## 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?