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No. Littlewood proved that $\theta(x) > x + c \sqrt{x} \log \log \log x$ holds for infinitely many integers $x$, for some $c > 0$. Cf the answer to this Mathoverflow question, noting that $\theta(x) = …

Your guess is correct. Since $2^{\omega(n)} = \sum_{d | n} \mu^2(d)$ one has $$ S = \sum_{(a,b,c) \in R(X)} 2^{\omega(4ac-b^2)} = \sum_{d \leq X} \mu^2(d) |R_d(X)| $$ where $R_d(X)$ is the set of inte …

One has $S_a(x) \sim C(a) x (\log x)^{2^a -1}$ where $$ C(a) = \Gamma(2^a)^{-1} \prod_p \left( 1 - \frac{1}{p} \right)^{2^a} \left( \sum_{k \geq 0} \frac{(k+1)^a}{p^k}\right). $$ This follows for exam …

Using the decomposition $$ H(x) := \sum_{a \leq x} \frac{1}{a} = \log(x) + \gamma - \frac{\psi(x)}{x} + \int_{x}^{+\infty} \frac{\psi(t) d t}{t^2}, $$ where $\psi(t) = \{ t \} - \frac{1}{2}$, one ge …

You can simplify Ramachandra's method by bounding the last sum of p.305 using Brun-Titchmarsh inequality (or Montgomery & Vaughan error-term free version of it) instead of Van der Corput's method + Se …

If you only need an upper bound of the right order of magnitude, I strongly recommend the use of Rankin's trick, which amounts to write $$ \sum_{r \leq R} f(r) \leq \prod_{p \leq R} \left( \sum_{\ell …

The answer is no, even if we assume there are no other poles than $1$ in $\sigma > 1- \epsilon_0$. I give an example below with $\epsilon_0=1$. This is a variant of an example given by Karamata in $19 …

As written above by Seva, one is led with exponential sums of the form $$ \sum_{d|\ell} \mu(d) e^{\frac{2 i \pi y}{d}} $$ where $y = hx$ is an integer multiple of $x$. If one wants to reduce the trivi …

Slightly off-topic : this article from Arxiv (in french) shows that zeros of functions f : s -> h(s) - h(1-s), where h is a meromorphic function satisfying appropriate growth conditions, are inclined …

I detail. It suffices to study the "tail" $P(X) = \prod_{p > X} (1- \frac{1}{p^s})^{-1} $. Using $-\log(1-y) = y + O(y^2)$, we get for real $s>1$ $$ \log P(X) = \sum_{p > X} \frac{1}{p^s} + O \left( …

I will answer a question slightly different from yours. Let $a(n)$ be the smallest integer $\geq n$ such that $a(n) = b c$ with $(b,c) = 1$ and $b \geq n^r$, $c \geq n^{1-r}$. Then : $a(n) - n \geq …

First note that $\prod_{\xi} (1 - \xi^n q^n)$ is equal to $(1- q^n)^5$ if $5 | n$ and to $1 - q^{5n}$ otherwise. Moreover $$ \varphi(q) = \prod_{n \geq 1} (1 - q^n)^{e_n} $$ where $e_n = 1,-2,3,$ or $ …

Two very nice answers have already been given, but I would like to add that Michel Balazard has a book in preparation on this topic (written for undergrads), in which he tries to give a deduction of t …