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\begin{center}
\vskip 1cm{\LARGE\bf 
A Generating Function for the Diagonal $T_{2n,n}$ 
\vskip .1in
in Triangles
}
\vskip 1cm
\large
Dmitry V. Kruchinin\\ 
Tomsk State University of Control Systems and Radioelectronics\\
and\\
National Research Tomsk Polytechnic University\\
Tomsk \\
Russian Federation\\ 
\href{mailto:KruchininDm@gmail.com}{\tt KruchininDm@gmail.com}
\\
\vskip 1cm
Vladimir V. Kruchinin\\
Tomsk State University of Control Systems and Radioelectronics\\
Tomsk \\
Russian Federation\\ 
\href{mailto:kru@2i.tusur.ru}{\tt kru@2i.tusur.ru}
\end{center}

\vskip .2 in

\begin{abstract}
We present techniques for obtaining a generating function for the
diagonal $T_{2n,n}$ of the triangle formed from the coefficients
of a generating
function $G(x)$ raised to the power $k$.  We obtain some relations between
central coefficients and coefficients of the diagonal $T_{2n,n}$, and we
also give some examples.
\end{abstract}

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\section{Introduction}

A triangle is a classic object of research in combinatorics.  For
instance, the Pascal triangle, the Bernoulli-Euler triangle, the
Catalan triangle, and the Motzkin triangle are discussed in many papers
and books \cite{ConcreteMath,Stanley_v2,Lando}.

Let $G(x)$ be an ordinary power series without a constant term, i.e., 
\begin{math}
G(x)=\sum\limits_{n&gt;0} g_n\,x^n,
\end{math}
where  \begin{math}g_0=0 \end{math} and \begin{math}g_1 \neq 0\end{math}.
In this paper we deal with the triangle $T_{n,k}$ defined as follows:
\begin{displaymath}
[G(x)]^k=\sum_{n\geq k}T_{n,k}x^n.
\end{displaymath}
Here we assume that $G(x)^0=T_{0,0}=1$. 

Then the generating function $G(x)$ raised to
the power $k$ gives the following triangle $T_{n,k}$
$$
\begin{array}{ccccccccccc}
&amp;&amp;&amp;&amp;&amp; T_{1,1}\\
&amp;&amp;&amp;&amp; T_{2,1} &amp;&amp; T_{2,2}\\
&amp;&amp;&amp; T_{3,1} &amp;&amp; T_{3,2} &amp;&amp; T_{3,3}\\
&amp;&amp; T_{4,1} &amp;&amp; T_{4,2} &amp;&amp; T_{4,3} &amp;&amp; T_{4,4}\\
&amp; \vdots &amp;&amp; \vdots &amp;&amp; \vdots &amp;&amp; \vdots &amp;&amp; \vdots\\
T_{n,1} &amp;&amp; T_{n,2} &amp;&amp; \ldots &amp;&amp; \ldots &amp;&amp; T_{n,n-1} &amp;&amp; T_{n,n}\\
\end{array}
$$

The  following notation  will  be  used  throughout  this  paper. 
The authors \cite{KruAppl,KruCompositae} introduced the notion of the \textit{composita} of a given ordinary generating function $G(x)=\sum_{n&gt;0}g(n)x^n$.  
\begin{definition}
The \textit{composita} is the function of two variables defined by
\begin{equation}
\label{Fnk0}G^{\Delta}(n,k)=\sum_{\pi_k \in C_n}{g(\lambda_1)g(\lambda_2) \cdots g(\lambda_k)},
\end{equation}
where $n,k$,$\lambda_i$ are integers that are greater than $0$, $C_n$ is the set of all compositions of $n$, and $\pi_k$ is the composition into $k$ parts exactly ($\sum_{i=1}^k\lambda_i=n$). 
\end{definition}

The generating function of the composita is equal to
\begin{equation}
[G(x)]^k=\sum_{n\geqslant k} G^{\Delta}(n,k)x^n=\sum_{n\geq k}T_{n,k}x^n.
\end{equation}

This notation coincides with the concept of Riordan array $(1,G(x))$ or \begin{math}
\left(\frac{G(x)}{x},G(x)\right)
\end{math},
which was given by Shapiro, Getu, Woan, and Woodson \cite{Shapiro1991}.

Recently, in \cite{KruchininTriangle2012}, we have shown how to find a generating function of the central elements of such triangles
\begin{equation}
C(x)=\sum_{n&gt;0}T_{2n-1,n}x^{n-1}=F'(x),
\label{central}
\end{equation}
where $F(x)$ is the solution of the equation
\begin{equation}
\label{LIF}
F(x)=xS(F(x))
\end{equation}
and
\begin{displaymath}
x\,S(x)=G(x).
\end{displaymath}

For solving (\ref{LIF}), one uses the Lagrange inversion formula (LIF), which was proved by Stanley \cite{Stanley_v2}. 
In \cite{KruchininTriangle2012}, we applied the LIF for the
generating functions raised to the $k$ power:
\begin{displaymath}
[G(x)]^k=\sum\limits_{n\geq k} G^{\Delta}(n,k)\,x^n
\end{displaymath}
and
\begin{displaymath}
[F(x)]^k=\sum\limits_{n\geq k} F^{\Delta}(n,k)\,x^n.
\end{displaymath}
We obtained the following relation between two triangles:
\begin{displaymath}
F^{\Delta}(n,k)=\frac{k}{n}T_{2n-k,n}.
\end{displaymath}


In this paper we present a method for obtaining the generating function for the diagonal $T_{2n,n}$ of a triangle $T_{n,k}$. The triangle is given by the following expression
\begin{displaymath}
[G(x)]^k=\sum\limits_{n\geq k} T_{n,k}\,x^n.
\end{displaymath}

\section{Main results}
The main result of this paper is given in the following theorem.
\begin{theorem}
\label{mainthm} Suppose we have the generating function
$
G(x)=\sum\limits_{n&gt;0} g_n\,x^n
$
that forms a triangle $T_{n,k}$:
$$
[G(x)]^k=\sum_{n\geq k}T_{n,k}x^n.
$$
Then the generating function 
$
A(x)=\sum_{n\geq 0} T_{2n,n}x^n
$
for the diagonal $T_{2n,n}$ of the triangle is defined by
\begin{equation}
A(x)=\frac{x\,F'(x)}{F(x)},
\end{equation}
where
$
F(x)=x\,S(F(x))
$
with 
$
S(x)=\frac{G(x)}{x}.
$
\end{theorem}


\begin{proof}
Suppose we have the following Laurent series
\begin{displaymath}
\Phi(z)=\varphi\,z+\varphi_0+\frac{\varphi_1}{z}+\cdots+\frac{\varphi_n}{z^n}+\cdots
\end{displaymath}
Then, raising this generating function to the power $k$, we get
\begin{displaymath}
[\Phi(z)]^k=\Phi_k(z)+E_k(z),
\end{displaymath}
where 
\begin{math}
\Phi_k(z)
\end{math}
contains the nonnegative powers of $z$ and 
\begin{math}
E_k(z)
\end{math}
contains the remaining powers of $z$.
According to Suetin \cite{Suetin}, \begin{math}
\Phi_k(z)
\end{math}
is the Faber polynomial.

Let us consider the generating function $G(z)$ in terms of $\Phi(z)$. That is,
\begin{displaymath}
[G(z)]^k=[z^2\,\Phi(1/z)]^k=\sum_{n\geq k} T_{n,k}\,z^n.
\end{displaymath}
Then we have
\begin{displaymath}
[\Phi(z)]^k=z^{2k}\sum_{n\geq k} T_{n,k}\,z^{-n}.
\end{displaymath}

After transformation, the Faber polynomial is equal to
\begin{equation}\label{Faber1}
\Phi_n(z)=\sum_{k=0}^n T_{2n-k,n}\,z^k,
\end{equation}

For the case  $z=0$, we have
\begin{equation}\label{z=0}
\Phi_n(0)=T_{2n,n}.
\end{equation}

According to Curtiss \cite{Curtiss} and Suetin \cite{Suetin},
the generating function for the Faber polynomials is equal to
\begin{displaymath}
\frac{t\phi '(t)}{\phi(t)-z}=\sum_{n \geq 0} \Phi_n(z)t^{-n},
\end{displaymath}
where $\phi(t)$ is the compositional inverse of $\Phi(t)$.
%% which is $x^2G(1/x)$.

Then the generating function for the case $z=0$ is equal to
$$
\frac{t\phi '(t)}{\phi(t)}=\sum_{n \geq 0} \Phi_n(0)t^{-n}.
$$

Next we set $t=\frac{1}{x}$. 
Taking into account that
$$
(\phi (1/x))'=\phi' (1/x)\,(1/x)'=\frac{-\phi' (1/x)}{x^2}
$$
or
$$
\phi' (1/x)=-x^2\,(\phi (1/x))'
$$
we get the generating function for $\Phi_n(0)$ 
\begin{equation}
\label{gen1/x}
A(x)=-\frac{x(\phi (1/x))'}{\phi(1/x)}=\sum_{n \geq 0} \Phi_n(0)x^{n}.
\end{equation}

Since $\phi(t)$ is the compositional inverse of $\Phi(t)$, the
following identity holds:
$$ \Phi(\phi(t))=t.  $$ 
If we substitute
$1/x$ for $t$, then we obtain the following relation: 
$$ \Phi(\phi(1/x))=1/x.  $$

Since $$
\Phi(x)=x^2G(1/x)=xS(1/x),
$$
we get
$$
\phi(1/x)S(1/\phi(1/x))=\frac{1}{x}.
$$
Then
$$\frac{1}{\phi(1/x)}=xS(1/\phi(1/x)) .$$
According to (\ref{LIF}), we have
$$
\frac{1}{\phi(1/x)}=F(x)
$$
Therefore, according to (\ref{z=0})  and (\ref{gen1/x}), the generating function for the diagonal $T_{2n,n}$ is equal to
$$
A(x)=-\frac{x(\frac{1}{F(x)})'}{\frac{1}{F(x)}}=\frac{x\,F'(x)}{F(x)}=\sum_{n \geq 0} T_{2n,n}\,x^n  .
$$
The theorem is thus proved.
\end{proof}

As applications of Theorem \ref{mainthm}, we give the following examples.

\begin{example} Let us consider the Pascal triangle.
This triangle can be defined by the generating function
\begin{displaymath}
[G(x)]^k=\left(\frac{x}{1-x}\right)^k=\sum_{n\geq k} {n-1 \choose n-k}x^n
\end{displaymath}

The solution of the equation
\begin{displaymath}
F(x)=\frac{x}{1-F(x)}
\end{displaymath} 
is the generating function
\begin{displaymath}
F(x)=\frac{1-\sqrt{1-4x}}{2}.
\end{displaymath}

Therefore,
the generating function for the diagonal $T_{2n,n}$ with the general term
\begin{math}
{2n-1  \choose n}
\end{math}
is
\begin{displaymath}
A(x)=\frac{x\,F'(x)}{F(x)}=\frac{2x}{\left(1-\sqrt{1-4x}\right)\,\sqrt{1-4x}}.
\end{displaymath}
\end{example}

\begin{example} 
Let us find the generating function
\begin{math}
A(x)=\sum_{n\geq 0}T_{2n,n}x^n
\end{math}
for the triangle defined by the following generating function
\begin{displaymath}
G(x)=x+x^2+x^3.
\end{displaymath}

Solving the equation
\begin{displaymath}
F(x)=x(1+F(x)+F(x)^2),
\end{displaymath}
we get the generating function for the Motzkin numbers (see the sequence 
\seqnum{A001006} in \cite{oeis})
\begin{displaymath}
F(x)={{-\sqrt{-3x^2-2x+1}-x+1}\over{2x}}.
\end{displaymath}
Then we have
\begin{displaymath}
\frac{x\,F'(x)}{F(x)}={{\sqrt{-3x^2-2x+1}+x-1}\over{\left(x-1\right)\,\sqrt{-3x^2-2 \,x+1}-3x^2-2x+1}}.
\end{displaymath}

After transformation, we obtain
\begin{displaymath}
A(x)=\frac{1}{\sqrt{-3x^2-2x+1}}.
\end{displaymath}
\end{example}

\begin{example} 
Let us find the generating function
\begin{math}
A(x)=\sum_{n\geq 0}T_{2n,n}x^n
\end{math}
for the triangle defined by the following expression
\begin{displaymath}
[G(x)]^k=\left[\frac{1-\sqrt{1-4x}}{2}\right]^k=
\sum\limits_{n\geq k}{{k\,{{2n-k-1}\choose{n-k}}}\over{n}}x^n.
\end{displaymath}

The solution of the functional equation (\ref{LIF}) for this case is the following generating function (see sequence \seqnum{A001764} in \cite{oeis})
\begin{displaymath}
F(x)=\frac{2}{\sqrt{3x}}{{\sin \left(\frac{1}{3}{{\arcsin \left({{\sqrt{27x}
 }\over{2}}\right)}}\right)}}.
\end{displaymath}

Therefore, the desired generating function has the form
\begin{displaymath}
A(x)=\frac{xF'(x)}{F(x)}=1+\sum_{n&gt;0}\frac{{3n-1\choose n}}{2}x^n={{\frac{\sqrt{3x}}{{2\sqrt{4-27x
 }}}\,\cot \left(\frac{1}{3}{{\arcsin \left({{\sqrt{27x
 }}\over{2}}\right)}}\right)}}+{{1}\over{2}}.
\end{displaymath}
\end{example}

\begin{example} 
Let  us consider the triangle defined by the expression
\begin{displaymath}
[G(x)]^m=[x^2\cot(x)]^m=\sum\limits_{n\geq m}T_{n,m}x^n,
\end{displaymath}
where
\begin{displaymath}
T_{n,m}=\left(-1\right)^{{{n-m}\over{2}}}\,\sum_{l=0}^{m}{\frac{2^{n-2m+l}}{{\left(n-2m+l\right)!}}{{{{m}\choose{l}}\,\sum\limits_{k=0}^{n-2m+l}{{{
  {s}\left(l+k , l\right)\,{S}\left(n-2m+l
   , k\right)}\over{k+l \choose l}}}}}}.
\end{displaymath}
Here $s(n,k)$ and $S(n,k)$ stand for the Stirling numbers of the first
and second kinds, respectively  \cite{Comtet_1974,ConcreteMath}.

This triangle forms the sequence \seqnum{A199542} in \cite{oeis}.
 Then we have
\begin{displaymath}
T_{2n,n}=\left(-1\right)^{{{n}\over{2}}}\,\sum_{l=0}^{n}{2^{l}\,\left(\sum_{
  k=0}^{l}{{{k!\,{S}\left(l , k\right)\,{s}
  \left(l+k , l\right)}\over{\left(l+k\right)!}}}\right)\,{{n}\choose{
  l}}}.
\end{displaymath} 
 For the equation 
 \begin{math}
 F(x)=x\,F(x)\cot(F(x))
 \end{math}, the solution is the generating function \begin{math}
 \arctan(x)
 \end{math}.
 
Hence,
 \begin{displaymath}
 \frac{xF'(x)}{F(x)}=\frac{x}{(1+x^2)\arctan(x)}=1-{{2x^2}\over{3}}+{{26x^4}\over{45}}-{{502x^6}\over{945}}+{{
  7102x^8}\over{14175}}+\cdots
 \end{displaymath}
Therefore, we obtain
 \begin{displaymath}
 A(x)=\frac{x}{(1+x^2)\arctan(x)}=\sum_{n\geq 0}\left(-1\right)^{{{n}\over{2}}}\,\sum_{l=0}^{n}{2^{l}\,\left(\sum_{
  k=0}^{l}{{{k!\,{S}\left(l , k\right)\,{s}
  \left(l+k , l\right)}\over{\left(l+k\right)!}}}\right)\,{{n}\choose{
  l}}}\,x^n.
 \end{displaymath}
\end{example} 

Next, we derive some interesting identities between coefficients in triangles.

\begin{theorem}
\label{cor1}
Suppose we have the triangle $T_{n,k}$, which is generated by $G(x)^k=\sum_{n\geqslant k}T_{n,k}x^n$.
Then the following identity holds for the central coefficients of the triangle
\begin{equation}\label{Coro1}
T_{2n-1,n}=\sum_{i=1}^n \frac{1}{i}T_{2i-1,i}\,T_{2(n-i),n-i}.
\end{equation}
\end{theorem}

\begin{proof}

The  result follows from Theorem \ref{mainthm} and the expression (\ref{central}).
We point out that
$$
F(x)=\sum_{n&gt;0} \frac{1}{n}T_{2n-1,n}x^n
$$
and
$$
\frac{x\,F'(x)}{F(x)}=\sum_{n\geqslant 0} T_{2n,n}x^n.
$$

Since 
$$
x\,F'(x)=\left(\frac{x\,F'(x)}{F(x)}\right)\,F(x),
$$
by applying the multiplication rule for formal power series, we obtain the desired result.
\end{proof}

\begin{example} Using Theorem \ref{cor1}, we obtain the identities for the Stirling numbers.

The Stirling numbers of the first kind $s(n,k)$ count the number of permutations of $n$ elements with $k$ disjoint cycles. The Stirling numbers of the first kind are defined by the following generating function \cite{Comtet_1974}:
$$
\psi_k(x)=\sum_{n\geq k}  s(n,k)  \frac{x^n}{n!}=\frac{1}{k!}\ln^k(1+x).
$$

With the help of (\ref{Coro1}), we find the following identity for the Stirling numbers of the first kind:
$$s(2n-1,n)=\sum_{i=1}^{n}\frac{{2n-1 \choose 2i-1}}{{n \choose i}}
\,{{{{s}\left(2i-1 , i\right)\,
{s}\left(2\left(n-i\right) , n-i
 \right)}\over{i\,\,}}}.$$ 

The Stirling numbers of the second kind $S(n,k)$ count the number of ways to partition a set of $n$ elements into $k$ nonempty subsets. The Stirling numbers of the second kind are defined by the following generating function \cite{Comtet_1974}:
$$
\Phi_k(x)=\sum_{n\geq k} S(n,k) \frac{x^n}{n!}=\frac{1}{k!}(e^x-1)^k.
$$

Using Theorem \ref{cor1}, we derive the following identity
$$S(2n-1,n)=\sum_{i=1}^{n}\frac{{2n-1 \choose 2i-1}}{{n \choose i}}
\,{{{{S}\left(2i-1 , i\right)\,
{S}\left(2\left(n-i\right) , n-i
 \right)}\over{i\,\,}}}.$$ 
\end{example} 

\begin{example}
Suppose we have the triangle defined by the following expression
$$(x\,e^x)^k=\sum_{n\geqslant k} \frac{k^{n-k}}{(n-k)!})x^n.$$
Then, using (\ref{Coro1}), we get
$$
\frac{n^{n-1}}{(n-1)!}=\sum_{i=1}^{n}{{{i^{i-2}\,\left(n-i\right)^{n-i}}\over{\left(i-1
 \right)!\,\left(n-i\right)!}}}
$$
or after simple manipulation
$$
n^{n-1}=\sum_{i=1}^{n}{{n-1 \choose i-1}{{i^{i-2}\,\left(n-i\right)^{n-i}}}}.
$$
\end{example} 

\begin{example} Suppose we have the triangle defined by the following expression 
$$\left(\frac{x}{(1-x)^m}\right)^k=\sum_{n\geqslant k} {n+(m-1)k-1 \choose n-k}x^n.$$
Then, according to (\ref{Coro1}), we obtain
 $${{\left(m+1\right)n-2}\choose{n-1}}=\sum_{i=1}^{n}\frac{1}{i}{{{{{im+i-2}\choose{i-1}}\,{{\left(m+1\right)n-im-i-1}\choose{n-i}}}}}.$$
If we put $m=2$, we derive the following identity
$${{3n-2}\choose{n-1}}=\sum_{i=1}^{n}{{{{{3i-2}\choose{i-1}}\,{{3n-3i-1}\choose{n-i
 }}}\over{i}}}.$$
\end{example} 

\begin{example} Suppose we have the triangle defined by the following expression  $$[G(x)]^k=\left(\frac{x^2}{e^x-1}\right)^k=\sum\limits_{n\geqslant k}T_{n,k}x ^n,$$
where
$$
T_{n,m}=\frac{m!}{(n-m)!}\sum_{k=0}^{n-m}\frac{k!\,S_1\left(m+k,m\right)\,{S_2}\left(n-m,k\right)}{(m+k)!}.
$$ 
The solution of the equation (\ref{LIF}) for this case, that is,  $F(x)=x\frac{F(x)}{e^{F(x)}-1}$, is the generating function $\ln(1+x)$ (see the sequence 
\seqnum{A191578} in \cite{oeis}).

Then, according to Theorem \ref{mainthm}, we have
$$
\frac{xF'(x)}{F(x)}=\frac{x}{(1+x)\ln(1+x)}=\sum\limits_{n \geqslant 0} T_{2n,n}\,x^n.
$$
where
$$
T_{2n,n}=\sum_{k=0}^{n}{{{k!\,{S_2}\left(n , k\right)\,
 {S_1}\left(n+k , n\right)}\over{\left(n+k\right)!}}}.
$$
This is reflected in the sequence A002208 in \cite{oeis}.

Using Theorem \ref{cor1}, we obtain the following identity
$$\sum_{m=0}^{n-1}\frac{\left(-1\right)^{m}}{n-m}\sum_{k=0}^{m}\frac{k!\,{S_2}\left(m , k\right)\,{S_1}\left(m+k , m
 \right)}{\left(m+k\right)!}=1.
 $$
 \end{example} 

\section{Acknowledgment}

The authors are grateful to the reviewer's valuable comments that improved the
manuscript. 

\begin{thebibliography}{99}
\bibitem{Comtet_1974} L. Comtet, 
\newblock \textit{Advanced Combinatorics}, D. Reidel Publishing Company, 1974.

\bibitem{Curtiss} J. H. Curtiss, Faber polynomials and the Faber series,
\newblock \emph{Amer. Math. Monthly}  \textbf{78} (1971), 577--596.

\bibitem{ConcreteMath} R.~L. Graham, D.~E. Knuth, and O.~Patashnik,
\newblock  \textit{Concrete Mathematics}, Addison-Wesley, 1989.

\bibitem{KruAppl} D. V.~Kruchinin and V. V. Kruchinin, Application of a composition of generating functions for obtaining explicit formulas of polynomials, 
\newblock \emph{J. Math. Anal. Appl.} \textbf{404} (2013), 161--171. 

\bibitem{KruCompositae} V. V.~Kruchinin and D. V. Kruchinin, Composita
and its properties, \newblock \emph{J. Analysis and Number
Theory} \textbf{2} (2014), 1--8 .

\bibitem{KruchininTriangle2012}
 D. V. Kruchinin and V. V. Kruchinin,
 \newblock A method for obtaining generating functions for  central coefficients of triangles,
 \newblock \emph{J. Integer Seq.}  \textbf{15} (2012), Article 12.9.3.

\bibitem{Lando}
 S.~K. Lando,
 \newblock {\em Lectures on Generating Functions},
 \newblock American Mathematical Society, 2003.

\bibitem{Shapiro1991} L. W. Shapiro,  S. Getu, W.-J. Woan,  and L. C. Woodson, The Riordan group,
\emph{Discr. Appl. Math.} \textbf{34} (1991), 229--239.

\bibitem{oeis}
N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences,
\url{http://oeis.org}.

\bibitem{Stanley_v2} R. P. Stanley, 
\newblock \textit{Enumerative Combinatorics 2},
Cambridge University Press, 1999.

\bibitem{Suetin} P. K. Suetin, 
\newblock {\em Series in Faber Polynomials},
\newblock Nauka, Moscow, 1984. (in Russian)
  
\end{thebibliography}

\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05A15; Secondary 11B75, 05A10.

\noindent \emph{Keywords: } 
generating function, triangle, diagonal, central coefficient, composita.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A001006},
\seqnum{A001764},
\seqnum{A002208},
\seqnum{A191578}, and
\seqnum{A199542}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received February 28 2014;
revised versions received November 10 2014; December 7 2014; February 
24 2015.
Published in {\it Journal of Integer Sequences},  May 13 2015.

\bigskip
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