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\begin{center}
\vskip 1cm{\LARGE\bf 
On Binomial Identities in Arbitrary Bases}
\vskip 1cm
\large
Lin Jiu\footnote{Author's current address:  
Research Institute for Symbolic Computation,
Johannes Kepler University, Linz, A-4040 Linz,
Austria; {\tt ljiu@risc.uni-linz.ac.at} .}\\
Department of Mathematics\\
Tulane University\\
New Orleans, LA 70118 \\
USA \\
\href{mailto:ljiu@tulane.edu}{\tt ljiu@tulane.edu} \\
\ \\
Christophe Vignat\\
Department of Mathematics\\
Tulane University\\
New Orleans, LA 70118 \\
USA \\
and\\
Department of Physics\\
Universit\'e Orsay Paris Sud\\
91405 Orsay Cedex\\
France\\
\href{mailto:cvignat@tulane.edu}{\tt cvignat@tulane.edu} \\

\end{center}

\vskip .2 in

\begin{abstract}
We first extend the digital binomial identity as given by Nguyen et
al.\ to an identity in an arbitrary base $b$, by introducing the
$b$-ary binomial coefficients. Then, we study the properties of these
coefficients such as their orthogonality, their link with Lucas'
theorem and their extension to multinomial coefficients. Finally, we
analyze the structure of the corresponding $b$-ary Pascal-like
triangles.
\end{abstract}

\section{Introduction}
In a series of recent articles,
Nguyen \cite{NguyenDigital,NguyenGeneralization},
and then Nguyen and Mansour \cite{Nguyenqdigital,Sheffer},
have introduced different versions of the binomial identity, in which
the usual integer powers are replaced by arithmetical functions that
count the digits of these powers in their representation in base $b$. Allouche and Shallit \cite[Chap.\ 3]{AllouchShallit}, for example, provide a variety of properties of these arithmetical functions.

Given a positive integer $n$, let $S_{b}^{\left(k\right)}\left(n\right)$
denote the number of $k$'s in the $b$-ary expansion of $n$:
\begin{equation}
n=\sum_{i}n_{i}b^{i}.\label{eq:nbaseb}
\end{equation}
Also let $S_{b}\left(n\right)$ denote the sum of all the digits, namely
\begin{equation}
S_{b}\left(n\right)=\sum_{i}n_{i}=\sum_{k=0}^{b-1}k S_{b}^{\left(k\right)}\left(n\right).\label{eq:Sbofn}
\end{equation}
The main extension of the binomial identity is 
\begin{equation}
\left(X+Y\right)^{S_{2}\left(n\right)}=\sum_{\genfrac{}{}{0pt}{}{0\leq k\leq n}{\left(k,n-k\right)\text{\thinspace\thinspace carry-free}}}X^{S_{2}\left(k\right)}Y^{S_{2}\left(n-k\right)},\label{eq:caseBquals2}
\end{equation}
which was originally given by Callan \cite{digialbinomial}
and also by Nguyen  \cite{NguyenDigital, NguyenGeneralization} in the case $b=2$.
Here, the sum on the right-hand side is over all values of $k$ such that the addition
of $k$ and $n-k$ in base $2$ is carry-free. Note that, in the general case of base $b$, this carry-free condition is equivalent to  $S_{b}\left(k\right)+S_{b}\left(n-k\right)=S_{b}\left(n\right)$.
Thus, identity \eqref{eq:caseBquals2} can be restated equivalently in the beautifully
symmetric form 
\begin{equation*}
\left(X+Y\right)^{S_{2}\left(n\right)}=\sum_{S_{2}\left(k\right)+S_{2}\left(n-k\right)=S_{2}\left(n\right)}X^{S_{2}\left(k\right)}Y^{S_{2}\left(n-k\right)}.
\end{equation*}
Nguyen \cite{NguyenDigital} proved identity \eqref{eq:caseBquals2} by using
a polynomial generalization of the Sierpinski triangle, which is the
Pascal triangle modulo $2$.

Nguyen \cite[Thm.\ 2]{NguyenGeneralization} also gives an extension of \eqref{eq:caseBquals2} to an arbitrary base $b$, 
in the form
\begin{equation}
\prod_{i=0}^{N-1}{x+y+n_{i}-1 \choose n_{i}}=\sum_{0\leq k\apprle_{b}n}\prod_{i=0}^{N-1}{x+k_{i}-1\choose k_{i}}{x+n_{i}-k_{i}-1\choose n_{i}-k_{i}}.\label{eq:product}
\end{equation}
Here, suppose $k$ and $n$ have the following expansions in base
$b$:
\begin{equation}
n=\underset{l=0}{\overset{N-1}{\sum}}n_{l}b^{l}\text{ and }k=\underset{l=0}{\overset{N-1}{\sum}}k_{l}b^{l}.\label{eq:digitalexpansionofnandk}
\end{equation}
Then, the summation range $0\leq k\apprle_{b}n$ means that the sum is over all positive
integers $k$ such that $k_{i}\leq n_{i}$, for all $i\in\left\{ 0,\ldots,N-1\right\}$. This summation range is also equivalent to summing over all values of $k$ such that the addition of $k$ and $n-k$ is carry free. 

The aim of this paper is to show that these results are a consequence
of an elementary identity stated in the next section. This approach
gives a more accessible form to Nguyen's results and provides
a number of generalizations. The paper is organized as follows.

In Section \ref{sec: AGeneralFormula}, we state a general formula on the sum over all the digits of
an integer number. It is an identity involving sequences of binomial
type, which lays the foundation for the remaining sections. In Section
\ref{sec: AGeneralizedBinomialIdentity}, we provide a new version of the binomial identity in base $b$
that involves newly defined $b$-ary binomial coefficients. Then, some
of the properties of these coefficients, such as their role in Lucas'
theorem and an orthogonality property, are presented in Section \ref{sec:Properties}.
Section \ref{sec:PascallikeTriangles} exhibits the construction rules for a Pascal-type triangle
built from these coefficients. Finally, Section \ref{sec:History} presents, besides an
historical approach, some open questions.

\section{A general formula}\label{sec: AGeneralFormula}
Nguyen proves \eqref{eq:product} by using a polynomial extension
of Sierpinski's matrices. We introduce here a more elementary approach.
First, we remark that \eqref{eq:product} can be restated equivalently
as 
\begin{equation}
\prod_{i=0}^{N-1}\frac{\left(x+y\right)_{n_{i}}}{n_{i}!}=\prod_{i=0}^{N-1}\sum_{0\leq k_{i}\leq n_{i}}\frac{\left(x\right)_{k_{i}}}{k_{i}!}\frac{\left(y\right)_{n_{i}-k_{i}}}{\left(n_{i}-k_{i}\right)!},\label{eq:digitalwiseproduct}
\end{equation}
where $\left(x\right)_{k}=\frac{\Gamma\left(x+k\right)}{\Gamma\left(x\right)}$
is the Pochhammer symbol. Next we realize that \eqref{eq:digitalwiseproduct} holds
in fact component-wise, i.e., for all $i\in\left\{ 0,\ldots,N-1\right\}$
and arbitrary positive integers $n_{i}$, we have
\begin{equation}
\frac{\left(x+y\right)_{n_{i}}}{n_{i}!}=\sum_{k_{i}=0}^{n_{i}}\frac{\left(x\right)_{k_{i}}}{k_{i}!}\frac{\left(y\right)_{n_{i}-k_{i}}}{\left(n_{i}-k_{i}\right)!},\label{eq:ChuVandermonde}
\end{equation}
which is actually the Chu-Vandermonde identity. Since \eqref{eq:ChuVandermonde}
is a consequence of the fact that the Pochhammer sequence $\left(\left(x\right)_{k}\right)_{k\geq 0}$
is of binomial type\footnote{A sequence of polynomials $p_{n}\left(x\right)$ is of binomial type \cite[Thm.\ 2.4.7, p.\ 26]{Roman} if it satisfies the convolution identity
\begin{equation*}
p_{n}\left(x+y\right)=\sum_{k=0}^{n}{n\choose k}p_{k}\left(x\right)p_{n-k}\left(y\right).
\end{equation*}}, we therefore have the following result.

\begin{proposition}
\label{prop:GeneralBinomialType}Assume that $\left(a_{n}\right)_{n\geq 0}$,
$\left(b_{n}\right)_{n\geq 0}$, and $\left(c_{n}\right)_{n\geq 0}$ are three
sequences related as
\begin{equation}
c_{n}=\sum_{k=0}^{n}{n\choose k}a_{k}b_{n-k}.\label{eq:Convolution}
\end{equation}
Using notation from \eqref{eq:digitalexpansionofnandk}, we have
\begin{equation}
\prod_{i=0}^{N-1}\frac{c_{n_{i}}}{n_{i}!}=\sum_{0\leq k\apprle_{b}n}\prod_{i=0}^{N-1}\frac{a_{k_{i}}}{k_{i}!}\frac{b_{n_{i}-k_{i}}}{\left(n_{i}-k_{i}\right)!},\label{eq:ProductConvolution}
\end{equation}
where the sum on the right-hand side is over all values of $k$ such
that $0\leq k\apprle_{b}n$, or equivalently such that the addition of $k$ and
$n-k$ is carry-free in base $b$.
\end{proposition}

We now apply formula \eqref{eq:ProductConvolution}
to obtain a generalization of the binomial identity in an arbitrary
base $b$.

\section{A generalized binomial identity}
\label{sec: AGeneralizedBinomialIdentity}
\begin{theorem}
\label{thm:MainTheorem}With the notation \eqref{eq:nbaseb}, \eqref{eq:Sbofn},
and \eqref{eq:digitalexpansionofnandk}, the identity 
\begin{equation}
\left(X+Y\right)^{S_{b}\left(n\right)}=\sum_{k=0}^{n}{n \choose k}_{b}X^{S_{b}\left(k\right)}Y^{S_{b}\left(n-k\right)}\label{eq:binomial}
\end{equation}
holds for all $X,Y\in\mathbb{C}$, where the $b$-ary binomial coefficients
${n\choose k}_{b}$ are given by 
\begin{equation}
{n\choose k}_{b}=\overset{N-1}{\underset{l=0}{\prod}}{n_{l} \choose k_{l}}.\label{eq:binomialcoefficients}
\end{equation}
\end{theorem}
\begin{proof}
By \eqref{eq:binomialcoefficients},  we have ${n \choose k}_{b}=0$ if $k_{l}>n_{l}$ holds for at least one value of $l$.  Therefore, it suffices to
assume $0\leq k_{l}\leq n_{l}$ for all $l\in\left\{ 0,\ldots,N-1\right\}$.
As a result, we have 
\begin{equation*}
n-k=\sum_{l=0}^{N-1}\left(n_{l}-k_{l}\right)b^{l},
\end{equation*}
which is equivalent to 
\begin{equation*}
S_{b}\left(k\right)+S_{b}\left(n-k\right)=S_{b}\left(n\right),
\end{equation*}
i.e., to the assumption that the addition of $k$ and $n-k$ is carry-free
in base $b$, as already mentioned.

We apply the identity \eqref{eq:ProductConvolution} in Proposition
\ref{prop:GeneralBinomialType} to obtain 
\begin{equation}
\prod_{l=0}^{N-1}c_{n_{l}}=\prod_{l=0}^{N-1}n_{l}!\sum_{k_{l}=0}^{n_{l}}\frac{a_{k_{l}}}{k_{l}!}\cdot\frac{b_{n_{l}-k_{l}}}{\left(n_{l}-k_{l}\right)!}=\sum_{S_{b}\left(k\right)+S_{b}\left(n-k\right)=S_{b}\left(n\right)}\left(\prod_{l=0}^{N-1}{n_{l} \choose k_{l}}a_{k_{l}}b_{n_{l}-k_{l}}\right).\label{eq:DigitsConvolution}
\end{equation}
The choice
\begin{equation*}
c_{n_{l}}=\left(X+Y\right)^{n_{l}},\ \ a_{k_{l}}=X^{k_{l}},\ \text{and}\ b_{k_{l}}=Y^{k_{l}},
\end{equation*}
which satisfies the convolution identity \eqref{eq:Convolution}, leads to
\begin{equation*}
\prod_{l=0}^{N-1}\left(X+Y\right)^{n_{l}}=\sum_{S_{b}\left(k\right)+S_{b}\left(n-k\right)=S_{b}\left(n\right)}\left(\prod_{l=0}^{N-1}{n_{l} \choose k_{l}}X^{k_{l}}Y^{n_{l}-k_{l}}\right).
\end{equation*}
Notice that the left-hand side reads 
\begin{equation*}
\prod_{l=0}^{N-1}\left(X+Y\right)^{n_{l}}=\left(X+Y\right)^{\overset{N-1}{\underset{l=0}{\sum}}n_{l}}=\left(X+Y\right)^{S_{b}\left(n\right)},
\end{equation*}
while the right-hand side can be written as 
\begin{equation*}
\prod_{l=0}^{N-1}\sum_{k_{l}=0}^{n_{l}}{n_{l} \choose k_{l}}X^{k_{l}}Y^{n_{l}-k_{l}}=\sum_{k=0}^{n}\left(\prod_{l=0}^{N-1}{n_{l} \choose k_{l}}\right)X^{S_{b}\left(k\right)}Y^{S_{b}\left(n-k\right)}.
\end{equation*}
This completes the proof.
\end{proof}
\begin{remark}
\label{rem:remark3p2}When $n<b$, the $b$-ary expansion of $n$
consists of only one digit, i.e., $n=n_{0}$. And since $k\leq n<b$, we also have $n-k<b$. Thus,
\begin{equation*}
S_{b}\left(n\right)=n_{0}=n,\ S_{b}\left(k\right)=k,\ S_{b}\left(n-k\right)=n-k,\ \text{and}\ {n \choose k}_{b}={n \choose k},
\end{equation*}
so that \eqref{eq:binomial} reduces to the usual binomial identity
\begin{equation*}
\left(X+Y\right)^{n}=\sum_{k=0}^{n}{n \choose k}X^{k}Y^{n-k}.
\end{equation*}
\end{remark}
\begin{corollary}
(I) In the binary case $b=2$, we recover the identity by Callan and Nguyen: 
\begin{equation*}
\left(X+Y\right)^{S_{2}\left(n\right)}=\sum_{S_{2}\left(k\right)+S_{2}\left(n-k\right)=S_{2}\left(n\right)}X^{S_{2}\left(k\right)}Y^{S_{2}\left(n-k\right)},
\end{equation*}
by noticing that the coefficients ${n_{l}\choose k_{l}}$ take the only values
${0\choose 0}={1\choose 0}={1\choose 1}=1$ and ${0\choose 1}=0$.

(II) In the case $b=3$, ${n\choose k}_{3}=0$ if $S_{3}\left(k\right)+S_{3}\left(n-k\right)\ne S_{3}\left(n\right)$. 
Otherwise, all ${n_{l}\choose k_{l}}=1$ except for the case $n_{l}=2$ and $k_{l}=1$ such that ${n_{l}\choose k_{l}}=2$.
Therefore, we have the equivalent expression 
\begin{equation*}
\left(X+Y\right)^{S_{3}\left(n\right)}=\sum_{S_{3}\left(k\right)+S_{3}\left(n-k\right)=S_{3}\left(n\right)}\left(2^{S_{3}^{\left(2\right)}\left(n\right)-S_{3}^{\left(2\right)}\left(k\right)-S_{3}^{\left(2\right)}\left(n-k\right)}\right)X^{S_{3}\left(k\right)}Y^{S_{3}\left(n-k\right)}.
\end{equation*}
\end{corollary}
\begin{remark}
Different choices of the sequences $\left(a_{n}\right)_{n\geq 0}$, $\left(b_{n}\right)_{n\geq 0}$,
and $\left(c_{n}\right)_{n\geq 0}$ lead to different identities. For instance,
another choice is the Pochhammer coefficients 
\begin{equation*}
c_{n}=\left(X+Y\right)_{n},\ \ a_{n}=\left(X\right)_{n},\ \text{and}\ b_{n}=\left(Y\right)_{n}.
\end{equation*}
The generating function
\begin{equation*}
\sum_{n\geq 0}\frac{\left(X\right)_{n}}{n!}z^{n}=\left(1-z\right)^{-X}
\end{equation*}
shows that the convolution property \eqref{eq:Convolution} holds.
Then from \eqref{eq:DigitsConvolution}, we deduce 
\begin{equation*}
\prod_{l=0}^{N-1}\frac{\left(X+Y\right)_{n_{l}}}{n_{l}!}=\sum_{S_{b}\left(k\right)+S_{b}\left(n-k\right)=S_{b}\left(n\right)}\left(\prod_{l=0}^{N-1}\frac{X_{k_{l}}}{k_{l}!}\cdot\frac{Y_{n_{l}-k_{l}}}{\left(n_{l}-k_{l}\right)!}\right),
\end{equation*}
or equivalently, we recover the result \cite[Thm.\ 2]{NguyenGeneralization}
\begin{equation*}
\prod_{l=0}^{N-1}{X+Y+n_{l}-1 \choose n_{l}}=\sum_{0\leq k\apprle_{b}n}\left(\prod_{l=0}^{N-1}{X+k_{l}-1 \choose k_{l}}{Y+n_{l}-k_{l}-1 \choose n_{l}-k_{l}}\right).
\end{equation*}
Then, for the
binary case $b=2$, $n_{l}\in\left\{ 0,1\right\} $ yields that
for all $l\in\left\{0,\ldots,N-1\right\}$,
\begin{displaymath}
\left(X+Y\right)_{n_{l}}=\begin{cases}
X+Y, & \text{if $n_{l}=1$;}\\
0, & \text{if $n_{l}=0$.}
\end{cases}
\end{displaymath}
Again, we recover
\begin{equation*}
\left(X+Y\right)^{S_{2}\left(n\right)}=\sum_{S_{2}\left(k\right)+S_{2}\left(n-k\right)=S_{2}\left(n\right)}X^{S_{2}\left(k\right)}Y^{S_{2}\left(n-k\right)}.
\end{equation*}

\end{remark}

\section{Properties of the $b$-ary binomial coefficients}
\label{sec:Properties}
We study in this section some properties of the $b$-ary binomial
coefficients defined by \eqref{eq:binomialcoefficients}.


\subsection{Generating function}
\begin{theorem}
A generating function for the $b$-ary binomial coefficients ${n\choose k}_{b}$
is 
\begin{equation}
\sum_{k=0}^{n}{n \choose k}_{b}x^{k}=\prod_{l=0}^{N-1}\left(1+x^{b^{l}}\right)^{n_{l}}.\label{eq:barybinomialgeneratingfunction}
\end{equation}
\end{theorem}
\begin{proof}
Define the right-hand side of \eqref{eq:barybinomialgeneratingfunction}
as $P\left(x\right)$, a polynomial of degree
\begin{equation*}
\deg P=\overset{N-1}{\underset{l=0}{\sum}}n_{l}b^{l}=n.
\end{equation*}
Let $p_{k}^{\left(n\right)}$ denote the coefficient of $x^{k}$ in $P\left(x\right)$
and remark that $p_{k}^{\left(n\right)}=0$ if and only if there is
a carry in the addition of $k$ and $n-k$ in base $b$. An elementary
enumeration shows
\begin{equation*}
p_{k}^{\left(n\right)}={n_{N-1} \choose k_{N-1}}{n_{N-2} \choose k_{N-2}}\cdots{n_{0} \choose k_{0}},
\end{equation*}
which gives the desired result.
\end{proof}

\subsection{Multinomial version}

The next property is the extension of \eqref{eq:binomial} to the multinomial
case.
\begin{theorem}
Define the $b$-ary multinomial coefficient by
\begin{equation*}
{n \choose k_{1},\ldots,k_{m}}_{b}=\prod_{l=0}^{N-1}{n_{l} \choose \left(k_{1}\right)_{l},\ldots,\left(k_{m}\right)_{l}},
\end{equation*}
where $\left(k_{i}\right)_{l}$ denotes the rank-$l$ digit in the
expansion $k_{i}=\underset{l}{\sum}\left(k_{i}\right)_{l}b^{l}$ of $k_{i}$ in base $b$. Then 
\begin{equation*}
\left(X_{1}+\cdots+X_{m}\right)^{S_{b}\left(n\right)}=\sum_{k_{1},\ldots,k_{m}}{n \choose k_{1},\ldots,k_{m}}_{b}X_{1}^{S_{b}\left(k_{1}\right)}\cdots X_{m}^{S_{b}\left(k_{m}\right)}.
\end{equation*}
\end{theorem}
\begin{proof}
The proof is straightforward and follows by induction on the value
of $m$, using the binomial expansion \eqref{eq:binomial} and the
definition \eqref{eq:binomialcoefficients} of the $b$-ary binomial
coefficients.
\end{proof}

\subsection{Chu-Vandermonde identity}

This section provides an extension to an arbitrary base $b$ of the
Chu-Vandermonde identity \cite[eq.\ (3), p.\ 8]{Riordan}. 
\begin{theorem}
Suppose that the addition of $m$ and $n$ is carry-free in base $b$,
then 
\begin{equation}
{m+n \choose r}_{b}=\sum_{0\leq k\apprle_{b}r}{m \choose k}_{b}{n \choose r-k}_{b}.\label{eq:bChuVandermonde}
\end{equation}
\end{theorem}
\begin{proof}
Let $N$ be the maximum number of digits among the numbers $m$, $n$,
and $r$. Also let
\begin{equation*}
m=\underset{l=0}{\overset{N-1}{\sum}}m_{l}b^{l},\ n=\underset{l=0}{\overset{N-1}{\sum}}n_{l}b^{l},\ \text{and}\ r=\underset{l=0}{\overset{N-1}{\sum}}r_{l}b^{l}.
\end{equation*}
Since the addition of $m$ and $n$ is carry-free, we have for all
$l\in\left\{ 0,\ldots,N-1\right\}$, 
\begin{equation*}
m_{l}+n_{l}<b\text{ and }m+n=\underset{l=0}{\overset{N-1}{\sum}}\left(m_{l}+n_{l}\right)b^{l}.
\end{equation*}
Thus, we first apply the ordinary Chu-Vandermonde identity digit-wise to get, for all $l\in\left\{ 0,\ldots,N-1\right\}$, 
\begin{equation*}
{m_{l}+n_{l}\choose r_{l}}=\sum_{k_{l}=0}^{r_{l}}{m_{l} \choose k_{l}}{n_{l} \choose r_{l}-k_{l}}.
\end{equation*}
Then we take the product over $l$ of both sides to obtain 
\begin{equation*}
{m+n \choose r}_{b}=\prod_{l=0}^{N-1}{m_{l}+n_{l} \choose r_{l}}=\prod_{l=0}^{N-1}\sum_{k_{l}=0}^{r_{l}}{m_{l} \choose k_{l}}{n_{l} \choose r_{l}-k_{l}}=\sum_{0\leq k\apprle_{b}r}{m \choose k}_{b}{n \choose r-k}_{b}.
\end{equation*}
\end{proof}
\begin{remark}
(I) Another proof could be obtained by considering generating
functions. It suffices to notice that if the addition $m+n$ is carry-free,
\begin{equation*}
\left(\prod_{l=0}^{N-1}\left(1+x^{b^{l}}\right)^{n_{l}}\right)\left(\prod_{l=0}^{N-1}\left(1+x^{b^{l}}\right)^{m_{l}}\right)=\prod_{l=0}^{N-1}\left(1+x^{b^{l}}\right)^{m_{l}+n_{l}}.
\end{equation*}


(II) In the case when the addition $m+n$ is not carry-free, the identity
fails. Take, for example, $b=2$ and 
\begin{equation*}
m=n=r=1\Rightarrow m+n=2=1\cdot2^{1}+0\cdot2^{0}.
\end{equation*}
Then,
\begin{equation*}
{m+n \choose r}_{b}={2 \choose 1}_{2}={1\cdot2^{1}+0\cdot2^{0} \choose 0\cdot2^{1}+1\cdot2^{0}}_{2}={1 \choose 0}{0 \choose 1}=0.
\end{equation*}
On the other hand,
\begin{equation*}
\sum_{0\leq k\apprle_{b}r}{m \choose k}_{b}{n \choose r-k}_{b}=\sum_{k=0}^{1}{1 \choose k}_{2}{1 \choose 1-k}_{2}={1 \choose 0}_{2}{1 \choose 1-0}_{2}+{1 \choose 1}_{2}{1 \choose 1-1}_{2}=2.
\end{equation*}


(III) The summation condition $0\leq k\apprle_{b}r$ in \eqref{eq:bChuVandermonde}
can not be replaced by the more natural condition $0\leq k\leq r$. Consider, for example, the case
\begin{equation*}
b=2,\ m=2,\ n=1,\ \text{and}\ r=2.
\end{equation*}
Then 
\begin{equation*}
{m+n \choose r}_{b}={3 \choose 2}_{2}={1\cdot2^{1}+1\cdot2^{0} \choose 1\cdot2^{1}+0\cdot2^{0}}_{2}={1 \choose 1}{1 \choose 0}=1.
\end{equation*}
On the other hand, considering all the values of $k$ that satisfy $0\leq k\leq 2$,
we have
\begin{displaymath}
{m \choose k}_{b}{n \choose r-k}_{b}={2 \choose k}_{2}{1 \choose 2-k}_{2}=\begin{cases}
0, & \text{if $k=0$;}\\
1, & \text{if $k=1$;}\\
1, & \text{if $k=2$.}
\end{cases}
\end{displaymath}
Note that $k=1=0\cdot2^{1}+1\cdot2^{0}$ is the only case for which $k\apprle_{b}r=1\cdot2^{1}+1\cdot2^{0}$ fails. Therefore, 
\begin{equation*}
\sum_{0\leq k\leq r}{m \choose k}_{b}{n \choose r-k}_{b}=2\ne1={m+n \choose r}_{b}=\sum_{0\leq k\apprle_{b}r}{m \choose k}_{b}{n \choose r-k}_{b}.
\end{equation*}
\end{remark}

\subsection{Symmetry and recurrence}
\begin{theorem}
The $b$-ary binomial coefficients satisfy
\begin{enumerate}
\item the symmetry property
\begin{equation*}
{n\choose k}_{b}={n\choose n-k}_{b},
\end{equation*}

\item when ${n\choose k}_b\neq 0$, the recurrence
\begin{equation}
{n\choose k}_{b}={n-1 \choose k-1}_{b}+{n-1 \choose k}_{b}, \label{eq:recurrence}
\end{equation}
\end{enumerate}
\end{theorem}

\begin{proof}
The symmetry property is easily deduced from definition \eqref{eq:binomialcoefficients}
and the invariance of each ${n_l\choose k_l}$ under the exchange $k_{l}\leftrightarrow n_{l}-k_{l}$ for each digit. For
the recurrence property, assume that both $k$ and $n$ have a non-zero
lowest rank digit, i.e., $k_{0}>0$ and $n_{0}>0$. Then, $\left(n-1\right)_{0}=n_{0}-1$,
$\left(k-1\right)_{0}=k_{0}-1$, 
\begin{equation*}
{n-1\choose k-1}_{b}={n_{N-1}\choose k_{N-1}}\cdots{n_{1}\choose k_{1}}{n_{0}-1\choose k_{0}-1},\text{ and }{n-1\choose k}_{b}={n_{N-1}\choose k_{N-1}}\cdots{n_{1}\choose k_{1}}{n_{0}-1\choose k_{0}}.
\end{equation*}
We deduce that
\begin{equation*}
{n-1\choose k-1}_{b}+{n-1\choose k}_{b}={n_{N-1}\choose k_{N-1}}\cdots{n_{1}\choose k_{1}}\left({n_{0}-1\choose k_{0}-1}+{n_{0}-1\choose k_{0}}\right)={n\choose k}_{b}.
\end{equation*}
This argument extends easily to the case where zero components appear in the pair
$\left(k_{0},n_{0}\right)$.
\end{proof}
\begin{remark}
As suggested by the referee, the recurrence \eqref{eq:recurrence}
can also be extended to higher digits, which leads to 
\begin{equation*}
{n\choose k}_{b}={n-b^{j} \choose k-b^{j}}_{b}+{n-b^{j} \choose k}_{b},
\end{equation*}
for all $j\in\left\{ 0,\ldots,N-1\right\}$, by a similar proof.
\end{remark}

\subsection{Link with Lucas' theorem}

The definition \eqref{eq:binomialcoefficients} will look familiar
to those readers who have already met Lucas' famous theorem \cite{FineBinomialModuloPrimep}, which we restate here.
\begin{theorem}
{[}Lucas{]} For a prime number $p$, the binomial coefficient ${n\choose k}$
satisfies the congruence
\begin{equation*}
{n\choose k}\equiv{n_{N-1}\choose k_{N-1}}\cdots{n_{0}\choose k_{0}}\pmod p.
\end{equation*}

\end{theorem}
\begin{remark}
(I) Note that, under the same condition, Lucas' theorem is concisely rephrased
in our notation as
\begin{equation*}
{n\choose k}\equiv{n\choose k}_{p}\pmod p.
\end{equation*} 
Lucas' theorem becomes obvious by considering the generating function \eqref{eq:barybinomialgeneratingfunction} and noting an elementary congruence:
\begin{equation*}
\sum_{k=0}^{n}{n \choose k}_{p}x^{k}=\prod_{l=0}^{N-1}\left(1+x^{p^{l}}\right)^{n_{l}}\equiv\left(1+x\right)^{n}=\sum_{k=0}^{n}{n\choose k}x^{k}\pmod p.
\end{equation*}


(II) The Sierpinski matrix, which Nguyen et al.\ use to prove their
results, contains the entries ${n\choose k}\bmod p$. These coefficients do not coincide with ${n\choose k}_{p}$ studied here, but are congruent to them.

\end{remark}

\subsection{Orthogonality relations}

There are many elementary identities involving the usual binomial
coefficients that can be extended to the case of the $b$-ary binomial
coefficients. Here, we show one example. 
\begin{example}
Assume, $\forall n\in\mathbb{N}$, two sequences $\left(a_{n}\right)_{n\geq 0}$ and $\left(c_{n}\right)_{n\geq 0}$
are related by
\begin{equation*}
a_{n}=\overset{n}{\underset{k=0}{\sum}}\left(-1\right)^{k}{n \choose k}c_{k}.
\end{equation*}
Then, 
\begin{equation*}
c_{n}=\overset{n}{\underset{k=0}{\sum}}\left(-1\right)^{k}{n \choose k}a_{k},
\end{equation*}
which is known as an inverse relation \cite[ex.\ 2, p.\ 4]{Riordan}. Note that this inverse relation is equivalent
to the orthogonality conditions 
\begin{displaymath}
\sum_{k=j}^{n}{n \choose k}{k \choose j}\left(-1\right)^{k+j}=\delta_{n,j}=\begin{cases}
1, & \text{if $j=n$;}\\
0, & \text{otherwise.}
\end{cases}
\end{displaymath}
The generalization to the $b$-ary case is as follows. 
\end{example}
\begin{theorem}
The $b$-ary binomial coefficients satisfy the orthogonality relations
\begin{equation*}
\sum_{k=0}^{n}{n \choose k}_{b}{k \choose j}_{b}\left(-1\right)^{S_{b}\left(k\right)+S_{b}\left(j\right)}=\delta_{n,j}.
\end{equation*}
As a consequence, if two sequences $\left(a_{n}\right)_{n\geq 0}$ and $\left(c_{n}\right)_{n\geq 0}$
satisfy 
\begin{equation*}
a_{S_{b}\left(n\right)}=\overset{n}{\underset{k=0}{\sum}}\left(-1\right)^{S_{b}\left(k\right)}{n \choose k}_{b}c_{S_{b}\left(k\right)},
\end{equation*}
then
\begin{equation*}
c_{S_{b}\left(n\right)}=\overset{n}{\underset{k=0}{\sum}}\left(-1\right)^{S_{b}\left(k\right)}{n \choose k}_{b}a_{S_{b}\left(k\right)}.
\end{equation*}
\end{theorem}
\begin{proof}
Since
\begin{equation*}
{n \choose k}_{b}=\prod_{l=1}^{N-1}{n_{j} \choose k_{l}},\ \ {k \choose j}_{b}=\prod_{l=1}^{N-1}{k_{l} \choose j_{l}},\ \ S_{b}\left(k\right)=\sum_{l=0}^{N-1}k_{l},\ \text{and}\ S_{b}\left(j\right)=\sum_{l=0}^{N-1}j_{l},
\end{equation*}
we deduce 
\begin{equation*}
\sum_{k=j}^{n}{n \choose k}_{b}{k \choose j}_{b}\left(-1\right)^{S_{b}\left(k\right)+S_{b}\left(j\right)}=\prod_{l=1}^{N-1}\sum_{k_{l}=j_{l}}^{n_{l}}{n_{j} \choose k_{l}}{k_{l} \choose j_{l}}\left(-1\right)^{k_{l}+j_{l}}=\prod_{l=1}^{N-1}\delta_{n_{l},j_{l}}=\delta_{n,j}.
\end{equation*}

\end{proof}

\section{Pascal-like triangles}
\label{sec:PascallikeTriangles}
In this section, we study the structure of Pascal-like triangles built
from the $b$-ary binomial coefficients ${n\choose k}_{b}$. Let us
start with two examples, where we systematically replace each null
binomial entry with a dot ``." symbol in order to make the structure
of the triangle more visible. Here we underline the fact that these
null entries correspond exactly to the pairs $\left(n,k\right)$ for
which the addition of $k$ and $n-k$ is not carry-free in base $b$.
\begin{example}
In base $b=3$, the top $9$ rows of the Pascal-like triangle
for the ternary binomial coefficients are\textbf{ }
\begin{equation*}
T_{2}^{\left(3\right)}=\begin{array}{ccccccccccccccccc}
 &  &  &  &  &  &  &  & 1\\
 &  &  &  &  &  &  & 1 &  & 1\\
 &  &  &  &  &  & 1 &  & 2 &  & 1\\
 &  &  &  &  & 1 &  & . &  & . &  & 1\\
 &  &  &  & 1 &  & 1 &  & . &  & 1 &  & 1\\
 &  &  & 1 &  & 2 &  & 1 &  & 1 &  & 2 &  & 1\\
 &  & 1 &  & . &  & . &  & 2 &  & . &  & . &  & 1\\
 & 1 &  & 1 &  & . &  & 2 &  & 2 &  & . &  & 1 &  & 1\\
1 &  & 2 &  & 1 &  & 2 &  & 4 &  & 2 &  & 1 &  & 2 &  & 1
\end{array}
\end{equation*}
Now, let 
\begin{equation*}
T_{1}^{\left(3\right)}=\begin{array}{ccccc}
 &  & 1\\
 & 1 &  & 1\\
1 &  & 2 &  & 1
\end{array}
\end{equation*}
denote the top $3$ rows of $T_{2}^{\left(3\right)}$ and also let 
\begin{equation*}
*=\begin{array}{ccc}
. &  & .\\
 & .
\end{array}=\begin{array}{ccc}
0 &  & 0\\
 & 0
\end{array}
\end{equation*}
be the elementary reverse triangle containing only zero entries. Then,
we remark that
\begin{equation*}
T_{2}^{\left(3\right)}=\begin{array}{ccccc}
 &  & T_{1}^{\left(3\right)}\\
 & T_{1}^{\left(3\right)} & * & T_{1}^{\left(3\right)}\\
T_{1}^{\left(3\right)} & * & 2T_{1}^{\left(3\right)} & * & T_{1}^{\left(3\right)}
\end{array}
\end{equation*}
and furthermore, introducing the notation
\begin{equation*}
T_{2}^{\left(3\right)}=T_{1}^{\left(3\right)}\otimes T_{1}^{\left(3\right)},
\end{equation*}
we deduce, for all $m\in\mathbb{N}$,
\begin{equation*}
T_{m}^{\left(3\right)}=\left(T_{1}^{\left(3\right)}\right)^{\otimes m}.
\end{equation*}
Here, the \emph{Kronecker product} $\otimes$ will be defined and discussed in Proposition \ref{prop:Thestructureof} below.
\end{example}

\begin{example}
In base $4$, the first $16$ rows of the Pascal-like triangle are as follows.
\end{example}
{\tiny{}
\begin{equation*}
T_{2}^{\left(4\right)}=\begin{array}{ccccccccccccccccccccccccccccccc}
 &  &  &  &  &  &  &  &  &  &  &  &  &  &  & 1\\
 &  &  &  &  &  &  &  &  &  &  &  &  &  & 1 &  & 1\\
 &  &  &  &  &  &  &  &  &  &  &  &  & 1 &  & 2 &  & 1\\
 &  &  &  &  &  &  &  &  &  &  &  & 1 &  & 3 &  & 3 &  & 1\\
 &  &  &  &  &  &  &  &  &  &  & 1 &  & . &  & . &  & . &  & 1\\
 &  &  &  &  &  &  &  &  &  & 1 &  & 1 &  & . &  & . &  & 1 &  & 1\\
 &  &  &  &  &  &  &  &  & 1 &  & 2 &  & 1 &  & . &  & 1 &  & 2 &  & 1\\
 &  &  &  &  &  &  &  & 1 &  & 3 &  & 3 &  & 1 &  & 1 &  & 3 &  & 3 &  & 1\\
 &  &  &  &  &  &  & 1 &  & . &  & . &  & . &  & 2 &  & . &  & . &  & . &  & 1\\
 &  &  &  &  &  & 1 &  & 1 &  & . &  & . &  & 2 &  & 2 &  & . &  & . &  & 1 &  & 1\\
 &  &  &  &  & 1 &  & 2 &  & 1 &  & . &  & 2 &  & 4 &  & 2 &  & . &  & 1 &  & 2 &  & 1\\
 &  &  &  & 1 &  & 3 &  & 3 &  & 1 &  & 2 &  & 6 &  & 6 &  & 2 &  & 1 &  & 3 &  & 3 &  & 1\\
 &  &  & 1 &  & . &  & . &  & . &  & 3 &  & . &  & . &  & . &  & 3 &  & . &  & . &  & . &  & 1\\
 &  & 1 &  & 1 &  & . &  & . &  & 3 &  & 3 &  & . &  & . &  & 3 &  & 3 &  & . &  & . &  & 1 &  & 1\\
 & 1 &  & 2 &  & 1 &  & . &  & 3 &  & 6 &  & 3 &  & . &  & 3 &  & 6 &  & 3 &  & . &  & 1 &  & 2 &  & 1\\
1 &  & 3 &  & 3 &  & 1 &  & 3 &  & 9 &  & 9 &  & 3 &  & 3 &  & 9 &  & 9 &  & 3 &  & 1 &  & 3 &  & 3 &  & 1
\end{array}.
\end{equation*}
}Starting from the elementary triangles:
\begin{equation*}
T_{1}^{\left(4\right)}=\begin{array}{ccccccc}
 &  &  & 1\\
 &  & 1 &  & 1\\
 & 1 &  & 2 &  & 1\\
1 &  & 3 &  & 3 &  & 1
\end{array}\text{ and }*=\begin{array}{ccccc}
. &  & . &  & .\\
 & . &  & .\\
 &  & .
\end{array}=\begin{array}{ccccc}
0 &  & 0 &  & 0\\
 & 0 &  & 0\\
 &  & 0
\end{array},
\end{equation*}
we similarly obtain that
\begin{equation*}
T_{2}^{\left(4\right)}=\begin{array}{ccccccc}
 &  &  & T_{1}^{\left(4\right)}\\
 &  & T_{1}^{\left(4\right)} & * & T_{1}^{\left(4\right)}\\
 & T_{1}^{\left(4\right)} & * & 2T_{1}^{\left(4\right)} & * & T_{1}^{\left(4\right)}\\
T_{1}^{\left(4\right)} & * & 3T_{1}^{\left(4\right)} & * & 3T_{1}^{\left(4\right)} & * & T_{1}^{\left(4\right)}
\end{array}
\end{equation*}
and more generally
\begin{equation*}
T_{m}^{\left(4\right)}=\left(T_{1}^{\left(4\right)}\right)^{\otimes m}.
\end{equation*}


These observations are now extended to an arbitrary base
$b$.
\begin{proposition}
\label{prop:Thestructureof}
The structure of the triangle built
from the coefficients ${n\choose k}_{b}$ satisfies 
\begin{enumerate}
\item \label{One} the top $b$ rows, denoted by $T_{1}^{\left(b\right)}$, coincide with the top $b$ rows of the usual Pascal triangle;
\item \label{Two} for all $m>1$, $T_{m}^{\left(b\right)}$ is obtained from $T_{m-1}^{\left(b\right)}$
by the Kronecker product
\begin{equation*}
T_{m}^{\left(b\right)}=T_{1}^{\left(b\right)}\otimes T_{m-1}^{\left(b\right)}
\end{equation*}
defined by 
\begin{equation*}
T_{m}^{\left(b\right)}=\begin{array}{ccccccc}
 &  &  & {0\choose 0}T_{m-1}^{\left(b\right)}\\
\\
 &  & {1\choose 0}T_{m-1}^{\left(b\right)} &  & {1\choose 1}T_{m-1}^{\left(b\right)}\\
 & \iddots &  &  &  & \ddots\\
{b-1\choose 0}T_{m-1}^{\left(b\right)} &  & \ldots & {b-1\choose j}T_{m-1}^{\left(b\right)} & \ldots &  & {b-1\choose b-1}T_{m-1}^{\left(b\right)}
\end{array}
\end{equation*}
so that 
\begin{equation*}
T_{m}^{\left(b\right)}=T_{1}^{\left(b\right)}\otimes\cdots\otimes T_{1}^{\left(b\right)}=\left(T_{1}^{\left(b\right)}\right)^{\otimes m};
\end{equation*}

\item \label{Three}  since 
\begin{equation*}
\#\text{rows}\left(T_{m}^{\left(b\right)}\right)=b\times\#\text{rows}\left(T_{m-1}^{\left(b\right)}\right),
\end{equation*}
the triangle $T_{m}^{\left(b\right)}$ has $b^{m}$ rows;
\item \label{Four} the bottom row of the triangle $T_{m}^{\left(b\right)}$ has $b^{m}$
entries.
\end{enumerate}
\end{proposition}

\begin{proof}
These properties are a direct consequence of \eqref{eq:binomialcoefficients}.
Since properties \ref{Three} and \ref{Four} are elementary, only properties \ref{One}  and
\ref{Two}  need to be verified. Here, we prove them by induction.

(i) For $m=1$, $T_{1}^{\left(b\right)}$ consists of the first $b$
rows of the triangle made of the $b$-ary binomial coefficients ${n\choose k}_{b}$ where
\begin{equation*}
0\leq k\leq n\leq b-1,
\end{equation*}
so that the $b$-ary expression of $n$ consists of a single digit:
$n=n_{0}$. In this case, ${n\choose k}_{b}$ coincides with the usual binomial coefficients ${n\choose k}$. Thus, the first $b$ rows coincide with those of the Pascal triangle
(see also Remark \ref{rem:remark3p2}). 

(ii) Consider $T_{m}^{\left(b\right)}$ by assuming that properties \ref{One}  and
\ref{Two} hold for $T_{1}^{\left(b\right)},\ldots, T_{m-1}^{\left(b\right)}$.
Then, the $b^{m}$ elements of $T_{m}^{\left(b\right)}$ correspond
exactly to the case
\begin{equation*}
0\leq n\leq b+\cdots+b^{m}-1,
\end{equation*}
which implies that $n$ has at most $m$ digits, namely
\begin{equation*}
n=n_{m-1}b^{m-1}+\cdots+n_{0}.
\end{equation*}
Thus, 
\begin{equation*}
{n \choose k}_{b}={n_{m-1} \choose k_{m-1}}\underset{\text{copy of }T_{m-1}^{\left(b\right)}}{\underbrace{\overset{m-2}{\underset{l=0}{\prod}}{n_{l} \choose k_{l}}}}.
\end{equation*}
Since $0\leq n_{m-1}\leq b-1$, the first factor ${n_{m-1} \choose k_{m-1}}$ gives a
copy of $T_{1}^{\left(b\right)}$ while the rest of the product gives
a copy of $T_{m-1}^{\left(b\right)}$. Hence, by induction,
\begin{equation*}
T_{m}^{\left(b\right)}=\left(T_{1}^{\left(b\right)}\right)^{\otimes m}.
\end{equation*}

\end{proof}

\section{Historical perspective and conclusion}
\label{sec:History}

Generalizations of the usual binomial coefficients have been extensively
studied in many different ways. In particular, Knuth and Wilf  \cite{Wilf} define
the \emph{generalized binomial coefficients} associated with a sequence of
numbers $\left(C_{n}\right)_{n\geq 0}$ by
\begin{equation*}
{m\choose n}_{C}=\frac{C_{m}C_{m-1}\cdots C_{n+1}}{C_{m-n}C_{m-n-1}\cdots C_{1}},
\end{equation*}
and study their divisibility by powers of a prime $p$. Ball et al.\ \cite{Ball} introduce the \emph{$b$-ary binomial coefficients} as those
of Knuth and Wilf built from a sequence $\left(C_{n}\right)_{n\geq 0}$
defined by
\begin{equation*}
C_{n}=b^{\nu_{b}\left(n\right)},
\end{equation*}
i.e., the highest power of $b$ that divides $n$, and study the arithmetic
properties of these coefficients.

In quantum physics, Gazeau et al.\ \cite{Gazeau} introduce the same generalized binomial coefficients as Knuth and Wilf to model
the influence of correlations between the states of quantum systems.

We note that the $b$-ary binomial coefficients
defined by \eqref{eq:binomialcoefficients} and studied here, look 
similar to the ones of Knuth
and Wilf. However, we failed to find a sequence $\left(C_{k}\right)_{k\geq0}$ such that 
\begin{equation*}
{n \choose k}_{b}={n \choose k}_{C}.
\end{equation*}
This may explain why we were unable to find a reference where these
$b$-ary binomial coefficients appear explicitly.

Nguyen \cite{NguyenGeneralization} has shown how the digital
binomial identity can be generalized to sequences of the Sheffer type
such as the Bernoulli or Hermite polynomials. An important question
that remains about these $b$-ary binomial coefficients is to know
if they satisfy identities which are not of the Sheffer type. The
example of the Chu-Vandermonde identity \eqref{eq:bChuVandermonde} already shows that even simple linear identities
do not extend automatically to the case of $b$-ary binomial coefficients,
indicating that extension to more difficult cases may be even trickier.
This will be the subject of future work.


\section{Acknowledgments} 

The work of the second author was partially supported by the iCODE
Institute, Research Project of the Idex Paris-Saclay. The authors
wish to thank Victor Hugo Moll for his support and discussions. The authors
also thank Hong Diep Nguyen for his valuable comments and the anonymous
reviewer for the careful analysis of the manuscript. Finally, the
authors thank Tanay Wakhare for his extension of the Chu-Vandermonde
identity.

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\bibitem{digialbinomial}D. Callan, Sierpinski\textquoteright s triangle
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\bibitem{Gazeau}E. M. F. Curado, J. P. Gazeau, and L. M. C. S. Rodrigues,
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\bibitem{FineBinomialModuloPrimep}N. J. Fine, Binomial coefficients
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\bibitem{Wilf}D. E. Knuth and H. S. Wilf, The power of a prime that
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\bibitem{Nguyenqdigital}T. Mansour and H. D. Nguyen, A q-digital
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\bibitem{NguyenDigital}H. D. Nguyen, A digital binomial theorem,
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\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 \emph{Mathematics Subject Classification:}
Primary 05C30; Secondary 05C78.

\noindent \emph{Keywords:} 
$b$-ary expansion, $b$-ary binomial coefficient, Lucas' theorem.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 12 2016;
revised versions received   April 15 2016; May 15 2016.
Published in {\it Journal of Integer Sequences}, June 2 2016.
Minor revision, November 13 2016.

This is the previous version, which
has been superseded by the corrected version, published September 25 2019.

\bigskip
\hrule
\bigskip

\noindent
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