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\begin{center}
    \vskip 1cm{\LARGE\bf Continued Fractions with Predictable\\
    \vskip .12in
    Patterns and Transcendental Numbers
    }
    \vskip 1cm
    \Large
Khalil Ayadi\footnote{Corresponding author.}, Chiheb Ben Bechir, and Maher Saadaoui\\
Department of Mathematics\\
Faculty of Sciences\\
Research Laboratory AGTS\\
Sfax University\\
Tunisia\\
\href{mailto:khalil.ayadi@isgis.usf.tn}{\tt khalil.ayadi@isgis.usf.tn}\\
\href{mailto:chihebbenbechir@gmail.com}{\tt chihebbenbechir@gmail.com}\\
\href{mailto:mahersaadaoui716@gmail.com}{\tt mahersaadaoui716@gmail.com}\\
\end{center}

\vskip .2 in
\begin{abstract} 
The continued fractions that are the subject of this paper display a particular kind of ``self-similar'' structure: that is, their partial quotients are created by manipulating and moving the denominators of their convergents. An alternating series characterizes the real numbers that these continued fractions represent. Using an application of Roth's theorem, we prove the transcendence of these numbers.
\end{abstract}


\section{Introduction}\label{sec1}

Let $x\in \mathbb{R}$. We write the continued fraction of $x$ as
\begin{equation}\label{x}\displaystyle x=[a_{0}, a_{1},\ldots]=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\ddots}}},\end{equation}
where the partial quotients $a_{n}$ are integers, positive for $n&gt;0$. 
We have 
\begin{equation*}x=\displaystyle\lim_{n\rightarrow \infty}[a_{0}, a_{1},\ldots,a_{n}],\end{equation*}
where
\begin{equation*}[a_{0}, a_{1},\ldots,a_{n}]=a_{0}+\frac{1}{a_{1}+\frac{1}{a_{2}+\frac{1}{\ddots+\frac{1}{a_{n}}}}}.\end{equation*}
Define $p_{n}$ and $q_{n}$ by
\begin{equation}\label{H}\begin{cases}
 p_{n}=a_{n}p_{n-1}+p_{n-2}, &amp; \text{for } n\geq1;\\
  q_{n}=a_{n}q_{n-1}+q_{n-2}, &amp; \text{for } n\geq1,
\end{cases}\end{equation}
with $p_{-1}=1$, $q_{-1}=0$, $p_{0}=a_{0}$ and $q_{0}=1$. Then $[a_{0}, a_{1},\ldots,a_{n}]=p_{n}/q_{n}$, which is called the $n$th convergent. 
We have the property
\begin{equation}\label{gcd}p_{n}q_{n-1}-p_{n-1}q_{n}=(-1)^{n+1},\ \ \text{ for } n\geq1.\end{equation}
 Using this property and induction, we obtain
\begin{equation}\label{eq1}
 \displaystyle\frac{p_{n}}{q_{n}}=
 a_{0}+\frac{1}{q_{0}q_{1}}-\frac{1}{q_{1}q_{2}}+\cdots+\frac{(-1)^{n-1}}{q_{n-1}q_{n}}.
 \end{equation}
Continued fractions are useful in providing the best rational approximations.

We recall that the irrationality exponent $\mu (x)$ is defined for real numbers $x$
 to be the supremum of the set of $\mu$ such that
$$\displaystyle 0&lt;\left|x-{\frac {p}{q}}\right|&lt;{\frac {1}{q^{\mu}}} $$
is satisfied by an infinite number of coprime integer pairs
$(p,q)$ with $q&gt;0$. Rational numbers have irrationality exponent $1$, while (as a consequence of Dirichlet's approximation theorem) every irrational number has irrationality exponent at least $2$.

Referring back to \cite{SO}, we recall a formula for the irrationality exponent of an irrational number in terms of its continued fraction expansion.
For a real number $x=[a_{0};a_{1},a_{2},\ldots]$ with convergents
$\displaystyle p_{i}/q_{i}$, it holds:
\begin{equation}\label{TT}\displaystyle \mu (x)=1+\limsup _{n\to \infty }{\frac {\ln q_{n+1}}{\ln q_{n}}}=2+\limsup _{n\to \infty }{\frac {\ln a_{n+1}}{\ln q_{n}}}.\end{equation}
A number $\displaystyle x\in \mathbb {R}$ with irrationality exponent
$\displaystyle \mu (x)=\infty$ is called a Liouville number.
In 1955, Roth \cite{R} published his famous theorem about the rational approximation of algebraic real numbers.

\begin{theorem} Let $x$ be a real algebraic number and let $\epsilon&gt;0$. 
Then the inequality
$$\displaystyle\bigg|x -\frac{p}{q}\bigg|&lt;\frac{1}{q^{2+\epsilon}}$$
has only a finite number of rational solutions $p/q$.
\end{theorem}

\noindent This theorem remains a very good tool to show the transcendence of real numbers. In fact, by this theorem, the irrationality exponent of any irrational algebraic number is exactly $2$.

The results presented here are similar in spirit to those in a paper by Davison and Shallit \cite{DS}, who found some continued fractions whose partial quotients are explicitly related to the denominators of their convergents, and used this to prove the transcendence of Cahen's constant:
\begin{equation*}C=\displaystyle\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}}{S_{n}-1}=\sum_{n=0}^{\infty}\displaystyle\frac{(-1)^{n}}{x_{n}}=0.64341\ldots,\end{equation*}
where $\{S_{n}\}$ is the Sylvester sequence defined by the nonlinear recurrence:
\begin{equation*}S_{0}=2, S_{n+1}=S_{n}^{2}-S_{n}+1 \hspace{5mm} (n\geq 0),\end{equation*}
and $x_{n}=S_{n}-1$ satisfies the recurrence \begin{equation*}x_{n+1}=x_{n}(x_{n} +1) \hspace{5mm}
 (n\geq 0).\end{equation*}
The partial quotients in continued fraction expansion of $C$ appear as sequence \seqnum{A006280} in the On-Line Encyclopedia of Integer Sequences (OEIS) \cite{OEIS}. Recently, Duverney and Shiokawa \cite{Duv} have proved that the exact value of its irrationality exponent is equal to 3.

The connections between infinite series representations for real numbers and continued fraction expansions are the subject of numerous studies. Several papers \cite{CL,HC2,DS,Duv,H0,S} looked into this, and we'll also present some more fascinating examples. More precisely, in this paper,
motivated by Davison and Shallit's concepts, we will reveal other analogous methods for displaying a real number as an infinite alternating series.

\section{Main results}\label{sec2}

\subsection{A first predictable pattern}

There are many numbers in the literature with a repeating pattern with a period of 2. For example
\begin{equation*}\tan(1/n)=[0,n-1,1,3n-2,1,5n-2,1,7n-2,\ldots],\end{equation*}
where $n$ is a positive integer (OEIS \seqnum{A019426}). We will see that there exist many other real numbers with a similar pattern which are transcendental numbers.

Let $x=[a_{0}, a_{1},\ldots]$ and $(p_{n}/q_{n})_{n}$ be the sequence of its convergents.
Let $\beta\in \mathbb{Z}_{&gt;0}$. Let $\omega_{0}$, $\omega_{1}$, $\omega_{2}$, $\ldots $ be a sequence of positive integers. Assume that
\begin{equation}\label{e2}\begin{cases}
   a_{0}=0,\\
a_{1}=\omega_{0},\\
a_{2n+2}=1,  &amp; \text{ for } n\geq0;\\
a_{2n+3}=\omega_{n+1}q_{2n+1}^{\beta},  &amp; \text{ for } n\geq0.
\end{cases}  \end{equation}
 Then $q_{0}=1$, $q_{1}=\omega_{0}$, and for $n\geq0$
 \begin{equation}\label{e22}\begin{cases}
   q_{2n+3}=\omega_{n+1}q_{2n+1}^{\beta}q_{2n+2}+q_{2n+1},\\
q_{2n +2}=q_{2n+1} +q_{2n}.
 \end{cases} \end{equation}
Using the equation \eqref{eq1}, we see
\begin{equation*}\displaystyle\frac{p_{2n+1}}{q_{2n+1}}=\sum_{0\leq i \leq 2n}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=[0,\omega_{0},1,\omega_{1}q_{1}^{\beta},1,\omega_{2}q_{3}^{\beta},1,\ldots,1,\omega_{n}q_{2n-1}^{\beta}]. \end{equation*}
 Hence, letting $n\rightarrow \infty$, we find the following way to express the continued fraction as an alternating series:

\begin{proposition}\label{p3}
 \begin{equation}\label{E3}
 x=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=[0,\omega_{0},1,\omega_{1}q_{1}^{\beta},1,\omega_{2}q_{3}^{\beta},1,\ldots,1,\omega_{n}q_{2n-1}^{\beta},\ldots].\ \end{equation}
 \end{proposition}

 \begin{theorem} The irrationality exponent of the number $x$ in Proposition \ref{p3} satisfies
 \begin{equation}\label{ll}
\mu(x)\geq 2+\beta +\limsup_{n \rightarrow \infty}\frac{\ln \omega_n}{\ln q_{2n}}\geq 2+\beta \geq 3.
\end{equation}
Hence $x$ is transcendental by Roth's theorem.
 \end{theorem}

 \begin{proof} Let $x$ be the number in equation \eqref{E3}, with continued fraction expansion $x=[a_{0}, a_{1}, \ldots]$. Let $\varepsilon&gt;0$ be arbitrary. From the equation \eqref{e22} we have $q_{2n}\leq 2q_{2n-1}$ for large $n$. So \begin{equation*}q_{2n}\leq q_{2n}^{\varepsilon}q_{2n-1},\end{equation*}
which yields $(1-\varepsilon)\ln q_{2n}\leq \ln q_{2n-1}$. Then by \eqref{TT}
 \begin{align*}\mu (x)&amp;=2+\displaystyle\limsup_{n\to \infty}\frac{\ln a_{n+1}}{\ln q_{n}}\\&amp;=2+\displaystyle\limsup _{n \rightarrow \infty}\frac {\ln a_{2n+1}}{\ln q_{2n}}\\&amp;=2+\displaystyle\limsup _{n \rightarrow \infty}\frac{\ln (\omega_{n}q_{2n-1}^{\beta})}{\ln q_{2n}}\\&amp;=
 2+\limsup _{n\to \infty }\frac {\ln \omega_{n}}{\ln q_{2n}}+\beta \frac{\ln q_{2n-1}}{\ln q_{2n}}
 \\&amp;\geq 2+ (1-\varepsilon)\beta+\displaystyle\limsup _{n \rightarrow \infty}\frac{\ln \omega_{n}}{\ln q_{2n}},\end{align*}
which yields \eqref{ll} since $\varepsilon$ is arbitrarily small.\end{proof}

 We now prove the converse to Proposition \ref{p3}.

 \begin{theorem} \label{t3}Let $(x_{i})$ be a sequence of positive integers such that $x_{0}=1$, and $x=\sum_{i\geq0}(-1)^{i}/x_{i}$ be an irrational number with continued fraction expansion
 $$x=[0,a_{1},a_{2},\ldots].$$
 Let $(p_{n}/q_{n})$ be the sequence of convergents of $x$. Assume that
 $$\displaystyle\sum_{i=0}^{n}\frac{(-1)^{i}}{x_{i}}=\frac{p_{n+1}}{q_{n+1}}$$
 for all $n\geq 0$, and that $x_{2n +1}\mid x_{2n+2}$ and $x_{2n+3}=\bigg(\displaystyle\frac{x_{2n+2}}{x_{2n+1}}\frac{x_{2n}}{x_{2n-1}}\frac{x_{2n-2}}{x_{2n-3}}\cdots\frac{x_{2}}{x_{1}}\bigg)^{2}+x_{2n+2}$ for all $n\geq 0$. Then  for $n\geq0$
  \begin{itemize}
 \item[(i)] $x_{n}=q_{n}q_{n+1}$.
 \item[(ii)] $q_{2n+1}\mid a_{2n+3}$.
 \item[(iii)] $a_{2n+2}=1$.
 \end{itemize}
 \end{theorem}

\begin{proof}
\leavevmode
\begin{itemize}
\item[(i)] Clearly verified for $n=0$. For $n\geq1$ we have
 $$\displaystyle\frac{p_{n+1}}{q_{n+1}}=\sum_{i=0}^{n}\frac{(-1)^{i}}{x_{i}}=\frac{p_{n}}{q_{n}}+\frac{(-1)^{n}}{x_{n}}.$$
 Hence
 $$\displaystyle\frac{p_{n+1}}{q_{n+1}}-\frac{p_{n}}{q_{n}}=\frac{(-1)^{n}}{x_{n}}.$$
 By \eqref{gcd}, we have
 $$\displaystyle\frac{p_{n+1}}{q_{n+1}}-\frac{p_{n}}{q_{n}}=\frac{(-1)^{n}}{q_{n+1}q_{n}},$$
 and hence $x_{n}=q_{n+1}q_{n}$.
\item[(ii)] Note that $x_{2n +1}\mid x_{2n+2}$ implies that $q_{2n+1}q_{2n+2}\mid q_{2n+2}q_{2n+3}$, and hence $q_{2n+1}\mid q_{2n+3}$. Since $q_{2n+3}=a_{2n+3}q_{2n+2}+q_{2n+1}$, so $q_{2n+1}\mid a_{2n+3}q_{2n+2}$. As $\gcd(q_{2n+1},q_{2n+2})=1$, so $q_{2n+1}\mid a_{2n+3}$.
\item[(iii)] We have \begin{align*}\bigg(\displaystyle\frac{x_{2n+2}}{x_{2n+1}}\frac{x_{2n}}{x_{2n-1}}\frac{x_{2n-2}}{x_{2n-3}}\cdots\frac{x_{2}}{x_{1}}\bigg)^{2}&amp;=
 \bigg(\displaystyle\frac{q_{2n+2}q_{2n+3}}{q_{2n+1}q_{2n+2}}\frac{q_{2n}q_{2n+1}}{q_{2n-1}q_{2n}}\cdots\frac{q_{2}q_{3}}{q_{1}q_{2}}\bigg)^{2}\\
 &amp;=q_{2n+3}^{2}\\&amp;=x_{2n+3}-x_{2n+2}=q_{2n+3}q_{2n+4}-q_{2n+2}q_{2n+3}.\end{align*}
 This gives that $q_{2n+4}=q_{2n+3}+q_{2n+2}$.
 \end{itemize}
\end{proof}

\begin{example} We put $\omega_{n}=1$ for $n\geq0$ and $\beta=1$. Then $q_{0}=1$, $q_{1}=1$, and $q_{2n+3}=q_{2n+1}(q_{2n+2}+1)$. Here are the first few values of the sequences $q_{n}$ and $x_{n}$:
\begin{center}
$\begin{array}{cccccccccc}
n &amp; 0 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp;7 &amp;\ldots \\
\hline
q_{n} &amp; 1 &amp; 1 &amp; 2 &amp; 3 &amp; 5 &amp; 18 &amp; 23 &amp; 432 &amp;\ldots \\
x_{n} &amp; 1 &amp; 2 &amp; 6 &amp; 15 &amp; 90 &amp; 414 &amp; 9936 &amp; 196560 &amp;\ldots
\end{array}$\end{center}
So we have 
\begin{align*}
 x=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}
 &amp; =1-\frac{1}{2}+\frac{1}{6}-\frac{1}{15}+\frac{1}{90}-\frac{1}{414}+\cdots
 \\[.05in]
 &amp;=[0,1,1,q_{1},1,q_{3},1,\ldots,1,q_{2n-1},\ldots]\\[.05in]
 &amp;=[0,1,1,1,1,3,1,18,1,432,\ldots],
 \end{align*}
 is a transcendental number. The continued fraction expansion of this series is referenced as \seqnum{A380013} in the OEIS \cite{OEIS}.
 \end{example}

\begin{example} We put $\beta=1$,
 $\omega_{0}=1$, and $\omega_{n+1}=q_{2n +1}$ for $n\geq0$. Then $q_{0}=1$, $q_{1}=1$, and $q_{2n+3}=q_{2n+1}^{2}q_{2n+2}+q_{2n+1}$.
So we have 
\begin{align*}
 x&amp;=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}
 =1-\frac{1}{2}+\frac{1}{6}-\frac{1}{15}+\frac{1}{240}+\cdots
 \\[.05in]
 &amp;=[0,1,1,1,1,q_{3}^{2},1,q_{5}^{2},1,\ldots,1,q_{2n-1}^{2},\ldots]\\[.05in]
 &amp;=[0,1,1,1,1,9,1,2304,1,14923065600,\ldots],
 \end{align*}
 is a transcendental number.
\end{example}

\begin{example} We put $\beta=1$,
 $\omega_{0}=1$, and $\omega_{n}=q_{2n}^{n}$ for $n\geq1$. Then $q_{0}=1$, $q_{1}=1$, and $q_{2n+3}=q_{2n+2}^{n+2}q_{2n+1}+q_{2n+1}=q_{2n+1}(q_{2n+2}^{n+2}+1)$.
So we have 
\begin{align*}
 x&amp;=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=
 1-\frac{1}{2}+\frac{1}{10}-\frac{1}{35}+\frac{1}{12040}-\cdots
 \\[.05in]
 &amp;=[0,1,1,q_{2}q_{1},1,q_{4}^{2}q_{3},1,q_{6}^{3}q_{5},\ldots,1,q_{2n+2}^{n+1}q_{2n+1},\ldots]\\[.05in]
 &amp;=[0,1,1,2,1,245,1,8859423442760,1,\ldots].
 \end{align*}
Since $\mu (x)\geq 2+\beta+\displaystyle\limsup _{n\rightarrow \infty }\frac {\ln \omega_{n}}{\ln q_{2n}}=+\infty$, then $x$ is a Liouville number.
\end{example}

\subsection{A second predictable pattern}

\noindent In this part, we construct many real numbers with a repeated pattern with a period $3$ analogous to those of $e^{\frac{1}{n}}$:
\begin{equation*}e^{\frac{1}{n}}=[1,n-1,1,1,3n-1,1,1,5n-1,1,1,\ldots],\end{equation*}
where $n$ is a positive integer (OEIS \seqnum{A003417}). We prove that these numbers are transcendental.

Let $x=[a_{0}, a_{1},\ldots]$ and $(p_{n}/q_{n})_{n}$ be the sequence of its convergents. As above, we let $\beta\in \mathbb{Z}_{&gt;0}$ and
$\omega_{0}$, $\omega_{1}$, $\omega_{2}$, $\ldots $ be a sequence of positive integers. Assume that
\begin{equation}\label{e3}\begin{cases}
   a_{0}=0,\\
a_{1}=\omega_{0},\\
a_{3n+2}=1  &amp; \text{ for } n\geq0;\\
a_{3n+3}=1  &amp; \text{ for } n\geq0;\\
a_{3n+4}=\omega_{n+1}q_{3n+2}^{\beta},  &amp; \text{ for } n\geq0.
\end{cases}
\end{equation}
 Then $q_{0}=1$, $q_{1}=\omega_{0}$, and for $n\geq0$
\begin{equation}\label{e33}\begin{cases}
 q_{3n +2}=q_{3n+1} +q_{3n},\\
 q_{3n +3}=q_{3n+2} +q_{3n+1},\\
 q_{3n+4}=\omega_{n+1}q_{3n+3}q_{3n+2}^{\beta}+q_{3n+2}.
\end{cases}
\end{equation}
 Using the equation \eqref{eq1}, we see
\begin{equation*}\displaystyle\frac{p_{3n+1}}{q_{3n+1}}=\sum_{0\leq i \leq 3n}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=[0,\omega_{0},1,1,\omega_{1}q_{2}^{\beta},1,1,\omega_{2}q_{5}^{\beta},1,1\ldots,1,\omega_{n}q_{3n-1}^{\beta}]. \end{equation*}
 Hence, letting $n\rightarrow \infty$, we also find the following way to express the real $x$ as an alternating series:

\begin{proposition}\label{p4}
 \begin{equation}\label{E4}
 x=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=
 [0,\omega_{0},1,1,\omega_{1}q_{2}^{\beta},1,1,\omega_{2}q_{5}^{\beta},1,1\ldots,1,1,\omega_{n}q_{3n-1}^{\beta},\ldots].
 \end{equation}
\end{proposition}

\begin{theorem} The irrationality exponent of the number $x$ in Proposition \ref{p4} satisfies
 \begin{equation}\label{lll}
\mu(x)\geq 2+\beta +\limsup_{n \rightarrow \infty}\frac{\ln \omega_n}{\ln q_{3n}}\geq 2+\beta \geq 3.
\end{equation}
Hence $x$ is transcendental by Roth's theorem.
\end{theorem}

 \begin{proof} Let $x$ be the number in equation \eqref{E4}, with continued fraction expansion $x=[a_{0}, a_{1}, \ldots]$. Let $\varepsilon&gt;0$ be arbitrary. From \eqref{e33}, we have $q_{3n}\leq 2q_{3n-1}$ for large $n$. So \begin{equation*}q_{3n}\leq q_{3n}^{\varepsilon}q_{3n-1},\end{equation*}
which yields $(1-\varepsilon)\ln q_{3n}\leq \ln q_{3n-1}$. Then by \eqref{TT}
 \begin{align*}\mu (x)&amp;=2+\displaystyle\limsup _{n \rightarrow \infty }\frac{\ln a_{n+1}}{\ln q_{n}}\\&amp;=2+\displaystyle\limsup _{n \rightarrow \infty}\frac {\ln a_{3n+1}}{\ln q_{3n}}\\&amp;=2+\displaystyle\limsup _{n \rightarrow \infty}\frac{\ln (\omega_{n}q_{3n-1}^{\beta})}{\ln q_{3n}}\\&amp;=
 2+\limsup _{n \to \infty }\frac {\ln \omega_{n}}{\ln q_{3n}}+\beta \frac{\ln q_{3n-1}}{\ln q_{3n}}
 \\&amp;\geq 2+ (1-\varepsilon)\beta+\displaystyle\limsup _{n \rightarrow \infty }\frac {\ln \omega_{n}}{\ln q_{3n}},\end{align*}
which yields \eqref{lll} since $\varepsilon$ is arbitrarily small.\end{proof}

 The proof of the converse to Proposition \ref{p4} is similar to the proof of Theorem \ref{t3}, so we omit it.

 \begin{theorem} Let $(x_{i})$ be a sequence of positive integers such that $x_{0}=1$, and $x=\sum_{i\geq0}(-1)^{i}/x_{i}$ be an irrational number with continued fraction expansion
 $$x=[0,a_{1},a_{2},\ldots].$$
 Let $(p_{n}/q_{n})$ be the sequence of convergents of $x$. Assume that
 $$\displaystyle\sum_{i=0}^{n}\frac{(-1)^{i}}{x_{i}}=\frac{p_{n+1}}{q_{n+1}}$$
 for all $n\geq 0$, and that $x_{3n +2}\mid x_{3n+3}$,
  \begin{equation*}x_{3n+2}-x_{3n+1}=\bigg(\displaystyle\frac{x_{3n+1}}{x_{3n}}\frac{x_{3n-1}}{x_{3n-2}}\frac{x_{3n-3}}{x_{3n-4}}\cdots\frac{x_{2}}{x_{1}}\bigg)^{2},\end{equation*} \begin{equation*}x_{3n+4}-x_{3n+3}=\bigg(\displaystyle\frac{x_{3n+3}}{x_{3n+2}}\bigg)^{2}(x_{3n+2}-x_{3n+1})\end{equation*} for all $n\geq 0$. Then for $n\geq0$ we have
   \begin{itemize}
 \item[(i)] $x_{n}=q_{n}q_{n+1}$.
 \item[(ii)] $q_{3n+2}\mid a_{3n+4}$.
 \item[(iii)] $a_{3n+2}=a_{3n+3}=1$.
 \end{itemize}
   \end{theorem}

\begin{example} We put $\beta=1$, $\omega_{0}=1$, and $\omega_{n+1}=q_{3n +2}$ for $n\geq0$. Then
\begin{equation*}\begin{cases}
   a_{0}=0,\\
a_{1}=1,\\
a_{3n+2}=a_{3n+3}=1, &amp; \text{ for } n\geq0;\\
a_{3n+4}=q_{3n+2}^{2}, &amp; \text{ for } n\geq0.
\end{cases}  \end{equation*}
We have 
\begin{align*}
 x&amp;=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=
 1-\frac{1}{2}+\frac{1}{6}-\frac{1}{42}+\frac{1}{238}-\cdots \\[0.05in]
 &amp;=[0,1,1,1,q_{2}^{2},1,1,q_{5}^{2},\ldots,1,1,q_{3n-1}^{2},\ldots]\\[0.05in]
 &amp;=[0,1,1,1,4,1,1,289,1,1,81126049,\ldots],
 \end{align*}
and this number is transcendental. The continued fraction expansion of this series is referenced as \seqnum{A380194} in the OEIS \cite{OEIS}.
\end{example}

\begin{example} We put $\beta=1$, $\omega_{0}=1$, and $\omega_{n}=q_{3n}^{n}$ for $n\geq1$. Then
\begin{equation*}\begin{cases}
   a_{0}=0,\\
a_{1}=1,\\
a_{3n+2}=a_{3n+3}=1, &amp; \text{ for } n\geq0;\\
a_{3n+4}=q_{3n+3}^{n+1}q_{3n+2}, &amp; \text{ for } n\geq0.
\end{cases}
\end{equation*}
We have 
\begin{align*}
 x&amp;=\sum_{i=0}^{\infty}\displaystyle\frac{(-1)^{i}}{q_{i}q_{i+1}}=1-\frac{1}{2}+\frac{1}{6}-\frac{1}{60}+\frac{1}{460}-\cdots \\[.05in]
 &amp;=[0,1,1,1,q_{2}q_{3},1,1,q_{6}^{2}q_{5},\ldots,1,1,q_{3n}^{n}q_{3n-1},\ldots]\\[.05in]
 &amp;=[0,1,1,1,6,1,1,42527,1,1,89468504340857877754313037,\ldots],
 \end{align*}
is a Liouville number.
\end{example}

\section{Acknowledgments}
The authors wish to thank the referee for several helpful comments and suggestions that significantly improved the quality of the manuscript.

\begin{thebibliography}{99}

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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A55; Secondary 11J70.\\
\noindent \emph{Keywords}: continued fraction, transcendental number.


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\noindent (Concerned with sequences
\seqnum{A003417},
\seqnum{A006280},
\seqnum{A019426},
\seqnum{A380013}, and
\seqnum{A380194}.)

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\vspace*{+.1in}
\noindent
Received August 11 2024;
revised versions received August 12 2024; October 8 2024; November 4 2024;
January 31 2025; February 7 2025.
Published in {\it Journal of Integer Sequences}, February 8 2025.

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