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\begin{center}
\vskip 1cm{\LARGE\bf 
Arithmetic Progressions of \\
\vskip .1in
$b$-Prodigious Numbers
}
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\begin{minipage}{3.1in}
\begin{center}
Michael~Gohn\\
Dept.\ of Mathematics and Computer Sci.\\
DeSales University\\2755 Station Ave.\\
Center Valley, PA 18034\\
USA\\
\href{mailto:mg7785@desales.edu}{\tt mg7785@desales.edu} 
\end{center}
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\quad
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\begin{center}
Joshua~Harrington\\
Department of Mathematics\\
Cedar Crest College\\
100 College Dr. \\
Allentown, PA 18104 \\
USA\\
\href{mailto:joshua.harrington@cedarcrest.edu}{\tt joshua.harrington@cedarcrest.edu}
\end{center}
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\bigskip

\begin{minipage}{3.1in}
\begin{center}
Sophia~Lebiere\\Department of Mathematics \\
Tufts University\\419 Boston Ave.\\
Medford, MA 02155 \\
USA\\
\href{mailto:slebie01@tufts.edu}{\tt slebie01@tufts.edu} 
\end{center}
\end{minipage}
\quad
\begin{minipage}{3.1in}
\begin{center}
Hani~Samamah\\Department of Mathematics \\
University of Florida\\Gainesville, FL 32611 \\
USA\\
\href{mailto:hanisamamah@ufl.edu}{\tt hanisamamah@ufl.edu}\\
\ \\
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\bigskip

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\begin{center}
Kyla~Shappell\\Department of Mathematics \\
Spring Hill College\\4000 Dauphin St.\\
Mobile, AL 36608\\
USA\\
\href{mailto:kyla.m.shappell@email.shc.edu}{\tt kyla.m.shappell@email.shc.edu}
\end{center}
\end{minipage}
\begin{minipage}{3.1in}
\begin{center}
Tony~W.~H.~Wong\\
Department of Mathematics\\
Kutztown University of Pennsylvania\\
15200 Kutztown Road \\
Kutztown, PA 19530 \\
USA\\
\href{mailto:wong@kutztown.edu}{\tt wong@kutztown.edu}
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\begin{abstract}
A positive integer $n$ is called a \emph{$b$-prodigious number} if $n$ is divisible by the product of its non-zero base-$b$ digits.  In this article, we investigate the maximum length of an arithmetic progression of $b$-prodigious numbers and the maximal length of consecutive sequences of $b$-prodigious numbers.
\end{abstract}

\section{Introduction}

In the {\it On-Line Encyclopedia of Integer Sequences}, sequence \seqnum{A055471} \cite{oeis} is the sequence of positive integers that are divisible by the product of their nonzero digits.  On the French website Diophante \cite{diophante}, this sequence is called \emph{nombres prodigieux}, which translates to \emph{prodigious numbers}.  In recognition of this translation, for an integer $b\geq 2$, we say that a positive integer $n$ is a \emph{$b$-prodigious number} if it is divisible by $p_b(n)$, which is defined as the product of the nonzero digits of $n$ in its base-$b$ representation.  Clearly, every positive integer is a $2$-prodigious number.  For $b=10$, De~Koninck and Luca \cite{dl2007} showed that for large enough values of $x$, the number of $b$-prodigious numbers less than $x$, denoted by $N_0(x)$, satisfies $x^{0.495}&lt;N_0(x)&lt;x^{0.901}$.  This result was later improved and generalized to arbitrary base by Sanna \cite{sanna}. 

In a similar manner, a positive integer $n$ is called a \emph{Niven number} if it is divisible by the sum of its digits.  More generally, for an integer $b\geq 2$, a \emph{$b$-Niven number} is a positive integer $n$ that is divisible by $s_b(n)$, which is defined as the sum of the digits of $n$ in its base-$b$ representation.  In 1992, Cooper and Kennedy \cite{ck} showed that there exist sequences of $20$ consecutive integers that are all Niven numbers and further proved that no sequences of $21$ consecutive Niven numbers exist. Their result was partially generalized in 1994 by Grundman \cite{grundman}, who showed that every sequence of consecutive integers of length at least $2b+1$ must contain a term that is not $b$-Niven.  Grundman's bound on the maximum length of consecutive Niven numbers was shown to be tight for $b\in\{2,3\}$ by Cai \cite{cai} and when $b\geq 4$ by Wilson \cite{wilson}.  In this paper, we establish analogous results for $b$-prodigious numbers.

Let $b\geq 3$ be a positive integer and let $\ell$ be the smallest positive integer that does not divide $b$.  In Section~\ref{sec:maximum} of this paper, we show that there exist sequences of $b+\ell$ consecutive integers that are all $b$-prodigious and we further show that no sequence of $b+\ell+1$ consecutive $b$-prodigious numbers exist.  In fact, we obtain this result by considering consecutive terms in an arithmetic progression.  In Section~\ref{sec:maximal}, we investigate maximal length of consecutive integers whose terms are all $b$-prodigious (here, \textit{maximal} stands for sequences that cannot be extended on either ends).








\section{The maximum length of arithmetic progressions of $b$-prodigious numbers}\label{sec:maximum}

We begin this section with the following lemma without proof.

\begin{lemma}\label{lem:digit}
Let $b\geq2$. Then for all integers $c&gt;0$, $1\leq j\leq b-1$, and $0\leq\beta&lt;\alpha$, we have
\begin{equation*}
p_b(cb^\alpha+jb^\beta)=p_b(c)\cdot j.
\end{equation*}
Furthermore, if $cb^\alpha+jb^\beta$ is a $b$-prodigious number, then
\begin{equation*}
j\mid(cb^\alpha+jb^\beta)\quad\text{and}\quad p_b(c)\mid(cb^\alpha+jb^\beta).
\end{equation*}
\end{lemma}

In view of Lemma~\ref{lem:digit}, the relation
$p_b(c)\mid(cb^\alpha+jb^\beta)$ is a necessary condition for $cb^\alpha+jb^\beta$ to be a $b$-prodigious number. For this reason, it will be useful to force $p_b(c)=1$ while also guaranteeing that $c$ satisfies certain congruence constraints. 

\begin{lemma}\label{lem:crt}
Let $\{x\mod{r_v}{m_v}:1\leq v\leq w\}$ be a consistent system of congruences such that $r_v=0$ if $\gcd(m_v,b)\neq1$. Then there exists a positive integer $c$ such that $c$ satisfies this system of congruences and $p_b(c)=1$.
\end{lemma}
\begin{proof}
For each $1\leq v\leq w$, let $m_v=m_{v,1}m_{v,2}$ such that $m_{v,1}$ divides a nonnegative power of $b$ and $\gcd(m_{v,2},b)=1$. Let $M_1=\lcm\{m_{v,1}:1\leq v\leq w\}$ and $M_2=\lcm\{m_{v,2}:1\leq v\leq w\}$. By the conditions on $r_v$, the system of congruences is equivalent to
$$\{x\mod{0}{M_1},\ x\mod{R}{M_2}\}$$
for some $1\leq R\leq M_2$. Then $c=\sum_{u=t}^{R+t-1}b^{u\varphi(M_2)}$ satisfies the desired properties, where $\varphi$ is the Euler's totient function and $t$ is a nonnegative integer such that $M_1\mid b^{t\varphi(M_2)}$.
\end{proof}

For a positive integer $d$, we say that a sequence $\mathcal{S}$ of integers is a $d$-AP if all consecutive terms of $\mathcal{S}$ have a common difference $d$.  When $d\mid b$ and $d&lt;b/2$, the following theorem establishes the maximum length of a $d$-AP whose terms are all $b$-prodigious numbers.  

\begin{theorem}\label{thm:dAP}
Let $b&gt;2$ and $d$ be positive integers such that $d\mid b$ and $d&lt;b/2$. Define $D_b=\{0\}\cup\{1\leq j\leq b-1:j\mid b\}$, and for each $j\in D_b$, let $k_j=\min\{\kappa\in\N:(j+\kappa d)\nmid b\}$. Then the maximum length of a $b$-prodigious $d$-AP is $b/d+k$, where $k=\max\{k_j:j\in D_b\}$.
\end{theorem}
\begin{proof}
Suppose $\mathcal{S}$ is a $b$-prodigious $d$-AP of length at least $b/d+k$. Let the first term of $\mathcal{S}$ be $ab+j$ for some nonnegative integers $a$ and $j$ with $0\leq j\leq b-1$. Then $\mathcal{S}$ contains $ab+j,ab+(j+d),\dotsc,(a+1)b+(j+(k-1)d)$.

If there exists $1\leq i\leq k-1$ such that $j+(i-1)d&lt; b\leq j+id$, then $j+(i-1)d\geq b-d&gt;b/2$. Thus, $(j+(i-1)d)\nmid b$. Since $j+(i-1)d$ is a factor of both $p_b(ab+(j+(i-1)d))$ and $p_b((a+1)b+(j+(i-1)d))$ by Lemma~\ref{lem:digit}, we deduce that $ab+(j+(i-1)d)$ and $(a+1)b+(j+(i-1)d)$ cannot be both $b$-prodigious, a contradiction. Hence $j+id\leq b-1$ for all $0\leq i\leq k-1$. Furthermore, since $ab+(j+id)$ and $(a+1)b+(j+id)$ are both $b$-prodigious, we have $j,j+d,\dotsc,j+(k-1)d\in D_b$. This implies that $j+(k-1)d\leq b/2$, so $j+kd&lt;b$. Therefore, $ab+(j+kd)$ and $(a+1)b+(j+kd)$ cannot be both prodigious since $(j+kd)\nmid b$ by the definition of $k$. In other words, $\mathcal{S}$ cannot be of length greater than $b/d+k$.

Finally, note that there exists a positive integer $c$ such that $c$ satisfies the congruence $x\mod{0}{(b-1)!}$ and $p_b(c)=1$ by Lemma~\ref{lem:crt}. As a result, if $j\in D_b$ satisfies $k_j=k$, then 
$$cb^2+j,cb^2+(j+d),\dotsc,cb^2+b+(j+(k-1)d)$$ is a $b$-prodigious $d$-AP of length $b/d+k$. This is because $p_b(cb^2+(j+id))=j+id$ by Lemma~\ref{lem:digit} when $0&lt;j+id&lt;b$; together with $(b-1)!\mid c$, we have $(j+id)\mid(cb^2+(j+id))$. Likewise, $p_b(cb^2+b+(j+id))=j+id$ when $0&lt;j+id&lt;j+kd$; together with $(j+id)\mid b$, we have $(j+id)\mid(cb^2+b+(j+id))$.
\end{proof}

As an example of how to use Theorem~\ref{thm:dAP}, consider the case $b=12$ and $d=4$.  Then $D_b=\{0,1,2,3,4,6\}$, $k_1=k_3=k_4=k_6=1$, and $k_0=k_2=2$.  Therefore, $k=2$ and the maximum length of a $12$-prodigious $4$-AP is $12/4+2=5$. Following the definitions in the proof of Lemma~\ref{lem:crt}, we have $m_1=11!$, $M_1=2^8\cdot3^4$, and $M_2=5^2\cdot7\cdot11$. Then $R=M_2$, $\varphi(M_2)=1200$, and $t=1$. Hence we have $c=\sum_{u=1}^{1925}12^{1200u}$, and 
$$144c,144c+4,144c+8,144c+12,144c+16$$
forms a $12$-prodigious $4$-AP of length $5$. Note that although our theorem provides a method to construct a $b$-prodigious $d$-AP of maximum length, the sequence constructed using this method may not be the simplest. For example, the sequence $2,6,10,14,18$ forms a $12$-prodigious $4$-AP of length $5$.

When $d=1$, the statement of Theorem~\ref{thm:dAP} simplifies to the following corollary.
\begin{corollary}\label{cor:ell}
Let $b&gt;2$. The maximum length of a sequence of consecutive $b$-prodigious numbers is $b+\ell$, where $\ell$ is the smallest positive integer that does not divide $b$.  
\end{corollary}
\begin{proof}
From Theorem~\ref{thm:dAP} we have $k_0=\ell\leq k$. On the other hand, for all $j\in D_b$, there always exists a multiple of $\ell$ in $\{j+\kappa:1\leq\kappa\leq\ell\}$; thus $k_j\leq \ell$. Combining both directions, our result follows from Theorem~\ref{thm:dAP}, since $k=\ell$.
\end{proof}

Since Theorem~\ref{thm:dAP} only applies when $d\mid b$ and $d&lt;b/2$, we end this section by addressing the maximum length of a $b$-prodigious $(b/2)$-AP.
\begin{theorem}
Let $b&gt;2$ be even. Then the maximum length of a $b$-prodigious $(b/2)$-AP is $6$.
\end{theorem}
\begin{proof}
Suppose $\mathcal{S}$ is a $b$-prodigious $(b/2)$-AP of length at least $6$. Then $\mathcal{S}$ contains $ab+(j+b/2)$ and $(a+1)b+(j+b/2)$ for some nonnegative integers $a$ and $j$ with $0\leq j&lt;b/2$. By Lemma~\ref{lem:digit}, $j+b/2$ divides both $ab+(j+b/2)$ and $(a+1)b+(j+b/2)$.  Hence $(j+b/2)\mid b$, which implies that $j=0$.

If $\mathcal{S}$ contains both $cb^2+ib$ and $cb^2+ib+b/2$ for some nonnegative integers $n$ and $1\leq i\leq b-1$, then $i\mid(cb^2+ib)$ and $(i\cdot b/2)\mid(cb^2+ib+b/2)$ by Lemma~\ref{lem:digit}. Hence $i$ divides $(cb^2+ib+b/2)-(cb^2+ib)=b/2$. Moreover, $cb^2+ib+b/2=mi\cdot b/2$ for some integer $m$. Dividing both sides of the last equation by $b/2$, we have $2cb+2i+1=mi$, so $i\mid(2cb+2i+1)$.

Combining the established observations that $i\mid(b/2)$ and $i\mid(2cb+2i+1)$, we have $i=1$. Therefore, $\mathcal{S}$ is of the form 
$$(c-1)b^2+(b-1)b+b/2,cb^2,cb^2+b/2,cb^2+b,cb^2+b+b/2,cb^2+2b$$ for some positive integer $c$. Note that this is a $b$-prodigious $(b/2)$-AP of length $6$ if $c=\sum_{u=0}^{b/2-1}b^u$.
\end{proof}



\section{Maximal lengths of consecutive $b$-prodigious numbers}\label{sec:maximal}

We say that the sequence $n,n+1,n+2,\ldots,n+t-1$ of consecutive $b$-prodigious numbers is \emph{maximal of length $t$} if $n-1$ and $n+t$ are not $b$-prodigious numbers.  Recall from Corollary~\ref{cor:ell} that $b+\ell$ is the maximum length of consecutive $b$-prodigious numbers, where $\ell$ is the smallest positive integer that does not divide $b$. In this section, we determine the existence of maximal sequences of consecutive $b$-prodigious numbers of lengths $b\leq t\leq b+\ell-1$.

We begin this section with the following three lemmas.
\begin{lemma}\label{lem:prod=1}
Let $b\geq2$. If $ab+j$ and $ab+(j+1)$ are both $b$-prodigious numbers, where $a$ is a positive integer and $0\leq j\leq b-2$, then $p_b(a)=1$.
\end{lemma}
\begin{proof}
If $ab+j$ and $ab+(j+1)$ are $b$-prodigious, then $p_b(a)\mid (ab+j)$ and $p_b(a)\mid (ab+(j+1))$ by Lemma~\ref{lem:digit}.  It follows that $p_b(a)\mid1$.  Thus  $p_b(a)=1$.
\end{proof}

\begin{lemma}\label{lem:minmax}
Let $b&gt;2$, let $\ell$ be the smallest positive integer that does not divide $b$, and let $\mathcal{S}$ be a maximal sequence of consecutive $b$-prodigious numbers of length at least $b$.  If the first term of $\mathcal{S}$ is of the form $cb^2$ for some positive integer $c$, then $\mathcal{S}$ is of length $b+\ell$. If the first term of $\mathcal{S}$ is of the form $cb^2+1$ for some nonnegative integer $c$, then $c=0$ and $\mathcal{S}$ is of length at $b+\ell-1$.
\end{lemma}

\begin{proof}
If the first term of $\mathcal{S}$ is of the form $cb^2$ for some positive integer $c$, then since $\mathcal{S}$ has at least $b$ terms, $p_b(c)=1$ by Lemma~\ref{lem:prod=1}. Thus, by Lemma~\ref{lem:digit}, we have
$p_b(cb^2+b+j)=j$ for $1\leq j\leq b-1$.  Hence, if $\ell$ is the smallest positive integer that does not divide $b$, then $cb^2,cb^2+1,\dotsc,cb^2+b+(\ell-1)$ are each $b$-prodigious.

If the first term of $\mathcal{S}$ is of the form $cb^2+1$ for some nonnegative integer $c$, then $c=0$ since $cb^2$ is not $b$-prodigious. Noting that $b+\ell$ is not $b$-prodigious since $p_b(b+\ell)=\ell$ by Lemma~\ref{lem:digit} and $\ell\nmid(b+\ell)$, our result follows.
\end{proof}

\begin{lemma}\label{lem:firstterms}
Let $b&gt;2$ and let $\mathcal{S}$ be a maximal sequence of consecutive $b$-prodigious numbers of length at least $b+1$. Then the first term of $\mathcal{S}$ is either $1$, of the form $cb^2$, or of the form $cb^2+b$ for some positive integer $c$.
\end{lemma}
\begin{proof}
Let $ab+j$ be the first term of $\mathcal{S}$ for some nonnegative integers $a$ and $j$ with $j\leq b-1$. Then the last term of $\mathcal{S}$ is at least $(a+1)b+j$. This implies that $j=0$ or $j\mid b$; thus $j\leq b/2$. If $1\leq j\leq b/2$, then Lemma~\ref{lem:prod=1} implies that $p_b(a)=p_b(a+1)=1$, and if $j=0$, then $p_b(a)=1$. Therefore, the first term of $\mathcal{S}$ is of the forms $cb^2+j$ or $cb^2+b$ for some nonnegative integers $c$ and $j$ with $0\leq j\leq b/2$.

If $c=0$, then since every integer $1\leq j\leq b-1$ is $b$-prodigious, the first term of the maximal sequence $\mathcal{S}$ of consecutive $b$-prodigious numbers is $1$. Hence it remains to consider $c&gt;0$, and we have $p_b(c)=1$. If $2\leq j\leq b/2$, then since $cb^2+2(j-1)\in\mathcal{S}$ is $b$-prodigious, we know that $j-1$ divides $cb^2+2(j-1)-(j-1)=cb^2+(j-1)$. As a result, $cb^2+(j-1)$ is also $b$-prodigious, contradicting that $cb^2+j$ is the first term of the maximal sequence $\mathcal{S}$. Finally, $j\neq1$ by Lemma~\ref{lem:minmax}.
\end{proof}

\begin{theorem}
Let $b&gt;2$ and let $\mathcal{S}$ be a maximal sequence of consecutive $b$-prodigious numbers of length $m$ for some $b+1\leq m\leq b+\ell-1$. Then
\begin{enumerate}[$(a)$]
\item\label{item:b+1&lt;m&lt;b+l-1} $m$ cannot be strictly between $b+1$ and $b+\ell-1$;
\item\label{item:m=b+l-1} $m=b+\ell-1$ if and only if the first term of $\mathcal{S}$ is $1$; and
\item\label{item:m=b+1} $m=b+1$ if and only if either $b$ is odd and the first term of $\mathcal{S}$ is $1$ or $b$ is a power of $2$.
\end{enumerate}
\end{theorem}
\begin{proof}
By Lemmas~\ref{lem:minmax} and \ref{lem:firstterms}, the first term of $\mathcal{S}$ is either $1$ or of the form $cb^2+b$ for some positive integer $c$. If the first term of $\mathcal{S}$ is $1$, then $m=b+\ell-1$ by Lemma~\ref{lem:minmax}. In the remainder of this proof, assume that the first term of $\mathcal{S}$ is $cb^2+b$.

If $m&gt;b+1$, then $\mathcal{S}$ contains both $cb^2+2b$ and $cb^2+2b+1$, contradicting Lemma~\ref{lem:prod=1}. This implies that $m=b+1$ and establishes part~$(\ref{item:b+1&lt;m&lt;b+l-1})$. Hence $m=b+\ell-1$ if and only if $\ell=2$, which is equivalent to $b$ being odd. However, $b$ cannot be odd since $2\mid(cb^2+b+2)$ and $2\mid(cb^2+2b)$. Therefore, if $m=b+\ell-1$, the first term of $\mathcal{S}$ cannot be $cb^2+b$. This establishes part~$(\ref{item:m=b+l-1})$.

If $b=2^\alpha m$ for some integer $\alpha\geq1$ and odd integer $m&gt;1$, then note that $2^{\alpha+1}\leq b-1$ and $2^{\alpha+1}\nmid(cb^2+b+2^{\alpha+1})$ since $2^{\alpha+1}\mid b^2$ but $2^{\alpha+1}\nmid b$. This contradicts that $\mathcal{S}$ contains $cb^2+b+2^{\alpha+1}$. Hence it remains to show that such a maximal sequence $\mathcal{S}$ exists when $b=2^\alpha$ for some integer $\alpha\geq2$. This can be achieved by finding a positive integer $c$ such that $j\mid(cb^2+b+j)$ for each $1\leq j\leq b-1$ and $p_b(c)=1$.

For each $1\leq j\leq b-1$, let $j=2^{\beta_j}j'$ for some nonnegative integer $\beta_j$ and odd integer $j'$. Note that $\beta_j&lt;\alpha$, so $j\mid(cb^2+b+j)$ is equivalent to $cb^2+b\mod{0}{j'}$. Since $\gcd(b,j')=1$, we have $c\mod{-b^{-1}}{j'}$. Such an integer $c$ exists by Lemma~\ref{lem:crt}, which completes the proof of part~$(\ref{item:m=b+1})$.
\end{proof}

\begin{lemma}\label{lem:b-1prime}
Let $b&gt;2$ and let 
$$ab+(b-1),(a+1)b,(a+1)b+1,\dotsc,(a+1)b+(b-2)$$ be a $b$-prodigious sequence. Then $b-1$ is a prime.
\end{lemma}
\begin{proof}
Suppose by way of contradiction that $b-1$ is not a prime. Then there exists $2\leq j\leq b-2$ such that $j\mid b-1$.  We deduce that $j$ divides $ab+(b-1)$ since $(b-1)\mid ab+(b-1))$ by Lemma~\ref{lem:digit}.  Since $j$ also divides $(a+1)b+j$ by Lemma~\ref{lem:digit}, it follows that $j$ divides $(a+1)b+j-(ab+(b-1))=j+1$, a contradiction.
\end{proof}

\begin{theorem}
Let $b&gt;2$. Then there exists a maximal sequence $\mathcal{S}$ of consecutive $b$-prodigious numbers of length $b$ if and only if either $b-1$ is a prime or $b$ is a power of an odd prime.
\end{theorem}
\begin{proof}

Let $\mathcal{S}$ be a maximal sequence of consecutive $b$-prodigious numbers of length $b$. By Lemma~\ref{lem:prod=1}, $\mathcal{S}$ is a subsequence of
$$(c-1)b^2+(b-1)b+(b-1),cb^2,cb^2+1,\dotsc,cb^2+2b$$ for some nonnegative integer $c$ such that $c=0$ or $p_b(c)=1$. Suppose by way of contradiction that $b-1$ is not a prime and $b$ is not a power of an odd prime. Then Lemma~\ref{lem:b-1prime} implies that the first term of $\mathcal{S}$ cannot be $(c-1)b^2+(b-1)b+(b-1)$ nor $cb^2+(b-1)$. Also, the first term of $\mathcal{S}$ cannot be $cb^2$ nor $cb^2+1$ since $cb^2+b$ and $cb^2+b+1$ are $b$-prodigious, and the first term of $\mathcal{S}$ cannot be $cb^2+b+1$ since $cb^2+b$ is $b$-prodigious.

If the first term of $\mathcal{S}$ is $cb^2+b$, then $b$ cannot be even; otherwise, $cb^2+2b$ is $b$-prodigious. Since $b$ is not a power of an odd prime, there exist distinct primes $p&lt;q$ such that $pq\mid b$. Let $\alpha$ be the greatest integer such that $p^\alpha\mid b$. Then $p^{\alpha+1}\leq b-1$, but $cb^2+b+p^{\alpha+1}$ is not $b$-prodigious, a contradiction. Hence $\mathcal{S}$ is given by 
$$cb^2+j,cb^2+(j+1),\dotsc,cb^2+b+(j-1)$$ for some $2\leq j\leq b-2$. Since $j-1$ divides both $cb^2+b+(j-1)$ and $cb^2+(j-1)$ by Lemma~\ref{lem:digit}, this implies that $(j-1)\nmid b$ since $cb^2+b+(j-1)$ is $b$-prodigious while $cb^2+(j-1)$ is not.

Since $cb^2+(b-1)$ is $b$-prodigious, we have by Lemma~\ref{lem:digit} that $(b-1)\mid c$. If there exists $2\leq\tilde{j}\leq j$ such that $\tilde{j}\mid b-1$, then $\tilde{j}\nmid b$; thus $\tilde{j}\nmid(cb^2+b+\tilde{j})$, a contradiction. Hence $j$ is less than every prime factor of $b-1$, implying that $j&lt;\sqrt{b-1}$ since $b-1$ is not a prime. Note that $b-1\geq4$, so $j&lt;\sqrt{b-1}\leq\frac{b-1}{2}$. As a result, $j\leq2(j-1)&lt;b-1$, so $cb^2+2(j-1)$ is $b$-prodigious. Together with $cb^2+b+(j-1)$ being $b$-prodigious, we deduce from Lemma~\ref{lem:digit} that $(j-1)\mid b$, a contradiction.

Conversely, we assume that $b-1$ is a prime or $b$ is a power of an odd prime. If $b-1$ is a prime, then let 
$$B=\max\{\beta\in\Z:p^\beta\mid j\text{ for some prime }p\mid b\text{ and }1\leq j\leq b-2\}$$
and
$$N=\max\{n\in\N:n\mid\lcm\{j\in\N:1\leq j\leq b-2\}\text{ and }\gcd(n,b)=1\}.$$
This construction ensures that $j\mid b^BN$ for all $1\leq j\leq b-2$. By Lemma~\ref{lem:crt}, there exists a positive integer $\overline{c}$ such that $\overline{c}\mod{b^{-2}(-b^2+1)-b^{B-2}+1}{(b-1)^B}$, $\overline{c}\mod{-b^{B-2}}{N}$, $\overline{c}\mod{0}{b^{B-1}}$, and $p_b(\overline{c})=1$. Define $c=\overline{c}+b^{B-2}$; thus $p_b(c)=1$. Note that \begin{align*}
p_b\big((c-1)b^2+(b-1)b+(b-1)\big)&amp;=p_b\big((\overline{c}+b^{B-2}-1)b^2+(b-1)b+(b-1)\big)\\
&amp;=p_b\big(\overline{c}b^2+b^B-1\big)\\
&amp;=p_b\left(\overline{c}b^2+\sum_{t=0}^{B-1}(b-1)b^t\right)\\
&amp;=(b-1)^B
\end{align*}
and $p_b(cb^2+j)=j$ for all $1\leq j\leq b-2$ by Lemma~\ref{lem:digit}. Furthermore,  \begin{align*}
(c-1)b^2+(b-1)b+(b-1)&amp;=(\overline{c}+b^{B-2}-1)b^2+(b-1)b+(b-1)\\
&amp;=\overline{c}b^2+b^B-1\\
&amp;\equiv(b^{-2}(-b^2+1)-b^{B-2}+1)b^2+b^B-1&amp;&amp;\pmod{(b-1)^B}\\
&amp;\equiv(-b^2+1)-b^B+b^2+b^B-1&amp;&amp;\pmod{(b-1)^B}\\
&amp;\equiv0&amp;&amp;\pmod{(b-1)^B}
\end{align*}
and $cb^2+j=(\overline{c}+b^{B-2})b^2+j=\overline{c}b^2+b^B+j$ is divisible by $j$ since $\overline{c}b^2+b^B\mod{0}{N}$ and $\overline{c}b^2+b^B\mod{0}{b^B}$. Hence 
$$(c-1)b^2+(b-1)b+(b-1),cb^2,cb^2+1,\dotsc,cb^2+(b-2)$$ forms a sequence of consecutive $b$-prodigious numbers of length $b$. This sequence is maximal since $(c-1)b^2+(b-1)b+(b-2)$ and $(c-1)b^2+(b-1)b+(b-1)$ cannot both be $b$-prodigious by Lemma~\ref{lem:prod=1} and $cb^2+(b-1)=\big((c-1)b^2+(b-1)b+(b-1)\big)+b$ is not divisible by $b-1$.

Lastly, if $b=p^\alpha$ for some odd prime $p$, then for each $1\leq j\leq b-1$, let $\beta_j$ be the maximum integer such that $p^{\beta_j}\mid j$ and $j=p^{\beta_j}j'$. By Lemma~\ref{lem:crt}, there exists a positive integer $c$ such that $c\mod{-p^{-\alpha}}{j'}$ for all $1\leq j\leq b-1$ and $p_b(c)=1$. Then
$$cb^2+b,cb^2+b+1,\dotsc,cb^2+b+(b-1)$$ forms a maximal sequence of consecutive $b$-prodigious numbers of length $b$, since $cb^2+b+(b-1)$ is even, and hence $cb^2+2b$ is not $b$-prodigious. 
\end{proof}






\section{Concluding remarks}
Although the main focus of our paper is on consecutive $b$-prodigious numbers, there are many other questions about $b$-prodigious numbers that could be investigated.  One such question is to consider the intersection of $b$-prodigious numbers with other integer sequences.  In this direction, we end our paper with a theorem that considers the $b$-prodigious-Niven numbers, i.e., the intersection of the set of $b$-prodigious numbers and the set of $b$-Niven numbers.

\begin{lemma}[\cite{ghw}]\label{lem:samesum}
Let $b\geq2$. If $x$ and $y$ are two $b$-Niven numbers such that $s_b(x)=s_b(y)$, then $s_b(x)\mid(x-y)$.
\end{lemma}

\begin{theorem}\label{thm:prodigiousNiven}
The maximum length of a sequence of consecutive $b$-prodigious-Niven numbers is exactly
\begin{enumerate}[$(a)$]
\item\label{item:b=2} $4$ if $b=2$;
\item\label{item:b&gt;2even} $b$ if $b&gt;2$ is even; and
\item\label{item:b&gt;2odd} $b+1$ if $b&gt;2$ is odd.
\end{enumerate}
\end{theorem}
\begin{proof}
Every positive integer is $2$-prodigious, so part~$(\ref{item:b=2})$ follows from the result by Cai \cite{cai} on $2$-Niven numbers as mentioned in the introduction. In the rest of the proof, we consider $b&gt;2$.

To prove parts~$(\ref{item:b&gt;2even})$ and $(\ref{item:b&gt;2odd})$, we first note that $1,2,\dotsc,b$ and $1,2,\dotsc,b+1$ are maximal sequences of consecutive $b$-prodigious-Niven numbers when $b$ is even and $b$ is odd, respectively. We call these the ``trivial" sequences. To search for nontrivial sequences, let $a$ and $j$ be positive integers with $1\leq j\leq b-1$ such that the sequence $ab+j,ab+j+1,\dotsc,(a+1)b+j-1$ is $b$-prodigious-Niven. Since $ab+(b-1)$ is $b$-prodigious, we have by Lemma~\ref{lem:digit} that $(b-1)\mid a$ and thus $(b-1)\mid s_b(a)$. If $j=b-1$, then $(a+1)b+b-1$ is not $b$-prodigious since $(b-1)\nmid(a+1)$. If $1\leq j\leq b-2$, then by Lemma~\ref{lem:prod=1} we see that
$p_b(a)=1$, and we deduce that $s_b(ab+j)=s_b((a+1)b+(j-1))=s_b(a)+j$. Thereforewe have $(s_b(a)+j)\mid (b-1)$ by Lemma~\ref{lem:samesum}, contradicting that $(b-1)\mid s_b(a)$. In other words, every nontrivial sequence of consecutive $b$-prodigious-Niven numbers with length at least $b$ is a subsequence of $ab,ab+1,\dotsc,ab+(b-1)$ or $ab+(b-1),(a+1)b,\dotsc,(a+1)b+(b-2)$ with maximum length $b$.
\end{proof}




\section{Acknowledgments}
These results are based on work supported by the National Science Foundation under grant numbered DMS-1852378.




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\bibitem{ck}
C.~Cooper and R. E.~Kennedy, On consecutive Niven numbers, \textit{Fibonacci Quart.\/} \textbf{31} (1993), 146--151.

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\bibitem{ghw}
H.~Grundman, J.~Harrington, and T.~W.~H.~Wong, Arithmetic progressions of $b$-Niven numbers, submitted.

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\end{thebibliography}



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\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A63; Secondary 11B25.

\noindent \emph{Keywords: } 
prodigious number, Niven number, arithmetic progression.

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\noindent (Concerned with sequence
\seqnum{A055471}.)

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\vspace*{+.1in}
\noindent
Received June 16 2022;
revised version received  October 18 2022.
Published in {\it Journal of Integer Sequences},
October 25 2022.

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