Fibonacci numbers and the nontrivial zeros of the Riemann zeta function Is this a mathematical coincidence? For $n=1,\dots,7$:
$$
\left\lfloor \prod_{k=1}^n \arg\left(\rho_k\right)\right\rfloor = F_{n+1},
$$
where $\arg$ is the complex argument, $\rho_n$ is the $n$th nontrivial zero of the Riemann zeta function, and $F_n$ is the $n$th Fibonacci number.  I've found this experimentally.
For larger $n$ values I've checked that for $n = 1,\dots,800$:
$$
F_{n+1} - \prod_{k=1}^n \arg\left(\rho_k\right) + 1 > 0. 
$$
I haven't found any counterexamples.
 A: This is not as surprising as it might seem. Let me address your second observation first and sketch a proof that it holds for all $n$. It is known that all non-trivial zeros $s$ lie in the "critical strip" $0<\Re(s)<1$ where $\Re$ is the real part of a complex number. Notice that the further away from the real axis we get, the closer the argument of any number in this strip will get to $\frac{\pi}2$, so we expect the product in your question be approximately $\left(\frac{\pi}2\right)^n$, but always strictly less. The Fibonacci numbers are asymptotic to $\varphi^n$ (where $\varphi$ is the golden ratio). Since $\varphi > \frac{\pi}2$ eventually the Fibonacci numbers will dominate the product in question, explaining your second observation. This may easily be extended into a formal proof of the fact.
This also gives us a hint about why the Fibonacci numbers are so close to the product in question - $\varphi$ and $\frac{\pi}2$ are very close to each other, so if the series are well-enough aligned to start with, they'll probably stay that way for a while, which is exactly what we observe. In particular, let us note that, $\frac{\pi}2$ is more than $F_2=1$, so the fact that our product will eventually be less than the Fibonacci sequence means things are being "pulled together" for the moment, allowing the coincidence to persist a while.
To really illustrate us, look at two bounds for your product. For $n=\{1,2,\ldots,7\}$ we have the following upper then lower bound:
$$\left\lfloor\left(\frac{\pi}2\right)^n\right\rfloor=\{1,2,3,6,9,15,23\}$$
$$\lfloor\operatorname{arg}(\rho_1)^n\rfloor=\{1,2,3,5,8,13,20\}$$
where the true product is somewhere between there. We're already pretty close to the Fibonacci sequence no matter what, and the true value just gets a little lucky and hits it right on.
In short: This is mostly a coincidence of how $\frac{\pi}2$ and $\varphi$ are close to each other, and the particular choice of $\operatorname{arg}(\rho_n)$  rather than $\frac{\pi}2$ tempers things "just enough" to make everything work out, but the coincidence doesn't rely on too many particulars of the Riemann zeta function.
