Linear independence of series related to the Thue–Morse sequence along powers

Michael Coons, Yohei Tachiya
{"title":"Linear independence of series related to the Thue–Morse sequence along powers","authors":"Michael Coons, Yohei Tachiya","doi":"10.4153/s0008439524000195","DOIUrl":null,"url":null,"abstract":"<p>The Thue–Morse sequence <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline1.png\"><span data-mathjax-type=\"texmath\"><span>$\\{t(n)\\}_{n\\geqslant 0}$</span></span></img></span></span> is the indicator function of the parity of the number of ones in the binary expansion of nonnegative integers <span>n</span>, where <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline2.png\"><span data-mathjax-type=\"texmath\"><span>$t(n)=1$</span></span></img></span></span> (resp. <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline3.png\"><span data-mathjax-type=\"texmath\"><span>$=0$</span></span></img></span></span>) if the binary expansion of <span>n</span> has an odd (resp. even) number of ones. In this paper, we generalize a recent result of E. Miyanohara by showing that, for a fixed Pisot or Salem number <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline4.png\"><span data-mathjax-type=\"texmath\"><span>$\\beta&gt;\\sqrt {\\varphi }=1.272019\\ldots $</span></span></img></span></span>, the set of the numbers <span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_eqnu1.png\"><span data-mathjax-type=\"texmath\"><span>$$\\begin{align*}1,\\quad \\sum_{n\\geqslant1}\\frac{t(n)}{\\beta^{n}},\\quad \\sum_{n\\geqslant1}\\frac{t(n^2)}{\\beta^{n}},\\quad \\dots, \\quad \\sum_{n\\geqslant1}\\frac{t(n^k)}{\\beta^{n}},\\quad \\dots \\end{align*}$$</span></span></img></span>is linearly independent over the field <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline5.png\"><span data-mathjax-type=\"texmath\"><span>$\\mathbb {Q}(\\beta )$</span></span></img></span></span>, where <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline6.png\"><span data-mathjax-type=\"texmath\"><span>$\\varphi :=(1+\\sqrt {5})/2$</span></span></img></span></span> is the golden ratio. Our result yields that for any integer <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline7.png\"><span data-mathjax-type=\"texmath\"><span>$k\\geqslant 1$</span></span></img></span></span> and for any <span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline8.png\"><span data-mathjax-type=\"texmath\"><span>$a_1,a_2,\\ldots ,a_k\\in \\mathbb {Q}(\\beta )$</span></span></img></span></span>, not all zero, the sequence {<span><span><img data-mimesubtype=\"png\" data-type=\"\" src=\"https://static.cambridge.org/binary/version/id/urn:cambridge.org:id:binary:20240321105855809-0602:S0008439524000195:S0008439524000195_inline9.png\"/><span data-mathjax-type=\"texmath\"><span>$a_1t(n)+a_2t(n^2)+\\cdots +a_kt(n^k)\\}_{n\\geqslant 1}$</span></span></span></span> cannot be eventually periodic.</p>","PeriodicalId":501184,"journal":{"name":"Canadian Mathematical Bulletin","volume":"156 1","pages":""},"PeriodicalIF":0.0000,"publicationDate":"2024-03-06","publicationTypes":"Journal Article","fieldsOfStudy":null,"isOpenAccess":false,"openAccessPdf":"","citationCount":"0","resultStr":null,"platform":"Semanticscholar","paperid":null,"PeriodicalName":"Canadian Mathematical Bulletin","FirstCategoryId":"1085","ListUrlMain":"https://doi.org/10.4153/s0008439524000195","RegionNum":0,"RegionCategory":null,"ArticlePicture":[],"TitleCN":null,"AbstractTextCN":null,"PMCID":null,"EPubDate":"","PubModel":"","JCR":"","JCRName":"","Score":null,"Total":0}
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Abstract

The Thue–Morse sequence Abstract Image$\{t(n)\}_{n\geqslant 0}$ is the indicator function of the parity of the number of ones in the binary expansion of nonnegative integers n, where Abstract Image$t(n)=1$ (resp. Abstract Image$=0$) if the binary expansion of n has an odd (resp. even) number of ones. In this paper, we generalize a recent result of E. Miyanohara by showing that, for a fixed Pisot or Salem number Abstract Image$\beta>\sqrt {\varphi }=1.272019\ldots $, the set of the numbers Abstract Image$$\begin{align*}1,\quad \sum_{n\geqslant1}\frac{t(n)}{\beta^{n}},\quad \sum_{n\geqslant1}\frac{t(n^2)}{\beta^{n}},\quad \dots, \quad \sum_{n\geqslant1}\frac{t(n^k)}{\beta^{n}},\quad \dots \end{align*}$$is linearly independent over the field Abstract Image$\mathbb {Q}(\beta )$, where Abstract Image$\varphi :=(1+\sqrt {5})/2$ is the golden ratio. Our result yields that for any integer Abstract Image$k\geqslant 1$ and for any Abstract Image$a_1,a_2,\ldots ,a_k\in \mathbb {Q}(\beta )$, not all zero, the sequence {Abstract Image$a_1t(n)+a_2t(n^2)+\cdots +a_kt(n^k)\}_{n\geqslant 1}$ cannot be eventually periodic.

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与 Thue-Morse 序列相关的序列沿幂级数的线性独立性
Thue-Morse 序列 $\{t(n)\}_{n\geqslant 0}$ 是非负整数 n 的二进制展开中 1 的个数奇偶性的指示函数,其中如果 n 的二进制展开中 1 的个数为奇数(或偶数),则 $t(n)=1$(或 $=0$)。在本文中,我们推广了宫之原(E. Miyanohara)最近的一个结果,证明对于一个固定的皮索特(Pisot)或萨利姆(Salem)数 $\beta>\sqrt {\varphi }=1.272019\ldots $, the set of the numbers $$\begin{align*}1,\quad \sum_{n\geqslant1}\frac{t(n)}{beta^{n}},\quad \sum_{n\geqslant1}\frac{t(n^2)}{beta^{n}}、\quad \dots, \quad \sum_{n\geqslant1}\frac{t(n^k)}{beta^{n}}, \quad \dots \end{align*}$$ 是线性独立于域 $\mathbb {Q}(\beta )$, 其中 $\varphi :=(1+\sqrt {5})/2$ 是黄金分割率。我们的结果表明,对于任意整数 $k\geqslant 1$,并且对于 \mathbb {Q}(\beta )$ 中的任意 $a_1,a_2,\ldots ,a_k\ 都不为零,序列 {$a_1t(n)+a_2t(n^2)+\cdots +a_kt(n^k)\}_{n\geqslant 1}$ 最终不可能是周期性的。
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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