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As Ofir says this shouldn't be a problem to prove directly using Perron. But let me describe another way to get the same result: we shall use the well-known result of Davenport that for any $A>0$ one …

We use the bound given by my previous answer: $$\sum_{n<x}\mu(kn)\ll_A \frac{kx}{(\log (kx))^A}.$$ Now we improve this bound by using that the sum $$\sum_{d \in \mathbb N\atop d\mid n, d\mid k}\mu(d)$ …

The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a …

I am assuming that explicit refers to the error term? In this case you can write $$Q_A(x)=\sum_{d\mid A}\mu(d)\sum_{k\leq \sqrt x \atop \gcd(A,k)=1}\mu(k)\left[\frac{x}{dk^2}\right],$$ where $[t]$ den …

One of Jacobi's theorems states that the number of representations of a positive integer $n$ as a sum of two squares of integers equals $$4(d_1(n)-d_3(n)),$$ where the function $d_i$ counts the number …