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\begin{center}
\vskip 1cm{\LARGE\bf Polygonal, Sierpi\'{n}ski, and Riesel Numbers}
\vskip 1cm
Daniel Baczkowski and Justin Eitner\\
Department of Mathematics\\
The University of Findlay\\
Findlay, OH 45840 \\
USA \\
\href{mailto:baczkowski@findlay.edu}{\tt baczkowski@findlay.edu}\\
\href{mailto:eitnerj@findlay.edu}{\tt eitnerj@findlay.edu}\\
\ \\
Carrie E. Finch\footnote{Partially supported by a Lenfest Grant.} and Braedon Suminski\footnote{Supported by the R. E. Lee Summer Scholar program.} \\
Department of Mathematics\\
Washington and Lee University\\
Lexington, VA 24450\\
USA\\
\href{mailto:finchc@wlu.edu}{\tt finchc@wlu.edu}\\
\href{mailto:suminskib14@mail.wlu.edu}{\tt suminskib14@mail.wlu.edu}\\
\ \\
Mark Kozek\\
Department of Mathematics \\
Whittier College\\
Whittier, CA 90608  \\
USA\\
\href{mailto:mkozek@whittier.edu}{\tt mkozek@whittier.edu}\\
\end{center}

\vskip .2in

\begin{abstract}
In this paper, we show that there are infinitely many Sierpi\'{n}ski
numbers in the sequence of triangular numbers, hexagonal numbers, and
pentagonal numbers. We also show that there are infinitely many Riesel
numbers in the same sequences. Furthermore, we show that there are
infinitely many $n$-gonal numbers that are simultaneously
Sierpi\'{n}ski and Riesel.
\end{abstract}


\section{Introduction}

Polygonal numbers are those that can be expressed geometrically by an
arrangement of equally spaced points. For example, a positive integer
$n$ is a triangular number if $n$ dots can be arranged in the form of
an equilateral triangle. Similarly, $n$ is a square number if $n$ dots
can be arranged in the form of a square. The diagram below represents
the first four hexagonal numbers, which are 1, 6, 15, and 28.

\begin{center}
\includegraphics[scale=.15]{hex.eps}
\end{center}

In 1960, Sierpi\'{n}ski \cite{sierp} showed that there are infinitely
many odd positive integers $k$ with the property that $k\cdot 2^n + 1$
is composite for all positive integers $n$. Such an integer $k$ is
called a Sierpi\'{n}ski number in honor of Sierpi\'{n}ski's work. Two
years later, Selfridge (unpublished) showed that $78557$ is a
Sierpi\'{n}ski number. To this day, this is the smallest known
Sierpi\'{n}ski number. As of this writing, there are six candidates
smaller than 78557 to consider: 10223, 21181, 22699, 24737, 55459,
67607. See \url{http://www.seventeenorbust.com} for the most up-to-date
information.

Riesel numbers are defined in a similar way: an odd positive integer $k$ is Riesel if $k\cdot 2^n - 1$ is composite for all positive integers $n$. These were first investigated by Riesel in 1956 \cite{riesel}. The smallest known Riesel number is 509203. As of this writing there are 50 remaining candidates smaller that 509203 to consider. See \url{http://www.prothsearch.net/rieselprob.html} for the most recent information.

The tool used to construct these numbers is a covering system. A collection of congruences 
\begin{align*}
r_1&  \pmod {m_1}\\
r_2&  \pmod {m_2}\\
&\vdots \\
r_t&  \pmod {m_t}
\end{align*}
is called a {\it covering system of congruences}, also called a {\it covering system}, if each integer $n$ satisfies $n \equiv r_i \pmod {m_i}$ for some $1 \leq i \leq t$. This technique was first introduced by Erd\H{o}s who later used the idea to show that there are infinitely many odd numbers that are not of the form $2^k + p$, where $p$ is a prime \cite{erdos}.

Previous work has been done to show an intersection between Sierpi\'{n}ski or Riesel numbers with familiar integer sequences such as the Fibonacci numbers \cite{ip,lucamejia} and the Lucas numbers \cite{bff}. Perfect power Sierpi\'{n}ski numbers and Riesel numbers have been studied in depth; in particular, there are infinitely many Sierpi\'{n}ski numbers of the form $k^r$ for any fixed positive integer $r$ \cite{chen,ffk}. For Riesel numbers, there are infinitely many $k$ such that $k^r$ is Riesel for values of $r$ with $\gcd(r,12) \leq 3$ \cite{chen} and for $\gcd(r,105) = 1$ \cite{fj}. In this paper we expand on these findings by considering the intersection of sequences of polygonal numbers with Sierpi\'{n}ski and Riesel numbers. As the $k$th polygonal number for an $n$-sided polygon is given by $\frac{1}{2}(k^2(n-2)-k(n-4))$, which is quadratic in $k$, we build on techniques for constructing perfect power Sierpi\'{n}ski numbers and binomial Sierpi\'{n}ski and Riesel numbers (cf.\ \cite{ffk,nonlin}). Through the use of coverings, we construct polygonal numbers that are also Riesel numbers and Sierpi\'{n}ski numbers. 

Cohen and Selfridge showed that there are infinitely many numbers that are simultaneously Sierpi\'{n}ski and Riesel \cite{cohenself}. The smallest Sierpi\'{n}ski-Riesel number that came from their construction has 26 digits. Several others also produced Sierpi\'{n}ski-Riesel numbers; for example, Brier (unpublished) produced an example with 41 digits in 1998, and Gallot (unpublished) produced an example with 27 digits in 2000. In 2008, an example with 24 digits was produced \cite{ffk}. Recently,
a Sierpi\'{n}ski-Riesel number that is 22 digits long was discovered by Clavier (unpublished). Entry A180247 in the Online Encyclopedia of Integer Sequences has more information about these results. (See \texttt{http://oeis.org/A180247}.) We use similar techniques in this paper to find polygonal numbers that are Sierpi\'{n}ski-Riesel.

All computations were performed using the computer algebra system Maple.

\section{Triangular and hexagonal numbers}

\subsection{Triangular-Sierpi\'{n}ski numbers}

Let $T_k$ denote the $k^{\textrm{th}}$ triangular number. That is, $T_k = \frac{k(k+1)}{2}$. Now, consider the implications in Table~\ref{TS} below.

\begin{table}[H]
\begin{align*}
&n \equiv 1 \nmod 2 && \& && k \equiv 1 \nmod {3} &\Longrightarrow && 3 \mid (T_k\cdot 2^n + 1)\\
&n \equiv 0 \nmod 3 && \& && k \equiv  3 \nmod {7} &\Longrightarrow && 7 \mid (T_k\cdot 2^n + 1)\\
&n \equiv 2 \nmod 4 && \& && k \equiv 1 \orr 3 \nmod {5} &\Longrightarrow && 5 \mid (T_k\cdot 2^n + 1)\\
&n \equiv 4 \nmod {8} && \& && k \equiv 1 \orr 15 \nmod {17} &\Longrightarrow &&17 \mid (T_k\cdot 2^n + 1)\\
&n \equiv 8 \nmod {12} && \& && k \equiv 4 \orr 8 \nmod {13} &\Longrightarrow && 13 \mid (T_k\cdot 2^n + 1)\\
&n \equiv 16 \nmod {24} && \& && k \equiv 53 \orr 187 \nmod {241} &\Longrightarrow && 241 \mid (T_k\cdot 2^n + 1)
\end{align*}
\caption{}
\label{TS}
\end{table}

The congruences for $n$ in Table~\ref{TS} form a covering. In each row, the congruence for $k$ results in $T_k\cdot 2^n + 1$ being divisible by one of the primes in $\mathcal{P}=\{3, 5, 7, 17, 13, 241\}$ for every positive integer in the congruence class for $n$.
Assume that our values of $k$ in each row will be chosen large enough from the congruence for $k$ such that $T_k\cdot 2^n+1$ is larger than its prime divisor in that row.  It follows that each of these $T_k\cdot 2^n + 1$ must be composite for every positive integer in this congruence class. To ensure that $T_k$ is odd (in order to satisfy the definition of a Sierpi\'{n}ski number), we also include the congruence
$k\equiv 1\orr 2$ (mod $4$). If $k = 4\ell + 1$, we have
$$T_k = \frac{1}{2}k(k+1) = \frac{1}{2}(4\ell + 1)(4\ell + 2) = (4\ell + 1)(2\ell + 1),$$
and if $k = 4\ell + 2$, we have
$$T_k = \frac{1}{2}k(k+1) = \frac{1}{2}(4\ell + 2)(4\ell + 3) = (2\ell + 1)(4\ell + 3),$$
 both of which are clearly odd.


Now, we find the intersection of all congruences for $k$ to find a $T_k$ that is a Sierpi\'{n}ski number.  Using the Chinese remainder theorem, we have the following result.

    \begin{theorem}\label{TriSierp}
    There are infinitely many Sierpi\'{n}ski numbers in the sequence of triangular numbers.
    \end{theorem}

    The smallest solution to the congruences for $k$ that we find using the Chinese remainder theorem is $698953$, and if $k$ is a natural number with $k \equiv 698953 \pmod {4 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 17 \cdot 241}$, then $T_k$ is a Sierpi\'{n}ski number. Thus, the smallest Triangular-Sierpi\'{n}ski number from this construction is 244267997581.

    \subsection{Triangular-Riesel numbers}

    Consider the implications in Table~\ref{TR} below.

\begin{table}[H]
\begin{align*}
&n \equiv 0 \nmod 2 && \& && k \equiv 1 \nmod {3} & \Longrightarrow && 3\mid (T_k\cdot 2^n - 1)\\
&n \equiv 0 \nmod 3 && \& && k \equiv 1 \orr 5 \nmod {7} & \Longrightarrow && 7\mid (T_k\cdot 2^n - 1)\\
&n \equiv 1 \nmod 4 && \& && k \equiv 2 \nmod {5} & \Longrightarrow && 5\mid (T_k\cdot 2^n - 1)\\
&n \equiv 7 \nmod {8} && \& && k \equiv 8 \nmod {17} & \Longrightarrow && 17\mid (T_k\cdot 2^n - 1)\\
&n \equiv 11 \nmod {12} && \& && k \equiv 5 \orr 7 \nmod {13} & \Longrightarrow && 13\mid (T_k\cdot 2^n - 1)\\
&n \equiv 16 \nmod {24} && \& && k \equiv  5 \orr 235 \nmod {241} & \Longrightarrow && 241\mid (T_k\cdot 2^n - 1)
\end{align*}
\caption{}
\label{TR}
\end{table}

We also include $k\equiv 1 \orr 2$ (mod $4$) to ensure $T_k$ is odd. Once again, the congruences for $n$ in Table~\ref{TR} form a covering. In each row, the congruences for $k$ result in $T_k\cdot 2^n - 1$ being divisible by one of the primes in the set $$\mathcal{P}=\{3, 7, 5, 17, 13, 241\}$$ for every positive integer $n$ in the same row. As before, assume that our values of $k$ will be chosen large enough from those in the congruence for $k$ so that $T_k$ is larger than any of the primes in the set $\mathcal{P}$.  It follows that each of these $T_k\cdot 2^n - 1$ must be composite for every positive integer $n$. We then find the intersection of all congruences for $k$ to find a $T_k$ that is a Riesel number.  By use of the Chinese remainder theorem, we see that the smallest residue class that is in the intersection of all congruences for $k$ is $k \equiv 888802 \pmod {4\cdot 3 \cdot 7 \cdot 5 \cdot 17 \cdot 13 \cdot 241}$. Thus, for every such value of $k$, the triangular number $T_k$ is Riesel. Hence, we have the following theorem.

\begin{theorem}\label{TriRies}
There are infinitely many Riesel numbers in the sequence of triangular numbers.
\end{theorem}

\subsection{Hexagonal-Sierpi\'{n}ski and Riesel numbers}

Let $H_k$ denote the $k^{\textrm{th}}$ hexagonal number. That is, $H_k = 2k^2-k$. Notice that if $k = 2\ell + 1$, we have $T_k = \frac{1}{2}k(k+1) = 2\ell^2 + 3\ell + 1 = H_{\ell + 1}$. That is, when $k$ is odd, the triangular number $T_k$ is also a hexagonal number. Thus, if we include the congruences from Table~\ref{TS} and $k \equiv 1$ (mod $2$), then we will have a subset of the triangular numbers that are also Sierpi\'{n}ski numbers in addition to hexagonal numbers. The smallest such $k$ is 698953, and all positive integers $k$ congruent to 698953 modulo $4 \cdot 3 \cdot 7 \cdot 5 \cdot 17 \cdot 13 \cdot 241$ also give $T_k$ which are also Sierpi\'{n}ski and hexagonal. Similarly, the positive integers $k$ that are congruent to 2916817 modulo $4 \cdot 3 \cdot 7 \cdot 5 \cdot 17 \cdot 13 \cdot 241$ yield $T_k$ which are Riesel and also hexagonal. Thus, we have the following corollaries.

\begin{corollary}
There are infinitely many Sierpi\'{n}ski numbers in the sequence of hexagonal numbers.
\end{corollary}

\begin{corollary}
There are infinitely many Riesel numbers in the sequence of hexagonal numbers.
\end{corollary}

\subsection{Triangular-Sierpi\'{n}ski-Riesel numbers}

Table~\ref{TSR} below gives congruences for $k$ to construct triangular numbers $T_k$ that are simultaneously Sierpi\'{n}ski and Riesel.

\begin{table}[H]
\begin{align*}
&n \equiv 1 \nmod 2 && \& && k \equiv 1 \nmod {3} & \Longrightarrow && 3\mid (T_k\cdot 2^n + 1)\\
&n \equiv 1 \nmod 3 && \& && k \equiv 2\orr 4 \nmod {7} & \Longrightarrow && 7\mid (T_k\cdot 2^n + 1)\\
&n \equiv 5 \nmod 9 && \& && k \equiv 23\orr 49 \nmod {73} & \Longrightarrow && 73\mid (T_k\cdot 2^n + 1)\\
&n \equiv 6 \nmod {12} && \& && k \equiv 1\orr 11 \nmod {13} & \Longrightarrow && 13\mid (T_k\cdot 2^n + 1)\\
&n \equiv 8 \nmod {18} && \& && k \equiv 6\orr 12 \nmod {19} & \Longrightarrow && 19\mid (T_k\cdot 2^n + 1)\\
&n \equiv 2 \nmod {36} && \& && k \equiv 15\orr 21 \nmod {37} & \Longrightarrow && 37\mid (T_k\cdot 2^n + 1)\\
&n \equiv 20 \nmod {36} && \& && k \equiv 24\orr 84 \nmod {109} & \Longrightarrow && 109\mid (T_k\cdot 2^n + 1)\\ 
&n \equiv 4 \nmod {5} && \& && k \equiv 13\orr 17 \nmod {31} & \Longrightarrow && 31\mid (T_k\cdot 2^n + 1)\\ 
&n \equiv 6 \nmod {10} && \& && k \equiv 3\orr 7 \nmod {11} & \Longrightarrow && 11\mid (T_k\cdot 2^n + 1)\\ 
&n \equiv 8 \nmod {20} && \& && k \equiv 9\orr 31 \nmod {41} & \Longrightarrow && 41\mid (T_k\cdot 2^n + 1)\\ 
&n \equiv 0 \nmod {15} && \& && k \equiv 69\orr 81 \nmod {151} & \Longrightarrow && 151\mid (T_k\cdot 2^n + 1)\\ 
&n \equiv 12 \nmod {60} && \& && k \equiv 20\orr 40 \nmod {61} & \Longrightarrow && 61\mid (T_k\cdot 2^n + 1)\\ 
\cline{0-9}
&n \equiv 0 \nmod 2 && \& && k \equiv 1 \nmod {3} & \Longrightarrow && 3\mid (T_k\cdot 2^n - 1)\\
&n \equiv 1 \nmod 4 && \& && k \equiv 2 \nmod {5} & \Longrightarrow && 5\mid (T_k\cdot 2^n - 1)\\
&n \equiv 7 \nmod {8} && \& && k \equiv 8 \nmod {17} & \Longrightarrow && 17\mid (T_k\cdot 2^n - 1)\\
&n \equiv 11 \nmod {16} && \& && k \equiv 128 \nmod {257} & \Longrightarrow && 257\mid (T_k\cdot 2^n - 1)\\
&n \equiv 11 \nmod {24} && \& && k \equiv 90\orr 150 \nmod {241} & \Longrightarrow && 241\mid (T_k\cdot 2^n - 1)\\
&n \equiv 3 \nmod {48} && \& && k \equiv 41\orr 55 \nmod {97} & \Longrightarrow && 97\mid (T_k\cdot 2^n - 1)\\
&n \equiv 19 \nmod {48} && \& && k \equiv 315\orr 357 \nmod {673} & \Longrightarrow && 673\mid (T_k\cdot 2^n - 1)
\end{align*}
\caption{}
\label{TSR}
\end{table}

In Table~\ref{TSR}, the congruences for $n$ above the horizontal line form a covering. This part of the table ensures that the congruences for $k$ yield a Sierpi\'{n}ski number $T_k$. In addition, the congruences for $n$ below the horizontal line also form a covering. Thus, the bottom part of the table ensures that the congruences for $k$ yield a Riesel number. Notice that the congruences for $k$ above the line and those below the line are compatible; the only modulus that is repeated in the two parts of the table is 3, and in both instances, we have $k~\equiv~1$~(mod~$3$).

Now we include the congruence $k \equiv 1 \orr 2$ (mod $4$) to ensure that $T_k$ is odd, and then use the Chinese remainder theorem to combine all of the congruences for $k$. The smallest solution to this set of congruences is $$k \equiv 92290397124858700233022 \pmod {270351155161021554764103899940}.$$ We conclude there are infinitely many Sierpi\'{n}ski-Riesel numbers in the sequence of triangular numbers, and the smallest example resulting from this construction is $$4258758700732063521204486546872386447899742753.$$ We state this result as a theorem below.

\begin{theorem} There are infinitely many triangular numbers that are simultaneously Sierpi\'{n}ski numbers and Riesel numbers.
\end{theorem}



\subsection{Hexagonal-Sierpi\'{n}ski-Riesel numbers}

If we again include the congruence $k \equiv 1$ (mod $2$) with the congruences in the previous subsection, we then have triangular numbers that are also hexagonal, in addition to being both Sierpi\'{n}ski and Riesel. Combining these congruences using the Chinese remainder theorem, we find the smallest solution to this set of congruences is $$k \equiv 24743267730877977274574137 \pmod{270351155161021554764103899940},$$ then $T_k$ is hexagonal, Sierpi\'{n}ski, and Riesel. We conclude with the following:

\begin{theorem} There are infinitely many hexagonal numbers that are simultaneously Sierpi\'{n}ski and Riesel numbers.
\end{theorem}


\section{Pentagonal numbers}

Let $P_k$ denote the $k^{\textrm{th}}$ pentagonal number. We then have $P_k = \frac{1}{2}k(3k-1)$. In this section, we show that there are infinitely many pentagonal-Sierpi\'{n}ski numbers, infinitely many pentagonal-Riesel numbers, and infinitely many pentagonal numbers that are simultaneously Sierpi\'{n}ski and Riesel.

\subsection{Pentagonal-Sierpi\'{n}ski numbers}

Consider the implications in Table~\ref{PS} below.

\begin{table}[H]
\begin{align*}
&n \equiv 1 \pmod 2 && \& && k \equiv 1 \pmod {3} & \Longrightarrow && 3\mid (P_k\cdot 2^n + 1)\\
&n \equiv 2 \pmod 3 && \& && k \equiv 2 \orr 3 \pmod {7} & \Longrightarrow && 7\mid (P_k\cdot 2^n + 1)\\
&n \equiv 2 \pmod 4 && \& && k \equiv 1 \pmod {5} & \Longrightarrow && 5\mid (P_k\cdot 2^n + 1)\\
&n \equiv 4 \pmod {8} && \& && k \equiv 1 \orr 5 \pmod {17} & \Longrightarrow && 17\mid (P_k\cdot 2^n + 1)\\
&n \equiv 0 \pmod {12} && \& && k \equiv 3 \orr 6 \pmod {13} & \Longrightarrow && 13\mid (P_k\cdot 2^n + 1)\\
&n \equiv 16 \pmod {24} && \& && k \equiv 189 \orr 213 \pmod {241} & \Longrightarrow && 241\mid (P_k\cdot 2^n + 1)
\end{align*}
\caption{}
\label{PS}
\end{table}

Observe that if $k \equiv 1 \orr 2$ (mod $4$), then $P_k$ is odd. To see this, notice that if $k = 4\ell + 1$, we have $$P_k = \frac{1}{2}k(3k-1) = \frac{1}{2}(4\ell + 1)(12\ell+3-1) = (4\ell + 1)(6\ell + 1),$$ and if $k = 4\ell + 2$ we have $$P_k = \frac{1}{2}k(3k-1) = \frac{1}{2}(4\ell + 2)(12\ell+6-1) = (2\ell+1)(12\ell + 5),$$ which are both clearly odd. Thus, we also include $k \equiv 1 \orr 2$ (mod $4$) in order to construct Sierpi\'{n}ski numbers in this sequence.

Once again, the congruences for $n$ in Table~\ref{PS} form a covering. In each row, the congruences for $k$ and $n$ result in $P_k\cdot 2^n + 1$ being divisible by one of the primes in the set $\mathcal{P}=\{3,5,7,13,17, 241\}$. As before, assume that our values of $k$ will be chosen large enough from those in the congruence classes for $k$ so that $P_k$ is larger than any of the primes in the set $\mathcal{P}$.  It follows that each of these $P_k\cdot 2^n - 1$ must be composite for every positive integer $n$. We deduce the intersection of all congruences for $k$ to find a $P_k$ that is a Sierpi\'{n}ski number.  Using the Chinese remainder theorem for the congruences for $k$, we find that there are infinitely many such $k$, and the smallest solution that arises out of these congruences is

$$k \equiv 56101 \pmod {22369620}.$$

We conclude the following:

\begin{corollary}
There are infinitely many pentagonal numbers that are Sierpi\'{n}ski numbers.
\end{corollary}

\subsection{Pentagonal-Riesel numbers}

In this section, we demonstrate the following result:

\begin{theorem}
There are infinitely many pentagonal numbers that are Riesel numbers.
\end{theorem}

We again prove this statement using a covering of the integers, shown in Table~\ref{PR} below.

\begin{table}[H]
\begin{align*}
&n \equiv 0 \pmod 2 && \& && k \equiv 1 \pmod {3} & \Longrightarrow && 3\mid (P_k\cdot 2^n - 1)\\
&n \equiv 2 \pmod 3 && \& && k \equiv 6 \pmod {7} & \Longrightarrow && 7\mid (P_k\cdot 2^n - 1)\\
&n \equiv 3 \pmod 4 && \& && k \equiv 3\orr 4 \pmod {5} & \Longrightarrow && 5\mid (P_k\cdot 2^n - 1)\\
&n \equiv 1 \pmod {8} && \& && k \equiv 10\orr 13 \pmod {17} & \Longrightarrow && 17\mid (P_k\cdot 2^n - 1)\\
&n \equiv 1 \pmod {12} && \& && k \equiv 11 \pmod {13} & \Longrightarrow && 13\mid (P_k\cdot 2^n - 1)\\
&n \equiv 21 \pmod {24} && \& && k \equiv 61 \orr 100 \pmod {241} & \Longrightarrow && 241\mid (P_k\cdot 2^n - 1)
\end{align*}
\caption{}
\label{PR}
\end{table}

The congruences for $n$ form a covering of the integers, so we again use the Chinese remainder theorem to combine the congruences for $k$ (including $k \equiv 1\orr 2$ (mod $4$)); the smallest solution for $k$ that satisfies all of the congruences in the table is $$k \equiv 590029 \pmod {22369620}.$$ For any of these solutions for $k$, the expression $P_k\cdot 2^n - 1$ is divisible by one of the primes in the set $\mathcal{P} = \{3,5,7,13,17, 241\}$.

\subsection{Pentagonal-Sierpi\'{n}ski-Riesel}

We show now that there are infinitely many pentagonal numbers that are simultaneously Sierpi\'{n}ski and Riesel. Consider the congruences in Table~\ref{PSR}.




In Table~\ref{PSR}, the congruences for $n$ above the horizontal line form a covering of the integers. Thus, the congruences for $k$ above this line yield pentagonal numbers $P_k$ that are Sierpi\'{n}ski. Similarly, the congruences for $n$ below the horizontal line also form a covering of the integers. Thus, the corresponding congruences for $k$ in the bottom part of the table yield pentagonal numbers $P_k$ that are also Riesel. Again, the congruences for $k$ above and below the line are compatible; the only repeated modulus is $3$, and in both parts of the table, we have $k \equiv 1$ (mod~$3$). When we also include $k \equiv 1 \orr 2$ (mod $4$) to make sure that the resulting $P_k$ is also an odd integer, we find that there are $2^{17}$ solutions for $k$ modulo $$M =4 \cdot 3 \cdot 5 \cdot 7 \cdot 11  \cdot 13 \cdot 17 \cdot 19 \cdot 31  \cdot 37  \cdot 41   \cdot 61 \cdot 73  \cdot 97  \cdot 109  \cdot 151 \cdot 241 \cdot 257 \cdot 673.$$ The smallest such solution is
$$k \equiv 180972518141277924651218 \pmod {M},$$ 
yielding the smallest pentagonal-Sierpi\'{n}ski-Riesel from this construction: $$49126578483592751315774667185145775331460999677.$$ Thus, we have shown the following result.

\begin{theorem}
There are infinitely many pentagonal numbers that are simultaneously Sierpi\'{n}ski and Riesel numbers.
\end{theorem}


\section{Polygonal numbers}

\begin{theorem}
For infinitely many values of $s$, there are infinitely many $s$-gonal numbers that are Sierpi\'{n}ski numbers.
\end{theorem}


\begin{table}[H]
\begin{align*}
&n \equiv 0 \nmod 2 && \& && k \equiv 2 \nmod {3} & \Longrightarrow && 3\mid (P_k\cdot 2^n + 1)\\
&n \equiv 1 \nmod 4 && \& && k \equiv 3 \orr 4 \nmod {5} & \Longrightarrow && 5\mid (P_k\cdot 2^n + 1)\\
&n \equiv 1 \nmod {10} && \& && k \equiv 2 \nmod {11} & \Longrightarrow && 11\mid (P_k\cdot 2^n + 1)\\
&n \equiv 7 \nmod {8} && \& && k \equiv 9 \orr 14 \nmod {17} & \Longrightarrow && 17\mid (P_k\cdot 2^n + 1)\\
&n \equiv 3 \nmod {18} && \& && k \equiv 15 \orr 17 \nmod {19} & \Longrightarrow && 19\mid (P_k\cdot 2^n + 1)\\ 
&n \equiv 11 \pmod {24} && \& && k \equiv 162 \orr 240 \pmod {241} & \Longrightarrow && 241\mid (P_k\cdot 2^n + 1)\\
&n \equiv 3 \pmod {16} && \& && k \equiv 140 \orr 203 \pmod {257} & \Longrightarrow && 257\mid (P_k\cdot 2^n + 1)\\
&n \equiv 43 \pmod {48} && \& && k \equiv 32 \orr 33 \pmod {97} & \Longrightarrow && 97\mid (P_k\cdot 2^n + 1)\\
&n \equiv 27 \pmod {48} && \& && k \equiv 112 \orr 337 \pmod {673} & \Longrightarrow && 673\mid (P_k\cdot 2^n + 1)\\ 
\cline{0-9}
&n \equiv 1 \nmod 2 && \& && k \equiv 2 \nmod {3} & \Longrightarrow && 3\mid (P_k\cdot 2^n - 1)\\
&n \equiv 6 \nmod {10} && \& && k \equiv 2 \nmod {11} & \Longrightarrow && 11\mid (P_k\cdot 2^n - 1)\\
&n \equiv 4 \nmod {12} && \& && k \equiv 4 \orr 5 \nmod {13} & \Longrightarrow && 13\mid (P_k\cdot 2^n - 1)\\
&n \equiv 12 \nmod {18} && \& && k \equiv 15 \orr 17 \nmod {19} & \Longrightarrow && 19\mid (P_k\cdot 2^n - 1)\\
&n \equiv 24 \nmod {36} && \& && k \equiv 29 \orr 33 \nmod {37} & \Longrightarrow && 37\mid (P_k\cdot 2^n - 1)\\
&n \equiv 10 \nmod {20} && \& && k \equiv 19 \orr 36 \nmod {41} & \Longrightarrow && 41\mid (P_k\cdot 2^n - 1)\\
&n \equiv 58 \nmod {60} && \& && k \equiv 50 \orr 52 \nmod {61} & \Longrightarrow && 61\mid (P_k\cdot 2^n - 1)\\
&n \equiv 6 \nmod {36} && \& && k \equiv 83 \orr 99 \nmod {109} & \Longrightarrow && 109\mid (P_k\cdot 2^n - 1)\\
&n \equiv 2 \nmod 3 && \& && k \equiv 6 \nmod {7} & \Longrightarrow && 7\mid (P_k\cdot 2^n - 1)\\
&n \equiv 0 \pmod {9} && \& && k \equiv 1 \orr 48 \pmod {73} & \Longrightarrow && 73\mid (P_k\cdot 2^n - 1)\\
&n \equiv 4 \nmod 5 && \& && k \equiv 22 \orr 30 \nmod {31} & \Longrightarrow && 31\mid (P_k\cdot 2^n - 1)\\
&n \equiv 7 \pmod {15} && \& && k \equiv 28 \orr 73 \pmod {151} & \Longrightarrow && 151\mid (P_k\cdot 2^n - 1)
\end{align*}
\caption{}
\label{PSR}
\end{table}


\begin{proof}
Observe that the $k^{\textrm{th}}$ $s$-gonal number is given by $$\P_k = \P_k(s) = \frac{1}{2}k\big((s-2)k-(s-4)\big).$$ Using the congruences in Table~\ref{TS}, if $s \equiv 3$ (mod $p$) for each $p$ in the set $$\mathcal{P}:=\{3, 5, 7, 13, 17, 241\},$$ then we have
\[
T_k \equiv \P_k \pmod{\prod_{p\in\mathcal{P}} p}
\]
and

\begin{table}[H]
\begin{align*}
&n \equiv 1 \nmod 2 && \& && k \equiv 1 \nmod {3} & \Longrightarrow && 3 \mid (S_k\cdot 2^n + 1)\\
&n \equiv 0 \nmod 3 && \& && k \equiv  3 \nmod {7} & \Longrightarrow && 7 \mid (S_k\cdot 2^n + 1)\\
&n \equiv 2 \nmod 4 && \& && k \equiv 1 \orr 3 \nmod {5} & \Longrightarrow && 5 \mid (S_k\cdot 2^n + 1)\\
&n \equiv 4 \nmod {8} && \& && k \equiv 1 \orr 15 \nmod {17} & \Longrightarrow && 17 \mid (S_k\cdot 2^n + 1)\\
&n \equiv 8 \nmod {12} && \& && k \equiv 4 \orr 8 \nmod {13} & \Longrightarrow && 13 \mid (S_k\cdot 2^n + 1)\\
&n \equiv 16 \nmod {24} && \& && k \equiv 53 \orr 187 \nmod {241} & \Longrightarrow && 241 \mid (S_k\cdot 2^n + 1)
\end{align*}
\caption{}
\label{InfPolyS}
\end{table}

This implies that the expression $\P_k\cdot 2^n + 1$ is composite for all positive integers $n$ if $k$ lies in the intersection of the congruence classes listed in the table above since the congruences for $n$ form a covering of the integers. If we also include the congruence $k \equiv 1$ (mod $4$), then the resulting polygonal number $\P_k$ is odd since $k = 4\ell + 1$ implies $$\P_k = (4\ell+1)(2\ell s - 4\ell +1).$$ The conclusion follows.
\end{proof}
Using the same technique, we also deduce the following results.

\begin{theorem}
For infinitely many values of $s$, there are infinitely many $s$-gonal numbers that are Riesel numbers.
\end{theorem}

\begin{theorem}
For infinitely many values of $s$, there are infinitely many $s$-gonal numbers that are Sierpi\'{n}ski-Riesel numbers.
\end{theorem}


\section {Acknowledgments}

The authors thank the referee for the valuable suggestions. The third
author wishes to thank the Department of Mathematics at The University
of Findlay for the hospitality during her visit. The fourth author is
grateful for the hospitality of the Mathematics Department at
Washington and Lee University during his visit. The authors gratefully
acknowledge the computational support from Washington and Lee
University student Elliot Emadian ('17) and Professor Gregg Whitworth
(Biology).

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\bibitem{cohenself} F. Cohen and J. L. Selfridge, Not every number is the sum or difference of two prime powers, {\it Math. Comp} {\bf 29} (1975), 79--81.

\bibitem{erdos} P. Erd\H{o}s, On integers of the form $2^k + p$ and some related problems, {\it Summa Brasil. Math.} {\bf 2} (1950), 113--123.

\bibitem{ffk} M. Filaseta, C. Finch, and M. Kozek, On powers associated with Sierpi\'nski numbers, Riesel numbers and Polignac's conjecture, {\it J.~Number Theory} {\bf 128} (2008), 1916--1940.

\bibitem{nonlin} C. Finch, J. Harrington, and L. Jones, Nonlinear Sierpi\'{n}ski and Riesel numbers, {\it J.~Number Theory} {\bf 133} (2013), 534--544.

\bibitem{fj} C. Finch and L. Jones, Perfect power Riesel numbers, {\it J.~Number Theory} {\bf 150} (2015), 41--46.

\bibitem{ip} D. Ismailescu and P. S. Park, On pairwise intersections of the Fibonacci, Sierpi\'{n}ski, and Riesel sequences. {\it J.~Integer Seq.} {\bf 16} (2013), Article 13.8.8.

\bibitem{lucamejia} F. Luca and V. J. Mej\'{i}a Huguet, Fibonacci-Riesel and Fibonacci-Sierpi\'{n}ski numbers, {\it Fibonacci Quart.} {\bf 46/47} (2008/09), 216--219.

\bibitem{riesel} H. Riesel, N\r{a}gra stora primtal, {\it Elementa} {\bf 39} (1956), 258--260.

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A07; Secondary 11Y55.

\noindent \emph{Keywords: }
Sierpi\'{n}ski number, Riesel number, triangular number, covering congruence.

\bigskip
\hrule
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\noindent (Concerned with sequences
\seqnum{A000217},
\seqnum{A000290},
\seqnum{A000326},
\seqnum{A000384}, and
\seqnum{A180247}.)

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\vspace*{+.1in}
\noindent
Received  November 25 2014;
revised version received  June 19 2015; July 9 2015.
Published in {\it Journal of Integer Sequences}, July 16 2015.

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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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