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On some problems of primes with the floor function
Number Theory · Submitted 18 Aug 2023
abstract compiled · 2026-09-23 · no language-model review
The authors' main claim In this paper, we obtain the asymptotic formula $$ \pi_{\theta}(x)=\frac{x^{\theta}}{(1-\theta) \log x}+O\left(x^{\theta}(\log x)^{-2}\right) $$ provide that $\frac{435}{923}<\theta<1$, and prove that $$ S_{\mathbb{P}}(x)=x\sum_{p} \frac{1}{p(p+1)}+O_{\varepsilon}\left(x^{435/923+\varepsilon}\right) $$ for $x \rightarrow \infty$.
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The abstract, sentence by sentence
Let $\left[x\right]$ be the largest integer not exceeding $x$. For $0<\theta \leq 1$, let $\pi_{\theta}(x)$ denote the number of integers $n$ with $1 \leq n \leq x^{\theta}$ such that $\left[\frac{x}{n}\right]$ is prime and $S_{\mathbb{P}}(x)$ denote the number of primes in the sequence $\left\{\left[\frac{x}{n}\right]\right\}_{n \geqslant 1}$. In this paper, we obtain the asymptotic formula $$ \pi_{\theta}(x)=\frac{x^{\theta}}{(1-\theta) \log x}+O\left(x^{\theta}(\log x)^{-2}\right) $$ provide that $\frac{435}{923}<\theta<1$, and prove that $$ S_{\mathbb{P}}(x)=x\sum_{p} \frac{1}{p(p+1)}+O_{\varepsilon}\left(x^{435/923+\varepsilon}\right) $$ for $x \rightarrow \infty$. Thus improve the previous result due to Ma, Wu and the author.
background contribution method result interpretation main claim
| # | sentence | role | the authors' own work | states the main conclusion | strength |
|---|---|---|---|---|---|
| s0 | Let $\left[x\right]$ be the largest integer not exceeding $x$. | background 1.00 | 0.05 | 0.01 | asserted 0.88 |
| s1 | For $0<\theta \leq 1$, let $\pi_{\theta}(x)$ denote the number of integers $n$ with $1 \leq n \leq x^{\theta}$ such that $\left[\frac{x}{n}\right]$ is prime and $S_{\mathbb{P}}(x)$ denote the number of primes in the sequence $\left\{\left[\frac{x}{n}\right]\right\}_{n \geqslant 1}$. | background 0.97 | 0.11 | 0.03 | asserted 0.51 |
| s2 | In this paper, we obtain the asymptotic formula $$ \pi_{\theta}(x)=\frac{x^{\theta}}{(1-\theta) \log x}+O\left(x^{\theta}(\log x)^{-2}\right) $$ provide that $\frac{435}{923}<\theta<1$, and prove that $$ S_{\mathbb{P}}(x)=x\sum_{p} \frac{1}{p(p+1)}+O_{\varepsilon}\left(x^{435/923+\varepsilon}\right) $$ for $x \rightarrow \infty$. | result 0.99 | 0.96 | 0.95 | asserted 1.00 |
| s3 | Thus improve the previous result due to Ma, Wu and the author. | interpretation 0.52 | 0.80 | 0.08 | asserted 0.98 |
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The paper
Original abstract
Let $\left[x\right]$ be the largest integer not exceeding $x$. For $0<\theta \leq 1$, let $\pi_{\theta}(x)$ denote the number of integers $n$ with $1 \leq n \leq x^{\theta}$ such that $\left[\frac{x}{n}\right]$ is prime and $S_{\mathbb{P}}(x)$ denote the number of primes in the sequence $\left\{\left[\frac{x}{n}\right]\right\}_{n \geqslant 1}$. In this paper, we obtain the asymptotic formula $$ \pi_{\theta}(x)=\frac{x^{\theta}}{(1-\theta) \log x}+O\left(x^{\theta}(\log x)^{-2}\right) $$ provide that $\frac{435}{923}<\theta<1$, and prove that $$ S_{\mathbb{P}}(x)=x\sum_{p} \frac{1}{p(p+1)}+O_{\varepsilon}\left(x^{435/923+\varepsilon}\right) $$ for $x \rightarrow \infty$. Thus improve the previous result due to Ma, Wu and the author.
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What it cites · 5 sentences without a confident label
Uses the method or tool of 1 paper
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We obtained this result by combine the methods in [ref] and [ref] .
On a variant of the prime number theorem · Introduction · 0.93
Cites as background 1 paper
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In 2021, Heyman [ref] proposed to study the number of primes or prime powers in the sequence ${[x/n]}_n ⩾1$ :
Primes in floor function sets · Introduction · 0.92
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