\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}}} \def\modd#1 #2{#1\ ({\rm mod}\ #2)} \newcommand{\lf}{\left\lfloor} \newcommand{\rf}{\right\rfloor} \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} \newtheorem*{notation*}{Notation} \begin{center} \vskip 1cm{\LARGE\bf Infinite Products Arising in Paperfolding } \vskip 1cm % first column \begin{minipage}[t]{0.5\textwidth} \begin{center} Leyda Almodovar \\ Department of Mathematics \\ University of Iowa\\ Iowa City, IA 52242 \\ USA \\ \href{mailto:leyda-almodovar@uiowa.edu}{\tt leyda-almodovar@uiowa.edu} \\ \ \\ Hadrian Quan \\ Department of Mathematics \\ UC Santa Cruz \\ Santa Cruz, CA 95064 \\ USA \\ \href{mailto:hquan1@ucsc.edu}{\tt hquan1@ucsc.edu} \\ \ \\ Eric Rowland\\ Department of Mathematics\\ University of Liege\\ Belgium \\ \href{mailto:rowland@lacim..edu}{\tt rowland@lacim.edu} \\ \end{center} \end{minipage} %second column \begin{minipage}[t]{0.4\textwidth} \begin{center} Victor H. Moll \\ Department of Mathematics \\ Tulane University \\ New Orleans, LA 70118 \\ USA\\ \href{mailto:vhm@tulane.edu}{\tt vhm@tulane.edu} \\ \ \\ Fernando Roman \\ Department of Mathematics \\ Kansas State University \\ Manhattan, KS 66506 \\ USA \\ \href{mailto:yahdiel@ksu.edu}{\tt yahdiel@ksu.edu} \\ \ \\ Michole Washington \\ Department of Mathematics \\ Georgia Institute of Technology \\ Atlanta, GA 30332 \\ USA \\ \href{mailto:mwashington9@gatech.edu}{\tt mwashington9@gatech.edu} \end{center} \end{minipage} \end{center} \vskip .2in \begin{abstract} J.-P.~Allouche recently examined two infinite products where the term is a rational function of the index $n$ raised to the term of the paperfolding sequence $\epsilon_{n}$. A closed form is given only for one of them. We discuss an attempt to produce the missing closed form. We give a detailed analysis of convergence and a closed form for the analogous question, where the paperfolding sequence is replaced by a periodic one. \end{abstract} \section{Introduction} The \textit{paperfolding sequence} $\epsilon_{n}$ is defined by the rules \begin{eqnarray} \epsilon_{2n} & = & (-1)^{n} \label{paper-fold-def0} \\ \epsilon_{2n+1} & = &\epsilon_{n}. \nonumber \end{eqnarray} The first few values are $\{ 1, \, 1, \, -1, \, 1, \, 1 \}$. For fixed $a \in \mathbb{N}$, the rules \eqref{paper-fold-def0} determine all subsequences of the form \begin{equation} \{\epsilon_{2^{a}n+b}: \, a \in \mathbb{N}, \, 0 \leq b < 2^{a} \} \end{equation} in terms of constants, $\{ \epsilon_{n} \}$ and $\{ (-1)^{n} \}$. For example, when $a=2$, \begin{equation} \epsilon_{4n} = 1, \, \epsilon_{4n+1} = \epsilon_{2n} = (-1)^{n}, \, \epsilon_{4n+2} = (-1)^{2n+1} = -1, \, \epsilon_{4n+3} = \epsilon_{2n+1} = \epsilon_{n}. \end{equation} The work presented here is motivated by results given by Allouche~\cite{allouche-2014a}. In particular, the evaluation \begin{equation} B = \prod_{n=1}^{\infty} \left( \frac{2n}{2n+1} \right)^{\epsilon_{n}} = \frac{1}{8 \sqrt{2 \pi}} \Gamma \left( \frac{1}{4} \right)^{2} \end{equation} is obtained using the auxiliary product \begin{equation} A = \prod_{n=0}^{\infty} \left( \frac{2n+1}{2n+2} \right)^{\epsilon_{n}}. \label{prod-A-11} \end{equation} Indeed, the identity \begin{equation} AB = \frac{1}{2} \prod_{n=1}^{\infty} \left( \frac{n}{n+1} \right)^{\epsilon_{n}} \end{equation} is split according to the parity of $n$ and \eqref{paper-fold-def0} yields \begin{equation} AB = \frac{1}{2} A \prod_{n=1}^{\infty} \left( \frac{2n}{2n+1} \right)^{(-1)^{n}}. \end{equation} The non-vanishing of $A$ gives \begin{equation} B = \frac{1}{2} \prod_{n=0}^{\infty} \frac{ (4n+4)(4n+3)}{(4n+5)(4n+2)}. \end{equation} A classical result expressing such products in terms of the gamma function gives the value of $B$. Observe that the value of $A$ does not come from this formulation. A search for a closed form for $A$ was the motivation for the results presented here. An early evaluation of an infinite product was produced by Wallis in his representation \begin{equation} \prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)} = \frac{\pi}{2}. \end{equation} The history of this discovery appears in Osler \cite{osler-2010b}. The literature contains a variety of infinite product evaluations. For instance, \begin{equation} \prod_{n=1}^{\infty} \left( 1 + \frac{1}{F_{2^{n}+1}} \right) = \frac{3}{\varphi} \text{ and } \prod_{n=1}^{\infty} \left( 1 + \frac{1}{L_{2^{n}+1}} \right) = 3 - \varphi, \end{equation} is given by Sondow~\cite{sondow-2011a}. Here $F_{n}, \, L_{n}$ are the Fibonacci (Lucas) numbers and $\varphi = \tfrac{1}{2}(\sqrt{5}+1)$ is the golden ratio. The value of infinite products usually involves classical constants of analysis. For instance, Borwein~\cite{borweinp-1993a} evaluates the function \begin{equation} D(x) = \lim\limits_{n \to \infty} \prod_{k=1}^{2n+1} \left( 1 + \frac{x}{k} \right)^{(-1)^{k+1}k} \end{equation} as a generalization of the values \begin{equation} \prod_{n=1}^{\infty} \left( 1 + \frac{2}{n} \right)^{(-1)^{n+1}n} = \frac{\pi}{2e} \quad \text{ and } \quad \prod_{n=1}^{\infty} \left( 1 + \frac{2}{n} \right)^{(-1)^{n}n} = \frac{6}{\pi e} \end{equation} established by Melzak~\cite{melzak-1961a}. Some exact evaluations are given in terms of the constant \begin{equation} A_{1} = \exp\!\left( \frac{1}{4} - \int_{0}^{\infty} \frac{e^{-s}}{s^{3}} \left( 1 - \frac{s}{2} + \frac{s^{2}}{12} - \frac{s}{e^{s}-1} \right) \, ds \right) \end{equation} and the Catalan constant \begin{equation} G = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}}. \end{equation} Examples include \begin{equation} D(1) = \frac{A_{1}^{6}}{2^{1/6} \sqrt{\pi}} \text{ and } D \left( \tfrac{1}{4} \right) = \frac{2^{1/6} \sqrt{\pi} A_{1}^{3}}{\Gamma \left( \tfrac{1}{4} \right)} e^{G/\pi}. \end{equation} Other types of products involving gamma factors have recently been analyzed by Chamberland and Straub~\cite{chamberland-2013a}. The question considered here deals with the evaluation of products of the form \begin{equation} \mathfrak{P}(R,s) = \prod_{n=1}^{\infty} R(n)^{s_{n}}. \label{gen-prod1} \end{equation} Here $R$ is a rational function and $s$ is an \textit{automatic sequence} (as studied by Allouche~\cite{allouche-2014a}). Examples include periodic sequences taking values in the alphabet $\{ +1, \, -1 \}$ or \textit{$k$-automatic sequences}: a sequence $\{ s_{n}: \, n \geq 0 \}$ is $k$-automatic if the set of subsequences $\{ s_{k^{j}n+\ell}: \, n \geq 0 \}$ with $j \geq 0, \, \ell \in [0, \, k^{j}-1 ]$ is finite. More information about such sequences appears in \cite{allouche-shallit-2003a}. The main example discussed here is the \textit{paperfolding sequence} $\epsilon_{n}$ defined in \eqref{paper-fold-def0}. Splitting the evaluation of a product into even and odd indices leads, in the special case of a rational function of degree $1$, to the identity \begin{equation} \prod_{n=0}^{\infty} \left( \frac{\alpha n+\beta}{\gamma n+\delta} \right)^{\epsilon_{n}} = \prod_{n=0}^{\infty} \left( \frac{2 \alpha n+\beta }{2 \gamma n+ \delta} \right)^{(-1)^{n}} \times \prod_{n=0}^{\infty} \left( \frac{2 \alpha n+ \alpha + \beta} {2 \gamma n+ \gamma + \delta} \right)^{\epsilon_{n}}. \label{folding-1} \end{equation} The exponent $(-1)^{n}$ appearing in the first product on the right is a periodic sequence of period length $2$. This motivates the evaluation of products with terms of the form $R(n)^{M_{n}}$ where $M_{n}$ is a periodic sequence. This is the topic of Sections~\ref{sec:convergence}--\ref{sec:con-period}. Section~\ref{sec:convergence} discusses the convergence of the product \begin{equation} \mathfrak{P}(R,1) = \prod_{n=0}^{\infty} R(n), \label{prod-M1} \end{equation} where \begin{equation} R(z) = \frac{(z+a_{1}) \cdots (z+a_{d})}{(z+b_{1}) \cdots (z+b_{d})}. \end{equation} This section reviews the elementary arguments showing that convergence in \eqref{prod-M1} is equivalent to $R(n) \to 1$ as $n \to \infty$ and $\mathfrak{S}(R) = 0$. Here \begin{equation} \mathfrak{S}(R) = \sum_{b \in R^{-1}(\infty)} b - \sum_{a \in R^{-1}(0)} a. \label{pzsum} \end{equation} The value of $\mathfrak{P}(R,1)$ is then given by \begin{equation} \prod_{n=0}^{\infty} \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{d})} = \prod_{k=1}^{d} \frac{\Gamma(b_{k})}{\Gamma(a_{k})}. \end{equation} Section~\ref{sec:simple} discusses the convergence of products $\mathfrak{P}(R,M)$, where $R$ is a rational function and $M$ is a periodic sequence of period length $2$. Section~\ref{sec:con-period} extends the results to any periodic sequence, with special emphasis on period lengths $3$ and $4$. Section~\ref{sec:paperfolding} considers some infinite products related to the paperfolding sequence, and Section~\ref{sec:automatic} considers a generalization to certain $k$-automatic sequences. An alternative proof of the evaluation of Allouche's product $B$ is presented and a new form of the product $A$ is given. The question of existence of a closed form for $A$ remains open. \section{Convergence of infinite products} \label{sec:convergence} This section considers the simplest type of product \eqref{gen-prod1}: $R$ is a given rational function and $s_{n} \equiv 1$. The data for the rational function is a sequence of complex numbers $\{ a_{k} \}$ and $\{b_{k} \}$ where $a_{k}, \, b_{k}$ are not $0$ nor a negative integer. The convergence of the partial finite products \begin{equation} \mathfrak{P}_{r}(R,1) = \prod_{n=1}^{r} \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{r})} \end{equation} is examined first. \begin{theorem} \label{thm-infprod} The infinite product \begin{equation} \mathfrak{P}(R,1) = \prod_{n=1}^{\infty} \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{r})} \end{equation} converges if and only if $d = r$ and $a_{1}+ \cdots + a_{d} = b_{1} + \cdots + b_{r}$; that is, $R(n) \to 1$ and $\mathfrak{S}(R) = 0$. \end{theorem} \begin{proof} The convergence of a product $\prod (1+ u_{k})$ is equivalent to the convergence of the series $\sum u_{k}$. Therefore $u_{k} \to 0$ is a necessary condition for convergence. This implies $d=r$. On the other hand \begin{equation} \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{r})} = 1 + \left( a_{1}+ \cdots + a_{d} - b_{1} - \cdots - b_{r} \right) \frac{1}{n} + O(1/n^{2}), \end{equation} and the second condition on the parameters $a_{k}, \, b_{k}$ is now clear. \end{proof} The next question is the evaluation of the limiting product. The motivation for the final result is this: consider the problem of producing a function $h(z)$ with zeros at a prescribed sequence $\{ z_{n} \}$. This is elementary if the sequence is finite: the solution is simply given as \begin{equation} P(z) = \prod_{n=1}^{N} \left( 1 - \frac{z}{z_{j}} \right) \end{equation} when $z_{j} \neq 0$. On the other hand, if the sequence is infinite, convergence issues might appear. For instance, if one would like to have a function that vanishes precisely at the negative integers, then the natural first attempt \begin{equation} P_{1}(z) = \prod_{n=1}^{\infty} \left( 1 + \frac{z}{n} \right) \end{equation} fails to converge. To fix this, introduce an exponential correction and form the partial products \begin{eqnarray} P_{2,N}(z) & = & e^{z \left(1+ \tfrac{1}{2} + \tfrac{1}{3} +\cdots + \tfrac{1}{N} \right) } \prod_{n=1}^{N} \left( 1 + \frac{z}{n} \right)e^{-z/n} \\ & = & e^{z \left(E_{1}(N) + \ln N \right) } \prod_{n=1}^{N} \left( 1 + \frac{z}{n} \right)e^{-z/n}, \nonumber \end{eqnarray} with \begin{equation} E_{1}(N) = 1 + \frac{1}{2} + \cdots + \frac{1}{N} - \ln N. \end{equation} The limit \begin{equation} \gamma = \lim\limits_{N \to \infty} E_{1}(N) \end{equation} is the famous \textit{Euler constant}. Therefore, the modified product \begin{equation} \frac{P_{2,N}(z)}{N^{z}} := e^{z E_{1}(N)} \prod_{n=1}^{N} \left( 1 + \frac{z}{n} \right)e^{-z/n} \end{equation} has a limit as $N \to \infty$. The infinite product has zeros at the negative integers. It turns out to be convenient to write an infinite product with poles at the negative integers and also to include $0$ as a pole. This yields the classical \textit{gamma function} $\Gamma(z)$. The functional equation $\Gamma(z+1) = z \Gamma(z)$ is used to simplify the result. \begin{theorem} The infinite product representation of the gamma function is given by \begin{equation} \prod_{n=1}^{\infty} \left(1 + \frac{z}{n} \right)^{-1} e^{z/n} = e^{\gamma z} \Gamma(z+1). \end{equation} \end{theorem} It is now easy to write the value of the infinite product \begin{equation} \mathfrak{P}(R,1) = \prod_{n=1}^{\infty} \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{r})} \label{product-0} \end{equation} in Theorem~\ref{thm-infprod}. Start with \begin{equation} \mathfrak{P}(R,1) = \prod_{n=1}^{\infty} \frac{(1+b_{1}/n)^{-1}e^{b_{1}/n} \cdots (1+b_{r}/n)^{-1} e^{b_{r}/n}} {(1+a_{1}/n)^{-1} e^{a_{1}/n}\cdots (1+a_{d}/n)^{-1} e^{a_{d}/n}} \label{product-1} \end{equation} and observe that the added exponential terms amount to $1$. Passing to the limit in \eqref{product-1} gives \begin{equation} \mathfrak{P}(R,1) = \prod_{k=1}^{d} \frac{\Gamma(b_{k}+1)}{\Gamma(a_{k}+1)}. \end{equation} To simplify the form of the result, shift $n$ to $n+1$ in \eqref{product-0} to produce the following result. \begin{theorem} \label{finite-gamma} Let $a_{k}, \, b_{k} \in \mathbb{C}$ none of which are $0$ or negative integers. Assume \begin{equation} a_{1} + \cdots + a_{d} = b_{1} + \cdots + b_{d}. \label{condition} \end{equation} Then \begin{equation} \prod_{n=0}^{\infty} \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{d})} = \prod_{k=1}^{d} \frac{\Gamma(b_{k})}{\Gamma(a_{k})}. \end{equation} \end{theorem} \section{The first example: Sequences of period length $2$} \label{sec:simple} This section considers products of the form \begin{equation} \mathfrak{P}(R,M) := \prod_{n=0}^{\infty} R(n)^{M_{n}} \end{equation} where $M_{n} = (-1)^{n}$. Start with the representation \begin{equation} \label{fact-R} R(z) = C \frac{(z+a_{1}) \cdots (z+a_{d})}{(z+b_{1}) \cdots (z+b_{r})}. \end{equation} The partial products of $\mathfrak{P}(R, s)$ are \begin{equation} \prod_{n=0}^{N} R(n)^{(-1)^{n}} = \prod_{n=0}^{\lf N/2 \rf} \frac{R(2n)}{R(2n+1)} \times \begin{cases} 1, & \text{if $N$ is odd;} \\ R(N+1), & \text{if $N$ is even}. \end{cases} \label{conv-pp} \end{equation} The first factor on the right in \eqref{conv-pp} is connected to the product $\mathfrak{P}(R_{1},1)$, where \begin{equation} R_{1}(z) = \frac{R(2z)}{R(2z+1)}. \label{prod-111} \end{equation} Its convergence is decided by Theorem~\ref{thm-infprod}. It is clear that the product on the left-hand side of \eqref{conv-pp} converges if and only if both factors on the right converge separately. In particular, if $\mathfrak{P}(R,M)$ converges, then $\begin{displaystyle} \lim\limits_{n \to \infty} R(n) = 1 \end{displaystyle}$ and it must be that $C=1$ in \eqref{fact-R}. To complete the discussion, it suffices to determine conditions under which $\mathfrak{P}(R_{1},1)$ is finite. The rational function \eqref{prod-111} factors as \begin{equation} R_{1}(z) = \frac{(2z+a_{1}) \cdots (2z+a_{d})}{(2z+b_{1}) \cdots (2z+b_{r})} \times \frac{(2z+1+b_{1}) \cdots (2z+1+b_{r})}{(2z+1+a_{1}) \cdots (2z+1+a_{d})}, \end{equation} with $d+r$ zeros at \begin{equation} -\tfrac{1}{2} a_{1}, \, \dots, -\tfrac{1}{2}a_{d}, \, -\tfrac{1}{2}(1+b_{1}), \dots, -\tfrac{1}{2}(1+ b_{r}) \end{equation} and $d+r$ poles at \begin{equation} -\tfrac{1}{2} b_{1}, \, \dots, -\tfrac{1}{2}b_{r}, \, -\tfrac{1}{2}(1+a_{1}), \dots, -\tfrac{1}{2}(1+ a_{d}). \end{equation} Since $R_{1}(z) \to 1$ as $z \to \infty$, convergence in \eqref{conv-pp} requires the relation \begin{equation} \sum_{k=1}^{d} a_{k} + \sum_{k=1}^{r} (1+ b_{k}) = \sum_{k=1}^{r} b_{k} + \sum_{k=1}^{d} (1+ a_{k}). \end{equation} This is equivalent to the condition $d=r$. The value of $\mathfrak{P}(R,M)$ is obtained from Theorem~\ref{finite-gamma} as \begin{equation} \mathfrak{P}(R,M) = \mathfrak{P}(R_{1},1) = \prod_{k=1}^{d} \frac{\Gamma(\frac{b_{k}}{2}) \Gamma(\frac{1 + a_{k}}{2}) }{\Gamma(\frac{1+b_{k}}{2}) \Gamma(\frac{a_{k}}{2})}. \end{equation} This is simplified using the duplication formula for the gamma function to obtain \begin{equation} \prod_{k=1}^{d} \frac{\Gamma(\frac{b_{k}}{2}) \Gamma(\frac{1 + a_{k}}{2}) }{\Gamma(\frac{1+b_{k}}{2}) \Gamma(\frac{a_{k}}{2})} = 2^{(b_{1}-a_{1}) + \cdots + (b_{d}-a_{d})} \prod_{k=1}^{d} \frac{\Gamma^{2}(\frac{b_{k}}{2}) \Gamma(a_{k})}{\Gamma^{2}(\frac{a_{k}}{2}) \Gamma(b_{k})}. \end{equation} The discussion above is summarized in the next statement. \begin{theorem} \label{exam-per2} Let $R(z)$ be a rational function and $M_{n} = (-1)^{n}$. Then $\mathfrak{P}(R,M)$ converges if and only if $R(z) \to 1$ as $z \to \infty$. If \begin{equation} R(z) = \prod_{k=1}^{d} \frac{(z+a_{k})}{(z+b_{k})} \text{ and } \mathfrak{S}(R) = \sum_{k=1}^{d} b_{k} - \sum_{k=1}^{d} a_{k}, \end{equation} then \begin{equation} \mathfrak{P}(R,M) = 2^{\mathfrak{S}(R)} \prod_{k=1}^{d} \frac{\Gamma^{2}(\frac{b_{k}}{2}) \Gamma(a_{k})}{\Gamma^{2}(\frac{a_{k}}{2}) \Gamma(b_{k})}. \end{equation} \end{theorem} \begin{example} Let $R(z) = (20z+5)/(20z+4)$. The convergence conditions are satisfied and Theorem~\ref{exam-per2} gives \begin{equation} \prod_{n=0}^{\infty} \left( \frac{20n+5}{20n+4} \right)^{(-1)^{n}} = \frac{\Gamma(\tfrac{1}{10}) \Gamma ( \tfrac{5}{8} )}{\Gamma( \tfrac{1}{8} ) \Gamma ( \tfrac{3}{5} )}. \label{example-00} \end{equation} \texttt{Mathematica} $9.0$ does not evaluate the original product, but it does give the right-hand side of \eqref{example-00} for \begin{equation} \mathfrak{P}(R_{1},1) = \prod_{n=0}^{\infty} \frac{80n^{2}+58n+6}{80n^{2}+58n+5}. \end{equation} \end{example} \begin{example} The infinite product \begin{equation} \mathfrak{P}(R,s) = \prod_{n=0}^{\infty} \left( \frac{2 \alpha n+\beta }{2 \gamma n+ \delta} \right)^{(-1)^{n}} \label{folding-2} \end{equation} encountered in the paperfolding product \eqref{folding-1} converges if and only if $\alpha = \gamma$. The product is then \begin{equation} \mathfrak{P}(R,s) = \prod_{n=0}^{\infty} \left( \frac{n+ 2v}{ n+ 2u} \right)^{(-1)^{n}} = 2^{2(u-v)} \frac{\Gamma^{2}(u) \Gamma(2v)}{\Gamma^{2}(v) \Gamma(2u)}, \label{folding-3} \end{equation} with $u = \delta/4 \alpha$ and $v = \beta/4 \alpha$. \end{example} \section{Convergence for periodic sequences} \label{sec:con-period} This section discusses the issue of convergence of the product \begin{equation} \mathfrak{P}(R,M) = \prod_{n=0}^{\infty} R(n)^{M_{n}} \end{equation} where $\{ M_{n} \}$ is a periodic sequence of period length $\ell$ of elements of the alphabet $\{ +1, \, -1 \}$. \begin{notation*} The results are expressed in terms of \begin{eqnarray} M^{+} & = & \{ i: \, M_{i} = +1 \text{ and } 0 \leq i \leq \ell -1 \} = \{ i_{1}, \, i_{2}, \dots, i_{|M^{+}|} \} \\ M^{-} & = & \{ j: \, M_{j} = -1 \text{ and } 0 \leq j \leq \ell -1 \} = \{ j_{1}, \, j_{2}, \dots, j_{|M^{-}|} \} , \nonumber \end{eqnarray} and the period length is $\ell = |M^{+}| + |M^{-}|$. \end{notation*} The rational function is written as \begin{equation} R(n) = C\frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{r})} \label{rational-fact0} \end{equation} with $a_{s}, \, b_{t} \not \in \{ 0, \, -1, \, 2, \dots \}$ and \begin{equation} \mathfrak{S}(R) = \sum_{t=1}^{r} b_{t} - \sum_{s=1}^{d} a_{s}. \end{equation} The partial product associated with $\mathfrak{P}(R,M)$ is \begin{eqnarray} \prod_{n=0}^{N} R(n)^{M_{n}} & = & \prod_{k=0}^{\lf N/\ell \rf} \prod_{i \in M^{+}} R(k \ell + i)^{M_{i}} \prod_{j \in M^{-}} R(k \ell + j)^{M_{j} } \prod_{n= \ell \lf N/\ell \rf +1 }^{N} R(n)^{M_{n}} \nonumber \\ & = & \prod_{k=0}^{\lf N/\ell \rf } \frac{\prod_{i \in M^{+}} R(k \ell + i)} {\prod_{j \in M^{-}} R(k \ell + j)} \prod_{n= \ell \lf N/\ell \rf + 1}^{N} R(n)^{M_{n}}, \label{prod-conv-11} \end{eqnarray} the last product being empty if $N$ is a multiple of the period length $\ell$. An elementary argument shows that the convergence of $\mathfrak{P}(R,M)$ requires the convergence of both products in \eqref{prod-conv-11}. The first product, which would lead to an expression of the form $\mathfrak{P}(R_{1},1)$ for a new rational function $R_{1}$ is labeled the \textit{main term}. The second product is called the \textit{tail product}. We analyze its convergence first. The tail product is defined by \begin{equation} P_{N,\ell}(M) = \prod_{n= \ell \lf N/\ell \rf + 1}^{N} R(n)^{M_{n}}. \label{prod-conv-12} \end{equation} Its convergence implies $R(n) \to 1$ as $n \to \infty$. Observe that $P_{N,\ell}(M) = 1$ if $N \equiv \modd{0} {\ell}$. On the other hand, in the case $N \equiv \modd{1} {\ell}$, one obtains $$P_{N,\ell}(M) = R(N)^{M_{N}} = R(N)^{M_{1}},$$ since $M_{N} = M_{1}$ by periodicity. Therefore, the convergence of $\mathfrak{P}(R,M)$ requires $R(N) \to 1$ for $N \equiv \modd{1} {\ell}$. Similarly, if $N \equiv \modd{2} {\ell}$, $$P_{N,\ell}(M) = R(N-1)^{M_{N-1}} R(N)^{M_{N}} = R(N-1)^{M_{1}}R(N)^{M_{2}}.$$ The convergence of $\mathfrak{P}(R,M)$ already implies $R(N-1) \to 1$ since $N-1 \equiv \modd{1} {\ell}$. This time it is required that $R(N) \to 1$. Iterating this argument it follows that $R(N) \to 1$ for $N \equiv \modd{j} {\ell}$ for any residue class $j$. This gives the next result. \begin{proposition} Assume $\mathfrak{P}(R,M)$ converges. Then $\begin{displaystyle} \lim\limits_{n \to \infty} R(n) = 1. \end{displaystyle}$ \end{proposition} The limiting value of the main term is $\mathfrak{P}(R_{1},1)$, where \begin{equation} R_{1}(n) = \frac{R(\ell n + i_{1}) \cdots R(\ell n + i_{|M^{+}|})} {R(\ell n + j_{1}) \cdots R(\ell n + j_{|M^{-}|})}. \end{equation} The ingredients entering into the convergence of $\mathfrak{P}(R_{1},1)$ are discussed in the next result. We assume the condition $R(n) \to 1$. \begin{proposition} \label{convergence1} Let $M_{*} = |M^{+}| - |M^{-}|$ and assume $\mathfrak{P}(R_{1},1)$ converges. Then \newline $\begin{displaystyle} \lim\limits_{n \to \infty} R_{1}(n) = 1 \end{displaystyle}$ and \begin{equation} \ell \mathfrak{S}(R_{1}) = M_{*} \mathfrak{S}(R). \label{zero-pole} \end{equation} \end{proposition} \begin{proof} The behavior of $R_{1}(n)$ as $n \to \infty$ comes directly from that of $R$. The identity \eqref{zero-pole} is a direct computation. \end{proof} Combining these propositions gives the following. \begin{theorem} \label{conv-R1} Let $R$ be a rational function satisfying $\begin{displaystyle} \lim\limits_{n \to \infty} R(n) = 1 \end{displaystyle}$ with zeros and poles of $R$ are outside $\{ 0, -1, -2, \dots \}$. There are two cases. \begin{enumerate} \item Assume $M_{*} \neq 0$. Then $\mathfrak{P}(R,M)$ converges if and only if $\mathfrak{S}(R) = 0$. \item Assume $M_{*} = 0$. Then $\mathfrak{P}(R,M)$ always converges. \end{enumerate} \end{theorem} For a general periodic sequence, the value of the product $\mathfrak{P}(R,M)$ is given by the following. \begin{theorem} \label{prod-form-0} Let $R(n)$ be a rational function written in the form \begin{equation} R(n) = \frac{(n+a_{1}) \cdots (n+a_{d})}{(n+b_{1}) \cdots (n+b_{d})} \label{rational-fact} \end{equation} with $a_{i}, b_{j} \not \in \{ 0, \, -1, \, -2, \dots \}$. Let $\{ M_{n} \}$ be a periodic sequence of $\pm 1$ with period length $\ell$. Assume the product \begin{equation} \mathfrak{P}(R,M) = \prod_{n=0}^{\infty} R(n)^{M_{n} } \end{equation} converges. Then \begin{equation} \label{ClosedFormPeriodicF} \mathfrak{P}(R,M) =\ell^{\mathfrak{S}(R)}\prod_{1\leq s \leq d} \frac{\Gamma(a_s)}{\Gamma(b_s)} \prod_{\substack{ i \in M^{+}}} \frac{\Gamma^2\left(\frac{b_s+i}{\ell}\right)}{\Gamma^2\left(\frac{a_s+i}{\ell}\right)}. \end{equation} \end{theorem} \begin{proof} Splitting the product according to its residues modulo $\ell$ gives \begin{eqnarray*} \prod_{n=1}^{\infty} R(n)^{M_{n}} & = & \prod_{n=0}^{\infty} \prod_{\stackrel{i \in M^{+}}{j \in M^{-}}} \frac{R(\ell n + i)}{R(\ell n + j)} \\ & = & \prod_{\stackrel{i \in M^{+}}{j \in M^{-}}} \prod_{n=0}^{\infty} \frac{ \left( n + \frac{a_{1}+i}{\ell} \right) \cdots \left( n + \frac{a_{d}+i}{\ell} \right) \left( n + \frac{b_{1}+j}{\ell} \right) \cdots \left( n + \frac{b_{d}+j}{\ell} \right) } { \left( n + \frac{b_{1}+i}{\ell} \right) \cdots \left( n + \frac{b_{d}+i}{\ell} \right) \left( n + \frac{a_{1}+j}{\ell} \right) \cdots \left( n + \frac{a_{d}+j}{\ell} \right) }. \end{eqnarray*} The products may be expressed in terms of the gamma function to obtain \begin{equation} \label{formula-gamma1} \prod_{n=1}^{\infty} R(n)^{M_{n}} = \prod_{\stackrel{i \in M^{+}}{j \in M^{-}}} \frac{ \Gamma \left( \frac{b_{1}+i}{\ell} \right) \cdots \Gamma \left( \frac{b_{d}+i}{\ell} \right) \Gamma \left( \frac{a_{1}+j}{\ell} \right) \cdots \Gamma \left( \frac{a_{d}+j}{\ell} \right) }{ \Gamma \left( \frac{a_{1}+i}{\ell} \right) \cdots \Gamma \left( \frac{a_{d}+i}{\ell} \right) \Gamma \left( \frac{b_{1}+j}{\ell} \right) \cdots \Gamma \left( \frac{b_{d}+j}{\ell} \right) } \end{equation} and the result is simplified using Gauss' multiplication formula \begin{equation} (2 \pi)^{\tfrac{\ell-1}{2}} \ell^{\tfrac{1}{2} - \ell z} \Gamma( \ell z) = \prod_{j=0}^{\ell-1} \Gamma \left( z + \frac{j}{\ell} \right). \end{equation} Take $z = a_{s}/\ell$ to produce \begin{eqnarray*} (2 \pi)^{\tfrac{\ell-1}{2}} \ell^{ 1/2 - a_{s}} \Gamma(a_{s}) & = & \Gamma \left( \frac{a_{s}}{\ell} \right) \Gamma \left( \frac{a_{s}+1}{\ell} \right) \cdots \Gamma \left( \frac{a_{s}+\ell-1}{\ell} \right) \\ & = & \Gamma \left( \frac{a_{s}+i_{1}}{\ell} \right) \cdots \Gamma \left( \frac{a_{s}+ i_{|M^{+}|}}{\ell} \right) \Gamma \left( \frac{a_{s}+j_{1}}{\ell} \right) \cdots \Gamma \left( \frac{a_{s}+ j_{|M^{+}|}}{\ell} \right) \end{eqnarray*} since every residue modulo $\ell$ appears exactly once in the sets $M^{+}$ and $M^{-}$. It follows that \begin{equation} \prod_{j \in M^{-}} \Gamma \left( \frac{a_{s}+j}{\ell} \right) = \frac{(2 \pi)^{(\ell-1)/2}\ell^{1/2 - a_{s}} \Gamma(a_{s})} {\prod_{i \in M^{+}} \Gamma \left( \frac{a_{s}+i}{\ell} \right)}, \end{equation} for $1 \leq s \leq d$. A similar result holds for $b_{s}$. Replacing in \eqref{formula-gamma1} concludes the proof. \end{proof} \begin{example} Consider the sequence $\overline{ \{ 1,-1,-1 \} }$, where the bar indicates the fundamental period; that is, \begin{equation} M_{n} = \begin{cases} \phantom{-}1, & \text{if $n \equiv \modd{0} {3}$;} \\ -1, & \text{if $n \equiv \modd{1, 2} {3}$.} \end{cases} \end{equation} Therefore $M^{+} = \{ 0 \}, \, M^{-} = \{ 1, \, 2 \}$ so that $M_{*} = -1$. Theorem~\ref{conv-R1} states that the convergence of $\mathfrak{P}(R_{1},1)$ is equivalent to $\mathfrak{S}(R) = 0$. Take $\begin{displaystyle} R(z) = \frac{(z+1)(z+3)}{(z+2)^{2}} \end{displaystyle}$. The conditions for convergence of $\mathfrak{P}(R,M)$ are satisfied, and its value is \begin{equation} \prod_{n=0}^{\infty} \left( \frac{(n+1)(n+3)}{(n+2)^{2}} \right)^{M_{n}} = \frac{\Gamma\left(1\right) \Gamma^2\left(\frac{2}{3}\right)}{\Gamma\left(2\right) \Gamma^2\left(\frac{1}{3}\right)} \frac{\Gamma\left(3\right) \Gamma^2\left(\frac{2}{3}\right)}{\Gamma\left(2\right) \Gamma^2\left(\frac{3}{3}\right)} = 2 \cdot \frac{\Gamma^4\left(\frac{2}{3}\right)}{\Gamma^2\left(\frac{1}{3}\right)} = \frac{3}{2 \pi^2} \Gamma^6\left(\tfrac{2}{3}\right). \end{equation} by Theorem~\ref{prod-form-0}. \end{example} \begin{example} \label{example-4b} Let $\displaystyle{R(z) = \frac{(z+2)(z+3)}{(z+1)(z+4)}}$ and $M = \overline{ \{ 1, \, 1, \, 1, \, -1 \}}$. Then $M^{+} = \{ 0, \, 1, \, 2\}$ and $M^{-} = \{ 3 \}$. Thus $M_{*} \neq 0$. The product $\mathfrak{P}(R,M)$ converges by Theorem~\ref{conv-R1}, and Theorem~\ref{prod-form-0} gives \begin{equation} \prod_{n=0}^{\infty} \left( \frac{(n+2)(n+3)}{(n+1)(n+4)} \right)^{M_{n}} = \frac{1}{24 \pi} \Gamma^{4} \left( \tfrac{1}{4} \right). \end{equation} \end{example} \section{The paperfolding sequence} \label{sec:paperfolding} The paperfolding sequence is defined by the rules \begin{equation} \epsilon_{2n} = (-1)^{n} \text{ and } \epsilon_{2n+1} = \epsilon_{n}. \label{rules-1} \end{equation} Allouche~\cite{allouche-2014a} considered the products \begin{equation} A = \prod_{n=0}^{\infty} \left( \frac{2n+1}{2n+2} \right)^{\epsilon_{n}} \label{value-A} \text{ and } B = \prod_{n=1}^{\infty} \left( \frac{2n}{2n+1} \right)^{\epsilon_{n}}, \end{equation} and proved \begin{equation} B = \frac{\Gamma \left( \tfrac{1}{4} \right)^{2}}{8 \sqrt{2 \pi}}. \label{value-B} \end{equation} The \textit{closed-form} evaluation of $A$ remains an open problem. The goal of this section is to present a new proof of \eqref{value-B} and to present an alternative product expression for $A$. Observe that \begin{equation} \prod_{n=0}^{\infty} \left( \frac{an+b}{cn+d} \right)^{\epsilon_{n}} = \prod_{n=0}^{\infty} \left( \frac{2an + b}{2cn+d} \right)^{(-1)^{n}} \times \prod_{n=0}^{\infty} \left( \frac{2an + (a+b)}{2cn+(c+d)} \right)^{\epsilon_{n}}. \label{prod-1a} \end{equation} The convergence of the first product requires $a=c$ and its value has been obtained in Theorem~\ref{exam-per2} as \begin{equation} \prod_{n=0}^{\infty} \left( \frac{2an + b}{2cn+d} \right)^{(-1)^{n}} = 2^{d/2c-b/2a} \frac{\Gamma^{2} \left( \frac{d}{4c} \right) \Gamma \left( \frac{b}{2a} \right)} {\Gamma^{2} \left( \frac{b}{4a} \right) \Gamma \left( \frac{d}{2c} \right) }. \end{equation} Iterating this procedure converts the second factor in \eqref{prod-1a} into \begin{equation} \label{form-56} \prod_{n=0}^{\infty} \left( \frac{2an + (a+b)}{2cn+(c+d)} \right)^{\epsilon_{n}} = \prod_{n=0}^{\infty} \left( \frac{4an + (a+b)}{4cn+(c+d)} \right)^{(-1)^{n}} \times \prod_{n=0}^{\infty} \left( \frac{4an + (3a+b)}{4cn+(3c+d)} \right)^{\epsilon_{n}}. \end{equation} The first product on the right-hand side of \eqref{form-56} converges and Theorem~\ref{exam-per2} gives \begin{equation} \label{period-100a} \prod_{n=0}^{\infty} \left( \frac{4an + (a+b)b}{4cn+(c+d)} \right)^{(-1)^{n}} = 2^{d/4c - b/4a} \frac{\Gamma^{2} \left( \frac{c+d}{8c} \right) \Gamma \left( \frac{a+b}{4a} \right)} {\Gamma^{2} \left( \frac{a+b}{8a} \right) \Gamma \left( \frac{c+d}{4c} \right) }. \end{equation} Now observe that \begin{equation} \frac{c+d}{8c} = \frac{1}{4} + \frac{d-c}{8c} \end{equation} so \eqref{period-100a} can be written as \begin{equation} \label{period-100b} \prod_{n=0}^{\infty} \left( \frac{4an + (a+b)b}{4cn+(c+d)} \right)^{(-1)^{n}} = 2^{d/4c - b/4a} \frac{\Gamma^{2} \left( \frac{1}{4} + \frac{d-c}{8c} \right) \Gamma \left(\frac{1}{2} + \frac{b-a}{2a} \right)} {\Gamma^{2} \left( \frac{1}{4} + \frac{b-a}{8a} \right) \Gamma \left( \frac{1}{2} + \frac{d-c}{4c} \right) }. \end{equation} Repeated application of this process gives \begin{multline} \prod_{n=0}^{\infty} \left( \frac{an+b}{cn+d} \right)^{\epsilon_{n}} = 2^{(d/c-b/a) \sum_{k=1}^{N} 1/2^{k}} \times {}\\ \prod_{k=2}^{N} \frac{ \Gamma^{2} \left( \frac{1}{4} + \frac{d-c}{c2^{k}} \right) \Gamma \left( \frac{1}{2} + \frac{b-a}{a 2^{k-1}} \right) } { \Gamma^{2} \left( \frac{1}{4} + \frac{b-a}{a2^{k}} \right) \Gamma \left( \frac{1}{2} + \frac{d-c}{c 2^{k-1}} \right) } \times {}\\ \prod_{n=0}^{\infty} \left( \frac{ 2^{N} an + a(2^{N}-1) + b}{2^{N} cn + c(2^{N}-1) + d} \right)^{\epsilon_{n}}. \end{multline} A direct argument shows that the last product converges to $1$ when $N \to \infty$. This completes the proof of the next statement. \begin{theorem} \label{nice-one} The infinite product associated with the paperfolding sequence is given by \begin{equation} \prod_{n=0}^{\infty} \left( \frac{an+b}{cn+d} \right)^{\epsilon_{n}} = 2^{(d/c-b/a)} \prod_{k=2}^{\infty} \frac{ \Gamma^{2} \left( \frac{1}{4} + \frac{d-c}{c2^{k}} \right) \Gamma \left( \frac{1}{2} + \frac{b-a}{a 2^{k-1}} \right) } { \Gamma^{2} \left( \frac{1}{4} + \frac{b-a}{a2^{k}} \right) \Gamma \left( \frac{1}{2} + \frac{d-c}{c 2^{k-1}} \right) }. \end{equation} \end{theorem} The product appearing in Theorem~\ref{nice-one} does not seem to admit a simple closed form for general choice of the parameters $a, \, b, \, d$ (recall that $a=c$ is required for the convergence of the product). Such a closed form is obtained in the special situation where the factors telescope. This occurs when $2d = a+b$. The next corollary (equivalent to a theorem of Allouche~\cite[Theorem~1]{allouche-2014a}) gives such a closed form, with $\alpha = d/a$. In that situation \begin{equation} \prod_{k=2}^{N} \frac{ \Gamma^{2} \left( \frac{1}{4} + \frac{d-c}{c2^{k}} \right) } { \Gamma^{2} \left( \frac{1}{4} + \frac{b-a}{a2^{k}} \right) } \to \frac{\Gamma^{2}(\tfrac{1}{4})}{\Gamma^{2}( \frac{\alpha}{2} - \frac{1}{4} )} \end{equation} and \begin{equation} \prod_{k=2}^{N} \frac{ \Gamma \left( \frac{1}{2} + \frac{b-a}{a2^{k-1}} \right) } { \Gamma \left( \frac{1}{2} + \frac{d-c}{a2^{k-1}} \right) } \to \frac{ \Gamma( \alpha -\tfrac{1}{2})}{\Gamma( \tfrac{1}{2})}. \end{equation} \begin{corollary} A special case of the paperfolding product is given by \begin{equation} \prod_{n=0}^{\infty} \left( \frac{n+ 2 \alpha -1}{n+ \alpha} \right)^{\epsilon_{n}} = 2^{1-\alpha} \, \frac{\Gamma^{2} \left( \tfrac{1}{4} \right)}{\Gamma \left( \tfrac{1}{2} \right)} \frac{\Gamma \left( \alpha - \tfrac{1}{2} \right) }{\Gamma^{2} \left( \frac{\alpha}{2} - \frac{1}{4} \right) }. \label{cor-11} \end{equation} \end{corollary} \begin{example} Take $\alpha = 3$ to obtain \begin{equation} \prod_{n=0}^{\infty} \left( \frac{n+ 5}{n+3} \right)^{\epsilon_{n}} = 3. \end{equation} \end{example} \begin{example} The infinite product $B$ in \eqref{value-A} comes by taking the limit as $\alpha \to \tfrac{1}{2}$. Indeed, write \eqref{cor-11} as \begin{equation} \prod_{n=1}^{\infty} \left( \frac{n+ 2 \alpha -1}{n+ \alpha} \right)^{\epsilon_{n}} = \frac{\alpha}{ \alpha-\tfrac{1}{2}} \, \frac{\Gamma^{2} \left( \tfrac{1}{4} \right)}{2^{\alpha} \Gamma \left( \tfrac{1}{2} \right)} \frac{\Gamma \left( \alpha - \tfrac{1}{2} \right) }{\Gamma^{2} \left( \frac{\alpha}{2} - \frac{1}{4} \right) }. \label{cor-11a} \end{equation} The limit \begin{equation} \lim\limits_{x \to 0} \frac{\Gamma(x)}{x \Gamma^{2}(x/2)} = \frac{1}{4} \end{equation} gives \begin{equation} \prod_{n=1}^{\infty} \left( \frac{2n}{2n+1} \right)^{\epsilon_{n}} = \frac{\Gamma^{2} \left( \tfrac{1}{4} \right)}{8 \sqrt{2 \pi}}, \end{equation} confirming \eqref{value-B}. \end{example} \begin{example} The method described above does not produce a closed form for the product $A$ in \eqref{value-A}. A direct use of the expression in Theorem~\ref{nice-one} gives \begin{equation} A = \prod_{n=0}^{\infty} \left( \frac{2n+1}{2n+2} \right)^{\epsilon_{n}} = \sqrt{2} \prod_{k=2}^{\infty} \left( \frac{ \Gamma \left( \tfrac{1}{4} \right) }{\Gamma \left( \tfrac{1}{4} - \frac{1}{2^{k+1}} \right) } \right)^{2} \times \frac{\Gamma \left( \tfrac{1}{2} - \frac{1}{2^{k}} \right) }{\Gamma \left( \tfrac{1}{2} \right)}. \label{formula-A1} \end{equation} Iterating the duplication formula for the gamma function yields the so-called Knar formula \cite[volume 1, page 6, formula 6]{erderly-1953a} \begin{equation} \Gamma(1+z) = 2^{2z} \prod_{k=1}^{\infty} \frac{1}{\sqrt{\pi}} \Gamma \left( \frac{1}{2} + \frac{z}{2^{k}} \right) \end{equation} and $z = - \tfrac{1}{2}$ gives \begin{equation} \prod_{k=2}^{\infty} \frac{\Gamma \left( \tfrac{1}{2} - \tfrac{1}{2^{k}} \right)}{\Gamma \left( \tfrac{1}{2} \right)} = 2 \sqrt{\pi}. \end{equation} Then \eqref{formula-A1} becomes \begin{equation} A = \prod_{n=0}^{\infty} \left( \frac{2n+1}{2n+2} \right)^{\epsilon_{n}} = 2 \sqrt{2 \pi} \prod_{k=3}^{\infty} \left( \frac{ \Gamma \left( \tfrac{1}{4} \right) }{\Gamma \left( \tfrac{1}{4} - \frac{1}{2^{k}} \right) } \right)^{2}. \end{equation} The authors have been unable to reduce this any further. \end{example} \section{Generalization to certain $k$-automatic sequences} \label{sec:automatic} This section extends the results on the paperfolding sequence to certain $k$-automatic sequences. As usual, let $R(z)$ be a rational function written in the form \begin{equation} R(z) = \frac{(z+a_{1}) \cdots (z+ a_{d}) }{(z+b_{1}) \cdots (z+b_{d}) } \end{equation} and assume that $a_{i}$ and $b_{j}$ are not in $\{ 0, \, -1, \, -2, \dots \}$. Consider the case in which $M_{n}$ is a $3$-automatic sequence defined by the rules \begin{eqnarray} M_{3n} & = & q_{0}(n), \label{3-automatic} \\ M_{3n+1} & = & q_{1}(n), \nonumber \\ M_{3n+2} & = & M_{n}, \nonumber \end{eqnarray} where $q_{j}$ takes values in $\{ +1, \, -1 \}$ and $q_{j}(n)$ is periodic of period length $\ell_{j}$. Now split the product according to residues modulo $3$ to produce \begin{eqnarray*} \prod_{n=0}^{\infty} R(n)^{M_{n}} & = & \prod_{n=0}^{\infty} R(3n)^{M_{3n}} \times \prod_{n=0}^{\infty} R(3n+1)^{M_{3n+1}} \times \prod_{n=0}^{\infty} R(3n+2)^{M_{3n+2} } \\ & = & \prod_{n=0}^{\infty} R(3n)^{q_{0}(n)} \times \prod_{n=0}^{\infty} R(3n+1)^{q_{1}(n)} \times \prod_{n=0}^{\infty} R(3n+2)^{M_{n}}. \end{eqnarray*} The convergence and values of the first two products are provided by Theorem~\ref{conv-R1} and Theorem~\ref{prod-form-0}. Assume the convergence of the product \begin{equation} \mathbb{P}_{0} = \prod_{n=0}^{\infty} R(3n)^{q_{0}(n)}. \end{equation} Theorem~\ref{conv-R1} shows that this happens if $|q_{0}|= 0$, where $|q_{0}|$ is the number of $+1$ minus the number of $-1$ in one period. In the remaining case, it is required that $\mathfrak{S}(R(3z)) = 0$, where $\mathfrak{S}(R)$ is defined in \eqref{pzsum}. The exact form of the product is obtained from Theorem~\ref{prod-form-0} which yields, with $R_{0}(z) = R(3z)$, \begin{equation} \label{ClosedFormPeriodicP0} \mathbb{P}_{0} = \mathfrak{P}(R_{0},q_{0}) =\ell_{0}^{\mathfrak{S}(R_{0})}\prod_{1\leq s \leq d} \frac{\Gamma(a_s/3)}{\Gamma(b_s/3)} \prod_{\substack{ i \in q_{0}^{+}}} \frac{\Gamma^2\left(\frac{b_s + 3i}{3\ell_{0}}\right)}{\Gamma^2\left(\frac{a_s+3i}{3\ell_{0}}\right)}. \end{equation} A similar process gives an analytic formula for the second product. Repeating the previous process yields a decomposition of the third product as \begin{eqnarray*} \prod_{n=0}^{\infty} R(n)^{M_{n}} & = & \prod_{n=0}^{\infty} R(9n+2)^{q_{0}(n)} \times \prod_{n=0}^{\infty} R(9n+5)^{q_{1}(n)} \times \prod_{n=0}^{\infty} R(9n+8)^{M_{n}}. \end{eqnarray*} As before, the first two products have an explicit analytic expression and the last one has to be split again. This process can be iterated to obtain a formula for the original product. For simplicity, the results are given for $R(z)$ a rational function of degree $1$ and only in the case in which all the periodic pieces $q_{i}(n)$ have a period length that is a power of a fixed even integer. In this situation, the final formula can be simplified. \begin{theorem}\label{closedformautomatic} Let $R(z)=\dfrac{z+b}{z+d}$, with $b, \, d\in \mathbb{R}^{+}$ and let $M_n$ be a $k$-automatic sequence satisfying the rules \begin{align*} M_{kn}&=q_0(n) \\ M_{kn+1}&=q_1(n) \\ &\,\,\,\vdots \\ M_{kn+k-2}&=q_{k-2}(n) \\ M_{kn+k-1}&=M_n. \end{align*} Assume there is an even integer $L$ such that each sequence $q_i(n)$ is a periodic sequence of period length $L_i = L^{\alpha_i}$ some power of $L$. In addition, assume that $|q_i^+|=|q_i^-|$ for all $0\leq i\leq k-2$. Then \begin{equation} \mathfrak{P}(R,M) = \prod_{n=0}^\infty R(n)^{M_n} \label{nice-prod} \end{equation} converges. Moreover, if $d=\dfrac{b+k-1}{k}$ the product in \eqref{nice-prod} can be evaluated as \begin{equation} \label{closedformautomatic2} \prod_{n=0}^\infty R(n)^{M_n}= \prod_{i=0}^{k-2}\left(L_{i}^{\frac{1-b}{k}} \frac{\Gamma(\frac{b+i}{k})}{\Gamma(\frac{i+1}{k})} \prod_{j\in q_i^+}\dfrac{\Gamma^2\left(\frac{i+1}{L_{i}k} +\frac{j}{L_{i} }\right)}{\Gamma^2\left(\frac{b+i}{L_{i} k} +\frac{j}{L_{i}}\right)}\right). \end{equation} \end{theorem} Note that the paperfolding sequence satisfies the hypothesis of the theorem. In this case $k=2$ and $q_0(n)=(-1)^n$, and $L = 2$. The rational function is $$R(n)=\dfrac{n+b}{n+\frac{b+1}{2}}$$ and \eqref{closedformautomatic2} reduces to the result of Allouche. The idea of the proof is the argument presented in the case of the $3$-automatic sequence above. Complete details may be found in \cite{quan-2014a}. \section{Acknowledgments} The second author acknowledges the partial support of NSF-DMS 1112656. This work was carried out during the $2014$ Mathematical Sciences Research Institute Undergraduate Program (MSRI-UP) which is funded by the National Science Foundation (grant No.~DMS-1156499) and the National Security Agency (grant No.~H-98230-13-1-0262). H.~Quan, F.~Roman and M.~Washington were undergraduates at this program, L.~Almodovar was a graduate assistant and E.~Rowland was a postdoctoral advisor at the MSRI-UP program. \begin{thebibliography}{1} \bibitem{allouche-2014a} J.-P. Allouche, Paperfolding infinite products and the gamma function, \textit{J. Number Theory} \textbf{148} (2015), 95--111. \bibitem{allouche-shallit-2003a} J.-P. Allouche and J.~Shallit, \textit{Automatic Sequences: Theory, Applications, Generalizations}, Cambridge University Press, 2003. \bibitem{borweinp-1993a} P.~Borwein and W.~Dykshoorn, An interesting infinite product, \textit{J. Math. Anal. 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Gaz.} \textbf{94} (2010), 430--437. \bibitem{quan-2014a} H.~Quan, F.~Roman, and M.~Washington, \href{http://www.msri.org/system/cms/files/81/files/original/Research_Reports_2014_MSRI-UP_(Single_File).pdf#page=63}{Infinite products and periodic sequences}, Technical report, MSRI-UP, Berkeley, California, 2014, pp.~60--90. \bibitem{sondow-2011a} J.~Sondow, Evaluation of Tachiya's algebraic infinite products involving Fibonacci and Lucas numbers, in \textit{Diophantine Analysis and Related Fields: Proc. DARF 2011}, American Institute of Physics, 2011, pp.~97--100. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B85; Secondary 33B15, 40A20. \noindent \emph{Keywords: } infinite product, gamma function, paperfolding sequence. \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A034947}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received May 29 2015; revised version received April 21 2016. Published in {\it Journal of Integer Sequences}, May 11 2016. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}