# Bounding an equivalent expression of Mertens function

Some months ago, I derived the following formula for the Merten's function $$M(n)$$ using the inclusion-exclusion principle:

$$M(n)=1-\pi\left(n\right)+\sum_{p_{i}\leq\frac{n}{p_i}}\left(\pi\left(\frac{n}{p_{i}}\right)-i\right)-\sum_{p_{i} This expression can be reformulated more elegantly (as proposed by @StevenClark in this previous question I posted some time ago) as $$M(x)=1 - \sum\limits_{Gpf(k)\leq \frac{x}{k}} \mu(k)\left(\pi\left(\frac{x}{k}\right)- \pi(Gpf(k)\right)$$

In both expressions we have that $$\pi(x)$$ is the prime counting function. In the second expression we have that $$Gpf(x)$$ is the greatest prime factor of $$x$$.

I have tried to derive something insightful from the above equivalence to bound $$M(x)$$ sharply, but unsuccessfully. I even do not know if either $$\sum\limits_{Gpf(k)\leq \frac{x}{k}} \mu(k) \pi\left(\frac{x}{k}\right)$$ or $$\sum\limits_{Gpf(k)\leq \frac{x}{k}} \mu(k) \pi(Gpf(k))$$ can be bounded sharply using Möbius inversion or some other technique.

I would appreciate any answer addressing (i) the possibility to extract valuable information from the expression in order to bound $$M(x)$$, and / or (ii) showing techniques that could potentially lead to bound the whole main expression or any of the sub-expressions.

I include a graph plotting both $$\sum\limits_{Gpf(k)\leq \frac{x}{k}} \mu(k) \pi\left(\frac{x}{k}\right)$$ and $$\sum\limits_{Gpf(k)\leq \frac{x}{k}} \mu(k) \pi(Gpf(k))$$, and including $$\sqrt{x}$$ and $$-\sqrt{x}$$: