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\begin{document}
\vspace*{-60pt} 
\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5(1) 
(2005), \#A31} 
\vskip 50pt 
\begin{center}
{\bf AN INFINITE FAMILY OF DUAL SEQUENCE IDENTITIES}
\vskip 20pt
{\bf Robin Chapman}\\
{\smallit Department of Mathematics,
University of Exeter, Exeter, EX4 4QE, UK}\\
{\tt rjc@maths.ex.ac.uk}
\end{center}
\vskip 30pt
\centerline{\smallit Received: 7/27/05, Accepted: 11/13/05,
 Published: 12/9/05}
\vskip 30pt

\centerline{\bf Abstract}

\noindent
We subsume three identities of Sun~[2] into an infinite family
of identities, and consider some special cases.

\pagestyle{myheadings}
\markright{\smalltt INTEGERS: \smallrm
 ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5(1) (2005),
\#A31\hfill}

\thispagestyle{empty}
\baselineskip=15pt
\vskip 30pt

\section*{\normalsize 1. Introduction}

Let $(a_n)_{n=0}^\infty$ be a sequence of complex numbers.
We call the sequence $(a_n^*)_{n=0}^\infty$ defined by
$$a_n^*=\sum_{i=0}^n{n\choose i}(-1)^i a_i$$
the \emph{dual sequence} of $(a_n)$. Repeating this operation
gives rise to the original sequence: $a_n^{**}=a_n$ for all
$n$~[1, 192--193].

A sequence $(a_n)$ satisfying $a_n^*=a_n$ is called \emph{self-dual}.
An important example of a self-dual sequence is $((-1)^n B_n)$ where
the Bernoulli numbers $B_n$ satisfy
$$\sum_{n=0}^\infty B_n\frac{t^n}{n!}=\frac{t}{\exp(t)-1}.$$
Sun~[2] considered the polynomials
\begin{equation}\label{A_n definition}
A_n(x)=\sum_{i=0}^n{n\choose i}(-1)^i a_i x^{n-i}
\quad\textrm{and}\quad
A_n^*(x)=\sum_{i=0}^n{n\choose i}(-1)^i a_i^* x^{n-i}.
\end{equation}
When $a_k=(-1)^k B_k$ then $A_n(x)=B_n(x)$, the classical Bernoulli
polynomial.
Sun's main theorem~[2, Theorem 1.1]
gives three identities involving the polynomials $A_n$ and $A_n^*$.
Using these identities he unifies and generalizes various identities
concerning Bernoulli numbers and polynomials in the recent literature.
The main theorem here gives an infinite family of identities
subsuming Sun's identities.

\vskip 30pt

\section*{\normalsize 2. The main theorem}

We begin with preliminaries concerning exponential generating functions.
Throughout we let $(a_n)_{n=0}^\infty$ be a sequence and $(a^*_n)_{n=0}^\infty$
be its dual. Also we let $A_n$ and $A_n^*$ be the polynomials defined
by~(\ref{A_n definition}).

We define
$$\al(t)=\sum_{n=0}^\infty a_n\frac{t^n}{n!}
\quad\textrm{and}\quad\al^*(t)=\sum_{n=0}^\infty a_n^*\frac{t^n}{n!}$$
as the exponential generating functions of the sequences $(a_n)$
and $(a_n^*)$.
Then
$$\al^*(t)=\sum_{n=0}^\infty\sum_{i=0}^n{n\choose i}(-1)^i a_i\frac{t^n}{n!}
=\sum_{i=0}^\infty a_i\frac{(-t)^i}{i!}\sum_{n=i}^\infty\frac{t^{n-i}}{(n-i)!}
=\exp(t)\al(-t).$$
As a consequence,
$$\exp(t)\al^*(-t)=\exp(t)\exp(-t)\al(t)=\al(t)$$
which recovers the fact that $a_n^{**}=a_n$.

It is well-known that the Bernoulli polynomials $B_n(x)$ are symmetric
or anti-symmetric about $1/2$ according to the parity of~$n$, to wit
$$B_n(1-x)=(-1)^n B_n(x).$$
The following lemma generalizes this to arbitrary sequences, relating
the polynomials $A_n$ and $A_n^*$.

\noindent
{\bf Lemma 1} \textbf{(Duality principle)}
\emph{
For each $n$, $A_n^*(x)=(-1)^n A_n(1-x).$}

\noindent
{\it Proof.}
First
\begin{eqnarray*}
\sum_{n=0}^\infty A_n(x)\frac{t^n}{n!}
&=&\sum_{n=0}^\infty\sum_{i=0}^n{n\choose i}(-1)^i a_i x^{n-i}\frac{t^n}{n!}\\
&=&\sum_{i=0}^\infty a_i\frac{(-t)^i}{i!}
\sum_{n=i}^\infty\frac{(xt)^{n-i}}{(n-i)!}=\exp(xt)\al(-t).
\end{eqnarray*}
Similarly
$$\sum_{n=0}^\infty A_n^*(x)\frac{t^n}{n!}=\exp(xt)\al^*(-t)=\exp((x-1)t)\al(t)
=\sum_{n=0}^\infty A_n(1-x)\frac{(-t)^n}{n!}.$$
Hence $A_n^*(x)=(-1)^n A_n(1-x)$ for all~$n$.
\epf

Sun's main theorem uses three variables $x$, $y$ and $z$
satisfying $x+y+z=1$ and expresses three sums featuring
polynomials $A_m(y)$ and $A_n^*(z)$ with coefficients in $x$ as
expressions depending only on~$x$. Our main theorem generalizes the first
two of these identities. We later see how to generalize the third.

\noindent
{\bf Theorem 2}\emph{
Let $x+y+z=1$. Then for all integers $k$, $l$, $r\ge0$,
\begin{eqnarray}\label{main identity}
&&(-1)^k\sum_{j=0}^k{k\choose j}x^{k-j}\frac{(l+j)!A_{l+j+r}(y)}{(l+j+r)!}\\
&&{}-(-1)^{j+r}\sum_{j=0}^l{l\choose j}x^{l-j}
\frac{(k+j)!A^*_{k+j+r}(z)}{(k+j+r)!}
\nonumber\\
&=&k!l!(-x)^{k+l+1}
\sum_{i=0}^{r-1}\frac{(-1)^i a_i}{i!(k+l+r-i)!}\nonumber\\
&&{}\times\sum_{j=0}^{r-1-i}
{k+r-i-1-j\choose k}{l+j\choose l}y^{r-i-1-j}(1-z)^j.\nonumber
\end{eqnarray}
}

\noindent
{\it Proof.}
We first recast (\ref{main identity}) in a more symmetric form.
We replace $z$ by $1-z$ and use the duality principle. Then $x=1-y-z$
must be replaced by $z-y$. After some rearrangement we find that
(\ref{main identity}) is equivalent to
\begin{eqnarray}\label{new main identity}
&&\sum_{j=0}^k{k\choose j}(z-y)^{k-j}\frac{(l+j)!A_{l+j+r}(y)}{(l+j+r)!}\\
&&{}-\sum_{j=0}^l{l\choose j}(y-z)^{l-j}\frac{(k+j)!A_{k+j+r}(z)}{(k+j+r)!}
\nonumber\\
&=&(-1)^k k!l!(y-z)^{k+l+1}
\sum_{i=0}^{r-1}\frac{(-1)^i a_i}{i!(k+l+r-i)!}\nonumber\\
&&{}\times\sum_{j=0}^{r-1-i}
{k+r-i-1-j\choose k}{l+j\choose l}y^{r-i-1-j}z^j.\nonumber
\end{eqnarray}

Define
$$\ca_r(k,l;y,z)=
\sum_{j=0}^k{k\choose j}(z-y)^{k-j}(l+j)!\frac{A_{l+j+r}(y)}{(l+j+r)!}$$
so that (\ref{new main identity}) is concerned with 
$\ca_r(k,l;y,z)-\ca_r(l,k;z,y)$.
We consider the exponential generating function of the $\ca_r(k,l;y,z)$.
We calculate
\begin{eqnarray*}
&&\sum_{k,l=0}^\infty\ca_r(k,l;y,z)\frac{t^ku^l}{k!l!}\\
&=&\sum_{k,l=0}^\infty\sum_{j=0}^k
{k\choose j}(z-y)^{k-j}(l+j)!\frac{A_{l+j+r}(y)}{(l+j+r)!}\frac{t^ku^l}{k!l!}\\
&=&\sum_{j,l=0}^\infty {l+j\choose j}t^ju^l\frac{A_{l+j+r}(y)}{(l+j+r)!}
\sum_{k=j}^\infty\frac{(z-y)^{k-j}t^{k-j}}{(k-j)!}\\
&=&\exp((z-y)t)\sum_{p=0}^\infty(t+u)^p\frac{A_{p+r}(y)}{(p+r)!}\\
&=&\frac{\exp((z-y)t)}{(t+u)^r}
\left[\exp(y(t+u))\al(-t-u)-\sum_{s=0}^{r-1}A_s(y)\frac{(t+u)^s}{s!}\right]\\
&=&\frac{\exp(zt+yu)\al(-t-u)}{(t+u)^r}-\frac{\exp((z-y)t)}{(t+u)^r}
\sum_{s=0}^{r-1}A_s(y)\frac{(t+u)^s}{s!}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
&&\sum_{k,l=0}^\infty[\ca_r(k,l;y,z)-\ca_r(l,k;z,y)]
\frac{t^ku^l}{k!l!}\\
&=&\frac{1}{(t+u)^r}\sum_{s=0}^{r-1}
[\exp((y-z)u)A_s(z)-\exp((z-y)t)A_s(y)]\frac{(t+u)^s}{s!}.
\end{eqnarray*}
As
$$\sum_{s=0}^{r-1}A_s(x)\frac{v^s}{s!}
=\sum_{s=0}^{r-1}\sum_{i=0}^s(-1)^i\frac{a_i}{i!}\frac{x^{s-i}}{(s-i)!}v^s
=\sum_{i=0}^{r-1}(-1)^i a_i\frac{v^i}{i!}
\sum_{s=i}^{r-1}\frac{(xv)^{s-i}}{(s-i)!}$$
then
\begin{eqnarray*}
&&\sum_{k,l=0}^\infty[\ca_r(k,l;y,z)-\ca_r(l,k;z,y)]
\frac{t^ku^l}{k!l!}\\
&=&\sum_{i=0}^{r-1}
\frac{(-1)^i a_i}{i!(t+u)^{r-i}}\sum_{s=i}^{r-1}
\frac{(t+u)^{s-i}}{(s-i)!}[z^{s-i}\exp((y-z)u)-y^{s-i}\exp((z-y)t)]\\
&=&\sum_{i=0}^{r-1}
\frac{(-1)^i a_i}{i!(t+u)^{r-i}}\sum_{s=0}^{r-1-i}
\frac{(t+u)^s}{s!}[z^s\exp((y-z)u)-y^s\exp((z-y)t)]\\
&=&\sum_{i=0}^{r-1}\frac{(-1)^i a_i}{i!}\cb_{r-i}(t,u,y,z)
\end{eqnarray*}
where
$$\cb_q(t,u,y,z)=
\frac{1}{(t+u)^q}\sum_{s=0}^{q-1}
\frac{(t+u)^s}{s!}[z^s\exp((y-z)u)-y^s\exp((z-y)t)].$$

We claim that
\begin{eqnarray}\label{evil identity}
&&\cb_q(t,u,y,z)\\
&=&\sum_{k,l=0}^\infty\frac{(y-z)^{k+l+1}}{(k+l+r)!}(-t)^ku^l
\sum_{j=0}^{q-1}{k+q-1-j\choose q-1-j}{l+j\choose j}y^{q-1-j}z^j
\nonumber\\
&=&\sum_{m=0}^\infty\frac{(y-z)^{m+1}}{(m+q)!}F_{m,q}(t,u,y,z)
\nonumber
\end{eqnarray}
where
$$F_{m,q}(t,u,y,z)=\sum_{k=0}^m\sum_{j=0}^{q-1}
{k+q-1-j\choose q-1-j}{m-k+j\choose j}(-t)^k u^{m-k}y^{q-1-j}z^j.$$
We remark that
$$F_{m,q}(t,u,y,z)=\sum_{j=0}^{q-1}G_{m,q-1-j,j}(t,u)y^{q-1-j}z^j$$
where
$$G_{m,i,j}(t,u)=\sum_{k=0}^m{k+i\choose i}{m-k+j\choose j}(-t)^ku^{m-k}.$$

We need to calculate $(t+u)G_{m,i,j}(t,u)$.
In doing this we adopt the convention
that ${x\choose i}$ is defined for arbitrary $x$ by
insisting that it be a polynomial of degree~$i$. Adopting this convention
ensures that ${-1\choose i}={i+1\choose i}=0$. We must take particular
care with the cases where $i=0$ or $j=0$.
If $i>0$ and $j>0$ then
\begin{eqnarray*}
&&(t+u)G_{m,i,j}(t,u)\\
&=&\sum_{k=0}^{m+1}{k+i\choose i}{m-k+j\choose j}(-t)^ku^{m+1-k}\\
&&{}-\sum_{k=-1}^m{k+i\choose i}{m-k+j\choose j}(-t)^{k+1}u^{m-k}\\
&=&\sum_{k=0}^{m+1}\left[{k+i\choose i}{m-k+j\choose j}
-{k-1+i\choose i}{m-k+1+j\choose j}\right]\\
&&{}\times(-t)^ku^{m+1-k}\\
&=&\sum_{k=0}^{m+1}\left[{k+i-1\choose i-1}{m-k+1+j\choose j}
-{k+i\choose i}{m-k+1+j-1\choose j-1}\right]\\
&&\times{}(-t)^ku^{m+1-k}\\
&=&G_{m+1,i-1,j}(t,u)-G_{m+1,i,j-1}(t,u).
\end{eqnarray*}
If $i>0$ then
\begin{eqnarray*}
(t+u)G_{m,i,0}&=&\sum_{k=0}^m{k+i\choose i}(-t)^ku^{m+1-k}
-\sum_{k=-1}^m{k+i\choose i}(-t)^{k+1}u^{m-k}\\
&=&\sum_{k=0}^m{k+i\choose i}(-t)^ku^{m+1-k}
-\sum_{k=0}^{m+1}{k-1+i\choose i}(-t)^k u^{m+1-k}\\
&=&\sum_{k=0}^{m+1}{k+i-1\choose i-1}(-t)^ku^{m+1-k}
-{m+1+i\choose i}(-t)^{m+1}\\
&&{}-{m+i\choose i}(-t)^{m+1}\\
&=&G_{m+1,i-1,0}(t,u)-{m+1+i\choose i}(-t)^{m+1}.
\end{eqnarray*}
Similarly, if $j>0$ then
$$(t+u)G_{m,0,j}(t,u)={m+1+j\choose j}u^{m+1}-G_{m+1,0,j-1}(t,u).$$
Finally,
$$(t+u)G_{m,0,0}(t,u)=(t+u)\sum_{k=0}^m(-t)^ku^{m+1-k}
=u^{m+1}-(-t)^{m+1}.$$

It follows that
$$(t+u)F_{m,1}(t,u,y,z)=(t+u)G_{m,0,0}(t,u)=u^{m+1}-(-t)^{m+1},$$
\begin{eqnarray*}
&&(t+u)F_{m,2}(t,u,y,z)\\
&=&(t+u)(G_{m,0,1}(t,u)z+G_{m,1,0}(t,u)y)\\
&=&(m+2)u^{m+1}z+G_{m,0,0}(t,u)(y-z)-(m+2)(-t)^{m+1}y,
\end{eqnarray*}
and for $r\ge3$,
\begin{eqnarray*}
&&(t+u)F_{m,r}(t,u,y,z)\\
&=&{m+r\choose r-1}u^{m+1}z^{r-1}-G_{m+1,0,r-2}(t,u)z^{r-1}\\
&&{}+G_{m+1,r-2,0}(t,u)y^{r-1}-{m+r\choose r-1}(-t)^{m+1}y^{r-1}\\
&&{}+\sum_{i=1}^{r-2}[G_{m+1,i-1,r-1-i}(t,u)-G_{m+1i,r-2-i}(t,u)]
y^i z^{r-1-i}\\
&=&(y-z)F_{m+1,r-1}(t,u,y,z)+{m+r\choose r-1}(u^{m+1}z^{r-1}
-(-t)^{m+1}y^{r-1}).
\end{eqnarray*}
Therefore
\begin{eqnarray*}
(t+u)^rF_{m,r}(t,u,y,z)&=&\sum_{l=1}^r{m+r\choose r-l}
(y-z)^{r-l}(u^{m+r}z^{r-l}-(-t)^{m+r}y^{r-l}).
\end{eqnarray*}
This establishes (\ref{evil identity}).

Consequently
\begin{eqnarray*}
&&\ca_r(k,l;y,z)-\ca_r(l,k;z,y)\\
&=&(-1)^k k!l!(y-z)^{k+l+1}\sum_{i=0}^{r-i-1}\frac{(-1)^ia_i}{i!(k+l+r-i)!}\\
&&{}\times\sum_{j=0}^{r-i-1}{k+r-1-i-j\choose r-1-i-j}
{m-k+j\choose j}y^{r-i-1-j}z^j
\end{eqnarray*}
as required.
\epf

\vskip 30pt

\section*{\normalsize 3. Examples and applications}

The identities (1.4) and (1.5) in [1] are respectively
the $r=1$ and $r=0$ cases of Theorem~2.

The next special case of Theorem~2, with $r=2$ yields, after
some rearrangement,
\begin{eqnarray*}
&&(-1)^k\sum_{j=0}^k{k\choose j}x^{k-j}\frac{A_{l+j+2}(y)}{(l+j+1)(l+j+2)}\\
&&{}-(-1)^j\sum_{j=0}^l{l\choose j}x^{l-j}
\frac{A^*_{k+j+2}(z)}{(k+j+1)(k+j+2)}\\
&=&k!l!(-x)^{k+l+1}\left(\frac{a_0}{(k+l+2)!}
\left[(k+1)y+(l+1)(1-z)\right]-\frac{a_1}{(k+l+1)!}\right)\\
&=&\frac{(-x)^{k+l+1}}{{k+l\choose k}}
\left(\frac{a_0\left[(k+1)y+(l+1)(1-z)\right]}{(k+l+1)(k+l+2)}
-\frac{a_1}{k+l+1}\right).
\end{eqnarray*}

Taking $l=k$ and $z=y$ in (\ref{main identity})
gives the following corollary.

\noindent
{\bf Corollary 3}\emph{
For all integers $k$, $r\ge0$,
\begin{eqnarray*}
&&\sum_{j=0}^k{k\choose j}(1-2y)^{k-j}
\frac{(k+j)![A_{k+j+r}(y)-(-1)^rA^*_{k+j+r}(y)]}{(k+j+r)!}\\
&=&(-1)^k k!^2(2y-1)^{2k+1}
\sum_{i=0}^{r-1}\frac{(-1)^i a_i}{i!(2k+r-i)!}\\
&&{}\times\sum_{j=0}^{r-1-i}
{k+r-i-1-j\choose k}{k+j\choose k}y^{r-i-1-j}z^j.
\end{eqnarray*}
}

\noindent
\emph{ In addition if $r$ is odd and $(a_n)$ is self-dual, or if $r$ is even
and $a_n^*=-a_n$ for all $n$ then 
\begin{eqnarray*}
&&2\sum_{j=0}^k{k\choose j}(1-2y)^{k-j}
\frac{(k+j)!A_{k+j+r}(y)}{(k+j+r)!}\\
&=&(-1)^k k!^2(2y-1)^{2k+1}
\sum_{i=0}^{r-1}\frac{(-1)^i a_i}{i!(2k+r-i)!}\\
&&{}\times\sum_{j=0}^{r-1-i}
{k+r-i-1-j\choose k}{k+j\choose k}y^{r-i-1-j}z^j.
\end{eqnarray*}
}
%\epf
%\hfill$\Box$

This corollary naturally gives rise to Bernoulli polynomial
identities by taking $a_n=(-1)^n B_n$. For example, when $r$ is
odd we have
\begin{eqnarray*}
&&2\sum_{j=0}^k{k\choose j}(1-2y)^{k-j}
\frac{(k+j)!B_{k+j+r}(y)}{(k+j+r)!}\\
&=&(-1)^k k!^2(2y-1)^{2k+1}
\sum_{i=0}^{r-1}\frac{B_i}{i!(2k+r-i)!}\\
&&{}\times\sum_{j=0}^{r-1-i}
{k+r-i-1-j\choose k}{k+j\choose k}y^{r-i-1-j}z^j.
\end{eqnarray*}

The following theorem which subsumes the second and
third identities [2, (1.5), (1.6)] in Sun's main theorem
follows from Theorem~2.

\noindent
{\bf Theorem 4}\emph{
Let $x+y+z=1$. Then
for all integers $k$, $l$, $r\ge0$,
\begin{eqnarray}\label{second identity}
&&(-1)^k\sum_{j=0}^k{k\choose j}{l+j\choose r}x^{k-j}A_{l+j-r}(y)\\
&=&(-1)^{l+r}\sum_{j=0}^l{l\choose j}{k+j\choose r}x^{l-j}A^*_{k+j-r}(z).
\nonumber
\end{eqnarray}
}

\noindent
{\it Proof.}
In this identity we interpret $A_m(x)/m!$ as zero whenever $m<0$.

Replacing $z$ by $1-z$ and using the duality principle we see that
(\ref{second identity}) is equivalent to
\begin{equation}\label{new second identity}
\sum_{j=0}^k{k\choose j}{l+j\choose r}(z-y)^{k-j}A_{l+j-r}(y)
=\sum_{j=0}^l{l\choose j}{k+j\choose r}(y-z)^{l-j}A_{k+j-r}(z)
\end{equation}

The $r=0$ case of (\ref{new second identity})
is the same as the $r=0$ case of~(\ref{new main identity}).
It is immediate from the definition of $A_m(x)$
that $A_m'(x)=mA_{m-1}(x)$. The case $r>0$ of
(\ref{new second identity}) follows from the $r=0$ case by applying the
partial differential operator
$$\frac{1}{r!}
\left(\frac{\partial}{\partial y}+\frac{\partial}{\partial z}\right)^r,$$
which annihilates all powers of $(y-z)$,
to both sides.
\hfill $\Box$

\vskip 30pt

\section*{\normalsize References}

\noindent[1]
R.L.\ Graham, D.E.\ Knuth \& O.\ Patashnik,
\emph{Concrete Mathematics},
Addison-Wesley, 1989.

\noindent[2]
Z.-W.\ Sun,
`Combinatorial identities in dual sequences',
\emph{European J.\ Combin.} \textbf{24} (2003) 709--718.


\end{document}