of primes goes to $0$ for large $n$, so even without working it out it is clear that the pattern must be short-lived.
(Note: Because the Fibonacci numbers diverge too quickly, the conjectured pattern is not itself sufficient to show that the density of primes is $\phi^{-3}$. But it is enough to prove that the upper density is at least this, which is enough to violate the density zero result. And this is easy to show - considering numbers not divisible by $2,3,5,7$ already gives density $8/35<\phi^{-3}$.)
A: The pattern does not seem to go on: there are only 17 primes between 144 and 233.
More generally, using the Prime number theorem $\pi(x) \sim_{x \to \infty} \frac x {\ln x}$ (where $\pi(x)$ is and the number of primes lower or equal to $x$), and the formula for the n-th Fibonacci number $F_n \sim_{n \to \infty}  \frac{\phi^n}{\sqrt 5}$, we can show that: $$\pi(F_n) \sim_{n \to \infty}  \frac{\phi^n}{n \sqrt 5 \ln \phi}$$
The number of primes between two consecutive Fibonacci numbers is therefore:
$$\pi(F_{n+1}) - \pi(F_n) \sim_{n \to \infty} \frac{\phi^n (\phi - 1)}{n \sqrt 5 \ln \phi} \sim_{n \to \infty} \frac{\phi - 1}{n\ln \phi} F_n$$.
Which contradicts what you would like to have by a factor of $\frac{\phi - 1}{n\ln \phi}$.
