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arXiv 2201.09287math.NT
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形式为$kf(k)$的数字

Numbers of the form $kf(k)$

Mikhail R. Gabdullin, Vitalii V. Iudelevich, Florian Luca

备注：
The error term in Theorem 1.2 is improved in this version of the paper
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摘要：

对于函数$f\colon \mathbb{N}\to\mathbb{N}$，定义$N^{\times}_{f}(x)=\#\{n\leq x: n=kf(k) \mbox{ for some $ k $} \}$。设$τ(n)=\sum_{d|n}1$为除数函数，$ω(n)=\sum_{p|n}1$为素除数函数，$φ(n)=\#\{1\leq k\leq n: (k,n)=1 \}$为欧拉函数。我们证明了\begin{gather*} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! 1) \quad N^{\times}_τ(x) \asymp \frac{x}{(\log x)^{1/2}}; \\ 2) \quad N^{\times}_ω(x) = (1+o(1))\frac{x}{\log\log x}; \\ \!\!\!\!\!\!\!\!\! 3) \quad N^{\times}_φ(x) = (c_0+o(1))x^{1/2}, \end{gather*}，其中$c_0=1.365...$ \，.

英文摘要：

For a function $f\colon \mathbb{N}\to\mathbb{N}$, define $N^{\times}_{f}(x)=\#\{n\leq x: n=kf(k) \mbox{ for some $k$} \}$. Let $τ(n)=\sum_{d|n}1$ be the divisor function, $ω(n)=\sum_{p|n}1$ be the prime divisor function, and $φ(n)=\#\{1\leq k\leq n: (k,n)=1 \}$ be Euler's totient function. We prove that \begin{gather*} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! 1) \quad N^{\times}_τ(x) \asymp \frac{x}{(\log x)^{1/2}}; \\ 2) \quad N^{\times}_ω(x) = (1+o(1))\frac{x}{\log\log x}; \\ \!\!\!\!\!\!\!\!\! 3) \quad N^{\times}_φ(x) = (c_0+o(1))x^{1/2}, \end{gather*} where $c_0=1.365...$\,.

机构、总结、中文摘要均由AI分析而来，可能存在错误，请以论文正文为准。

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