what is a good estimation of $\rho(x)$.
$\rho(x)$ is related to $\pi(x)$ the prime counting function?
$\omega(x)$ is the number of prime factors of $x$ and
$\sigma_0(x)$ is the number of factors of $x$
$$\rho(x)=\sum_{n=2}^{x} \frac{\omega(x)}{(\sigma_0(x)-1)^2}$$
$$|(2,3,5)|=p_{30},|(1,2,3,5,6,10,15,30)|-1=q_{30},\frac{p_{30}}{(q_{30}-1)^2}=\frac{3}{49}$$
I've heard of good approximations of $\pi(x)$ like $\pi(x)\sim\frac{x}{\ln(x)}$
I wonder how to find one for $\rho(x)$
First we investigate \begin{align*} \sum_{2\le n\le x} \frac{\omega(n)}{\sigma_0(n)^2} &= \sum_{2\le n\le x} \frac{1}{\sigma_0(n)^2} \sum_{p\mid n} 1 \\ &= \sum_{p\le x} \sum_{\substack{2\le n\le x \\ p\mid n}} \frac{1}{\sigma_0(n)^2} \\ &= \frac14 \sum_{p\le x} \sum_{m\le x/p} \frac{\sigma_0(p)^2}{\sigma_0(mp)^2}. \end{align*} The function $\frac{\sigma_0(p)^2}{\sigma_0(mp)^2}$ is nonnegative, multiplicative in $m$, and has value $\frac14$ on all primes except $p$ itself. General tools, such as the Wirsing–Odoni method, then give (I'm skipping many technical steps) \begin{align*} \sum_{2\le n\le x} \frac{\omega(n)}{\sigma_0(n)^2} &\sim \frac14 \prod_q \bigg( 1-\frac1q \bigg)^{1/4} \bigg( 1 + \frac1{4q} + \frac1{9q^2} + \cdots \bigg) \sum_{p\le x} \frac{x/p}{(\log(x/p))^{3/4}} \frac{1 + \frac1{9p} + \frac1{16p^2} + \cdots}{1 + \frac1{4p} + \frac1{9p^2} + \cdots} \\ &\sim \frac14 \prod_q \bigg( 1-\frac1q \bigg)^{1/4} \bigg( 1 + \frac1{4q} + \frac1{9q^2} + \cdots \bigg) \frac{x\log\log x}{(\log x)^{3/4}} \end{align*} (where the product is over all primes $q$). A similar argument gives \begin{align*} \sum_{2\le n\le x} \frac{\omega(n)}{\sigma_0(n)^3} &\sim \frac18 \prod_q \bigg( 1-\frac1q \bigg)^{1/8} \bigg( 1 + \frac1{8q} + \frac1{27q^2} + \cdots \bigg) \frac{x\log\log x}{(\log x)^{7/8}} = o \bigg( \frac{x\log\log x}{(\log x)^{3/4}} \bigg). \end{align*} Now the inequalities $\frac1{t^2} \le \frac1{(t-1)^2} \le \frac1{t^2}+\frac6{t^3}$, valid for $t\ge2$, yield finally \begin{align*} \sum_{2\le n\le x} \frac{\omega(n)}{(\sigma_0(n)-1)^2} &= \sum_{2\le n\le x} \frac{\omega(n)}{\sigma_0(n)^2} + o \bigg( \frac{x\log\log x}{(\log x)^{3/4}} \bigg) \\ &\sim \frac14 \prod_q \bigg( 1-\frac1q \bigg)^{1/4} \bigg( 1 + \frac1{4q} + \frac1{9q^2} + \cdots \bigg) \frac{x\log\log x}{(\log x)^{3/4}}. \end{align*}