TITLE: Moebius / Zeta function connections
QUESTION [5 upvotes]: Following on from this question, I include a plot of the slightly less clear, but far simpler mathematically Mertens function against $x$ to the power of Zeta Zero 1, where the correlation between the two is fairly convincing - the largest scale changes in frequency imitating those of $x^{\rho_{1}}/e$:

It appears that
$$M(x)\equiv\sum_{k=1}^{x}\mu(k)\sim \frac{\Im(x^{\rho_{1}})}{e}$$
I have not tried it yet, but am wondering whether
$$\frac{1}{n}\sum_{\rho_{n=1}}^{\rho_\infty}\frac{\Im(x^{\rho_{n}})}{e}$$
(or something similar) will yield the precise fluctuations of the Mertens function.
I realise that
$$\frac{1}{\zeta(s)}=\sum_{n=0}^{\infty}\frac{\mu(n)}{n^{s}}$$
but cannot make the leap to the connection with the fluctuations of the zeros. Sorry, but I am a bit out of my depth here, and would greatly appreciate some pointers in the right direction.
Update
A plot of
$$M(x)\equiv\sum_{k=1}^{x}\mu(k)\text{ against } \bigg(\frac{\Im(x^{\rho_{1}})}{e}+\frac{\Im(x^{\rho_{2}})}{e}\bigg)/2$$
for comparison:

Update 2
It appears that to divide by $e$ is not necessary once enough Zeros have been added. A plot of
$$M(x)\equiv\sum_{k=1}^{x}\mu(k)\text{ against } \bigg(\Im(x^{\rho_{1}}+x^{\rho_{2}}+x^{\rho_{3}}+\dots x^{\rho_{9}})\bigg)/9:$$

REPLY [3 votes]: You're on to something.  I recently found a reference to something similar here (see the bottom of p. 141)-Odlyzko and Te Riele's disproof of the Mertens Conjecture.  In his 1951 text on the zeta function, Titchmarsh derived, on condition of the RH:
$M_0(x) = \underset{k\rightarrow \infty}{\lim}\sum_{Im(\rho)<T_k}{\frac{x^{\rho}}{\mid \rho\zeta '(\rho)\mid}} -2+ f(x)$
where f(x) is given as an alternating power series in $\frac{1}{x}$ with very small coefficents (so tending to 0 very fast), and $k\leq T_k\leq k+1$.  $M_0(x)$ is just real-analytic version of $M(n)$ interpolating to half-integer values at the jump discontinuities.
So instead of taking the simple average, try weighting the $x^\rho$ terms as above, then subtract 2 and see how that turns out!  (please share a graph if you do.  I always love a good convincing visual to keep me believing in the RH.)