# Tag Info

### abscissa of convergence of $\sum_{n = 1}^\infty \frac{\mu(n)}{n^s}$

I will outline how those properties can be related to each other. It is through an intermediate integral involving $M(x) = \sum_{1 \leq n \leq x} \mu(n)$. For a sequence $\{a_n\}$ in $\mathbf C$, ...
1 vote
Accepted

### Unable to Understand Notations in Davenport's MNT

In the formula $$2\pi N(T)=\Delta_R {\rm arg} \xi(s)$$ the term on the RHS stands for the increment of the argument of $\xi(s)$ along the boundary of the rectangle $R$, i.e., the change in the ...

### Call $n\in\Bbb N$ "balanced" if the sum of its digits equals the count of its divisors. How many "balanced" numbers are there up to $m$?

Upto $10^9$ , we have the following values and counts : ...

### How is the logarithm of an integer analogous to the degree of a polynomial?

Here's a concrete way in which they're analogous: $\mathrm{deg}\,(f \times g) = \mathrm{deg}\,f + \mathrm{deg}\,g$ $\log |xy| = \log |x| + \log |y|$ Also: the degree of the constant zero polynomial ...
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### Probability for finding product?

By "first billion natural numbers", I'm going to assume you mean all the numbers from $1$-$1,000,000,000$ (so 0 is not included). The trick here is to look at the numbers in terms of ...
1 vote
Accepted

### What is $17$ adic value of $3+2\sqrt{-2}\in \Bbb{Q}_{17}$?

After knowing $\sqrt{-2}=7+a_1p+\cdots\ \ (p=17)$, we see $$\nu_{17}(3+2\sqrt{-2})=\nu_{17}(17+2a_1p+\cdots)\geqq 1$$ and $$\nu_{17}(3-2\sqrt{-2})=\nu_{17}(-11-2a_1p-\cdots)=0,$$ but these two numbers ...

### For every matrix $A$ mod $r$, is there a nonzero vector $\vec x$ so that $A \vec x$ has 0 mod $r$ nonzero entries?

So with the author's comment in OP I get to understand that it intends to prove that for any given $A\in\mathbb Z_r^{m\times n}$ with $n\ge 2$, either there is some $y\in\mathbb Z_r^n$ so that Ay=0 or ...
1 vote
Accepted

### Call $n\in\Bbb N$ "balanced" if the sum of its digits equals the count of its divisors. How many "balanced" numbers are there up to $m$?

The number of divisors of some positive integer can only be determined by factoring it. But the count is routine for , say , $m\le 10^{10}$ I do not think that we can prove that there are infinite ...
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### Alternating Dirichlet series involving the Möbius function.

It is indeed true. The function $(-1)^{n+1}$ is a multiplicative function (really! confirm this), and therefore the Dirichlet series can be expanded into an Euler product as usual (in its half-plane ...
1 vote

### How to prove the p-adic expansion of negative p-adic integers ends with p-1?

(Edit: Replaced earlier answer with different approach.) Your expansion of $-2$ is incorrect. I would think if you see the expansions of negative numbers, you'll see what is happening. Let's take $p=5$...
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### What is the decomposition of global units $1+\mathfrak{p}$?

EDIT: As clarified by the OP, the intended question is the decomposition of $U_{K, \mathfrak p} = \{x \in \mathcal O_K^\times: x \equiv 1 \pmod{\mathfrak p}\}$ instead of $1 + \mathfrak p$. So we ...

### the generators of the modular group are S and T

As I see, the problem is the induction step. Consider this Hint: $X \in G \implies T^{a}S^{b}X \in G$ Solution: $$ST^mA \in G \implies T^{-m}S^{-1}(ST^mA) \in G \implies A \in G$$
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### Is it true that an irreducible polynomial is only divisible by a fixed set of primes finitely many times?

Let $f \in \mathbb{Z}[x]$ be nonzero with at least two distinct complex roots (that is, $f$ is not of the form $a(x-b)^n$). Let's prove that the statement holds for $f$. Let $S$ be a finite set of ...

### Are there infinitely many $c$-th perfect powers having a constant congruence speed of $c$?

Ok, I have just found a way to prove (a generalization) of the present statement. I am planning to write a preprint in a few of weeks, then I will be happy to share it here. Basically, it is just a ...
Accepted

1 vote

### Does Euler product formula give any hints about asymptotics of primes?

Intuitively, the limit $s \to \infty$ shouldn't be expected to give asymptotic information, because the contributions from the tail of the series $\sum_{n} n^{-s}$ and $\prod_{p} \frac{1}{1 - p^{-s}}$ ...
We can use exponential generating functions. To preface, we want to assign, in some form or another, a value $c_i$ to each $1 \leq i \leq N$ such that $c_i = S_j - 1$ for some $j$ and \$\sum_{i = 1}^N ...