32

This problem was asked by Sierpinski in 1958 and answered by Hasse in the 1960s.
For each nonzero rational number $a$ (take $a \in \mathbf Z$ if you wish) and each prime $\ell$, let $S_{a,\ell}$ be the set of primes $p$ not dividing the numerator or denominator of $a$ such that $a \bmod p$ has multiplicative order divisible by $\ell$. When $a = \pm 1$, $S_{a,...

18

This isn't really a full answer, but it's too long for a comment, and perhaps it's informative all the same.
Your sum $S_k[\mathcal{O}]$ can be written as the value at $s = k$ of the sum
$$\sum_{0 \ne \lambda \in \mathcal{O}} \frac{\lambda^k}{Nm(\lambda)^s} = \sum_{n \ge 1} a^{(k)}_n n^{-s},$$
where $a^{(k)}_n := \sum_{N(\lambda) = n} \lambda^k$.
Now, I ...

12

The following explicit version of the Prime Number Theorem was proved by Trudgian:
$$ |\pi(x)-\mathrm{li}(x)|<x e^{-0.39\sqrt{\ln x}},\qquad x\geq 229.\tag{$\ast$}$$
In fact Trudgian's Theorem 2 is somewhat stronger than $(\ast)$, and with Mathematica it is straightforward to extend the validity of $(\ast)$ to $x\geq 2$.

7

Zeros are always counted with multiplicity, both in $N_\chi(\alpha,T)$ and in sums over zeros. This becomes clear when you look at how this quantity is estimated. Note also that the multiplicity of each zero $s$ of $L(s,\chi)$ is small, namely $O(\log q(2+|s|))$ by Jensen's formula.
For example, in Gallagher's paper the crucial step to look at is on p.336: &...

6

I will write here an approach which gives some interesting upper bounds on $p$ and $q$. The trivial lower bounds are $p \geq n$ and $q \geq p (\geq n)$. The idea shown here does not give an effective method for evaluating some lower bounds, so this is only a partial answer to your question. This approach is based on the following result and on some of its ...

6

Following some discussion with Harald (offline)... It appears that the answer is YES, the set is indeed infinite.
First ingredient:
We take the idea of adding upper densities, but use logarithmic densities instead of natural densities (logarithmic density meaning one sums reciprocals to $x$, divides by $\log{x}$, and sends $x$ to infinity). A short ...

answered Sep 23 at 21:59

so-called friend Don

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6

An integral representation:$$\sum (\zeta(n)^2-1)=\sum_{n\geqslant 2, (a,b)\ne (1,1)}\frac1{a^nb^n}=\sum_{(a,b)\ne (1,1)}\frac1{ab(ab-1)}\\=
\sum_{(a,b)\ne (1,1)}\int_0^1 x^{ab-2}(1-x)dx=\int_0^1 (1-x)\left(\sum_{b\geqslant 2}x^{b-2}+x^{-2}\sum_{a=2}^\infty \sum_{b=1}^\infty x^{ab}\right)\\=1+
\int_0^1 \sum_{a=2}^\infty(1-x)\frac{x^{a-2}}{1-x^a}dx=
1+
\int_0^...

reference-request analytic-number-theory sequences-and-series riemann-zeta-function binomial-coefficients

5

If for every $q>1 $ the function $b(n)^q$ has average $O_q(1)$ then by H"{o}lder one can get $$ \sum_{n<x} \lambda^{\omega(n)} b(n) \ll_\epsilon x (\log x)^{\lambda-1+\epsilon}$$ for every fixed $\epsilon >0$. Apart from the $(\log x)^\epsilon$ term, this is close to the best one can hope in general.
If you know the stronger (compared to the ...

4

The Dirichlet series of the indicator function of $k$-free numbers is $\zeta(s)/\zeta(ks)$. Hence any exponent less than $1/k$ in the error term implies a quasi-Riemann Hypothesis. More precisely, if the number of $k$-free numbers is $x/\zeta(k)+O(x^c)$, then $s=1$ is the only pole of $\zeta(s)/\zeta(ks)$ in the half-plane $\Re(s)>c$, whence all the zeros ...

4

There are now at least two instances of such an explicit result on the arxiv.
Theorem 1.1 of Bordignon: https://arxiv.org/abs/2001.05114
Theorem 1.1 (and Corollary 1.2) of Jain-Sharma, Khale, and Liu: https://arxiv.org/abs/2010.09530.
Bordignon's result applies to convoluted Dirichlet characters; however, applying the result with $\psi$ identically $1$ (...

4

(Just a quick aside: the set of solutions to the equation $\zeta(s) = a$ when $a \neq 0$ are usually called the $a$-points of the Riemann zeta function in the literature, in case you want to look up the state of the art)
As Steve Huntsman mentions in a comment, it is indeed possible to use the universality of the zeta function to prove that $\zeta(s)$ is ...

2

You can try to follow the standard theta argument for general $k$, to get a series expansion, except not one that relates the function to itself.
Consider
$$G_{2k}(z) = \sum_{n=1}^{\infty} e^{-n^{2k} z}$$
Take the Mellin transform of both sides
$$\int_0^\infty G_{2k}(z)z^{s-1}dz = \Gamma(s) \zeta(2ks)$$
Replace $z$ with $1/z$, and $s$ with $-s$
$$\int_0^\...

2

Note that this is a correct answer to the original question, so I will leave it here, even though the question has now been changed. (The original question is recoverable by going back to the previous versions.)
In fact, every complex manifold has such an atlas.
Let $(M,J)$ be a (finite-dimensional) complex $n$-manifold and let $\mathscr{U}$ be an open cover ...

ag.algebraic-geometry nt.number-theory dg.differential-geometry complex-geometry analytic-number-theory

2

Q3: The first identity (1) is not correct, it should read
\begin{align} \sum_{n=2}^{\infty}(\zeta(n)^{2} -1) &= \frac{7}{4} - \zeta(2) + 2\sum_{m=2}^{\infty} \frac{H_{m-1-\frac{1}{m}} - H_{-\frac{1}{m}} - H_{m-1}}{m} . \end{align}
The final identity (*) then becomes
$$
\sum_{m=2}^{\infty} \frac{H_{-\frac{1}{m}}}{m} = \frac{3}{4} - \zeta(2) + 2 \sum_{m=2}^...

answered Oct 9 at 11:57

Carlo Beenakker

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1

I don't know if the following willy full answer your question, since I'm not sure if you attach a huge importance to have integral coefficients/primitive solutions. Apologies if it doesn't.
Let $Q$ be any nondegenerate (I assume this is the case of interest) ternary quadratic form such that $Q(\mathbb{Q})\neq 0$.
Then the theory of quadratic forms says that $...

1

1. We have
$$\sum_{\substack{dd'\leq X\\q\mid d+d'\\d\leq d'}}1
=\sum_{d\leq\sqrt{X}}\sum_{\substack{d'\leq X/d\\q\mid d+d'\\d\leq d'}}1
\leq\sum_{d\leq\sqrt{X}}\sum_{\substack{c\leq 2X/d\\q\mid c}}1
\leq\sum_{d\leq\sqrt{X}}\frac{2X}{qd}<\frac{2X(1+\log\sqrt{X})}{q}.$$
We get the same bound when the roles of $d$ and $d'$ are interchanged,
hence in the end
...

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