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\begin{center}
	\vskip 1cm{\LARGE\bf 
		The 2-Pascal Triangle and\\
		\vskip .1in
		a Related Riordan Array 
	}
	\vskip 1cm
	\large
	 Yassine Otmani \\
	Faculty of Mathematics\\
	USTHB, RECITS Laboratory\\
	P.O. Box 32 \\El Alia, 16111\\ Bab Ezzouar\\  Algiers\\ Algeria\\
	 \href{mailto:yotmani@usthb.dz}{\tt yotmani@usthb.dz}  and 
	\href{mailto:otmani.yassine@yahoo.fr}{\tt otmaniyassine93@gmail.com}
	
	
	
\end{center}

\vskip .2 in
\begin{abstract}
In this paper, we determine the structure of the $2$-Pascal triangle and prove that it consists of even rows within a specific proper Riordan array. We provide the generating functions for two identities involving the coefficients of the triangle. As a consequence, we derive some combinatorial identities, including the generating functions for the sums of diagonal elements along a finite ray through the $2$-Pascal triangle. Finally, we determine the inverse transform of the trinomial transform.
\end{abstract}

\section{Introduction}
Let $\mathcal{F}:=\mathbb{R}[ t] $ denote the set of formal power series over $\mathbb{R}$ in the indeterminate $t$. We let $\mathcal{F}_{0}$ denote the set of all formal power series $f(t) \in\mathcal{F}$ in which $f( 0) \neq0$, and by $\mathcal{F}_{1}=t\mathcal{F}_{0}$ the set of formal power series $f( t) \in\mathcal{F}$ with $f( 0) = 0$ and  $f^\prime( 0)  \neq 0$. A Riordan array $D:=( d_{n,k}) _{n,k \geq 0}$ is an infinite lower triangular matrix in which every entry is given by 
\begin{equation}\label{1.1}
d_{n,k}:=[ t^{n}] d( t) h(t) ^{k},
\end{equation}
where $d(t)\in\mathcal{F}$ and $h(t) \in t\mathcal{F}$ (see, for instance, \cite{rogers,sprugnoli}). The notation $[t^{k}]f (t)$ indicates the extraction coefficient of $t^k$ in $f (t)$ for every integer $k$. The operator $[t^{k}]$  has many properties that are illustrated in \cite[Section 1.2]{sprugnolibook}  and \cite{MSV07}; some of them are given below. Let $f(t):=\sum_{k\geq0}f_{k}t^{k} $ and $g(t):=\sum_{k\geq0}g_{k}t^{k} $, where $f$ and $g$ are formal power series. The following statements hold.
 \begin{center}
\begin{tabular}{ c c c }
 $( R1) $ & Linearity & $[ t^{n}](  \alpha f(t)+ \beta g(t)) =\alpha[ t^{n}] f(t)+\beta [ t^{n}] g(t)$; \\
 $( R2)$ & Shifting & $[ t^{n}]t^{m} f(t)=[ t^{n-m}] f(t)$; \\
  $( R3) $ & Convolution &   $[ t^{n}] f(t)g(t)=\sum_{k=0}^{n}f_{k}g_{n-k}$;
\end{tabular}
\end{center}
where $m$ is an integer. According to (R2), for $k>n$, we have $d_{n,k}=0$.

In addition, we use the notation $D:=\mathcal{R}( d(t) ,h(t)) $. A good example of a Riordan array is the Pascal triangle $P=\left(\binom{n}{k} \right)_{n,k\geq 0}$  (\seqnum{A008277}), in which
\begin{equation}\label{1.2}
\binom{n}{k}=[ t^{n}] \frac{1}{1-t}\left( \frac{t}{1-t}\right) ^{k},
\end{equation}
with the convention $\binom{n}{k}=0$, for $k<0$ and $n>0$. It follows from \eqref{1.1} that $d(t)h(t)^{k}$ is the generating function of $k$-column of the Riordan array $D$. It can also be characterized by the following bivariate generating function \cite{sprugnoli},
\begin{equation*}\label{1.3}
d( t,z) =\sum_{k\geq0}d(t)h(t)^{k}z^{k}=\frac{d(t) }{1-zh(t)}.
\end{equation*}
If $d(t) \in\mathcal{F}_{0}$ and $h(t)\in\mathcal{F}_{1}$ the Riordan array $D$ is called proper. In the literature, there are numerous articles dealing with the theory of Riordan arrays. Let $D_1=\mathcal{R}( d_{1}(t),h_{1}(t))$ and $D_2=\mathcal{R}( d_{2}(t),h_{2}(t))$ be two Riordan arrays. We let $D_1*D_2$ denote the row-by-column product of two Riordan arrays. Then we have
\begin{equation*}\label{1.4}
D_1*D_2=\mathcal{R}( d_{1}(t)d_{2}\left( h_{1}(t)\right),h_{2}\left( h_{1}(t)\right)  ).
\end{equation*}

Let $\mathbb{A}$ denote the set of proper Riordan arrays. The valuable algebraic property of proper Riordan arrays is the fact that the set $\mathbb{A}$ with the usual matrix row-by-column product form a group called the Riordan group \cite{shapiro}, where the identity element of the group is $\mathcal{R}( 1,t) $, and for all $\mathcal{R}( d(t) ,h(t)) \in\mathbb{A}$ the inverse element is given as $\mathcal{R}^{-1}( d\left( 1/\bar{h}(t)\right) ,\bar{h}(t))$, in which $\bar{h}(t)$ is the compositional inverse of $h(t)$.

Rogers \cite{rogers} discovered a fundamental characterization of Riordan arrays; he showed that for every proper Riordan array $( d_{n,k}) _{n,k\geq 0}$ there exists a sequence $A=( a_{n}) _{n\geq0}$ (where $a_0\neq0$) such that for every $n,k\in\mathbb{N}\cup \{ 0\} $, we have
\begin{align*}\label{14}
d_{n+1,k+1}=\sum_{j=0}^{\infty}a_{j}d_{n,k+j}.
\end{align*}
This is called the $A$-sequence of the Riordan array. Let $A( t) $ be the generating function of the sequence $A$. It follows that $$h( t) =tA( h( t)) ,$$ as shown in \cite[Theorem 1.3]{sprugnoli} and \cite{MRSV97}.

 An important property pointed out by Sprugnoli \cite[Theorem 1.1]{sprugnoli} is that the sums involving the product of rows of a Riordan array $\mathcal{R}( d(t) ,h(t))$ by a sequence $( f_{n})_{n\geq0}$ can describe the coefficient of a formal power series relative to $d(t)$, $h(t)$ and $f(t) $ (where $f(t)$ is the generating function of the sequence $( f_{n})_{n\geq0}$). This property can be stated as follows:
 \begin{theorem}[Sprugnoli \cite{sprugnoli}]\label{theorem1} Let $d(t) $ and $h(t) $ be formal power series with $h\in t\mathcal{F}$. Let $\mathcal{R}( d(t), h(t) )=( d_{n,k})_{n,k\geq0}$ be a Riordan array, and $f(t) :=\sum_{k\geq0}f_{k}t^{k}$. We have
 \begin{equation}\label{1.5}
 \sum_{k=0}^{n}d_{n,k}f_{k}=[ t^{n}] d( t) f(h( t)) .
\end{equation}
 \end{theorem}
A common example of this result is the Euler transformation
\begin{equation*}\label{1.6}
\sum_{k=0}^{n}\binom{n}{k}f_{k}=[ t^{n}]\frac{1}{1-t}f\left( \frac{t}{1-t}\right).
\end{equation*}
Sprugnoli \cite[Section 3]{sprugnoli} illustrated many  applications of the Riordan array transformation \eqref{1.5}, proving  almost all identities in Gould's book \cite{gould}. In particular, he proved the following generalizations that allow the treatment of many combinatorial formulas.
\begin{theorem}\label{corollary1}
Let $n,m$ be nonnegative integers and $x,a,b$ are integers and $y$ is a complex number. If $b-a\in\mathbb{N}$, then
 \begin{equation}\label{1.7}
 \sum_{k}\binom{n+ak}{x+bk}f_{k}=[t^{n}]\frac{t^{x}}{(1-t )^{x+1} }f\left(\frac{t^{b-a}}{(1-t )^{b} } \right).
 \end{equation}
If $b<0$, then
 \begin{equation}\label{1.8}
 \sum_{k}\binom{y+ak}{m+bk}f_{k}=[ t^{m}] (1+t ) ^{y}f\left(t^{-b}(1+t)^{a}  \right).
 \end{equation}
 \end{theorem}

Let $n$ be a nonnegative integer with $s\in\mathbb{N}$. The bi$^s$nomial coefficient $\binom{n}{k}_{s}$ is defined by
\begin{align}\label{1.12}
\binom{n}{k}_{s}&:=\begin{cases}[ t^k]  ( 1+t+\cdots+t^s) ^{n},& \text{if $k\in \{ 0,1,\ldots,sn\}$;} \\0,& \text{otherwise;} \end{cases}
\end{align}
see Belbachir et al.\ \cite{benm} and Comtet \cite[Page 77]{comtet}. The bi$^s$nomial coefficient is a natural extension of the binomial coefficient. It appears that there are several representations and denominations of the bi$^s$nomial coefficients in the literature \cite[Table 1]{belbachir4}. Moreover, many combinatorial interpretations for the bi$^s$nomial coefficients are known. Freund \cite{freund} showed that the term $\binom{n}{k}_s$ counts the number of different ways of distributing $k$ objects among $n$ cells where each cell contains at most $s$ objects. In addition, one can see \cite{andre,Iguerofa2020}.

 We list below some of the important properties related to the bi$^s$nomial coefficient.
\begin{itemize}
\item The symmetric relation:
\begin{equation*}
\binom{n}{k}_s=\binom{n}{sn-k}_s.
\end{equation*}
  \item The longitudinal recurrence relation:
  \begin{equation*}\label{recurrence}
  \binom{n}{k}_s= \binom{n-1}{k}_s+ \binom{n-1}{k-1}_s+\cdots+\binom{n-1}{k-s}_s.
  \end{equation*}
  \item The explicit form of bi$^s$nomial coefficient in the term of binomial coefficients:
  \begin{equation}\label{terminbinomial}
  \binom{n}{k}_s=\sum_{j_1+j_2+\cdots+j_s=k}\binom{n}{j_1}\binom{j_1}{j_2}\cdots\binom{j_{s-1}}{j_s}.
  \end{equation}
\end{itemize}
For more properties we refer to \cite{belbachir}. For $s=2$ in \eqref{1.12}, we obtain the bi$^2$nomial coefficients. Let $z$ be a complex number, Comtet \cite[Page 77]{comtet} (see also \cite[Section 4]{belbachir3}) extends the bi$^s$nomial coefficient as
\begin{equation}\label{3.300}
\binom{z}{k}_{s}=[ t^{k}] ( 1+t+\cdots+t^s) ^{z}.
\end{equation}
Explicitly, we have
\begin{equation}\label{4.6}
\binom{z}{k}_{s}= \sum_{\substack{j_1+j_2+\cdots+j_s=k\\ j_1\geq j_2 \geq\cdots\geq j_s}}\frac{z( z-1) \cdots( z-j_1+1) }{( j_1-j_2) !( j_2-j_3)!\cdots(j_{s-1}-j_s ) !j_s !}.
\end{equation}
Relation \eqref{3.300} suggests the convention $\binom{z}{k}_{s}=0$, for $k\notin\mathbb{N}\cup \{ 0 \} $.
 The $s$-Pascal triangle (that is, the bi$^s$nomial coefficients arranged in triangle form) is an extension of Pascal's triangle which, by the longitudinal recurrence relation \ref{recurrence}, is constructed by the sum of $s$ adjacent values in the preceding row with the constant term $1$ in the first row (for $s=2 $ see Table \ref{table:nonlin+}). For $s=3,4,5$ see
\seqnum{A008287}, \seqnum{A035343}, \seqnum{A063260}, respectively.

\begin{table}[ht]
\centering 
\begin{tabular}{c c c c c c c c c c c c c c} 
\hline
$n\backslash k$ & 0 & 1 & 2  & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12\\ [0.5ex] 
\hline 
0 & 1  \\ 
1 & 1 & 1 & 1 \\
2 & 1 & 2 & 3 & 2 & 1 \\
3 & 1 & 3 & 6 & 7 & 6 & 3 & 1 \\
4 & 1 & 4 & 10 & 16 & 19 & 16 & 10 & 4 & 1\\
5 & 1 & 5 & 15 & 30 & 45 & 51 & 45 & 30 & 15 & 5 & 1           \\
6 & 1 & 6 & 21 & 50 & 90 & 126 & 141 & 126 & 90 & 50 & 21 & 6 & 1             \\[1ex] 
\hline 
\end{tabular}
\caption{The $2$-Pascal triangle (\seqnum{A027907}), also called the trinomial triangle.}
\label{table:nonlin+} 
\end{table}
The aim of the present paper is to study the structure of the $2$-Pascal triangle. In particular, we provide the column generating functions of the $2$-Pascal triangle, and furthermore, we give formulas similar to Eqs.~\eqref{1.7} and \eqref{1.8} involving the bi$^2$nomial coefficients. Several other results are established.

The paper is organized as follows: in Section \ref{section2}, we examine the bi$^2$nomial coefficient and determine a relationship between the $2$-Pascal triangle and a proper Riordan array. In Section \ref{section3}, we demonstrate two formulas similar to 
Eqs.~\eqref{1.7} and \eqref{1.8} for the bi$^2$nomial coefficients. In Section \ref{section4}, we investigate the formulas from Section \ref{section3} to derive several combinatorial identities. Sections \ref{section55} and \ref{section5} are devoted to studying particular cases of the formulas mentioned in Section \ref{section3}. Finally, in Section \ref{section6}, we provide the inverse transform of the trinomial transform \cite{nem}.
\section{The characterization of $2$-Pascal triangle}
\label{section2} In this section, we primarily derive a new generating function for the bi$^2$nomial coefficients. Furthermore, by investigating this generating function, we apply the Lagrange inversion formula to establish a connection between the $2$-Pascal triangle and a proper Riordan array.
\begin{theorem}\label{theorem21} Let $n$ be a nonnegative integer and $k$ be an integer. Then
\begin{equation}\label{2.1}
\binom{n}{k}_{2}=[ t^{2n}] (1-t+t^2) ^{n}\frac{1}{1-t  }\left(\frac{t}{1-t} \right)^{k} .
\end{equation} 
\end{theorem}
\begin{proof}Let $n$ be a nonnegative integer and $k$ be an integer. Setting 
$$a( n,k) = [ t^{2n}] (1-t+t^2) ^{n}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k} .$$
Note that $a(0,0)=a(n,2n)=1$. Additionally, for $k>2n$ or $k<0$, we have $a( n,k)=0$. Setting $n\geq1$ with $k\leq 2n$, we obtain
\begin{align*}
&a( n-1,k)+a( n-1,k-1)+a( n-1,k-2)\\
&=[ t^{2( n-1)}] (1-t+t^2) ^{n-1}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k}+
[ t^{2( n-1) }](1-t+t^2) ^{n-1}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k-1}\\&\quad +[ t^{2( n-1) }](1-t+t^2) ^{n-1}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k-2}\\
&=[ t^{2( n-1) }](1-t+t^2) ^{n-1}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k-2}\left( \left(\frac{t}{1-t} \right)^{2}+\frac{t}{1-t}+1\right)\\
&=[ t^{2n }] (1-t+t^2) ^{n}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k}\\
&=a( n,k) .
\end{align*}
One can observe that $a( n,k) $ verified the same recurrence relation of bi$^2$nomial coefficient over $\mathbb{Z}$. Hence, the proof is done.
\end{proof}
\begin{remark}Note that
\begin{align*}
a( n,k)&=[ t^0] \left(1+\dfrac{1-t}{t^2}\right) ^{n}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{k}\\
&=\sum_{j=0}^{n}\binom{n}{j}[ t^j]\frac{t^{k-j}}{( 1-t) ^{k-j+1}}.
\end{align*}
Now from Eq.~\eqref{1.2}, it follows that
\begin{align*}
a( n,k) =\sum_{j=0}^{\min( n,k) }\binom{n}{j}\binom{j}{k-j}.
\end{align*}
In view of Eq.~\eqref{terminbinomial}, for $s=2$, we conclude
\begin{align*}
a( n,k) =\binom{n}{k}_2.
\end{align*}
\end{remark}
The Lagrange inversion formula \cite[Thm.\ 1.2.4]{goulden} is a remarkable tool for solving certain functional equations arise in combinatorial enumeration. The theorem stated as follows:
\begin{theorem}[Lagrange inversion formula]\label{lemma11}
Let $F( t) $ be a formal power series and $\phi( t) \in\mathcal{F}_{0}$. We have
\begin{equation}\label{3.4}
[ t^{n}] F(t) \phi( t) ^{n}=[ t^{n}]\left[ \frac{F( w) }{1-t\phi'( w) }\middle\vert \quad w=t\phi( w) \right] .
\end{equation}
\end{theorem}
The notation $[ f( y)  \vert \quad y=g( t)  ]$ denotes the composition $f( g( t))$.

Let $\alpha( t)\in\mathcal{F}_{0} $ and $\beta(t)\in \mathcal{F}_{1}$   defined as follows: 
\begin{equation*}
\alpha( t) :=-\frac{2\left(2\sqrt{-t\sqrt{4-3t^2}-t^2+2}+t^2\sqrt{8-6t^2}-t\sqrt{2}\right)}{\sqrt{-t\sqrt{4-3t^2}-t^2+2}\left(3t^4+2t\sqrt{4-3t^2}-t^2-4\right)},
\end{equation*}
\begin{equation*}
\beta(t):=-\frac{t(t+\sqrt{4-3t^2})}{2t^2-2}.
\end{equation*}
\begin{theorem}\label{theorem3}
Let $n$ and $k$ be nonnegative integers. Then
\begin{equation}\label{theorem320}
\binom{n}{k}_{2}=[ t^{2n}] \alpha( t) \left( \beta( t) \right) ^{k}.
\end{equation}
\end{theorem}
\begin{proof}
From Theorem \ref{theorem21}, the formula \eqref{2.1} can be rewritten as
follows:
\begin{align*}
\binom{n}{k}_{2}=[ t^{2n}]\left( \sqrt{1-t+t^2}\right) ^{2n}\frac{1}{1-t}\left( \frac{t}{1-t}\right) ^{k}.
\end{align*}
Setting $\phi( t)=\sqrt{1-t+t^2}$ and $F(t) =\frac{1}{1-t}\left( \frac{t}{1-t}\right) ^{k}$. Solving the function equation $w=t\phi( w) $, we obtain
\begin{align*}
w( t) =-\frac{t(-t+\sqrt{4-3t^2})}{2t^2-2}.
\end{align*}
Applying Eq.~\eqref{3.4}, and after simplifications, we deduce Eq.~\eqref{theorem320}.
\end{proof}
Obviously, the $2$-Pascal triangle is a collection of even rows of the proper Riordan array $\mathcal{R}( \alpha( t) ,\beta( t))$ (see Table \ref{table:nonlin++}).
\begin{table}[ht]
\centering 
\begin{tabular}{c c c c c c c c c c } 
\hline
$n\backslash k$ & 0 & 1 & 2  & 3 & 4 & 5 & 6 & 7 & 8  \\ [0.5ex] 
\hline 
$0$ & ${1}$  \\ 
$1$ & $\frac{1}{2}$ & $1$  \\
$2$ & ${1}$ & ${1}$ & ${1}$  \\
$3$ & $\frac{11}{16}$ & $\frac{15}{8}$ & $\frac{3}{2}$ & 1 \\
$4$ & ${1}$ & ${2}$ & ${3}$ & ${2}$ & ${1}$\\ 
$5$ & $\frac{203}{256}$ & $\frac{355}{128}$ & $\frac{65}{16}$ & $\frac{35}{8}$ & $\frac{5}{2}$ & $1$           \\
$6$   & ${1}$ & ${3}$ & ${6}$ & ${7}$ & ${6}$ & ${3}$ & ${1}$  \\
$7$ & $\frac{1759}{2048}$ & $\frac{3787}{1024}$ &$\frac{1953}{256}$ &$\frac{1435}{128}$ & $\frac{175}{16}$ &$\frac{63}{8}$ &$\frac{7}{2}$ &$1$
\\
$8$ & ${1}$ & ${4}$ & ${10}$ &${16}$ & ${19}$ & ${16}$ & ${10}$ & ${4}$ & ${1}$
\\[1ex] 
\hline 
\end{tabular}
\caption{The Riordan array $ \mathcal{R}( \alpha( t) ,\beta( t))  $.} 
\label{table:nonlin++} 
\end{table}
\section{Combinatorial bi\texorpdfstring{$^2$}{}nomial identities }\label{section3}
We demonstrated in the previous section how the generating function in Eq.~\eqref{2.1} plays an important role in determining the relationship between the $2$-Pascal triangle and the proper Riordan array $\mathcal{R}( \alpha( t) ,\beta( t))$. In this section, we further examine the generating function in Eq.~\eqref{2.1}, through which we establish formulas for bi$^2$nomial coefficients, as shown in Theorem \ref{corollary1}. Additionally, we derive several new combinatorial identities.

Let $n,m$ and $k$ be nonnegative integers. Define
\begin{align*}
d_{n,k}:=\binom{n+ak}{x+bk}_{2} \text{and } z_{m,k}:=\binom{y+ak}{m+bk}_{2},
\end{align*}
where $a,b,x$ are integers and $y$ is a complex number. For $b-2a\in\mathbb{N}$, the triangle $(d_{n,k}) _{n,k\geq 0}$ is an extension of $2$-Pascal triangle. In addition, for $b<0$, the triangle $(z_{m,k}) _{m,k\geq 0}$  is a Riordan array, in both cases we can apply the Riordan array transformation Eq.~\eqref{1.5}, in which we establish the following formulas.
\begin{theorem}
\label{theorem4} Let $f(t) $ be the generating function of the sequence $( f_{n}) _{n\geq0}$.
\begin{itemize}
\item If $b-2a\in\mathbb{N}$, then
\begin{align}\nonumber
&\sum_{k}\binom{n+ak}{x+bk}_{2}z^{2n-x+\left(2a-b \right) k}f_{k}\\&=[t^{2n}]( 1-zt+( zt) ^2) ^{n}\frac{t^{x}}{(1-zt )^{x+1} }f\left(\frac{t^{b-2a}( 1-zt+( zt) ^2) ^{a}}{(1-zt )^{b} } \right) \label{2.3}.
\end{align}
\item If $b<0$, then
\begin{align}
\label{2.4}
\sum_{k}\binom{y+ak}{m+bk}_{2}z^{m+bk}f_{k}&=[ t^{m}] (1+zt+( zt) ^2 ) ^{y}f\left(t^{-b}(1+zt+( zt) ^2)^{a}  \right).
 \end{align}
\end{itemize}
\end{theorem}

\begin{proof} Recall that for all formal power series $g( t) $ (see \cite[(0TR), Page 7]{sprugnolibook}), we have
\begin{align*}
[ t^n] g( zt)=z^{n} [ t^n] g(t).
\end{align*}
From the formula \eqref{2.1}, we obtain
\begin{align*}
\binom{n+ak}{x+bk}_{2}z^{2n-x+(2a-b ) k}&=z^{2n-x+(2a-b ) k}[ t^{2(n+ak)}] (1-t+t^2) ^{n+ak}\frac{1}{1-t}\left(\frac{t}{1-t} \right)^{x+bk}\\
&=[ t^{2n}] (1-zt+( zt) ^2) ^{n}\ \frac{t^x}{(1-zt) ^{x+1}}\left(\frac{t^{b-2a}}{(1-zt+( zt) ^2)^{-a}( 1-zt)^b } \right)^{k}.
\end{align*}
Assume that $b-2a\in\mathbb{N}$. Then the $n$-th row of the triangle formed by bi$^2$nomial coefficients  $\binom{n+ak}{x+bk}_{2}$ is the $2n$-th row of the Riordan array $\mathcal{R}\left(\frac{(1-zt+( zt) ^2) ^{n}}{1-zt}\left( \frac{t}{1-zt}\right) ^{x},\frac{t^{b-2a}}{(1-zt+( zt) ^2)^{-a}( 1-zt)^b} \right) $. Therefore, applying Theorem \ref{theorem1}, we deduce the formula \eqref{2.3}.
 
Now, suppose that $b<0$. Hence
\begin{align*}
\binom{y+ak}{m+bk}_{2}z^{m+bk}&=[ t^{m+bk}] (1+zt+( zt)^2) ^{y+ak}\\
&=[ t^{m}](1+zt+( zt) ^2) ^{y}\left( t^{-b}( 1+zt+( zt) ^2) ^{a}\right) ^{k}.
\end{align*}
Thus, the triangle formed by the bi$^2$nomial coefficients $\binom{y+ak}{m+bk}_{2}$ is exactly the Riordan array $\mathcal{R}\left((1+zt+( zt) ^2) ^{y},t^{-b}( 1+zt+( zt) ^2) ^{a} \right) $. Again, using Theorem \ref{theorem1}, we deduce the formula \eqref{2.4}. 
\end{proof}
\begin{example}Taking $b=3$, $x=0$ and $f_{k}=1$ in \eqref{2.3}, we get
\begin{align*}
\sum_{k=0}^{\lfloor2n/3\rfloor}\binom{n}{3k}_{2}&=[t^{2n} ](1-t+t^2) ^{n}\left[\frac{1}{1-y} \middle\vert \quad y= \frac{t^{3}}{(1-t )^{3}}  \right]\\
&=[t^{2n} ](1-t+t^2) ^{n}\frac{( 1-t) ^{2}}{( 1-2t) ( 1-t+t^2) }\\
&=[t^{2n-2} ](1-t+t^2) ^{n-1}\frac{1}{1-2t}\\
&=\sum_{k=0}^{2(n-1)}\binom{n-1}{k}_{2}( -1) ^{k}2^{2(n-1)-k}\\
&=3^{n-1}.
\end{align*}
It follows that
\begin{equation*}
\sum_{k=0}^{\lfloor2n/3\rfloor}\binom{n}{3k}_{2}=3^{n-1}.
\end{equation*}
Similarly, we deduce that
\begin{align*}
\sum_{k=0}^{\lfloor ( 2n-1) /3\rfloor}\binom{n}{3k+1}_{2}&=\sum_{k=0}^{\lfloor ( 2n-2) /3\rfloor}\binom{n}{3k+2}_{2}=3^{n-1}.
\end{align*}
\end{example}
More generally, we establish the following result.
\begin{theorem}Let $x\in\mathbb{Z}^{+}$, $b\in\mathbb{N}$ and $z$ be a complex number. If $0\leq x<b$, then
\begin{equation}\label{2.6-1}
\sum_{k=0}^{\lfloor(2n-x)/b\rfloor}\binom{n}{x+bk}_{2}z^{x+bk}=z^n\frac{1}{b}\sum_{j=1}^{b}w_{b}^{j(n-x)}\left( 1+zw_{b}^{j}+( zw_{b}^j) ^{-1}\right) ^{n},
\end{equation}
where $w_{b}=\exp{( 2i\pi/b) }$ with $i^2=-1$.
\end{theorem}
\begin{proof} Applying \eqref{2.3}, for $a=0$ and $0\leq x<b$, substituting $z$ with $1$, and taking $f_{k}=z^{x+bk}$, we obtain
\begin{align*}
\sum_{k\geq0}\binom{n}{x+bk}_{2}z^{x+bk}&=[t^{2n} ]( 1-t+t^2) ^{n}\frac{t^{x}}{(1-t )^{x+1}} \left[\frac{z^x}{1-z^by} \middle\vert \quad y= \frac{t^{b}}{(1-t)^{b}}  \right]\\
&=[t^{2n} ]( 1-t+t^2) ^{n}\frac{( zt) ^{x}(1-t )^{b-x-1}}{(1-t )^{b}-( zt) ^b}.
\end{align*}
Recall that the formal power series ${\textstyle g(t) =\frac{( zt) ^{x}(1-t)^{b-x-1}}{(1-t )^{b}-( zt) ^b}}$ is the generating function of lacunary sums of binomial coefficients (see \cite[(1.53)]{gould} and \cite[Section 4]{belbachir2}). This implies that
\begin{align*}
\sum_{k\geq0}\binom{n}{x+bk}_{2}z^{x+bk}&=\frac{1}{b}\sum_{k=0}^{2n}\binom{n}{k}_{2}( -1)^{k}\sum_{j=1}^{b}( w_{b}^j) ^{-x}( 1+zw_{b}^{j}) ^{2n-k}.
\end{align*} 
 Thus,
\begin{align*}
\sum_{k\geq0}\binom{n}{x+bk}_{2}z^{x+bk}&=\frac{1}{b}\sum_{j=1}^{b}w_{b}^{-jx}\left( 1-( 1+zw_{b}^{j})+( 1+zw_{b}^{j})^2\right) ^{n}\\
&=\frac{1}{b}\sum_{j=1}^{b}w_{b}^{-jx}\left(  1+zw_{b}^{j}+( zw_{b}^j) ^{2}\right) ^{n}.
\end{align*}
This concludes the proof of the theorem.
\end{proof}
Substituting $z=1$ into \eqref{2.6-1} and using the fact $2\cos\theta=( e^{i\theta}+e^{-i\theta})$, we conclude that
\begin{align}\label{2.7-1}
\sum_{k\geq0}\binom{n}{x+bk}_{2}=\frac{1}{b}\sum_{j=1}^{b} \left( 1+2\cos\left( \frac{2j\pi}{b}\right) \right) ^{n}\cos\left( \frac{2j\left( n-x\right)\pi }{b}\right).
\end{align}
The lacunary sums of bi$^2$nomial coefficients \eqref{2.7-1} have been studied in various papers. Hoggatt et al.\ \cite{hoggatt} examined the case $b=5$. In addition, Hoggatt et al.\ \cite{hoggatt} derived the lacunary sums for all bi$^{s}$nomial coefficients. Andrews \cite[Eq. (2.18)]{andrews} determined the case $b=10$ for $x=0,1,\ldots,5$. Ni et al.\ \cite[Thm.\ 1.1]{ni} obtained a new explicit form of \eqref{2.7-1} where they treated the case $b=12$. To our knowledge, no one has dealt with the alternative case of lacunary sums of bi$^2$nomial coefficients. Let us examine some cases.
\begin{corollary}Let $n$ be a nonnegative integer. Then
\begin{equation}\label{3k+x}
3\sum_{k\geq0}\binom{n}{3k+x}_{2}( -1) ^{k+x}=\begin{cases}( -1) ^{n}2^{n+1}+1, &\text{if $n\equiv x \pmod3$;} \\ ( -1) ^{n+1}2^{n}+1 ,&\text{if $n\equiv x\pm1\pmod3$,}    \end{cases}
\end{equation}
and
\begin{equation}\label{3.17+}
5\sum_{k\geq0}\binom{n}{5k+x}_{2}( -1) ^{k+x}=\begin{cases}2(-1)^nL_{2n}+1, &\text{if $n\equiv x \pmod5$;} \\ (-1)^{n+1}L_{2n+1}+1,&\text{if $n\equiv x\pm1\pmod5$;}\\(-1)^{n}L_{2n-1} +1,&\text{if $n\equiv x\pm2\pmod5$,}    \end{cases}
\end{equation}
where $( L_{n}) _{n\geq0}$ is the Lucas sequence \emph{(\seqnum{A000032})}.
\end{corollary}
\begin{proof} We prove only \eqref{3.17+}, since the same steps can be followed to prove the identity in \eqref{3k+x}. Taking $b=5$ and $z=-1$ in \eqref{2.6-1}, we obtain

\begin{align*}
&\sum_{k}\binom{n}{5k+x}_{2}( -1) ^{k+x}=\frac{1}{5}\sum_{j=1}^{5} \left(2\cos\left( \frac{2j\pi}{5}\right)-1 \right) ^{n}\cos\left( \frac{2j(n-x)\pi }{5}\right).
\end{align*}
Note that $\cos(2\pi/5)=\cos(8\pi/5)=( \sqrt{5}-1) /4 $ and $\cos(4\pi/5)=\cos(6\pi/5)=-( \sqrt{5}+1) /4 $. Now, if $n-x\equiv 0\pmod5$, we get
\begin{align*}
\sum_{k}\binom{n}{5k+x}_{2}( -1) ^{k+x}&=\frac{2}{5}\left(\frac{\sqrt{5}-1}{2}-1 \right) ^{n}+\left(-\frac{\sqrt{5}+1}{2}-1 \right) ^{n}+\dfrac{1}{5}\\
 &=\frac{1}{5}( 2( -1) ^nL_{2n}+1).
\end{align*}
Following the same steps we can easily prove the other cases modulo 5.
\end{proof}
Let us now examine the case when $b$ is an even number and $z=-1$ in \eqref{2.6-1}.
\begin{corollary}Let $b$ be an even number. Then
\begin{equation}\label{2.10}
\sum_{k\geq0}\binom{n}{x+bk}_{2}( -1) ^{k}=\frac{1}{b}\sum_{j=1}^{b}\cos\left(\frac{\pi(n-x)( 1+2j) }{b} \right) \left( 1+2\cos\left(\frac{\pi(2j+1) }{b} \right) \right) ^{n} .
\end{equation}
\end{corollary}
\begin{proof} Setting $z=\exp{(i\pi/b) }$, where $b$ is an even nonnegative integer. It follows that
\begin{align*}
\sum_{k\geq0}\binom{n}{x+bk}_{2}( -1) ^{k}&=z^{-x}\sum_{k\geq0}\binom{n}{x+bk}_{2}z^{bk+x}.
\end{align*}
In view of \eqref{2.6-1}, we obtain
\begin{align*}
\sum_{k\geq0}\binom{n}{x+bk}_{2}( -1) ^{k}&=\frac{z^{n-x}}{b}\sum_{j=1}^{b}w_{b}^{j(n-x)}( 1+zw_{b}^{j}+( zw_{b}^j) ^{-1}) ^{n},\\
&=\frac{z^{n-x}}{b}\sum_{j=1}^{b}w_{b}^{j(n-x)}\left( 1+2\cos\left(\frac{\pi(2j+1) }{b} \right) \right) ^{n}.
\end{align*} 
By simplifying the last line of the above equation, the proof is completed.
\end{proof}
\begin{example}The following identity holds:
\begin{equation}\label{pell-lucas-equation}
\sum_{k\geq0}\binom{n}{4k+x}_{2}( -1) ^{k}=\begin{cases}( -1) ^{( n-x) /4}Q_{n}/2, &\text{if $n\equiv x \pmod4$;} \\( -1) ^{( n-x\pm1) /4}P_{n},&\text{if $n\equiv x\mp1\pmod4$;}\\0,&\text{if $n\equiv x+2\pmod4$,}   \end{cases}
\end{equation}
where $( Q_{n}) _{n\geq0}$ is the Pell-Lucas sequence (\seqnum{A001333}), and $( P_n)_{n\geq0} $ is the Pell sequence (\seqnum{A000129}).

Moreover, we find
\begin{equation}
\sum_{k\geq0}\binom{n}{6k+x}_{2}( -1) ^{k}=\begin{cases}( -1) ^{( n-x) /6}a_{n}, &\text{if $n\equiv x \pmod6$;} \\( -1)^{( n-x\mp1) /6}b_{n},&\text{if $n\equiv x\pm1\pmod6$;}\\( -1) ^{( n-x\mp2) /6}c_{n-2},&\text{if $n\equiv x\pm2\pmod6$;}\\0,&\text{if $n\equiv x+3\pmod6$,}   \end{cases}
\end{equation}
where the sequences $( a_{n}) _{n\geq0}$, $( b_{n}) _{n\geq0}$, and $( c_{n})_{n\geq0} $ correspond exactly to \seqnum{A052948}, \seqnum{A002605}, and \seqnum{A077846}, respectively.
\end{example}
One can see that some amazing cases can be obtained from \eqref{2.10}. Let us now examine the case when $b=2$, $x=a=0$ as in \eqref{2.3}, as follows:
\begin{corollary}
Let $z$ be an indeterminate and $n$ be a nonnegative integer. Then
\begin{equation}\label{2.13}
\sum_{k\geq0}\binom{n}{2k}_{2}z^{k}=\frac{\left( 1+\sqrt{z}+z\right) ^{n}+\left(1-\sqrt{z}+z\right) ^{n}}{2}.
\end{equation}
\end{corollary}
\begin{proof}
With the help of \eqref{2.3}, we get
\begin{align*}
\sum_{k\geq0}\binom{n}{2k}_{2}z^{k}&=[ t^{2n}] ( 1-t+t^2) ^{n}\frac{1-t}{1-2t+( 1+z)t}\\
&=\frac{1}{2}[ t^{2n}] ( 1-t+t^2) ^{n}\left(\frac{1}{1-( 1-\sqrt{z}) t}+\frac{1}{1-( 1+\sqrt{z}) t} \right)\\
&=\frac{\left( 1-( 1+\sqrt{z})+( 1+\sqrt{z})^2\right) ^{n}+\left( 1-( 1-\sqrt{z})+( 1-\sqrt{z})^2\right) ^{n}}{2}.
\end{align*}
Simplifying the last line in the above equation completes the proof.
\end{proof}
We let $p_{n}( z)$ denote the left-hand side of identity \eqref{2.13}. The polynomial $p_{n}(z)$ verifies the following recurrence relation
\begin{equation}\label{2.14}
p_{n}(z)=2( z+1)p_{n-1}(z)-(z^2+z+1 )p_{n-2}(z),
\end{equation}
where $p_{0}(z)=1$ and $p_{1}(z)=z+1$. For $z=2,3,4,5$ we get \seqnum{A083878}, \seqnum{A083882}, \seqnum{A081336}, \seqnum{A152107}, respectively. The proof of \eqref{2.14} can be obtained easily by applying the right-hand side of \eqref{2.13}.
\begin{remark}
Note that for all complex numbers $\lambda$ and $w$, we have
\begin{align*}
[ t^{2n}] ( 1-wt+( wt) ^2) ^{n}\frac{1}{1-\lambda t}=( w^2-\lambda w+\lambda^2) ^{n}.
\end{align*}
\end{remark}
By applying the same steps, we derive the following result.
\begin{corollary}Let $z$ be an indeterminate and $n$ be a nonnegative integer. Then
\begin{equation*}\label{2.15}
q_{n}(z):=\sum_{k\geq0}\binom{n}{2k+1}_{2}z^{k}=\frac{\left( 1+\sqrt{z}+z\right) ^{n}-\left(1-\sqrt{z}+z\right) ^{n}}{2\sqrt{z}}.
\end{equation*}
\end{corollary}
The polynomial $q_{n}( x) $ satisfies the same recurrence \eqref{2.14} where the initial values are $q_{0}(z)=0$ and $q_{1}(z)=1$. The value $z=2$ gives \seqnum{A081179}.
\section{Some combinatorial identities}\label{section4}
In this section, we focus on a distinct category of identities. We examine a number of cases where $f_k$ relates to the binomial coefficients. Our first identify is as follows:
\begin{theorem}
Let $n$ and $m$ be nonnegative integers. Then
\begin{equation*}
\sum_{k=m}^{2n}\binom{n}{k}_{2}\binom{k}{m}( -1) ^{k}=\binom{n}{m}_{2}.
\end{equation*}
\end{theorem}
\begin{proof}Applying \eqref{2.3}, for $a=x=0$, $b=z=1$, and taking $f_{k}=\binom{k}{m}( -1) ^{k}$, we find
\begin{align*}
\sum_{k=m}^{2n}\binom{n}{k}_{2}\binom{k}{m}( -1) ^{k}&=[ t^{2n}] ( 1-t+t^2)^n\frac{1}{1-t}\left[\frac{(-y) ^{m}}{( 1+y) ^{m+1}} \middle\vert \quad y=\frac{t}{1-t}\right]\\
&=( -1) ^{m}[ t^{2n-m}] ( 1-t+t^2)^n.
\end{align*}
This concludes the proof of the theorem.
\end{proof} 
\begin{theorem}
Let $n$ and $m$ be nonnegative integers. Then
\begin{equation*}
\sum_{k=0}^{m}\binom{n}{k}_{2}\binom{m}{k}=\sum_{k=0}^{2n}\binom{n}{k}_{2}\binom{m+k}{k}( -1)^k .
\end{equation*}
\end{theorem}
\begin{proof}
 Applying \eqref{2.3}, for $a=x=0$, $b=z=1$, and taking $f_{k}=\binom{m}{k}$, we get
\begin{align*}
\sum_{k=0}^{m}\binom{n}{k}_{2}\binom{m}{k}&=[ t^{2n}] ( 1-t+t^2)^n\frac{1}{1-t}\left[(1+y) ^{m}  \middle\vert \quad y=\frac{t}{1-t}\right]\\
&=[ t^{2n}]( 1-t+t^2)^n\frac{1}{(1-t) ^{m+1}}.
\end{align*}
Finally, by convolution product the proof is done.
\end{proof}

\begin{theorem}
Let $n$ and $m$ be nonnegative integers. If $n\leq m$, then
\begin{equation*}
\sum_{k=0}^{n}\binom{n-k}{m}_{2}\binom{n}{k}( -1) ^{k}=\binom{n}{m-n}.
\end{equation*}
\end{theorem}

\begin{proof}In view of \eqref{2.3}, for $a=-z=1$, $b=0$, and substituting $x$ with $m$, let $f_{k}=\binom{n}{k}$, we obtain
\begin{align*}
\sum_{k=0}^{n}\binom{n-k}{m}_{2}\binom{n}{k}( -1) ^{k}&=[ t^{2n}] ( 1-t+t^2)^n\frac{t^{m}}{( 1-t)^{m+1} }\left[( 1-y) ^{n} \middle\vert \quad y=\frac{t^2}{1-t+t^2}\right]\\
&=[ t^{2n}] \frac{t^{m}}{( 1-t)^{m-n+1} }.
\end{align*}
Finally, applying $(R2)$ (see Page 2) and using the formula \eqref{1.2}, we complete the proof.
\end{proof}
\begin{theorem}
Let $n$ be a nonnegative integer. Then
\begin{equation*}
\sum_{k=0}^{n}\binom{2n-2k}{n}_{2}\binom{n}{k}( -1) ^{k}=2^n.
\end{equation*}
\end{theorem}
\begin{proof}In view of \eqref{2.3}, for $a=-2$, $z=1$, $b=0$, and substituting $x$ with $n$, let $f_{k}=\binom{n}{k}(-1)^k$, we find
\begin{align*}
\sum_{k=0}^{n}\binom{2n-2k}{n}_{2}\binom{n}{k}( -1) ^{k}&=[ t^{4n}] ( 1-t+t^2)^{2n}\frac{t^{n}}{( 1-t)^{n+1} }\left[( 1-y) ^{n} \middle\vert \quad y=\frac{t^4}{( 1-t+t^2) ^{2}}\right]\\
&=[ t^{3n}] \frac{\left( ( 1-t)^n ( 1-t+2t^2)\right) ^{n}}{( 1-t)^{n+1} }\\
&=[ t^{3n}] \frac{( 1-t+2t^2) ^{n}}{1-t} .
\end{align*}
By convolution product, we obtain
\begin{align*}
\sum_{k=0}^{n}\binom{2n-2k}{n}_{2}\binom{n}{k}( -1) ^{k}&=\sum_{k=0}^{3n}\sum_{j=0}^{n}\binom{n}{j}\binom{j}{k-j}2^{k-j}( -1) ^{k}\\
&=\sum_{j=0}^{n}\binom{n}{j}\sum_{k=j}^{3n}\binom{j}{k-j}2^{k-j}( -1) ^{k}.
\end{align*}
Observe that for $j=0,1,\ldots,n$, we have $3n>2j$. Hence
\begin{align*}
\sum_{k=j}^{3n}\binom{j}{k-j}2^{k-j}( -1) ^{k}&=\sum_{k=j}^{2j}\binom{j}{k-j}2^{k-j}( -1) ^{k}\\
&=\sum_{k=0}^{j}\binom{j}{k}2^{k}( -1) ^{k+j}.
\end{align*}
Now, in view of binomial theorem, we conclude
\begin{align*}
\sum_{k=0}^{n}\binom{2n-2k}{n}_{2}\binom{n}{k}( -1) ^{k}=\sum_{j=0}^{n}\binom{n}{j}=2^n.
\end{align*}
The proof is done.
\end{proof}
In what follows, we present a theorem similar to Vandermonde's convolution identity \cite[(3.20)]{gould}.
\begin{theorem}
Let $n$ and $m$ be nonnegative integers and $l$ be an integer. Then
\begin{equation}\label{(37)}
\sum_{k=0}^{2m}\binom{n}{l+k}_{2}\binom{m}{k}_{2}=\binom{n+m}{2m+l}_{2}.
\end{equation}
\end{theorem}
\begin{proof}Applying \eqref{2.3}, for $a=0$, $b=z=1$, substituting $x$ with $l$, and taking $f_{k}=\binom{m}{k}_{2}$, we find
\begin{align*}
\sum_{k=0}^{2m}\binom{n}{l+k}_{2}\binom{m}{k}_{2}&=[ t^{2n}]( 1-t+t^2)^n\frac{t^l}{( 1-t) ^{l+1}}\left[( 1+y+y^2) ^{m} \middle\vert \quad y=\frac{t}{1-t}\right]\\
&=[ t^{2n}] ( 1-t+t^2)^{n+m}\frac{t^l}{(1-t) ^{2m+l+1}}.
\end{align*}
Finally, multiplying and dividing by $t^{2m}$, and then using \eqref{2.1}, we conclude \eqref{(37)}.
\end{proof}
 Let us now examine some cases of the formula \eqref{2.4}. We obtain the results below.
\begin{theorem} Let $x$ and $y$ be two complex numbers and $n$ be a nonnegative integer. Then the following identities hold:
\begin{equation}\label{2.31}
\sum_{k=0}^{n}\binom{2x}{n-k}_{2}\binom{y-2x}{k}( -1) ^{n-k}=\sum_{k=0}^{\lfloor n/3\rfloor}\binom{x}{k}\binom{y-3x}{n-3k}=\sum_{k=0}^{n}\binom{x}{k}\binom{y-2k}{n-k}( -3) ^{k},
\end{equation}
\begin{equation}\label{2.32}
\sum_{k=0}^{\lfloor n/3\rfloor}\binom{-x}{n-3k}_{2}\binom{x}{k}( -1) ^{k}=( -1) ^{n}\binom{x}{n}.
\end{equation}
\end{theorem}
\begin{proof}In view of \eqref{2.4}, for $a=0$, $b=z=-1$, substituting $y$ with $2x$ and $m$ with $n$, and taking $f_{k}=\binom{y-2x}{k}$, we obtain
\begin{align*}
\sum_{k=0}^{n}\binom{2x}{n-k}_{2}\binom{y-2x}{k}( -1) ^{n-k}&=[ t^{n}]( 1-t+t^2) ^{2x}\left[( 1+w) ^{y-2x} \middle\vert \quad w=t\right]\\
&=[ t^{n}]( 1-t+t^2) ^{2x}( 1+t) ^{y-2x}.
\end{align*}
For the right-hand side of \eqref{2.31}, we use the formula \eqref{1.8}, which gives the same generating function as the one in the above identity.

 Similarly, using \eqref{2.4}, for $a=0$, $b=-3$, $z=1$, substituting $y$ with $-x$ and $m$ with $n$, and taking $f_{k}=\binom{x}{k}(-1)^k$, we find
\begin{align*}
\sum_{k=0}^{\lfloor n/3\rfloor}\binom{-x}{n-3k}_{2}\binom{x}{k}( -1) ^{k}&=[ t^n] ( 1+t+t^2) ^{-x}\left[( 1-y) ^{x} \middle\vert \quad y=t^3\right]\\
&=[ t^n]( 1-t) ^{x}.
\end{align*}
Hence, the proof of \eqref{2.32} is finished.
\end{proof}
\begin{remark}
The right-hand side of \eqref{2.31} is the identity \cite[(3.51)]{gould}.
\end{remark}
The next identity represents the extended form of the identity \cite[(3.42)]{gould} to the bi$^2$nomial coefficients.
\begin{theorem}
Let $x$ be a complex number and $n$ be a nonnegative integer. Then
\begin{equation*}
\sum_{k=0}^{n}\binom{2n-x}{n-k}_{2}\binom{x}{k}_{2}( -1) ^{k}=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{2n-2x}{n-2k}_{2}\binom{x}{k}_{2}.
\end{equation*}
\end{theorem}
\begin{proof}
Applying \eqref{2.4}, for $a=0$, $b=-z=-1$, substituting $y$ with $2n-x$ and $m$ with $n$, and taking $f_{k}=\binom{x}{k}_2(-1)^k$, we obtain
\begin{align*}
\sum_{k=0}^{n}\binom{2n-x}{n-k}_{2}\binom{x}{k}_{2}( -1) ^{k}&=[ t^{n}] ( 1+t+t^2) ^{2n-x}\left[( 1-y+y^2) ^{x} \middle\vert \quad y=t\right]\\
&=[ t^{n}]( 1+t+t^2) ^{2n-x}( 1-t+t^2) ^{x}.
\end{align*}
Alternatively, we find
\begin{align*}
\sum_{k=0}^{\lfloor n/2\rfloor}\binom{2n-2x}{n-2k}_{2}\binom{x}{k}_{2}&=[ t^{n}] ( 1+t+t^2) ^{2n-2x}\left[( 1+y+y^2) ^{x} \middle\vert \quad y=t^2\right]
\\&=[ t^{n}] ( 1+t+t^2) ^{2n-2x}( 1+t^2+t^4) ^{x}\\
&=[ t^{n}] ( 1+t+t^2) ^{2n-2x}( 1+t+t^2) ^{x}( 1-t+t^2) ^{x}\\
&=[ t^{n}] ( 1+t+t^2) ^{2n-x}( 1-t+t^2) ^{x}.
\end{align*}
This concludes the proof of the theorem.
\end{proof}
\begin{theorem}
Let $n$ be a nonnegative integer. Then
\begin{equation}\label{3.40}
\sum_{k=0}^{2n}\binom{n}{k}^2_2( -1) ^{k}=\binom{n}{n}_2.
\end{equation}
\end{theorem}
\begin{proof}
Using \eqref{2.4}, for $a=0$, $b=-z=-1$, substituting $y$ with $n$ and $m$ with $2n$, and taking $f_{k}=\binom{n}{k}_2(-1)^k$, we find
\begin{align*}
\sum_{k=0}^{2n}\binom{n}{k}^2_2( -1) ^{k}&=\sum_{k=0}^{2n}\binom{n}{2n-k}_2\binom{n}{k}_2( -1) ^{k}\\
&=[ t^{2n}]( 1+t+t^2) ^{n}\left[( 1-y+y^2) ^{n} \middle\vert \quad y=t\right]\\
&=[ t^{2n}]( 1+t^2+t^4) ^{n}.
\end{align*}
Then the proof is done.
\end{proof}
\begin{remark}
The identity \eqref{3.40} may be shown as an extended form of the identity (3.81) in Gould's book \cite{gould}.
\end{remark}
Finally, we extend the Identities (3.70) and (3.72) in Gould's book \cite{gould} to bi$^2$nomial coefficients as next.
\begin{theorem}
If $n$ be a nonnegative integer, then the following identities hold:
\begin{equation}\label{2.36}
\sum_{k=0}^{n}\binom{n}{2k}_{2}^{2}=\frac{1}{2}\binom{2n}{2n}_{2}+\frac{1}{2}\binom{n}{n}_{2},
\end{equation}
\begin{equation}\label{2.37}
\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{n-2k}_{2}\binom{n}{2k}_2=\frac{1}{2}\binom{2n}{n}_{2}+\frac{1}{2}\binom{n}{n/2}_{2}.
\end{equation}
\end{theorem}

\begin{proof} One can observe that
\begin{align*}
\sum_{k=0}^{n}\binom{n}{2k}_{2}^{2}=\sum_{k=0}^{n}\binom{n}{2n-2k}_{2}\binom{n}{2k}_{2}.
\end{align*}
Using the generating function \eqref{2.13} and the formula \eqref{2.4}, for $a=0$, $b=-2$, $z=1$, substituting $y$ with $n$ and $m$ with $2n$, and taking $f_{k}=\binom{n}{2k}_2$, we obtain
\begin{align*}
\sum_{k=0}^{n}\binom{n}{2k}_{2}^{2}&=[ t^{2n}] ( 1+t+t^2) ^{n}\left[\frac{\left( 1+\sqrt{y}+y\right) ^{n}+\left(1-\sqrt{y}+y\right) ^{n}}{2} \middle\vert \quad y=t^2\right]\\
&=\frac{1}{2}[ t^{2n}] ( 1+t+t^2) ^{2n}+\frac{1}{2}[ t^{2n}] ( 1+t^2+t^4) ^{n}.
\end{align*}
Hence, the proof of \eqref{2.36} is done. The identity \eqref{2.37} follows by the same steps.
\end{proof}

\section{On the combinatorial sums involving the diagonal elements of the $2$-Pascal triangle}\label{section55}
The purpose of this section is to study some combinatorial identities involving the diagonal elements of the $2$-Pascal triangle by investigating the formula \eqref{2.3}.

Let $( y_{n}) _{n\geq0} $ be a sequence (\seqnum{A077943}) defined by 
\begin{align*}
y_{n}=2y_{n-1}-2y_{n-2}+2y_{n-3}& \text{, where }y_{0}=y_{1}=0 \text{ and } y_{2}=1.
\end{align*}
The characteristic polynomial $p$ of the sequence $\left( y_{n}\right) _{n\geq0} $ gives $p\left( t\right) =t^3-2t^2+2t-2$. Let $\alpha, \beta,\overline{\beta}$ be the zeros of $p$, where $\overline{\beta}$ is the complex conjugate of $\beta$. By applying Cardan's formulas to $p$, we obtain
\begin{align*}
\alpha&=\dfrac{1}{3} \left( 2 +( 17-3\sqrt{33})^{1/3}- ( -17+3\sqrt{33})^{1/3}\right),\\
\beta&=\dfrac{1}{6}\left(4+( 1+i\sqrt{3})( -17+3\sqrt{33})^{1/3} -( 1-i\sqrt{3})( 17+3\sqrt{33})^{1/3} \right).
\end{align*}
 Therefore
 \begin{equation}\label{2.16}
 y_{n}=\dfrac{\alpha^n}{( \alpha-\beta)( \alpha-\overline{\beta}) }+\dfrac{\beta^n}{( \beta-\alpha)( \beta-\overline{\beta}) }+\dfrac{\overline{\beta}^n}{( \overline{\beta}-\alpha)( \overline{\beta}-\beta) }.
 \end{equation}
The formula \eqref{2.16} is called the Binet formula of the sequence $( y_{n}) _{n\geq0} $.

 It is well-known that the Tribonacci sequence $( T_{n}) _{n\geq0}$ (\seqnum{A000073})  defined by
\begin{align*}
T_{n}=T_{n-1}+T_{n-2}+T_{n-3}& \text{, where }T_{0}=T_1=0 \text{ and }T_2=1.
\end{align*}
Furthermore, the Tribonacci numbers appear as the sums of the diagonals of the $2$-Pascal triangle. In the formula \eqref{2.3}, the Tribonacci numbers follow the direction $b=-a=1$ for $f_{k}=1$, yielding the following identity.
\begin{theorem}Let $n$ be a nonnegative integer. Then
\begin{equation*}
T_{n+2}=\frac{1}{2^{n+1}}y_{3( n+1) }.
\end{equation*}
\end{theorem}
\begin{proof}From \eqref{2.3}, setting $x = 0$, $b = -a = 1$, and taking $f_k = 1$, it follows that
\begin{align*}
T_{n+2}&=\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\\&=[ t^{2n}] ( 1-t+t^2) ^{n}\frac{1}{1-t}\left[\frac{1}{1-y} \middle\vert \quad y=\frac{t^3}{( 1-t) (1-t+t^2) }\right]\\
&=[ t^{2n}] ( 1-t+t^2) ^{n+1}\frac{1}{( 1-t) (1-t+t^2)-t^3}\\
&=[ t^{2(n+1)}] ( 1-t+t^2) ^{n+1}\frac{t^2}{1-2t+2t^2-2t^3}.
\end{align*} 
One can observe that the formal power series $\frac{t^2}{1-2t+2t^2-2t^3}$ is the generating function of the sequence $( y_{n}) _{n\geq0}$. Thus, by $(R3)$ we get
\begin{align*}
T_{n+2}=\sum_{k=0}^{2n}\binom{n}{k}_{2}( -1) ^{2n-k}y_{k}.
\end{align*}
Now, using \eqref{2.16}, we find
\begin{align*}
T_{n+2}=\dfrac{( 1-\alpha+\alpha^2) ^{n+1}}{( \alpha-\beta)( \alpha-\overline{\beta}) }+\dfrac{( 1-\beta+\beta^2) ^{n+1}}{( \beta-\alpha)( \beta-\overline{\beta}) }+\dfrac{( 1-\overline{\beta}+\overline{\beta}^2) ^{n+1}}{( \overline{\beta}-\alpha)( \overline{\beta}-\beta) }.
\end{align*}
In addition, we have $\alpha^3=2(\alpha^2-\alpha+1 )$ (where also $\beta$ and $\overline{\beta}$ verify the same equality). This concludes the proof of the theorem.
\end{proof}
More generally, consider the case when $f_{k}=z^k$, $b=-a=1$, and let $w$ be a complex number. Define the sequence $( Y_{n}) _{n\geq0}$ as
\begin{align*}
Y_{n}=2wY_{n-1}-2w^2Y_{n-2}+( z+w^3) Y_{n-3}& \text{, where }Y_{0}=Y_{1}=0  \text{ and } Y_{2}=1 .
\end{align*}
Let $\alpha_{z,w},\beta_{z,w}$ and $\overline{\beta}_{z,w}$ be the zeros of the polynomial $p_{z,w}(t ) =t^3-2wt^2+2w^2t-( z+w^3) $, in which
\begin{align*}
\alpha_{z,w}&=\frac{1}{6}u( z,w) -\frac{4w^2}{3u( z,w)}+\dfrac{2w}{3}\\
\beta_{z,w}&=-\frac{1}{12}u( z,w) +\frac{2w^2}{3u( z,w)}+\dfrac{2w}{3}+\frac{i\sqrt{3}}{2}\left(\frac{1}{6}u( z,w)+\frac{4w^2}{3u( z,w)} \right),
\end{align*}
where
\begin{align*}
u( z,w)=\left(28w^3 + 108z + 12\sqrt{81z^2+42zw^3+ 9w^6} \right) ^{1/3}
\end{align*}
and $\overline{\beta}_{z,w}$ is the complex conjugate of $\beta_{z,w}$. Applying the same steps as in the previous proof, we deduce the following result.
\begin{theorem}Let $n$ be a nonnegative integer. Then
\begin{align}\label{2.21}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}w^{2n-3k}z^k&=\dfrac{( w^2-w\alpha_{z,w}+\alpha_{z,w}^2) ^{n+1}}{( \alpha_{z,w}-\beta_{z,w})( \alpha_{z,w}-\overline{\beta}_{z,w}) }+\dfrac{( w^2-w\beta_{z,w}+\beta_{z,w}^2) ^{n+1}}{( \beta_{z,w}-\alpha_{z,w})( \beta_{z,w}-\overline{\beta}_{z,w}) }\nonumber+\\&\quad\quad \dfrac{( w^2-w\overline{\beta}_{z,w}+\overline{\beta}_{z,w}^2) ^{n+1}}{( \overline{\beta}_{z,w}-\alpha_{z,w})( \overline{\beta}_{z,w}-\beta_{z,w}) }.
\end{align}
\end{theorem}

\begin{example}
Substituting $z=3$ and $w=1$ (resp., $z=-1$ and $w=1$) into \eqref{2.21}, we obtain
\begin{equation*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}3^k=\frac{3^n}{2}+\frac{( -1) ^{n}}{4}\left(( 1+i\sqrt{2})^n- ( 1-i\sqrt{2} )^n\right)
\end{equation*}
and
\begin{equation*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}( -1) ^k=\begin{cases}1,&\text{if $n\equiv 0,1\pmod 4$;}\\0,&\text{if $n\equiv2,3\pmod 4$. } \end{cases}
\end{equation*}
\end{example}
\begin{remark}\label{remark}
Putting $w=1$, the cases $z=-1$, $z=2$, and $z=3$ are recorded in OEIS \cite{oeis} as \seqnum{A133872}, \seqnum{A102001}, and \seqnum{A103770}, respectively. Setting $z=w$, the cases $z=\sqrt{2}$, $z=\sqrt{3}$, and $z=2$ are recorded in OEIS \cite{oeis} as \seqnum{A077835}, \seqnum{A077828}, and \seqnum{A103771}, respectively.
\end{remark}
\begin{lemma}\label{lemma2} 
Let $n$ and $k$ be nonnegative integers with $n>0$. Then
\begin{align*}
\frac{k}{n}\binom{n}{k}_{2}=\binom{n-1}{k-1}_{2}+2\binom{n-1}{k-2}_{2}.
\end{align*}
\end{lemma}
\begin{proof}The proof follows directly from \cite[Thm.\ 2.2]{belbachir4}.
\end{proof}
Now, we turn our attention to another identity, which we establish as follows:
\begin{theorem}Let $n>0$ be an integer, $f( t) $ be the generating function of the sequence $( f_{n}) _{n\geq0}$. Then 
\begin{align}\label{2.27}
\sum_{k=0}^{\lfloor 2n/3\rfloor }&\binom{n-k}{k}_{2}\frac{n}{n-k}f_{k}=[ t^{2n}]\frac{( 1-t+t^2) ^{n-1}( 3-4t+2t^2) }{1-t} f\left( \frac{t^3}{( 1-t+t^2)( 1-t)}\right).
\end{align}
\end{theorem}
\begin{proof}Note that
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}f_{k}&=\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{((n-k)+k)}{n-k}f_{k}\\
&=\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}f_k+\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{k}{n-k}f_k.
\end{align*}
In view of Lemma \ref{lemma2}, we get
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}f_{k}&=\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}f_k+\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-1-k}{k-1}_{2}f_k+\\& \quad 2\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-1-k}{k-2}_{2}f_k.
\end{align*}
Applying \eqref{2.3}, we find
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}f_{k}&=[ t^{2n}] ( 1-t+t^2) ^{n}\frac{1}{1-t}f\left( \frac{t^3}{( 1-t+t^2)( 1-t)}\right) \\&+[ t^{2(n-1)}] ( 1-t+t^2) ^{n-1}t^{-1}f\left( \frac{t^3}{( 1-t+t^2)( 1-t)}\right)\\
&+2[ t^{2(n-1)}] ( 1-t+t^2) ^{n-1}\frac{t^{-2}}{( 1-t) ^{-1}}f\left( \frac{t^3}{( 1-t+t^2)( 1-t)}\right).
\end{align*}
Therefore
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}&\binom{n-k}{k}_{2}\frac{n}{n-k}f_{k}=\\&[ t^{2n}] ( 1-t+t^2) ^{n-1}f\left( \frac{t^3}{( 1-t+t^2)( 1-t)}\right)\left(\frac{1-t+t^2}{1-t}+t+2( 1-t)  \right).
\end{align*}
By simplifying the last expression, the proof is done.
\end{proof}
Let us examine some cases of the identity \eqref{2.27}. Let $( \lambda_{n})_{n\geq 0}$ be a sequence (\seqnum{A075115}) defined as follows:
\begin{align*}
 \lambda_{n}=2 \lambda_{n-1}-2 \lambda_{n-2}+2 \lambda_{n-3},\quad \lambda_{0}=3, \lambda_1 =2,\lambda_2 =0.
\end{align*}
It is clear that
\begin{align}\label{2.29}
\lambda_{n}=\alpha^{n}+\beta^{n}+\overline{\beta}^n.
\end{align}

\begin{corollary}Let $n>0$ be an integer. Then
\begin{equation*}\label{2.30+}
K_{n}:=\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}=\frac{1}{2^n}\lambda_{3n}.
\end{equation*}
\end{corollary}
\begin{proof}
From \eqref{2.27}, putting $f_{k}=1$, we obtain
\begin{align*}
K_{n}&=[ t^{2n}]\frac{( 1-t+t^2) ^{n-1}( 3-4t+2t^2) }{1-t} \left[\frac{1}{1-y} \middle\vert \quad y=\frac{t^3}{( 1-t) (1-t+t^2) }\right]\\
&=[ t^{2n}]( 1-t+t^2) ^{n}\frac{3-4t+2t^2}{1-2t+2t^2-2t^3}.
\end{align*}
Observe that the formal power series $\frac{3-4t+2t^2}{1-2t+2t^2-2t^3}$ is the generating function of the sequence $\left( \lambda_{n}\right)_{n\geq 0}$. Using the Binet formula \eqref{2.29}, we conclude
\begin{align*}
K_{n}=( 1-\alpha+\alpha^2) ^{n}+( 1-\beta+\beta^2) ^{n}+( 1-\overline{\beta}+\overline{\beta}^2) ^{n}.
\end{align*}
Finally, we have $\alpha^3=2(\alpha^2-\alpha+1 )$ (where also $\beta$ and $\overline{\beta}$ verify the same equality). Thus, the proof is done.
\end{proof}
\begin{remark}
The sequence $( K_{n}) _{n\geq0}$ is known as the Tribonacci-Lucas numbers (\seqnum{A001644}).
\end{remark}
Recall that the sequence (\seqnum{A146559}) is defined by
\begin{equation*}
a_{n}:=[ t^n] \frac{1-t}{1-2t+2t^2},
\end{equation*}
in which $a_{n}$ has the following explicit form
\begin{align*}
a_{n}=\frac{1}{2}( 1+i) ^{n}+\frac{1}{2}( 1-i) ^{n},
\end{align*}
where $i^2=-1$.
\begin{corollary}Let $n>0$ be an integer. Then
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}( -1)^k=\begin{cases}1,& \text{if $n\equiv \pm1\pmod4$; }\\-1,& \text{if $n\equiv 2\pmod4$;}\\3,& \text{if  $n\equiv 0\pmod4$.}\end{cases}
\end{align*}

\end{corollary}
\begin{proof}Applying \eqref{2.27}, for $f_{k}=-1$, we obtain
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}( -1) ^{k}&=[ t^{2n}]( 1-t+t^2) ^{n}\frac{3-4t+2t^2}{1-2t+2t^2}\\
&=1+2[ t^{2n}]( 1-t+t^2) ^{n}\frac{1-t}{1-2t+2t^2}.
\end{align*}
This implies that
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}( -1) ^{k}&=[ t^{2n}]( 1-t+t^2) ^{n}\frac{3-4t+2t^2}{1-2t+2t^2}\\
&=1+\left( 1-( 1-i)+( 1-i) ^{2}\right) ^{n} +\left( 1-( 1+i)+( 1+i) ^{2}\right) ^{n}.
\end{align*}
Therefore
\begin{align*}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}( -1) ^{k}=1+i^{n}\left(( -1) ^{n}+1\right) .
\end{align*}
Finally, we examine the cases of $n$ modulo $4$.
\end{proof}
More generally, we establish the following identity.
\begin{corollary}Let $n>0$ be an integer, $w$ and $z$ be complex numbers. Then
\begin{align}\label{2.30++}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}w^{2n-3k}z^k&=( w^2-w\alpha_{z,w}+\alpha_{z,w}^2) ^{n}+( w^2-w\beta_{z,w}+\beta_{z,w}^2) ^{n}\nonumber+\\&\quad \quad( w^2-w\overline{\beta}_{z,w}+\overline{\beta}_{z,w}^2) ^{n}.
\end{align}
\end{corollary}
\begin{example}Taking $z=3$ and $w=1$ in \eqref{2.30++}, we get
\begin{equation}
\sum_{k=0}^{\lfloor 2n/3\rfloor}\binom{n-k}{k}_{2}\frac{n}{n-k}3^k=3^n+( -1) ^{n}\left( ( 1+i\sqrt{2}) ^{n}+( 1-i\sqrt{2}) ^{n}\right). 
\end{equation}
\end{example} 

\section{On the bi$^2$nomial coefficient of the form  
${{{n+ak} \choose {n+bk}}}_2$}

\label{section5}

If $x$ depends on $n$, or $y$ depends on $m$, the formulas \eqref{2.3} and \eqref{2.4} are only applicable in limited cases. In this section, we examine the bi$^2$nomial coefficients of the form $\binom{n+ak}{n+bk}_{2}$. We evaluate the cases of $a$ and $b$ in order to find out the nature of the arrays $\left( \binom{n+ak}{n+bk}_{2}\right) _{_{n,k\geq0}}$. We also derive some combinatorial identities involving these kinds of bi$^2$nomial coefficients.

To establish the generating function of some cases of the bi$^2$nomial coefficients $\binom{n+ak}{n+bk}_{2}$, we use the Lagrange inversion formula (Theorem \ref{lemma11}). Then we obtain the result below.
\begin{theorem}Let $n$ be a nonnegative integer and $a,b$ are integers. If $b-2a\geq0$, then
\begin{equation}\label{56+}
\binom{n+ak}{n+bk}_{2}=[ t^{n}]\frac{1}{\sqrt{1-2t-3t^2}}\left(\frac{( 1-t-\sqrt{1-2t-3t^2}) ^{b-a}}{2^{b-a}t^{b}} \right) ^{k} .
\end{equation}
\end{theorem}
\begin{proof} Let $b-2a\geq0$. In view of \eqref{2.1}, we get
\begin{align*}
\binom{n+ak}{n+bk}_{2}&=[ t^{2n}] ( 1-t+t^2)^{n}\frac{t^{n}}{( 1-t) ^{n}}\frac{1}{1-t}\left( \frac{t^{b-2a}( 1-t+t^2) ^{a}}{( 1-t) ^{b}}\right) ^{k}\\
&=[ t^{n}] ( 1-t+t^2)^{n}\frac{1}{( 1-t) ^{n}}\frac{1}{1-t}\left( \frac{t^{b-2a}( 1-t+t^2) ^{a}}{( 1-t) ^{b}}\right) ^{k}.
\end{align*}
Setting $\phi( t) =( 1+t+t^2) /( 1-t) $. We obtain
\begin{align*}
w( t)=\frac{t - \sqrt{1-2t-3t^2} + 1}{2( t + 1) }.
\end{align*}
With the help of \eqref{3.4} and, after some simplifications, we arrive at \eqref{56+}.
\end{proof}
Let $ D:=\left( \binom{n+ak}{n+bk}_{2}\right) _{n,k\geq0}$.  Setting $a\geq0$ and $b\geq0$, we distinguish the following cases.
\begin{itemize}
  \item If $0\leq b\leq2a$, then $D$ is not a triangular array;
  \item If $b=2a+1$, then $D$ is a proper Riordan array;
  \item If $b>1+2a$, then $D$ is an improper Riordan array.
\end{itemize}

Now, suppose that $b<0$. According to symmetric relation, we have $\binom{n+ak}{n+bk}_{2}=\binom{n+ak}{n+( 2a-b) k}_{2}$. Take $c=2a-b$. Note that $c\geq 2a$, then this letter reduces to the cases above.

Now, consider $a<0$. If $b\geq0$, one can observe that for all values of $k$ we have $n+bk\geq0$. Hence, according to \eqref{4.6} $D$ is not a triangular array. If $b<0$, according to the convention on page 4, we have $\binom{n+ak}{n+bk}=0$ for $k>\lfloor n/b\rfloor$. Here we get
\begin{equation*}
\binom{n+ak}{n-ck}_2=[ t^{n}]\frac{1}{\sqrt{1-2t-3t^2}}\left(\frac{( 1-t-\sqrt{1-2t-3t^2}) ^{c+a}}{2^{c+a}t^{c+2a}} \right) ^{k},
\end{equation*}
where $b=-c$. Hence
\begin{itemize}
  \item If $c=1$, then $D$ is a proper Riordan array;
  \item If $c>1$, then $D$ is an improper Riordan array.
\end{itemize}

\begin{remark}Substituting $a=0$ and $b=1$ into \eqref{56+}, we get
$\mathcal{R}\left(\frac{1}{\sqrt{1-2t-3t^2}},\frac{1-t-\sqrt{1-2t-3t^2}}{2t} \right)$,
which is called the Motzkin triangle (\seqnum{A094531}). It was studied in \cite{LMMS14}.
\end{remark}
Let us now derive some identities.
\begin{corollary}Let $n$ be a nonnegative integer and $w$ be a complex number. Then
\begin{align}\label{51}
\sum_{k=0}^{n}\binom{n}{n-k}_{2}w^{n+k}&=\frac{( 1+w+w^2) ^n}{2}+\frac{1}{2}\binom{n}{n}_{2}w^n+\frac{( w^2-1)}{2}\sum_{k=0}^{n-1}\binom{k}{k}_{2}( 1+w+w^2) ^{n-1-k}w^k.
\end{align}
\end{corollary}
\begin{proof}
Substituting $a=0$ and $b=-1$ into \eqref{56+},
then applying \eqref{1.5} for $f_k=w^k$, we find
\begin{align*}
\sum_{k=0}^{n}\binom{n}{n-k}_{2}w^k&=[ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\left[\frac{1}{1-wy} \middle\vert \quad y=\frac{1-t-\sqrt{1-2t-3t^2}}{2t}\right]\\
&= [ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\frac{2t}{( 2+w) t-w+w\sqrt{1-2t-3t^2}}\\
&=[ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\frac{2t\left( w\sqrt{1-2t-3t^2}-( 2+w) t+w\right)}{(-4w^2t^2- 4t^2-4wt^2+4wt)}.
\end{align*}
Simplifying the last expression, we get
\begin{align*}
\sum_{k=0}^{n}\binom{n}{n-k}_{2}w^k&=\frac{1}{2}[ t^{n}]\frac{w}{w-( w^2+w+1) t}+\frac{1}{2}[ t^{n}]\frac{1}{\sqrt{1-2t-3t^2} }\\&+\frac{1}{2}[ t^{n-1}]\frac{w^2-1}{\sqrt{1-2t-3t^2}(w-( w^2+w+1) t)}.
\end{align*}
Therefore, the proof is done.
\end{proof}
\begin{remark}
Setting $w=1$ in \eqref{51} yields \seqnum{A027914}, and setting $w=-1$ in \eqref{51} yields \seqnum{A246437}.
\end{remark}

\begin{corollary}Let $n$ be a nonnegative integer. Then
\begin{equation}\label{3.8}
\sum_{k}\binom{n+k}{n-k}_{2}=\frac{1}{2}T_{2(n+1)}+\frac{1}{2}\binom{n}{n}_{2}+\frac{1}{2}\sum_{k=0}^{n-1}\binom{k}{k}_{2}( T_{2( n-k)} +T_{2( n-1-k) }) ,
\end{equation}
where $T_{2n}$ is the bisection of Tribonacci numbers \emph{(\seqnum{A099463})}.
\end{corollary}
\begin{proof}
Substituting $a=1$ and $b=-1$ into \eqref{56+}, then applying \eqref{1.5}, for $f_k=1$, we obtain
\begin{align*}
\sum_{k=0}^{n}\binom{n+k}{n-k}_{2}&=[ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\left[\frac{1}{1-y} \middle\vert \quad y=\frac{( 1-t-\sqrt{1-2t-3t^2}) ^{2}}{4t^{3}}\right]\\
&=[ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\frac{4t^3}{4t^3-( 1-t-\sqrt{1-2t-3t^2}) ^{2}}\\
&=[ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\frac{4t^3}{\left( 4t^3 + 2t^2+4t-2 - 2\sqrt{-3t^2 - 2t + 1}(t-1)\right) }.
\end{align*}
Simplifying the last expression, we get
\begin{align*}
\sum_{k=0}^{n}\binom{n+k}{n-k}_{2}&=\frac{1}{2}[ t^{n}]\frac{(t-1)}{t^3 + t^2 + 3t -1}+\frac{1}{2}[ t^{n}] \frac{1}{\sqrt{1-2t-3t^2}}\\&\quad +\frac{1}{2}[ t^{n-1}] \frac{t^2-1}{\sqrt{1-2t-3t^2}( t^3 + t^2 + 3t -1) }.
\end{align*}
Then the proof is done.
\end{proof}
Following the same steps of precedent proof, we get the alternating sign version of \eqref{3.8};
\begin{corollary}Let $n$ be a nonnegative integer. Then
\begin{equation*}
\sum_{k}\binom{n+k}{n-k}_{2}( -1) ^{k}=\frac{1+( -1) ^n}{4}+\binom{n}{n}_{2}-\frac{w_n}{2},
\end{equation*}
where $( w_n) _{n\geq0}$ is \emph{\seqnum{A097893}}.
\end{corollary}
\section{Trinomial transform and its inverse relation}
\label{section6} In this final section, we introduce the trinomial transform concept and we determine its inverse relation.
\begin{definition}
Let $\gamma$ be a real number. Recall that the Gegenbauer polynomial $C^{(\gamma) }_{n}(x) $ (cf.\ \cite[Chapter 22]{abr}) is defined by
\begin{align*}
C^{(\gamma) }_{n}(x)=[ t^{n}] \frac{1}{( 1-2xt+t^2) ^{\gamma}}.
\end{align*}
\end{definition}
Beside to important application of the Riordan array transformation \eqref{1.5}, to solve combinatorial identities, we can also use the algebraic property of proper Riordan arrays to find the sums inversion, i.e., for all proper Riordan array  $D=(d_{n,k}) _{n,k\geq0}$ and its inverse $D^{-1}=( \bar{d}_{n,k}) _{n,k\geq0}$, we have
\begin{equation}\label{5.1}
\sum_{k=0}^{n}d_{n,k}f_{k}=g_{n}\Longleftrightarrow f_{n}=\sum_{k=0}^{n}\bar{d}_{n,k}g_{k},
\end{equation}
where $( f_{n}) _{n\geq0}$ and $( g_{n})_{n\geq0}$ are two sequences \cite{merlini}. In particular, we have
\begin{equation}\label{5.2}
\sum_{k=0}^{n}\binom{n}{k}f_{k}=g_{n}\Longleftrightarrow f_{n}=\sum_{k=0}^{n}\binom{n}{k}( -1) ^{n-k}g_{k}.
\end{equation}
The right-hand side of the formula \eqref{5.2} is known as the binomial transform \cite{riordan}.

N{\'e}meth \cite{nem} gave a new transform called the trinomial transform of a sequence. His definition inspired from the well-known binomial transform, replacing binomial coefficients with bi$^2$nomial coefficients. 

 Examining Theorem \ref{theorem3} we get the following result.
\begin{theorem}
Let $( f_{n}) _{n\geq0}$ and $( g_{n}) _{n\geq0}$ be sequences. Then
\begin{equation}\label{6.3}
\sum_{k=0}^{2n}\binom{n}{k}_{2}f_{k}=g_{2n}\Longleftrightarrow \sum_{k=0}^{2n}\bar{a}_{2n,k}g_{k}=f_{2n},
\end{equation}
where $a_{0,0}=a_{n,n}=1$, and for $n\geq1$, we have
\begin{align*}
\bar{a}_{n,k}= C^{(k/2+1) }_{n-k}( -1/2) + \frac{1}{2}C^{(k/2+1) }_{n-k-1}( -1/2).
\end{align*}
\end{theorem}
\begin{proof}
Let $\mathcal{R}(\alpha( t) ,\beta( t) )=( a_{n,k}) _{n,k\geq0}$. One can show that
 $$\mathcal{R}^{-1}(\alpha( t) ,\beta( t) )=\mathcal{R}\left(\frac{2+t}{2+2t+2t^2},\frac{t}{\sqrt{1+t+t^2}}\right),$$ see Page 2. Define $B=( \bar{a}_{n,k}) _{n,k\geq0}$, where
\begin{align*}
\bar{a}_{n,k}=\left[ t^{n}\right] \frac{2+t}{2+2t+2t^2}\left( \frac{t}{\sqrt{1+t+t^2}}\right) ^{k}.
\end{align*}
This gives 
\begin{align*}
\bar{a}_{n,k}=C^{(k/2+1) }_{n-k}( -1/2) + \frac{1}{2}C^{(k/2+1)}_{n-k-1}( -1/2).
\end{align*}
In view of \eqref{5.1}, we have 
\begin{align*}
\sum_{k=0}^{n}a_{n,k}f_{k}=g_{n}\Longleftrightarrow f_{n}=\sum_{k=0}^{n}\bar{a}_{n,k}g_{k}.
\end{align*}
By changing $n$ to $2n$ and applying Theorem \ref{theorem3}, the proof is completed.
\end{proof}
\begin{remark} 
Obviously, the inverse transform of the trinomial transform gives only the even-index terms of the sequence $( f_{n}) _{n\geq0}$, because it is mainly related to the position of bi$^2$nomial rows over the Riordan array $\mathcal{R}(\alpha( t) ,\beta( t) )$. 
\end{remark}
Using the Lagrange inversion formula, Merlini et al.\ \cite[Thm.\ 6]{merlini} established that.
\begin{theorem}
Let us suppose that the sum $a_n=
\sum_{k=0}^n d_{n,k}b_k$ corresponds to the implicit Riordan array $D =(  f( t) \psi( t)^n , h( t) ) $, that is, to the relation $a_n =[ t^n]f( t) \psi( t)^nb (h( t)) $, with $\psi( 0)\neq0$, $h( 0)=0$ and $h'( 0)\neq0$. Then the inverse sum is
\begin{equation*}
b_{n}=\sum_{k=0}^{n}\left( [ t^{n-k}] \frac{\left( \psi( t) -t\psi'( t) \right)h'( t)  }{f( t) \psi( t)^{k+1}}\frac{t^{n+1}}{h( t)  ^{n+1}}\right) a_{k}.
\end{equation*}
\end{theorem}
\begin{remark}
The concept of implicit Riordan arrays is given in \cite[Page 2]{merlini}.
\end{remark}
According to \eqref{2.1} the bi$^2$nomial transform can be written as
\begin{equation*}
g_{2n}=[ t^{2n}] ( 1-t+t^2) ^{n}\frac{1}{1-t}f\left( \frac{t}{1-t}\right) ,
\end{equation*}
where $f( t) $ is the generating function of the sequence $( f_{n}) _{n\geq0}$. Let $g( t)=\alpha( t) f( \beta( t) )  $. Then we obtain.
\begin{corollary}Let $n$ be a nonnegative integer. Then
\begin{equation*}
f_{2n}=[ t^{2n}]\frac{( 2-t) ( 1-t) ^{2n}}{2( 1-t+t^2) }g\left(\frac{t}{\sqrt{1-t+t^2}} \right).
\end{equation*}
\end{corollary}

\section{Acknowledgment}
Our sincere thanks to the reviewers for their valuable feedback and constructive suggestions, which have significantly strengthened this paper. This paper is partially supported by DGRSDT grant $n^{\circ}$ C0656701.


\begin{thebibliography}{99}
\bibitem{abr}M. Abramowitz and I. A. Stegun, (Eds.), {\it Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables} (Vol. 55), US Government printing office, 1948.

\bibitem{andre}D. Andr{\'e}, M{\'e}moire sur les combinaisons r{\'e}guli{\`e}res et leurs applications, \textit{Ann. Sci. {\'E}c. Norm. Sup{\'e}r.} \textbf{5} (1876), 155--198.

\bibitem{andrews} G. Andrews, {Euler's `exemplum memorabile inductionis fallacis' and $q$-trinomial coefficients}, {\it J. Amer. Math. Soc.} \textbf{3} (1990), 653--669. 

\bibitem{belbachir3}H. Belbachir, Determining the mode for convolution powers for discrete uniform distribution, {\it Probab. Engrg. Inform. Sci.} \textbf{25} (2011), 469--475. 

\bibitem{benm}H. Belbachir and A. Benmezai, A $q$-analogue for ${\rm bi}^s$nomial coefficients and generalized Fibonacci sequences, \textit{C. R. Math. Acad. Sci. Paris} \textbf{352} (2014), 167--171.

\bibitem{belbachir}H. Belbachir, S. Bouroubi, and A. Khelladi, Connection between ordinary multinomials, Fibonacci numbers, Bell polynomials and discrete uniform distribution, {\it Ann. Math. Inform.} \textbf{35} (2008), 21--30.

\bibitem{belbachir4}H. Belbachir and O. Igueroufa, Congruence properties for bi$^s$nomial coefficients and like extended Ram and Kummer theorems under suitable hypothesis, {\it Mediterr. J. Math.} \textbf{17} (2020), 1--14.

\bibitem{Iguerofa2020}H. Belbachir and O. Igueroufa, Combinatorial interpretation of bi$^s$nomial coefficients and generalized Catalan numbers, {\it Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020)}, 47--54.

\bibitem{belbachir2} H. Belbachir, T. Komatsu, and L. Szalay, Linear recurrences associated to rays in Pascal's triangle and combinatorial identities, {\it Math. Slovaca} \textbf{64} (2014), 287--300.

\bibitem{cheon}G. S. Cheon and S. T. Jin, The group of multidimensional Riordan arrays, {\it Linear Algebra Appl.} \textbf{524} (2017), 263--277.

\bibitem{comtet}L. Comtet, {\it Advanced Combinatorics: The Art of Finite and Infinite Expansions}, Reidel, 1974.

\bibitem{freund} J. E. Freund, Restricted occupancy theory---a generalization of Pascal's triangle, {\it Amer. Math. Monthly} \textbf{63} (1956), 20--27.

\bibitem{ni}H. X. Ni and H. Pan, On the lacunary sum of trinomial coefficients, {\it Appl. Math. Comput.} \textbf{339} (2018), 286--293.

\bibitem{hoggatt} V. E. Hoggatt Jr and G. L. Alexanderson, Sum of partition sets in generalized Pascal triangles, {\it Fibonacci Quart.} \textbf{14} (1976), 117--125.

\bibitem{gould}H. W. Gould, {\it Combinatorial Identities}, Morgantown, W. Va., 1972.

\bibitem{goulden}I. P. Goulden and D. M. Jackson, {\it Combinatorial Enumeration}, John Wiley and Sons, 1983. 

\bibitem{LMMS14}A. Luz{\'o}n, D. Merlini, M. A. Mor{\'o}n, and R. Sprugnoli, Complementary Riordan arrays, \textit{Discrete Appl. Math.} \textbf{172} (2014), 75--87.

\bibitem{merlini}D. Merlini, R. Sprugnoli, and M. Verri, Combinatorial sums and implicit Riordan arrays, {\it Discrete Math.} \textbf{309} (2009), 475--486.

\bibitem{MRSV97} D. Merlini, D. G. Rogers, R. Sprugnoli, and M. C. Verri, On some alternative characterizations of Riordan arrays, \textit{Canad. J. Math.} \textbf{49} (1997), 301--320.

\bibitem{MSV07}D. Merlini, R. Sprugnoli, and M. C. Verri, The method of coefficients, \textit{Amer. Math. Monthly} \textbf{114} (2007), 40--57.

\bibitem{nem}L. N{\'e}meth, The trinomial transform triangle, \textit{J. Integer Seq.} \textbf{21} (2018), \href{https://cs.uwaterloo.ca/journals/JIS/VOL21/Nemeth/nemeth6.pdf}{Article 18.7.3}.

\bibitem{rogers}D. G. Rogers, Pascal triangles, Catalan numbers and renewal arrays, {\it Discrete Math.} \textbf{22} (1978), 301--310.

\bibitem{riordan}J. Riordan, {\it Combinatorial Identities}, John Wiley and Sons, 1968.

\bibitem{shapiro}L. V. Shapiro, S. Getu, W.-J. Woan, and L. Woodson, The Riordan group, {\it Discrete Appl. Math.} \textbf{34} (1991), 229--239.

\bibitem{oeis}N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, \url{https://oeis.org}, 2024.

\bibitem{sprugnolibook}R. Sprugnoli, {\it Riordan Array Proofs of Identities in Gould's Book}, University of Florence, Italy, 2006.

\bibitem{sprugnoli}R. Sprugnoli, Riordan arrays and combinatorial sums, {\it Discrete Math.} \textbf{132} (1994), 267--290.

\end{thebibliography}


\hrule
\vspace*{+.1in}

\noindent 2020 {\em Mathematics Subject Classification}: Primary 11B65; Secondary 05A10, 05A15, 05A19.\\
{\em Keywords}:
Riordan array, $2$-Pascal triangle, generating function, bi$^2$nomial coefficient, combinatorial identity.
\vspace*{+.1in}
\hrule
\vspace*{+.1in}

\noindent (Concerned with sequences 
\seqnum{A000032},
\seqnum{A000073},
\seqnum{A000129},
\seqnum{A001333},
\seqnum{A001644},
\seqnum{A002605},
\seqnum{A008277},
\seqnum{A008287},
\seqnum{A027907},
\seqnum{A027914},
\seqnum{A035343},
\seqnum{A052948},
\seqnum{A063260},
\seqnum{A075115},
\seqnum{A077828},
\seqnum{A077835},
\seqnum{A077843},
\seqnum{A077846},
\seqnum{A077943},
\seqnum{A081179},
\seqnum{A081336},
\seqnum{A083878},
\seqnum{A083882},
\seqnum{A094531},
\seqnum{A097893},
\seqnum{A099463},
\seqnum{A102001},
\seqnum{A103770},
\seqnum{A103771},
\seqnum{A133872},
\seqnum{A146559},
\seqnum{A152107}, and
\seqnum{A246437}.)

\bigskip
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\vspace*{+.1in}
\noindent
Received September  17 2023;
revised versions received  November 9 2024; November 16 2024; December 3 2024.
Published in {\it Journal of Integer Sequences}, April 19 2025.

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