Asymptotic estimation of $A_n$

Let $ A_n$ represent the number of integers that can be written as the product of two element of $ [[1,n]]$ .

I am looking for an asymptotic estimation of $ A_n$ .

First, I think it’s a good start to look at the exponent $ \alpha$ such that :

$ $ A_n = o(n^\alpha)$ $

I think we have : $ 2 < alpha $ . To prove this lower bound we use the fact that the number of primer numbers $ \leq n$ is about $ \frac{n}{\log n}$ . Hence we have the trivial lower bound (assuming $ n$ is big enough) :

$ $ \frac{n}{\log n} \cdot \binom{ E(\frac{n}{\log n})}{2} = o(n^3)$ $

Now is it possible to get a good asymptotic for $ A_n$ and not just this lower bound ? Is what I’ve done so far correct ?

Thank you !

Asymptotic formula for number of square-free numbers congruent to $1$ modulo $p$


Knowing that $ \sum_{n\leq x}\mu^2(n)=\frac{6}{\pi^2}x+O(\sqrt{x})$ , prove that:

$ $ \sum_{n\leq x,\,\,n\equiv 1(\text{mod }p)}\mu^2(n)=\frac{6}{\pi^2(p-1)}x+O(\sqrt{x})$ $

Using Dirichlet charaters, we have: \begin{align*} \sum_{n\leq x,\,\,n\equiv 1(\text{mod }p)}\mu^2(n)&=\frac{1}{\varphi(p)}\sum_{n\leq x}\sum_{\chi(\text{mod }p)}\mu^2(n)\chi(n)\ &=\frac{1}{\varphi(p)}\sum_{\chi(\text{mod }p)}\sum_{n\leq x}\mu^2(n)\chi(n)\ &=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)\chi_0(n)+\underbrace{\frac{1}{\varphi(p)}\sum_{\chi\neq \chi_0}\sum_{n\leq x}\mu^2(n)\chi(n)}_{=:\Delta(x)} \end{align*} Where $ \chi_0$ is the principal character. I was able to prove that $ \Delta(x)=O(\sqrt{x})$ basically by using the fact that $ \mu^2(n)=\sum_{d|n}\mu(d)$ , and that $ \left|\sum_{n\leq x}\chi(n)\right|\leq \varphi(p)$ . My problem is the main term, which doesn’t match my calculation.

By definition, $ \chi_0(n)=1$ if $ p\not| n$ and $ \chi_0(n)=0$ if $ p|n$ . Therefore: \begin{align*} \frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)\chi_0(n)&=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)-\frac{1}{\varphi(p)}\sum_{n\leq x,\,\,p|n}\mu^2(n)\ &=\frac{1}{\varphi(p)}\sum_{n\leq x}\mu^2(n)-\frac{1}{\varphi(p)}\sum_{n\leq \frac{x}{p}}\mu^2(n)\ &=\frac{1}{\varphi(p)}\left(\frac{6}{\pi^2}x-\frac{6}{\pi^2}\frac{x}{p}\right)+O(\sqrt{x})\ &=\frac{6x}{\pi^2\varphi(p)}\left(1-\frac{1}{p}\right)+O(\sqrt{x})\ &=\frac{6}{\pi^2p}x+O(\sqrt{x}) \end{align*} What am I missing?