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\begin{center}
\vskip 1cm{\LARGE\bf
Appending Digits to Generate an Infinite \\
\vskip .1in
Sequence of Composite Numbers
}
\vskip 1cm
\large
Lenny Jones and Daniel White\\
Department of Mathematics \\
Shippensburg University \\
Shippensburg, PA 17257 \\
USA\\
\href{mailto:lkjone@ship.edu}{\tt lkjone@ship.edu}\\
\href{mailto:dw9878@ship.edu}{\tt dw9878@ship.edu}\\
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\begin{abstract}
Let $D$ be a list of single digits, and let $k$ be a positive integer. We construct an infinite sequence of positive integers by repeatedly appending, in order, one at a time, the digits from the list $D$ to the integer $k$, in one of four ways: always on the left, always on the right, alternating and starting on the left, or alternating and starting on the right. In each of these four situations, we investigate, for various lists $D$, how to find infinitely many positive integers $k$ such that every term of the sequence is composite.
\end{abstract}
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\section{Introduction}\label{Intro}
In a previous paper \cite{L}, the first author proved that, for any digit $d\in \{1,3,7,9\}$, there exist infinitely many positive integers $k$, with $\gcd(k,d)=1$, such that appending $d$ on the right of $k$, any number of times, always gives a composite number. Using similar methods, this paper generalizes this result, with tweaks and alterations along the way as each variation requires. We describe this process more precisely as follows.
Given a list $D=\left[d_1,\ldots, d_t\right]$, where $d_i\in \left\{0,1,\ldots ,9\right\}$, and a positive integer $k$, we construct an infinite sequence $\left\{s_n\right\}_{n=1}^{\infty}$ of positive integers by repeatedly appending, in order, one at a time, the digits from the list $D$ to the integer $k$, in one of four ways: always on the left, always on the right, alternating and starting on the left, or alternating and starting on the right.
For example, if $D=\left[1,7,9\right]$ and $k=a_1a_2\cdots a_q$, where $a_i$ is the $i$th digit of $k$ (reading from left to right), then the sequence $\left\{s_n\right\}_{n=1}^{\infty}$ generated by appending the digits from $D$ to $k$ in an alternating manner, starting on the left, is
\[s_1=1a_1a_2\cdots a_q, \quad s_2=1a_1a_2\cdots a_q7, \quad s_3=91a_1a_2\cdots a_q7, \quad s_4=91a_1a_2\cdots a_q71, \ldots .\] In each of these four situations, we investigate, for various lists $D$, how to find infinitely many positive integers $k$ such that every term of the sequence $\left\{s_n\right\}_{n=1}^{\infty}$ is composite. We can also assume that the list $D=\left[d_1,\ldots, d_t\right]$ is minimal in the sense that there is no $j10$, certain computational roadblocks arise, which we discuss in Section \ref{Section:t larger than 10} and Section \ref{Section: Final Remarks}.
\subsection{The Cases $t\le 10$}\label{Section: t<=10}
\begin{table}[h]
\caption{Right (or Left) $t\le 6$}
\begin{center}
\begin{tabular}{c|c}%\hline
$t$ & $\C$\\ \hline %\hline
1&$\left\{(1,2,11),(0,3,37),(0,4,101),(4,6,7),(2,6,13)\right\}$ \\ \hline
2&$\left\{(1,4,11),(3,4,101),(0,6,7),(2,6,13),(4,6,37)\right\}$ \\ \hline
&$\left\{(1,6,7),(2,6,11),(3,6,13),(2,9,333667),(6,12,101),(4,12,9901),\right.$\\
3&$(5,18,19),(17,18,52579),(0,24,73),(10,24,137),(22,24,99990001),$\\
&$\left.(12,36,999999000001),(60,72,3169),(36,72,98641)\right\}$
\\ \hline
4& Given in Section \ref{Case:t=4R}\\ \hline
5& Given in Section \ref{SubSectionL5}\\ \hline
&$\left\{ (1,12,101),(5,12,9901),(3,18,19),(15,18,333667),\right.$\\
&$(14,18,52579),(11,24,73),(19,24,137),(23,24,99990001),$\\
&$(22,30,31),(0,30,41),(12,30,211),(18,30,241),(4,30,271),$\\
&$(6,30,2906161),(24,30,9091),(16,30,2161),(8,36,999999000001),$\\
6&$(32,42,43),(20,42,127),(14,42,239),(2,42,10838689),$\\
&$(8,42,459691),(26,42,4649),(38,42,909091),(22,42,1933)$\\
&$(10,42,2689),(31,48,17),(7,48,9999999900000001),$\\
&$(45,48,5882353),(27, 54, 757),(9, 54, 440334654777631),$\\
&$(45, 54, 14175966169),(10, 54, 70541929),(58, 60, 61),$\\
&$\left.(40, 60, 39526741),(28, 60, 4188901),(10, 60, 27961)\right\}$\\ \hline
\end{tabular}
\end{center}
\label{Table:1}
\end{table}
\begin{table}[h]
\caption{Right (or Left) $7\le t \le 10$}
\begin{center}
\begin{tabular}{c|ccc}%\hline
$t$ & $|\C|$ & $M$ & $L$\\ \hline %\hline
7 & 63 & 126 & 360360\\ \hline
8& 73 & 120 & 2882880\\ \hline
9& 80 & 144 & 6486480\\ \hline
10 & 80 & 140& 3603600\\ \hline
\end{tabular}
\end{center}
\label{Table:2}
\end{table}
In Table \ref{Table:1} we give explicit coverings for values of $t\le 6$ that can be used in each of these cases, and to illustrate the techniques used to produce these coverings, we provide the details for the single case $t=4$ in Section \ref{Case:t=4R}. Because of the large number of congruences in $\C$ for values of $t$ with $7\le t\le 10$, Table \ref{Table:2} gives only the number of congruences in $\C$, denoted $|C|$, the largest modulus in $\C$, denoted $M$, and the least common multiple of the moduli in $\C$, denoted $L$, in these cases.
\subsubsection{A Detailed Example: The Case $t=4$}\label{Case:t=4R}
Let $D=[d_1,d_2,d_3,d_4]$.
Because we are appending four digits, the form of the term $s_n$ will vary according to the value of $n$ modulo 4. For example, when $n\equiv 1 \pmod{4}$, we have
\begin{align*}
s_n&=10^nk+d_110^{n-1}+d_210^{n-2}+d_310^{n-3}+d_410^{n-4}+\\
&\hspace*{.21in}\cdots +d_110^4+d_210^3+d_310^2+d_410+d_1\\
&=10^nk+d_1\left(\frac{10^{n+3}-1}{10^4-1}\right)+\left(10^3d_2+10^2d_3+10d_4\right)\left(\frac{10^{n-1}-1}{10^4-1}\right).
\end{align*}
Therefore, when $n\equiv 0,1,2,3 \pmod{4}$, we have respectively:
\[s_n=\left\{\begin{array}{l}
10^nk+\left(10^3d_1+10^2d_2+10d_3+d_4\right)\left(\frac{10^n-1}{10^4-1}\right)\\
10^nk+d_1\left(\frac{10^{n+3}-1}{10^4-1}\right)+\left(10^3d_2+10^2d_3+10d_4\right)\left(\frac{10^{n-1}-1}{10^4-1}\right)\\
10^nk+\left(10d_1+d_2\right)\left(\frac{10^{n+2}-1}{10^4-1}\right)+\left(10^3d_3+10^2d_4\right)\left(\frac{10^{n-2}-1}{10^4-1}\right)\\
10^nk+\left(10^2d_1+10d_2+d_3\right)\left(\frac{10^{n+1}-1}{10^4-1}\right)+10^3d_4\left(\frac{10^{n-3}-1}{10^4-1}\right).\\
\end{array} \right.\]
We use a table \cite{K} of prime factors of numbers of the form $10^m-1$ to construct, within the constraints of Conditions \ref{Conditions}, a list of possible moduli for our covering $\C$. We illustrate this procedure by determining the first few moduli in $\C$. According to Conditions \ref{Conditions}, the minimum modulus in $\C$ must be at least $2t=8$, each modulus must be divisible by 4, and we cannot use the primes 3, 11 and 101, since they divide $10^4-1$. Therefore, we start with $m_1=8$. There are two prime divisors of $10^8-1$ that do not divide $10^4-1$, namely 73 and 137. Thus, we can use the modulus 8 twice in our covering, with the corresponding primes 73 and 137. The next possible modulus in $\C$ is 12, which we can use four times since there are four prime divisors of $10^{12}-1$ available to us that we have not yet used: 7, 13, 37, 9901. The next possible modulus in $\C$ is 16. We can use it twice since there are two available primes divisors of $10^{16}-1$, namely 17 and 5882353.
Continuing in this manner with consecutive multiples of 4, we eventually reach a list of moduli for which a covering actually exists (verified by Magma):
\[
\M=[8,8,12,12,12,12,16,16,20,20,20,20,20,24,28,28,28,28,28,28,32,32,32,32,32].
\] We can, in many cases, judiciously prune the list of moduli, and still construct a covering. For example, adding the modulus $28$ in the construction of $\M$ here increases the least common multiple of the moduli by a factor of 7. However, skipping over 28 and moving directly to 32 after 24 only increases the least common multiple of the moduli by a factor of 2, and therefore creates fewer ``holes" to fill. Consequently, we are able to construct a covering using the moduli in the list $\M$ without 28, and using one less 32. One advantage of this method is that it reduces the size of the smallest value of $k$ that we seek. In order to conserve space, we have attempted to employ such an optimizing strategy for our detailed examples, although such optimization is, in general, of no concern to us here.
One covering $\C$ generated by Magma is
\begin{align*}
\C=\left\{\right.& (1,8,73),(0,8,137),(2,12,7),(6,12,13),(10,12,37),(3,12,9901),(12,16,17),\\
& (5,16,5882353),(19,20,41),(11,20,271),(15,20,3541),(7,20,9091),(3,20,27961),\\
& (5,24,99990001),(20,32,353),(4,32,449),(29,32,641),(13,32,1409)\left.\right\}.
%& \left.\right\}.
\end{align*}
To illustrate the use of $\C$, we give a specific example.
\begin{example}
Suppose that $D=[1,3,5,7]$. We start with $(1,8,73)\in \C$. We want to find $k$ such that $s_n\equiv 0 \pmod{73}$. Since $n\equiv 1 \pmod{8}$, we have that $n\equiv 1 \pmod{4}$. Thus, we take the form for $s_n$ corresponding to $n\equiv 1 \pmod{4}$, set it equal to zero, and solve for $k$. Finally, we substitute in $n=1$ (since $n\equiv 1 \pmod{8}$), and reduce modulo 73 to get
\begin{align*}
k&=\frac{-d_1\left(\frac{10^{n+3}-1}{10^4-1}\right)-\left(10^3d_2+10^2d_3+10d_4\right)\left(\frac{10^{n-1}-1}{10^4-1}\right)}{10^n}\\
&\equiv 51 \pmod{73}, \quad \mbox{ when $n=1$}.\\
%&\equiv 2 \pmod{7}.
\end{align*}
For each element in $\C$, we get a corresponding congruence for $k$, and so we generate a system of congruences in $k$ modulo the primes from the ordered triples in $\C$. By the Chinese remainder theorem, the smallest positive solution to this system is \[k=329487380848404895573199266357097656655612892130353175.\]
\end{example}
%\subsubsection{The Cases $t\le 6$, $t\ne 4$}\label{Section: t<=6}
\subsection{The Cases $t\ge 11$}\label{Section:t larger than 10}
The minimum modulus in any covering $\C$ here is at least $2t$, and so Remark \ref{Rem:Covering necessity} tells us that the largest modulus in $\C$ must grow accordingly. There are then two implications. The first implication is that for large $t$, we will eventually run out of prime factors of $10^m-1$ in our database. The second implication is that, even if we do have the prime factors of $10^m-1$ in our database, the number of moduli needed to construct a covering is growing so rapidly that we very quickly reach a point in which the Magma program runs out of memory before it can generate a covering. Although, theoretically, we see no obstruction to the existence of a covering satisfying Conditions \ref{Conditions} when $t\ge 11$, we are unable to prove or disprove the existence of such a covering. For more discussion, see Section \ref{Section: Final Remarks}.
\section{Appending the Digits Only on the Left}\label{Section:OnlyLeft}
Again we let $D=[d_1,d_2,\ldots, d_t]$, where $d_i\in \left\{0,1,2,\ldots ,9\right\}$. Let $k$ be a positive integer, and suppose that $k=a_1a_2\cdots a_q$, where $a_i$ is the $i$th digit of $k$, reading from left to right. Then we construct the sequence of positive integers $\left\{s_n\right\}_{n=1}^{\infty}$ defined by
\[s_1=d_1a_1a_2\cdots a_q, \quad s_2=d_2d_1a_1a_2\cdots a_q, \quad s_3=d_3d_2d_1a_1a_2\cdots a_q, \quad\ldots \]
\[s_{t+1}=d_1d_td_{t-1}\cdots d_2d_1a_1a_2 \cdots a_q, \quad\ldots .\]
As before, we wish to determine values of $k$ such that $s_n$ is composite for all $n\ge 1$. However, because we are appending the digits on the left, there are two minor complications that arise. First, each of the expressions for $s_n$ now involves $q$, the number of digits of $k$. Second, assuming we can solve the system of congruences for $k$ that is produced by our methods, we would like to ensure that there is a solution $k$ that is not divisible by 2 or 5, since such values of $k$ constitute a trivial situation. Fortunately, we can overcome both of these concerns. We illustrate how to deal with these complications in the case when $t=5$. Despite these complications, we see that, as in the situation when we were appending the digits on the left of $k$, the number of distinct formulas for $s_n$ required here is $n$, one for each congruence class modulo $n$. Therefore, since the coverings we seek must satisfy Conditions \ref{Conditions}, the same covering can be used for a particular $t$, regardless of whether we are appending the digits on the left or right of $k$. See Tables \ref{Table:1} and \ref{Table:2}.
\subsection{A Detailed Example: The Case $t=5$}\label{SubSectionL5}
Let $D=[d_1,d_2,d_3,d_4,d_5]$.
We begin by formulating the term $s_n$. As before,
we see here that the form of $s_n$ is dependent on the congruence class of $n$ modulo the number of elements in our list $D$, which in this example is $5$. But here, $s_n$ is also a function of $q$, the number of digits in $k$. When $n\equiv 0,1,2,3,4 \pmod{5}$, we have respectively:\\ %. Thus, when $n\equiv 0,1,2,3,4 \pmod{5}$, we have respectively\\
\[s_n=\left\{\begin{array}{l}
k+\left(10^qd_1+10^{q+1}d_2+10^{q+2}d_3+10^{q+3}d_4+10^{q+4}d_5\right)\left(\frac{10^n-1}{10^5-1}\right)\\\\
k+10^qd_1\left(\frac{10^{n+4}-1}{10^5-1}\right)\\
\quad\quad+\left(10^{q+1}d_2+10^{q+2}d_3+10^{q+3}d_4+10^{q+4}d_5\right)\left(\frac{10^{n-1}-1}{10^5-1}\right)\\\\
k+\left(10^qd_1+10^{q+1}d_2\right)\left(\frac{10^{n+3}-1}{10^5-1}\right)\\
\quad\quad+\left(10^{q+2}d_3+10^{q+3}d_4+10^{q+4}d_5\right)\left(\frac{10^{n-2}-1}{10^5-1}\right)\\\\
k+\left(10^qd_1+10^{q+1}d_2+10^{q+2}d_3\right)\left(\frac{10^{n+2}-1}{10^5-1}\right)\\
\quad\quad+\left(10^{q+3}d_4+10^{q+4}d_5\right)\left(\frac{10^{n-3}-1}{10^5-1}\right)\\\\
k+\left(10^qd_1+10^{q+1}d_2+10^{q+2}d_3+10^{q+3}d_4\right)\left(\frac{10^{n+1}-1}{10^5-1}\right)\\
\quad\quad+10^{q+4}d_5\left(\frac{10^{n-4}-1}{10^5-1}\right).\\\\
\end{array} \right.\]
Here we wish to construct a covering and a set of corresponding primes $\P$ that satisfy Conditions \ref{Conditions} with $t=5$. We proceed, as in Section \ref{Section:OnlyRight}, to construct the covering
\begin{align*}
\C=\left\{\right.& (0,10,11),(5,10,9091),(13,15,31),(2,15,37),(7,15,2906161),(6,20,101),\\
& (14,20,3541),(19,20,27961),(12,30,7),(18,30,13),(1,30,211),(11,30,241),\\
& (3,30,2161),(4,40,73),(29,40,137),(16,40,1676321),(36,40,5964848081),\\
& (8,45,238681),(23,45,333667),(38,45,4185502830133110721),(57,60,61),\\
& (27,60,9901),(51,60,4188901),(21,60,39526741),(9,80,17),\\
& (49,80,5070721),(24,80,5882353),\\
& (64,80,19721061166646717498359681)\left.\right\}.
\end{align*}
As in Section \ref{Section:OnlyRight}, each of the elements in $\C$ corresponds to a particular congruence in a system of congruences for $k$. To avoid the trivial situations that $k$ is divisible by 2 or 5 when we solve the system, we add to the system the additional congruence $k\equiv d \pmod{10}$, where $d\in \{1,3,7,9\}$. We must still contend with the fact that each of the other congruences in the system is dependent on $q$, the number of digits of $k$. However, this problem is easily rectified. We introduce some notation. We let $\ell=10 \prod_{p_i\in \P}p_i$, the least common multiple of the moduli in our system of congruences for $k$, and we let $\delta(z)$ denote the number of digits in the positive integer $z$. For any positive integer $w$, we let $v_w=\delta(\ell)+w$. For any value of $w$, we can substitute $v_w$ in for $q$ into each of the congruences in the system. This allows us to solve for $k$ using the Chinese remainder theorem. If $k_0$ is the smallest positive integer solution to this system, then all solutions to the system satisfy the congruence $x\equiv k_0 \pmod{\ell}$. Note that the number of digits in $k_0$ is at most $\delta(\ell)$. It is then easy to see that there is a positive integer $a$ such that $\delta(k_0+a \ell)=v_w$. In other words, $k_0+a \ell$ is a solution to our system with exactly $q$ digits, and since $w$ can be any positive integer, we get infinitely many such solutions. To illustrate these ideas and the use of $\C$, we give an example.
\begin{example}\label{Ex:Left}
Suppose that $D=[3,7,2,5,4]$. We wish to find $k$ such that $s_n\equiv 0 \pmod{p}$ for all $p \in \P$. For each ordered triple $(r,m,p)$ in $\C$, we substitute $r$ in for $n$ into the formula for $s_n$, where $n\equiv r \pmod{5}$. Then we set each of these expressions equal to zero, and solve for $k$. Here we have that $\delta(\ell)=146$, and we choose to let $q=147$. We reduce each of the congruences modulo the corresponding prime $p\in \P$, and we add the additional congruence $k\equiv 1 \pmod{10}$ to avoid certain trivial situations.
By the Chinese remainder theorem, the smallest positive solution to this system is
\begin{center}%\vspace{1pc}
{\normalsize $k_0$=10982121324843319893051990697602742531038520281080212527033380031702058677\\429664543284836432559714302725330265818303246372107452182645651819050681.}
\end{center}
However, $\delta(k_0)=146$, and so we must add a multiple of $\ell$ to $k_0$ to get an integer that has exactly 147 digits, which will then be a solution to our problem. The least value of $a$ such that $\delta(k_0+a\ell)=147$ is $a=9$. Thus, the smallest solution produced by our methods is
\begin{center}%\vspace{1pc}
{\normalsize $k$=110983121334843519896052020698002747531098520881087212607034180039702148678\\329672543364837232566714362725930270818343246672110452202645751820050691.}
\end{center}
\end{example}
\begin{remark}
We should mention that of course the same computational obstacles for larger values of $t$ that we faced in Section \ref{Section:OnlyRight} also plague us here.
\end{remark}
\section{Appending the Digits in an Alternating Manner}
\label{Section:Alternating}
Let $D=[d_1,d_2,\ldots, d_t]$, where $d_i\in \left\{0,1,\ldots ,9\right\}$. Let $k$ be a positive integer, and suppose that $k=a_1a_2\cdots a_q$, where $a_i$ is the $i$th digit of $k$, reading from left to right. By first starting on the right and appending the elements of list $D$ in an alternating manner, we construct the sequence of positive integers $\left\{s_n\right\}_{n=1}^{\infty}$ defined by
\[s_1=a_1a_2\cdots a_qd_1, \quad s_2=d_2a_1a_2\cdots a_qd_1, \quad s_3=d_2a_1a_2\cdots a_qd_1d_3, \quad\ldots \]
As before, we wish to determine values of $k$ such that $s_n$ is composite for all $n\ge 1$. The methods we use here are similar to the methods used in the previous sections, and they are valid regardless of whether we start appending on the left or on the right of $k$. Only the formulas for $s_n$ will be affected. In either situation, we require ${\rm lcm}(2,t)$ versions of $s_n$, which implies that we have two subcases: $t$ even and $t$ odd. Accordingly, we alter Conditions \ref{Conditions} as follows.
\begin{Con}\label{ConditionsA} Let $m_1\le m_2\le \cdots \le m_s$ be the moduli in the covering, and let $\P=\{p_1,p_2,\ldots ,p_s\}$ be a set of corresponding primes. Let $a={\rm lcm}(2,t)/2$ and $b={\rm lcm}(2,t)$.
\begin{enumerate}
\item $m_i\equiv 0 \pmod{b}$ for all $i$.\label{Item:1A}
\item No prime in $\P$ is a divisor of $10^a-1$.\label{Item:2A}
\item The prime $p_i\in \P$ divides $10^{m_i/2}-1$.
\end{enumerate}
\end{Con}
The variances in Conditions \ref{ConditionsA} from Conditions \ref{Conditions} are due to the period of occurrence of $d_i \in D$ on either side of $k$, and the number of variations of $s_n$. Note that Conditions \ref{ConditionsA} imply that the minimum modulus in $\C$ must be at least $2t$ when $t$ is even, and at least $4t$ when $t$ is odd. One additional implication that follows from Conditions \ref{ConditionsA} is that if we are able to find a covering and a set of corresponding primes satisfying Conditions \ref{ConditionsA} for some particular odd value of $t$, then the same covering and set of primes can be used for the case of $2t$. In the alternating situation when $t\ge 7$ is odd, we are faced with the same computational difficulties that we encountered when appending on a single side in the cases $t\ge 11$. These obstructions are discussed in Section \ref{Section:t larger than 10}. To illustrate the general methods used in the alternating situation, we give a detailed analysis of the case $t=4$ in Section \ref{Section: t=4A}.
\subsection{The Cases $t\le 10$}\label{Section: t<=10A}
Although an explicit covering is given only in the case of $t=4$ in Section \ref{Section: t=4A}, we give, as we gave in Section \ref{Section: t<=10} in the situation of appending on a single side, some basic information in all other cases when $t\le 10$, except $t=7$ and $t=9$. This general information is given in Table \ref{Table:3}, where $|C|$ is the number of congruences in the covering $\C$, $M$ is the largest modulus in $\C$, and $L$ is the least common multiple of the moduli in $\C$.
\begin{table}[ht]
\caption{Alternating $t\le 10$}
\begin{center}
\begin{tabular}{c|ccc}%\hline
$t$ & $|\C|$ & $M$ & $L$\\ \hline %\hline
1 & 31 & 84 & 840\\ \hline
2& \multicolumn{3}{c}{Same as $t=1$}\\ \hline
3& 79 & 420 & 15120\\ \hline
4& \multicolumn{3}{c}{Given in Section \ref{Section: t=4A}}\\ \hline
%4 & 43 & 168 & 3360\\ \hline
5 & 331 & 15840 & 7207200\\ \hline
6 & \multicolumn{3}{c}{Same as $t=3$}\\ \hline
7& \multicolumn{3}{c}{Unknown}\\ \hline
8 & 177 & 1584 & 221760\\ \hline
9& \multicolumn{3}{c}{Unknown}\\ \hline
10 & \multicolumn{3}{c}{Same as $t=5$}\\ \hline
\end{tabular}
\end{center}
\label{Table:3}
\end{table}
\subsubsection{A Detailed Example: The Case $t=4$}\label{Section: t=4A}
We give the ${\rm lcm}(2,t)=4$ formulas for $s_n$ for the situation when we start appending on the right. The formulas for $s_n$, and the methods used, when we start appending on the left are similar. When $n\equiv 0,1,2,3 \pmod{4}$, we have respectively:\\
\[s_n=\left\{\begin{array}{l}
10^{n/2}k+\left(10d_1+10^{(n/2)+q}d_2+d_3+10^{(n+2)/2+q}d_4\right)\left(\frac{10^{n/2}-1}{10^2-1}\right)\\\\
10^{(n+1)/2}k + d_1\left(\frac{10^{(n+3)/2}-1}{10^2-1}\right)\\
\quad\quad+\left(10^{(n+1)/2+q}d_2+10d_3+10^{(n+3)/2+q}d_4\right)\left(\frac{10^{(n-1)/2}-1}{10^2-1}\right)\\\\
10^{n/2}k+\left(10d_3+10^{(n+2)/2+q}d_4\right)\left(\frac{10^{(n-2)/2}-1}{10^2-1}\right)\\
\quad\quad+\left(d_1+10^{(n/2)+q}d_2\right)\left(\frac{10^{(n+2)/2}-1}{10^2-1}\right)\\\\
10^{(n+1)/2}k+10^{(n+3)/2+q}d_4\left(\frac{10^{(n-3)/2}-1}{10^2-1}\right)\\
\quad\quad+ \left(10d_1+10^{(n+1)/2+q}d_2+d_3\right)\left(\frac{10^{(n+1)/2}-1}{10^2-1}\right).\\\\
\end{array} \right.\]
The covering we use here is
\begin{align*}
\C=\left\{\right.& (1, 8, 101),(2, 12, 7),(10, 12, 13),(6, 12, 37),(13, 16, 73),(3, 16, 137),\\
& (8, 20, 41), (4, 20, 271), (16, 20, 9091), (15, 24, 9901), (23, 28, 239),\\
& (7, 28, 4649), (11, 28, 909091), (5, 32, 17), (21, 32, 5882353), (0, 40, 27961),\\
& (12, 40, 3541), (31, 44, 23), (39, 44, 513239), (23, 44, 21649), (27, 44, 8779),\\
& (35, 44, 4093), (15, 56, 29), (31, 56, 281), (12, 60, 31), (32, 60, 211),\\
& (20, 60, 241), (40, 60, 2906161), (52, 60, 2161), (11, 64, 353), (27, 64, 449),\\
& (43, 64, 641), (59, 64, 1409), (31, 64, 69857), (60, 80, 5964848081), (20, 80, 1676321),\\
& (47, 84, 43),(83, 84, 127), (55, 84, 1933), (19, 84, 2689) \left.\right\}.
\end{align*}
\begin{example}
Suppose that $D=[7,1,4,9]$. We follow the methods used in Section \ref{SubSectionL5}. Here $\delta(\ell)=131$, and we choose $v=132$. Then the smallest value of $k$ with 132 digits produced by this process, such that every term of the sequence $\left\{s_n\right\}_{n=1}^{\infty}$ is composite, is
\begin{center}
{\normalsize $k$=13096455548520391404136872903624576303901893826181040740105459909726\\1712572158731534146652935794771514600650784321669697928418283852.}
\end{center}
\end{example}
\section{Final Remarks}\label{Section: Final Remarks}
As mentioned in Section \ref{Section:t larger than 10}, there are certain computational restrictions that prohibit us from extending the results in Theorem \ref{Thm:Main} to larger values of $t$. Theoretically, however, it might be possible to prove the existence of a covering satisfying Conditions \ref{Conditions} or Conditions \ref{ConditionsA}, which in turn, would prove the existence of the desired values of $k$.
One possible approach is the following. Although we might not know the explicit prime divisors of numbers of the form $10^m-1$, there are theorems \cite{Bang, BV, S, Z} that guarantee the existence of certain prime divisors of such numbers. For every modulus $m$ in the covering, these theorems guarantee the existence of at least one prime divisor of $10^m-1$ that does not divide $10^n-1$ for any integer $n