Recall that Merten's function is defined as:
$$M(x) = \sum_{n\le x}\mu(n)$$
Using the following prime counting functions to represent the count of integers less than $x$ with $k$ prime divisors:
$$\pi_k(x) \sim \frac{x}{\log(x)}\cdot \frac{(\log(\log(x))^{k-1}}{(k-1)!}$$
I obtained:
$$M(x) \sim \sum_{k=1}^\infty (-1)^k \ \frac{x\cdot(\log(\log(x))^{k-1}}{\log(x)\cdot(k-1)!} - \left\lfloor\left(\frac{x}{\log^2(x)}\right)^\frac{1}{k}\right\rfloor$$
When I calculate the value of this equation and compare it to Merten's function it seems to be accurate, however, the value within the floor braces is off by some factor of x. This part of the equation is meant to correct for values where the mobius function is equal to 0 (when n contains a squared prime factor). I can't seem to figure out what I am doing wrong and any corrections would be greatly appreciated.
According to the Riemann Hypothesis, this equation should exist within $O(x^{\frac{1}{2}+\epsilon})$ if that helps.
Thanks!
EDIT:
I worked out a new bound using ideas from Sieve Theory as follows:
$$M(x)\lt\lt -\pi(x)\cdot\left(1+\sum_{n=1}^{O\left(\frac{\log(x)}{\log(\log(x))}\right)}(-1)^n\cdot\prod_{k=1}^n{\left(\pi\left(\frac{x}{k\#}\right)-\pi(k\#)\right)}\right) \\\sim -\pi(x)\cdot\left(1+\sum_{n=1}^{O\left(\frac{\log(x)}{\log(\log(x))}\right)}(-1)^n\cdot\prod_{k=1}^n{\left(\frac{x\theta(k)-(k\#)^2(\log(x)-\theta(k))}{\theta(k)(k\#)(\log(x)-\theta(k))}\right)}\right)$$
Where $\theta$ is the first Chebyshev function defined by:
$$\theta(n)=\sum_{p\le x}\log(p)$$
Not sure what to do with this yet, but it is certainly more accurate than my previous bound. Hopefully I can get $\mu$ back involved in the relation. Any ideas are welcome.