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\newtheorem{lemma}[theorem]{Lemma}
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\begin{document}

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\title{A SEQUENCE OF BINOMIAL COEFFICIENTS RELATED TO LUCAS AND FIBONACCI NUMBERS}

\maketitle

\centerline{Moussa Benoumhani}

\begin{center}
Mathematical Department \\
Sana'a University\\
P. O. Box 14026 \\
Sana'a \\
Yemen\\
E-mail: benoumhani@yahoo.com \\
\end{center}

\date{}
\maketitle

\begin{abstract}
Let $L(n,k)={n \over {n-k}} {{n-k}\choose k} $.
We prove that all the zeros of the polynomial 
$L_n(x)=\sum\limits_{k\geq 0}L(n,k)x^k$ 
are real.
The sequence $L(n,k)$ is thus strictly log-concave, and
hence unimodal with at most two consecutive maxima. We determine those
integers where the maximum is reached. In the last section we prove that 
$L(n,k)$ satisfies a central limit theorem as well as a local limit theorem.
\end{abstract}

\section{Introduction}

A positive real sequence $(a_k)_{k=0}^n$ is said to be {\it unimodal} if there
exist integers $k_0,k_1,0\leq k_0\leq k_1\leq n$ such that 
\begin{equation*}
a_0\leq \,a_1\leq \cdots \leq a_{k_0}=a_{k_0+1}= \cdots =a_{k_1}\geq a_{k_1+1}\geq
\cdots \geq a_n.
\end{equation*}
The integers $l,$ $k_0\leq l\leq k_1$ are called the {\it modes} of the sequence.
If $k_0&lt;k_1$ then $(a_k)_{k=0}^n$ is said to have a {\it plateau} of $%
k_1-k_0+1 $ elements; if $k_0=k_1$ then it is said to have a {\it peak}. A real
sequence is said to be {\it logarithmically concave} (log-concave for short) if 
\begin{equation*}
a_k^2\geq a_{k-1}a_{k+1},\text{\thinspace \thinspace \thinspace \thinspace
\thinspace }1\leq k\leq n-1\text{ \qquad }(1)
\end{equation*}
If the inequalities in (1) are strict, then $(a_k)_{k=0}^n$ is said to be
{\it strictly log-concave} (SLC for short). A sequence is said to be have {\it no
internal zeros} if $i&lt;j\,\,$, $a_i\neq 0$ and $a_j\neq 0$, then $a_k\neq 0$
for $i\leq k\leq j$. A log-concave sequence with no internal zeros is
obviously unimodal, and if it is SLC, then it has at most two consecutive
modes. The following result is sometimes useful in proving log-concavity.
For a proof of this theorem, see Hardy and Littlewood \cite{5}.

\begin{theorem}
(I. Newton) Let $(a_k)_{k=0}^n$ be a real sequence. Assume that the
polynomial $P(x)=\sum\limits_{k=0}^na_kx^k$ has only real zeros. Then 
\begin{equation*}
a_k^2\geq \frac{n-k+1}{n-k}\cdot \frac{k+1}ka_{k+1}a_{k-1},\,\,\,1\leq k\leq
n-1.\,\,\,\,\,\,\,\,\,\,\,\,(2)
\end{equation*}
\end{theorem}

\vfill\eject

If the sequence $(a_k)_{k=0}^n$ is positive and satisfies the hypothesis of
the previous theorem, then it is SLC. The two possible values of the modes
are given by the next theorem.

\begin{theorem}
Let $(a_k)_{k=0}^n$ be a real sequence satisfying the hypothesis of the
previous theorem. Then every mode of the sequence $(a_k)_{k=0}^n$ satisfies 
\begin{equation*}
\left\lfloor \frac{\sum\limits_{k=1}^nka_k}{\sum\limits_{k=0}^na_k}%
\right\rfloor \leq k_0\leq \left\lceil \frac{\sum\limits_{k=0}^nka_k}{%
\sum\limits_{k=0}^na_k}\right\rceil ,
\end{equation*}
where $\left\lfloor x\right\rfloor $ and $\left\lceil x\right\rceil $ are
respectively the floor and the ceiling of $x$.
\end{theorem}

For a proof of this theorem, see Benoumhani \cite{2,3}.

Let $g(n,k)= {{n-k} \choose k} $.   This sequence was
been investigated by S. Tanny and M. Zuker \cite{8}; they proved that it is SLC,
and determined its modes. If $r_n$ is the smallest mode of $g(n,k)$, then 
\begin{equation*}
r_n=\left\lceil \frac{5n-3-\sqrt{5n^2+10n+9}}{10}\right\rceil
\,\,\,.\,\,\,\,\,\,\,\,\,\,(3)
\end{equation*}
They proved that there are infinitely many integers where a double maximum
occurs. The integers where this happen are given by: $n_j=F_{4j}-1$, where $%
F_{k\text{ }}$ is the $k^{th}$ Fibonacci number. The smallest mode
corresponding to $n_j$ is given by $r_j=\frac 15\left( L_{4j-1}-4\right) $,
where $L_j$ is the $j$'th Lucas number.

In this paper we consider the sequence $L(n,k)=\frac n{n-k} {{n-k} \choose k}$,
$0\leq k\leq \left\lfloor \frac
n2\right\rfloor$, $n\geq 1$. It is known that $L(n,k)$ counts the number of
ways of choosing $k$ points, no two consecutive, from a collection of $n$
points arranged in a circle; see Stanley \cite[p.\ 73, Lemma 2.3.4]{7} and Sloane
\cite[A034807]{6}.

In Section 2, for the sake of completeness, we prove that all
zeros of the polynomials $P_n(x)=\sum\limits_{k\geq 0}g(n,k)x^k$ are real.
The explicit formula for $P_n(x)$ allows us to derive some identities. Also
it enables us to rediscover a result of S. Tanny and M. Zuker. In the third
section, we consider the polynomials $L_n(x)=\sum\limits_{k\geq 0}L(n,k)x^k.$
We prove that all zeros of $L_n(x)$ are real and negative. In this case, too,
the explicit formula for $L_n(x)$ gives some identities. The SLC of the
sequence $L(n,k)$ is deduced from the fact that $L_n(x)$ has real zeros. We
determine the modes, and the integers $n$ where $L(n,k)$ has a double
maximum. In the last section we prove that the sequence $L(n,k)$ is
asymptotically normal, and satisfies a local limit theorem on $R$.

\section{The polynomials $P_n(x)$}

It is well known that the sequence $g(n,k)=
{{n-k} \choose k}$,
$0\leq k\leq \left\lfloor \frac n2\right\rfloor $, is related
to the Fibonacci numbers by the relation 
$\sum\limits_{k\geq 0}
{{n-k} \choose k}
=F_{n+1}$ .
Recall that the sequence $(F_n)$ is
defined as follows: 
\begin{equation*}
F_n=F_{n-1}+F_{n-2},\,n\geq 2,
\end{equation*}
with $F_0=0,\,F_1=1$. Also we have the explicit formula
$$ F_n=\frac 1{\sqrt{%
5}}\left( \left( \frac{1+\sqrt{5}}2\right) ^n-\left( \frac{1-\sqrt{5}}%
2\right) ^n\right) .$$

It is straightforward to see that $P_n(x)$ satisfies the recursion
\begin{equation*}
P_n(x)=P_{n-1}(x)+xP_{n-2}(x),\text{ \thinspace \thinspace \thinspace
\thinspace \thinspace \thinspace \thinspace \thinspace \thinspace \thinspace
\thinspace \thinspace \thinspace \thinspace \thinspace \thinspace \thinspace
\thinspace \thinspace \thinspace \thinspace }(4)
\end{equation*}
with initial conditions $P_0(x)=P_1(x)=1.$ Using the relation (4) we prove

\begin{proposition}
For all $n\geq 0,$ all zeros of
the polynomials $P_n(x)$ are real. More
precisely, we have
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;P_n(x)=\frac 1{\sqrt{4x+1}}\left( \left( 
\frac{1+\sqrt{4x+1}}2\right) ^{n+1}-\left( \frac{1-\sqrt{4x+1}}2\right)
^{n+1}\right) .\,\,\,\,\,\,\,\,(5)$
\end{proposition}

%TCIMACRO{\TeXButton{Proof}{\proof }}
%BeginExpansion
\proof %
%EndExpansion
Write the relation (4) in matrix form, as follows:
$\left( 
\begin{array}{l}
P_n(x) \\ 
P_{n-1}(x)
\end{array}
\right) =\left( 
\begin{array}{ll}
1 &amp; x \\ 
1 &amp; 0
\end{array}
\right) \left( 
\begin{array}{l}
P_{n-1}(x) \\ 
P_{n-2}(x)
\end{array}
\right) $.
We deduce 
\begin{equation*}
\left( 
\begin{array}{l}
P_n(x) \\ 
P_{n-1}(x)
\end{array}
\right) =\left( 
\begin{array}{ll}
1 &amp; x \\ 
1 &amp; 0
\end{array}
\right) ^{n-1}\left( 
\begin{array}{l}
P_1(x) \\ 
P_0(x)
\end{array}
\right) =\left( 
\begin{array}{ll}
1 &amp; x \\ 
1 &amp; 0
\end{array}
\right) ^{n-1}\left( 
\begin{array}{l}
1 \\ 
1
\end{array}
\right) .
\end{equation*}
The eigenvalues of the matrix $A=\left( 
\begin{array}{ll}
1 &amp; x \\ 
1 &amp; 0
\end{array}
\right) $ are 
\begin{equation*}
\lambda _1=\frac{1+\sqrt{4x+1}}2\text{ },\text{ }\lambda _2=\frac{1-\sqrt{%
4x+1}}2,
\end{equation*}
and two eigenvectors of $A$ are $V_1=$ $\left( 
\begin{array}{l}
\lambda _1 \\ 
1
\end{array}
\right) $and $V_2=$ $\left( 
\begin{array}{l}
\lambda _2 \\ 
1
\end{array}
\right) $. Now the matrix $A$ may be written 
\begin{equation*}
\left( 
\begin{array}{ll}
1 &amp; x \\ 
1 &amp; 0
\end{array}
\right) =\left( 
\begin{array}{ll}
\lambda _1 &amp; \lambda _2 \\ 
1 &amp; 1
\end{array}
\right) \left( 
\begin{array}{ll}
\lambda _1 &amp; 0 \\ 
0 &amp; \lambda _2
\end{array}
\right) \left( 
\begin{array}{ll}
\lambda _1 &amp; \lambda _2 \\ 
1 &amp; 1
\end{array}
\right) ^{-1}.
\end{equation*}
From this, we obtain 
\begin{eqnarray*}
\left( 
\begin{array}{ll}
1 &amp; x \\ 
1 &amp; 0
\end{array}
\right) ^{n-1} &amp;=&amp;\frac 1{\lambda _1-\lambda _2}\left( 
\begin{array}{ll}
\lambda _1 &amp; \lambda _2 \\ 
1 &amp; 1
\end{array}
\right) \left( 
\begin{array}{ll}
\lambda _1^{n-1} &amp; 0 \\ 
0 &amp; \lambda _2^{n-1}
\end{array}
\right) \left( 
\begin{array}{ll}
1 &amp; -\lambda _2 \\ 
-1 &amp; \lambda _1
\end{array}
\right) \\
&amp;=&amp;\frac 1{\lambda _1-\lambda _2}\left( 
\begin{array}{ll}
\lambda _1^n-\lambda _2^n &amp; -\lambda _1^n\lambda _2+\lambda _1\lambda _2^n
\\ 
\lambda _1^{n-1}-\lambda _2^{n-1} &amp; -\lambda _1^{n-1}\lambda _2+\lambda
_1\lambda _2^{n-1}
\end{array}
\right) .
\end{eqnarray*}

The vector $\left( 
\begin{array}{l}
P_n(x) \\ 
P_{n-1}(x)
\end{array}
\right) $ is now 
\begin{equation*}
\left( 
\begin{array}{l}
P_n(x) \\ 
P_{n-1}(x)
\end{array}
\right) =\frac 1{\lambda _1-\lambda _2}\left( 
\begin{array}{ll}
\lambda _1^n-\lambda _2^n &amp; -\lambda _1^n\lambda _2+\lambda _1\lambda _2^n
\\ 
\lambda _1^{n-1}-\lambda _2^{n-1} &amp; -\lambda _1^{n-1}\lambda _2+\lambda
_1\lambda _2^{n-1}
\end{array}
\right) \left( 
\begin{array}{l}
1 \\ 
1
\end{array}
\right) .
\end{equation*}

So, 
\begin{equation*}
P_n(x)=\frac 1{\lambda _1-\lambda _2}\left( \lambda _1^n-\lambda
_2^n-\lambda _1^n\lambda _2+\lambda _1\lambda _2^n\right) .
\end{equation*}
Since $\lambda _1+\lambda _2=1$ and $\lambda _1-\lambda _2=\sqrt{4x+1},$ we
finally obtain
$$P_n(x)=\frac 1{\sqrt{%
4x+1}}\left( \left( \frac{1+\sqrt{4x+1}}2\right) ^{n+1}-\left( \frac{1-\sqrt{%
4x+1}}2\right) ^{n+1}\right) .$$
This is the desired result.

For the roots of $P_n(x),$ we have
$$P_n(x)=0\Longleftrightarrow \left( \frac{1+\sqrt{4x+1}}{1-\sqrt{%
4x+1}}\right) ^{n+1}=1\Longleftrightarrow \left( \frac{1+\sqrt{4x+1}}{1-%
\sqrt{4x+1}}\right) =\varepsilon _k,1\leq k\leq \left\lfloor \frac
n2\right\rfloor \;$$
where the $\varepsilon _k$ are the $(n+1)^{th}$ roots of unity. Thus, 
\begin{equation*}
P_n(x)=0\Longleftrightarrow \sqrt{4x+1}=\frac{\varepsilon _k-1}{\varepsilon
_k+1}\Longleftrightarrow 4x=-1+\left( \frac{\varepsilon _k-1}{\varepsilon
_k+1}\right) ^2.
\end{equation*}
Furthermore, we obtain $P_n(x)=0\Longleftrightarrow x=-\frac 14\left( 1+\tan
^2\left( \frac{k\pi }{n+1}\right) \right) ,1\leq k\leq \left\lfloor \frac
n2\right\rfloor .$ This proves that the roots of $P_n(x)$ are real and
negative.%
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

\noindent{\bf Remark.}
In the sequel, we need Lucas numbers.  Let us recall their definition: 
\begin{equation*}
L_n=L_{n-1}+L_{n-2},\;\;\;L_0=2,\,L_1=1.
\end{equation*}
It is not hard to see that
$$L_n=\left( \frac{1+\sqrt{5}}2\right) ^n+\left( \frac{1-\sqrt{5}}2\right)
^n\,\,\, \text{and} \thinspace \thinspace \thinspace \thinspace \thinspace
\thinspace L_n=F_n+F_{n-2},\;$$
holds.

\begin{corollary}
We have the following identities:

\medskip

$1.\,\sum\limits_{n\geq 0}P_n(x)z^n=\frac 1{1-z-xz^2}.$

\medskip

$2.\,\sum\limits_{k\geq 0}(-1)^k
{{n-k} \choose k}
=\left\{ 
\begin{array}{c}
0,\;{\text\rm if}\;n=6k+2,\;6k+5; \\ 
1,\;{\text\rm if}\;n=6k,\;6k+1;\;\;\;\; \\ 
-1,\;{\text \rm if}\;n=6k+3,\;6k+4.
\end{array}
\right. $

\medskip

$3.\,\sum\limits_{k\geq 0}k
{{n-k} \choose k}
=\sum\limits_{k=0}^{n-2}F_kF_{n-k-2}=\frac{(n+1)L_n-2F_n}5=\frac{%
(n-1)F_n+(n+1)F_{n-2}}5.$

\medskip

$4.\,(n+1)L_n-2F_n=(n-1)F_n+(n+1)F_{n-2}\equiv 0$ {\rm (mod} $5${\rm )}.

\bigskip

$5.\,\sum\limits_{k\geq 0}(-1)^k k
{{n-k} \choose k}
=\left\{ 
\begin{array}{c}
\frac 23n,\;\;\;\;\;\text{\rm if}\;n=6k; \\ 
\frac{n-1}3,\;\text{\rm if}\;n=6k+1; \\ 
-\frac{n+1}3,\;\text{\rm if}\;n=6k+2; \\ 
-\frac{2n}3,\;\text{\rm if}\;n=6k+3; \\ 
-\frac{(n-1)}3,\;\text{\rm if}\;n=6k+4; \\ 
\frac{n+1}3,\;\text{\rm if}\;n=6k+5.\;\;
\end{array}
\right. $
\end{corollary}

%TCIMACRO{\TeXButton{Proof}{\proof }}
%BeginExpansion
\proof %
%EndExpansion
The first is known and easy to establish using $(4).$ For $(2),$ put $x=-1$
in $(5).$ For the third, differentiate the generating function of $P_n(x)$
with respect to $x$, and compare the coefficients, and then put $x=1.$
Relation $4$ is immediate from $3.$ For the last one, put $x=-1$ in the
derivative of $P_n(x).$%
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

According to Theorem 2, every mode $r_n$ of the sequence 
${{n-k} \choose k}$
satisfies the relation
\begin{equation*}
\left\lfloor \frac{\sum\limits_{k=1}^n k
{{n-k} \choose k}
}{F_n}\right\rfloor \leq r_n\leq \left\lceil \frac{%
\sum\limits_{k=0}^n k 
{{n-k} \choose k}
}{F_n}%
\right\rceil .
\end{equation*}

S. Tanny and M. Zuker gave an exact formula for $r_n$, but this is somewhat
opaque. So they used another method to give a more explicit one; but it is
less precise. Namely, they proved that
$r_n=\left\lfloor \frac n2\left( 1-\frac{\sqrt{5}}%
5\right) \right\rfloor $\ or$\;r_n=\left\lceil \frac n2\left( 1-\frac{\sqrt{5%
}}5\right) \right\rceil $. We give another proof of this result.\newline

\begin{proposition}
(S. Tanny, M. Zuker \cite{8})

The modes of the sequences 
${{n-k} \choose k}$
are
given by $\;r_n=\left\lfloor \frac n2\left( 1-\frac{\sqrt{5}}5\right)
\right\rfloor $ or\ $r_n=\left\lceil \frac n2\left( 1-\frac{\sqrt{5}}%
5\right) \right\rceil .$
\end{proposition}

%TCIMACRO{\TeXButton{Proof}{\proof }}
%BeginExpansion
\proof %
%EndExpansion
Since all zeros of the polynomial $P_n(x)$ are real, it suffices to compute $%
\frac{\sum\limits_{k=1}^n k {{n-k} \choose k}
}
{% \sum\limits_{k\geq 0} {{n-k} \choose k}
}=\frac{%
\sum\limits_{k=1}^n k {{n-k} \choose k}}
{F_n}\,.$
The last corollary gives 
\begin{equation*}
\mu _n=\frac{\sum\limits_{k=1}^n k {{n-k} \choose k}
}{F_n}=\frac{(n+1)L_n-2F_n}{5F_n}=\frac{(n+1)L_n}{5F_n}-\frac 25.
\end{equation*}
Using the explicit formula for the Lucas and Fibonacci numbers; we obtain

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mu _n=\frac{(n+1)}2\left(
1-\frac{\sqrt{5}}5\right) \frac{1+a^n}{1-a^{n+1}},\;a=-\frac{3-\sqrt{5}}2.$

Now consider the sequence$\;$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\mu _n=\frac{(n+1)}2\left( 1-\frac{\sqrt{5}}%
5\right) \frac{1+a^n}{1-a^{n+1}}-\frac 25=\frac{(n+1)}2\left( 1-\frac{\sqrt{5%
}}5\right) A_n-\frac 25,$

where 
\begin{equation*}
A_n=\frac{1+a^n}{1-a^{n+1}}.
\end{equation*}
Also, observe that for every $n$ we have 
\begin{equation*}
A_{2n+1}&lt;1&lt;A_{2n}.
\end{equation*}
So$\;\;\;\;\;$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,%
\,\,\,\,\mu _{2n}=\frac{2n+1}2\left( 1-\frac{\sqrt{5}}5\right) A_{2n}-\frac
25\geq \frac{2n+1}2\left( 1-\frac{\sqrt{5}}5\right) -\frac 25,\;$

and$\;$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mu _{2n+1}=\frac{2n+2}2\left( 1-%
\frac{\sqrt{5}}5\right) A_{2n+1}-\frac 25\leq \frac{2n+2}2\left( 1-\frac{%
\sqrt{5}}5\right) -\frac 25.$

Thus

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\frac{2n+1}2\left( 1-\frac{\sqrt{5}}%
5\right) -\frac 25\leq \mu _{2n}\leq \mu _{2n+1}\leq \frac{2n+2}2\left( 1-%
\frac{\sqrt{5}}5\right) -\frac 25.$

We deduce that for every $n\geq 2,$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\frac n2\left( 1-\frac{\sqrt{5}}%
5\right) -\frac 25\leq \mu _n\leq \frac{n+2}2\left( 1-\frac{\sqrt{5}}%
5\right) -\frac 25.$

Since the difference between the two bounds is $\left( 1-\frac{\sqrt{5}}%
5\right) &lt;1;$ there is a unique integer $r_n$ in the interval $\left( \frac
n2\left( 1-\frac{\sqrt{5}}5\right) -\frac 25,\frac{n+2}2\left( 1-\frac{\sqrt{%
5}}5\right) -\frac 25\right) $ and of course \ \ $r_n=\left\lfloor \frac
n2\left( 1-\frac{\sqrt{5}}5\right) \right\rfloor $ or $r_n=\left\lceil \frac
n2\left( 1-\frac{\sqrt{5}}5\right) \right\rceil .$ 
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

\section{The polynomials $L_n(x)$}

In this section, we consider the sequence $L(n,k)=\frac n{n-k}
{{n-k} \choose k}
$. 
We prove that all zeros of the polynomials $%
L_n(x)=\sum\limits_{k\geq 0}L(n,k)x^k$ are real.

\begin{proposition}
For all $n\geq 2,$ all zeros of the polynomials $L_n(x)$ are real.
We have \thinspace $\,\,\,\,L_n(x)=\left( \frac{1+\sqrt{4x+1}}2\right)
^n+\left( \frac{1-\sqrt{4x+1}}2\right) ^n.\;\;\;\;(6)$.
\end{proposition}

\textit{Proof. }Since the polynomials satisfy the recursion 
\begin{equation*}
L_n(x)=L_{n-1}(x)+xL_{n-2}(x);
\end{equation*}
\thinspace with $L_0=2,$ $L_1=1,$\thinspace \thinspace the proof is exactly
the same as for $P_n(x)$ .%
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

\begin{corollary}
We have the following identities:

$1.\,\sum\limits_{n\geq 0}L_n(x)z^n=\frac{2-z}{1-z-xz^2}.$

\medskip

$2.\,\sum\limits_{k\geq 0}\frac n{n-k} {{n-k} \choose k}
=L_n.$

\medskip

$3.\,\sum\limits_{k\geq 0}(-1)^k\frac n{n-k} {{n-k} \choose k}
=\left\{ 
\begin{array}{c}
1,\,{\text\rm if}\,\;n=6k+1\text{ or }\;6k+5;\,\,\,\,\,\, \\ 
-1,\,\,{\text\rm if}\,\,\,\,n=6k+2\text{ or }\;6k+4;\,\,\,\,\,\,\,\,\,\, \\ 
\,\,\,2,\,\,\,{\text\rm if}\,\,\,\,n=6k\,;\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,%
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \\ 
-2,\;{\text\rm if}\;\;n=6k+3.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,%
\,\,\,\,\,\,\,\,\,\,
\end{array}
\right. $

$4.\,\sum\limits_{k=0}^{n-1}L_kF_{n-k-1}=nF_n.$
\end{corollary}

%TCIMACRO{\TeXButton{Proof}{\proof }}
%BeginExpansion
\proof %
%EndExpansion
Relation (1) is immediate, for the second one, it suffices to put $x=1$ in $%
(6).$ For the third one, put $x=-1$ again in (6). The last one is obtained
by differentiating the generating function of $L_n(x)\;$with respect to $x\;$%
and then equating the coefficients of $z^n\;$in both sides. 
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

Since all zeros of the polynomials $L_n(x)$ are real, it follows that the
sequence $L(n,k)$ is SLC. We follow S. Tanny and M. Zuker to give the modes.

\begin{theorem}
The smallest mode of the sequence $L(n,k)$ is given by 
\begin{equation*}
k_n=\left\lceil \frac{5n-4-\sqrt{5n^2-4}}{10}\right\rceil .
\end{equation*}
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof }}
%BeginExpansion
\proof %
%EndExpansion
The integer $k_n$ satisfies 
\begin{equation*}
\left\{ 
\begin{array}{l}
L(n,k_n-1)&lt;L(n,k_n)\text{ \qquad }(a) \\ 
L(n,k_n)\geq L(n,k_n+1)\text{ \qquad }(b)
\end{array}
\right.
\end{equation*}
\thinspace \thinspace \thinspace Let\thinspace \thinspace \thinspace
\thinspace \thinspace 
\begin{equation*}
f(x)=5x^2-(5n+6)x+n^2+3n+2,
\end{equation*}
and 
\begin{equation*}
g(x)=5x^2-(5n-4)x+n^2+2n+1.
\end{equation*}
We have 
\begin{eqnarray*}
(a) &amp;\Longleftrightarrow &amp;f(k_n)&gt;0; \\
(b) &amp;\Longleftrightarrow &amp;g(k_n)\leq 0.
\end{eqnarray*}
The roots of the first equation are $\frac{5n+6\pm \sqrt{5n^2-4}}{10}$, and
those of the second one are $\frac{5n-4\pm \sqrt{5n^2-4}}{10}.\,$ The
desired integer satisfies 
\begin{equation*}
\frac{5n-4-\sqrt{5n^2-4}}{10}\leq k_n&lt;\frac{5n+6-\sqrt{5n^2-4}}{10}.
\end{equation*}
Which is what we wanted.%
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

The previous formula for $k_n$ is not as explicit as expected. We give a
more explicit one.

\begin{corollary}
The integer $k_n$ satisfies the following

\ \ \ \ \ \ \ \ $k_n=\left\lfloor \frac n2\left( 1-\frac{\sqrt{5}}5\right)
\right\rfloor $ or $k_n=\left\lceil \frac n2\left( 1-\frac{\sqrt{5}}5\right)
\right\rceil .$
\end{corollary}

%TCIMACRO{\TeXButton{Proof}{\proof } }
%BeginExpansion
\proof %
%EndExpansion
The proof is the same as for $r_n$ .%
%TCIMACRO{\TeXButton{End Proof}{\endproof }}
%BeginExpansion
\endproof %
%EndExpansion

In the next result, the integers $n,$ such that the sequence $L(n,k)$ has a
double maximum will be determined. Before determining these integers, we
need the following lemmas:

\begin{lemma}
For every $n\geq 0,$ $5F_n^2+4(-1)^n=L_n^2.$
\end{lemma}

%TCIMACRO{\TeXButton{Proof}{\proof}}
%BeginExpansion
\proof%
%EndExpansion
This is known, and straightforward using the explicit formulas of $F_n$ and $%
L_n$ .%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion

\begin{lemma}
For every $n\geq 0,$ $5F_{4n+1}-L_{4n+1}-4\equiv 0$ {\rm (mod 10)}.
\end{lemma}

%TCIMACRO{\TeXButton{Proof}{\proof}}
%BeginExpansion
\proof%
%EndExpansion
Again, the explicit formulas of $F_n$ and $L_n$ give easily the wanted
result.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion

\begin{theorem}
The sequence $L(n,k)$ has a double maximum if and only if $n=F_{4j+1},$ and
in this case the smallest mode is given by $k_n=F_{2j}^2.$
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof }}
%BeginExpansion
\proof %
%EndExpansion
If $l$ is the smallest mode of $L(n,k)$ then it satisfies 
\begin{equation*}
L(n,l)=L(n,\,l+1),
\end{equation*}
which is equivalent to 
\begin{equation*}
f(n,l)=5l^2-(5n-4)l+n^2-2n+1=0\,.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,%
\,\,\,\,\,\,\,\,\,(7)
\end{equation*}
Equation (7) has two roots in $l$%
\begin{equation*}
l_{1,\,2}=\frac{5n-4\pm \sqrt{5n^2-4}}{10}.
\end{equation*}
The solution greater than $\frac n2$ is rejected, since the modes of $L(n,k)$
are less than $\frac n2$. The smallest one remains, i.e., 
\begin{equation*}
l=\frac{5n-4-\sqrt{5n^2-4}}{10}\,\,.\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,%
\,\,\,\,\,\,\,\,\,\,\,\,\,(8)
\end{equation*}
So, we are looking for all pairs of integers $(n_j,k_j),\,0\leq k_j\leq 
\frac{n_j}2\,$, satisfying (7) (or (8)). We may transform (8) to an equation
related to Pell's equation as in Tanny and Zuker \cite{8}, and then use some
classical facts about units (invertible elements) in quadratic fields (see
Cohn \cite{4} for details). But we proceed differently: by Lemma $10$, $%
5F_{2j+1}^2-4=L_{2n+1}^2$, and by Lemma $11$,
$5F_{4j+1}-4-\sqrt{5F_{4j+1}^2-4 }\equiv 5F_{4j+1}-4-L_{4j+1}\equiv 0$ (mod $10$),
that is, $k_j=\frac{55F_{4j+1}-4\pm \sqrt{5F_{4j+1}^2-4}}{10}=\frac{5F_{4j+1}-4-L_{4j+1}}{10}
=F_{2j}^2\leq \frac{F_{4j+1}}2.$ So, some of the Fibonacci numbers are
certainly among the $n_j$. Now let $(n_0,k_0)=(1,0),\,(n_1,k_1)=(5,1),%
\,(n_2,k_2)=(34,9),\,(n_3,k_3)=(233,64),\,...,$with $n_j=F_{4j+1},$ $k_j=$ $%
F_{2j}^2$. The following recursions are easily derived: 
\begin{equation*}
\left\{ 
\begin{array}{l}
n_{j+1}=7n_j-n_{j-1}; \\ 
k_{j+1}=7k_j-k_{j-1}+2\,.
\end{array}
\right.
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,%
\,\,\,\,\,\,\,\,\,\,(9)
\end{equation*}

Now, we prove that all solutions of (7) are in fact $(n_j=F_{4j+1},$ $k_j=$ $%
F_{2j}^2)_{j\geq 0}$. We will show that if $(n_j,k_j)$ is a solution of (7),
then 
\begin{equation*}
(n_{j+1},k_{j+1})=(7n_j-n_{j-1},7k_j-k_{j-1}+2)
\end{equation*}
is another one. Indeed 
\begin{eqnarray*}
f(n_{j+1},k_{j+1}) &amp;=&amp;5k_{j+1}^2-(5n_{j+1}-4)k_{j+1}+n_{j+1}^2-2n_{j+1}+1 \\
&amp;=&amp;5(7k_j-k_{j-1}+2)^2-(5(7n_j-n_{j-1})-4)(7k_j-k_{j-1}+2) \\
&amp;&amp;+(7n_j-n_{j-1})^2-2(7n_j-n_{j-1})+1 \\
&amp;=&amp;0
\end{eqnarray*}
since $f(n_i,k_i)=5k_i^2-(5n_i-4)k_i+n_i^2-2n_i+1=0$ for $0\leq i\leq j.$
Suppose that $(n,k)$ is another one, $\,0\leq k\leq \frac n2;$ different
from those $(n_j,k_j)$. There is a unique $(n_i,k_i)$ such that $%
n_i&lt;n&lt;n_{i+1}.$ We verify easily that $f(7n-n_{i-1},7k-k_{i-1}+2)=0.$ This
means that $(n,k)=(n_i,k_i),$ and proves that all the solutions of (7) are
given by the recursions (9).This ends the proof.%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion

\noindent{\bf Remarks.}
1. There is a relation between the modes of the sequence $g(n,k)$ and those
of $L(n,k)$. Let $(m_j,r_j)$ be the sequence of integers such that $%
g(m_j,r_j)=g(m_j,r_j+1).$ Since $m_j=F_{4j}-1,$and $r_j=\frac 15(L_{4j-1}-4),
$ it is easy to establish (by direct calculations,
or generating functions of $r_j$ ), that
\begin{equation*}
\,\,\left\{ 
\begin{array}{l}
n_j=r_{j+1}-r_j; \\ 
k_j=m_j-2r_j-1.
\end{array}
\right. 
\end{equation*}

2. Note that our relation for $k_j$ was derived by
S. Tanny and M. Zuker \cite[p.\ 301]{9}.
There, the initial conditions for the Fibonacci numbers are: $%
F_0=\,F_1=1.$

3. Using the recursions $(9)$, we obtain the generating functions: 
\begin{equation*}
g(x)=\sum\limits_{j=0}^\infty n_jx^j=\frac{1-2x}{1-7x+x^2}\text{ \thinspace
and}\,\,\,\,\,\,h(x)=\sum\limits_{j=1}^\infty k_jx^j=\frac{x+x^2}{%
(1-x)(1-7x+x^2)}.
\end{equation*}

\section{A central and a local theorem for $L(n,k)$}

A positive real sequence $a(n,k)_{k=0}^n,$ with $A_n=\sum%
\limits_{k=0}^na(n,k)\neq 0,$ is said to satisfy a central limit theorem
(or is {\it asymptotically normal}) with mean $\mu _n$ and variance $\sigma _n^2$
if 
\begin{equation*}
\lim_{n\longrightarrow +\infty }\sup_{x\in R}\left| \sum\limits_{0\leq k\leq
\mu _n+x\sigma _n}\frac{a(n,k)}{A_n}-(2\pi )^{-1/2}\int\limits_{-\infty
}^xe^{-\frac{t^2}2}dt\right| =0.
\end{equation*}

The sequence satisfies a local limit theorem on $B\subseteq R$ ; with mean $%
\mu _n$ and variance $\sigma _n^2$ if 
\begin{equation*}
\lim_{n\longrightarrow +\infty }\sup_{x\in B}\left| \frac{\sigma _na(n,\mu
_n+x\sigma _n)}{A_n}-(2\pi )^{-1/2}e^{-\frac{x^2}2}\right| =0.
\end{equation*}

Recall the following result (see Bender \cite{1}).

\begin{theorem}
Let $(P_n)_{n\geq 1}$ be a sequence of real polynomials; with only real
negative zeros. The sequence of the coefficients of the $(P_n)_{n\geq 1}$
satisfies a central limit theorem; with $\mu _n=\frac{P_n^{"}(1)}{P_n(1)}$
and $\sigma _n^2=$ $\left( \frac{P_n^{"}(1)}{P_n(1)}+\frac{P_n^{^{\prime
}}(1)}{P_n(1)}-\left( \frac{P_n^{^{\prime }}(1)}{P_n(1)}\right) ^2\right) $
provided that $\lim\limits_{n\longrightarrow +\infty }\sigma _n^2=+\infty .$
If, in addition, the sequence of the coefficients of each $P_n$ is with no
internal zeros; then the sequence of the coefficients satisfies a local
limit theorem on $R$.
\end{theorem}

The fact that the zeros of the sequence $L_n(x)$ are real implies
the following result.

\begin{theorem}
The sequence $(L(n,k))_{k\geq 0}$ satisfies a central limit and a local
limit theorem on $R$  with $\mu _n=\frac{L_n^{"}(1)}{L_n(1)}\thicksim \frac
n2\left( 1-\frac{\sqrt{5}}5\right) $and $\sigma _n^2=\frac{L_n^{"}(1)}{L_n(1)%
}+\frac{L_n^{^{\prime }}(1)}{L_n(1)}-\left( \frac{L_n^{^{\prime }}(1)}{L_n(1)%
}\right) \thicksim 5^{-\frac 34}n$
\end{theorem}

%TCIMACRO{\TeXButton{Proof}{\proof}}
%BeginExpansion
\proof%
%EndExpansion
We have
\begin{equation*}
\sigma _n^2=\frac{L_n^{"}(1)}{L_n(1)}+\frac{L_n^{^{\prime }}(1)}{L_n(1)}%
-\left( \frac{L_n^{^{\prime }}(1)}{L_n(1)}\right) ^2=\frac{%
n^2L_{n-2}L_n-5n^2F_{n-1}}{5L_n^2}+\frac{3nF_{n-1}-nL_{n-2}}{5L_n}.
\end{equation*}

\bigskip Let $\alpha =\frac{1+\sqrt{5}}2,$ $\beta =\frac{1-\sqrt{5}}2.$
Using the explicit formulas of $L_n$ and $F_n$, we obtain
\begin{equation*}
\sigma _n^2=\frac{(-1)^nn^2}{\alpha ^{2n}+\beta ^{2n}+2(-1)^n}+\frac{\alpha
^{n-2}\left( \frac{3\sqrt{5}\alpha }5-1\right) n-\beta ^{n-2}\left( \frac{3%
\sqrt{5}\beta }5+1\right) n}{5\left( \alpha ^n+\beta ^n\right) }\thicksim
5^{-\frac 34}n.
\end{equation*}
So,$\lim\limits_{n\longrightarrow +\infty }\sigma _n=+\infty .$ The local
limit theorem is then easily seen to be satisfied; since $L(n,k)\neq 0,$ for 
$0\leq k\leq \left\lfloor \frac n2\right\rfloor .$%
%TCIMACRO{\TeXButton{End Proof}{\endproof}}
%BeginExpansion
\endproof%
%EndExpansion

As a consequence of the local limit theorem, we have

\begin{corollary}
Let $L=\max \{L(n,k),\,0\leq k\leq \frac n2\}.$ Then 
\begin{equation*}
L\thicksim \frac{5^{\frac 34}\left( \frac{1+\sqrt{5}}2\right) ^n}{\sqrt{2\pi
n}}.
\end{equation*}
\end{corollary}

\thinspace \thinspace \thinspace Acknowledgments: My sincere thanks to
Andreas Dress and Jean-Louis Nicolas for their valuable corrections and
comments.

\begin{thebibliography}{9}
\bibitem{1}  E. A. Bender, Central and local limit theorems applied to
asymptotic enumeration, \textit{\ J. Combin. Theory}, Ser. A $\mathbf{15}$
(1973), 91--111.

\bibitem{2}  M. Benoumhani, Polyn\^{o}mes \`{a} racines r\'{e}elles et
applications combinatoires, Th\`{e}se de doctorat, Universit\'{e} Claude
Bernard, Lyon 1, Lyon, France.1993.

\bibitem{3}  M. Benoumhani, Sur une propri\'{e}t\'{e} des polyn\^{o}mes
\`{a} racines n\'{e}gatives, \textit{J. Math. Pures Appl, }$\mathbf{75}$
(1996), 85--105.

\bibitem{4}  H. Cohn, \textit{A Classical Invitation to Algebraic Numbers
and Class Fields}, Springer-Verlag, 1978.

\bibitem{5}  G. H. Hardy, J. E. Littlewood and G. P\'olya, \textit{Inequalities},
Cambridge Univ. Press, 1956.

\bibitem{6}  N. Sloane, Online Encyclopedia of Integer Sequences,
{\tt www.research.att.com/\symbol{126}njas/sequences/index.html}.

\bibitem{7}  R. Stanley, \textit{Enumerative Combinatorics}, Wadsworth \&amp;
Brooks / Cole, Monterey, California 1986.

\bibitem{8}  S. Tanny and M. Zuker, On a unimodal sequence of binomial
coefficients, \textit{Discrete Math}. $\mathbf{9}$ (1974), 79--89.

\bibitem{9}  S. Tanny and M. Zuker, Analytic methods applied to a sequence of
binomial coefficients, \textit{Discrete Math}. $\mathbf{24}$ (1978),
299--310.
\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords:  
Fibonacci number, log--concave sequence, limit theorems, Lucas number,
polynomial with real zeros, unimodal sequence.}


\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A034807}.)


\bigskip
\hrule
\bigskip



\vspace*{+.1in}
\noindent
Received December 21, 2002;
revised version received  April 25, 2002.
Published in {\it Journal of Integer Sequences},
June 5, 2003.
\bigskip
\hrule
\bigskip

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



\end{document}

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