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\begin{center}
\vskip 1cm {\LARGE\bf A Common Generating Function for Catalan Numbers and Other Integer
Sequences}
\vskip 1cm
\large
G. E. Cossali\\
Universit\`{a} di Bergamo\\
24044 Dalmine\\
Italy\\
\href{mailto:cossali@unibg.it}{\tt cossali@unibg.it}\\
\vskip .5cm
\end{center}

\date{}
\pagestyle{myheadings}

\begin{abstract}
Catalan numbers and other integer sequences (such as the triangular numbers) are
shown to be particular cases of the same sequence array $g(n,m)=\frac{\left(
2n+m\right) !}{m!n!(n+1)!}$ . Some features of the sequence array are
pointed out and a unique generating function is proposed.
\end{abstract}

\section{Introduction}

Catalan numbers can be found in many different combinatorial problems, as
shown by Stanley \cite{1}, and exhaustive information about this sequence can be found in
\cite{2}. In this note I show that the Catalan numbers (A000108) and other
known sequences (triangular numbers A000217, A034827, A001700, A002457,
A002802, A002803, A007004, A024489) can be derived by the same generating
function and are related to the same polynomial set.

\section{The polynomials $j_m(y)$}

Consider the following recurrence relation defining the polynomials $%
j_{m}(y) $: 
\begin{equation}
\begin{array}{c}
j_{0}(y)=1; \\ 
\\
j_{m+1}\left( y\right) =y j_{m}\left( y\right)
+\sum_{s=0}^{m}j_{s}\left( y\right) j_{m-s}\left( y\right).
\end{array}
\label{polyrec}
\end{equation}

It may immediately be noticed that for $y=0$ this formula coincides with the
recursive definition of the Catalan numbers: 
\begin{equation}
C_{s+1}=\sum_{m=0}^{s}C_{m}C_{s-m}  \label{catarec}
\end{equation}%
where
\begin{equation}
C_{n}=\frac{2n!}{n!(n+1)!}=\frac{1}{n+1}\binom{2n}{n}.  \label{catadef}
\end{equation}

This means that $C_{n}$ is the zero-order coefficient of the $n$th-order
polynomial $ j_{n}\left( y\right) $, i.e.,
\begin{equation}
j_{m}(y)=\sum_{q=0}^{m}e(m,q)\ y^{q}  \label{jn}
\end{equation}%
and $e(m,0)=C_{m}$. The first few values of $e(m,q)$ are shown in Table 1.

\begin{center}
\begin{eqnarray*}
&amp;&amp;%
\begin{tabular}{|c|r|r|r|r|r|r|r|r|}
\hline
$m\;\diagdown \;q$ &amp; \textbf{0} &amp; \textbf{1} &amp; \textbf{2} &amp; \textbf{3} &amp; 
\textbf{4} &amp; \textbf{5} &amp; \textbf{6} &amp; \textbf{7} \\ 
\hline
\textbf{0} &amp; 1 &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\ 
\textbf{1} &amp; 1 &amp; 1 &amp;  &amp;  &amp;  &amp;  &amp;  &amp;  \\ 
\textbf{2} &amp; 2 &amp; 3 &amp; 1 &amp;  &amp;  &amp;  &amp;  &amp;  \\ 
\textbf{3} &amp; 5 &amp; 10 &amp; 6 &amp; 1 &amp;  &amp;  &amp;  &amp;  \\ 
\textbf{4} &amp; 14 &amp; 35 &amp; 30 &amp; 10 &amp; 1 &amp;  &amp;  &amp;  \\ 
\textbf{5} &amp; 42 &amp; 126 &amp; 140 &amp; 70 &amp; 15 &amp; 1 &amp;  &amp;  \\ 
\textbf{6} &amp; 132 &amp; 462 &amp; 630 &amp; 420 &amp; 140 &amp; 21 &amp; 1 &amp;  \\ 
\textbf{7} &amp; 429 &amp; 1716 &amp; 2772 &amp; 2310 &amp; 1050 &amp; 252 &amp; 28 &amp; 1 \\
\hline
\end{tabular}
\\
\end{eqnarray*}
\centerline{\text{Table 1: Some of the coefficients } $e(m,q)$}
\end{center}

Equation (\ref{jn}) can be introduced into (\ref{polyrec}) to obtain
\begin{equation*}
\sum_{q=0}^{m+1}e(m+1,q)\ y^{q}=\sum_{q=0}^{m}e(m,q)\
y^{q+1}+\sum_{s=0}^{m}\sum_{p=0}^{m-s}e(m-s,p)\ \sum_{r=0}^{s}e(s,r)\ y^{p+r}
\end{equation*}%
and transformed as follows: 
\begin{equation*}
\begin{array}{c}
\sum_{q=0}^{m+1}e(m+1,q)\ y^{q}=\sum_{q=1}^{m+1}e(m,q-1)\
y^{q}+\sum_{s=0}^{m}\sum_{p=0}^{m-s}e(m-s,p)\ \sum_{r=0}^{s}e(s,r)\ y^{p+r}=
\\ 
=\sum_{q=1}^{m+1}e(m,q-1)\ y^{q}+\sum_{p=0}^{m}\ \sum_{r=0}^{m-p}\ y^{p+r}\ 
\left[ \sum_{s=r}^{m-p}e(m-s,p)\ e(s,r)\right] = \\ 
=\sum_{q=1}^{m+1}e(m,q-1)\ y^{q}+\sum_{p=0}^{m}\ \sum_{q=p}^{m}\ y^{q}\ 
\left[ \sum_{s=q-p}^{m-p}e(m-s,p)\ e(s,q-p)\right] = \\ 
=\sum_{q=1}^{m+1}e(m,q-1)\ y^{q}+\sum_{q=0}^{m}\ \sum_{p=0}^{q}\ \ \left[
\sum_{s=q-p}^{m-p}e(m-s,p)\ e(s,q-p)\right] y^{q}= \\ 
=\sum_{q=1}^{m+1}e(m,q-1)\ y^{q}+\sum_{q=0}^{m}\ \sum_{p=0}^{q}\ \ \left[
\sum_{l=q}^{m}e(m-l+p,p)\ e(l-p,q-p)\right] y^{q}.
\end{array}
\end{equation*}
The following set of equations can then be obtained for any natural
number $m$:

(1) for $q=0$: 
\begin{equation*}
e(m+1,0)=\sum_{s=0}^{m}e(m-s,0)\ e(s,0),
\end{equation*}
whose solution is
\begin{equation}
e(m,0)=C_{m}=\frac{1}{m+1}\binom{2m}{m}.  \label{qzero}
\end{equation}

(2) for $0&lt;q\leq m$: 
\begin{equation}
e(m+1,q)=e(m,q-1)+\sum_{p=0}^{q}\ \ \sum_{l=q}^{m}e(m-l+p,p)\ e(l-p,q-p) .
\label{enkeq}
\end{equation}

(3) for $q=m+1$: 
\begin{equation}
e(m+1,m+1)=e(m,m)=\cdots =1 . \label{qmax}
\end{equation}

It is useful to introduce the modified matrix $g(n,k)$ defined as follows:
\begin{equation}
\begin{array}{c}
g(n,k)=e(n+k,k) \\ 
e(n,k)=g(n-k,k)%
\end{array}
\label{geconv}
\end{equation}
Table 2 reports the first few values: 

\begin{center}
\begin{eqnarray*}
&amp;&amp;%
\begin{tabular}{|c|r|r|r|r|r|r|r|r|}
\hline
$m\;\diagdown \;q$ &amp; \textbf{0} &amp; \textbf{1} &amp; \textbf{2} &amp; \textbf{3} &amp; 
\textbf{4} &amp; \textbf{5} &amp; \textbf{6} &amp; \textbf{7} \\ 
\hline
\textbf{0} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\ 
\textbf{1} &amp; 1 &amp; 3 &amp; 6 &amp; 10 &amp; 15 &amp; 21 &amp; 28 &amp; 36 \\ 
\textbf{2} &amp; 2 &amp; 10 &amp; 30 &amp; 70 &amp; 140 &amp; 252 &amp; 420 &amp; 660 \\ 
\textbf{3} &amp; 5 &amp; 35 &amp; 140 &amp; 420 &amp; 1050 &amp; 2310 &amp; 4620 &amp; 8580 \\ 
\textbf{4} &amp; 14 &amp; 126 &amp; 630 &amp; 2310 &amp; 6930 &amp; 18018 &amp; 42042 &amp; 90090 \\ 
\textbf{5} &amp; 42 &amp; 462 &amp; 2772 &amp; 12012 &amp; 42042 &amp; 126126 &amp; 336336 &amp; 816816 \\ 
\textbf{6} &amp; 132 &amp; 1716 &amp; 12012 &amp; 60060 &amp; 240240 &amp; 816816 &amp; 2450448 &amp; 6651216
\\ 
\textbf{7} &amp; 429 &amp; 6435 &amp; 51480 &amp; 291720 &amp; 1312740 &amp; 4988412 &amp; 16628040 &amp; 
49884120 \\
\hline
\end{tabular}
\end{eqnarray*}
\centerline{\text{Table 2: \ Some values of the coefficients } $g(m,q)$}
\end{center}

\noindent Equations (\ref{qzero}), (\ref{enkeq}), (\ref{qmax}) then become:

(1) for $q=0$: 
\begin{equation}
g(n,0)=C_{n}.  \label{gqzero}
\end{equation}

(2) for $0&lt;q\leq n+1$:
\begin{equation}
g(n+1,q)=g(n+1,q-1)+\sum_{p=0}^{q}\ \ \sum_{l=0}^{n}g(n-l,p)\ q(l,q-p),
\label{gnkeq}
\end{equation}%
where $m-q$ was replaced by $n=m-q$. Moreover, from (\ref{qmax}) we get 
$g(0,n)=1.$

The solution of (\ref{gnkeq}) can be written as follows:
\begin{equation}
g(n,q)=\frac{\left( 2n+q\right) !}{q!n!(n+1)!}=C_{n}\dbinom{2n+q}{q} .
\label{gnksol}
\end{equation}
In fact, (\ref{gqzero}) is satisfied, and substituting (\ref{gnksol}) into (%
\ref{gnkeq}), we get
\begin{eqnarray}
C_{n+1}\dbinom{2(n+1)+q}{q} &amp;=&amp;C_{n+1}\dbinom{2(n+1)+q-1}{q-1}+
\label{solution} \\
&amp;&amp;+\sum_{l=0}^{n}C_{n-l}C_{l}\left[ \sum_{p=0}^{q}\ \ \dbinom{2(n-l)+p}{p}%
\dbinom{2l+q-p}{q-p}\right] \nonumber
\end{eqnarray}

It is possible to show that (see appendix)
\begin{equation*}
\sum_{p=0}^{q}\ \ \dbinom{2(n-l)+p}{p}\dbinom{2l+q-p}{q-p}=\dbinom{2n+q+1}{q} ,
\end{equation*}%
and 
\begin{equation*}
\begin{array}{c}
\dbinom{2(n+1)+q}{q}-\dbinom{2(n+1)+q-1}{q-1}=\dbinom{2(n+1)+q-1}{q-1}\left[ 
\dfrac{2(n+1)+q}{q}-1\right]  \\ 
\\
=\left\{ \dfrac{\left[ 2(n+1)+q-1\right] !}{\left[ q-1\right] !\left[ 2(n+1)%
\right] !}\right\} \left[ \dfrac{2(n+1)}{q}\right] =\left\{ \dfrac{\left(
2n+q+1\right) !}{q!\left( 2n+1\right) !}\right\} =\dbinom{2n+q+1}{q}.
\end{array}%
\end{equation*}
By using the recursive definition (\ref{catarec}) of Catalan numbers, (\ref%
{solution}) becomes an indentity.

\section{Some features of the array $g(n,q)$}

The sequence
\begin{equation*}
g(n,q)=\frac{2n+q!}{q!n!(n+1)!}=C_{n}\dbinom{2n+q}{q}
\end{equation*}%
can be seen as a generalization of the Catalan sequence, as it reduces to the
Catalan sequence for $q=0$. There are also some other interesting features.
In Table 3 I report the known names of the integer sequences, referenced
in \textit{The On-line Encyclopedia of Integer Sequences} \cite{2}, that can be
extracted from the matrix $g(n,q)$. 
\begin{eqnarray*}
&amp;&amp;%
\begin{tabular}{|c|r|r|r|r|r|r|r|}
\hline
&amp;  &amp; A000108 &amp; A001700 &amp; A002457 &amp; A002802 &amp; A002803 &amp; none \\ 
\hline
&amp; $m\;\diagdown \;q$ &amp; \textbf{0} &amp; \textbf{1} &amp; \textbf{2} &amp; \textbf{3} &amp; 
\textbf{4} &amp; \textbf{5} \\ 
\hline
- &amp; \textbf{0} &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 &amp; 1 \\ 
A000217 &amp; \textbf{1} &amp; 1 &amp; 3 &amp; 6 &amp; 10 &amp; 15 &amp; 21 \\ 
A034827 &amp; \textbf{2} &amp; 2 &amp; 10 &amp; 30 &amp; 70 &amp; 140 &amp; 252 \\ 
none &amp; \textbf{3} &amp; 5 &amp; 35 &amp; 140 &amp; 420 &amp; 1050 &amp; 2310 \\ 
none &amp; \textbf{4} &amp; 14 &amp; 126 &amp; 630 &amp; 2310 &amp; 6930 &amp; 18018 \\ 
- &amp; \textbf{5} &amp; 42 &amp; 462 &amp; 2772 &amp; 12012 &amp; 42042 &amp; 126126 \\ 
- &amp; \textbf{6} &amp; 132 &amp; 1716 &amp; 12012 &amp; 60060 &amp; 240240 &amp; 816816 \\
\hline
\end{tabular}
\\
\end{eqnarray*}%
\centerline{\text{Table 3:} $g(m,q)$\text{ numbers and names of known sequences}}
\newline

The first five columns correspond to the known sequences: A000108 (Catalan),
A001700, A002457, A002802, A002803. The first two rows correspond to the
sequences A000217 ($g(1,q)$, triangular numbers) and A034827. For the other
rows and columns no reference was found by the author. Also the sequence on
the main diagonal $g(k,k)$ (1,3,30,420,6930,126126,$\ldots $) is known as
A007004 and the sequence on the diagonal $g(k,k+1)$ (1,6,70,1050,18018, $%
\ldots $) is known as A024489.

\section{Generating function}

Consider the algebraic equation in $J$:
\begin{equation}
-xJ^{2}+(1-yx)J-1=0,   \label{opera}
\end{equation}
and its solutions
\begin{equation}  \label{generating}
J(x,y)=\frac{(1-yx)\pm \sqrt{\left( 1-yx\right) ^2-4x}}{2x} .
\end{equation}
Let now suppose that $J(x,y)$ admits a Taylor expansion in $x$ (which
excludes the solution $J(x,y)=\frac{(1-yx)+\sqrt{\left( 1-yx\right) ^2-4x}}{2x%
}$ unlimited in $x=0$) 
\begin{equation}  \label{taylor}
J(x,y)=\sum_{m=0}^\infty j_m\left( y\right) x^m.
\end{equation}
Substituting (\ref{taylor}) into equation (\ref{opera}) we get
\begin{equation*}
\begin{array}{c}
0=-x\sum_{m=0}^{\infty }j_{m}\left( y\right) x^{m}\sum_{m=0}^{\infty
}j_{n}\left( y\right) x^{n}+\sum_{m=0}^{\infty }j_{m}\left( y\right)
x^{m}-yx\sum_{m=0}^{\infty }j_{m}\left( y\right) x^{m}-1= \\ 
=-x\sum_{s=0}^{\infty }j_{s}\left( y\right) \sum_{m=s}^{\infty
}j_{s-m}\left( y\right) x^{s}+\sum_{m=0}^{\infty }j_{m}\left( y\right)
x^{m}-y\sum_{m=0}^{\infty }j_{m}\left( y\right) x^{m+1}-1= \\ 
=-\sum_{s=0}^{\infty }\sum_{m=0}^{s}j_{s}\left( y\right) j_{s-m}\left(
y\right) x^{s+1}+\sum_{s=0}^{\infty }j_{s}\left( y\right)
x^{s}-y\sum_{s=0}^{\infty }j_{s}\left( y\right) x^{s+1}-1= \\ 
=j_{0}(y)-1+\sum_{s=0}^{\infty }\left[ -\sum_{m=0}^{s}j_{s}\left( y\right)
j_{s-m}\left( y\right) +j_{s+1}\left( y\right) -yj_{s-1}\left( y\right) %
\right] x^{s+1}.
\end{array}%
\end{equation*}%
Then
\begin{equation}
\begin{array}{c}
j_{0}(y)=1 \\ 
j_{s+1}\left( y\right) =y\ j_{s}\left( y\right) +\sum_{m=0}^{s}j_{s}\left(
y\right) j_{s-m}\left( y\right),
\end{array}
\label{recursive1}
\end{equation}%
which is the recursive definition given by (\ref{polyrec}). This means that
\begin{equation*}
j_{m}\left( y\right) =\underset{x\rightarrow 0}{\lim }\frac{1}{m!}\frac{%
d^{m}J(x,y)}{dx^{m}},
\end{equation*}%
and for $y=0$ the function $J(x,0)$ is the generating function of the
Catalan sequence
\begin{equation*}
\begin{array}{c}
j_{m}(0)=C_{m} \\ 
J(x,0)=\frac{1-\sqrt{1-4x}}{2x}=C_{a}(x).
\end{array}%
\end{equation*}%
Now, using Equations (\ref{jn}) and (\ref{taylor}), we get
\begin{equation}
\begin{array}{c}
J(x,y)=\sum_{m=0}^{\infty }\sum_{q=0}^{m}e(m,q)\
y^{q}x^{m}=\sum_{q=0}^{\infty }\sum_{m=q}^{\infty }e(m,q)\ y^{q}x^{m}= \\ 
=\sum_{q=0}^{\infty }y^{q}\sum_{m=0}^{\infty }e(m+q,q)\
x^{m+q}=\sum_{q=0}^{\infty }\left( yx\right) ^{q}\sum_{m=0}^{\infty }g(m,q)\
x^{m}.
\end{array}
\label{Jxy}
\end{equation}%
Then, the function
\begin{equation*}
L(x,z)=\frac{(1-z)-\sqrt{\left( 1-z\right) ^{2}-4x}}{2x}=J(x,z/x)
\end{equation*}%
can be expanded to get (see equation (\ref{Jxy}))
\begin{equation}
L(x,z)=\sum_{q=0}^{\infty }\sum_{m=0}^{\infty }g(m,q)\ x^{m}z^{q}, 
\label{Lxz}
\end{equation}%
and this can be seen to be the generating function of $g(m,q)$:
\begin{equation*}
g(m,q)=\underset{x,z\rightarrow 0}{\lim }\ \frac{1}{m!q!}\frac{\partial
^{m+q}L(x,z)}{\partial x^{m}\partial z^{q}}.
\end{equation*}

It is interesting to observe that
\begin{eqnarray*}
L(x,z) &amp;=&amp;\sum_{q=0}^{\infty }\sum_{m=0}^{\infty }g(m,q)\
x^{m}z^{q}=\sum_{m=0}^{\infty }C_{m}\sum_{q=0}^{\infty }\dbinom{2m+q}{q}%
z^{q}x^{m}= \\
&amp;=&amp;\sum_{m=0}^{\infty }C_{m}T_{m}(z)x^{m}
\end{eqnarray*}%
with 
\begin{equation*}
T_{m}(z)=\sum_{q=0}^{\infty }\dbinom{2m+q}{q}z^{q}.
\end{equation*}%
It can be proven that
\begin{equation*}
T_{m}(z)=\frac{1}{(1-z)^{2m+1}}
\end{equation*}%
as 
\begin{equation*}
\frac{1}{q!}\frac{d^{q}T_{m}(z)}{dz^{q}}=\frac{(2m+1)\cdots (2m+q)}{q!\
(1-z)^{2m+1+q}}
\end{equation*}%
and 
\begin{equation*}
\underset{z\rightarrow 0}{\lim }\frac{1}{q!}\frac{d^{q}T_{m}(z)}{dz^{q}}=%
\frac{(2m+q)!}{q!\ 2m!}=\dbinom{2m+q}{q} .
\end{equation*}

The generating function $L(x,z)$ can then be written also in the form
\begin{equation*}
L(x,z)=\frac{1}{(1-z)}\sum_{n=0}^{\infty }C_{n}\left[ \frac{x}{(1-z)^{2}}%
\right] ^{n}=\frac{1}{(1-z)}C_{a}\left[ \frac{x}{(1-z)^{2}}\right]
\end{equation*}
that better shows the strong link existing between the sequence array $%
g(n,q) $ and the Catalan numbers.

\section{Appendix}

The following binomial identity: 
\begin{equation}  \label{idbas}
\sum_{p=0}^q\dbinom{m+p}m\dbinom{n+q-p}n=\dbinom{m+n+q+1}q
\end{equation}
holds for any non-negative integers $m,n,q$.
\bigskip

\begin{proof}
We define
\begin{equation*}
M(m,n,q)=\sum_{p=0}^{q}\dbinom{m+p}{m}\dbinom{n+q-p}{n}.
\end{equation*}%
Then the proposition (\ref{idbas}) is equivalent to
\begin{equation*}
M(m,n,q)=\dbinom{m+n+q+1}{q}
\end{equation*}%
The proof is based on the use of the binomial identity
\begin{equation}
\sum_{k=r}^{n}\dbinom{k}{r}=\dbinom{n+1}{r+1}  \label{idbin}
\end{equation}%
that can also be written as
\begin{equation*}
\sum_{k=0}^{m}\dbinom{r+k}{r}=\dbinom{m+r+1}{r+1}.
\end{equation*}%
The identity (\ref{idbas}) holds for $n=0$ and any $m,q$. In fact,
\begin{equation*}
M(m,0,q)=\sum_{p=0}^{q}\dbinom{m+p}{m}\dbinom{q-p}{0}=\sum_{p=0}^{q}\dbinom{%
m+p}{m}=\dbinom{m+q+1}{m+1}=\dbinom{m+q+1}{q},
\end{equation*}%
where the binomial identity (\ref{idbin}) was used.

For any $n\neq 0$, using (\ref{idbin}) again, and with $q,m$ natural numbers,
\begin{equation*}
\begin{array}{c}
M(m,n,q)=\sum_{p=0}^{q}\dbinom{m+p}{m}\dbinom{n+q-p}{n}=\sum_{p=0}^{q}%
\dbinom{m+p}{m}\dbinom{\left( n-1\right) +q-p+1}{\left( n-1\right) +1}= \\ 
=\sum_{p=0}^{q}\dbinom{m+p}{m}\sum_{k=0}^{q-p}\dbinom{n-1+k}{n-1}%
=\sum_{k=0}^{q}\sum_{p=0}^{q-k}\dbinom{m+p}{m}\dbinom{n-1+k}{n-1}= \\ 
=\sum_{k=0}^{q}\dbinom{q-k+m+1}{m+1}\dbinom{n-1+k}{n-1}=M(m+1,n-1,q).
\end{array}%
\end{equation*}%
By repeatedly applying the rule $M(m,n,q)=M(m+1,n-1,q)$, it is easy to
obtain 
\begin{equation*}
M(m,n,q)=M(m+n,0,q)=\dbinom{m+n+q+1}{q}.
\end{equation*}
\end{proof}

\begin{thebibliography}{9}
\bibitem{1} R. P. Stanley. 
\newblock {\textit{Enumerative Combinatorics},
Vol. 2}. \newblock Cambridge University Press, 1999.

\bibitem{2} N. J. A. Sloane. 
\newblock {\em On-Line Encyclopedia of Integer
Sequences}. \newblock published electronically at
\href{http://www.research.att.com/~njas/sequences}{\tt http://www.research.att.com/$\sim$njas/sequences}.
\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 05A15, 11Y55, 11B65.\ \

\noindent \emph{Keywords:  Generating function, Catalan numbers, binomial identity, polynomials}



\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000108}
\seqnum{A000217}
\seqnum{A034827}
\seqnum{A001700}
\seqnum{A002802}
\seqnum{A002803}
\seqnum{A007004}
\seqnum{A024489}
\seqnum{A002457}.)


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\noindent
Received October 6, 2002;
revised version received April 17, 2003.
Published in {\it Journal of Integer Sequences} May 2, 2003.
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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.

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