论文标题
$ kf（k）$的形式的数字
Numbers of the form $kf(k)$
论文作者
论文摘要
对于功能，$ f \ colon \ mathbb {n} \ to \ mathbb {n} $，定义$ n^{\ times} _ {f} _ {f}（x）= \＃\ {n \ leq x：n = kf（kf（kf（kf（kf（kf））令$τ（n）= \ sum_ {d | n} 1 $为除数函数，$ω（n）= \ sum_ {p | n} 1 $是素数divisor函数，$φ（n）= \＃\ {1 \ leq k \ leq k \ leq n：（k，n）= 1 \ n）= 1 \ \ \ \ e（k，n）= 1 \ \ \ be euler'soffiction。我们证明了这一点 \ begin {chater*} \！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！\！ 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 \ log x}; \\ \！\！\！\！\！\！\！\！\！\！\！ 3）\ quad n^{\ times}_φ（x）=（c_0+o（1））x^{1/2}，\ end {change*}其中$ c_0 = 1.365 ... $ \,。
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 $c_0=1.365...$\,.
