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This is fairly classical, and the answer is zero for any $n>1$, and $1$ for $n=1$. The reason basically comes down to the fact that asymptotic-wise, almost all prime powers are prime. Here is the prop …

Let $\theta(x)=\sum_{p\leq x}\log p$ be first Chebyshev function. Then we have $A(n)=e^{\theta(n)}$, and $$B(n)=\prod_k A(n^{1/k})^k=e^{\sum_k k\theta(n^{1/k})}.$$ One can easily show that $\sum_k k\t …

$\newcommand{\li}{\operatorname{li}}$Yes, those values of $x$ are unbounded. As JoshuaZ indicates in a comment, the key here is Littlewood's result on sign changes, and because of the way $\pi(x)$ gro …

It is standard that all prime factors of Fermat number $F_m$ are of the form $2^{m+2}k+1$, in particular they are all at least $2^m$. It is then also clear that $F_m$ can only be divisible by at most …

I would expect there are only finitely many such $n$. A numerical search suggests the complete set of such integers is $\{1,2,3,4,5,7,16,17,19,23\}$ - there are no others below 10000. Here is an argum …

Assuming the strengthened version of Hardy-Littlewood conjecture I discuss here (which follows from Dickson's conjecture), the following much stronger result holds: let $a_0,\dots,a_m$ be any sequence …

The answer is no. A set $S$ to be piecewise syndetic iff there is an integer $d$ such that there exist intervals $I$ of arbitrary length such that distances between elements of $S\cap I$ are bounded b …

I figured the remarks I gave in the comments deserve to be gathered up into a more coherent form as an answer. One thing I will start with is that comparing $L(n)$ (or $M(n)$, the Mertens function) to …

Bratus and Pak have devised an algorithm which, given a black box (or "gray box") group $G$ which is known to be isomorphic to $S_n$ or $A_n$, actually explicitly finds such an isomorphism. The algori …

The asymptotic $\sigma(n)\ll n\omega(n)$ is true and has nothing to do with RH or Robin's criterion. Let us rewrite it as $\frac{\sigma(n)}{n}\ll\omega(n)$. By writing $n=\prod_{i=1}^k p_i^{e_i}$, whe …

There is a completely elementary way to see that the answer to a) is negative - if $f$ had only finitely many Fourier coefficients, it would extend to an entire function on $\mathbb C$. But for any $\ …

Assuming the abc conjecture, there are only finitely many solutions with $r\geq 5$. Indeed, more generally, abc conjecture implies there are only finitely many sums of the form $a+b=c$ in which $a,b,c …

Summary: I consider the limit $\lim_{n\to\infty}\prod_{i=1}^n i^{\chi(i)}$ (let me drop the $\mod s\mathbb Z$ for brevity). If we restrict to $n$ divisible by $s$, then the limit will always be equal …

Suppose $k^4+4=L_{k-1}L_{k+1}$ has fewer than four prime divisors. Since the factors are relatively prime, this implies that one of them is a prime power. But since $L_{k\pm 1}$ is one larger than a s …

It doesn't seem to me that either article follows the line of reasoning as you have presented it. Indeed, we do not take the integral over the union of integrals $(x_n,(1+\varepsilon)x_n)$. We do get …