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\begin{center}
\vskip 1cm{\LARGE\bf A Generalization of the \\
\vskip .12in
Digital Binomial Theorem}
\vskip 1cm
\large
Hieu D. Nguyen\\
Department of Mathematics\\
Rowan University\\
Glassboro, NJ 08028\\
USA \\
\href{mailto:nguyen@rowan.edu}{\tt nguyen@rowan.edu}
\end{center}

\vskip .2 in

\begin{abstract}
We prove a generalization of the digital binomial theorem by
constructing a one-parameter subgroup of generalized Sierpi\'{n}ski
matrices.  In addition, we derive new formulas for the coefficients of
Prouhet-Thue-Morse polynomials and describe group relations satisfied
by generating matrices defined in terms of these Sierpi\'{n}ski
matrices.
\end{abstract}

\section{Introduction}
The classical binomial theorem describes the expansion of $(x+y)^N$ in terms of binomial coefficients.  More precisely, for any non-negative integer $N$, we have
\begin{equation}\label{eq:binomial-theorem}
(x+y)^N = \sum_{k=0}^N \binom{N}{k} x^k y^{N-k}, 
\end{equation}
where the binomial coefficients
$\binom{N}{k}$ are defined in terms of factorials:
\[
\binom{N}{k}=\frac{N!}{k!(N-k)!}.
\]

The author \cite{N1} introduced a digital version of this theorem (based on Callan \cite{C}) where the exponents appearing in (\ref{eq:binomial-theorem}) are viewed as sums of digits.  To illustrate this, consider the binomial theorem for $N=2$:
\begin{equation} \label{eq:binomial-theorem-n=2}
(x+y)^2  =x^2y^0+x^1y^1+x^1y^1+x^0y^2.
\end{equation}
It is easy to verify that (\ref{eq:binomial-theorem-n=2}) is equivalent to
\begin{equation} \label{eq:sum-of-digit-expansion-n=2}
(x+y)^{s(3)}  = x^{s(3)}y^{s(0)}+x^{s(2)}y^{s(1)}+x^{s(1)}y^{s(2)}+x^{s(0)}y^{s(3)},
\end{equation}
where $s(k)$ denotes the sum of digits of $k$ expressed in binary.  For example, 
\[
s(3)=s(1\cdot 2^1+1\cdot 2^0)=2.
\]  
More generally, we have


\begin{theorem}[digital binomial theorem \cite{N1}] \label{th:digital-binomial-theorem}
 Let $n$ be a non-negative integer.  Then
\begin{equation} \label{eq:digital-binomial-theorem-carry-free}
(x+y)^{s(n)}  = \sum_{\substack{0\leq m \leq n \\  (m,n-m) \ 
\text{\rm carry-free}}}x^{s(m)}y^{s(n-m)}.
\end{equation}
\end{theorem}
\noindent Here, a pair of non-negative integers $(j,k)$ is said to be {\em carry-free} if their addition involves no carries when performed in binary.  For example, $(8,2)$ is carry-free since $8+2=(1\cdot 2^3+0\cdot 2^2+0\cdot 2^1+0\cdot 2^0)+1\cdot 2^1=10$ involves no carries.  It is clear that $(j,k)$ is carry-free if and only if $s(j)+s(k)=s(j+k)$ \cite{BEJ, N1}.  Also, observe that if $n=2^N-1$, then (\ref{eq:digital-binomial-theorem-carry-free}) reduces to (\ref{eq:binomial-theorem}). 

In this paper we generalize Theorem~\ref{th:digital-binomial-theorem} to any base $b\geq 2$.  To state this result, we shall need to introduce a {\em digital dominance} partial order on $\mathbb{N}$ as defined by Ball, Edgar, and Juda \cite{BEJ}.  Let
\begin{align*}
n & =n_0b^0+n_1b^1+\dots + n_{N-1}b^{N-1}
\end{align*}
represent the base-$b$ expansion of $n$ and denote $d_{b,i}(n):=n_i$ to be the $i$-th digit of $n$ in base $b$.  We shall say that $m$ is {\em digitally} less than or equal to $n$ if $d_{b,i}(m)\leq d_{b,i}(n)$ for all $i=0,1,\ldots, N-1$.  In that case, we shall write $m\preceq_b n$.  We are now ready to state our result.

\begin{theorem} \label{th:digital-binomial-theorem-non-binary}
Let $n$ be a non-negative integer.  Then for any base $b\geq 2$, we have
\begin{equation} \label{eq:digital-binomial-theorem-non-binary}
\prod_{i=0}^{N-1}\binom{x+y+d_{b,i}(n)-1}{d_{b,i}(n)} =\sum_{0\leq m\preceq_b n}\left( \prod_{i=0}^{N-1}\binom{x+d_{b,i}(m)-1}{d_{b,i}(m)} \prod_{i=0}^{N-1}\binom{y+d_{b,i}(n-m)-1}{d_{b,i}(n-m)} \right).
\end{equation}
\end{theorem}

Let $\mu_j(n):=\mu_j^{(b)}(n)$ denote the multiplicity of the digit $j>0$ in the base-$b$ expansion of $n$, i.e.,
\[
\mu_j(n)=|\{i: d_{b,i}(n)=j\}|.
\]
As a corollary, we obtain

\begin{corollary} \label{co:digital-binomial-theorem-non-binary}
Let $n$ be a non-negative integer.  Then for any base $b\geq 2$, we have
\begin{align*}
\prod_{j=1}^{b-1}\binom{x+y+j-1}{j}^{\mu_j(n)} & =\sum_{0\leq m\preceq_b n}\left( \prod_{j=1}^{b-1}\binom{x+j-1}{j}^{\mu_j(m)} \prod_{j=1}^{b-1}\binom{y+j-1}{j}^{\mu_j(n-m)} \right).
\end{align*}
\end{corollary}
\noindent Observe that if $b=2$, then Corollary~\ref{co:digital-binomial-theorem-non-binary} reduces to Theorem~\ref{th:digital-binomial-theorem}.

The source behind Theorem~\ref{th:digital-binomial-theorem} is a one-parameter subgroup of Sierpi\'nski matrices, investigated by Callan \cite{C}, which encodes the digital binomial theorem.  To illustrate this, define a sequence of lower-triangular matrix functions $S_N(x)$ of dimension $2^N\times 2^N$ recursively by
\begin{equation}
S_1(x)=\left(
\begin{array}{rr}
1 & 0 \\
x & 1 \\
\end{array}
\right), \ \ S_{N+1}(x)=S_1(x)\otimes S_N(x),
\end{equation}
where $\otimes$ denotes the Kronecker product of two matrices.  For example, $S_2(x)$ and $S_3(x)$ can be computed as follows:
\begin{align*}
S_2(x) & =S_1(x)\otimes S_1(x) = \left(\begin{array}{rr} 1\cdot S_1(x) & 0\cdot S_1(x) \\ x\cdot S_1(x) & 1\cdot S_1(x) \end{array} \right)
= \left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
x & 1 & 0 & 0 \\
x & 0 & 1 & 0 \\
x^2 & x & x & 1
\end{array}
\right),
\\
\\
S_3(x) & =S_1(x)\otimes S_2(x) = \left(\begin{array}{rr} 1\cdot S_2(x) & 0\cdot S_2(x) \\ x\cdot S_2(x) & 1\cdot S_2(x) \end{array} \right)
=\left(
\begin{array}{llllllll}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0  \\
x & 1 & 0 & 0 & 0 & 0 & 0 & 0   \\
x & 0 & 1 & 0 & 0 & 0 & 0 & 0   \\
x^2 & x & x & 1 & 0 & 0 & 0 & 0  \\
x & 0 & 0 & 0 & 1 & 0 & 0 & 0  \\
x^2 & x & 0 & 0 & x & 1 & 0 & 0  \\
x^2 & 0 & x & 0 & x & 0 & 1 & 0  \\
x^3 & x^2 & x^2 & x & x^2 & x & x & 1
\end{array}
\right).
\end{align*}
Observe that when $x=1$, the infinite matrix $S=\lim_{N\rightarrow \infty}S_N(1)$ contains Sierpi\'nski's triangle.  

Callan \cite{C} gives a formula for the entries of $S_N(x)=(\alpha_N(j,k,x))$, $0\leq j,k\leq 2^N-1$, in terms of the sum-of-digits function:
\begin{equation} \label{eq:formula-for-entries}
\alpha_N(j,k,x)=\left\{
\begin{array}{cl}
x^{s(j-k)}, & \mathrm{if} \ 0\leq k\leq j \leq 2^N-1 \ \mathrm{and} \  (k, j-k) \ \textrm{are} \ \textrm{carry-free;}  \\ 
0, & \mathrm{otherwise}.
\end{array}
\right.
\end{equation}
Moreover, Callan proved that $S_N(x)$ forms a one-parameter subgroup of $SL(2^N,\mathbb{R})$, i.e., the group of $2^N\times 2^N$ real matrices with determinant one.  Namely, we have
\begin{equation} \label{eq:one-parameter-matrix-product}
S_N(x)S_N(y)=S_N(x+y).
\end{equation}
If we denote the entries of $S_N(x)S_N(y)$ by $t_N(j,k)$, then the equality 
\[
t_N(j,k)=\sum_{i=0}^{2^N-1}\alpha_N(j,i,x)\alpha_N(i,k,y)=\alpha_N(j,k,x+y)\]
corresponds precisely to Theorem~\ref{th:digital-binomial-theorem} with $j=0$ and $k=0$.  For example, if $N=2$, then (\ref{eq:one-parameter-matrix-product}) becomes
\begin{align*}
\left(
\begin{array}{llll}
1 & 0 & 0 & 0 \\
x & 1 & 0 & 0 \\
x & 0 & 1 & 0 \\
x^2 & x & x & 1 
\end{array}
\right)
\left(
\begin{array}{llll}
1 & 0 & 0 & 0 \\
y & 1 & 0 & 0 \\
y & 0 & 1 & 0 \\
y^2 & y & y & 1 
\end{array}
\right) & =
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
x+y & 1 & 0 & 0 \\
x+y & 0 & 1 & 0 \\
(x+y)^2 & x+y & x+y & 1 
\end{array}
\right).
\end{align*}

The rest of this paper is devoted to generalizing Callan's construction of 
Sierpi\'nski matrices to arbitrary bases and considering two applications of them.  In Section~\ref{two}, we use these generalized Sierpi\'nski matrices to prove Theorem~\ref{th:digital-binomial-theorem-non-binary}.  In Section~\ref{three}, we demonstrate how these matrices arise in the construction of Prouhet-Thue-Morse polynomials defined by the author \cite{N2}.  In Section~\ref{four}, we describe a group presentation in terms of generators defined through these matrices and show that these generators satisfy a relation that generalizes the three-strand braid relation found by Ferrand \cite{F1}.  This relation suggests that Sierpi\'nski matrices encode not only a digital binomial theorem but also an interesting group structure.

%%%%%%%%%%%%%

\section{Sierpi\'nski triangles}
\label{two}

To prove Theorem~\ref{th:digital-binomial-theorem-non-binary}, we consider the following generalization of the Sierpi\'nski matrix $S_N(x)$ in terms of binomial coefficients.  Define lower-triangular matrices $S_{b,N}(x)$ of dimension $b^N\times b^N$ recursively by
\begin{align*}
S_{b,1}(x) & =\left(
\begin{array}{cccll}
1 & 0 & 0 & \cdots & 0 \\
\binom{x}{1} & 1 & 0 & \cdots & 0 \\
\binom{x+1}{2} & \binom{x}{1} & 1 & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots  & \vdots \\
\binom{x+b-2}{b-1} & \binom{x+b-3}{b-2} & \binom{x+b-4}{b-3} & \cdots &  1
\end{array}
\right)=
\left\{
\begin{array}{cl}
\binom{x+j-k-1}{j-k}, & \textrm{if } 0\leq k \leq j \leq b-1; \\
0, & \textrm{otherwise},
\end{array}
\right. \\
\end{align*}
and for $N>1$,
\[
S_{b,N+1}(x) =S_{b,1}(x)\otimes S_{b,N}(x).
\]

\begin{example} To illustrate our generalized Sierpi\'nski matrices, we calculate $S_{3,1}(x)$ and $S_{3,2}(x)$:
\begin{align*}
S_{3,1}(x)
& =\left(
\begin{array}{ccc}
1 & 0 & 0 \\
\binom{x}{1} & 1 & 0  \\
\binom{x+1}{2} & \binom{x}{1} & 1
\end{array} 
\right),
\\
S_{3,2}(x) & =S_{3,1}(x)\otimes S_{3,1}(x) \\
& =\left(
\begin{array}{ccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0  \\
\binom{x}{1} & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0  \\
\binom{x+1}{2} & \binom{x}{1} & 1 & 0 & 0 & 0 & 0 & 0 & 0  \\
\binom{x}{1} & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0  \\
\binom{x}{1}\binom{x}{1} & \binom{x}{1} & 0 & \binom{x}{1} & 1 & 0 & 0 & 0 & 0  \\
\binom{x}{1}\binom{x+1}{2} & \binom{x}{1}\binom{x}{1} & \binom{x}{1} & \binom{x+1}{2} & \binom{x}{1} & 1 & 0 & 0 & 0  \\
\binom{x+1}{2} & 0 & 0 & \binom{x}{1} & 0 & 0 & 1 & 0 & 0  \\
\binom{x+1}{2}\binom{x}{1} & \binom{x+1}{2} & 0 & \binom{x}{1}\binom{x}{1} & \binom{x}{1} & 0 & \binom{x}{1} & 1 & 0 \\
\binom{x+1}{2}\binom{x+1}{2} & \binom{x+1}{2}\binom{x}{1} & \binom{x+1}{2} & \binom{x}{1}\binom{x+1}{2} & \binom{x}{1}\binom{x}{1} & \binom{x}{1} & \binom{x+1}{2} & \binom{x}{1} & 1 \\
\end{array}
\right).
\end{align*}
\end{example}

We now generalize Callan's result for $S_N(x)$ by presenting a formula for the entries of $S_{b,N}(x)$; see \cite{IM} for a similar generalization but along a different vein.

\begin{theorem} \label{th:formula-for-entries-arbitrary-b}
Let $\alpha_N(j,k):=\alpha_N(j,k,x)$ denote the $(j,k)$-entry of $S_{b,N}(x)$.
Then
\begin{equation} \label{eq:formula-for-entries-arbitrary-b}
\alpha_N(j,k)=\left\{
\begin{array}{cl}
\binom{x+d_0-1}{d_0} \binom{x+d_1-1}{d_1} \cdots \binom{x+d_{N-1}-1}{d_{N-1}}, & \mathrm{if} \ 0\leq k\leq j \leq b^N-1 \ \mathrm{and} \  k \preceq_b j;  \\ 
\\
0, & \mathrm{otherwise},
\end{array}
\right.
\end{equation}
where $j-k=d_0b^0+d_1b^1+\ldots + d_{N-1}b^{N-1}$ is the base-$b$ expansion of $j-k$, assuming $j\geq k$.
\end{theorem}

\begin{proof} We argue by induction on $N$.  It is clear that (\ref{eq:formula-for-entries-arbitrary-b}) holds for $S_{b,1}(x)$.  Next, assume that (\ref{eq:formula-for-entries-arbitrary-b}) holds for $S_{b,N}(x)$ and let $\alpha_{N+1}(j,k)$ be an arbitrary entry of $S_{b,N+1}(x)$, where $pb^N\leq j \leq (p+1)b^N-1$ and $qb^N \leq k \leq (q+1)b^N-1$ for some non-negative integers $p,q\in \{0,1,\ldots,b-1\}$.  Set $j'=j-pb^N$ and $k'=k-qb^N$.  We consider two cases:

\vskip 6pt
\noindent Case 1. $p<q$.  Then $j< k$ and $\alpha_{N+1}(j,k)=0\cdot \alpha_N(j',k')=0$.

\vskip 6pt
\noindent Case 2. $p\geq q$.  Then $j\geq k$ and
\begin{equation}\label{eq:case-2-part-1}
\alpha_{N+1}(j,k)=\binom{x+p-q-1}{p-q}\alpha_N(j',k').
\end{equation}
Let $j-k=d_0b^0+d_1b^1+\dots + d_{N}b^{N}$, where $d_N=p-q$.  Then $j'-k'=d_0b^0+d_1b^1+\dots +d_{N-1}b^{N-1}$.  By assumption,
\begin{equation} \label{eq:case-2-part-2}
\alpha_{N}(j',k')=\left\{
\begin{array}{cl}
\binom{x+d_0-1}{d_0} \binom{x+d_1-1}{d_1} \cdots \binom{x+d_{N-1}-1}{d_{N-1}} & \mathrm{if} \ 0\leq k' \leq j' \leq b^N-1 \ \mathrm{and} \  k' \preceq_b j';  \\ 
\\
0 & \textrm{otherwise.}
\end{array}
\right.
\end{equation}
Since $k \preceq_b j$ if and only if $k' \preceq_b j'$, it follows from (\ref{eq:case-2-part-1}) and (\ref{eq:case-2-part-2}) that
\begin{equation}
\alpha_{N+1}(j,k)=\left\{
\begin{array}{cl}
\binom{x+d_0-1}{d_0} \binom{x+d_1-1}{d_1} \cdots \binom{x+d_N-1}{d_N} & \mathrm{if} \ 0\leq k\leq j \leq b^{N+1}-1 \ \mathrm{and} \  k \preceq_b j;  \\ 
\\
0 & \textrm{otherwise.}
\end{array}
\right.
\end{equation}
Thus,  (\ref{eq:formula-for-entries-arbitrary-b})  holds for $S_{b,N+1}$.
\end{proof}

Next, we show that $S_{b,N}(x)$ forms a one-parameter subgroup of $SL(b^N,\mathbb{R})$.  To prove this, we shall need the following two lemmas; the first is due to Gould \cite{G} and the second follows easily from the first through an appropriate change of variables.

\begin{lemma}[Gould \cite{G}] \label{le:binomial-sum-of-product}
\begin{equation}\label{eq:binomial-sum-formula}
\sum_{k=0}^n \binom{x+k}{k}\binom{y+n-k}{n-k} = \binom{x+y+n+1}{n}.
\end{equation}
\end{lemma}

\begin{proof} Gould \cite{G} derives (\ref{eq:binomial-sum-formula}) as a special case of a generalization of Vandemonde's convolution formula.  We shall prove  (\ref{eq:binomial-sum-formula}) more directly using a combinatorial argument. Let $A$, $B$, and $C=\{0,1,\ldots,n\}$ denote three sets containing $x$, $y$, and $n+1$ elements (all distinct), respectively, where $n$ is a positive integer.  For any non-negative integer $k$, define $A_k=A\cup \{0,\dots,k-1\}$ and $B_k=B\cup \{k+1,\dots,n\}$.  Then given any $n$-element subset $S$ of $A\cup B\cup C$, there exists a unique integer $k_S$ in $C-S$, called the index of $S$ with respect to $A$ and $B$, such that $|S\cap A_{k_S}|=k_S$ and $|S\cap B_{k_S}|=n-k_S$.  To see this, define $S_A=A\cap S$, $S_B=B\cap S$, and $T=C-S$.  We begin by deleting $|S_A|$ consecutive elements from $T$, in increasing order and beginning with its smallest element, to obtain a subset $T'$.  We then delete $|S_B|$ consecutive elements from $T'$, in decreasing order and beginning with its largest element, to obtain a subset $T''$, which must now contain a single element denoted by $k_S$.  It is now clear that $|S\cap A_{k_S}|=k_S$ and $|S\cap B_{k_S}|=n-k_S$.

To prove  (\ref{eq:binomial-sum-formula}), we count the $n$-element subsets $S$ of $A\cup B \cup C$ in two different ways.  On the one hand, since $|A\cup B \cup C|=x+y+n+1$, the number of such subsets is given by $\binom{x+y+n+1}{n}$.  On the other hand, we partition all such $n$-element subsets into equivalence classes according to each subset's index value.  Since $|S\cap A_{k}|=k$ and $|S\cap B_{k}|=n-k$, it follows that the number of $n$-element subsets $S$ having the same index $k$ is given by $\binom{x+k}{k}\binom{y+n-k}{n-k}$ and total number of $n$-element subsets is given by
\[
\sum_{k=0}^n \binom{x+k}{k}\binom{y+n-k}{n-k}.
\]
Lastly, we equate the two answers to obtain  (\ref{eq:binomial-sum-formula}).

\end{proof}

\begin{lemma}\label{le:binomial-sum-of-product-2} Let $p$ and $q$ be positive integers with $q \leq p$.  Then
\begin{equation}\label{eq:binomial-sum-formula-2}
\sum_{v=q}^{p} \binom{x+p-v-1}{p-v} \binom{y+v-q-1}{v-q} =\binom{x+y+p-q-1}{p-q} .
\end{equation}
\end{lemma}

\begin{proof}
Set $k=v-q$ and $n=p-q$.  Then (\ref{eq:binomial-sum-formula-2}) can be rewritten as
\[
\sum_{w=0}^{p-q} \binom{x+p-q-w-1}{p-q-w} \binom{y+w-1}{w} =\binom{x+y+p-q-1}{p-q},
\]
which follows from Lemma~\ref{le:binomial-sum-of-product}.
\end{proof}

\begin{theorem} For all $N\in \mathbb{N}$, 
 \begin{equation}\label{eq:one-parameter}
 S_{b,N}(x)S_{b,N}(y)=S_{b,N}(x+y).
 \end{equation}
\end{theorem}

\begin{proof}  We argue by induction on $N$.  Lemma~\ref{le:binomial-sum-of-product} proves that (\ref{eq:one-parameter}) holds for $S_{b,1}(x)$.  Next, assume that (\ref{eq:one-parameter}) holds for $S_{b,N}(x)$.   Let $t_{N+1}(j,k)$ denote the $(j,k)$-entry of $S_{b,N+1}(x)S_{b,N+1}(y)$.  Then
\begin{align*}
t_{N+1}(j,k) &=\sum_{m=0}^{b^{N+1}-1} \alpha_{N+1}(j,m,x)\alpha_{N+1}(m,k,y) \\
&=\sum_{v=0}^{b-1}\sum_{r=0}^{b^{N}-1} \alpha_{N+1}(j,vb^N+r,x)\alpha_{N+1}(vb^N+r,k,y).
\end{align*}
As before, assume $pb^N\leq j \leq (p+1)b^N-1$ and $qb^N \leq k \leq (q+1)b^N-1$ for some non-negative integers $p,q\in \{0,1,\ldots,b-1\}$.  Set $j'=j-pb^N$ and $k'=k-qb^N$.  Again, we consider two cases:

\vskip 6pt
\noindent Case 1: $j<k$.  Then $\alpha_{N+1}(j,k,x+y)=0$ by definition.  On the other hand, we have
\begin{align*}
t_{N+1}(j,k) & =\sum_{m=0}^{j} \alpha_{N+1}(j,m,x)\alpha_{N+1}(m,k,y)+\sum_{m=j+1}^{b^{N+1}-1} \alpha_{N+1}(j,m,x)\alpha_{N+1}(m,k,y) \\
& =\sum_{m=0}^{j} \alpha_{N+1}(j,m,x)\cdot 0+\sum_{m=j+1}^{b^{N+1}-1} 0\cdot \alpha_{N+1}(m,k,y) \\
& = 0
\end{align*}
and thus (\ref{eq:one-parameter}) holds.

\vskip 6pt
\noindent Case 2: $j\geq k$.  Since $S_{b,N+1}(x)  = S_{b,1}(x)\otimes S_{b,N}(x)$, we have
\[
\alpha_{N+1}(j,vb^N+r,x)=\left\{\begin{array}{cl}
\binom{x+p-v-1}{p-v}\alpha_{N}(j',r,x) & \textrm{if} \ j\geq vb^N+r; \\
0 & \textrm{if} \ j< vb^N+r. \\
\end{array}
\right.
\]
Similarly, we have
\[
\alpha_{N+1}(vb^N+r,k,y)=\left\{\begin{array}{cl}
\binom{y+v-q-1}{v-q}\alpha_{N}(r,k',y) & \textrm{if} \  k\leq vb^N+r; \\
0 & \textrm{if} \ k > vb^N+r. \\
\end{array}
\right.
\]
It follows that
\begin{align*}
t_{N+1}(j,k) 
&=\sum_{v=q}^{p}\sum_{r=0}^{b^{N}-1} \binom{x+p-v-1}{p-v} \binom{y+v-q-1}{v-q} \alpha_{N}(j',r,x)\alpha_{N}(r,k',y) \\
&=\left(\sum_{v=q}^{p} \binom{x+p-v-1}{p-v} \binom{y+v-q-1}{v-q}\right) \sum_{r=0}^{b^{N}-1} \alpha_{N}(j',r,x)\alpha_{N}(r,k',y) \\
& = \binom{x+y+p-q-1}{p-q}  \alpha_{N}(j',k',x+y) \\
& = \alpha_{N+1}(j,k,x+y),
\end{align*}
where we have made use of the inductive assumption and Lemma~\ref{le:binomial-sum-of-product-2}.  This proves that (\ref{eq:one-parameter}) holds.
\end{proof}

As a corollary, we obtain Theorem~\ref{th:digital-binomial-theorem-non-binary}, which we now prove.

\begin{proof}[Proof of Theorem~\ref{th:digital-binomial-theorem-non-binary}]
Let $j=n$ and $k=0$.  Then the identity
\[
\sum_{m=0}^{b^{N}-1} \alpha_{N}(j,m,x)\alpha_{N}(m,k,y) = \alpha_{N}(j,k,x+y),
\]
which follows from (\ref{eq:one-parameter}), is equivalent to (\ref{eq:digital-binomial-theorem-non-binary}).
\end{proof} 

We end this section by describing the infinitesimal generator of $S_{b,N}(x)$.  Define $X_{b,1}(x)=(\chi_{j,k})$ to be a strictly lower-triangular matrix whose entries $\chi_{j,k}$ are given by
\begin{equation}
\chi_{j,k}=\left\{
\begin{array}{ll}
x/(j-k), & \textrm{if } j\geq k+1; \\
0, & \textrm{otherwise.} \\
\end{array}
\right.
\end{equation}
For $N>1$, we define matrices
\[X_{b,N+1}(x)=X_{b,1}(x)\oplus X_{b,N}(x)=X_{b,1}(x)\otimes I_{b^N} +I_b\otimes X_{b,N}(x),
\]
where $\oplus$ denotes the Kronecker sum and $I_{b^N}$ denotes the $b^N\times b^N$ identity matrix.  Observe that $X_{b,1}(x)$ has the following matrix form:
\begin{equation}
X_{b,1}(x)=\left(
\begin{array}{ccccc}
0 & 0 & 0 & \cdots & 0 \\
x & 0 & 0 & \cdots & 0 \\
x/2 & x & 0 & \cdots & 0 \\
x/3 & x/2 & x & \cdots & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x/(b-1) & x/(b-2) & x/(b-3) & \cdots & 0
\end{array}
\right)
\end{equation}

The following lemmas will be needed.  The first states a useful identity involving the unsigned Stirling numbers of the first kind, $c(n,k)$, defined by the generating function
\[
x(x+1)...(x+n-1)=\sum_{k=0}^n c(n,k) x^k.
\]
It is well known that $c(n,k)$ counts the number of $n$-element permutations consisting of $k$ cycles.

\begin{lemma} \label{le:stirling-number-formula}
Let $l$ and $n$ be positive integers with $l\geq n$.  Then
\begin{equation} \label{eq:stirling-number-formula}
\sum_{i=1}^{l-n+1}(i-1)!\binom{l}{i}c(l-i,n-1)=n c(l,n).
\end{equation}
\end{lemma}

\begin{proof} We give a combinatorial argument.  Let $A=\{1,2,...,l\}$.  We count in two different ways the number of permutations $\pi=\sigma_1\sigma_2 \cdots \sigma_n$ of $A$ consisting of $n$ cycles where we distinguish one of the cycles $\sigma_j$ of $\pi$.  On the one hand, since there are $c(l,n)$ such permutations $\pi$ and $n$ ways to distinguish a cycle of $\pi$, it follows that the answer is given by $n c(l,n)$.  On the other hand, we can construct $\pi$ by first choosing our distinguished cycle $\sigma_1$ consisting of $i$ elements.  The number of possibilities for $\sigma_1$ is $\binom{l}{i}(i-1)!$ since there are $\binom{l}{i}$ ways to choose $i$ elements from $A$ and $(i-1)!$ ways to construct a cycle from these $i$ elements.  It remains to construct the remaining cycles $\sigma_2,\dots, \sigma_n$, which we view as a permutation $\pi'=\sigma_2\cdots \sigma_n$ on $l-i$ elements consisting of $n-1$ cycles.  Since there are $c(l-i,n-1)$ such possibilities for $\pi'$, it follows that the number of permutations $\pi$ with a distinguished cycle is given by 
\[\sum_{i=1}^{l-n+1}(i-1)!\binom{l}{i}c(l-i,n-1).
\]
Equating the two answers yields (\ref{eq:stirling-number-formula}) as desired.
\end{proof}

\begin{lemma} \label{le:infinitesimal-generator-power} Let $n$ be a positive integer with $1\leq n \leq b-1$.  Then
\begin{equation} \label{eq:infinitesimal-generator-power}
X_{b,1}^n(x)=(\chi_n(j,k)),
\end{equation}
where the entries $\chi_n(j,k)$ are given by
\begin{equation}  \label{eq:infinitesimal-generator-entry}
\chi_n(j,k)=\left\{
\begin{array}{cl}
 \frac{n!}{(j-k)!}c(j-k,n)x^n, & \textrm{if } j\geq k+n; \\
0, & \textrm{otherwise.} \\
\end{array}
\right.
\end{equation}

\end{lemma}

\begin{proof} We argue by induction on $n$.  It is clear that (\ref{eq:infinitesimal-generator-entry}) holds when $n=1$.  Suppose $n>1$.  If $j< k+n$, then $\chi_n(j,k)=0$ because $X_{b,1}^m$ is a power of strictly lower-triangular matrices.  Therefore, assume $j\geq k+n$.  Then
\begin{align*}
\chi_n(j,k) & =\sum_{i=0}^{b-1}\chi_{n-1}(j,i)\chi_1(i,k) = \sum_{i=k+1}^{j-n+1}\chi_{n-1}(j,i)\chi_1(i,k)  \\
& =x^n \sum_{i=k+1}^{j-n+1}\frac{(n-1)!}{(j-i)!}c(j-i,n-1)\frac{1}{i-k} \\
& =x^n \sum_{i=k+1}^{l+k-n+1}\frac{(n-1)!}{(l+k-i)!}c(l+k-i,n-1)\frac{1}{i-k} \\
& = \frac{(n-1)! x^n}{l!}  \sum_{m=1}^{l-m}\frac{l!}{m(l-m)!}c(l-m,n-1) \\
& = \frac{(n-1)! x^n}{l!} \sum_{m=1}^{l-n+1}(m-1)!\binom{l}{m}c(l-m,n-1),
\end{align*}
where $l=j-k$ and $m=i-k$.  It follows from Lemma~\ref{le:stirling-number-formula} that
\begin{align*}
\chi_n(j,k)=\frac{(n-1)!x^n}{l!}\cdot n c(l,n)=\frac{n!}{(j-k)!}c(j-k,n)x^n
\end{align*}
as desired.
\end{proof}

\begin{lemma}  \label{le:infinitesimal-generator-base-case} We have
\begin{equation} \label{eq:infinitesimal-generator-base-case}
\mathrm{exp}(X_{b,1}(x))=S_{b,1}(x).
\end{equation}
\end{lemma}

\begin{proof} Denote the entries of $\mathrm{exp}(X_{b,1}(x))$ by $\xi(j,k)$.  It is clear that $\xi(j,k)=0$ for $j<k$ and $\xi(j,k)=1$ for $j=k$ since $X_{b,1}$ is strictly lower triangular.
Therefore, assume $j\geq k+1$.  Since $X_{b,1}^{n}=0$ for $n\geq b$, we have
\begin{align*}
\xi(j,k) & =\sum_{n=1}^{b-1}\frac{\chi_n(j,k)}{n!} \\
&  =\frac{1}{(j-k)!}\sum_{n=1}^{b-1}c(j-k,n) x^n \\
& =\frac{ x(x+1)\cdots (x+j-k-1) }{(j-k)!} \\
& =\binom{x+j-k-1}{j-k} \\
& = \alpha_1(j,k).
\end{align*}
Thus, (\ref{eq:infinitesimal-generator-base-case}) holds.
\end{proof}

\begin{theorem} Let $N$ be a positive integer.  Then
\begin{equation}\label{eq:infinitesimal-generator}
\mathrm{exp}(X_{b,N}(x))=S_{b,N}(x).
\end{equation}
\end{theorem}

\begin{proof} We argue by induction on $N$.
Lemma~\ref{le:infinitesimal-generator-base-case} shows that (\ref{eq:infinitesimal-generator}) is true for $N=1$.  Then since $\mathrm{exp}(A\oplus B)=\mathrm{exp}(A)\otimes \mathrm{exp}(B)$ for any two matrices $A$ and $B$, it follows that
\[
\mathrm{exp}(X_{b,N}(x))= \mathrm{exp}(X_{b,1}(x) \oplus X_{b,N-1}(x))=S_{b,1}(x)\otimes S_{b,N-1}(x)=S_{b,N}(x),
\]
which proves (\ref{eq:infinitesimal-generator}).
\end{proof}

%%%%%%%%%%%%%%%%%

\section{Prouhet-Thue-Morse polynomials}
\label{three}

In this section we demonstrate how the generalized Sierpi\'nski matrices $S_{b,N}(1)$ arise in the study of Prouhet-Thue-Morse polynomials, first investigated by the author \cite{N2}.   These polynomials were used in the same paper to give a new proof of the well-known Prouhet-Tarry-Escott problem, which seeks $b\geq 2$ sets of non-negative integers $S_0$, $S_1$, \dots, $S_{b-1}$ that have equal sums of like powers up to degree $M\geq 1$, i.e.,
\[
\sum_{n\in S_0} n^m =\sum_{n\in S_1} n^m= \cdots =\sum_{n\in S_{b-1}} n^m 
\]
for all $m=0,1,\dots,M$.  In 1851, E. Prouhet \cite{P} gave a solution (but did not publish a proof; see Lehmer \cite{L} for a proof) by partitioning the first $b^{M+1}$ non-negative integers into the sets $S_0,S_1,\ldots,S_{b-1}$ according to the assignment
\[
n\in S_{u_b(n)}.
\]
Here, $u_b(n)$ is the generalized Prouhet-Thue-Morse sequence 
\href{http://oeis.org/A010060}{A010060}, defined as the residue of  the sum of digits of $n$ (base $b$):
\[
u_b(n)=\sum_{j=0}^d n_j \bmod b,
\]
where $n=n_0d^0+\dots + n_db^d$ is the base-$b$ expansion of $n$.  When $b=2$, $u(n):=u_2(n)$ generates the classical Prouhet-Thue-Morse sequence: $0,1,1,0,1,0,0,1,\ldots$. 

Let $A=(a_0,a_1,\dots,a_{b-1})$ be a \textit{zero-sum vector}, i.e., an ordered collection of $b$ arbitrary complex values that sum to zero:
\[
a_0+a_1+\dots+a_{b-1}=0.
\]
We define $F_N(x;A)$ to be the \textit{Prouhet-Thue-Morse} (PTM) polynomial of degree $b^N-1$ whose coefficients belong to $A$ and repeat according to $u_b(n)$, i.e.,
\begin{equation} \label{de:F}
F_N(x;A)=\sum_{n=0}^{b^{N}-1} a_{u_b(n)}x^n.
\end{equation}
In the case where $b=2$, $a_0=1$, and $a_1=-1$, we obtain the classic product generating function formula
\begin{equation} \label{eq:classic-product-generating-function}
 \prod_{m=0}^{N} (1-x^{2^m}) =\sum_{n=0}^{2^{N+1}-1} (-1)^{u(n)}x^n.
\end{equation}
Lehmer generalized this formula to the case where $A=(1,\omega,\omega^2,\ldots, \omega^{b-1})$ consists of all $b$-th roots of unity with $\omega=e^{i2\pi/b}$.  The following theorem, proven by the author \cite{N2}, extends this factorization to $F_N(x;A)$ for arbitrary zero-sum vectors.

\begin{theorem}[\cite{N2}] \label{th:factor-F}
Let $N$ be a positive integer and $A$ a zero-sum vector.  There exists a polynomial $P_N(x)$ such that
\begin{equation} \label{eq:ptm-polynomial}
F_N(x;A)=P_{N}(x)\prod_{m=0}^{N-1}(1-x^{b^m}).
\end{equation}
\end{theorem}
Theorem~\ref{th:factor-F} is useful in that it allows us to establish that the polynomial $F_N(x,A)$ has a zero of order $N$ at $x=1$, from which Prouhet's solution follows easily by setting $N=M+1$ and differentiating $F_N(x;A)$ $m$ times \cite{N2}.  

We now derive formulas for the coefficients of $P_N(x)$ in terms of generalized Sierpi\'nski triangles.  Towards this end, let
\[
P_N(x;\mathbf{c}_N)=\sum_{n=0}^{b^N-1}c_nx^n
\]
denote a polynomial whose coefficients are given by the column vector $\mathbf{c}_N=(c_0,...,c_{b^N-1})^T$.  Also, let 
\[
\mathbf{a}_n=(a_{u_b(0)},a_{u_b(1)},\ldots,a_{u_b(b^N-1)})^T
\]
 be a column vector consisting of elements of $A$ generated by the PTM sequence $u_b(n)$.  Next, define a sequence of $b^N \times b^N$ matrices $M_N$ recursively as follows.  Set
\[
M_1=(m_{j,k}) =\left(
\begin{array}{rrrrrr}
1 & 0 & 0 & \cdots & 0 & 0 \\
-1 & 1 & 0 & \cdots & 0 & 0 \\
& & & \ddots \\
0 & 0 & 0 & \cdots & -1 & 1
\end{array}
\right)
\]
where 
\[
m_{j,k} = \begin{cases}
1, & \mbox{if } j=k; \\
-1, & \mbox{if } j-k=1; \\
0, & \mbox{otherwise.}
\end{cases}
\]
Then for $N>1$, define
\begin{equation}
M_{N+1}= M_1\otimes M_N =
\left(
\begin{array}{rrrrrr}
M_N & 0_N & 0_N &  \cdots & 0_N &  0_N \\
-M_N & M_N & 0 &  \cdots & 0_N & 0_N \\
& & &  \ddots  \\
0_N & 0_N  & 0_N & \cdots & -M_N & M_N \\
\end{array}
\right),
\end{equation}
where $M_1 \otimes M_N$ denotes the Kronecker product. 
The following theorem establishes a matrix relationship between the vectors $\mathbf{a}_N$ and $\mathbf{c}_N$.

\begin{theorem} \label{th:matrix-equation}
Let $A=(a_0,\ldots, a_{b-1})$ be a zero-sum vector.  The polynomial equation 
\begin{equation} \label{eq:polynomial-factorization}
F_N(x;A)=P_N(x;\mathbf{c}_N)\prod_{m=0}^{N-1}(1-x^{b^m})
\end{equation}
 is equivalent to the matrix equation
\begin{equation}\label{eq:matrix-equation}
\mathbf{a}_N=M_N\mathbf{c}_N
\end{equation}
together with the condition $c_n=0$ for any $n$ that contains the digit $b-1$ in its base-$b$ expansion, where $0\leq n\leq b^N-1$.
\end{theorem}

To prove Theorem~\ref{th:matrix-equation}, we shall need the following two lemmas, which we state without proof since their results are easy to verify.

\begin{lemma} \label{le:system-equations-base-case}
Let $A=(a_0,a_1,\ldots, a_{b-1})$ and $\mathbf{c}_1=(c_0,\ldots,c_{b-1})^T$.  Then the polynomial equation
\[
\sum_{n=0}^{b-1}a_nx^n=\left(\sum_{n=0}^{b-1}c_nx^n\right)(1-x)
\]
is equivalent to the system of equations
\begin{align*}
a_0 & =c_0 \\
a_1 & =-c_0+c_1 \\
& \ \vdots \\
a_{b-1} & =-c_{b-2} +c_{b-1}
\end{align*}
together with the condition $c_{b-1}=0$.
\end{lemma}

\begin{lemma} \label{le:matrix-equation-base-case}
The system of equations in Lemma~\ref{le:system-equations-base-case}
can be expressed in matrix form as
\[
\mathbf{a}_1=M_1\mathbf{c}_1.
\]
\end{lemma}

\begin{proof}[Proof of Theorem~\ref{th:matrix-equation}] We argue by induction on $N$.  It is clear that (\ref{eq:matrix-equation}) holds for $N=1$ since the polynomial equation $F_1(x;A)=P_1(x;\mathbf{c}_1)(1-x)$ is equivalent to $\mathbf{a}_1=M_1\mathbf{c}_1$ because of Lemmas~\ref{le:system-equations-base-case} and \ref{le:matrix-equation-base-case}.  Next, assume that (\ref{eq:matrix-equation}) holds for case $N$.  We shall prove that (\ref{eq:matrix-equation}) holds for case $N+1$.  Define 
\[
P_N(x;C_N(p))=\sum_{n=0}^{b^N-1}c_{n+pb^N}x^{n+pb^N}
\]
to be a polynomial whose coefficients are given by the entries of $C_N(p)=(c_{pb^N},\dots,c_{(p+1)b^N-1})$, i.e., those elements in $\mathbf{c}_{N+1}$ from position $pb^N$ to position $(p+1)b^N$, so that
\[
P_{N+1}(x;\mathbf{c}_{N+1}) = P_{N-1}(x;C_N(0))+\dots+P_{N-1}N(x;C_N(b-1)).
\]
We then expand the right-hand side of (\ref{eq:ptm-polynomial}) for case $N+1$ as follows:
\begin{align*}
F_{N+1}(x;A) & = \left(P_N(x;C_N(0))+\dots+P_N(x;C_N(b-1))\right)\left(\prod_{m=0}^{N-1}(1-x^{b^m})\right)(1-x^{b^N}) \\
& = \left(Q_N(x;C_N(0))+\dots+Q_N(x;C_N(b-1))\right)(1-x^{b^N}) \\
&= Q_N(x;C_N(0))+\left(Q_N(x;C_N(1))-x^{b^N}Q_N(x;C_N(0))\right)+\dots + \\
& \ \ \ \ \left(Q_N(x;C_N(b-1))-x^{b^N}Q_N(x;C_N(b-2))\right) - x^{b^N}Q_N(x;C_N(b-1)),
\end{align*}
where we define
\[
Q_N(x;C_N(p))=P_N(x;C_N(p))\prod_{m=0}^{N-1}(1-x^{b^m}).\]
Next, we equate this result with the definition of $F_{N+1}(x;A)$:
\[
 F_{N+1}(x;A)=\sum_{n=0}^{b^{N+1}-1}a_{u_b(n)} x^n=F_N(x;A_N(0))+\dots+F_N(x;A_N(b-1)),
 \]
 where we define
 \[
 F_N(x;A_N(p))=\sum_{n=0}^{b^N-1}a_{u(n+pb^N)}x^{n+pb^N}.
 \]
This leads to the system of polynomial equations
\begin{align*}
F_N(x;A_N(0)) & =Q_N(x;C_N(0)) \\
F_N(x;A_N(1)) & =Q_N(x;C_N(1))-x^{b^N}Q_N(x;C_N(0)) \\
& \vdots \\
F_N(x;A_N(b-1)) & =Q_N(x;C_N(b-1))-x^{b^N}Q_N(x;C_N(b-2))
\end{align*}
together with the condition $Q_N(x;C_N(b-1))=0$, i.e. $c_n=0$ for all $(b-1)b^N \leq n \leq b^{N+1}-1$.
It follows from Lemmas~\ref{le:system-equations-base-case} and \ref{le:matrix-equation-base-case} that this system is equivalent to the system of matrix equations
\begin{align*}
\mathbf{a}_N(0) & =M_N\mathbf{c}_N(0) \\
\mathbf{a}_N(1) & = -M_N\mathbf{c}_N(0) + M_N\mathbf{c}_N(1) \\
& \vdots \\
\mathbf{a}_N(b-1) & = -M_N\mathbf{c}_N(b-2) + M_N\mathbf{c}_N(b-1),
\end{align*}
where the first matrix equation by assumption satisfies the condition $c_n=0$ for any $0\leq n\leq b^N-1$ that contains the digit $b-1$ in its base-$b$ expansion.
This in turn is equivalent to the matrix equation $\mathbf{a}_{N+1}=M_{N+1}\mathbf{c}_{N+1}$ together with the condition that $c_n=0$ for any $0\leq n \leq b^{N+1}-1$ that contains the digit $b-1$ in its base-$b$ expansion.  This establishes the theorem for case $N+1$ and completes the proof.
\end{proof}

\begin{lemma}
The matrix $M_N$ has inverse $S_N=M_N^{-1}$, where $S_N$ is given recursively by
\begin{equation}
S_1=
\left(
\begin{array}{rrrrr}
1 & 0 & 0 &  \cdots & 0 \\
1 & 1 & 0 &  \cdots & 0 \\
& & &  \ddots  \\
1 & 1  & 1 & \cdots & 1 \\
\end{array}
\right)
\end{equation}
and for $N>1$,
\begin{equation}
S_{N+1}=S_1\otimes S_N=
\left(
\begin{array}{rrrrr}
S_N & 0_N & 0_N &  \ldots &  0_N \\
S_N & S_N & 0_N &  \ldots &  0_N \\
& & &  \ddots  \\
S_N & S_N  & S_N & \ldots & S_N \\
\end{array}
\right).
\end{equation}
Thus, if $\mathbf{a}_N=M_N\mathbf{c}_N$, then
\begin{equation} \label{eq:matrix-equation-case-N}
\mathbf{c}_N=S_N\mathbf{a}_N.
\end{equation}
\end{lemma}

\begin{proof}
It is straightforward to verify directly that $S_1=M_1^{-1}$.  Since $S_{N+1}=S_1\otimes S_{N}$ and $M_{N+1}=M_1\otimes M_N$, it follows that $S_N=M_N^{-1}$.
  \end{proof}

Observe that $S_N=S_{b,N}(1)$ is the generalized Sierpi\'nski triangle defined in the previous section. 
We now present a formula for the coefficients $c_n$ in terms of the elements of $A$.

\begin{theorem} \label{th:formula-coefficients}
Let $A=(a_0,\ldots,a_{b-1})$ be a zero-sum vector.  Then the polynomial equation (\ref{eq:polynomial-factorization}) has solution $P_N(x;C_N)$ whose coefficients $c_n$, $0\leq n \leq b^N-1$, are given by
\begin{equation} \label{eq:formula-coefficients}
c_n=\sum_{k\in I_b(n)}a_{u_b(k)},
\end{equation}
where $I_b(n)=\{k\in \mathbb{N}: k\preceq_b n \}$.  Moreover, $c_n=0$ for all $n$ whose base-$b$ expansion contains the digit $b-1$.
\end{theorem}

\begin{proof} From Theorem~\ref{th:formula-for-entries-arbitrary-b}, we know that the non-zero entries in the $n$-th row of $S_N$, which are all equal to 1, are located at $(n,k)$ where $k\preceq_b n$.  Formula (\ref{eq:formula-coefficients}) now follows from (\ref{eq:matrix-equation-case-N}).  It remains to show that (\ref{eq:formula-coefficients}) yields $c_n=0$ for all $n$ whose base-$b$ expansion contains the digit $b-1$.  Let $n=n_0b^0+\dots+n_Lb^L+\dots + n_{N-1}b^{N-1}$ where $n_L=b-1$.  Denote $I_b(n;L)$ to be the subset of $I_b(n)$ consisting of integers whose base-$b$ expansion has digit 0 at position $L$, i.e.,
\[
I_b(n;L)=\{k\in \mathbb{N}:k\preceq_b n, k=k_0b^0+\dots+k_Nb^N, k_L=0\}.
\]
Then using the fact that $A$ is a zero-sum vector, i.e., $a_0+\dots + a_{b-1}=0$, we have
\begin{align*}
c_n & =\sum_{k\in I_b(n;L)}a_{u_b(k)} + \sum_{k\in I_b(n;L)}a_{u_b(k+b^L)}+\dots+\sum_{k\in I_b(n;L)}a_{u_b(k+(b-1)b^L))} \\
& =\sum_{k\in I_b(n;L)}\left(a_{u_b(k)} + a_{(u_b(k)+1)_b}+\dots+a_{(u_b(k)+(b-1))_b}\right) \\
& = 0
\end{align*}
as desired.
\end{proof}

Next, we specialize Theorem~\ref{th:formula-coefficients} to base $b=3$.  Define $w(n)$ to be the sum of the digits of $n$ in its base-$3$ representation modulo 2.

\begin{corollary} Suppose $b=3$.  Let $n\in \mathbb{N}$ be such that its base-3 representation does not contain the digit 2.  Then 
\begin{equation} \label{eq:formula-coefficient-base-3}
c_n=(-1)^{w(n)}a_{u_3(2n)}.
\end{equation}
\end{corollary}

\begin{proof} Let $n=n_03^0+\dots+n_{N-1}3^{N-1}$ be the base-3 representation of $n$ with no digit equal to 2 and leading digit $n_{N-1}=1$.  We argue by induction on $N$.  It is clear that (\ref{eq:formula-coefficient-base-3}) holds when $N=1$ since from (\ref{eq:formula-coefficients}) we have
\[
c_1=a_0+a_1=-a_2=(-1)^{w(n)}a_{u_3(2n)}.
\]
Next, assume (\ref{eq:formula-coefficient-base-3}) holds for a given $N$.  To prove that (\ref{eq:formula-coefficient-base-3}) holds for $N+1$, we consider two cases by decomposing $n=n'+3^{N}$.
\\

\noindent Case 1: $n_i=0$ for all $i=0,\ldots,N-1$.  Then $n=3^n$ and
\begin{align*}
c_{n} & =\sum_{k\in I_3(n)}a_{u_3(k)}=a_{u_3(0)}+a_{u_3(n)}=a_0+a_1 = -a_2 \\
& = (-1)^{w(n)}a_{u_3(2n)}.
\end{align*}

\noindent Case 2: $n_L=1$ for some $0\leq L \leq N-1$.  Then set $n''=n'-3^L$.  It follows that
\begin{align*}
c_{n} & =\sum_{k\in I_3(n)}a_{u_3(k)}=\sum_{k\in I_3(n,L)}a_{u_3(k)}+\sum_{k\in I_3(n,L)}a_{u_3(k+3^N)} \\
& =\sum_{k\in I_3(n')}a_{u_3(k)}+\sum_{k\in I_3(n')}a_{u_3(k+3^L)} \\
& =\sum_{k\in I_3(n')}a_{u_3(k)}+\sum_{k\in I_3(n',L)}a_{u_3(k+3^L)}+\sum_{k\in I_3(n',L)}a_{u_3(k+2\cdot 3^L)} +\sum_{k\in I_3(n',L)}a_{u_3(k)}-\sum_{k\in I_3(n',L)}a_{u_3(k)} \\
& =\sum_{k\in I_3(n')}a_{u_3(k)}+\sum_{k\in I_3(n',L)}\left( a_{u_3(k)}+a_{(u_3(k)+1)_3} +a_{(u_3(k)+2)_3}\right) -\sum_{k\in I_3(n'')}a_{u_3(k)} \\
& = (-1)^{w(n')}a_{u_3(2n')}- (-1)^{w(n'')}a_{u_3(2n'')} \\
& = (-1)^{w(n')}a_{u_3(2n')}+(-1)^{w(n''+3^L)}a_{u_3(2(n'-3^L))} \\
& = (-1)^{w(n')}\left( a_{u_3(2n')}+a_{(u_3(2n')-2)_3} \right) = (-1)^{w(n')}\left(-a_{(u_3(2n')-1)_3}\right) \\
& = (-1)^{w(n'+3^N)}a_{(u_3(2n')+2)_3))} = (-1)^{w(n)}a_{u_3(2n'+2\cdot 3^L)} \\
& = (-1)^{w(n)}a_{u_3(2n)}.
\end{align*}
Thus, (\ref{eq:formula-coefficient-base-3}) holds for $N+1$.
\end{proof}

%%%%%%%%%%%%%%%%

\section{Group generators and relations}
\label{four}

In this section, we reveal further evidence of the rich structure of Sierpi\'nski matrices by describing group generators and relations defined by the matrices $S_N$ and $M_N$.  Recall that $S_N$ and $M_N$ are matrices of dimension $b^N\times b^N$ for a given base $b$.  Define $T_N=M_N^t$ to be the transpose of $M_N$ and $U_N=S_NT_N$, $V_N=T_NS_N$.  
The following lemma gives a recursive construction of $U_N$ and$V_N$.

\begin{lemma} We have
\begin{align*}
U_{N+1} & =U_1\otimes U_N \\
V_{N+1} & = V_1\otimes V_N
\end{align*}
\end{lemma}

\begin{proof} The result follows from the mixed-product property of the Kronecker product.  We demonstrate this for $U_{N+1}$:
\begin{align*}
U_{N+1}=S_{N+1}T_{N+1}=(S_1\otimes S_N)(T_1\otimes T_N)=(S_1T_1)\otimes (S_NT_N)=U_1\otimes U_N.
\end{align*}
The calculation is the same for $V_{N+1}$.
\end{proof}

Observe that $U_N=(u_{i,j})$ and $V_N=(v_{i,j})$ are skew-triangular and that $V_N$ is the skew-transpose of $U_N$, i.e., $v_{i,j}=u_{b-1-j,b-1-i}$ for $i,j=0,1,\ldots,b-1$.  Here are some examples of $U_N$ and $V_N$ when $b=2$:
\[
U_1=\left(
\begin{array}{rr}
1 & -1 \\
1 & 0
\end{array}
\right), \ \
U_2=\left(
\begin{array}{rrrr}
1 & -1 & -1 & 1 \\
1 & 0 & -1 & 0 \\
1 & -1 & 0 & 0 \\
1 & 0 & 0 & 0
\end{array}
\right), 
\]

\[
V_1=\left(
\begin{array}{rr}
0 & -1 \\
1 & 1
\end{array}
\right), \ \
V_2=\left(
\begin{array}{rrrr}
0 & 0 & 0 & 1 \\
0 & 0 & -1 & -1 \\
0 & -1 & 0 & -1 \\
1 & 1 & 1 & 1
\end{array}
\right).
\]

The next lemma establishes that the eigenvalues of $U_N$ and $V_N$ are $(b+1)$-th roots of 1 or $-1$.

\begin{lemma} \label{le:eigenvalues}
The set of eigenvalues of $U_1$ and $V_1$ are exactly the same and consist of all roots of the polynomial equation
\[
-1+r-r^2+\dots+(-1)^{b+1} r^b=0.
\]
\end{lemma}

\begin{proof}
Let $r$ be a root of $-1+r-r^2+\dots+(-1)^{b+1} r^b=0$.  We claim that $r$ is an eigenvalue of $U_1$ with eigenvector $\mathbf{v}=(v_1,\ldots, v_b)^T$, where 
\[
v_k=\sum_{j=1}^k (-1)^{j+1}r^j
\]
for $k=1,\ldots ,b$.  Observe that $v_b=1$.
Denote $\mathbf{w}=(U_1-rI_b)\mathbf{v}=(w_1,\ldots,w_b)^T$.  It suffices to show $\mathbf{w}=0$.  It is straightforward to verify that
\[
U_1=(u_{j,k})=
\begin{cases}
1, & \mbox{if } k=1; \\
-1, & \mbox{if } k=j+1; \\
0, & \mbox{otherwise.}
\end{cases}
\]
We now calculate $w_j$ by considering three cases:

\vskip 6pt
\noindent Case 1: $k=1$.  Then
\[
w_1=(u_{1,1}-r)v_1+u_{1,2}v_2=(1-r)r-(r-r^2)=0.
\]
Case 2: $1<k<b$.  Then
\[
w_k=u_{k,1}v_1+(u_{k,k}-r) v_k+u_{k,k+1}v_{k+1}=r-r\sum_{j=1}^k (-1)^{j+1}r^j-\sum_{j=1}^{k+1} (-1)^{j+1}r^j=0.
\]
Case 3: $k=b$.  Then
\[
w_b=u_{b,1}v_1+(u_{b,b}-r)v_b = r - r =0.
\]
Thus, $\mathbf{w}=0$.  It can be shown by a similar argument that the eigenvalues of $V_1$ are exactly the same as those of $U_1$.
\end{proof}

\begin{theorem} \label{th:group-relation}
The matrices $U_N$ and $V_N$ satisfy the relation
\begin{equation}\label{eq:group-relation}
U_N^{b+1}=V_N^{b+1}=(-1)^{N(b+1)}I_{b^N},
\end{equation}
where $I_{b^N}$ is the $b^N\times b^N$ identity matrix.
\end{theorem}

\begin{proof}
We shall only prove (\ref{eq:group-relation}) for $U_N$ since the proof for $V_N$ is the same.  We argue by induction on $N$.  When $N=1$, we know from Lemma~\ref{le:eigenvalues} that all eigenvalues of $U_1$ are roots of 
$-1+r-r^2+\dots+(-1)^{b+1} r^b=0$.
It follows that every eigenvalue $r$ satisfies $r^{b+1}=(-1)^{b+1}$; moreover, since they are all distinct, the corresponding eigenvectors are all linearly independent.  Thus, $U_1^{b+1}=(-1)^{b+1}I_b$.

Next, assume that (\ref{eq:group-relation}) holds for $U_{N-1}$.  It follows from the mixed-product property of the Kronecker product that
\begin{align*}
U_N^{b+1} & = (U_1\otimes U_{N-1})^{b+1} = U_1^{b+1}\otimes U_{N-1}^{(N-1)(b+1)} \\
& =(-1)^{b+1}I_b\otimes (-1)^{(N-1)(b+1)}I_{b^{N-1}} \\
& =(-1)^{N(b+1)}I_{b^N}.
\end{align*}
Thus,  (\ref{eq:group-relation}) holds for $N$.
\end{proof}

We note that for $b=2$,  Ferrand \cite{F1} proved that the matrices $S_N$ and $T_N$ satisfy the three-strand braid relation
\begin{equation} \label{eq:braid-relation}
S_NT_NS_N=T_NS_NT_N.
\end{equation}
We give an alternate proof of (\ref{eq:braid-relation}) based on Theorem~\ref{th:group-relation}.  Define $Q_N=S_NT_NS_N$ and $R_N=T_NS_NT_N$.  Then $Q_NR_N=U_N^3=(-1)^{3N}I_{2^N}=(-1)^NI_{2^N}$ because of (\ref{eq:group-relation}).  Moreover, we claim that
 $Q_N^2=R_N^2=(-1)^NI_{2^N}$.  This follows by induction on $N$, which we demonstrate for $Q_N$ by again using the mixed-product property of the Kronecker product:
 \begin{align*}
Q_N^2 &=(S_NT_NS_N^2T_NS_N) \\
& = (S_1T_1S_1^2T_1S_1)\otimes(S_{N-1}T_{N-1}S_{N-1}^2T_{N-1}S_{N-1}) \\
& = (-I_2)\otimes((-1)^{N-1}I_{2^{N-1}}) \\
& = (-1)^N I_{2^N}
\end{align*}
Thus, $Q_NR_N=Q_N^2$, which implies the braid relation $Q_N=R_N$.   However, we find that the braid relation fails to hold for $b>2$. 




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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11C20; Secondary 05A10.

\noindent \emph{Keywords: } 
binomial theorem, Sierpi\'nski triangle, Prouhet-Thue-Morse sequence.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A010060}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 1 2015;
revised version received April 30 2015. 
Published in {\it Journal of Integer Sequences}, May 26 2015.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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