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\begin{center}
\vskip 1cm{\LARGE\bf Unimodality of Certain Sequences Connected With 
Binomial Coefficients
}
\vskip 1cm
\large
Hac\`ene Belbachir and Farid Bencherif\\
Department of Mathematics\\
USTHB\\
Po.~Box 32 \\
El Alia, Algiers\\
Algeria\\
\href{mailto:hbelbachir@usthb.dz}{\tt hbelbachir@usthb.dz}  \\
\href{mailto:hacenebelbachir@gmail.com}{\tt hacenebelbachir@gmail.com} \\
\href{mailto:fbencherif@usthb.dz}{\tt fbencherif@usthb.dz} \\
\href{mailto:fbencherif@yahoo.fr}{\tt fbencherif@yahoo.fr}\\ 
\ \\
L\'aszl\'o Szalay\footnote{This
research is supported by the J\'anos Bolyai Scholarship of
HAS, and by the Hungarian National Foundation for Scientific
Research Grant No.~T 048954 MAT, No.~K 61800 FT2.}\\
Institute of Mathematics and Statistics\\
University of West Hungary\\
Erzs\'ebet utca 9 \\
H-9400 Sopron\\
Hungary\\
\href{mailto:laszalay@ktk.nyme.hu}{\tt laszalay@ktk.nyme.hu}\\
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\vskip .2 in

\begin{abstract}
This paper is devoted to the study of certain unimodal sequences
related to binomial coefficients. Although the paramount purpose is to
prove unimodality, in a few cases we even determine the maxima of
the sequences. Our new results generalize some earlier theorems on
unimodality. The proof techniques are quite varied.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
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\newtheorem{corollary}{Corollary}[section]
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%\date{\today}

\section{Introduction}


Let $\{H_n\}_{n=0}^\infty$ denote a binary recurrence sequence defined by the initial values $H_0\in\R$ and $H_1\in\R$, not both zero,
and the recurrence relation
\begin{equation*} \label{linrec}
H_n=AH_{n-1}+BH_{n-2},\quad\quad (n\ge2),
\end{equation*}
where the coefficients $A$ and $B$ are also real numbers. Moreover, let
\begin{equation}\label{Hh}
h_{n,k}=A^{n-2k}B^k {n-k\choose k}\left( H_1+\frac{k}{n-2k+1}AH_0\right),\quad\quad k=0,\dots,\left\lfloor \frac{n}{2}\right\rfloor.
\end{equation}
The sequences $\{H_n\}_n$ and $\{h_{n,k}\}_k$ are strongly linked since
\begin{equation*} \label{bin}
H_{n+1}=(n\bmod 2)\cdot B^{\frac{n+1}{2}}H_0+\sum_{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor}h_{n,k},\quad(n\ge0).
\end{equation*}
In what follows, for the natural numbers $n$ and $k$ we denote the
floor $\left\lfloor \frac{n}{k}\right\rfloor$ by $n_k$.

For instance,
if $A=B=H_1=1$ and $H_0=0$ then the Fibonacci sequence $\{F_n\}_{n=0}^\infty$,
and the well-known
$F_{n+1}=\sum_{k=0}^{n_2}{n-k \choose k}$ identity are obtained.

One of the purposes of this paper is to investigate the unimodality of the
sequence $\{h_{n,k}\}_k$. A finite sequence of real numbers
$\{a_k\}_{k=0}^m$ ($m\ge1$) is {\it unimodal} if there exists an integer
$l\in\{0,\dots,m\}$  such that the subsequence $\{a_k\}_{k=0}^l$
increases, while $\{a_k\}_{k=l}^m$ decreases. 
If $a_0\le a_1\le\dots\le a_{l_0-1}< a_{l_0}=\dots=a_{l_1}>a_{l_1+1}\ge\dots\ge a_m$ then the integers
$l_0,\dots,l_1$ are the {\it modes} of $\{a_k\}_{k=0}^m$. In case of $l_0=l_1$ we talk about a {\it peak}, otherwise the set of modes is called a {\it plateau}.
Naturally, these definitions can even be extended to
infinite sequences. For po\-sitive sequences, unimodality is implied by strict log-concavity.
A sequence $\{a_k\}_{k=0}^m$ is said to be {\it strictly log-concave} (SLC for short) if $a_l^2>a_{l-1}a_{l+1}$, $1\le l\le m-1$.

Theorems \ref{t:1}-\ref{t:3} discuss three classes of (\ref{Hh}) and
generalize most of the earlier results on unimodality of the sequences
related to binomial coefficients. Gene\-rally, $\{h_{n,k}\}_k$ is not
unimodal, not even if $A$ and $B$ are positive. For
example, let $A=B=1$, further $H_0=-8$, $H_1=5$. Now
\begin{equation*}
h_{8,k}=5,~27,~27,~-30,~-27,\quad(k=0,\dots,4)
\end{equation*}
is not unimodal.

The first result dealing with unimodality of the elements of the Pascal
triangle is due to Tanny \& Zuker \cite{TZ1}, who showed
that $n-k \choose k$ $(k=0,\dots,n_2)$ is unimodal. They
\cite{TZ1fel,TZ2} also investigated  the unimodality of $n-\alpha k
\choose k$. In this work we also treat certain cases of the binomial
sequence $n+\alpha k \choose \beta k$.

Benoumhani \cite{B} proved the unimodality of the
sequence $\frac{n}{n-k}{n-k \choose k}$ connected to Lucas
numbers. Recently, Belbachir and Bencherif 
%\cite{BB1},
\cite{BB2} proved that
the sequences $2^{n-2k}{n-k \choose k}$ and $2^{n-2k}\frac{n}{n-k}{n-k \choose k}$ linked to Pell sequence and its
companion sequence are unimodal. In all the aforesaid cases the
authors descibe the peaks and the plateaus with two elements, the
elements in which the monotonity changes.


\section{Results}

\begin{theorem} \label{t:1}
Suppose that $A$ and $B$ are given real numbers with $A>0$,
moreover $H_0=0$ and $H_1=1$. Assuming that $n\ge2$, the sequence
\begin{equation*}
h_{n,k}=A^{n-2k}B^k {n-k\choose k},\quad  (k=0,\dots, n_2)
\end{equation*} 
is unimodal if and only if $B\ge0$. In this case the peak $k=p_n$ of
$\{h_{n,k}\}$ satisfies
\begin{equation*}
p_n\in\left\{
\left\lfloor \frac{n}{2}\left(1-\frac{1}{\sqrt{4c+1}}\right)\right\rfloor,\left\lceil \frac{n}{2}\left(1-\frac{1}{\sqrt{4c+1}}\right) \right\rceil
\right\},
\end{equation*}
and the plateau with two elements may occur at the places $p_n$ and
$p_n+1$.
\end{theorem}


Obviously, $h_{n,k}$ cannot be unimodal with negative $B$, and
trivially unimodal when $B=0$. Considering $A$ and $B$ as two natural
numbers, a combinatorial interpretation of $A^{n-2k}B^k {n-k\choose k}
$ is the number of words formed with the letters $R,S_1,\dots,
S_A,T_1,\dots, T_B$ of length $n$, beginning vith $k$ consecutive $R$'s
and containing exactly $k$ letters choosing among $\{T_1,\dots,T_B\}$.

A similar theorem to Theorem \ref{t:1} is true for the companion sequence
of $H_n$.

\begin{theorem} \label{t:2}
Let $A>0$ and $B$ denote real numbers, $H_0=2$, $H_1=A$
and let $n\ge2$ be an integer. The sequence 
\begin{equation*}
h_{n-1,k}=A^{n-2k}B^k\frac{n}{n-k}{n-k\choose k},\quad (k=0,\dots ,n_2)
\end{equation*} 
is unimodal if and
only if $B\ge0$. The description of peaks and plateaus coincide as we have in
Theorem \ref{t:1}.

\end{theorem}

The choice $H_0=0$ and $H_1=1$ makes the formula (\ref{Hh}) as simple
as possible. The following theorem does not fix the initial values, but
only the coefficients.

\begin{theorem} \label{t:3}
Assume that the initial values $H_0$ and $H_1$ are positive,
$A=B=1$ and $n\ge2$. Under these conditions the sequence
\begin{equation*}
h_{n,k}={n-k\choose k}\left(H_1+\frac{k}{n-2k+1}H_0\right),\quad
(k=0,\dots ,n_2)
\end{equation*} 
is also unimodal.
\end{theorem}

Another direction is to investigate the unimodality of the sequence
$n+\alpha k \choose \beta k$, where $\alpha$ and $\beta\ge0$ are
integers. If either $\alpha$ or $\beta$ is zero then unimodality is
trivial, as well as with $(\alpha,\beta)=(1,1)$. The case
$(\alpha,\beta)=(-1,1)$ has been treated in \cite{TZ1}, while
$\alpha<-1$, $\beta=1$ in \cite{TZ1fel} and \cite{TZ2}.

The first pair worth considering is $(\alpha,\beta)=(-1,2)$, when
$h_{n,k}={n-k\choose 2k}$. Note that if
$a_n=\sum_{k=0}^{n_3}{n-k\choose 2k}$ then the terms of $\{a_n\}$
satisfy the recurrence relation $a_n=a_{n-1}+ a_{n-2}+a_{n-4}$
with the initial values $a_0=a_1=a_2=1$ and $a_3=2$ (Sloane: A005251).

\begin{theorem} \label{t:4}
The sequence $h_{n,k}={n-k\choose 2k}$  is unimodal ($k=0,\dots,
n_3$).
\end{theorem}

A more interesting pair is $(\alpha,\beta)=(1,2)$. Now
$F_{2n+1}=\sum_{k=0}^{n}{n+k\choose 2k}$ (Sloane: A001519). Note that the sequence $a_n=F_{2n+1}$
satisfies $a_n=3a_{n-1}-a_{n-2}$ $(n\ge2)$.

\begin{theorem} \label{t:5}
The sequence $h_{n,k}={n+k\choose 2k}$  is unimodal ($k=0,\dots,
n$). A plateau exists if and only if $n=(F_{12u+4}-1)/2$ or
$n=(F_{12u+8}-1)/2$ with $k=(L_{12u+4}-7)/10$ or
$k=(L_{12u+8}-7)/10$ ($u\in\N$), respectively.
\end{theorem}

\noindent For numerical examples, see the following two
tables. \vskip 0.2cm


{\renewcommand{\arraystretch}{1.5}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
  \hline 

  $u$ & $F_{12u+4}$ & $n$ & $L_{12u+4}$ & $k$ &   ${n+k\choose 2k}={n+k+1\choose 2k+2}$ 
  \\  \hline
  ${ 0 }$ & 3 & 1 & 7 & 0 & ${1\choose 0}={2\choose 2}$ \\ \hline
  1 & 987 & 493 & 2207 & 220 & ${713\choose 440}={714\choose 442}$
  \\ \hline
  2 & 317811 & 158905 & 710647 & 71064 & ${229969\choose 142128}={229970\choose 142130}$ \\ \hline
\end{tabular}
\end{center}
\vskip 0.2cm

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}
  \hline
  $u$ & $F_{12u+8}$ & $n$ & $L_{12u+8}$ & $k$ & ${n+k\choose 2k}={n+k+1\choose 2k+2}$
  \\ \hline
  0 & 21 & 10 & 47 & 4 & ${14\choose 8}={15\choose 10}$ \\ \hline
  1 & 6765 & 3382 & 15127 & 1512 & ${4894\choose 3024}={4895\choose 3026}$
  \\ \hline
  2 & 2178309 & 1089154 & 4870847 & 487084 & ${1576238\choose974168 }={1576239\choose 974170}$ \\ \hline
\end{tabular}
\end{center}
}


Finally, let us consider the case $(1,\beta)$. For instance, if
$\beta=3$ then the numbers $\sum_{k=0}^{n_2}{n+k\choose 3k}$ give every
third term of the Padovan sequence $\{p_n\}$ determined by
$p_n=p_{n-2}+p_{n-3}$ and $p_0=1$, $p_1=p_2=0$ (Sloane: A003522).
Theorem \ref{t:6} partially generalizes Theorem \ref{t:5}.


\begin{theorem} \label{t:6}
The sequence $h_{n,k}={n+k\choose \beta k}$  is unimodal ($2\le\beta\in\N$, $k=0,\dots,
n_{\beta-1}$).
\end{theorem}

The strict log-concavity is only utilized in the proof of Theorem \ref{t:6}, because it really simplifies the treatment. In the rest of the cases we start from the definition of unimodality, which is useful mainly where the peaks and plateaus are also aimed to determined.  Finally, we draft two conjectures on Pascal triangles.


\begin{conj}
Let ${n\choose k}$ be a fixed element of the Pascal triangle crossed by a ray.
The sequence of binomial coefficents located along this ray is unimodal.
\end{conj}

\begin{conj}
Take a generalized Pascal triangle
corresponded to the Tribonacci sequence given by $T_n=T_{n-1}+T_{n-2}+T_{n-3}$ ($n\ge3$) and $T_0=T_1=0$, $T_2=1$. This triangle contains the elements
\begin{equation*}
{n\choose k}_2=\sum_{i=m}^{{\left\lfloor
\frac{k}{2}\right\rfloor}}\frac{n!}{i!\cdot(k-2i)!\cdot(n-k+i)!}=
\sum_{i=m}^{{\left\lfloor \frac{k}{2}\right\rfloor}}{n\choose
n-i}{n-i\choose k-2i}
\end{equation*}
with $m=\max\{0\;,\;k-n\}$. We conjecture that the sequence ${n-k\choose k}_2$ is
unimodal.
\end{conj}


\section{Proofs}

{\bf Proof of Theorem \ref{t:1}.} There remains only the case $B>0$ to
consider. The inequality
\begin{equation*} \label{e11}
A^{n-2k}B^k {n-k\choose k}\le A^{n-2(k+1)}B^{k+1} {n-(k+1)\choose
k+1}, \quad\quad (k=0,\dots,{n_2}-1)
\end{equation*}
is equivalent to
\begin{equation}\label{e12}
0\le (4c+1)k^2-(4c+1)kn+cn^2+(2c+1)k-(c+1)n
\end{equation}
with $c=B/A^2>0$. Multiplying (\ref{e12}) by $4(4c+1)$ it can be rewritten as
\begin{equation}\label{e13}
(4c+1)(n+1)^2+4c^2\le \left((4c+1)(2k-n)+(2c+1)\right)^2.
\end{equation}
It is easy to check that $(4c+1)(2k-n)+(2c+1)<0$. Consequently, by
(\ref{e13}) we have
\begin{equation}\label{e14}
k\le t_n=\frac{(4c+1)n-(2c+1)-\sqrt{(4c+1)(n+1)^2+4c^2}}{ 2(4c+1)}.
\end{equation}
Put $t^\star_n=(n/2)(1-1/\sqrt{4c+1})$. From $-1<t_n-t^\star_n<0$ it follows that
the peak $p_n$ of $\{h_{n,k}\}$ is one of $\left\lfloor t^\star_n\right\rfloor$ and $\left\lceil t^\star_n \right\rceil$.
When $t_n$ is a natural number then $\{h_{n,k}\}$ possesses a plateau with two elements at $k=p_n$ and $k=p_n+1$.
Obviously, plateaus can be identified  by the positive integer solutions $x,y$ of the Pell equation $(4c+1)(x+1)^2+4c^2=y^2$.

\begin{rem}
{\rm Formula (\ref{e14}) returns with
\begin{equation*}%label{e14spec1}
k\le \frac{5n-3-\sqrt{5n^2+10n+9}}{10} \quad {\rm and} \quad k\le
\frac{4n-3-\sqrt{8n^2+16n+9}}{8}
\end{equation*}
in the particular case of Fibonacci and Pell sequences (see
\cite{TZ1} and \cite{BB2}), respectively.}
\end{rem}

\noindent{\bf Proof of Theorem \ref{t:2}.} Suppose that $B>0$ and follow the steps of the proof of Theorem \ref{t:1}. Thus we can transform  $h_{n,k}\le h_{n,k+1}$  into
\begin{equation}\label{e22}
0\le (4c+1)k^2-(4c+1)kn+cn^2+(2c+2)k-(c+1)n+1,
\end{equation}
($c=B/A^2$) and by a suitable multiplication of (\ref{e22}) we have
\begin{equation*}\label{e23}
(4c+1)n^2+4c(c-2)\le \left((4c+1)(2k-n)+(2c+2)\right)^2.
\end{equation*}
Hence
\begin{equation}\label{e24}
k\le \frac{(4c+1)n-(2c+2)-\sqrt{(4c+1)n^2+4c(c-2)}}{2(4c+1)}.
\end{equation}
Now we find again that $p_n$ is $\left\lfloor t^\star_n\right\rfloor$ or $\left\lceil t^\star_n \right\rceil$ with
the same $t^\star_n$ we had in Theorem \ref{t:1}. For plateaus the diophantine equation
$(4c+1)x^2+4c(c-2)=y^2$ should be investigated.


\begin{rem}
{\rm The inequality (\ref{e24}) provides formulae
\begin{equation*}\label{e14spec2}
k\le \frac{5n-4-\sqrt{5n^2-4}}{10} \quad {\rm and} \quad k\le
\frac{4n-5-\sqrt{8n^2-7}}{8}
\end{equation*}
in case of Lucas numbers and the companion sequence of Pell numbers (see
\cite{B} and \cite{BB2}), respectively.}
\end{rem}

\noindent{\bf Proof of Theorem \ref{t:3}}. Now $A=B=1$ and the
inequality $h_{n,k}\le h_{n,k+1}$ provides
\begin{eqnarray*} 
\lefteqn{(n-k)(k+1)\left(H_1(n-2k+1)+H_0k\right)\le}  \\
 & & (n-2k+1)(n-2k)\left(H_1(n-2k-1)+H_0(k+1)\right),  \\
\end{eqnarray*}
which is equivalent to $0\le f_n(k)=\sum_{i=0}^3e_i(n)k^i$, where
\begin{eqnarray*}
e_0(n)&=&n^3+(h-1)n^2+(h-2)n,\\
e_1(n)&=&(h-7)n^2+(-4h+2)n+(-2h+3),\\
e_2(n)&=&(-5h+15)n+(3h-1),\\
e_3(n)&=&5h-10,
\end{eqnarray*}
and $h=H_0/H_1>0$. The special case $h=2$ has essentially been
treated by Benoumhani \cite{B}. Therefore, by the sign of $e_3(n)$, we must
distinguish two cases.

Firstly, we assume that $h>2$, which entails positive leading coefficient in
the polynomial $f_n(k)$ of degree three. Since $f_n(0)=e_0(n)>0$
and $f_n(n/2)=-(hn^3+(2h+2)n^2+4n)/8<0$, it follows that $f_n(k)$
possesses exactly one zero in the interval $[0;n/2]$. Hence
$h_{n,k}$ is unimodal.

Now, the assertion $0<h<2$ implies $e_3(n)<0$. The reader
can easily verify that $f_n(0)=n(n^2+(h-1)n+(h-2))$ is positive if
$n\ge2$ and $f_n(n/2)<0$ as it has already been occurred previously. But the negative leading
coefficient $e_3(n)$ causes ambiguous structure for the zeros of
$f_n(k)$. Therefore a deeper analysis is necessary to clarify the
situation.

The polynomial
\begin{eqnarray*} 
\lefteqn{f_n^\prime(k)=(15h-30)k^2+\left((-10h+30)n+(6h-2)\right)k+}  \\
& & +\left((h-7)n^2+(-4h+2)n+(-2h+3)\right) \\
\end{eqnarray*}
has two distinct real zeros because its discriminant
\begin{equation*}
D=20(2h^2-3h+3)n^2+40(3h^2-5h+3)n+4(39h^2-111h+91)
\end{equation*}
is positive when $0<h<2$. Indeed, $2h^2-3h+3>0$ and the
discriminant
\begin{equation*}
D_1=-960(h-2)^2(11h^2-19h+19)
\end{equation*}
of $D$ takes negative values on the interval $]0,2[$.

 To finish the proof it is sufficient to show that the larger real
zero $f_2$ of $f^\prime_n(k)$ satisfies $n/2<f_2$, more exactly
%The left hand side of
%\begin{equation} \label{f1}
%0<f_1={(10h-30)n+(-6h+2)+\sqrt{D}\over 30h-60}<{n\over2}
%\end{equation}
%leads to
%\begin{equation}
%(-h+7)n^2+(4h-2)n+(2h-3)>0,
%\end{equation}
%which is fulfilled when $0<h<2$. The right hand side of (\ref{f1}) is equivalent to
%\begin{equation} \label{f11}
%(h^2-4h+4)n^2+(4h^2-12h+8)n+(8h^2-28h+24)>0.
%\end{equation}
%But all the coefficents of the powers of $n$ in (\ref{f11}) are positive if $h$ is in the given interval.
%Finally, we prove that
\begin{equation*} \label{f2}
f_2=\frac{(10h-30)n+(2-6h)-\sqrt{D}}{30h-60}>\frac{n}{2}.
\end{equation*}
To do this it is enough to see that $2-6h-5hn<\sqrt{D}$, which is trivially true when $h\ge1/3$. Contrary, if $h<1/3$,
it suffices to confirm
\begin{equation} \label{f21}
0<(-h+2)n^2+(-4h+4)n+(-8h+12).
\end{equation}
And (\ref{f21}) is fulfilled since its right hand side has no real
zero under the given conditions.
\bigskip

\noindent{\bf Proof of Theorem \ref{t:4}}. Starting with
${n-k\choose 2k}\le{n-k-1\choose 2k+2} $, we obtain
\begin{equation*}
0\le f(k)=-23k^3+(23n-21)k^2+(-9n^2+12n-4)k+(n^3-3n^2).
\end{equation*}
Since $f(0)=n^2(n-3)\ge0$ if $n\ge3$, further
$f(n/3)=-4n(n+3)(2n+3)/27<0$, it is sufficient to verify that
$f(k)$ is strictly monotone decreasing. And, really, if $n\ge3$
then $f^\prime(k)\ne0$ implies the required monotonity.
\bigskip

\noindent{\bf Proof of Theorem \ref{t:5}}. Assuming 
$h_{n,k}\le h_{n,k+1}$, it provides
\begin{equation*}
0\le f(k)=-5k^2-7k+(n^2+n-2).
\end{equation*}
Since $k=-0.7$ is the only solution of $f^\prime(k)=0$, it follows
that $f(k)$ is strictly decreasing in the interval
$[0;n-1]$. Further, the zeros of the polynomial $f(k)$ are
\begin{equation} \label{e51}
f_{1,2}=\frac{7\pm\sqrt{20n^2+20n+9}}{-10},
\end{equation}
the smaller $f_1$ is negative, the larger $f_2$ is approximately $0.45n$ since
\begin{equation*}
f\left(\frac{n}{\sqrt{5}}-1\right)\cdot  f\left(\frac{n}{\sqrt{5}}\right)=
-\frac{4}{5}(4+\sqrt{5})n^2-\frac{2}{5}(5+3\sqrt{5})n<0
\end{equation*}
implies $n/\sqrt{5}-1<f_2<n/\sqrt{5}$.
Thus $h_{n,k}$ is unimodal, and its smallest mode $k_n=\left\lceil f_2\right\rceil$ satisfies $\left\lfloor n/\sqrt{5}\right\rfloor\le k_n\le\left\lceil n/\sqrt{5}\right\rceil$.

Now we inspect the existence of plateaus. By (\ref{e51}), we
need to solve the Pell equation $20x^2+20x+9=y^2$. It is well
known that
\begin{equation*}
y^2-5(2x+1)^2=4
\end{equation*}
implies $y=L_{2v}$ and $2x+1=F_{2v}$. A Fibonacci number with even
suffix is odd if and only if 3 does not divide $v$. Put
$v=3u\pm1$. A simple verification shows that
$L_{6u\pm2}\equiv\pm7\;\;(\!\!\!\mod 10)$ depending on the parity
of $u$. Choosing the appropriate cases we obtain $f_2$ as positive
integer.

\begin{rem} {\rm
Another proof can be obtained by introducing the function 
$G_{n}\left( x\right) =\sum_{0\leq k\leq n}\binom{n+k}{2k}x^{k},$ for which 
\begin{equation*}
G_{n}\left( x\right) =\frac{1}{x^{n}\sqrt{x^{2}+4x}}\left( \left( 
\frac{x+\sqrt{x^{2}+4x}}{2}\right) ^{2n+1}-\left( \frac{x-\sqrt{x^{2}+4x}}{2}%
\right) ^{2n+1}\right) .
\end{equation*}
It suffices to see that 
\begin{equation*}
G_{n}\left( x\right) =\left( x+2\right)
G_{n-1}\left( x\right) -G_{n-2}\left( x\right) ,\ G_{0}\left( x\right) =1,\
G_{1}\left( x\right) =x+1.
\end{equation*}
We deduce that $G_{n}\left( x\right) $ admits $n$ distinct real zeros given
by 
\begin{equation*}
x_{k}=-2\left( 1+\cos \frac{2k\pi }{2n+1}\right) ,\ k=1,2,\ldots ,n,
\end{equation*}
and then the sequence $\binom{n+k}{2k}$ is SLC, which gives unimodality
with a peak or a plateau with two elements.

For $m\ge3$ let denote $\left\{ \alpha _{m}\right\} _{m\geq -1}$ and $\left\{ \beta _{m}\right\}
_{m\geq -1}$\ sequences defined by%
\[
\left\{ 
\begin{array}{l}
\left( \alpha _{-1},\alpha _{0},\alpha _{1},\alpha _{2}\right) =\left(
-11,-2,1,10\right) \ \text{and }\alpha _{m}=322\alpha _{m-2}-\alpha
_{m-4}+160, \\ 
\left( \beta _{-1},\beta _{0},\beta _{1},\beta _{2}\right) =\left(
4,0,0,4\right) \ \ \ \text{and }\beta _{m}=322\beta _{m-2}-\beta _{m-4}+224.%
\end{array}%
\right. 
\]


The integers $n$ for which the sequences $\binom{n+k}{2k},\ (k=0,1,\ldots ,n)$
admit a plateau with two elements $\left\{ k_{n},k_{n}+1\right\} $ are
exactly the integers $\alpha _{s},\ s\geq 1$ such that $k_{\alpha
_{s}}=\beta _{s}$ for $s\geq 1$.

\begin{center}
\begin{tabular}{|c|c|c|}
\hline
$s$ & $n=\alpha _{s}$ & $k_{n}=\beta _{s}$ \\ \hline
{\small 1} & {\small 1} & {\small 0} \\ \hline
{\small 2} & {\small 10} & {\small 4} \\ \hline
{\small 3} & {\small 493} & {\small 220} \\ \hline
{\small 4} & {\small 3382} & {\small 1512} \\ \hline
{\small 5} & {\small 158905} & {\small 71064} \\ \hline
{\small 6} & {\small 1089154} & {\small 487084} \\ \hline
\end{tabular}%
\end{center}
}
\end{rem}

\bigskip

\noindent{\bf Proof of Theorem \ref{t:6}} \ We show that $h_{n,k}$ is strictly log-concave, or more precisely
\begin{equation} \label{e61}
{n+k\choose \beta k}^2>{n+k-1\choose \beta k-\beta}{n+k+1\choose \beta k+\beta}.
\end{equation}
Put $b=\beta-1\ge1$. Thus (\ref{e61}) is equivalent to
\begin{equation*}
\frac{(n+k)(n-bk+1)}{(n+k+1)(n-bk)}\cdot{\prod_{i=0}^{\beta-1}\left(1+\frac{\beta}{\beta k-i}\right)}
\cdot{\prod_{j=1}^{b-1}\left(1+\frac{b+1}{n-bk-j}\right)}>1.
\end{equation*}
Since $(n+k)(n-bk+1)>(n+k+1)(n-bk)>0$, further the other multipliers are trivially greater than 1, the statement is proved.









\begin{thebibliography}{29}

%\bibitem{BB1} Belbachir, H. and Bencherif, F., Unimodality of
%sequences with binomial coefficients, preprint.

\bibitem{BB2} H.~Belbachir and F.~Bencherif, Unimodality of
sequences associated to Pell numbers, preprint.

\bibitem{B} M.~Benoumhani,
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Benoumhani/benoumhani8.html}{A sequence of binomial coefficients related to Lucas and Fibonacci numbers},
{\it J.~Integer Seq.}, {\bf 6} (2003), Article 03.2.1.

\bibitem{TZ1} S.~Tanny and M.~Zuker, On a unimodality sequence of binomial coefficients, {\it Discrete Math.}, {\bf 9} (1974), 79--89.

\bibitem{TZ1fel} S.~Tanny and M.~Zuker, On a unimodal sequence of binomial coefficients,
{\it J.~Combin.~Inform.~System Sci.}, {\bf 1} (1976), 81--91.

\bibitem{TZ2} S.~Tanny and M.~Zuker, Analytic methods applied to a sequence of binomial coefficients, {\it Discrete Math.}, {\bf 24} (1978), 299--310.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B65; Secondary 05A10, 11B39 .

\noindent \emph{Keywords: } binomial coefficients, unimodality, log-concavity, linear recurrences.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences 
\seqnum{A001519},
\seqnum{A003522}, and
\seqnum{A005251}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 27 2006;
revised version received  January 16 2007.
Published in {\it Journal of Integer Sequences}, January 17 2007.

\bigskip
\hrule
\bigskip

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


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