The argument of a finite Euler product at a non-trivial zero of $\zeta(s)$. With $p_n$ is the $n$-th prime number, we know that:
$$\arg\left(\zeta(s)\right)=\arg\prod_{n=1}^{\infty} \frac{1}{1-\frac{1}{p_n^s}}=\sum_{n=1}^{\infty} \arg\left(\frac{1}{1-\frac{1}{p_n^s}}\right)\qquad \Re(s) > 1$$
Now define the finite series:
$$f(s,N)=\sum_{n=1}^{N} \arg\left(\frac{1}{1-\frac{1}{p_n^s}}\right) \qquad s \in \mathbb{C}$$
and plot it for values at and near a non-trivial zero ($\rho$) of $\zeta(s)$ for subsequent $N$. This gives for instance:



For $N \rightarrow \infty$ these oscillating series show a reducing frequency and increasing amplitude ('spiral waves'). Oscillating series do neither converge or diverge. however it seems that only when $s=\rho$ (middle graph) the oscillation occurs along a horizontal line (the same pattern emerges for all $\rho$s I tested). 
Is there any possible way of calculating the location of such a line? 

The Euler product diverges for $\Re(s) < 1$, you need to replace it by the regularized version (valid for $\Re(s)\in (1/2,1)$ assuming the RH is true)
$$\lim_{x\to \infty} -\sum_{p \le x} \log(1-p^{-s}) - pv\int_0^x \frac{t^{-s}}{\log t}dt + \frac{\log(s-1)}{s}$$
If there are infinitely many zeros of real part $\ge \Re(s) $ then there is no regularized version at $s$.
