\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Transcendence of Digital Expansions \\ \vskip .02in Generated by a Generalized Thue-Morse \\ \vskip .10in Sequence } \vskip 1cm \large Eiji Miyanohara \\ Graduate School of Fundamental Science and Engineering\\ Waseda University\\ 3-4-1 Okubo, Shinjuku\\ Tokyo 169-8555 \\ Japan\\ \href{mailto:j1o9t5acrmo@fuji.waseda.jp}{\tt j1o9t5acrmo@fuji.waseda.jp} \end{center} \vskip .2 in \begin{abstract} In this article, first we generalize the Thue-Morse sequence by means of a cyclic permutation and the $k$-adic expansion of non-negative integers, giving a sequence $(a(n))_{n=0}^\infty$, and consider the condition that $(a(n))_{n=0}^\infty$ is non-periodic. Next, we show that, if a generalized Thue-Morse sequence $(a(n))_{n=0}^\infty$ is not periodic, then no subsequence of the form $(a(N+nl))_{n=0}^\infty$ (where $N \ge 0$ and $l>0$) is periodic. We apply the combinatorial transcendence criterion established by Adamczewski, Bugeaud, Luca, and Bugeaud to find that, for a non-periodic generalized Thue-Morse sequence taking its values in $\{0,1,\ldots,\beta-1\}$ (where $\beta$ is an integer greater than $1$), the series $\sum_{n=0}^\infty a(N+nl) {\beta}^{-n-1}$ gives a transcendental number. Furthermore, for non-periodic generalized Thue-Morse sequences taking positive integer values, the continued fraction $[0, a(N), a(N+l),\ldots,a(N+nl ), ... ]$ gives a transcendental number. \end{abstract} \section{Introduction}\label{s:1} First we introduce the Thue-Morse sequence, defined by digit counting. Let $k$ be an integer greater than $1$. We define the $k$-adic expansion of a non-negative integer $n$ as follows: \begin{align}\label{eq1.1} n=\sum_{q=1}^{{\rm finite}} s_{n,q} k^{w_n(q)}, \end{align} where $1\le s_{n,q}\le k-1, 0\le w_n(q) 0$ $)$ is a transcendental number. \end{theorem} By Theorem \ref{th.1.2}, one can find an uncountable quantity of transcendental numbers. Moreover, if $\sum_{n=0}^\infty \frac{a(n)}{ {\beta}^{n+1}}$ is a transcendental number, then $\sum_{n=0}^\infty \frac{a(N+nl)}{ {\beta}^{n+1}}$ $($ for all $N \ge 0$ and for all $l >0$ $)$ is also a transcendental number. The proof of Theorem \ref{th.1.2} does not rely on the pure periodicity of the periodic generalized Thue-Morse sequence $(a(n))_{n=0}^\infty$ and the Cobham conjecture. Here we study non-periodicity of the subsequence $(a(N+nl))_{n=0}^\infty$ $($ for all $N \ge 0$ and for all $l >0$ $)$ of a generalized Thue-Morse sequence $(a(n))_{n=0}^\infty$. See also Morgenbesser, Shallit, and Stoll \cite{MSS}. Almost no generalized Thue-Morse sequence $(a(n))_{n=0}^\infty$ is $k$-automatic (see Proposition \ref{p.5.1} in Section \ref{s:5}). Therefore, the proof of Theorem \ref{th.1.2} is different from the proof of Theorem \ref{MM-AB}. We prove Theorem \ref{th.1.2} by combining Theorem \ref{th.3.1} in Section \ref{s:3} with the combinatorial transcendence criterion established by Adamczewski, Bugeaud, and Luca \cite{ABL}. This paper is organized as follows. In Section \ref{s:2}, we review the basic concepts of the periodicity of sequences, and give the formal definition of the generalized Thue-Morse sequences. For a sequence $(a(n))_{n=0}^\infty$, we set its generating function $g(z) \in \mathbb{C}[[z]]$ to be \[ g(z):=\sum_{n=0}^\infty a(n) z^n. \] For a generalized Thue-Morse sequence, one can prove that the generating function is convergent on the open unit disk and that it has an infinite product expansion. In Section \ref{s:3}, first we prove the key lemma on the $k$-adic expansion of non-negative integers. Next, we use this lemma and the infinite product expansion of the generating function of a generalized Thue-Morse sequence to prove a necessary-sufficient condition for the non-periodicity of the generalized Thue-Morse sequence. Furthermore, we prove that if the generalized Thue-Morse sequence is not periodic, then no subsequence $(a(N+nl))_{n=0}^\infty$ ( for all $N \ge 0$ and for all $l >0$) of the generalized Thue-Morse sequences is periodic. In Section \ref{s:4}, we introduce the concept of the stammering sequence, introduced by Adamczewski, Bugeaud, and Luca \cite{ABL}, and the combinatorial transcendence criterion, established by Adamczewski, Bugeaud, Luca \cite{ABL} and Bugeaud \cite{Bu1}. By applying this combinatorial transcendence criterion to the generalized non-periodic Thue-Morse sequence $(a(n))_{n=0}^\infty$, which takes its values from $\{0,1,\ldots,\beta-1\}$, we show that $\sum_{n=0}^\infty a(N+nl){\beta}^{-n-1}$ is a transcendental number. Furthermore by applying this combinatorial transcendence criterion to the generalized non-periodic Thue-Morse sequence $(a(n))_{n=0}^\infty$, which takes its values in bounded positive integers, we show that the continued fraction $[0, a(N), a(N+l),\ldots,a(N+nl ), \ldots]$ is also transcendental number. This result includes Theorem \ref{th.1.2}. In Section \ref{s:5}, we consider the necessary-sufficient condition that a generalized Thue-Morse sequence is a $k$-automatic sequence. Then we find many transcendental numbers whose irrationality exponent is finite in all arithmetical subsequences of the corresponding generalized Thue-Morse sequence by applying the Adamczewski and Cassaigne result on $k$-automatic irrational numbers \cite{AC}. Furthermore, we consider the transcendence of the value at the algebraic point of the generating function $\sum_{n=0}^\infty a(N+nl)z^{-n-1}$ by applying Becker's result on $k$-automatic power series. \section{Generalized Thue-Morse sequences and their generating functions}\label{s:2} Let $(a(n))_{n=0}^\infty$ be a sequence with values in $\mathbb{C}$. The sequence $(a(n))_{n=0}^\infty$ is called {\it ultimately periodic} if there exist non-negative integers $N$ and $l>0$ such that \begin{align}\label{eq2.1} a(n)=a(n+l) \qquad (\forall n \ge N). \end{align} An $arithmetical$ $subsequence$ of $(a(n))_{n=0}^\infty$ is defined to be a subsequence such as $(a(N+tl))_{t=0}^\infty$, where $N \ge 0$ and $l >0$. \begin{definition}\label{de.2.1} Let $(a(n))_{n=0}^\infty$ be a sequence with values in $\mathbb{C}$. The sequence $(a(n))_{n=0}^\infty$ is called $everywhere$ $non$-$periodic$ if no arithmetical subsequence of $(a(n))_{n=0}^\infty$ takes on only one value. \end{definition} Now we present some lemmas about the everywhere non-periodic sequences. \begin{lemma}\label{Le.2.1} If $(a(n))_{n=0}^\infty$ is everywhere non-periodic, then $(a(n))_{n=0}^\infty$ is not ultimately periodic. \end{lemma} \begin{proof} We prove the contrapositive. Assume that $(a(n))_{n=0}^\infty$ is ultimately periodic. From the definition of everywhere non-periodic, there exist non-negative integers $N$ and $l>0$ such that \begin{align}\label{eq2.2} a(n)=a(n+l) \qquad (\forall n \ge N). \end{align} It follows from \eqref{eq2.2} that the arithmetical subsequence $(a(N+tl))_{t=0}^\infty$ takes on only one value. \end{proof} \begin{lemma}\label{Le.2.2} If $(a(n))_{n=0}^\infty$ is everywhere non-periodic, then all arithmetical subsequences of $(a(n))_{n=0}^\infty$ are everywhere non-periodic. \end{lemma} \begin{proof} We prove contraposition. If $(a(N+tl))_{t=0}^\infty$ is not everywhere non-periodic, then there exist non-negative integers $k$ and $J>0$ such that $(a(N+kl+mJl))_{m=0}^\infty$ takes on only one value. The subsequence $(a(N+kl+mJl))_{m=0}^\infty$ is also an arithmetical subsequence of $(a(n))_{n=0}^\infty$. Therefore, $(a(n))_{n=0}^\infty$ is not everywhere non-periodic. \end{proof} \begin{corollary}\label{Co.2.1} $(a(n))_{n=0}^\infty$ is everywhere non-periodic if and only if no arithmetical subsequence of $(a(n))_{n=0}^\infty$ is ultimately periodic. \end{corollary} \begin{proof} Assume $(a(n))_{n=0}^\infty$ is everywhere non-periodic. By Lemma \ref{Le.2.1} and Lemma \ref{Le.2.2}, no arithmetical subsequence of the sequence $(a(n))_{n=0}^\infty$ is periodic. We show the sufficient condition. Assume $(a(n))_{n=0}^\infty$ is not everywhere non-periodic. Then there exist non-negative integers $N$ and $l>0$ such that $(a(N+tl))_{t=0}^\infty$ takes on only one value. This sequence is ultimately periodic. \end{proof} Next, we generalize the Thue-Morse sequence of Emmanuel \cite{Emma}. \begin{definition}\label{de.2.2} Let $L$ be an integer greater than $1$, and let $a_0,a_1, \ldots ,a_{L-1}$ be $L$ distinct complex numbers. We let ${\{a_0,a_1 \ldots a_{L-1}\}}^*$ denote the free monoid generated by $\{a_0,a_1 \ldots a_{L-1}\}$. We define a morphism $f$ from ${\{a_0,a_1 \ldots a_{L-1}\}}^*$ to ${\{a_0,a_1 \ldots a_{L-1}\}}^*$ as follows: \begin{align}\label{eq2.3} f(a_i) = a_{i+1}, \end{align} where the index $i$ is computed modulo $L$. Let $f^j$ be the $j$ times composed mapping of $f$, and let $f^0$ be the identity mapping. Let $A$ and $B$ be two finite words on $\{a_0,a_1, \ldots ,a_{L-1}\}$, and let $AB$ denote the concatenation of $A$ and $B$. Let $A_0=a_0$, $k$ be an integer greater than $1$, and let $\kappa$ be a map $\kappa$:$\{1, \ldots ,k-1\} \times\mathbb{N} \rightarrow\ \{0, \ldots ,L-1\}$. For a non-negative integer $m$, we define a space of words $W_m$ by \begin{align}\label{eq2.4} W_{m}:=\{ a_{i_1} a_{i_2} \ldots a_{i_m} \; | a_{i_1}, a_{i_2}, \cdots ,a_{i_m}\in \{ a_0,a_1, \ldots , a_{L-1}\} \} . \end{align} We define $A_{n+1}\in W_{k^{n+1}}$ recursively as \begin{align}\label{eq2.5} A_{n+1}:=A_n f^{\kappa(1,n)}(A_n)\cdots \cdots f^{\kappa (k-1,n)}(A_n), \end{align} and we let \begin{align}\label{eq2.6} A_{\infty}:=\lim_{n \to \infty} A_n \end{align} denote the limit of $A_n$. The sequence (or infinite word) $A_{\infty}$ is called the $generalized$ $Thue$-$Morse$ $sequence$ $of$ $type$ $(L,k,\kappa)$, abbreviated as the $(L,k,\kappa)$-TM sequence. \end{definition} \begin{example}[ \cite{Emma}]\label{exa1} Let $L=2$, $a_0=0$, $a_1=1$ and $\kappa(1,y)=1$ for all $y\in \mathbb{N}$. The $(2,2,1)$-TM sequence is as follows: \begin{align*} &\qquad \qquad A_0=0,\; A_1=01,\; A_2=0110,\; A_3=01101001,\\ & A_{\infty}=0110100110010110100101100110100110010110011010010110100110\cdots. \end{align*} This example is the Thue-Morse sequence of Emmanuel \cite{Emma}. \end{example} \begin{example}\label{exa2} Let $L=2$, $a_0=0$, $a_1=1$ and \[ \kappa (1, y)= \begin{cases} 1, & \text{ $y$ is a prime number};\\ 0, & \text{ otherwise}. \end{cases} \] The $(2,2, \kappa)$-TM sequence is \begin{align*} &A_0=0,\; A_1=00,\; A_2=0000,\; A_3=00001111,\\ & A_{\infty}=0000111100001111 1111000011110000\cdots. \end{align*} \end{example} \begin{example}\label{exa3} Let $L=2$, $a_0=0$, $a_1=1$ and \[ \kappa (1, y)= \begin{cases} 1, & \text{ $y$ is a square number and $s=2$};\\ 0, & \text{ otherwise} \end{cases}. \] The $(2,3,\kappa)$-TM sequence is \begin{align*} &\qquad \qquad A_0=0,\; A_1=001,\; A_2=001001001,\\ & A_{\infty}=001001001001001001001001001 001001001001001001001001001 110110\cdots. \end{align*} \end{example} Let $(a(n))_{n=0}^\infty$ be a sequence with values in $\mathbb{C}$. The $generating$ $function$ of $(a(n))_{n=0}^\infty$ is the formal power series $g(z) \in \mathbb{C}[[z]]$, defined as \[ g(z):=\sum_{n=0}^\infty a(n) z^n. \] The following lemma clarifies the meaning of an $(L,k,\kappa)$-TM sequence. \begin{lemma}\label{Le.2.3} Let $A_{\infty}=(b(n))_{n=0}^\infty$ be a $(L,k,\kappa)$-TM sequence with $a_j=\exp \frac{2\pi \sqrt{-1} j}{L}$ $($ for all $j$ with $0\le \; j \le L-1$ $)$. Let $ G_{A_{\infty}}(z)$ be the generating function of $(b(n))_{n=0}^\infty$, \[ G_{A_{\infty}}(z):=\sum_{n=0}^\infty b(n) z^n. \] The generating function $G_{A_{\infty}}(z)$ will have the infinite product on $|z|<1$, \begin{align}\label{eq2.7} G_{A_{\infty}}(z)=\prod_{y=0}^\infty (1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,y) } {L}z^{s k^y}). \end{align} \end{lemma} \begin{proof} From the assumption $a_j=\exp \frac{2\pi \sqrt{-1} j}{L}$ for all $j$ with $0\le \; j \le L-1$, we have \begin{align} \label{eq2.8} f(a_j)=\exp \frac{2\pi \sqrt{-1} }{L} a_j \end{align} for all $j$ with $0\le \; j \le L-1$. The $(L,k,\kappa)$-TM sequence takes on only finite values, and by the Cauchy-Hadamard theorem, $G_{A_{\infty}}(z)$ and $\prod_{y=0}^\infty(1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,y) } {L}z^{s k^y})$ converge absolutely on the unit disk. Let $G_{A_n}(z)$ be the generating function of $A_n$; We identify the infinite word $A_n 0\cdots0\cdots=:A_n 0^{\infty}$ with $A_n$. We will show by induction that the following equality holds for $n$, \begin{align}\label{eq2.9} G_{A_n}(z)=\prod_{y=0}^{n-1}(1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,y)}{L}z^{s k^y}). \end{align} First, we check the case $n=1$. From the definition of $A_1$, we have \begin{align}\label{eq2.10} G_{A_1}(z)=1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,0)}{L}z^s. \end{align} Thus, the $n=1$ case is true. By the induction hypothesis we may assume that \begin{align}\label{eq2.11} G_{A_j}(z)=\prod_{y=0}^{j-1}(1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,y)}{L}z^{s k^y}). \end{align} Therefore, we have \begin{align}\label{eq2.12} G_{A_{j+1}}(z)=G_{A_j}(z)+\sum_{s=1}^{k-1} G_{f^{\kappa(s,j)}(A_j)}(z)z^{s k^j}. \end{align} Alternatively, \begin{align}\label{eq2.13} G_{f^{\kappa(s,j)}(A_j)}(z)=\exp \frac{2\pi \sqrt{-1} \kappa(s,j)}{L}G_{A_j}(z). \end{align} From \eqref{eq2.11}-\eqref{eq2.13}, we get \begin{align}\label{eq2.14} &G_{A_{j+1}}(z)=G_{A_j}(z)(1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,y)}{L}z^{s k^j})\nonumber \\ &\qquad\qquad =\prod_{y=0}^j(1+\sum_{s=1}^{k-1}\exp \frac{2\pi \sqrt{-1} \kappa(s,y)}{L}z^{s k^y}). \end{align} Therefore \eqref{eq2.7} is true. Finally, we will compare the coefficients of $z^j$ on both sides of \eqref{eq2.7}. On the right-hand side of \eqref{eq2.7}, the coefficient of $z^j$ are determined by $G_{A_N}(z)$ for sufficiently large $N$. From the definition of $A_{\infty}$, the prefix word, $p^N$, of $A_{\infty}$ is $A_N$. From the above argument and \eqref{eq2.9}, the coefficients of $z^j$ on both sides of \eqref{eq2.7} must coincide. \end{proof} \begin{proposition}\label{p.2.1} Let $A_{\infty}=(b(n))_{n=0}^\infty$ be a $(L,k,\kappa)$-TM sequence with $a_j=\exp \frac{2\pi \sqrt{-1} j}{L}$ $($for all $j$ with $0\le \; j \le L-1$ $)$. Let $(a(n))_{n=0}^\infty$ be a sequence defined by \eqref{eq1.5}. Then \begin{align}\label{eq2.15} \frac{L}{2\pi \sqrt{-1}} \log b(n)\equiv a(n) \; \pmod{L}. \end{align} \end{proposition} \begin{proof} Let the $k$-adic expansion of $n$ be as follows: \begin{align}\label{eq2.16} n=\sum_{q=1}^{n(k)} s_{n,q} k^{w_n(q)}, \end{align} where $1\le s_{n,q}\le k-1$, $0 \le w_n(q)1$ and $l>0$ be integers and $t$ be a non-negative integer, then there exists an integer $x$ such that \begin{align}\label{eq3.1} xl=\sum_{q=1}^{{\rm finite}} s_{xl,q} k^{w_{xl}(q)}, \end{align} where $s_{{xl},1}=1, w_{xl}(2)-w_{xl}(1)>t,w_{xl}(q+1)>w_{xl}(q)\ge 0$. Furthermore, if $t'$ be other non-negative integer, then there exists an integer $X$ such that \begin{align}\label{eq3.2} Xl=\sum_{q=1}^{{\rm finite}} s_{Xl,q} k^{w_Xl(q)}, \end{align} where $s_{{Xl},1}=1, w_{Xl}(2)-w_{Xl}(1)>t', w_{Xl}(q+1)>w_{Xl}(q)\ge 0, w_{xl}(1)=w_{Xl}(1)$. \end{lemma} \begin{proof} Let us assume the factorization of $k$ into prime factors is \begin{align}\label{eq3.3} k=\prod_{t=1}^N {p_t}^{y_t}, \end{align} where $p_1,p_2,\cdots p_N$ are $N$ distinct prime numbers and $y_t$\;for $p_t$ $\;(1\le t \le N\;)$ are $N$ positive integers. Let $l$ be represented as \begin{align}\label{eq3.4} l=G\prod_{u=1}^n {p_{t_u}}^{x_u}, \end{align} where $G$ and $k$ are coprime, ${p_{t_u}} \in \{p_t|1\le t \le N\}$ and $x_u$ are $n$ positive integers. As $G$ and $k$ are coprime, there exist integers $D$ and $E$ such that \begin{align}\label{eq3.5} DG=1-k^{t+1}E. \end{align} We set $$F= \max \{A | x_u=y_{t_u} A +H, \; 0 \leq H < y_{t_u}, \; 1 \leq u \leq n \}.$$ From the definition of $F$, $k^{F+1} \prod_{u=1}^n {p_{t_u}}^{-x_u}$ is a non-negative integer. Thus we have \begin{align}\label{eq3.6} lD^2G k^{F+1} \prod_{u=1}^n {p_{t_u}}^{-x_u}=k^{F+1} D^2G^2. \end{align} On the other hand, by \eqref{eq3.5} we have \begin{align}\label{eq3.7} D^2G^2=1+k^{t+1}E(k^{t+1}E-2). \end{align} Thus $E(k^{t+1}E-2)$ is a non-negative integer. If $E(k^{t+1}E-2)>0$, it follows from the $k$-adic expansion of $E(k^{t+1}E-2)$ that $k^{F+1}D^2G^2$ satisfies the Lemma. If $E(k^{t+1}E-2)=0$, then $G=1$. The integer $k^{F+1}(1+k^{t+1})$ also satisfies the Lemma. As $F+1$ is independent of $t$, the second claim is trivial. \end{proof} Now we will show the everywhere non-periodic result by the previous lemma. \begin{proposition}\label{p.3.1} Let $A_{\infty}=(a(n))_{n=0}^\infty$ be a sequence with values in $\mathbb{C}$, and let $G_{A_{\infty}}(z)$ denote the generating function of $(a(n))_{n=0}^\infty$. Assume that $G_{A_{\infty}}(z)$ has the following infinite product expansion for an integer $k$ greater than $1$ and $t_{s,y} \neq0$ for all $s$ with $1\le s\le k-1$ and for all $y\in\mathbb{N}$, \begin{align}\label{eq3.8} G_{A_{\infty}}(z)=\prod_{y=0}^\infty (1+\sum_{s=1}^{k-1} t_{s,y} z^{s k^y}). \end{align} If there exists a periodic arithmetical subsequence of $(a(n))_{n=0}^\infty$, then $G_{A_{\infty}}(z)$ has the following infinite product expansion \begin{align}\label{eq3.9} G_{A_{\infty}}(z)=(\sum_{n=0}^{k^A -1} a(n) z^n) \prod_{y=0}^\infty(1+\sum_{s=1}^{k-1} h^{s k^y} z^{s k^{A+y}}), \end{align} where $A$ is a non-negative integer and $h$ is a complex number. \end{proposition} \begin{proof} Let $n$ and $m$ be two non-negative integers and their respective $k$-adic expansions are as follows: \begin{align}\label{eq3.10} n=\sum_{q}^{{\rm finite}} s_{n,q} k^{w_n(q)}, \quad m=\sum_{p}^{{\rm finite}} s_{m,p} k^{w_m(p)}, \end{align} where $1\le s_{n,q}, s_{m,p}\le k-1$, $ 0\le w_n(q) 0$, such that \begin{align}\label{eq3.12} a(N)=a(N+tl) \qquad (\forall t \in \mathbb{N}). \end{align} Let the $k$-adic expansion of $N$ be \begin{align}\label{eq3.13} N=\sum_{q=1}^{N(k)} s_{N,q} k^{w_N(q)} \qquad where\; 1\le s_{N,q}\le k-1, 0 \le w_N(q) w_N(N(k))). \\&\label{eq3.15}\qquad \qquad \qquad \qquad a(N) \neq 0. \end{align} From \eqref{eq3.14} and \eqref{eq3.15}, we get \begin{align}\label{eq3.16} a(k^r t l) = 1\qquad (\forall r > w_N(N(k))). \end{align} By Lemma \ref{Le.3.1}, there exists an integer $x$ greater than zero such that \begin{align}\label{eq3.17} x l=\sum_{q=1}^{x l(k)} s_{xl,q} k^{w_{xl}(q)}, \end{align} where $s_{xl,1}=1$ and $w_{xl}(2)-w_{xl}(1)>1$. Moreover, there exists an integer $X$ greater than zero such that \begin{align}\label{eq3.18} X l=\sum_{q=1}^{X l(k)} s_{{Xl}_q} k^{w_{Xl}(q)}, \end{align} where $s_{Xl,1}=1$, $w_{Xl}(2)-w_{Xl}(1)>w_{xl}(xl(k))$ and $w_{Xl}(1)=w_{xl}(1)$. Let $xl k^{-w_{xl}(1)}$ and $Xl k^{-w_{xl}(1)}$ be replaced by $xl$ and $Xl$, respectively. Let $r$ be an integer greater than $w(N(k))+w_{xl}(1)$ and $s$ be an integer in $\{1,\ldots, k-1\}$. By the definition of $Xl$ and \eqref{eq3.11}, we have \begin{align}\label{eq3.19} a(k^r sX l)=a(sk^r )a(k^rsXl-sk^r). \end{align} From \eqref{eq3.15}, we get \begin{align}\label{eq3.20} &1=a(k^r x l), \\&\label{eq3.21}1=a(k^r sX l), \\&\label{eq3.22}1=a(k^r x l+k^r sX l). \end{align} By \eqref{eq3.11}, \eqref{eq3.19}-\eqref{eq3.22} and the definitions of $xl$ and $Xl$, we have \begin{align}\label{eq3.23} &a(k^r)a(k^r x l-k^r)=1, \\& \label{eq3.24}a(lsk^r)a(sX lk^r-s k^r)=1, \\& \label{eq3.25}a(k^r (s+1))a(x l k^r-k^r)a(sX lk^r-s k^r)=1. \end{align} From \eqref{eq3.23}-\eqref{eq3.25}, we get \begin{align}\label{eq3.26} a(k^r (s+1))=a(k^r)a(k^r s). \end{align} Put $h:=a(k^{w(N(k))+w_{xl}(1)+1})$. By \eqref{eq3.11}, we have \begin{align}\label{eq3.27} a(sk^y)=t_{s,y}, \end{align} for all $s$ with $1\le s\le k-1$ and for all $y \in \mathbb{N}$. By \eqref{eq3.15}, \eqref{eq3.26}, \eqref{eq3.27} and inductive computation, we get the relations \begin{align}\label{eq3.28} t_{s,w(N(k))+w_{xl}(1)+1+y}=h^{s k^y}, \end{align} for all $s$ with $1\le s\le k-1$ and for all $y \in \mathbb{N}$. From the assumption of $G_{A_{\infty}}(z)$, the proof is complete. \end{proof} \qquad Finally, we prove the main theorem in Section \ref{s:3}. \begin{theorem}\label{th.3.1} Let $A_{\infty}=(a(n))_{n=0}^\infty$ be an $(L,k,\kappa)$-TM sequence. The sequence $A_{\infty}=(a(n))_{n=0}^\infty$ is ultimately periodic if and only if there exists an integer $A$ such that \begin{align}\label{eq3.29} \kappa(s,A+y) \equiv \kappa(1,A) s k^y \pmod{L}, \end{align} for all $s$ with $1\le s\le k-1$ and for all $ y \in \mathbb{N}$. Moreover, if the $(L,k,\kappa)$-TM sequence is not ultimately periodic, then no arithmetical subsequence of $(L,k,\kappa)$-TM sequence is ultimately periodic. \end{theorem} \begin{proof} We assume, without loss of generality, that $A_{\infty}=(a(n))_{n=0}^\infty$ is an $(L,k,\kappa)$-TM sequence with $a_j=\exp \frac{2\pi \sqrt{-1} j}{L}$ $($for all $0\le j \le L-1$ $)$. From this assumption and Lemma \ref{Le.2.3}, $(a(n))_{n=0}^\infty$ satisfies the assumption of Proposition \ref{p.3.1}. Therefore, \eqref{eq3.29} is the necessary condition. Now, we show the sufficient condition. Let $G_{A_{\infty}}(z)$ be the generating function of $(a(n))_{n=0}^\infty$. Notation is the same as for Proposition \ref{p.3.1}. If we assume that $(a(n))_{n=0}^\infty$ satisfies \eqref{eq3.29}, then there exists a non-negative integer $A$ such that \begin{align}\label{eq3.30} t_{s,A+y}=h^{s k^y} \qquad (\forall y \in \mathbb{N}). \end{align} Thus $G_{A_{\infty}}(z)$ has the infinite product expansion \begin{align}\label{eq3.31} G_{A_{\infty}}(z)=(\sum_{n=0}^{k^A -1} b(n) z^n) \prod_{y=0}^\infty(1+\sum_{s=1}^{k-1} (h z^{k^A})^{s k^y}). \end{align} Let $Z=h z^{k^A}$. As $h$ is the $L$-th root of $1$ and $\kappa$ is a zero map in Lemma \ref{Le.2.3}, we find \begin{align}\label{eq3.32} \prod_{y=0}^\infty(1+\sum_{s=1}^{k-1} Z^{s k^y})=\sum_{n=0}^\infty Z^n \qquad on \; |Z|<1. \end{align} We put $G(z)=\sum_{n=0}^{k^A -1} a(n) z^n$. From \eqref{eq3.31} and \eqref{eq3.32}, \begin{align}\label{eq3.33} G_{A_{\infty}}(z)=G(z)(\sum_{n=0}^\infty (h z^{k^A})^n). \end{align} As $h$ is the $L$-th root of $1$, \begin{align}\label{eq3.34} G_{A_{\infty}}(z)=(G(z) (\sum_{n=0}^{L-1} (h z^{k^A})^n))(1+\sum_{s=1}^\infty z^{s L k^A})=\frac{G(z) (\sum_{n=0}^{L-1} (h z^{k^A})^n)}{1-z^{L k^A}}. \end{align} As the degree of $G(z)$ is $k^A -1$, and using \eqref{eq3.34}, we find that the sequence $(a(n))_{n=0}^\infty$ that satisfies \eqref{eq3.29} has a period $Lk^A$. Moreover, if the $(L,k,\kappa)$-TM sequence is not ultimately periodic, then no arithmetical sequence of $(L,k,\kappa)$-TM sequence is ultimately periodic by the above argument and by Proposition \ref{p.3.1}. \end{proof} If an $(L,k,\kappa)$-TM sequence satisfies \begin{align}\label{eq3.35} \kappa (s,y)=\kappa (s,y+1) \end{align} for all $s$ with $1\le s\le k-1$ and for all $ y \in \mathbb{N}$, then $(\kappa(1),\kappa(2),\ldots ,\kappa(k-1))$-$L$ will denote the $(L,k,\kappa)$-TM sequence. The weak version of the corollary that follows is given as Theorem $2$ in Morton and Mourant \cite{MM}. See also Allouche and Shallit \cite{AllSha2}, Frid \cite{Fr}. \begin{corollary}\label{Co.3.1} The sequence $(\kappa(1),\kappa(2),\ldots ,\kappa(k-1))$-$L$ is periodic if and only if $\kappa(s)$ $($ for all $s$ with $1\le s\le k-1$ $)$ satisfies \begin{align}\label{eq3.36} s \kappa(1) \equiv \kappa(s) ,\kappa(k-1)\equiv 0 \pmod{L}. \end{align} Moreover, if $(\kappa(1),\kappa(2),\ldots ,\kappa(k-1))$-$L$ is not periodic, then no arithmetical subsequence of $(\kappa(1),\kappa(2),\ldots ,\kappa(k-1))-L$ is periodic. \end{corollary} \begin{proof} By Theorem \ref{th.3.1}, the necessary-sufficient condition for the periodicity of $(\kappa(1),\kappa(2),\ldots ,\kappa(k-1))$-$L$ comprises the following relations: \begin{align}\label{eq3.37} \kappa(1,A+1) \equiv \kappa(1,A)k \pmod{L}, \kappa(k-1)\equiv(k-1)\kappa(1) \equiv 0 \pmod{L}. \end{align} . \end{proof} \section{Transcendence results of the generalized Thue-Morse sequences}\label{s:4} Adamczewski, Bugeaud, and Luca \cite{ABL} introduced a new class of sequences, as follows. For any positive number $y$,$\lfloor y\rfloor $ and $\lceil y\rceil $ are the floor and ceiling functions. Let $W$ be a finite word on $\{a_0,a_1, \ldots, a_{L-1}\}$ and let $|W|$ be the length of $W$. For any positive number $x$, we let $W^x$ defined the word $W^{ \lfloor x \rfloor} W^`$, where $W^`$ is a prefix of $W$ of length $\lceil(x-\lfloor x\rfloor)|W|\rceil$. \begin{definition}\label{de.4.1} $(a(n))_{n=0}^\infty$ is called a $stammering$ $sequence$ if $(a(n))_{n=0}^\infty$ satisfies the following conditions: $(1)$ The sequence $(a(n))_{n=0}^\infty$ is a non-periodic sequence. $(2)$ There exist two sequences of finite words, $(U_m)_{m\ge 1}$ and $(V_m)_{m\ge 1}$, such that, $(A)$ there exists a real number $w >1$ independent of $n$ such that the word \qquad \qquad \qquad\qquad $U_m {V_m}^w$ is a prefix of the word $(a(n))_{n=0}^\infty$, $(B)$ $\lim_{m \to \infty} |U_m|/|V_m|\;< \;+\infty$, and $(C)$ $\lim_{m \to \infty} |V_m|=+\infty$. \end{definition} Let $(a(n))_{n=0}^\infty$ be a sequence of positive integers. We define the $continued$ $fraction$ of $(a(n))_{n=0}^\infty$ as \begin{align}\label{eq4.1} [0, a(0),a(1),\ldots, a(n),\ldots ]:=\cfrac{1}{a(0)+ \cfrac{1}{a(1)+ \cfrac{1}{\cdots + \cfrac{1}{a(n)+ \cfrac{1}{ \ddots}}}}}. \end{align} Adamczewski, Bugeaud, Luca \cite{ABL} and Bugeaud \cite{Bu1} proved the result that follows by the Schmidt subspace theorem. \begin{theorem}[\cite{ABL,Bu1}]\label{ABL-Bu1} If $\beta$ is an integer greater than $1$ and $(a(n))_{n=0}^\infty$ is a stammering sequence on $\{0,1,\ldots,\beta-1\}$, then $\sum_{n=0}^\infty \frac{a(n)}{ {\beta}^{n+1}}$ is a transcendental number. Moreover, if $(a(n))_{n=0}^\infty$ is a stammering sequence on bounded positive integers, then the continued fraction $[0, a(0), a(1)\ldots,a(n) \ldots]$ is also a transcendental number. \end{theorem} We will prove the next theorem using Theorems and \ref{th.1.2}, \ref{th.3.1} and \ref{ABL-Bu1}. \begin{theorem}\label{th.4.2} Let $A_{\infty}=(a(n))_{n=0}^\infty$ be an $(L,k,\kappa)$-TM sequence and $\beta$ be an integer greater than $1$. We assume that $(a(n))_{n=0}^\infty$ takes its input from $\{0,1,\ldots,\beta-1\}$. If there is no integer $A$ such that \begin{align}\label{eq4.2} \kappa(s,A+y) \equiv \kappa(1,A) s k^y \pmod{L} \end{align} for all $s$ with $1\le s\le k-1$ and for all $ y \in \mathbb{N}$, then $\sum_{n=0}^\infty \frac{a(N+nl)}{ {\beta}^{n+1}}$ $($ for all $N \ge 0$ and for all $l>0$ $)$ is a transcendental number. Moreover, if we assume that $(a(n))_{n=0}^\infty$ takes its input from the positive integers, and if there is no integer $A$ such that \begin{align}\label{eq4.3} \kappa(s,A+y) \equiv \kappa(1,A) s k^y \pmod{L} \end{align} for all $s$ with $1\le s\le k-1$ and for all $ y \in \mathbb{N}$, then $[0, a(N), a(N+s),\ldots,a(N+nl) \ldots]$ $($ for all $N \ge 0$ and for all $l>0$ $)$ is a transcendental number. \end{theorem} \begin{proof} Let $N$ and $l>0$ be positive integers. By Theorem \ref{th.3.1}, $(a(N+nl))_{n=0}^\infty$ is non-periodic. Therefore, we only have to prove that $(L,k,\kappa)$-TM satisfies the condition $(2)$ of Definition \ref{de.4.1}. We choose an integer $M$ such that $k^M> 2(N+l)$, and assume that $m>M$. As $f$ is a cyclic permutation of order $L$ and by Definition \ref{de.2.2}, the $(Ll+1)k^m$ prefix word of $(a(n))_{n=0}^\infty$ is as follows \begin{align}\label{eq4.4} A_{\infty}=(a(n))_{n=0}^\infty =A_m f^{i_1}(A_m) \cdots f^{i_{Ll}}(A_m)\cdots, \end{align} where $A_m$ is the $k^m$ prefix word of $(a(n))_{n=0}^\infty$, $i_j (1\le j\le Ll) \in \{0,\dots, L-1 \}$. By \eqref{eq4.4}, we have \begin{align}\label{eq4.5} f^{i_{tl}}(a(n))=a(n+k^m tl ) \end{align} for all $0\le n\le k^m-1$ and for all $1\le t\le L$. As $f$ is a cyclic permutation of order $L$, by \eqref{eq4.4}, \eqref{eq4.5} and the Dirichlet schubfachprinzip, we have \begin{align}\label{eq4.6} (a(N+n l))_{n=0}^\infty =W_{1,m} W_{2,m} W_{3,m} W_{2,m} \cdots , \end{align} where $W_{i,m}$ $(i\in\{1,2,3\})$ are finite words such that \begin{align}\label{eq4.7} &|W_{1,m}|\le ((Ll+1)k^m -N)/l +1, \\&\label{eq4.8}|W_{2,m}| \ge (k^m-N)/l-1, \\& \label{eq4.9}|W_{2,m}|+|W_{3,m}|\le ((Ll+1)k^m -N)/l +1. \end{align} We put $U_m:=W_{1,m}$, $V_m:=W_{2,m}W_{3,m}$ and $w:=1+\frac{1}{2 Ll+3}$. By \eqref{eq4.7}-\eqref{eq4.9} and the assumption of $m$, we obtain \begin{eqnarray}\label{eq4.10} &\lceil (w-1)|V_m|\rceil =\lceil \frac{1}{ 2 Ll+3}(|W_{2,m}|+|W_{3,m}|)\rceil \le \nonumber \\ &\frac{1}{ 2 Ll+3}((Ll+1)k^m-N+l)/l\le \frac{k^m}{2 l} \; < |W_{2,m}|. \end{eqnarray} From \eqref{eq4.10}, $(a(n))_{n=0}^\infty$ satisfies Condition $(A)$. Furthermore, \begin{eqnarray}\label{eq4.11} &|U_m|/|V_m|=|W_{1,m}|/|W_{2,m}W_{3,m}|\le \nonumber \\ &((Ll+1)k^m-N+l)/l \times l /( k^m-N-l) \le 2 Ll+3. \end{eqnarray} From \eqref{eq4.11}, $(a(n))_{n=0}^\infty$ satisfies Condition $(B)$. It follows directly that $(V_m)_{m\ge 1}$ satisfies Condition $(C)$. \end{proof} \begin{corollary}\label{Co.4.1} Let $(a(n))_{n=0}^\infty$ be an $(L,k,\kappa)$-TM sequence and $\beta$ be an integer greater than $1$. If $(a(n))_{n=0}^\infty$ takes its input from $\{0,1,\ldots,\beta-1\}$ and there is no integer $A$ such that \begin{align}\label{eq4.12} \kappa(s,A+y) \equiv \kappa(1,A) s k^y \pmod{L} \end{align} for all $s$ with $1\le s\le k-1$ and for all $y \in \mathbb{N}$, then the generating function $f(z):=\sum_{n=0}^\infty \frac{a(N+nl)}{ z^{n+1}}$ $($ for all $N \ge 0$ and for all $l>0$ $)$ is transcendental over $\mathbb{C}(z)$. \end{corollary} \begin{proof} We assume $f(z)$ is algebraic over $\mathbb{C}(z)$. As $f(z)$ is algebraic over $\mathbb{Q}(z)$ if and only if $f(z)$ is algebraic over $\mathbb{C}(z)$ $($see the Remark in Theorem $1.2$ in Nishioka \cite{Ni} $)$, then $f(z)$ satisfies the equation \begin{align}\label{eq4.13} c_n(z) f^n(z)+c_{n-1}(z)f^{n-1}(z)+ \cdots + c_0(z)=0, \end{align} where $c_i(z) \in \mathbb{Q}[z] $ $( 0 \le i \le n)$, $c_n(z) c_0(z) \neq 0$ and $c_i(z)$ $(0 \le i \le n)$ are coprime. From Theorem \ref{th.4.2}, $f(\frac{1}{\beta})$ is a transcendental number. From the above argument and by \eqref{eq4.13}, $c_i(\frac{1}{\beta}) =0$ $($ for all $0 \le i \le n)$. This contradicts the assumption that $c_i(z)$ $( 0 \le i \le n)$ are coprime. \end{proof} \section{ $k$-Automatic generalized Thue-Morse sequences and some results}\label{s:5} First, we introduce some definitions. \begin{definition}\label{de.5.1} Let $\alpha$ be an irrational real number. The irrationality exponent $\mu (\alpha)$ of $\alpha$ is the supremum of the real numbers $\mu$ such that the inequality \begin{align}\label{eq5.1} \left| \alpha - \frac{p}{q} \right| < \frac{1}{q^{\mu}} \end{align} has infinitely many solutions in non-zero integers $p$ and $q$. \end{definition} \begin{definition}\label{de.5.2} The $k$-$kernel$ of $(a(n))_{n=0}^\infty$ is the set of all subsequences of the form $(a(k^e n+j))_{n=0}^\infty$, where $e\ge 0$ and $0\le j\le k^e-1$. \end{definition} \begin{definition}\label{de.5.3} The sequence$(a(n))_{n=0}^\infty$ is called a $k$-$automatic$ $sequence$ if the $k$-kernel of $(a(n))_{n=0}^\infty$ is the finite set. \end{definition} \begin{definition}\label{de.5.4} The power series $\sum_{n=0}^\infty a(n) z^n \in\mathbb{C}[[x]]$ is called a $k$-$automatic$ $power$ $series$ if $(a(n))_{n=0}^\infty$ is a $k$-automatic sequence. \end{definition} \begin{definition}\label{de.5.5} An $(L,k,\kappa)$-TM sequence is called $y$-$periodic$ if there exist non-negative integers $N$ and $t(0w(q)\ge 0. \end{align} Let $l(t)\equiv l \;\pmod {t}$, where $0\le l(t) \le t-1$. By Lemma \ref{Le.2.3} with \eqref{eq3.11}, we have \begin{align}\label{eq5.8} a(k^{e+l} n)=a(\sum_{q=1}^{N(n)} s_q k^{w(q)+e+l} )= \prod_{q=1}^{N(n)}a(s_q k^{w(q)+e+l}). \end{align} From \eqref{eq5.7}, \eqref{eq5.8}, and Lemma \ref{Le.2.3} with \eqref{eq3.11}, we get \begin{align}\label{eq5.9} &a(k^{e+l} n)= \prod_{q=1}^{N(n)}a(s_q k^{w(q)+e+l})= \prod_{q=1}^{N(n)}a(s_q k^{{w(q)+e+l(t)} })\nonumber \\ &=a(\sum_{q=1}^{N(n)} s_q k^{w(q)+e+l(t)} )= a(k^{e+l(t)} n). \end{align} As $a(j)$ takes on only finitely many values, and by \eqref{eq5.6} and \eqref{eq5.9}, it follows that the $k$-kernel of $(a(n))_{n=0}^\infty$ is a finite set. \end{proof} \begin{theorem}\label{th.5.3} Let $(a(n))_{n=0}^\infty$ be an $(L,k,\kappa)$-TM and $\beta$ be an integer greater than $1$. If $(a(n))_{n=0}^\infty$ takes on the values $\{0,1,\ldots,\beta-1\}$, is $y$-periodic and there is no integer $A$ such that \begin{align}\label{eq5.11} \kappa(s,A+y) \equiv \kappa(1,A) s k^y \pmod{L} \end{align} for all $s$ with $1\le s\le k-1$ and for all $y \in \mathbb{N}$, then $\mu(\sum_{n=0}^\infty \frac{a(N+nl)}{ {\beta}^{n+1}})$ $($ for all $N \ge 0$ and for all $l>0$ $)$ is finite. \end{theorem} \begin{proof} By the previous proposition, $(a(n))_{n=0}^\infty$ is a $k$-automatic sequence. As the arithmetical subsequence of a $k$-automatic sequence is $k$-automatic, see Theorem $2.3$ and Theorem $2.6$ in Allouche and Shallit \cite{AllSha1}, and by Theorems \ref{th.4.2} and \ref{th.5.1}, $\mu(\sum_{n=0}^\infty \frac{a(N+nl)}{ {\beta}^{n+1}})$ is finite. \end{proof} \begin{theorem} Let $(a(n))_{n=0}^\infty$ be an $(L,k,\kappa)$-TM, $\beta$ be an integer greater than $1$, $f(z):= \sum_{n=0}^\infty \frac{a(N+nl)}{ z^{n+1}}$ $($ for all $N \ge 0$ and for all $l>0$ $)$, and $00$ $)$ is a $k$-automatic sequence. Therefore, $f(z)$ is a $k$-automatic power series. Theorem \ref{th.5.2} implies that $f(\alpha)$ is transcendental for all but finitely many algebraic numbers $\alpha$ with $|\alpha| \le R$. \end{proof} \section{Acknowledgments} I would like to thank Dr. Hajime Kaneko for useful advice and discussions, Prof. Kimio Ueno for his support, and the members of the geometric laboratories for their instructive comments. \begin{thebibliography}{99} \bibitem{AB1} B. 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Sci.} {\bf 4 }(2001), 357--362. \bibitem{MM}P. Morton and W. J. Mourant, Digit patterns and transcendental numbers, {\it J. Australian. Math. Soc} {\bf 51} (1991), 216--236. \bibitem{MSS}J. F. Morgenbesser, J. Shallit, and T. Stoll, Thue-Morse at multiples of an integer, {\it J. Number Theory} {\bf 131 } (2011), 1498--1512. \bibitem{Ni}K. Nishioka, \emph{Mahler Functions and Transcendence}, Lecture Notes in Mathematics, Springer, 1996. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A63; Secondary 11J99. \noindent \emph{Keywords: } digit counting, stammering sequence. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received August 9 2014; revised version received October 10 2014; April 1 2015; July 21 2015; July 31 2015. Published in {\it Journal of Integer Sequences}, July 31 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}