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\begin{center}
\vskip 1cm{\LARGE\bf 
Several Generating Functions for \\
\vskip .1in
Second-Order Recurrence Sequences
}
\vskip 1cm
\large
Istv\'an Mez\H{o}\\
Institute of Mathematics\\
University of Debrecen\\
Hungary\\
\href{mailto:imezo@math.klte.hu}{\tt imezo@math.klte.hu} \\
\end{center}

\vskip .2 in

\begin{abstract}Carlitz and Riordan began a study on closed form of
generating functions for powers of second-order recurrence sequences.
This investigation was completed by St\u{a}nic\u{a}. In this paper we
consider exponential and other types of generating
functions for such sequences. Moreover, an extensive table of
generating functions is provided.
\end{abstract}

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\section{Introduction}

The Fibonacci sequence, which is
is sequence \seqnum{A000045} in Sloane's {\it Encylopedia},
\cite{Sloane} is defined recursively as follows:
\[F_{n}=F_{n-1}+F_{n-2}\quad(n\ge2),\]
with initial conditions
\[F_0=0,\quad F_1=1.\]
The Lucas numbers $L_n$, which comprise Sloane's
sequence \seqnum{A000032}, are defined by the same manner but with
initial conditions
\[L_0=2,\quad L_1=1.\]


In 1962, Riordan \cite{Riordan} determined the generating functions for 
powers of Fibonacci numbers:
\[f_k(x)=\sum_{n=0}^\infty F_n^kx^n.\]
This question had been suggested by Golomb \cite{Golomb} in 1957.
Riordan found the recursive solution
\begin{equation}
(1-L_kx+(-1)^kx^2)f_k(x)=1+x\sum_{j=1}^{[k/2]}(-1)^jA_{kj}f_{k-2j}(x(-1)^j),\label{eq:Riordan}
\end{equation}
with initial functions
\[f_0(x)=\frac{1}{1-x},\quad\mbox{and}\quad f_1(x)=\frac{1}{1-x-x^2}.\]
We mention that in his paper Riordan used the $F_0=F_1=1$ condition.
In the result above, the coefficients $A_{kj}$ have a complicated definition and cannot be handled easily.

In the same journal and volume, Carlitz \cite{Carlitz} made the following generalization. Let
\[u_n=pu_{n-1}-qu_{n-2}\quad(n\ge2),\]
with initial conditions
\[u_0=1,\quad u_1=p.\]
He computed the generating functions for the sequences $u_n^k$. They have the same form as Eq.~\eqref{eq:Riordan}.

In his recent paper \cite{Stanica}, St\u{a}nic\u{a} gave the most general and simple answer for the questions above (see Theorem \ref{thm:Stanica}) with an easy proof. Namely, let the so-called second-order recurrence sequence be given by
\begin{equation}
u_{n}=pu_{n-1}+qu_{n-2}\quad(n\ge2),\label{eq:undef}
\end{equation}
where $p,q,u_0$ and $u_1$ are arbitrary numbers such that we eliminate the degenerate case $p^2+4q=0$. Then let
\begin{equation}
\alpha=\frac{1}{2}(p+\sqrt{p^2+4q}),\;\beta=\frac{1}{2}(p-\sqrt{p^2+4q}),\label{eq:alphabeta}
\end{equation}
\begin{equation}
A=\frac{u_1-u_0\beta}{\alpha-\beta},\;B=\frac{u_1-u_0\alpha}{\alpha-\beta}.\label{eq:AB}
\end{equation}
It is known that $u_n$ can be written in the form
\[u_n=A\alpha^n-B\beta^n\quad\mbox{(Binet formula)}.\]

Many famous sequences have this shape. A comprehensive table can be found at the end of the paper.

To present St\u{a}nic\u{a}'s result, we need to introduce the sequence $V_n$ given by its Binet formula:
\[V_n=\alpha^n+\beta^n,\;V_0=2,V_1=p.\]

\begin{Theorem}[St\u{a}nic\u{a}]\label{thm:Stanica}The generating function for the $r$th power of the sequence $u_n$ is
\[\sum_{n=0}^\infty u_n^rx^n=\]
\[\sum_{k=0}^\frac{r-1}{2}(-1)^kA^kB^k\binom{r}{k}\frac{A^{r-2k}-B^{r-2k}+(-b)^k(B^{r-2k}\alpha^{r-2k}-A^{r-2k}\beta^{r-2k})x}{1-(-b)^kV_{r-2k}-x^2},\]
if $r$ is odd, and
\[\sum_{n=0}^\infty u_n^rx^n=\]
\[\sum_{k=0}^{\frac{r}{2}-1}(-1)^kA^kB^k\binom{r}{k}\frac{B^{r-2k}+A^{r-2k}-(-b)^k(B^{r-2k}\alpha^{r-2k}+A^{r-2k}\beta^{r-2k})x}{1-(-b)^kV_{r-2k}x+x^2}\]
\[+\binom{r}{\frac{r}{2}}\frac{A^{\frac{r}{2}}(-B)^{\frac{r}{2}}}{1-(-1)^{\frac{r}{2}}x},\]
if $r$ is even.
\end{Theorem}

In the spirit of this result we present the same formulas for even and odd indices, exponential generating functions for powers, product of such sequences and so on.

\section{Non-exponential generating functions}

The result in this and the following sections yield rich and varied examples which are collected in separate tables at the end of the paper.

First, the generating function for $u_n$ is given:
\begin{Proposition}\label{prop:ordgf}We have
\[\sum_{n=0}^\infty u_nx^n=\frac{u_0+(u_1-pu_0)x}{1-px-qx^2}.\]
\end{Proposition}

For the sake of a more readable presentation, the proof of this statement and all of the others will be collected in a separate section. We remark that 
Proposition~\ref{prop:ordgf} is not new but the proof is easy and typical.

Sequences with even and odd indices appear so often that it is worth to construct the general generating function of this type.
%We use the abbreviations ``e'' and ``o'' to the words ``even'' and ``odd'', respectively.

\begin{Theorem}\label{thm:oddgf}The generating function for the sequence $u_{2n}$ is
\[\sum_{n=0}^\infty u_{2n}x^n=\frac{u_0+(u_2-u_0(p^2+2q))x}{1-(p^2+2q)x+q^2x^2},\]
while
\[\sum_{n=0}^\infty u_{2n+1}x^n=\frac{u_1+(u_0pq-u_1q)x}{1-(p^2+2q)x+q^2x^2}.\]
\end{Theorem}

\begin{Example}\label{ex:f2n}As a consequence, we can state the following identity which we use later.
\[\sum_{n=0}^\infty F_{2n}x^n=\frac{x}{1-3x+x^2}.\]
See the paper of Johnson \cite{Johnson}, for example.
\end{Example}

Generating functions for powers of even and odd indices are interesting. The following theorem contains these results.

\begin{Theorem}\label{thm:evenoddStanica}Let $u_n=pu_{n-1}+qu_{n-2}$ be a sequence with initial values $u_0$ and $u_1$. Then
\[\sum_{n=0}^\infty u_{2n}^rx^n=\]
\[\sum_{k=0}^\frac{r-1}{2}(-1)^kE^kF^k\binom{r}{k}\frac{E^{r-2k}-F^{r-2k}+q^{2k}(F^{r-2k}\rho^{r-2k}-E^{r-2k}\sigma^{r-2k})x}{1-q^{2k}V_{r-2k}-x^2},\]
if $r$ is odd, and
\[\sum_{n=0}^\infty u_{2n}^rx^n=\]
\[\sum_{k=0}^{\frac{r}{2}-1}(-1)^kE^kF^k\binom{r}{k}\frac{F^{r-2k}+E^{r-2k}-q^{2k}(F^{r-2k}\rho^{r-2k}+E^{r-2k}\sigma^{r-2k})x}{1-q^{2k}V_{r-2k}x+x^2}\]
\[+\binom{r}{\frac{r}{2}}\frac{E^{\frac{r}{2}}(-F)^{\frac{r}{2}}}{1-(-1)^{\frac{r}{2}}x},\]
if $r$ is even. For odd indices we have to make the substitution $E\leadsto G$ and $F\leadsto H$. Here
\begin{eqnarray*}
\rho&amp;=&amp;\frac{1}{2}\left(p^2+2q+p\sqrt{p^2+4q}\right),\\
\sigma&amp;=&amp;\frac{1}{2}\left(p^2+2q-p\sqrt{p^2+4q}\right),\\
E&amp;=&amp;\frac{u_2-u_0\sigma}{\rho-\sigma},\quad F=\frac{u_2-u_0\rho}{\rho-\sigma},\\
G&amp;=&amp;\frac{u_3-u_1\sigma}{\rho-\sigma},\quad H=\frac{u_3-u_1\rho}{\rho-\sigma},\\
V_n&amp;=&amp;\rho^n+\sigma^n,\;V_0=2,\;V_1=p^2+2q.
\end{eqnarray*}

%For the sequence with odd indices we have the same formulas but in this case
%\begin{eqnarray*}
%G&amp;=&amp;\frac{u_3-u_1\sigma}{\rho-\sigma},\\
%H&amp;=&amp;\frac{u_3-u_1\rho}{\rho-\sigma}.
%\end{eqnarray*}

\end{Theorem}

\begin{Remark}\label{rem:efgh}These constants are calculated for the named sequences:
\renewcommand{\arraystretch}{1.4}
\begin{center}
\begin{tabular}{|l|c|c|c|c|c|c|}\hline
Sequence&amp;$\rho$&amp;$\sigma$&amp;$E$&amp;$F$&amp;$G$&amp;$H$\\\hline\hline
$F_n$&amp;$\frac{3+\sqrt{5}}{2}$&amp;$\frac{3-\sqrt{5}}{2}$&amp;$\frac{\sqrt{5}}{5}$&amp;$\frac{\sqrt{5}}{5}$&amp;$\frac{\sqrt{5}}{5}\phi$&amp;$\frac{\sqrt{5}}{5}\ophi$\\\hline
$L_n$&amp;$\frac{3+\sqrt{5}}{2}$&amp;$\frac{3-\sqrt{5}}{2}$&amp;1&amp;$-1$&amp;$\frac{\sqrt{5}}{10}(5+\sqrt{5})$&amp;$\frac{\sqrt{5}}{10}(5-\sqrt{5})$\\\hline
$P_n$&amp;$3+2\sqrt{2}$&amp;$3-2\sqrt{2}$&amp;$\frac{\sqrt{2}}{4}$&amp;$\frac{\sqrt{2}}{4}$&amp;$\frac{\sqrt{2}}{4}(1+\sqrt{2})$&amp;$\frac{\sqrt{2}}{4}(1-\sqrt{2})$\\\hline
$Q_n$&amp;$3+2\sqrt{2}$&amp;$3-2\sqrt{2}$&amp;1&amp;$-1$&amp;$\frac{\sqrt{2}}{2}(2+\sqrt{2})$&amp;$\frac{\sqrt{2}}{2}(2-\sqrt{2})$\\\hline
$J_n$&amp;4&amp;1&amp;$\frac{1}{3}$&amp;$\frac{1}{3}$&amp;$\frac{2}{3}$&amp;$\frac{-1}{3}$\\\hline
$j_n$&amp;4&amp;1&amp;1&amp;$-1$&amp;2&amp;1\\\hline
\end{tabular}
\end{center}
\renewcommand{\arraystretch}{1}
\end{Remark}

The product of the sequences $u_n$ and $v_n$ has a simple generating function
as given in the following proposition.

\begin{Proposition}\label{prop:unvn}Let $u_n$ and $v_n$ be two second-order recurrence sequences given by their Binet formulae:
\[u_n=A\alpha^n-B\beta^n,\quad v_n=C\gamma^n-D\delta^n,\]
where $A,B,C,D,\alpha,\beta,\gamma,\delta$ are defined as in Eqs.~\eqref{eq:alphabeta} and \eqref{eq:AB}.
Then the generating function for $u_nv_n$ is
\[\sum_{n=0}^\infty u_nv_nx^n=\frac{AC}{1-\alpha\gamma x}-\frac{AD}{1-\alpha\delta x}-\frac{BC}{1-\beta\gamma x}+\frac{BD}{1-\beta\delta x}.\]
\end{Proposition}

We mention that a similar statement can be obtained for the products $u_nv_{2n}$, $u_{2n}v_{2n}$, $u_{2n+1}v_{2n}$, $u_{2n+1}v_{2n+1}$, etc.

\begin{Remark}As a special case, let $u_n=F_n$ and $v_n=L_n$. Then it is well known (from Binet formula, for example) that
\[A=B=\frac{1}{\sqrt{5}},\quad\alpha=\frac{1+\sqrt{5}}{2}\quad\beta=\frac{1-\sqrt{5}}{2},\]
\[C=1,\quad D=-1,\quad\gamma=\frac{1+\sqrt{5}}{2}\quad\delta=\frac{1-\sqrt{5}}{2}.\]
The quantity $\frac{1+\sqrt{5}}{2}$ is called the golden ratio (or golden mean, or golden section). For further use we apply the standard notation $\phi$ for this, and $\ophi$ for $\frac{1-\sqrt{5}}{2}$. We remark that $\phi\ophi=-1$ and $\phi-\ophi=\phi^2-\ophi^2=\sqrt{5}$.

Therefore
\[\sum_{n=0}^\infty F_nL_nx^n=\frac{1}{\sqrt{5}}\left(\frac{1}{1-\phi^2x}+\frac{1}{1+x}-\frac{1}{1+x}-\frac{1}{1-\ophi^2x}\right)\]
\[=\frac{1}{\sqrt{5}}\frac{(\phi^2-\ophi^2)x}{(1-\phi^2x)(1-\ophi^2x)}=\frac{x}{x^2-3x+1}.\]
Comparing the result obtained in Example \ref{ex:f2n}, this yields the known identity
\[F_{2n}=F_nL_n.\]
See Mordell's book \cite[pp.\ 60--61]{Mordell}.
\end{Remark}

\begin{Remark}The variation $u_n=F_n$ and $v_n=P_n$ (where $P_n$ are the Pell numbers \seqnum{A000129}) also have combinatorial sense. See the paper of Sellers \cite{Sellers}. 
The generating function for $F_nP_n$ is known \seqnum{A001582},
but it can be deduced using the proposition above:
\[\sum_{n=0}^\infty F_nP_nx^n=\frac{x-x^3}{x^4-2x^3-7x^2-2x+1}.\]
Using Theorem \ref{thm:eqfunvn}, the exponential generating function for $F_nP_n$ is derived:
\[\sum_{n=0}^\infty F_nP_n\frac{x^n}{n!}=\frac{1}{4}\sqrt{\frac{2}{5}}\left[e^{\phi(1+\sqrt{2})x}-e^{\phi(1-\sqrt{2})x}-e^{\ophi(1+\sqrt{2})x}+e^{\ophi(1-\sqrt{2})x}\right].\]
\end{Remark}

\begin{Remark} As the author realized, the sequence $(J_nj_n)$ appears in the on-line encyclopedia \cite{Sloane} but not under this identification ($J_n$ and $j_n$ are called Jacobsthal \seqnum{A001045}
and Jacobsthal-Lucas \seqnum{A014551} numbers). The sequence \seqnum{A002450} has the generating function as $(J_nj_n)$.
Thus, the definition of \seqnum{A002450}
gives the (otherwise elementary but not depicted) observation
\[J_nj_n=\frac{4^n-1}{3}.\]
\end{Remark}

Let us turn the discussion's direction to the determination of generating functions with coefficients $\frac{u_n}{n^q}$. (We do not restrict ourselves
to the case of positive $q$.) To do this, we present the notion of polylogarithms which are themselves generating functions, having coefficients $\frac{1}{n^q}$. Concretely,
\[\Li_q(x)=\sum_{n=1}^\infty\frac{x^n}{n^q}.\]
Because of the coefficients $\frac{1}{n^q}$, it is extremely difficult to find closed forms of these sums but the situation changes when we take negative powers:
\[\sum_{n=1}^\infty n^qx^n=\Li_{-q}(x)=\frac{1}{(1-x)^{q+1}}\sum_{i=0}^{q-1}\euler{q}{i}x^{q-i},\]
where the symbol $\euler{a}{b}$ denotes the Eulerian numbers; that is,
$\euler{a}{b}$ is the number of permutations on the set $1,\dots,a$ in which exactly $b$ elements are greater than the previous element \cite{GKP}.

After these introductory steps, we state the following.

\begin{Proposition}\label{prop:un/nq}For any $u_n$ second-order recurrence sequence and for any $q\in\ZZ$,
\[\sum_{n=1}^\infty\frac{u_n}{n^q}x^n=A\Li_q(\alpha x)-B\Li_q(\beta x).\]
In particular, if $q=1$ then
\begin{equation}
\sum_{n=1}^\infty\frac{u_n}{n}x^n=-A\ln(1-\alpha x)+B\ln(1-\beta x),\label{eq:un/n}
\end{equation}
while for $q=-1$
\[\sum_{n=1}^\infty nu_nx^n=A\frac{x}{(1-\alpha x)^2}-B\frac{x}{(1-\beta x)^2}.\]
\end{Proposition}

Applications can be found at the end of the paper. We mention that the special case $u_n=F_n$ and $x=\frac{1}{2}$ was investigated by Benjamin et al.\ \cite{Benjamin} from a probabilistic point of view. Moreover, we can easily formulate the parallel results for even and odd indices:
\[\sum_{n=1}^\infty\frac{u_{2n}}{n^q}x^n=E\Li_q(\rho x)-F\Li_q(\sigma x),\]
\[\sum_{n=1}^\infty\frac{u_{2n+1}}{n^q}x^n=G\Li_q(\rho x)-H\Li_q(\sigma x).\]



\begin{Remark}In their paper on transcendence theory, Adhikari et al.\ \cite{Adhikari} noted the beautiful fact that the sum
\[\sum_{n=1}^\infty\frac{F_n}{n2^n}\]
is transcendental.

Possessing the results above, we are able to take a closer look at this sum. Let $u_n=F_n$ and $x=\frac{1}{2}$ in Eq.~\eqref{eq:un/n}. Then
\[\sum_{n=1}^\infty\frac{F_n}{n2^n}=-\frac{1}{\sqrt{5}}\ln\left(1-\frac{\phi}{2}\right)+\frac{1}{\sqrt{5}}\ln\left(1-\frac{\ophi}{2}\right)\]
\[=\frac{1}{\sqrt{5}}\left(-\ln\left(\frac{3-\sqrt{5}}{4}\right)+\ln\left(\frac{3+\sqrt{5}}{4}\right)\right)=\frac{1}{\sqrt{5}}\ln\left(\frac{3+\sqrt{5}}{3-\sqrt{5}}\right)\]
\[=\frac{1}{\sqrt{5}}\ln\left(\frac{2-\ophi}{2-\phi}\right).\]
So, this value is a transcendental number.

A similar calculation shows that
\[\sum_{n=1}^\infty\frac{L_n}{n2^n}=2\ln(2),\]
which is again a transcendental number. In addition, we present an interesting example for series whose members' denominators and the sum are the same but the numerators are different. Namely,
\[\sum_{n=1}^\infty\frac{L_n}{n2^n}=2\ln(2)=\sum_{n=1}^\infty\frac{2}{n2^n}.\]

Finding closed form for different arguments of polylogarithms is an
intensively investigated and very hard topic. Fortunately, some
functional equations gives the chance to find a closed form for the sum
of certain series involving Fibonacci and Lucas numbers. In the book
\cite[pp.\ 6--7, 137--139]{Lewin} of Lewin, these are all the
known special values:

\renewcommand{\arraystretch}{1.4}
\begin{longtable}{lll}
$\Li_2(1)$&amp;=&amp;$\frac{\pi^2}{6}$,\\
$\Li_2(-1)$&amp;=&amp;$-\frac{\pi^2}{12}$,\\
$\Li_2\left(\frac{1}{2}\right)$&amp;=&amp;$\frac{\pi^2}{12}-\frac{1}{2}\log^2(2)$,\\
$\Li_2(\ophi)$&amp;=&amp;$\frac{1}{2}\log^2(-\ophi)-\frac{\pi^2}{15}$,\\
$\Li_2(-\ophi)$&amp;=&amp;$-\log^2(-\ophi)+\frac{\pi^2}{10}$,\\
$\Li_2(-\phi)$&amp;=&amp;$\frac{1}{2}\log^2(\phi)-\frac{\pi^2}{10}$,\\
$\Li_2\left(\frac{1}{\phi^2}\right)$&amp;=&amp;$\frac{\pi^2}{15}-\frac{1}{4}\log^2\left(\frac{1}{\phi^2}\right)$,\\
$\Li_3(-1)$&amp;=&amp;$-\frac{3}{4}\zeta(3)$,\\
$\Li_3\left(\frac{1}{2}\right)$&amp;=&amp;$\frac{7}{8}\zeta(3)-\frac{\pi^2}{12}\log(2)+\frac{1}{6}\log^3(2)$,\\
$\Li_3\left(\frac{1}{\phi^2}\right)$&amp;=&amp;$\frac{4}{5}\zeta(3)+\frac{\pi^2}{15}\log\left(\frac{1}{\phi^2}\right)-\frac{1}{12}\log^3\left(\frac{1}{\phi^2}\right)$.
\end{longtable}
\renewcommand{\arraystretch}{1}
Here $\zeta(3)=\Li_3(1)$ is the Ap\'ery's constant without known closed form.

With these identities, we deduce the following beautiful sums:

\begin{eqnarray*}
\sum_{n=1}^\infty\frac{(-1)^nF_n}{\phi^nn^2}&amp;=&amp;\frac{1}{\sqrt{5}}\left(\log^2(\phi)-\frac{3\pi^2}{20}\right),\\
\sum_{n=1}^\infty\frac{(-1)^nL_n}{\phi^nn^2}&amp;=&amp;-\log^2(\phi)-\frac{\pi^2}{60},\\
\sum_{n=1}^\infty\frac{(-1)^nF_n}{\phi^nn^3}&amp;=&amp;\frac{1}{\sqrt{5}}\left(\frac{2\pi^2}{15}\log(\phi)-\frac{2}{3}\log^3(\phi)-\frac{31}{20}\zeta(3)\right),\\
\sum_{n=1}^\infty\frac{(-1)^nL_n}{\phi^nn^3}&amp;=&amp;\frac{1}{20}\zeta(3)-\frac{2\pi^2}{15}\log(\phi)+\frac{2}{3}\log^3(\phi).
\end{eqnarray*}
Using Proposition \ref{prop:un/nq},
\begin{eqnarray*}
\sum_{n=1}^\infty\frac{(-1)^nF_n}{\phi^nn^2}&amp;=&amp;\sum_{n=1}^\infty\frac{F_n}{n^2}\ophi^n=\frac{1}{\sqrt{5}}\Li_2(\phi\ophi)-\frac{1}{\sqrt{5}}\Li_2(\ophi\ophi)\\
&amp;=&amp;\frac{1}{\sqrt{5}}\left(\Li_2(-1)-\Li_2\left(\frac{1}{\phi^2}\right)\right),
\end{eqnarray*}
since $\ophi=\frac{-1}{\phi}$. Using the table of polylogarithms above, an elementary calculation shows the result. The same approach can be applied to derive the other sums (with data from the table with respect to $A,B,\alpha,\beta$).

We can rewrite these sums in a more curious form, because
\[\frac{\sqrt{5}-1}{2}=2\sin\left(\frac{\pi}{10}\right).\]
That is,
\[\frac{-1}{\phi}=\ophi=-2\sin\left(\frac{\pi}{10}\right).\]
Whence, for example,
\[\sum_{n=1}^\infty\frac{(-1)^nF_n}{\phi^nn^2}=\sum_{n=1}^\infty\frac{(-2)^nF_n}{n^2}\sin^n\left(\frac{\pi}{10}\right).\]
\end{Remark}

\section{Exponential generating functions}

The results in the section above can also have exponential versions, which
we give next. Since such expressions often cannot be simplified and finding the exponential generating function is only a substitution of constants, we omit the tables.

\begin{Theorem}\label{thm:expgford_even_odd}The recurrence sequence $u_n$ has the exponential generating function
\[\sum_{n=0}^\infty u_n\frac{x^n}{n!}=Ae^{\alpha x}-Be^{\beta x},\]
while for even and odd indices
\begin{eqnarray*}
\sum_{n=0}^\infty u_{2n}\frac{x^n}{n!}&amp;=&amp;Ee^{\rho x}-Fe^{\sigma x},\\
\sum_{n=0}^\infty u_{2n+1}\frac{x^n}{n!}&amp;=&amp;Ge^{\rho x}-He^{\sigma x},
\end{eqnarray*}
where $E,F,G,H,\rho,\sigma$ are defined in Theorem \ref{thm:evenoddStanica}.
\end{Theorem}

We phrase the exponential version of St\u{a}nic\u{a}'s theorem in a wider sense.

\begin{Theorem}\label{thm:eqfofStanica}We have
\begin{eqnarray*}
\sum_{n=0}^\infty u_n^r\frac{x^n}{n!}=\sum_{k=0}^r\binom{r}{k}A^k(-B)^{r-k}e^{\alpha^k\beta^{r-k}x},\\
\sum_{n=0}^\infty u_{2n}^r\frac{x^n}{n!}=\sum_{k=0}^r\binom{r}{k}E^k(-F)^{r-k}e^{\rho^k\sigma^{r-k}x},\\
\sum_{n=0}^\infty u_{2n+1}^r\frac{x^n}{n!}=\sum_{k=0}^r\binom{r}{k}G^k(-H)^{r-k}e^{\rho^k\sigma^{r-k}x}.
\end{eqnarray*}
\end{Theorem}


The exponential generating function for product of recurrence sequences is presented in the following
\begin{Theorem}\label{thm:eqfunvn}Under the hypotheses of Proposition \ref{prop:unvn}, we have
\[\sum_{n=0}^\infty u_nv_n\frac{x^n}{n!}=ACe^{\alpha\gamma x}-ADe^{\alpha\delta x}-BCe^{\beta\gamma x}+BDe^{\beta\delta x}.\]
Again, the same statement can be obtained for the products $u_nv_{2n}$, $u_{2n}v_{2n}$, $u_{2n+1}v_{2n}$, $u_{2n+1}v_{2n+1}$, etc.
\end{Theorem}

\section{Proofs}

\textit{Proof of Proposition \ref{prop:ordgf}.}
Let the generating function be $f(x)$. Then
\[f(x)-pxf(x)-qx^2f(x)=u_0+u_1x-pu_0x+\sum_{n=2}^\infty(u_n-pu_{n-1}-qu_{n-2})x^n\]
\[=u_0+u_1x-pu_0x,\]
by Eq.~\eqref{eq:undef}. The result follows.\hfill\qed

\textit{Proof of Theorem \ref{thm:oddgf}.}
In order to reach our aim, we need the following identity:
\begin{equation}
u_{2n}=(p^2+2q)u_{2n-2}-q^2u_{2n-4}.\label{eq:u2nlemma}
\end{equation}
Since
\[u_{2n-1}=pu_{2n-2}+qu_{2n-3},\quad\mbox{and}\quad u_{2n-2}=pu_{2n-3}+qu_{2n-4},\]
we get that
\[u_{2n-3}=\frac{1}{q}(u_{2n-1}-pu_{2n-2})=\frac{1}{p}(u_{2n-2}-qu_{2n-4}).\]
If we express $u_{2n-1}$ and consider the identity
\[u_{2n-1}=\frac{1}{p}(u_{2n}-qu_{2n-2}),\]
we will arrive at Eq.~\eqref{eq:u2nlemma}.

Let $f_e(x)$ be the generating function for $u_{2n}$ (``e'' abbreviates the word ``even''). Then
\[q^2x^2f_e(x)-(p^2+2q)xf_e(x)+f_e(x)\]
\[=q^2\sum_{n=2}^\infty u_{2n-4}x^n-(p^2+2q)\sum_{n=1}^\infty u_{2n-2}x^n+\sum_{n=0}^\infty u_{2n}x^n\]
\[=\sum_{n=2}^\infty (q^2u_{2n-4}-(p^2+2q)u_{2n-2}+u_{2n})x^n-(p^2+2q)u_0x+u_0+u_2x.\]
\[=u_0+(u_2-u_0(p^2+2q))x.\]
We get the result.

Let $f_o(x)$ be the generating function for the sequence $u_{2n+1}$.
\[pf_o(x)+qf_e(x)=\sum_{n=0}^\infty(pu_{2n+1}+qu_{2n})x^n=\sum_{n=0}^\infty u_{2n+2}x^n\]
\[=\frac{1}{x}\sum_{n=1}^\infty u_{2n}x^n=\frac{1}{x}\left(\sum_{n=0}^\infty u_{2n}x^n-u_0\right)=\frac{1}{x}(f_e(x)-u_0).\]
Thus
\[f_o(x)=\frac{1}{p}\left(f_e(x)\left(\frac{1}{x}-q\right)-\frac{u_0}{x}\right).\]
If we consider the closed form of $f_e(x)$ this formula can be transformed into the wanted form.\hfill\qed

\textit{Proof of Theorem \ref{thm:evenoddStanica}.}
We know (see Eq.~\eqref{eq:u2nlemma}), that
\[u_{2n}=(p^2+2q)u_{2n-2}-q^2u_{2n-4}.\]
This allows us to construct a second-order recurrence sequence $v_n$ from $u_n$ with the property
\[v_n=u_{2n},\]
namely,
\begin{equation}
v_n:=(p^2+2q)v_{n-1}-q^2v_{n-2},\quad v_0:=u_0,\;v_1:=u_2.\label{eq:vn}
\end{equation}
Therefore
\[\rho=\frac{1}{2}\left(p^2+2q+\sqrt{(p^2+2q)^2+4(-q^2)}\right)=\frac{1}{2}\left(p^2+2q+p\sqrt{p^2+4q}\right),\]
\[\sigma=\frac{1}{2}\left(p^2+2q-\sqrt{(p^2+2q)^2+4(-q^2)}\right)=\frac{1}{2}\left(p^2+2q-p\sqrt{p^2+4q}\right),\]
\[E=\frac{v_1-v_0\sigma}{\rho-\sigma}=\frac{u_2-u_0\sigma}{\rho-\sigma},\quad F=\frac{v_1-v_0\rho}{\rho-\sigma}=\frac{u_2-u_0\rho}{\rho-\sigma}\]
with respect to the sequence $v_n$. That is,
\[v_n=E\rho^n-F\sigma^n.\]
If we apply St\u{a}nic\u{a}'s theorem for $v_n=u_{2n}$, we get the first statement. Secondly, we find the corresponding identity of Eq.~\eqref{eq:u2nlemma}.
\[u_{2n-1}=pu_{2n-2}+qu_{2n-3},\quad\mbox{and}\quad u_{2n}=pu_{2n-1}+qu_{2n-2}.\]
We express $u_{2n-2}$ from these:
\[u_{2n-2}=\frac{1}{p}(u_{2n-1}-qu_{2n-3})=\frac{1}{q}(u_{2n}-pu_{2n-1}),\]
whence
\[u_{2n}=\frac{q}{p}(u_{2n-1}-qu_{2n-3})+pu_{2n-1}.\]
On the other hand,
\[u_{2n}=\frac{1}{p}(u_{2n+1}-qu_{2n-1}).\]
Putting together the last two equalities we get the wanted formula:
\begin{equation}
u_{2n+1}=(p^2+2q)u_{2n-1}-q^2u_{2n-3}.\label{eq:u2n+1lemma}
\end{equation}
Again, we are able to construct the sequence $w_n$ for which
\[w_n=u_{2n+1}.\]
We are in the same situation as before. The only thing we should care about is that
\[w_0=u_1,\;w_1=u_3.\]
\hfill\qed

\textit{Proof of Proposition \ref{prop:unvn}.}
If $u_n$ and $v_n$ have the form as in the proposition, then we see that
\[u_nv_n=AC(\alpha\gamma)^n-AD(\alpha\delta)^n-BC(\beta\gamma)^n+BD(\beta\delta)^n.\]
Thus
\[\sum_{n=0}^\infty u_nv_nx^n\]
\[=AC\sum_{n=0}^\infty(\alpha\gamma x)^n-AD\sum_{n=0}^\infty(\alpha\delta x)^n-BC\sum_{n=0}^\infty(\beta\gamma x)^n+BD\sum_{n=0}^\infty(\beta\delta x)^n.\]
The result follows. In addition, we mention that there are too many parameters, so it is not worth to look for an expression with parameters $u_0,u_1,v_0,v_1,p,q,r,s$ directly. However, the remains can be completed easily, as the author calculated for the standard sequences.\hfill\qed

\textit{Proof of Proposition \ref{prop:un/nq}.}
It is straightforward from Binet formula and the definition of polylogarithms.\hfill\qed

\textit{Proof of Theorem \ref{thm:expgford_even_odd}.} This proof is again straightforward,
\[\sum_{n=0}^\infty u_n\frac{x^n}{n!}=A\sum_{n=0}^\infty\frac{(\alpha x)^n}{n!}-B\sum_{n=0}^\infty\frac{(\beta x)^n}{n!}=Ae^{\alpha x}-Be^{\beta x}.\]
Finally, we choose $v_n$ and $w_n$ as in the proof of Theorem \ref{thm:evenoddStanica}, and follow the usual argument.\hfill\qed

\textit{Proofs of Theorems \ref{thm:eqfofStanica} and \ref{thm:eqfunvn}.}
The binomial theorem, the same approach as described in the proof of Theorem \ref{thm:evenoddStanica} and the Binet formula immediately gives the results:
\begin{eqnarray*}
\sum_{n=0}^\infty u_n^r\frac{x^n}{n!}&amp;=&amp;\sum_{n=0}^\infty(A\alpha^n-B\beta^n)^r\frac{x^n}{n!}\\
&amp;=&amp;\sum_{n=0}^\infty\sum_{k=0}^r\binom{r}{k}A^k(\alpha^k)^n(-B)^{r-k}(\beta^{r-k})^n\frac{x^n}{n!}\\
&amp;=&amp;\sum_{k=0}^r\binom{r}{k}A^k(-B)^{r-k}\sum_{n=0}^\infty(\alpha^k)^n(\beta^{r-k})^n\frac{x^n}{n!}\\
&amp;=&amp;\sum_{k=0}^r\binom{r}{k}A^k(-B)^{r-k}e^{\alpha^k\beta^{r-k}x}.
\end{eqnarray*}
The rest can be proven by the same approach.\hfill\qed

\newpage

\section{Tables}

\vspace{5mm}

\begin{center}
\textbf{Standard parameters for the named sequences}\\\vspace{2mm}
\begin{longtable}{|l|c|c|c|c|c|c|}\hline
Name&amp;Notation&amp;$u_0$&amp;$u_1$&amp;$p$&amp;$q$&amp;First few values\\\hline\hline
Fibonacci&amp;$F_n$&amp;0&amp;1&amp;1&amp;1&amp;0,\,1,\,1,\,2,\,3,\,5,\,8,\,13,\,21\\\hline
Lucas&amp;$L_n$&amp;2&amp;1&amp;1&amp;1&amp;2,\,1,\,3,\,4,\,7,\,11,\,18,\,29,\,47\\\hline
Pell&amp;$P_n$&amp;0&amp;1&amp;2&amp;1&amp;0,1,2,5,12,29,70,169,408\\\hline
Pell-Lucas&amp;$Q_n$&amp;2&amp;2&amp;2&amp;1&amp;2,2,6,14,34,82,198,478\\\hline
Jacobsthal&amp;$J_n$&amp;0&amp;1&amp;1&amp;2&amp;0,\,1,\,1,\,3,\,5,\,11,\,21,\,43,\,85\\\hline
Jacobsthal-Lucas&amp;$j_n$&amp;2&amp;1&amp;1&amp;2&amp;2,1,5,7,17,31,65,127,257\\\hline
\end{longtable}
\end{center}

\renewcommand{\arraystretch}{1.4}

\begin{center}
\begin{longtable}{|c|c|c|c|c|}\hline
Sequence&amp;$A$&amp;$B$&amp;$\alpha$&amp;$\beta$\\\hline\hline
$F_n$&amp;$\frac{1}{\sqrt{5}}$&amp;$\frac{1}{\sqrt{5}}$&amp;$\frac{1+\sqrt{5}}{2}$&amp;$\frac{1-\sqrt{5}}{2}$\\\hline
$L_n$&amp;1&amp;$-1$&amp;$\frac{1+\sqrt{5}}{2}$&amp;$\frac{1-\sqrt{5}}{2}$\\\hline
$P_n$&amp;$\frac{\sqrt{2}}{4}$&amp;$\frac{\sqrt{2}}{4}$&amp;$1+\sqrt{2}$&amp;$1-\sqrt{2}$\\\hline
$Q_n$&amp;1&amp;$-1$&amp;$1+\sqrt{2}$&amp;$1-\sqrt{2}$\\\hline
$J_n$&amp;$\frac{1}{3}$&amp;$\frac{1}{3}$&amp;2&amp;$-1$\\\hline
$j_n$&amp;1&amp;$-1$&amp;2&amp;$-1$\\\hline
\end{longtable}
\end{center}

\begin{center}
\textbf{Ordinary generating functions}\\\vspace{2mm}
\begin{longtable}{|c|c|}\hline
Coefficient of $x^n$&amp;Generating function\\\hline\hline
$F_n$&amp;$\frac{x}{1-x-x^2}$\\
$L_n$&amp;$\frac{2-x}{1-x-x^2}$\\
$P_n$&amp;$\frac{x}{1-2x-x^2}$\\
$Q_n$&amp;$\frac{2-2x}{1-2x-x^2}$\\
$J_n$&amp;$\frac{x}{1-x-2x^2}$\\
$j_n$&amp;$\frac{2-x}{1-x-2x^2}$\\\hline
\end{longtable}
\end{center}

\newpage

\begin{center}
\textbf{Generating functions of even and odd indices}\\\vspace{2mm}
\begin{longtable}{|c|c||c|c|}\hline
$x^n$&amp;Generating function&amp;$x^n$&amp;Generating function\\\hline\hline
$F_{2n}$&amp;$\frac{x}{1-3x+x^2}$&amp;$F_{2n+1}$&amp;$\frac{1-x}{1-3x+x^2}$\\
$L_{2n}$&amp;$\frac{2-3x}{1-3x+x^2}$&amp;$L_{2n+1}$&amp;$\frac{1+x}{1-3x+x^2}$\\
$P_{2n}$&amp;$\frac{2x}{1-6x+x^2}$&amp;$P_{2n+1}$&amp;$\frac{1-x}{1-6x+x^2}$\\
$Q_{2n}$&amp;$\frac{2-6x}{1-6x+x^2}$&amp;$Q_{2n+1}$&amp;$\frac{2+2x}{1-6x+x^2}$\\
$J_{2n}$&amp;$\frac{x}{1-5x+4x^2}$&amp;$J_{2n+1}$&amp;$\frac{1-2x}{1-5x+4x^2}$\\
$j_{2n}$&amp;$\frac{2-5x}{1-5x+4x^2}$&amp;$j_{2n+1}$&amp;$\frac{1+2x}{1-5x+4x^2}$\\\hline
\end{longtable}
\end{center}

\vspace{10mm}

\begin{center}
\textbf{Generating functions for products of sequences}\\\vspace{2mm}
\begin{longtable}{|c|c|}\hline
Coefficient of $x^n$&amp;Generating function\\\hline\hline
$F_nL_n$&amp;$\frac{x}{1-3x+x^2}$\\\hline
$F_nP_n$&amp;$\frac{x-x^3}{1-2x-7x^2-2x^3+x^4}$\\\hline
$F_nQ_n$&amp;$\frac{2x+2x^2+2x^3}{1-2x-7x^2-2x^3+x^4}$\\\hline
$F_nJ_n$&amp;$\frac{1-2x^2}{1-x-7x^2-2x^3+4x^4}$\\\hline
$F_nj_n$&amp;$\frac{x+4x^2+2x^3}{1-x-7x^2-2x^3+4x^4}$\\\hline
$L_nP_n$&amp;$\frac{x+4x^2+x^3}{1-2x-7x^2-2x^3+x^4}$\\\hline
$L_nQ_n$&amp;$\frac{4-6x-14x^2-2x^3}{1-2x-7x^2-2x^3+x^4}$\\\hline
$L_nJ_n$&amp;$\frac{x+2x^2+2x^3}{1-x-7x^2-2x^3+4x^2}$\\\hline
$L_nj_n$&amp;$\frac{4-3x-14x^2-2x^3}{1-x-7x^2-2x^3+4x^4}$\\\hline
$P_nQ_n$&amp;$\frac{2x}{1-6x+x^2}$\\\hline
$P_nJ_n$&amp;$\frac{x-2x^3}{1-2x-13x^2-4x^3+4x^4}$\\\hline
$P_nj_n$&amp;$\frac{x+8x^2+2x^3}{1-2x-13x^2-4x^3+4x^4}$\\\hline
$Q_nJ_n$&amp;$\frac{2x+2x^2+4x^3}{1-2x-13x^2-4x^3+4x^4}$\\\hline
$Q_nj_n$&amp;$\frac{4-6x-26x^2-4x^3}{1-2x-13x^2-4x^3+4x^4}$\\\hline
$J_nj_n$&amp;$\frac{x}{1-5x+4x^2}$\\\hline
\end{longtable}
\end{center}

\vspace{10mm}

\begin{center}
\textbf{Generating functions for squares}\\\vspace{2mm}
\begin{longtable}{|c|c||c|c||c|c|c}\hline
$x^n$&amp;Gen. function&amp;$x^n$&amp;Gen. function&amp;$x^n$&amp;Gen. function\\\hline\hline
$F_n^2$&amp;$\frac{x-x^2}{1-2x-2x^2+x^3}$&amp;$F_{2n}^2$&amp;$\frac{x+x^2}{1-8x+8x^2-x^3}$&amp;$F_{2n+1}^2$&amp;$\frac{1-4x+x^2}{1-8x+8x^2-x^3}$\\\hline
$L_n^2$&amp;$\frac{4-7x-x^2}{1-2x-2x^2+x^3}$&amp;$L_{2n}^2$&amp;$\frac{4-23x+9x^2}{1-8x+8x^2-x^3}$&amp;$L_{2n+1}^2$&amp;$\frac{1+8x+x^2}{1-8x+8x^2-x^3}$\\
$P_n^2$&amp;$\frac{x-x^2}{1-5x-5x^2+x^3}$&amp;$P_{2n}^2$&amp;$\frac{4x+4x^2}{1-35x+35x^2-x^3}$&amp;$P_{2n+1}^2$&amp;$\frac{1-10x+x^2}{1-35x+35x^2-x^3}$\\\hline
$Q_n^2$&amp;$\frac{4-16x-4x^2}{1-5x-5x^2+x^3}$&amp;$Q_{2n}^2$&amp;$\frac{4-104x+36x^2}{1-35x+35x^2-x^3}$&amp;$Q_{2n+1}^2$&amp;$\frac{4+56x+4x^2}{1-35x+35x^2-x^3}$\\\hline
$J_n^2$&amp;$\frac{x-2x^2}{1-3x-6x^2+8x^3}$&amp;$J_{2n}^2$&amp;$\frac{x+4x^2}{1-21x+84x^2-64x^3}$&amp;$J_{2n+1}^2$&amp;$\frac{1-12x+16x^2}{1-21x+84x^2-64x^3}$\\\hline
$j_n^2$&amp;$\frac{4-11x-2x^2}{1-3x-6x^2+8x^3}$&amp;$j_{2n}^2$&amp;$\frac{4-59x+100x^2}{1-21x+84x^2-64x^3}$&amp;$j_{2n+1}^2$&amp;$\frac{1+28x+16x^2}{1-21x+84x^2-64x^3}$\\\hline
\end{longtable}
\end{center}

\vspace{10mm}

\begin{center}
\textbf{Generating functions for sequences $(n\cdot u_{(2)n})$}\\\vspace{2mm}
\begin{longtable}{|c|c||c|c|c|}\hline
$x^n$&amp;Gen. function&amp;$x^n$&amp;Gen. function\\\hline\hline
$nF_n$&amp;$\frac{x+x^3}{1-2x-x^2+2x^3+x^4}$&amp;$nF_{2n}$&amp;$\frac{x-x^3}{1-6x+11x^2-6x^3+x^4}$\\
$nL_n$&amp;$\frac{x+4x^2-x^3}{1-2x-x^2+2x^3+x^4}$&amp;$nL_{2n}$&amp;$\frac{3x-4x^2+3x^3}{1-6x+11x^2-6x^3+x^4}$\\
$nP_n$&amp;$\frac{x+x^3}{1-4x+2x^2+4x^3+x^4}$&amp;$nP_{2n}$&amp;$\frac{2x-2x^3}{1-12x+38x^2-12x^3+x^4}$\\
$nQ_n$&amp;$\frac{2x+4x^2-2x^3}{1-4x+2x^2+4x^3+x^4}$&amp;$nQ_{2n}$&amp;$\frac{6x-4x^2+6x^3}{1-12x+38x^2-12x^3+x^4}$\\
$nJ_n$&amp;$\frac{x+2x^3}{1-2x-3x^2+4x^3+4x^4}$&amp;$nJ_{2n}$&amp;$\frac{x-4x^3}{1-10x+33x^2-40x^3+16x^4}$\\
$nj_n$&amp;$\frac{x+8x^2-2x^3}{1-2x-3x^2+4x^3+4x^4}$&amp;$nj_{2n}$&amp;$\frac{5x-16x^2+20x^3}{1-10x+33x^2-40x^3+16x^4}$\\\hline
\end{longtable}
\end{center}

\vspace{10mm}

\begin{center}
\textbf{Generating functions for sequences $(n\cdot u_{2n+1})$}\\\vspace{2mm}
\begin{longtable}{|c|c|}\hline
Coefficient of $x^n$&amp;Generating function\\\hline\hline
$nF_{2n+1}$&amp;$\frac{2x-2x^2+x^3}{1-6x+11x^2-6x^3+x^4}$\\
$nL_{2n+1}$&amp;$\frac{4x-2x^2-x^3}{1-6x+11x^2-6x^3+x^4}$\\
$nP_{2n+1}$&amp;$\frac{5x-2x^2+x^3}{1-12x+38x^2-12x^3+x^4}$\\
$nQ_{2n+1}$&amp;$\frac{14x-4x^2-2x^3}{1-12x+38x^2-12x^3+x^4}$\\
$nJ_{2n+1}$&amp;$\frac{3x-8x^2+8x^3}{1-10x+33x^2-40x^3+16x^4}$\\
$nj_{2n+1}$&amp;$\frac{7x-8x^2-8x^3}{1-10x+33x^2-40x^3+16x^4}$\\\hline
\end{longtable}
\end{center}

\renewcommand{\arraystretch}{1}

\newpage

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\bibitem[2]{Benjamin}A. T. Benjamin, J. D. Neer, D. E. Otero, J. A.
Sellers, A probabilistic view of certain weighted Fibonacci sums,
\emph{Fib. Quart.} \textbf{41} (4) (2003), 360--364.

\bibitem[3]{Carlitz}L. Carlitz, Generating functions for powers of
certain sequences of numbers, \emph{Duke Math. J.} \textbf{29} (1962),
521--537.

\bibitem[4]{Golomb}S. Golomb, Problem 4270, \emph{Amer. Math. Monthly}
\textbf{64}(1) (1957), p. 49.

\bibitem[5]{GKP}R. L. Graham, D. E. Knuth and O. Patashnik,
\emph{Concrete Mathematics}, Addison Wesley, 1993.

\bibitem[6]{Johnson}R. C. Johnson, \emph{Matrix method for Fibonacci
and related sequences}, notes for undegraduates, 2006. Available online
at \href{http://www.dur.ac.uk/bob.johnson/fibonacci/}{\tt http://www.dur.ac.uk/bob.johnson/fibonacci/}.

\bibitem[7]{Lewin}L. Lewin, \emph{Dilogarithms and Associated
Functions}, Macdonald, London, 1958.

\bibitem[8]{Mordell}L. J. Mordell, Diophantine Equations, Academic
Press, London and New York, 1969.

\bibitem[9]{Riordan}J. Riordan, Generating functions for powers of
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\end{thebibliography}

\bigskip
\hrule
\bigskip

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

\noindent \emph{Keywords: } 
Recurrence sequences, Fibonacci sequence, Lucas sequence, Pell
sequence, Pell-Lucas sequence, Jacobsthal sequence, Jacobsthal-Lucas
sequence, generating function, exponential generating function.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A000129},
\seqnum{A001045},
\seqnum{A001582},
\seqnum{A002450}, and
\seqnum{A014551}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received July 17 2007;
revised version received  April 24 2009.
Published in {\it Journal of Integer Sequences}, April 29 2009.

\bigskip
\hrule
\bigskip

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


\end{document}

                                                                                

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