If $\rho (n)$ are the zeros of the Riemann zeta function, it seems that:
$$\rho (n)=\frac{1}{2} +i \frac{2 \pi n}{ln(n)}+ i \frac{2 \pi ln(ln(n)) n}{ln(n)^2}+ o(\frac{ln(ln(n))n}{ln(n)^2})$$
In particular, there can't be double zeros of the Riemann zeta function for $n$ large.