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\begin{center}
\vskip 1cm{\LARGE\bf Some Properties of Hyperfibonacci \\
\vskip .1in
and Hyperlucas Numbers } \vskip 1cm \large Ning-Ning
Cao\footnote{This work supported by the Science Research
Foundation of Dalian University of Technology
(2008).} and Feng-Zhen Zhao$^1$ \\
School of Mathematical Sciences\\
Dalian University of Technology \\
Dalian 116024 \\
China \\
\href{mailto:caoning0928@163.com}{\tt caoning0928@163.com} \\
\href{mailto:fengzhenzhao@yahoo.com.cn.}{\tt fengzhenzhao@yahoo.com.cn}\\
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\vskip .2 in

\begin{abstract}
In this paper, we discuss the properties
of hyperfibonacci numbers and hyperlucas numbers. We derive some identities for hyperfibonacci and
hyperlucas numbers by the method of coefficients. Furthermore, we give
asymptotic expansions of certain sums involving hyperfibonacci and
hyperlucas numbers by Darboux's method.
\end{abstract}

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%************************* section 1 *******************************************


\section{Introduction}
Dil and Mez\H{o} \cite{ref1} introduced the definitions of
``hyperfibonacci'' numbers $F_n^{(r)}$ and ``hyperlucas'' numbers
$L_n^{(r)}$
\begin{eqnarray*}
F_n^{(r)}=\sum_{k=0}^nF_k^{(r-1)}, \quad {\rm with} \quad
F_n^{(0)}=F_n, \quad F_0^{(r)}=0, \quad F_1^{(r)}=1,
\end{eqnarray*}
\begin{eqnarray*}
L_n^{(r)}=\sum_{k=0}^nL_k^{(r-1)}, \quad {\rm with} \quad
L_n^{(0)}=L_n, \quad L_0^{(r)}=2, \quad L_1^{(r)}=2r+1,
\end{eqnarray*}
where $r$ is a positive integer, and $F_n$ and $L_n$ are Fibonacci
and Lucas numbers, respectively. The generating functions of
$F_n^{(r)}$ and $L_n^{(r)}$ are as follows:
\begin{eqnarray*}
\sum_{n=0}^{\infty}F_n^{(r)}t^n&=&\frac{t}{(1-t-t^2)(1-t)^r}, \\
\sum_{n=0}^{\infty}L_n^{(r)}t^n&=&\frac{2-t}{(1-t-t^2)(1-t)^r}.
\end{eqnarray*}
The first few values of $F_n^{(r)}$ and $L_n^{(r)}$ are as follows:\\
\begin{displaymath}
F_n^{(1)}: 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, \ldots;
\end{displaymath}
\begin{displaymath}
F_n^{(2)}:
0, 1, 3, 7, 14, 26, 46, 79, 133, 221,
364, 596, 972, 1581, 2567, 4163, 6746,\ldots;
\end{displaymath}
\begin{displaymath}
L_n^{(1)}:
2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, \ldots;
\end{displaymath}
\begin{displaymath}
L_n^{(2)}: 2, 5, 11, 21, 38, 66, 112, 187, 309, 507,
828, 1348, 2190, 3553, 5759, 9329, 15106,\ldots \ .
\end{displaymath}
There are some elementary identities for $F_n^{(r)}$ and $L_n^{(r)}$ when $r=1,2$.
For example,
\begin{eqnarray*}
F^{(1)}_n&=&F_{n+2}-1,\\
F^{(2)}_n&=&F_{n+4}-n-3\\
          &=&\sum_{k=0}^n (n-k)F_k,\\
L^{(1)}_n&=&L_{n+2}-1 \\
          &=&F_n+F_{n+2}-1, \\
L^{(2)}_n&=&4 (F_{n+1}-1)+3 F_n-n \\
          &=&L_{n+3}-(n+4).
\end{eqnarray*}
For the above values and elementary identities of $F
_n^{(r)}$ and $L
_n^{(r)}$, see \cite{ref4} (\seqnum{A000071}, \seqnum{A001924}, \seqnum{A001610}, \seqnum{A023548}).


Hyperfibonacci numbers and hyperlucas numbers are useful, and Dil
and Mez\H{o} \cite{ref1} derived some equalities for Fibonacci and Lucas
numbers by applying them. Hence, hyperfibonacci numbers and
hyperlucas numbers deserve to be investigated. In this paper, we
discuss properties of $F_n^{(r)}$ and $L_n^{(r)}$. We
establish some identities for $F_n^{(r)}$ and $L_n^{(r)}$.
Furthermore, we give asymptotic expansions of certain sums related
to $F_n^{(r)}$ and $L_n^{(r)}$.

For convenience, we first recall some notation. Let
$\alpha=(1+\sqrt 5)/2$. It is well known that
\begin{eqnarray*}
F_n=\frac{\alpha^n-(-1)^n\alpha^{-n}}{\sqrt 5}, \quad
L_n=\alpha^n+(-1)^n\alpha^{-n},
\end{eqnarray*}
and $F_n$ and $L_n$ satisfy the following recurrence
relation
\begin{eqnarray}
W_{n+1}=W_n+W_{n-1}, \quad n\geq 1. \label{eq:E1}
\end{eqnarray}
 As usual, the binomial coefficient ${n\choose m}$ is defined by
\begin{eqnarray*}
{n\choose m}=\begin{cases} \displaystyle\frac{n!}{m!(n-m)!}, \text{ if } \quad n\geq m; \\
0, \quad \quad \quad \quad \quad \text{ if } n<m;
\end{cases}
\end{eqnarray*}
where $n$ and $m$ are nonnegative integers. Throughout, $[z^n]f(z)$
denotes the coefficient of $z^n$ in $f(z)$, where
$$
f(z)=\sum_{n=0}^{\infty}f_nz^n.
$$
If $f(t)$ and $g(t)$ are formal power series, the following
relations hold \cite{ref2}
\begin{eqnarray}
&&[t^n](af(t)+bg(t))=a[t^n]f(t)+b[t^n]g(t),
\label{eq:E2}\\
&&[t^n]tf(t)=[t^{n-1}]f(t), \label{eq:E3}\\
&&[t^n]f(t)g(t)=\sum_{k=0}^n[y^k]f(y)[t^{n-k}]g(t). \label{eq:E4}
\end{eqnarray}
The above relations \eqref{eq:E2}--\eqref{eq:E4} will be used later
on.

Now we recall the notation of the binomial transform of a sequence,
the inverse binomial transform of a sequence and Euler-Seidel
infinite matrix \cite{ref1, ref5}. Let
$\{a_k\}_{k=0}^\infty$ be a sequence. The binomial transform of
$\{a_k\}$ is given by $\sum^n_{k=0}{n\choose k}a_k$, the inverse
binomial transform of $\{a_k\}$ is given by $\sum^n_{k=0}{n\choose
k}(-1)^k a_k$, and the Euler-Seidel infinite matrix corresponding to
the sequence $\{a_k\}$ is determined by the following formulas
\begin{eqnarray*}
&&a_n^{[0]}=a_n \quad (n\geq0),\\
&&a^{[k]}_n=a_n^{[k-1]}+a_{n+1}^{[k-1]} \quad (n\geq0,\quad k\geq0),
\end{eqnarray*}
where $a^{[k]}_n$ is the element at the $(k+1)th$ row and the
$(n+1)th$ column. The sequences $\{a_n^{[0]}\}$ and $\{a_0^{[n]}\}$ satisfy
\cite{ref1}
\begin{eqnarray}
&&a_0^{[n]}=\sum_{k=0}^n{n\choose k}a_k^{[0]} ,\label{eq:E5}\\
&&a_n^{[0]}=\sum_{k=0}^n{n\choose
k}(-1)^{n-k}a_0^{[k]}.\label{eq:E6}
\end{eqnarray}
Let $a(t)$ and $\bar{a}(t)$ be the generating function of
$\{a_n^{[0]}\}$ and $\{a_0^{[n]}\}$, respectively,
\begin{eqnarray*}
a(t)=\sum_{n=0}^\infty
a_n^{[0]}t^n,\quad\bar{a}(t)=\sum_{n=0}^\infty a_0^{[n]} t^n.
\end{eqnarray*}
The functions $a(t)$ and $\bar{a}(t)$ satisfy \cite{ref1}
\begin{eqnarray}
\bar{a}(t)&=&\frac{1}{1-t}a\bigg(\frac{t}{1-t}\bigg),\label{eq:E7}\\
a(t)&=&\frac{1}{t+1}\bar{a}\bigg(\frac{t}{t+1}\bigg).\label{eq:E8}
\end{eqnarray}


\section{Main Results}
\label{sec2}
In this section, we derive some identities for $F_n^{(r)}$ and
$L_n^{(r)}$. Later, we give asymptotic expansions of
certain sums involving $F_n^{(r)}$ and $L_n^{(r)}$.

Various identities involving Fibonacci and Lucas numbers were
established. The following sums were
investigated \cite{ref3, ref7, ref8}
\begin{eqnarray*}
\sum_{j_1+j_2+\cdots+j_k=n}F_{j_1}F_{j_2}\cdots
F_{j_k}, \quad
\sum_{j_1+j_2+\cdots+j_k=n}L_{j_1}L_{j_2}\cdots
L_{j_k}.
\end{eqnarray*}
For example,
\begin{eqnarray*}
\sum_{j_1+j_2=n}F_{j_1}F_{j_2}&=&\frac{(n-1)L_n+2F_{n-1}}{5},\\
\sum_{j_1+j_2=n}L_{j_1}L_{j_2}&=&(n+1)L_n+2F_{n+1}.
\end{eqnarray*}
Now we derive some identities for $F_n^{(r)}$ and $L_n^{(r)}$. Denote
\begin{eqnarray*}
A_{n, k, r}=\sum_{j_1+j_2+\cdots+j_k=n}F^{(r)}_{j_1}F^{(r)}_{j_2}\cdots F^{(r)}_{j_k}, \quad
B_{n, k, r}=\sum_{j_1+j_2+\cdots+j_k=n}L^{(r)}_{j_1}L^{(r)}_{j_2}\cdots L^{(r)}_{j_k}.
\end{eqnarray*}
These sums are interesting because they can help us to find some new
convolution properties. For $A_{n, k, r}$ and $B_{n, k, r}$, we have
\begin{theorem}
\label{thm:1} {Let $k$, $n\geq 1$ and $r\geq 1$ be positive integers.
For $A_{n, k, r}$ and $B_{n, k, r}$, we have
\begin{eqnarray}
A_{n, 2, 1}&=&n+5-2F_{n+4}+\frac{(n+1)L_{n+4}-2F_{n+1}}{5},  \label{eq:E9}\\
B_{n, 2, 1}&=&n+9-10F_{n+4}-2F_{n+1}+\frac{(5n+9)L_{n+4}+4L_{n+6}}{5},\label{eq:E10}\\
A_{n, k+1, r}&=&\sum_{j=0}^nA_{n, k, r}F_j^{(r)}, \label{eq:E11}\\
B_{n, k+1, r}&=&\sum_{j=0}^nB_{n, k, r}L_j^{(r)}. \label{eq:E12}
\end{eqnarray}}
\end{theorem}
\begin{proof}
Let
$$
F_r(t)=\frac{t}{(1-t-t^2)(1-t)^r},\quad L_r(t)=\frac{2-t}{(1-t-t^2)(1-t)^r}.
$$
Clearly,
\begin{eqnarray*}
F_1(t)&=&\bigg(\frac{\alpha^2-\alpha^{-2}}{t-1}-\frac{\alpha}{t-\alpha^{-1}}
-\frac{\alpha^{-1}}{t+\alpha}\bigg)\frac{1}{\sqrt 5}\\
&=&\bigg(\sum_{n=0}^{\infty}F_{n+2}t^n-\sum_{n=0}^{\infty}t^n\bigg)t,\\
A_{n, 2, 1}&=&[t^n]F^2_1(t),\\
B_{n, 2, 1}&=&[t^n] L^2_1(t)\\
&=&[t^n]F^2_1(t)-4[t^{n+1}]F^2_1(t)+4[t^{n+2}]F^2_1(t)\\
&=&A_{n, 2, 1}-4A_{n+1, 2, 1}+4A_{n+2, 2, 1}.
\end{eqnarray*}
Then we get
\begin{eqnarray*}
\sum_{j_1+j_2=n}F_{j_1}^{(1)}F_{j_2}^{(1)}&=&[t^n]F_1^2(t)\\
&=&[t^n]\frac{1}{5}\bigg[\frac{(\alpha^2-\alpha^{-2})^2}{(t-1)^2}+\frac{\alpha^2}{(t-\alpha^{-1})^2}+\frac{\alpha^{-2}}{(t+\alpha)^2}-\frac{2\alpha
(\alpha^2-\alpha^{-2})}{(t-1)(t-\alpha^{-1})}\\
&&-\frac{2\alpha^{-1}(\alpha^2-\alpha^{-2})}{(t-1)(t+\alpha)}+\frac{2}{(t-\alpha^{-1})(t+\alpha)}\bigg]\\
&=&[t^n]\frac{1}{5}\bigg\{\frac{(\alpha^2-\alpha^{-2})^2}{(t-1)^2}+\frac{\alpha^4}{(1-\alpha t)^2}+\frac{\alpha^{-4}}{(1+\alpha^{-1}t)^2}\\
&&+\frac{2(\alpha^2-\alpha^{-2})(\alpha^3+\alpha^{-3})}{1-t}
-2 \bigg[ \alpha^4 (\alpha^2-\alpha^{-2})+\frac{\alpha}{\alpha+\alpha^{-1}}\bigg]\frac{1}{1-\alpha t}\\
&&+2 \bigg[ \alpha^{-4} (\alpha^2-\alpha^{-2})-\frac{\alpha^{-1}}{\alpha+\alpha^{-1}}\bigg]\frac{1}{1+\alpha^{-1} t}
\bigg\}\\
&=&[t^n]\frac{1}{5}\bigg\{(\alpha^2-\alpha^{-2})^2\sum_{n=0}^{\infty} (n+1)t^n + \sum_{n=0}^{\infty}[\alpha^{n+4}+(-1)^n \alpha^{-n-4}](n+1)t^n \\
&&+2(\alpha^2-\alpha^{-2})(\alpha^3+\alpha^{-3})\sum_{n=0}^ {\infty} t^n \\
&&-2 \bigg [\alpha^4 (\alpha^2-\alpha^{-2})+\frac{\alpha}{\alpha+\alpha^{-1}}\bigg]
 \sum_{n=0}^{\infty} \alpha^n t^n \\
&&+2\bigg[\alpha^{-4} (\alpha^2-\alpha^{-2})-\frac{\alpha^{-1}}{\alpha+\alpha^{-1}}\bigg]
 \sum_{n=0}^{\infty} (-1)^n \alpha^{-n} t^n
\bigg\}\\
&=&[t^n]\bigg\{
\sum_{n=0}^{\infty}(n+1)t^n + \sum_{n=0}^{\infty}\frac{\alpha^{n+4}+(-1)^n\alpha^{-n-4}}{5}(n+1)t^n + 2F_3\sum_{n=0}^{\infty}t^n\\
&&-\frac{2}{5}(\alpha^2-\alpha^{-2})\sum_{n=0}^{\infty}[\alpha^{n+4}-(-1)^n\alpha^{-n-4}]t^n \\
&&- \frac{2}{5(\alpha+\alpha^{-1})}\sum_{n=0}^{\infty}[\alpha^{n+1}+(-1)^n\alpha^{-n-1}]t^n
\bigg\}\\
&=&[t^n]\sum_{n=0}^{\infty}(n+1)t^n + [t^n]\sum_{n=0}^{\infty}\frac{(n+1)L_{n+4}}{5}t^n + [t^n]4\sum_{n=0}^{\infty}t^n \\
&& -[t^n]2\sum_{n=0}^{\infty}F_{n+4}t^n -[t^n] \frac{2}{5}\sum_{n=0}^{\infty}F_{n+1}t^n.
\end{eqnarray*}
By applying \eqref{eq:E3} and the definitions of $F_n$ and $L_n$,
we have \eqref{eq:E9}. Naturally, we deduce that
\begin{eqnarray*}
B_{n, 2,
1}&=&n+9-10F_{n+4}+\frac{4(n+3)L_{n+6}-4(n+2)L_{n+5}+(n+1)L_{n+4}}{5}-2F_{n+1}.
\end{eqnarray*}
By using \eqref{eq:E1}, we prove that \eqref{eq:E10} holds. By
means of \eqref{eq:E4}, we can show that
\eqref{eq:E11} and \eqref{eq:E12} hold.
\end{proof}

It is interesting that we can get congruence relations from \eqref{eq:E9} and \eqref{eq:E10}
\begin{eqnarray*}
A_{n,2,1}\equiv \bigg(n-2F_{n+4}+\frac{(n+1)L_{n+4}-2F_{n+1}}{5}\bigg)\pmod 5,\\
B_{n,2,1}\equiv \bigg(n-10F_{n+4}-2F_{n+1}+\frac{(5n+9)L_{n+4}+4L_{n+6}}{5}\bigg)\pmod 9.
 \end{eqnarray*}

When $k$ or $r$ gets large, it is difficult to compute the closed
forms of $A_{n, k, r}$ and $B_{n, k, r}$. However, we can give their
asymptotic values. Now we recall a lemma \cite{ref6}.
\begin{lemma}
\label{lem:1} {
Assume that $f(t)=\sum_{n\geq 0}a_nt^n$ is an analytic
function for $|t|<r$ and with a finite number of algebraic
singularities on the circle $|t|=r$. $\alpha_1$, $\alpha_2$,
$\cdots$, $\alpha_l$ are singularities of order $\omega$, where
$\omega$ is the highest order of all singularities. Then
\begin{eqnarray}
a_n=(n^{\omega-1}/\Gamma(\omega))\times\bigg(\sum_{k=1}^lg_k(\alpha_k)\alpha^{-n}_k
+\mbox{o}(r^{-n})\bigg),
\label{eq:E13}
\end{eqnarray}
where $\Gamma(\omega)$ is the gamma function, and
$$
g_k(\alpha_k)=\lim_{t\rightarrow
\alpha_k}(1-(t/\alpha_k))^{\omega}f(t).
$$}
\end{lemma}

\begin{theorem}
\label{thm:3} {Suppose that $k$ and $r$ are fixed positive integers.
For $A_{n, k, r}$ and $B_{n, k, r}$, when $n\to\infty$,
\begin{eqnarray}
A_{n, k, r}&=&\frac{n^{k-1}}{(k-1)!}\bigg[\bigg(\frac{\alpha}{\alpha^2+1}\bigg)^k(1+\alpha)^{kr}\alpha^n+\mbox{o}(
\alpha^n))\bigg],\label{eq:E14}\\
B_{n, k,
r}&=&\frac{n^{k-1}}{(k-1)!}\bigg[\bigg(\frac{2\alpha^2-\alpha}{\alpha^2+1}\bigg)^k(1+\alpha)^{kr}\alpha^n+\mbox{o}(
\alpha^n))\bigg]. \label{eq:E15}
\end{eqnarray}}
\end{theorem}

\begin{proof}
Let
$$
f_{k, r}(t)=\frac{t^k}{(1-t-t^2)^k(1-t)^{kr}}.
$$
We know that $f_{k, r}(t)$ is analytic for $|t|<1/\alpha$, and with
one algebraic singularity on the circle $|t|=1/\alpha$. The order of
$1/\alpha$ is $k$. One can compute that
\begin{eqnarray*}
\lim_{t\to1/\alpha}(1-\alpha t)^kf_{k,
r}(t)=\bigg(\frac{\alpha}{\alpha^2+1}\bigg)^k(1+\alpha)^{kr}.
\end{eqnarray*}
By using Lemma \ref{lem:1}, we can prove that \eqref{eq:E14} holds. Using
the same method, we can prove that \eqref{eq:E15} holds.
\end{proof}

In addition, we give asymptotic expansions of other sums for
$F_n^{(r)}$ and $L_n^{(r)}$.

\begin{theorem}
\label{thm:4} {
Let $n$ be a positive integer. When $n\rightarrow\infty$,
\begin{eqnarray}
\sum_{k=0}^n{n\choose k}
F_k^{(r)}&=&\frac{\alpha(1+\alpha)^r}{(\alpha^2+1)(2-\alpha)^n}+o((2-\alpha)^{-n}), \label{eq:E16}\\
\sum_{k=0}^n{n\choose
k}L_k^{(r)}&=&\frac{(1+\alpha)^r}{(2-\alpha)^n}+o((2-\alpha)^{-n}),
\label{eq:E17} \\
\sum_{k=0}^n{n\choose k}(-1)^k
F_k^{(r)}&=&\frac{-\alpha^{n+1}(2-\alpha)^r}{\alpha^2+1}+o((-\alpha)^n),
\label{eq:E18}\\
\sum_{k=0}^n{n\choose k}(-1)^k
L_k^{(r)}&=&(2-\alpha)^r\alpha^n+o((-\alpha)^n).\label{eq:E19}
\end{eqnarray}
}
\end{theorem}

\begin{proof}
We only give the proofs of \eqref{eq:E16} and \eqref{eq:E18}. The proofs of
\eqref{eq:E17} and \eqref{eq:E19} are similar to those of
\eqref{eq:E16} and \eqref{eq:E18}, respectively, and they are
omitted here. Let $F_k^{(r)}$ be the first row, and we get the
Euler-Seidel infinite
 matrix A. Then let $F_k^{(r)}$ be the first column, and we get Euler-Seidel infinite
 matrix B. The elements of A and B are denoted by $a_n^{[k]}$ and $b_n^{[k]}$. By using \eqref{eq:E5} and \eqref{eq:E6}, we obtain
\begin{eqnarray*}
a_0^{[n]}&=&\sum_{k=0}^n{n\choose k}F_k^{(r)},\\
b_n^{[0]}&=&\sum_{k=0}^n{n\choose k}(-1)^{n-k}F_k^{(r)}.
\end{eqnarray*}
By means of \eqref{eq:E7} and \eqref{eq:E8}, we get
\begin{eqnarray*}
\sum_{k=0}^n{n\choose k}F_k^{(r)}&=&a_0^{[n]}\\
&=&[t^n]\frac{t(1-t)^r}{(t^2-3t+1)(1-2t)^r},\\
\sum_{k=0}^n{n\choose k}(-1)^{n-k}F_k^{(r)}&=&b_n^{[0]}\\
&=&[t^n]\frac{-t(1+t)^r}{t^2-t-1}.
\end{eqnarray*}
We know that the function
$\bar{a}(t)=\displaystyle\frac{t(1-t)^r}{(t^2-3t+1)(1-2t)^r}$ is
analytic for $|t|<2-\alpha$ with one algebraic
singularity $\alpha_1=2-\alpha$ on the circle
 $|t|=2-\alpha$, and $a(t)=\displaystyle\frac{-t(1+t)^r}{t^2-t-1}$ is analytic for
$|t|<\alpha^{-1}$ with one algebraic singularity
$\beta_1=-\alpha^{-1}$ on the circle
 $|t|=\alpha^{-1}$. The
orders of $\alpha _1$ and $\beta_1$ are $1$. We can compute that
\begin{eqnarray*}
\lim_{t\to
{2-\alpha}}(1-\frac{t}{2-\alpha})\bar{a}(t)&=&\frac{\alpha(1+\alpha)^r}{\alpha^2+1},\\
\lim_{t\to-\alpha^{-1}}\bigg(1+\frac{t}{\alpha^{-1}}\bigg)a(t)
&=&-\frac{\alpha(2-\alpha)^r}{\alpha^2+1}.
\end{eqnarray*}
By using Lemma \ref{lem:1}, we prove that \eqref{eq:E16}  and \eqref{eq:E18}
hold.
\end{proof}

\begin{theorem}
\label{thm:5} {
Suppose that $m$ and $r$ are fixed positive integers.
When $n\to\infty$,
\begin{eqnarray}
\sum_{j=0}^n F_{n-j}^{(r)}{j+m-1\choose
j}&=&\frac{\alpha^{n+1}(1+\alpha)^{r+m}}{\alpha^2+1}+o(\alpha^n),\label{eq:E20}\\
\sum_{j=0}^n L_{n-j}^{(r)}{j+m-1\choose
j}&=&(1+\alpha)^{r+m}\alpha^n+o(\alpha^n).
\label{eq:E21}
\end{eqnarray}
}
\end{theorem}
\begin{proof}
 We can verify that
\begin{eqnarray*}
\sum_{j=0}^nF_{n-j}^{(r)}{j+m-1\choose
j}&=&\sum_{j=0}^n[t^{n-j}]\frac{t}{(1-t-t^2)(1-t)^r}[t^j]\frac{1}{(1-t)^m}\\
&=&[t^n]\frac{t}{(1-t-t^2)(1-t)^{m+r}},\\
\sum_{j=0}^n L_{n-j}^{(r)}{j+m-1\choose
j}&=&[t^n]\frac{2-t}{(1-t-t^2)(1-t)^{m+r}}.
\end{eqnarray*}
By using Lemma \ref{lem:1}, we have \eqref{eq:E20} and \eqref{eq:E21}.
\end{proof}

In the final of this section, we compare the accurate values with
the asymptotic values. In Theorem \ref{thm:4}, let
\begin{eqnarray*}
X_n=\sum_{k=0}^n{n\choose k}F_k^{(r)},\quad
Y_n=\frac{\alpha(1+\alpha)^r}{(\alpha^2+1)(2-\alpha)^n}.
\end{eqnarray*}

\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|}\hline
$n$ & $X_n$ & $Y_n$ & $|X_n-Y_n|/X_n$\\
\hline
50 & $9.2737156629317196\times10^{20} $& $9.2737269219308562\times10^{20}$ & $1.2141\times10^{-6}$\\
\hline
100 & $7.345448671565505\times10^{41}$ & $7.3454486715782857\times10^{41}$ & $1.7399\times10^{-12}$\\
\hline
150 & $5.818115698360039\times10^{62}$ &$ 5.8181156983601641\times10^{62}$ &$ 2.1352\times10^{-14}$\\
\hline
\end{tabular}
\end{center}
\caption{some values of $X_n$ and $Y_n$}
\end{table}

From the above table, we find that the value of $|X_n-Y_n|/X_n$ gets
smaller and smaller with the increasing of $n$.

\section{Remarks}
\label{sec3}
Consider the sequences $\{u_n^{(r)}\}$ and $\{v_n^{(r)}\}$ defined by
\begin{eqnarray*}
\sum_{n=0}^\infty u_n^{(r)}z^n=\frac{z}{(1-pz-z^2)(1-z)^r}, \\
\sum_{n=0}^\infty v_n^{(r)}z^n=\frac{2-pz}{(1-pz-z^2)(1-z)^r},
\end{eqnarray*}
where $p>0$. It is clear that $u_n^{(r)}=F_n^{(r)}$ and $v_n^{(r)}=L_n^{(r)}$ when $p=1$. The conclusions of $F_n^{(r)}$ and $L_n^{(r)}$ can be generalized to the cases of $\{u_n^{(r)}\}$ and $\{v_n^{(r)}\}$.
For example, put
\begin{eqnarray*}
U_{n,k,r}=\sum_{j_1+j_2+\cdots+j_k=n}u_{j_1}^{(r)} u_{j_2}^{(r)}\cdots u_{j_k}^{(r)},  \quad
V_{n,k,r}=\sum_{j_1+j_2+\cdots+j_k=n}v_{j_1}^{(r)} v_{j_2}^{(r)}\cdots v_{j_k}^{(r)}.
\end{eqnarray*}
Then we have
\begin{eqnarray}
U_{n,2,1}&=&\frac{n-1}{p^2(p^2+4)}(v_n^{(0)}+2v_{n+1}^{(0)}+v_{n+2}^{(0)})
              -\frac{2}{p^3}(u_{n+1}^{(0)}+2u_n^{(0)}+u_{n-1}^{(0)})  \nonumber \\
&& +\frac{2}{p(p^2+4)}u_{n-1}^{(0)}+\frac{np+p+4}{p^3},
         \label{eq:E22}  \\
U_{n,k+1,r}&=&\sum_{j=0}^n U_{n,k,r}u_j^{(r)}.     \label{eq:E23}
\end{eqnarray}


We can verify that
\begin{eqnarray*}
U_{n,2,1}&=&[t^n]\frac{t^2}{(1-pt-t^2)^2(1-t)^2}\\
&=&[t^n]\bigg\{\frac{1}{\Delta (1+\tau)^2}\frac{t^2}{(t+\tau)^2} + \frac{1}{\Delta} \bigg(\frac{1}{1-\tau^{-1}}-\frac{1}{1+\tau}\bigg)^2 \frac{t^2}{(1-t)^2} \\
&&+\frac{1}{\Delta} \frac{1}{(1-\tau^{-1})^2} \frac{t^2}{(t-\tau^{-1})^2} - \frac{2}{\Delta(1+\tau)}\bigg(\frac{1}{1-\tau^{-1}}-\frac{1}{1+\tau}\bigg)\frac{t^2}{\tau+1}\bigg(\frac{1}{t+\tau}+\frac{1}{1-t}\bigg)\\
&&-\frac{2t^2}{\Delta (1+\tau)(1-\tau^{-1})} \bigg(\frac{-1}{t+\tau}+\frac{1}{t-\tau^{-1}}\bigg)\frac{1}{\sqrt{\Delta}} \\
&&+\frac{2t^2}{\Delta(1-\tau^{-1})}\bigg(\frac{1}{1-\tau^{-1}}-\frac{1}{1+\tau}\bigg) \bigg(\frac{1}{t-\tau^{-1}}+\frac{1}{1-t}\bigg) \frac{1}{1-\tau^{-1}}
\bigg\}\\
&=&[t^n] \bigg\{
\frac{1}{p^2\Delta}\sum_{n=0}^\infty (n+1) t^{n+2}(v_{n+2}^{(0)}+2v_{n+3}^{(0)}+v_{n+4}^{(0)})+ \frac{1}{p^2}\sum_{n=0}^\infty (n+1) t^{n+2} \\
&& -\frac{2}{p^3}\sum_{n=0}^\infty t^{n+2}(u_{n+3}^{(0)}+2u_{n+2}^{(0)}+u_{n-1}^{(0)})
+\frac{2}{p\Delta}\sum_{n=0}^\infty t^{n+2}u_{n+1}^{(0)}+\frac{4+2p}{p^3}\sum_{n=0}^\infty t^{n+2}
\bigg\},
\end{eqnarray*}
where $\Delta=p^2+4$, $\tau=\frac{p+\sqrt{\Delta}}{2}$. Hence \eqref{eq:E22} holds. The proof of \eqref{eq:E23} is similar to that of \eqref{eq:E11}, and it is omitted here.
The identities about $\{v_n^{(0)}\}$ can be found in the same way.



\section{Acknowledgments}

The authors would like to thank the anonymous referees for their
criticism and useful suggestions.

\begin{thebibliography}{99}
\bibliographystyle{plain}


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\end{thebibliography}

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\noindent 2000 {\it Mathematics Subject Classification}:  Primary
11B37, 11B39; Secondary 05A16, 05A19.

\noindent {\it Keywords}: Fibonacci numbers, Lucas numbers, generating function, asymptotic expansion,
Darboux's method.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000048},
\seqnum{A000071},
\seqnum{A001610},
\seqnum{A001924}, and
\seqnum{A023548}.)

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received June 6 2010;
revised version received  September 8 2010.
Published in {\it Journal of Integer Sequences}, September 10 2010.
Corrected, March 26 2011.

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