Search Results

Here is a more explicit version of Noam Elkies's comment. Theorem. Let $p>2$ be a prime number. Let $\mathcal{U},\mathcal{V}\subseteq\{1,2,\dots,p-1\}$ be two intervals, and let $r\in\{1,2,\dots,p-1\ …

I think the standard approach to this problem is to estimate the sum $$ S(x) := \sum_{m=1}^x\prod_{k=1}^n\left(1-\left(\frac{m}{p_k}\right)\right), $$ where $p_k$ is the $k$-th odd prime. Indeed, we a …

For a given $n$, let $P\subset\{2,3\dots,n\}$ be the set of primes up to $n$, and let $S\subset\{2,3\dots,n\}$ be any subset with pairwise coprime elements. Consider the function $f:S\to P$ that assig …

This answer is an elaboration of Lucia's comment, all mistakes are mine. Consider the primes $7\leq p\leq\sqrt{n}$ with $p\equiv 3\pmod{4}$. For each such prime $p$, the number of solutions of the co …

I deleted my original answer, because I realized (thanks to user alpoge, see the comments below the original post) that I have already answered the same question earlier, with a much better analysis. …

This is certainly possible. I will construct an example with $\alpha=\sqrt{N}$. That is, I will exhibit a square number $N$ with more than $(\log N)^4$ divisors lying in $[\tfrac{1}{2}\sqrt{N},\sqrt{N …

The answer is probably no. Assume that $u_{k+1}<u_k$ for all $k$. Then $$ \frac{1}{p_{k+1}}<\frac{\log\log N_{k+1}}{\log\log N_k}-1=\frac{\log\frac{\log N_{k+1}}{\log N_k}}{\log\log N_k}<\frac{\frac{\ …

The affirmative answer to your question is equivalent to the Riemann Hypothesis, and this was observed by Solé and Planat as a consequence of Nicolas's earlier work. I recall their argument briefly. …

Here is a counterexample assuming the Riemann Hypothesis. Let $a(n):=\Lambda(n)-1$, where $\Lambda$ is the von Mangoldt function. Then $A(x)=\psi(x)-[x]$, where $\psi$ is the Chebyshev function, and $ …

It is easy to see that $\theta_s\leq 1/2$, hence $EH(\theta_{s})$ holds by the Bombieri-Vinogradov theorem. To see the claim $\theta_s\leq 1/2$, assume that $1/2<\theta<1$, and let $1\leq q\leq x^\th …

Estimating this sum is much the same as estimating the error term in the prime number theorem for arithmetic progressions. See Exercises 7-8 in Section 11.3 of Montgomery-Vaughan: Multiplicative numbe …

The last display in your original post is used in (2.5) of Selberg: An elementary proof of the prime-number theorem, Annals of Math. 50 (1949), 305-313. It is a simple consequence of Chebyshev's estim …

The answer is probably no. Your inequality implies $$ \frac{\log p_{k+1}}{\log N_k} > \log\left(1+\frac{\log p_{k+1}}{\log N_k}\right)>\frac{\log\log N_k}{p_{k+1}-1}>\frac{\log\log N_k}{p_{k+1}}.$$ In …

The Chebyshev function $\psi(x)$ is constant zero for $0<x<1$. It does not get more explicit than that.

The proof of Theorem 1.2 relies on Propositions 3.4 and 3.5. In particular, if $h$ is sufficiently large but fixed, the bound you quote is true for a positive proportion of $x\in\mathbb{N}\cap[X,2X]$. …