{"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>\\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}
引用次数: 0
Abstract
The Thue–Morse sequence $\{t(n)\}_{n\geqslant 0}$ is the indicator function of the parity of the number of ones in the binary expansion of nonnegative integers n, where $t(n)=1$ (resp. $=0$) 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 $\beta>\sqrt {\varphi }=1.272019\ldots $, the set of the numbers $$\begin{align*}1,\quad \sum_{n\geqslant1}\frac{t(n)}{\beta^{n}},\quad \sum_{n\geqslant1}\frac{t(n^2)}{\beta^{n}},\quad \dots, \quad \sum_{n\geqslant1}\frac{t(n^k)}{\beta^{n}},\quad \dots \end{align*}$$is linearly independent over the field $\mathbb {Q}(\beta )$, where $\varphi :=(1+\sqrt {5})/2$ is the golden ratio. Our result yields that for any integer $k\geqslant 1$ and for any $a_1,a_2,\ldots ,a_k\in \mathbb {Q}(\beta )$, not all zero, the sequence {$a_1t(n)+a_2t(n^2)+\cdots +a_kt(n^k)\}_{n\geqslant 1}$ cannot be eventually periodic.