On the multiplicative group generated by \Big\{{[\sqrt {2}n]\over n}~\mid~n\in\mathbb{N} \Big\}. V

IF 0.4 Q4 MATHEMATICS
I. Kátai, B. M. Phong
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引用次数: 0

Abstract

Let $f,g$ be completely multiplicative functions, $\vert f(n)\vert=\vert g(n)\vert =1 (n\in\mathbb{N})$. Assume that $${1\over {\log x}}\sum_{n\le x}{\vert g([\sqrt{2}n])-Cf(n)\vert\over n}\to 0 \quad (x\to\infty).$$ Then $$f(n)=g(n)=n^{i\tau},\quad C=(\sqrt{2})^{i\tau}, \tau\in \mathbb{R}.$$
在由\Big {{[\sqrt 2n{] }\over n }\mid n \in\mathbb{N}\Big}生成的乘法组上。v
设$f,g$是完全乘法函数$\vert f(n)\vert=\vert g(n)\vert =1 (n\in\mathbb{N})$。假设$${1\over {\log x}}\sum_{n\le x}{\vert g([\sqrt{2}n])-Cf(n)\vert\over n}\to 0 \quad (x\to\infty).$$那么 $$f(n)=g(n)=n^{i\tau},\quad C=(\sqrt{2})^{i\tau}, \tau\in \mathbb{R}.$$
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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