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\title[Explicit Formul{\ae} for some Numerical Triangles of Z.~W.~Trzaska]
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{\Large{\bf
{Explicit Formul{\ae} for some Numerical Triangles of Z.~W.~Trzaska}
}}


{\sc Helmut Prodinger\\
Department for Algebra and Discrete Mathematics \\
TU Wien \\
Wiedner Hauptstrasse 8--10/118 \\
A-1040 Wien\\
 Austria \\
\rm email: Helmut.Prodinger@@tuwien.ac.at}
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\date{\today}





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In \cite{Trzaska3}, the following modified numerical triangle {\sc Mnt1} is considered.
The polynomials $T_{k}(q)$ are given by $T_0(q)=1$, $T_1(q)=1+q$ and the recursion formula
\begin{equation}\label{eq1}   
T_{k+1}(q)=(2+q)T_k(q)-T_{k-1}(q)~.
\end{equation}   
The numbers $A(k,m)$ of the modified numerical triangle {\sc Mnt1} are defined
as
\begin{equation*}   
A(k,m)=[q^m]T_k(q)~.
\end{equation*}
(Read: ``coefficient of $q^m$ in $T_k(q)$.'')
\begin{theorem}\label{first}
\begin{equation*}   
A(k,m)=\binom{k+m}{2m}~. 
\end{equation*}
\end{theorem} 

The second modified numerical triangle {\sc Mnt2} is the following.
The polynomials $P_{k}(q)$ are given by $P_0(q)=0$, $P_1(q)=1$ and the recursion formula
\begin{equation*}   
P_{k+1}(q)=(2+q)P_k(q)-P_{k-1}(q)~.
\end{equation*}   
The numbers $B(k,m)$ of the modified numerical triangle {\sc Mnt2} are defined
as
\begin{equation*}   
B(k,m)=[q^m]P_k(q)~.
\end{equation*}

\begin{theorem}
\begin{equation*}   
B(k,m)=\binom{k+m}{2m+1}~. 
\end{equation*}
\end{theorem} 


The paper gives  a backward pointer to \cite{Trzaska2}, which also has {\sc Mnt1},
but also another one, $g(k,m)$, which is a little bit clumsy to describe, but for which I
found the formula 
\begin{theorem}
\begin{equation*}   
g(k,m)=\frac{k^2-k+m^2}{m(2m-1)}\binom{k+m-2}{2m-2}~. 
\end{equation*}
\end{theorem} 

The paper  \cite{Trzaska1} is also referenced, but it does not give more triangles than
the ones just described.

\medskip

We only give a proof of Theorem \ref{first}, the other ones being similar.
The formula is evidently true for $m=0$ and also for $k=0$ and $k=1$. Now, proceeding
by induction and reading off the coefficient of $q^m$ in equation (\ref{eq1})  we get
for $m, k\ge1$ 
\begin{equation*}   
A(k+1,m)=2A(k,m)+A(k,m-1)-A(k-1,m)~.
\end{equation*}
It is a routine matter to check that $A(k,m)=\binom{k+m}{2m} $ fulfills this.






\bibliographystyle{plain} 
\begin{thebibliography}{1}

\bibitem{Trzaska1}
Z.~W. Trzaska.
\newblock Numerical triangle, {F}ibonacci sequence and ladder networks: some
  further results.
\newblock {\em Appl. Math. Lett.}, 6.4:55--61, 1993.

\bibitem{Trzaska2}
Z.~W. Trzaska.
\newblock Modified numerical triangle and the {F}ibonacci sequence.
\newblock {\em The Fibonacci Quarterly}, 32:124--129, 1994.

\bibitem{Trzaska3}
Z.~W. Trzaska.
\newblock On {F}ibonacci hyperbolic trigonometry and modified numerical
  triangles.
\newblock {\em The Fibonacci Quarterly}, 34:129--138, 1996.

\end{thebibliography}

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\bigskip 
\noindent 1991 \textit{Mathematics Subject Classification.}\ {05A10.}


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