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\begin{document}

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\theoremstyle{plain}
\newtheorem{theorem}{Theorem}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}

\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}
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\begin{center}
\vskip 1cm{\LARGE\bf Log-Concavity and LC-Positivity for \\
\vskip .1in
Generalized Triangles
}
\vskip 1cm
{\large
Moussa Ahmia \\
LMAM Laboratory \\
Department of Mathematics \\
UMSB \\
Jijel 18000 \\
Algeria \\
\href{mailto:ahmiamoussa@gmail.com}{\texttt{ahmiamoussa@gmail.com}} \\ 
\ \\
Hac\`{e}ne Belbachir \\
USTHB\\
Faculty of Mathematics\\
RECITS Laboratory\\
BP 32\\
El Alia 16111\\
Bab Ezzouar\\
Algiers\\
Algeria\\
\href{mailto:hacenebelbachir@gmail.com}{\texttt{hacenebelbachir@gmail.com}}}
\end{center}

\begin{abstract}
In this paper, we propose the generalized triangles called
$s$-\textit{triangles} for $s$ a given positive integer, as a {\it bi}-indexed
sequence of nonnegative numbers $\left\{a_{s}(n,k)  \right\}_{0\leq
k\leq ns}$ satisfying $a_{s}(n,k)=0$ for $k<0$. We extend some results of
Wang and Yeh, and show that if the $s$-triangle is LC-positive
(resp., doubly LC-positive) then it preserves (resp., it doubly preserves)
the log-concavity of the sequences. Applications related to bi$^{s}$nomial
coefficients are given.
\end{abstract} 

\vskip .2 in

\section{Introduction}
A sequence of nonnegative numbers $(x_{k})_{k}$ is \textit{log-concave} (LC for short) if $x_{i-1}x_{i+1}\leq x_{i}^{2}$ for all
$i>0$, which is equivalent to $x_{i-1}x_{j+1}\leq x_{i}x_{j}$ for all $j\geq i\geq1$; see \cite{fb}. Log-concave sequences arise often in combinatorics, algebra, geometry, analysis, probability and statistics and have been extensively investigated; see Stanley \cite{St} and Brenti \cite{fb} for details.

For two polynomials with real coefficients $A(q)$ and $B(q)$, we write
$A(q)$ $\geq _{q}B(q)$ if the difference $A(q)-B(q)$ has only nonnegative
coefficients. A polynomial sequence $\left(A_{n}(q)\right)_{n\geq
0}$ is called $q$-log-concave (as introduced by Sagan \cite{sg1})
if 
\begin{equation*} 
A_{n-1}(q)A_{n+1}(q)\leq _{q}A_{n}(q)^{2}
\end{equation*}
for $n\geq 1$.

It is easy to see that if the sequence $\left(A_{n}(q)\right)_{n\geq 0}$ is $q$-log-concave, then for each fixed nonnegative number $q$, the sequence $\left(f_{n}(q)\right)_{n\geq 0}$ is log-concave. The $q$-log-concavity of polynomials have been extensively studied; see Butler \cite{But}, Krattenthaler \cite{Krat}, Leroux \cite{Ler} and Sagan \cite{sg1,sg2}, for instance.

Let $\{a_{s}(n,k)\}_{0\leq k\leq ns}$ be a $s$-triangle of nonnegative numbers with $s\geq1$.
We illustrate a $4$-triangle as follows:
$$\
\begin{array}
[c]{cccccccccccccccccc}
n\backslash k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 &
14 & 15 & 16\\
0 & {\star} &  &  &  &  &  &  &  &  &  &  &  &  &  &  &  & \\
1 & {\star} & {\star} & {\star} & {\star} & {\star} &  &  &  &  &  &  &  &  &
&  &  & \\
2 & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} &
{\star} & {\star} &  &  &  &  &  &  &  & \\
3 & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} &
{\star} & {\star} & {\star} & {\star} & {\star} & {\star} &  &  &  & \\
4 & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} &
{\star} & {\star} & {\star} & {\star} & {\star} & {\star} & {\star} & {\star}
& {\star} & {\star}
\end{array}
$$
\begin{center}
Table 1. The $4$-triangle.
\end{center}

A nice example of these $s$-triangles is the triangle given by the ordinary multinomials or bi$^{s}$nomial coefficients \cite{bsb}: let $s\geq1$ and $n\geq0$ be two integers, and $k=0,1,\ldots,sn,$ the bi$^{s}$nomial number $\binom{n}{k}_{s}$ is defined as the $k$-th coefficient in the expansion
\begin{equation}
(1+x+x^{2}+\cdots+x^{s})^{n}=\sum_{k\geq0} \binom{n}{k}_{s}x^{k}\label{eq1}.
\end{equation}

Below we list some related identities for the bi$^s$nomial coefficients. For
more details see \cite{bsb} and references therein.
\begin{itemize}
 \item Expression of bi$^s$nomial coefficients in terms of binomial coefficients,
\begin{equation}
\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}}\label{eq2}.
\end{equation}
  \item The symmetry relation
\begin{equation}
\binom{n}{k}_{s}=\binom{n}{sn-k}_{s}\label{eq3}.
\end{equation}
\item The longitudinal recurrence relation
\begin{equation}
\binom{n}{k}_{s}=
\sum_{j=0}^{s} \binom{n-1}{k-j}_{s}\label{eq}.
\end{equation}
\end{itemize}
These coefficients, as for usual binomial coefficients, are defined as
in the Pascal triangle known as the ``$s$-Pascal triangle''.
One can find the first values of the $s$-Pascal triangle in the 
{\it On-Line Encyclopedia of Integer Sequences} (OEIS) \cite{slo} as 
\seqnum{A027907} for $s=2$,
as \seqnum{A008287} for $s=3$,
and as \seqnum{A035343} for $s=4$.
\small{
$$
\begin{array}
[c]{ccccccccccccccccc}
n\backslash k & \text{{0}} & \text{{1}} & \text{{2}} &
\text{3} & \text{{4}} & \text{{5}} & \text{{6}}
& \text{7} & \text{{8}} & \text{9} & \text{{10}} &
\text{11} & \text{12}  \\
\text{{0}} & \text{1} &  &  &  &  &  &  &  &  &  &  &  &  &  &     \\
\text{{1}} & \text{1} & \text{1} & \text{1} &  &  &  &  &  &  &  &  &
&  &  &   \\
\text{2} & \text{1} & \text{2} & \text{3} & \text{2} & \text{1} &  &
&  &  &  &  &  &  &  &     \\
\text{3} & \text{1} & \text{3} & \text{6} & \text{7} & \text{6} &
\text{3} & \text{1} &  &  &  &  &  &  &  &     \\
\text{4} & \text{1} & \text{4} & \text{10} & \text{16} & \text{19} &
\text{16} & \text{10} & \text{4} & \text{1} &  &  &  &  &  &     \\
\text{5} & \text{1} & \text{5} & \text{15} & \text{30} & \text{45} &
\text{51} & \text{45} & \text{30} & \text{15} & \text{5} & \text{1} &  &  &  &
  \\
\text{6} & \text{1} & \text{6} & \text{21} & \text{50} & \text{90} &
\text{126} & \text{141} & \text{126} & \text{90} & \text{50} & \text{21} &
\text{6} & \text{1} &  &     \\
\text{7} & \text{1} & \text{7} & \text{28} & \text{77} & \text{161} &
\text{266} & \text{357} & \text{393} & \text{357} & \text{266} & \text{161} &
\text{77} &  $\ldots$  \\
\text{8} & \text{1} & \text{8} & \text{36} & 112 & 266 & 504 & 784 & 1016 &
1107 & 1016 & 784 & 504 &  $\ldots$
\end{array}
$$
}
\begin{center}
Table 2. Triangle of trinomial coefficients: $s=2$.
\end{center}
\normalsize

Brondarenko \cite{bro} gives a combinatorial interpretation of the bi$^{s}$nomial coefficient $\binom{n}{k}_{s}$ as \textit{the number of different ways of distributing} ``$k$'' \textit{balls among} ``$n$'' \textit{cells where each cell contains at most} ``$s$'' \textit{balls}.
Using this combinatorial argument, one can easily establish the following
relation
\begin{equation*}
\binom{n}{k}_{s}=\sum_{n_{1}+2n_{2}+\cdots+sn_{s}=k}\binom{n}
{n_{0},n_{1},\ldots,n_{s-1}}.
\end{equation*}


These coefficients are also naturally linked to generalized 
Fibonacci sequence: the ``multibonacci'' sequence, given for $s\geq1$, by
\begin{displaymath}
\begin{cases}
\Phi_{0}=\Phi_{1}=\cdots=\Phi_{s-1}=0,\ \ \Phi_{s}=1,\\
\Phi_{n}=\Phi_{n-1}+\Phi_{n-2}+\cdots+\Phi_{n-s-1} \text{  }\left(n\geq1\right).
\end{cases}
\end{displaymath}

We have the following identity \cite{bsb}
\begin{equation*}
\Phi_{n+1}=\sum_{k}\binom{n-k}{k}_{s}.
\end{equation*}

The case $s=1$ provides a nice identity for Fibonacci numbers
(sequence \seqnum{A000045}):
\begin{equation*}
F_{n+1}=\sum_{k}{\binom{n-k}{k}}.
\end{equation*}

One of the extensions of binomial coefficients are $q$-binomial
coefficients. Several works and applications were done in this area.
For Fibonacci sequences, see Carlitz \cite{Ca} and Cigler \cite{Cig}. For
Lucas sequences, see Belbachir and Benmezai \cite{bbm2}. For a variant of
$q$-bi$^{s}$nomials, see Belbachir and Benmezai \cite{bbm} or our paper
\cite{Baz}, and for a recent application to the determinant, see Arikan and
Kili\c{c} \cite{TK}.

Let us consider the following two \textit{linear transformations} of sequences:
\begin{equation}
t_{n}=\sum_{k=0}^{ns}a_{s}(n,k)x_{k},\ \ (n\geq 0),\label{1}
\end{equation}
\begin{equation}
z_{n}=\sum_{k=0}^{ns}a_{s}(n,k)x_{k}y_{sn-k},\ \ (n\geq 0).\label{2}
\end{equation}
We say that the linear transformation \eqref{1} (resp., \eqref{2}) has the PLC (resp., double PLC) property if it preserves log-concavity of sequences, i.e., the log-concavity of $(x_{n})$ (resp., $(x_{n})$ and $(y_{n})$) implies that of $(t_{n})$ (resp., $(z_{n})$). The corresponding $s$-triangle $\{a_{s}(n,k)\}$ is also called PLC (resp., double PLC).

This is a good way to obtain log-concavity by linear transformations or some operators. For instance, Menon \cite{Mn} demonstrated that log-concavity is preserved under the ordinary convolution. Walkup in \cite{walk},
and later, Wang and Yeh \cite{42} also proved that log-concavity is preserved under the binomial convolution. It is also established that the $q$-binomial convolution preserves log-concavity; see \cite{ZH}. In \cite{MH, MH1, AH}, we established the preserving log-convexity and log-concavity properties, respectively, for the bi$^{s}$nomial coefficients and the $p,q$-binomial coefficients.

In this paper, we generalize the aforementioned results for the generalized triangles like the $s$-Pascal triangle. In $\S$ 2, we give the necessary conditions to establish the PLC (resp., double PLC) property of the generalized triangles $\{a_{s}(n,k)\}$. In $\S$ 3, some examples of the both properties are given include the $s$-Pascal triangle.
\section{LC-positivity and preservation of log-concavity}
In this section, we give a relation between LC-positivity (resp., double LC-positivity) and the PLC property (resp., the double PLC property) for generalized triangles. We start with the concept of LC-positivity introduced by Wang and Yeh \cite{42}.
\begin{definition}\label{df1}
Let $s\geq1$ and $n\geq0$ be two integers. For $0\leq r \leq sn$, define the polynomial
\begin{equation*}
\mathcal{A}_{s,r}(n;q):=\sum_{k=r}^{ns}a_{s}(n,k)q^{k}.
\end{equation*}
We say that the $s$-triangle $\{a_{s}(n,k)\}$ has the LC-positive property if for each $r\geq0$, the sequence of polynomials $\left (\mathcal{A}_{s,r}(n;q)\right)_{n\geq r}$ is $q$-log-concave in $n$.
\end{definition}

\begin{definition}\label{df2}
Let $s\geq1$ and $n\geq0$ be two integers. For $0\leq k \leq sn$, define the reciprocal triangle $\{a^{\ast}_{s}(n,k)\}$ of $\{a_{s}(n,k)\}$ by
\begin{equation*}
a^{\ast}_{s}(n,k)=a_{s}(n,sn-k)
\end{equation*}
and for $0\leq r \leq sn$, the polynomial
\begin{equation*}
\mathcal{A}^{\ast}_{s,r}(n;q):=\sum_{k=r}^{ns}a^{\ast}_{s}(n,k)q^{k}.
\end{equation*}
We say that the $s$-triangle $\{a_{s}(n,k)\}$ has the double LC-positive property if for each $r\geq0$, the sequence of polynomials $\left (\mathcal{A}_{s,r}(n;q)\right)_{n\geq r}$ and $\left (\mathcal{A}^{\ast}_{s,r}(n;q)\right)_{n\geq r}$ are $q$-log-concave in $n$.
\end{definition}

We shall need the following lemma due to Wang and Yeh \cite{42}.
\begin{lemma}\label{lemm1}
Let $h\in\mathbb{N}$. Suppose that two sequences $a_{0},\ldots,a_{h}$ and $X_{0},\ldots,X_{h}$ of real numbers satisfy the following two conditions:
\begin{description}
\item[1] $\sum_{k=r}^{h}a_{k}\geq0$\ \ ($0\leq r\leq h$);

\item[2] $0\leq X_{0}\leq \cdots\leq X_{h}$.
\end{description}
Then
 $$\sum_{k=0}^{h}a_{k}X_{k}\geq X_{0}\sum_{k=0}^{h}a_{k}\geq0.$$
\end{lemma}

Let $\{a_{s}(n,k)\}_{0\leq k\leq ns}$ be a $s$-triangle of nonnegative numbers and $(x_{k})_{k\geq0}$ be a log-concave sequence. Let $(z_{n})_{n\geq0}$
be the sequence defined by \eqref{1} and let us consider the difference
\begin{equation}
\triangle_{n}:=\left(  \sum_{k=0}^{ns}a_{s}(n,k)x_{k}\right)  ^{2}-\left(
\sum_{k=0}^{ns-s}a_{s}(n-1,k)x_{k}\right)  \left(  \sum_{k=0}^{ns+s}
a_{s}(n+1,k)x_{k}\right). \label{3}
\end{equation}
Then $\triangle_{n}$ is a quadratic form in $ns+s+1$ variables $x_{0},x_{1},\ldots,x_{ns+s}$.

Let $S_{t}$ be the sum of terms $x_{k}x_{t-k}$\ in $\triangle_{n}.$ For $0\leq k\leq\lfloor t/2\rfloor$ with $0\leq t\leq2ns$, let $a_{s,k}(n,t)$ be the
coefficient of the term $x_{k}x_{t-k}$\ in $\triangle_{n}.$ Then
\begin{equation}
\triangle_{n}=\sum_{t=0}^{2ns}S_{t}\text{ \ with }S_{t}=\sum_{k=0}^{\lfloor
t/2\rfloor}a_{s,k}(n,t)x_{k}x_{t-k}.\label{e4}
\end{equation}

Thus, it suffices to show that $S_{t}$ $\geq0$ ($0\leq t\leq2ns $). We have the following inequalities $x_{0}x_{t}\leq x_{1}x_{t-1}\leq x_{2}x_{t-2}\leq\cdots$. Hence by Lemma \ref{lemm1}, it suffices to establish that 
\begin{equation}
A_{s,r}(n,t):=\sum_{k=r}^{\lfloor t/2\rfloor}a_{s,k}(n,t)\geq0,\ \ \left(  0\leq
r\leq\lfloor t/2\rfloor\right).\label{e5}
\end{equation}

Using relation \eqref{3}, for $k<t/2,$ we obtain
\begin{align}
 a_{s,k}(n,t) &= 2a_{s}(n,k)a_{s}(n,t-k)-a_{s}(n-1,k)a_{s}(n+1,t-k)\notag\\
&\ \ \ -a_{s}(n+1,k)a_{s}(n-1,t-k),\label{eq4}
\end{align}
and for $t$ even and $k=t/2$, we have
\begin{equation}
a_{s,k}(n,t)=a_{s}(n,k)^{2}-a_{s}(n-1,k)a_{s}(n+1,k).\label{e6}
\end{equation}

Let us remark that $A_{s,r}(n,t)$ is precisely the coefficient of $q^{t}$ in the polynomial $\mathcal{A}_{s,r}^{2}(n;q)-\mathcal{A}_{s,r}(n-1;q)\mathcal{A}
_{s,r}(n+1;q)$, i.e.,
\begin{equation}
\mathcal{A}_{s,r}^{2}(n;q)-\mathcal{A}_{s,r}(n-1;q)\mathcal{A}_{s,r}
(n+1;q)=\sum_{t=2r}^{2ns}A_{s,r}(n,t)q^{t}.\label{5}
\end{equation}

Hence, the following characterization of positivity holds:
\begin{lemma}\label{l1}
The $s$-triangle $\{a_{s}(n,k)\}_{0\leq k\leq ns}$ is LC-positive if and only if $A_{s,r}(n,t)\geq0$ for all $2r\leq
t\leq2ns.$
\end{lemma}
Now, from the discussion above, we obtain the following:
\begin{theorem}\label{t1}
The LC-positive $s$-triangles are PLC.
\end{theorem}

The relation between double LC-positivity and the double PLC property is given by the following proposition.
\begin{proposition}\label{p1}
Given a $s$-triangle $\{a_{s}(n,k)\}_{0\leq k\leq ns}$ of nonnegative numbers and two log-concave sequences $(x_{k})_{k\geq 0}$ and $(y_{k})_{k\geq 0}$.

Define three $s$-triangles $\{b_{s}(n,k)\}$, $\{c_{s}(n,k)\}$ and $
\{d_{s}(n,k)\}$ by
\begin{equation*}
b_{s}(n,k)=a_{s}(n,k)x_{k},\text{ \ \ \ }c_{s}(n,k)=a_{s}(n,k)y_{ns-k},\text{
\ \ }d_{s}(n,k)=a_{s}(n,k)x_{k}y_{ns-k}.
\end{equation*}
For $2r\leq t\leq 2ns$, define $B_{s,r}(n,t)$, $C_{s,r}(n,t)$ and $D_{s,r}(n,t)$ similar to $A_{s,r}(n,t)$ in \eqref{5}.

\begin{enumerate}
\item If the $s$-triangle $\{a_{s}(n,k)\}$ is LC-positive, then the $s$-triangle $\{b_{s}(n,k)\}$ is LC-positive and $B_{s,r}(n,t)\geq A_{s,r}(n,t)x_{r}x_{t-r}$.

\item If the $s$-triangle $\{a_{s}(n,k)\}$ is double LC-positive, then the $s$-triangle $\{c_{s}(n,k)\}$ is LC-positive and $C_{s,r}(n,t)\geq
A_{s,r}(n,t)y_{ns-t+r}y_{ns-r}$ for $t\leq ns+r$.

\item If the $s$-triangle $\{a_{s}(n,k)\}$ is double LC-positive, then the $s$-triangle $\{d_{s}(n,k)\}$ is LC-positive and $D_{s,r}(n,t)\geq
A_{s,r}(n,t)x_{r}x_{t-r}y_{ns-t+r}y_{ns-r}$ for $t\leq ns+r$.
\end{enumerate}
\end{proposition}

\begin{proof}
\leavevmode
\begin{enumerate}
\item Let $0\leq t\leq 2ns.$ It is easy to see by definition that $b_{s,k}(n,t)=a_{s,k}(n,t)x_{k}x_{t-k}$ for $0\leq k\leq \lfloor t/2\rfloor
.$ Hence for $0\leq r\leq \lfloor t/2\rfloor $
\begin{equation*}
B_{s,r}(n,t):=\sum_{k=r}^{\lfloor t/2\rfloor}b_{s,k}(n,t)=\sum_{k=r}^{\lfloor t/2\rfloor }a_{s,k}(n,t)x_{k}x_{t-k},
\end{equation*}

Now $\{a_{s}(n,k)\}$ is LC-positive and $x_{0}x_{t}\leq x_{1}x_{t-1}\leq\cdots$ by the log-concavity of $(x_{k})$. From Lemma \ref{lemm1} it
follows that
\begin{equation*}
B_{s,r}(n,t)\geq x_{r}x_{t-r}\sum_{k=r}^{\lfloor t/2\rfloor}a_{s,k}(n,t)=A_{s,r}(n,t)x_{r}x_{t-r}\geq 0,
\end{equation*}

So the $s$-triangle $\{b_{s}(n,k)\}$ is LC-positive.

\item Let $2r\leq t\leq 2ns.$ We need to prove $C_{s,r}(n,t)\geq 0$. For brevity, we do this only for the case $t$ odd since the same technique is
still valid for the case where $t$ is even.

Let $t=2l+1$ for $0\leq k\leq l$.  Then we define
\begin{align*}
\alpha _{k} &=a_{s}(n,k)a_{s}(n,t-k), \\
\beta _{k} &=a_{s}(n-1,k)a_{s}(n+1,t-k), \\
\gamma _{k} &=a_{s}(n+1,k)a_{s}(n-1,t-k), \\
Y_{k} &=c_{ns-t+k}y_{ns-k}.
\end{align*}
Then
\begin{equation*}
a_{s,k}(n,t)=2\alpha _{k}-\beta _{k}-\gamma _{k},
\end{equation*}
and
\begin{equation*}
c_{s,k}(n,t)=2\alpha _{k}Y_{k}-\beta _{k}Y_{k+s}-\gamma _{k}Y_{k-s}
\end{equation*}
by definition. It follows that
\begin{align*}
C_{s,r}(n,t) &=\sum_{k=r}^{l}(2\alpha _{k}Y_{k}-\beta _{k}Y_{k+s}-\gamma
_{k}Y_{k-s}) \\
&=\sum_{k=r}^{l}(2\alpha _{k}-\beta _{k-s}-\gamma
_{k+s})Y_{k}+\sum_{j=1}^{s}\beta _{r-j}Y_{r+s-j} \\
&\ \ \ -\sum_{j=1}^{s}\gamma _{r+j-1}Y_{r-s+j-1}-\sum_{j=1}^{s}\beta
_{l-j+1}Y_{l+s-j+1}+\sum_{j=1}^{s}\gamma _{l+j}Y_{l-s+j},
\end{align*}
where we use the fact that $Y_{l+s-j+1}=Y_{l-s+j}$ and $\beta_{l-j+1}=\gamma _{l+j}$ for $j=\overline{1,s}$. Note that $(Y_{k})$ is nondecreasing by the log-concavity of $(y_{k})$ and
\begin{align*}
2\alpha _{k}-\beta _{k-s}-\gamma _{k+s} &=2a_{s}^{\star
}(n,ns-k)a_{s}^{\star }(n,np-t+k) \\
&\ \ \ -a_{s}^{\star }(n-1,ns-k)a^{\star }(n+1,ns-t+k) \\
&\ \ \ -a_{s}^{\star }(n+1,np-k)a_{s}^{\star }(n-1,ns-t+k) \\
&=a_{s,ns-t+k}^{\star }(n,2ns-t).
\end{align*}

Hence by the LC-positivity of $\{a_{s}^{\star }(n,k)\}$, we have
\begin{align*}
C_{s,r}(n,t) &=\sum_{j=ns-t+r}^{\lfloor (2ns-t)/2\rfloor }a_{s,j}^{\star
}(n,2ns-t)Y_{j-ns+t}+\sum_{j=1}^{s}\beta _{r-j}Y_{r+s-j}  \\
&\ \ \ -\sum_{j=1}^{s}\gamma _{r+j-1}Y_{r-s+j-1}   \\
&\geq Y_{r}\sum_{j=ns-t+r}^{\lfloor (2ns-t)/2\rfloor }a_{s,j}^{\star
}(n,2ns-t)+Y_{r}\sum_{j=1}^{s}\beta _{r-j}-Y_{r-s}\sum_{j=1}^{s}\gamma
_{r+j-1}   \\
&=Y_{r}\sum_{k=r}^{s}(2\alpha _{k}-\beta _{k-s}-\gamma
_{k+s})+Y_{r}\sum_{j=1}^{s}\beta _{r-j}-Y_{r-s}\sum_{j=1}^{s}\gamma _{r+j-1}
 \\
&=Y_{r}\sum_{k=r}^{s}(2\alpha _{k}-\beta _{k}-\gamma
_{k})+(Y_{r}-Y_{r-s})\sum_{j=1}^{s}\gamma _{r+j-1}   \\
&=A_{s,r}(n,t)Y_{r}+(Y_{r}-Y_{r-s})\sum_{j=1}^{s}\gamma _{r+j-1}.
\end{align*}

Thus $C_{s,r}(n,t)\geq A_{s,r}(n,t)y_{ns-t+r}y_{ns-r}$.

\item We have $d_{s}(n,k)=a_{s}(n,k)x_{k}y_{ns-k}=c_{s}(n,k)x_{k}$ and
\begin{equation*}
D_{s,r}(n,t)=\sum_{k=r}^{\lfloor t/2\rfloor}d_{s,k}(n,t)=\sum_{k=r}^{\lfloor t/2\rfloor }c_{s,k}(n,t)x_{k}x_{t-k},
\end{equation*}
by 1 and 2, so
\begin{equation*}
D_{s,r}(n,t)\geq C_{s,r}(n,t)x_{r}x_{t-r}\geq A_{s,r}(n,t)x_{r}x_{t-r}y_{ns-t+r}y_{ns-r}.
\end{equation*}
\end{enumerate}
\end{proof}

Now we establish the second result.
\begin{theorem}\label{t2}
The double LC-positive $s$-triangles are double PLC.
\end{theorem}

\begin{proof}
Let the $s$-triangle $\{a_{s}(n,k)\}$ be doubly LC-positive. Suppose that both $(x_{k})$ and $(y_{k})$ are log-concave. Then the $s$-triangle $
\{a_{s}(n,k)x_{k}y_{ns-k}\}$ is LC-positive by Proposition \ref{p1} (3) and is therefore PLC by Theorem \ref{t1}. Thus the row-sum sequence
\begin{equation*}
z_{n}=\sum_{k=0}^{ns}a_{s}(n,k)x_{k}y_{ns-k},\hspace{1cm}n=0,1,2,\ldots
\end{equation*}
is log-concave. In other words, the $s$-triangle $\{a_{s}(n,k)\}$ is double
PLC.
\end{proof}

By Lemma \ref{l1}, $\{a_{s}(n,k)\}$ is LC-positive if and only if the inequality $\sum_{k=r}^{\lfloor t/2\rfloor }a_{s,k}(n,t)\geq 0$ for all $2r\leq t\leq 2ns$, so the following corollary is immediate.
\begin{corollary}\label{cc1}
Suppose that the following two conditions hold:

\begin{description}
\item[A] There exists an index $m=m(n,t)$ such that $a_{k}(n,t)<0$ for $k<m$ and $a_{s,k}(n,t)\geq 0$ for $k\geq m$;

\item[B] The sequence $\left(\mathcal{A}_{s,0}(n;q)\right)_{n\geq 0}$ is $q$-log-concave.
\end{description}
Then the $s$-triangle $\{a_{s}(n,k)\}$ is LC-positive and therefore PLC.
\end{corollary}

\begin{corollary}
Suppose that $s$-triangle $\{a_{s}(n,k)\}$ satisfies Conditions (A) and (B) in Corollary \ref{cc1} and $\{a_{s}^{\ast }(n,k)\}$ satisfies Condition (A).
Then $\{a_{s}(n,k)\}$ is doubly LC-positive and therefore double PLC.
\end{corollary}

\begin{proof}
It suffices to show that $\left(\mathcal{A}_{s,0}^{\star }(n;q)\right)$ is $q$-log-concave. We have
\begin{equation*}
\mathcal{A}_{s,0}^{\star
}(n;q)=\sum_{k=0}^{ns}a_{s}(n,ns-k)q^{k}=\sum_{k=0}^{ns}a_{s}(n,k)q^{ns-k}=q^{ns}\mathcal{A}_{s,0}(n;q^{-1})
\end{equation*}

It follows that
\begin{equation*}
\mathcal{A}_{s,0}^{\star 2}(n;q) -\mathcal{A}_{s,0}^{\star }(n-1;q)\mathcal{A}_{s,0}^{\star }(n+1;q)
=q^{2ns}\left(\mathcal{A}_{s,0}^{2}(n;q^{-1})-\mathcal{A}_{s,0}(n-1;q^{-1})
\mathcal{A}_{s,0}(n+1;q^{-1})\right)
\end{equation*}
which has nonnegative coefficients by the $q$-log-concavity of $\left(\mathcal{A}_{s,0}(n;q)\right)$.
\end{proof}

\section{Application to linear operators of finite order}

In this section, for selected examples of $s$-triangles we show their LC-positivity leading to the PLC property.

Let $\mathfrak{S}$ denote the set of sequences $(u_{k})_{k\in\mathbb{Z}}$  of nonnegative numbers. Given $(s+1)$ nonnegative numbers $\lambda_{0},\lambda_{1},\ldots,\lambda_{s}$, define the linear operator $L=L[\lambda_{0},\lambda_{1},\ldots,\lambda_{s}]$, on $\mathfrak{S}$ by
\begin{equation*}
L(u_{k})=\sum_{j=0}^{s}\lambda_{j}u_{k-j}\ \ (k\in\mathbb{Z}).
\end{equation*}


For $n\geq2,$ define $L^{n}:=L(L^{n-1})$ by induction. It is convenient to view $L^{0}$ as the identity operator. Let $(u_{k}
)_{k\in\mathbb{Z}}$ be a log-concave sequence.
\begin{lemma}\label{llm}
If the sequence $\left(\lambda_{0},\lambda_{1},\ldots,\lambda_{s}\right)$ is log-concave, then so is the sequence $\left(L^{n}(u_{k})\right)_{k\in\mathbb{Z}}$.
\end{lemma}

\begin{proof}
In fact
\begin{align*}
&\left( L(u_{k})\right)^{2}-L(u_{k-1})L(u_{k+1})=\left(  \sum_{j=0}^{s}\lambda_{j}u_{k-j}\right)  ^{2}-\sum_{j=0}^{s}\lambda_{j}u_{k-j-1}\sum_{j=0}^{s}\lambda_{j}u_{k-j+1}\\
&=\sum_{j=0}^{s}\lambda_{j}^{2}(u_{k-j}^{2}-u_{k-j-1}u_{k-j+1})
+\sum_{
0\leq l<j\leq s}\lambda_{j}\lambda_{l}(u_{k-j}u_{k-l}-u_{k-j-1}u_{k-l+1})\\
&\ \ \ +\sum_{0\leq l<j\leq s}\lambda_{j}\lambda_{l}u_{k-j}u_{k-l}-\sum_{
0\leq l<j\leq 0}\lambda_{j}\lambda_{l}u_{k-j+1}u_{k-l-1}\\
&=T_1+ T_2+T_3,
\end{align*}
with
\begin{align*}
T_1 &=\sum_{j=0}^{s}\lambda_{j}^{2}(u_{k-j}^{2}-u_{k-j-1}u_{k-j+1}),\\
T_2 &= \sum_{0\leq l<j\leq s}\lambda_{j}\lambda_{l}(u_{k-j}u_{k-l}-u_{k-j-1}u_{k-l+1})
\end{align*}
and
\begin{align*}
T_3 &= -\sum_{1\leq l+1<j\leq s}\lambda_{j}\lambda_{l}(u_{k-j+1}u_{k-l-1}-u_{k-j}u_{k-l}).
\end{align*}

It follows that
\begin{align*}
&  \left( L(u_{k})\right)^{2}-L(u_{k-1})L(u_{k+1})=\sum_{j=1}^{s-1}\left(\lambda_{j}^{2}
-\lambda_{j-1}\lambda_{j+1}\right)
\times\left(u_{k-j}^{2}-u_{k-j-1}u_{k-j+1}\right)\\
&+\lambda_{0}^{2}\left(u_{k}^{2}-u_{k-1}u_{k+1}\right)
+\lambda_{s}^{2}\left(u_{k-s}^{2}-u_{k-s-1}u_{k-s+1}\right)\\
&+\sum_{2\leq l+2<j\leq s-1}\left(\lambda_{l}\lambda_{j}-\lambda_{l-1}\lambda_{j+1}\right)
 \times\left(u_{k-j}u_{k-l}-u_{k-j-1}u_{k-l+1}\right)\\
&+\sum_{l=0}^{s-1}\lambda_{l}\lambda_{l+1}\left(u_{k-l-1}u_{k-l}
-u_{k-l-2}u_{k-l+1}\right)\\
&+\sum_{l=0}^{s-2}\lambda_{l}\lambda_{l+2}\left(u_{k-l-2}u_{k-l}
-u_{k-l-3}u_{k-l+1}\right)\\
& \geq 0.
\end{align*}
By induction, the polynomial sequence $\left(L^{n}(u_{k})\right)_{k\in\mathbb{Z}}$ is also log-concave for $n\geq0$.
\end{proof}

This brings us to the following theorem.
\begin{theorem}\label{th3}
Given $(s+1)$ nonnegative numbers $\lambda_{0},\lambda_{1},\ldots,\lambda_{s}$ and a log-concave sequence $(u_{k})_{k\in\mathbb{Z}}$, define
\begin{equation*}
a_{s}(n,k)=L^{n}(u_{k}),\ \left(  0\leq k\leq ns\right).
\end{equation*}
If $\left(\lambda_{0},\lambda_{1},\ldots,\lambda_{s}\right)$ is log-concave. Then the $s$-triangle $\{a_{s}(n,k)\}$ is doubly LC-positive and therefore double PLC.
\end{theorem}

\begin{proof}
Denote $a_{k}=L^{n-1}(u_{k})$ for $k\in\mathbb{Z}$ and $\mathcal{A}_{s,r}(n-1;q)=\sum_{k=r}^{ns-s}a_{k}q^{k}$. If $\left(\lambda_{0},\lambda_{1}, \ldots,\lambda_{s}\right)$ is log-concave, then by Lemma \ref{llm} so is the sequence $(a_{k})_{k\in\mathbb{Z}}$. We have
\begin{align*}
\mathcal{A}_{s,r}(n;q) & =\lambda_{0}\sum_{k=r}^{ns}a_{k}q^{k}+\lambda_{1}\sum_{k=r}^{ns}a_{k-1}
q^{k}+\cdots+\lambda_{s}\sum_{k=r}^{ns}a_{k-s}q^{k}\\
& =\mathcal{A}_{s,r}(n-1;q)\sum_{j=0}^{s}\lambda_{j}q^{j}+\sum_{j=1}
^{s}\lambda_{j}\sum_{l=1}^{j}a_{r-l}q^{r+j-l}\\
&\ \ \ +\sum_{j=0}^{s-1}\lambda_{j}\sum_{l=1}^{s-j}a_{ns-s+l}q^{ns-s+l+j},
\end{align*}
thus
\begin{align*}
& \mathcal{A}_{s,r}(n;q)^{2}-\mathcal{A}_{s,r}(n-1;q)\mathcal{A}
_{s,r}(n+1;q)=\\
& \sum_{j=1}^{s}\sum_{l=1}^{j}\sum_{f=0}^{s} \lambda_{j}\lambda_{f}\left(\sum_{k=r}^{ns}
[a_{r-l}a_{k-f}-a_{r-f-l}a_{k}]q^{k+r+j-l}\right.\\
&\left. +\sum_{k=ns-s+1}^{ns}a_{r-f-l}a_{k}q^{k+r+j-l}\right) \\
& +\sum_{j=0}^{s-1}\sum_{l=1}^{s-j}\sum_{f=0}^{s} \lambda
_{j}\lambda_{f}\left(\sum_{k=r}^{ns}\left(  a_{ns-s+l}a_{k-f}-a_{ns+l-f}a_{k-s}\right)
q^{k+ns-s+l+j}\right.\\
&\left. +\sum_{k=r}^{r+s-1}a_{ns+l-f}a_{k}q^{k+ns-s+l+j}\right),
\end{align*}
which has nonnegative coefficients by the log-concavity of the sequence $(a_{k})$. Hence the $s$-triangle $\{a_{s}(n,k)\}_{0\leq k\leq ns}$ is LC-positive.


On the other hand, let $u_{k}^{\star }=u_{-k}$ for $k\in \mathbb{Z}$. Then the sequence $(u_{k}^{\star })_{k\in \mathbb{Z}}$ is log-concave and $
a_{s}^{\star }(n,k)=L^{n}[\lambda ](u_{k}^{\star })$. Thus the $s$-triangle $\{a_{s}^{\star }(n,k)\}_{0\leq k\leq ns}$ is also LC-positive, and the $s$-triangle $\{a_{s}(n,k)\}_{0\leq k\leq ns}$ is therefore doubly LC-positive.
\end{proof}

\begin{corollary}\label{cr1}
Let $a$ and $b$ be two nonnegative integers with $a\geq b$. If the sequences $(x_{k})$ and $(y_{k})$ are log-concave, then so is the sequence
\begin{equation*}
z_{n}=\sum_{k=0}^{ns}\binom{a+n}{b+k}_{s}x_{k}y_{sn-k},\ \ (n\geq0).
\end{equation*}
\end{corollary}

\begin{proof}
Using relation \eqref{eq}, we have  $\binom{a+n}{b+k}_{s}=\sum_{j=0}^{s}\binom{a+n-1}{b+k-j}_{s}$, and taking $u_{k}=\binom{a}{b+k}_{s}$ with $\lambda_{j}=1$, ($1\leq j\leq s$) in Theorem \ref{th3}, we obtain the result.
\end{proof}
When $s=1$, we obtain the result of Y. Wang \cite[Corollary 3.4]{39}.
Taking $a=b=0$ in Corollary \ref{cr1}, we obtain the following nice result.
\begin{corollary}\label{cr2}
If the sequences $(x_{k})$ and $(y_{k})$ are log-concave, then
so is
\begin{equation*}
z_{n}=\sum_{k=0}^{ns}\binom{n}{k}_{s}x_{k}y_{sn-k}\ \ (n\geq0).
\end{equation*}

\end{corollary}

The following theorem is in a sense dual to Theorem \ref{th3}.

\begin{theorem}\label{thh4}
Let $\lambda_{0},\lambda_{1},\ldots,\lambda_{s}, (s+1)$ nonnegative numbers and $\{a_{s}(n,k)\}$ an $s$-triangle of nonnegative numbers. Suppose that each row of $\{a_{s}(n,k)\}$ is log-concave and satisfies the following recurrence relation
\begin{equation}
a_{s}(n,k)=\sum_{j=0}^{s}\lambda_{j}a_{s}(n+1,k+j),\ \ (0\leq k\leq ns).\label{6}
\end{equation}
Then the $s$-triangle $\{a_{s}(n,k)\}$ is LC-positive and therefore double PLC.
\end{theorem}

\begin{proof}
Denote $a_{s}(n+1,k)=v_{k}$\ \ ($0\leq k\leq ns+s$). Then the sequence $(v_{k})$ is log-concave and $\mathcal{A}_{s,r}(n+1;q)=\sum_{k=r}
^{ns+s}v_{k}q^{k}$. By the recurrence relation \eqref{6} we have
\begin{align*}
\mathcal{A}_{s,r}(n;q)  & =\mathcal{A}_{s,r}(n+1;q)\sum_{j=0}^{s}\lambda_{j}q^{-j}-\sum_{j=0}
^{s-1}\sum_{l=1}^{s-j}\lambda_{j}v_{ns+j+l}q^{ns+l}-\sum_{j=1}^{s}\sum_{l=0}^{j-1}\lambda_{j}v_{r+l}q^{r+l-j}.
\end{align*}


It follows that
\begin{align*}
&  \mathcal{A}_{s,r}^{2}(n;q)-\mathcal{A}_{s,r}(n-1;q)\mathcal{A}
_{s,r}(n+1;q)\\
&=  \mathcal{A}_{s,r}(n;q)\left(  \mathcal{A}_{s,r}(n+1;q)\sum_{j=0}^{s}
\lambda_{j}q^{-j}-\sum_{j=0}^{s-1}\sum_{l=1}^{s-j}\lambda_{j}v_{ns+j+l}
q^{ns+l}\right.\\
&\left.  -\sum_{j=1}^{s}\sum_{l=0}^{j-1}\lambda_{j}
v_{r+l}q^{r+l-j}\right)  \\
& -\mathcal{A}_{s,r}(n+1;q)\left(  \mathcal{A}_{s,r}(n;q)\sum_{j=0}
^{s}\lambda_{j}q^{-j}\right.  -\sum_{j=0}^{s-1}\sum_{l=1}^{s-j}\sum_{f=0}
^{s}\lambda_{j}\lambda_{f}v_{ns-s+j+l+f}q^{ns-s+l}\\
&\left.  -\sum_{j=1}^{s}\sum_{l=0}^{j-1}\sum_{f=0}^{s}\lambda_{j}\lambda
_{f}v_{r+l+f}q^{r+l-j}\right)  \\
&  =S_{1}+S_{2}+S_{3},
\end{align*}
with
\begin{align*}
S_{3}&= \sum_{j=0}^{s-1}\sum_{f=0}^{s}\lambda_{j}\lambda_{f}\left(  \sum
_{l=1}^{s-j}\sum_{k=r}^{r+s-1}v_{ns-s+j+l+f}v_{k}q^{k+ns-s+l}+\sum_{l=0}
^{j-1}\sum_{k=ns+1}^{ns+s}v_{r+l+f}v_{k}q^{k+r+l-j}\right),
\end{align*}
and
\begin{align}
  S_{1} & =\sum_{j=0}^{s-1}\sum_{f=0}^{s}\sum_{l=1}^{s-j}\lambda_{j}\lambda
_{f}\left( \left(  \sum
_{k=r+s}^{ns}+\sum_{k=ns+1}^{ns+j}+\sum_{k=ns+j+1}^{ns+s}\right)\right.
(v_{ns-s+j+l+f}v_{k}\nonumber\\
&\ \ \ -v_{ns+j+l}v_{k+f-s})\left.q^{k+ns-s+l}\right) \nonumber  \\
&  =\sum_{j=0}^{s-1}\sum_{f=0}^{s}\sum_{l=1}^{s-j}q^{ns-s+l}\lambda_{j}
\lambda_{f}\left(  \sum_{k=r+s}^{ns}(v_{ns-s+j+l+f}v_{k}-v_{ns+j+l}
v_{k+f-s})q^{k}  \nonumber\right.\\
&\ \ \  \left.
+\sum_{k=ns+j}^{ns+s}(v_{ns-s+j+l+f}v_{k}-v_{k+(ns+j+l-k)}
v_{ns-s+j+l+f-(ns+j+l-k)})q^{k}\right) \label{eqq}
\end{align}
since
\begin{align*}
&  \sum_{j=0}^{s-1}\sum_{f=0}^{s}\sum_{l=1}^{s-j}\sum_{k=ns+j+1}^{ns+s}
\lambda_{j}\lambda_{f}(v_{ns-s+j+l+f}v_{k}-v_{ns+j+l}v_{k+f-s})q^{k+ns-s+l}\\
&  =\sum_{j=0}^{s-1}\sum_{f=0}^{s}\lambda_{j}\lambda_{f}\sum_{l=1}
^{s-j}\left(  \left(  \sum_{k=ns+j+1}^{ns+j+l}+\sum_{k=ns+j+l}^{ns+s}\right)\right.\\
&\ \ \ \left.(v_{ns-s+j+l+f}v_{k}-v_{ns+j+l}v_{k+f-s})q^{k+ns-s+l}\right)  \\
&  =0,
\end{align*}
by setting, $l^{\prime}=k-ns-j$ and $k^{\prime}=l+j+ns$ in the
second term. The sum \eqref{eqq} has nonnegative coefficients by log-concavity of $(v_{k})_{k},$ and the first term of \eqref{eqq} gives the following:
if $ns-s+j+l+f\leq k,$ then $$v_{ns-s+j+l+f}v_{k}-v_{ns+j+l}v_{k+f-s}=
v_{ns-s+j+l+f}v_{k}-v_{k+(ns+j+l-k)}v_{ns-s+j+l+f-(ns+j+l-k)} \geq0;$$
and otherwise, 
$$v_{ns-s+j+l+f}v_{k}-v_{ns+j+l}v_{k+f-s}=v_{ns-s+j+l+f}v_{k}-v_{ns-s+j+f+(s-f)}v_{k-(s-f)}\geq0.$$

\begin{align}
S_{2}&=\sum_{f=1}^{s}\lambda_{f}\left(  \sum_{k=r+1}^{ns}\lambda
_{j}(v_{r+f}v_{k}-v_{r}v_{k+f})q^{k+r-j}\right. \nonumber \\
&\ \ \ +\sum_{j=2}^{s}\lambda_{j}\left(
\sum_{k=r+1}^{r+j-1}(v_{r+f}v_{k}-v_{r}v_{k+f})q^{k+r-j}  \right.
\nonumber\\
&\ \ \ +\sum_{k=r+j}^{ns}(v_{r+f}v_{k}-v_{r}v_{k+f})q^{k+r-j}
+\sum_{l=1}^{j-1}\left(  (v_{r+l+f}v_{r}-v_{r+l}{}v_{r+f})q^{2r+l-j}\right.
\nonumber\\
&\ \ \ +\sum_{k=r+1}^{r+l}(v_{r+l+f}v_{k}-v_{r+l}v_{k+f})q^{k+r+l-j}+\sum
_{k=r+l}^{r+j-1}(v_{r+l+f}v_{k}-v_{r+l}v_{k+f})q^{k+r+l-j}\nonumber\\
&\ \ \ \left.  \left. \left. +\sum_{k=r+j}^{ns}  (v_{r+l+f}v_{k}-v_{r+l}
v_{k+f})q^{k+r+l-j}\right)  \right)  \right)  \nonumber\\
 & =\sum_{f=1}^{s}\lambda_{f}\left(  \sum_{k=r+1}^{ns}\lambda_{j}(v_{r+f}
v_{k}-v_{r}v_{k+f})q^{k+r-j}+\sum_{j=2}^{s}\sum_{k=r+j}^{ns}\lambda
_{j}(v_{r+f}v_{k}-v_{r}v_{k+f})q^{k+r-j} \right. \nonumber\\
&\ \ \ \left.  +\sum_{j=2}^{s}\sum_{l=1}^{j-1}\sum_{k=r+j}^{ns}\lambda
_{j}(v_{r+l+f}v_{k}-v_{r+l}v_{k+f})q^{k+r+l-j}\right)  \label{eqqq}
\end{align}
since, by setting $k^{\prime}=l+r$ in the second term
\begin{align*}
&\sum_{j=2}^{s}\sum_{f=1}^{s}\lambda_{j}\lambda_{f}\left(  \sum_{k=r+1}
^{r+j-1}(v_{r+f}v_{k}-v_{r}v_{k+f})q^{k+r-j}+\sum_{l=1}^{j-1}(v_{r+l+f}v_{r}-v_{r+l}{}v_{r+f})q^{2r+l-j}\right)  =0,
\end{align*}
also, by setting $k^{\prime}=l+r$ and $l^{\prime}=k-r$ in second term
\begin{equation*}
\sum_{j=2}^{s}\sum_{f=1}^{s}\sum_{l=1}^{j-1}\lambda_{j}\lambda_{f}\left(
\left(  \sum_{k=r+1}^{r+l}+\sum_{k=r+l}^{r+j-1}\right)  (v_{r+l+f}
v_{k}-v_{r+l}v_{k+f})q^{k+r+l-j}\right)  =0.
\end{equation*}

The sum \eqref{eqqq} has nonnegative coefficients by the log-concavity of $(v_{k})_{k}$. Hence the polynomial $\mathcal{A}_{s,r}^{2}(n;q)-\mathcal{A}
_{s,r}(n-1;q)\mathcal{A}_{s,r}(n+1;q)$ has nonnegative coefficients. So the triangle $\{a_{s}(n,k)\}$ is LC-positive.

Clearly, the reciprocal $s$-triangle $\{a_{s}^{\ast }(n,k)\}$ possesses the same property as $\{a_{s}(n,k)\}$ does. Hence $\{a_{s}^{\ast }(n,k)\}$ is
also LC-positive. Thus the $s$-triangle $\{a_{s}^{{}}(n,k)\}$ is doubly LC-positive and therefore double PLC.
\end{proof}

In Theorem \ref{thh4}, the choice $\lambda_{j}=1$ ($1\leq j\leq s$) and $a_{s}(n,k)=\binom{a-n}{b-k}_{s}$ \ ($0\leq k\leq ns$), leads to the following:

\begin{corollary}\label{cr3}
Let $a,b\in\mathbb{N}$ with $a\geq b$. If the sequences $(x_{k})$ and $(y_{k})$ are log-concave, then so is the sequence
\[
z_{n}=\sum_{k=0}^{ns}\binom{a-n}{b-k}_{s}x_{k}y_{sn-k},\ \ (n\geq0).
\]
\end{corollary}
\noindent By setting $s=1$ in the above result, we obtain the result of Wang \cite[Corollary\ 3.9]{39}.

We conclude this paper with the following.
\begin{conjecture}
The $s$-triangle $\bigl(\binom{n}{k}_{s}\binom{a-n}{b-k}_{s}\bigr)_{k}$ is double PLC.
\end{conjecture}

\section{Acknowledgments}
The authors would like to thank the anonymous referees for their careful reading and helpful comments which have hopefully led to a clearer paper.

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\end{thebibliography}

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\noindent 2000 \textit{Mathematics Subject Classification}: Primary 05A20;
Secondary 05A10, 15A04, 11B65.

\noindent \textit{Keywords}: log-concavity, LC-positivity,
$q$-log-concavity, ordinary multinomial, linear transformation,
convolution.

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\noindent (Concerned with sequences
\seqnum{A000045},
\seqnum{A008287},
\seqnum{A027907}, and
\seqnum{A035343}.)
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\vspace*{+.1in}
\noindent
Received May 30 2019;
revised versions received January 2 2020; January 3 2020.
Published in {\it Journal of Integer Sequences}, May 11 2020.

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