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\begin{center}
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{\LARGE\bf  Dickson  Polynomials, Chebyshev \\
\vskip .1in
Polynomials, 
and Some Conjectures of Jeffery }
\vskip 1cm
\large
Stefano Barbero\\
Department of Mathematics \\
University of  Turin \\
via Carlo Alberto 10 \\
10123 Turin \\
Italy\\
\href{mailto:stefano.barbero@unito.it}{\tt stefano.barbero@unito.it}\\
\end{center}

\begin{abstract}
By using Dickson polynomials in several variables and Chebyshev
polynomials of the second kind, we derive the explicit expression of
the entries in the array defining the sequence \seqnum{A185095} .  As a
result, we  obtain a straightforward proof of some conjectures of
Jeffery concerning this sequence  and other related ones.
\end{abstract}

\section{Proof of main conjectures}

As we can find in the {\it Encyclopedia of Integer Sequences} \cite{oeis}, the sequence \seqnum{A185095}  is  a rectangular array read by antidiagonals, where the $n$-th row  has generating function 
\begin{equation}\label{rowgf}
F_n(z) =\frac{ \sum_{r=0}^{n} {(n+1-r)(-1)^r {2(n+1)-r \choose r}z^r} }{ \sum_{r=0}^{n+1}{ (-1)^r {2(n+1)-r \choose r}z^r}}
\end{equation}
for $n=0,1,2,\ldots$. 
In this section we find the explicit expression for the entries of this rectangular array and we determine the ordinary generating functions for the columns.

Let us consider the Dickson polynomials of the first kind in several indeterminates, as defined in the book of Lidl, Mullen,  and Turnwald \cite{LMT}.

\begin{definition}\label{Dickdef}
The Dickson polynomials of the first kind in $n$ indeterminates of total degree $k$
 $$D_k^{(i)}(x_1,\ldots,x_n,a), \quad i=1,\ldots,n$$ 
satisfy the functional equations 
\begin{equation} D_k^{(i)}(x_1,\ldots,x_n,a)=S_i(u_1^k,\ldots,u_{n+1}^k), \quad i=1,\ldots,n
\end{equation}
where  $x_i=S_i(u_1,\ldots,u_{n+1})$, $S_i(y_1,\ldots,y_{n+1})$ is the $i$-th symmetric function of $y_1, \ldots,y_{n+1}$, and  $u_1\cdot \ldots \cdot u_{n+1}=a$. 
\end{definition}
In particular when $i=1$ we have $D_k^{(1)}(x_1,\ldots,x_n,a)=S_1(u_{1}^k,\ldots,u_{n+1}^k)=\sum_{j=1}^{n+1}u_{j}^k$. Moreover if we pose $x_0=1$ and $x_{n+1}=a$, the Dickson polynomials of the first kind  $D_k^{(1)}(x_1,\ldots,x_n,a)$ have the following generating function (see  Lidl, Mullen, and Turnwald \cite [Lemma 2.23]{LMT} ):
\begin{equation}\label{Dgf}
\sum_{k=0}^{+\infty} D_k^{(1)}(x_1,\ldots,x_n,a) z^k=\frac{\sum_{r=0}^{n}(n+1-r)(-1)^{r}x_r z^r}{\sum_{r=0}^{n+1}(-1)^{r}x_r z^r}, \quad k\geq0.
\end{equation}
\begin{proposition} 
The $k$-th entry  in the $n$-th row $R_{n,k}$, $k,n \geq 0$, of the rectangular array which defines the sequence \seqnum{A185095},  corresponds to $D_k^{(1)}(x_1,\ldots,x_n,1)$ where 
$x_r={2(n+1)-r \choose r}$ for $ r=0,\ldots,n+1$. Furthermore we have that
 \begin{equation}\label{Rnk}
R_{n,k}=2^{2k}\sum_{j=1}^{n+1}\cos^{2k}\left(\frac{j\pi}{2n+3}\right).
\end{equation}
\end{proposition}
\begin{proof} After a comparison between the two generating functions (\ref{rowgf}) and (\ref{Dgf}),  we immediately find that $R_{n,k}=D_k^{(1)}(x_1,\ldots,x_n,1)$.  Moreover, by definition (\ref{Dickdef}),  the role played by symmetric functions when $x_r={2(n+1)-r \choose r}$ for $ r=0\ldots,n+1$, allow us to prove that
$$R_{n,k}=D_k^{(1)}(x_1,\ldots,x_n,1)=\sum_{j=1}^{n+1}\alpha_{j}^k,$$ where 
$\alpha_{j}$, for  $j=1,\ldots,n+1$ are the zeros of the polynomial
\begin{equation}
P_n(x)=\sum_{j=0}^{n+1} (-1)^{j}{2(n+1)-j \choose j}x^{n+1-j}.
\end{equation}
Now let us recall the definition of the Chebyshev polynomials of the second kind $U_{h}\left(\frac{x}{2}\right)$ (see for references the books of Rivlin \cite{Riv}  and of Mason and Hascomb \cite{MH}):
\begin{equation}
U_{h} \left(\frac{x}{2}\right)=\sum_{j=0}^{\lfloor \frac{h}{2}\rfloor}(-1)^{j}
{h-j \choose j} x^{h-2j}.
\end{equation}
It is clear  that $P_n(x)=U_{2n+2}\left(\frac{\sqrt{x}}{2}\right)$. So  $\alpha_{j}$ is a zero of $P_n(x)$ if and only if $\frac{\sqrt{\alpha_{j}}}{2}$ is a positive zero for $U_{2n+2}(x)$.  Since the zeros of $U_{2n+2}(x)$ are $x_{j}=\cos\left(\frac{j\pi}{2n+3}\right)$, $j=1,\ldots,2n+3$, a simple trigonometric consideration allow us to find the positive ones when $j=1,\ldots, n+1$, showing that $\alpha_{j}=2^2\cos^2\left(\frac{j\pi}{2n+3}\right)$. The thesis immediately follows.
\end{proof}
\begin {corollary}
The entry  $R_{n,k}$ for $n\geq1$  has the alternative expression
\begin{equation}\label{Rnkal}
\begin{cases}
R_{n,k}={2k \choose k}\left(n+\frac{3}{2}\right)-2^{2k-1}, \quad 1 \leq k &lt;2n+3\\
R_{n,k}={2k \choose k}\left(n+\frac{3}{2}\right)-2^{2k-1}+(2n+3)\sum_{i=1}^{\lfloor \frac{k}{2n+3}\rfloor}{2k \choose k-i(2n+3)}, \quad k\geq 2n+3
\end{cases}
\end{equation}
where $R_{n,0}=n+1$, $R_{0,k}=1$.
\begin{proof}
Obviously if $k=0$ we obtain $R_{n,0}=n+1$ and if $n=0$ we have $R_{0,k}=1$.  When  $k\geq 1$, by (\ref{Rnk})
and thanks to the summation formulas (see Gradshteyn and Ryzhik \cite{GR}):
$$\cos^{2k}(x)=\frac{1}{2^{2k}}\left\lbrace{2k \choose k}+2\sum_{h=0}^{k-1}{2k \choose h}\cos(2(k-h)x)\right\rbrace$$
and
$$ \sum_{j=0}^{n}\cos(jx)=\frac{1}{2}\left[ 1+\frac{\sin\left(\frac{(2n+1)x}{2}\right)}{\sin\left(\frac{x}{2}\right)}\right],$$
we get
\begin{align*}
R_{n,k}&amp;=\sum_{j=1}^{n+1}\left({2k \choose k}+2\sum_{h=0}^{k-1}{2k \choose h}\cos\left(2(k-h)\frac{j\pi}{2n+3}\right)\right)=&amp;
\\ &amp;={2k \choose k}(n+1)+2\sum_{h=0}^{k-1}{2k \choose h}\left(\sum_{j=0}^{n+1}\cos\left(2(k-h)\frac{j\pi}{2n+3}\right)-1\right)=&amp; \\
&amp;={2k \choose k}(n+1)+2\sum_{h=0}^{k-1} {2k \choose h}
\left(\sum_{j=0}^{n+1}\cos\left(2(k-h)\frac{j\pi}{2n+3}\right)\right)-2\sum_{h=0}^{k-1}{2k\choose h}\\
\end{align*}

Now we have 
\begin{equation}
\begin{cases}
\sum_{j=0}^{n+1}\cos\left(2(k-h)\frac{j\pi}{2n+3}\right)=\frac{1}{2}\left[1+\frac{\sin\left(\frac{2n+3}{2}\frac{2(k-h)\pi}{2n+3}\right)}{\sin\left(\frac{(k-h)\pi}{2n+3}\right)}\right]=\frac{1}{2}\quad \text{if} \quad k\neq i(2n+3)+h\\
\sum_{j=0}^{n+1}\cos\left(2(k-h)\frac{j\pi}{2n+3}\right)=n+2 \quad \text{if}\quad k=i(2n+3)+h
\end{cases}
\end{equation}
for a certain positive integer $i$.

%\left(\frac{1}{2}\left[ 1+\frac{\sin\left(\frac{(2n+3)}{2}\frac{2(k-h)\pi}{2n+3}\right)}{\sin\left(\frac{(k-h)\pi}{2n+3}\right)}\right]-1\right)=&amp; \\
%&amp;={2k \choose k}(n+1)-\sum_{h=0}^{k-1}{2k \choose h}.
%\end{align*}
Recalling the following well known identity
$$\sum_{h=0}^{k-1}{2k \choose h}=2^{2k-1}-\frac{1}{2}{2k \choose k},$$ if $1\leq k&lt;2n+3$ we clearly obtain 
$$R_{n,k}={2k \choose k}\left(n+\frac{3}{2}\right)-2^{2k-1},$$
on the other hand, when $k\geq 2n+3$ we find
\small
$$ R_{n,k}={2k \choose k}(n+1)+2\left(\sum_{h=0}^{k-1} {2k \choose h}\left(-\frac{1}{2}\right)-\frac{1}{2}\sum_{i=1}^{\lfloor \frac{k}{2n+3}\rfloor}{2k \choose k-i(2n+3)}
+(n+2)\sum_{i=1}^{\lfloor \frac{k}{2n+3}\rfloor}{2k \choose k-i(2n+3)}\right)=$$
$$={2k \choose k}\left(n+\frac{3}{2}\right)-2^{2k-1}+(2n+3)\sum_{i=1}^{\lfloor \frac{k}{2n+3}\rfloor}{2k \choose k-i(2n+3)}.$$
\normalsize
\end{proof}
\end{corollary}
Considering the rectangular array in the definition of sequence \seqnum{A 185095} we can also study its columns, finding their generating functions. Thanks to the previous result we  have the following 
\begin{proposition}
The ordinary generating function $G_k(z)$ for the $k$-th column of  the rectangular array in the definition of \seqnum{A 185095} is
\begin{equation}
G_{k}(z)=\frac{\left(2^{2k-1}-\frac{3}{2}{2k \choose k}+1\right)z^2+\left(\frac{5}{2}{2k \choose k}-2^{2k-1}-2\right)z+1}{(1-z)^2}+P_{n_0}(z),\quad k\geq0,
\end{equation}
where $n_0=\lfloor\frac{k-3}{2}\rfloor$ and
\begin{equation}\label{pn0}
P_{n_0}(z)=\sum_{n=1}^{n_0}\left((2n+3)\sum_{i=1}^{\lfloor\frac{k}{2n+3}\rfloor}{2k \choose k-i(2n+3)}\right)z^n
\end{equation}
with the convention that the summation in (\ref{pn0}) is equal to $0$ when
$n_0&lt;1$.
\end{proposition}
\begin{proof}
Using the alternative expression determined in  Corollary \ref{Rnkal}, and the well--known formulas $\sum_{n=0}^{+\infty}z^n=\frac{1}{1-z}$ and $\sum_{n=0}^{+\infty}(n+1)z^n=\frac{1}{(1-z)^2},$
 we can compute the ordinary generating function $G_k(z)$ as follows:
 \begin{align*}
G_{k}(z)&amp;=\sum_{n=0}^{+\infty}R_{n,k}z^n=1+{2k \choose k}\left(\sum_{n=0}^{+\infty}(n+1)z^n-1\right)-\left(2^{2k-1}-\frac{1}{2}{2k \choose k}\right)\left(\sum_{n=0}^{+\infty}z^n-1\right)+P_{n_0}(z)=&amp; \\
&amp;=1+{2k \choose k}\frac{(2z-z^2)}{(1-z)^2}-\frac{z}{1-z}\left(2^{2k-1}-\frac{1}{2}{2k \choose k}\right)+P_{n_0}(z)=&amp; \\
&amp;=\frac{\left(2^{2k-1}-\frac{3}{2}{2k \choose k}+1\right)z^2+\left(\frac{5}{2}{2k \choose k}-2^{2k-1}-2\right)z+1}{(1-z)^2}+P_{n_0}(z),
\end{align*}
where the term $P_{n_0}(z)$ described in equation (\ref{pn0}) appears because when $2n+3\leq k$, or equivalently $n\leq\lfloor\frac{k-3}{2}\rfloor=n_0$, we use the second expression of $R_{n,k}$ in (\ref{Rnkal}).
\end{proof}
\section{Concluding remarks}
In this section we  discuss some consequences of the previous results concerning  the sequence \seqnum{A185095}. We show that the proof of the remaining conjectures and the relations with other sequences are immediate.
First of all we observe that the $k$-th column of the array defining the sequence \seqnum{A185095} satisfies the recurrence relation 
$$ R_{n+1,k}=2R_{n,k}-R_{n-1,k},\ \text{for}\quad  n\geq\lfloor\frac{k-3}{2}\rfloor+1,$$ which is a straightforward consequence of  (\ref{Rnkal}).
The rectangular array which generates \seqnum{A186740}  clearly is the transpose of the rectangular array that we have considered. In fact, if we start by  numbering by $0$ the columns of the rectangular array which defines \seqnum{A186740}, we find that the entry in the $k$-th row and $n$-th column  corresponds to $R_{n,k}$ by definition.  Moreover by  (\ref{Rnkal}) and using  our numeration for the columns we have 
\begin{description}
\item $R_{n,0}=n+1, n\geq0$:  the column $0$ corresponds to the natural numbers \seqnum{A000027} ;
\item $R_{n,1}=2n+1, n\geq0$ : the column $1$ is the sequence of odd integers \seqnum{A005408} ;
\item $R_{n,2}=6n+1, n\geq0$ : the column $2$ is \seqnum{A016921} ;
\item $R_{n,3}=20n-2, n\geq1$ : the column $3$ is \seqnum{A114698} ;
\item $R_{n,4}=70n-93, n\geq2$: the column $4$ is \seqnum{A114646} , 
\end{description}
and so on .

Finally another interesting consequence of the formula for $R_{n,k}$ concerns the sequence \seqnum{A198632}. By considering the array $w(h,2k)$, as defined in the page related to \seqnum{A198632} (see OEIS \cite{oeis}),  we have immediately  $w(2(n+1),2k)=2R_{n,k}$. In fact  $$w(h,l)=\text{Tr}(J_{h}^l)=\sum_{j=1}^{h}\lambda_{j}^{l},$$
where $\lambda_j$, $j=1,\ldots,h$,  are the eigenvalues of the adjacency matrix $J_{h}$, a Jacobi matrix, whose characteristic polynomial is $U_{h}(\frac{x}{2})$. Thus
$$w(2(n+1),2k)=\sum_{j=1}^{2n+2}\left(2\cos\left(\frac{j\pi}{2n+3}\right)\right)^{2k}=2^{2k+1}\sum_{j=1}^{n+1}\left(\cos\left(\frac{j\pi}{2n+3}\right)\right)^{2k}=2R_{n,k}.$$


\section{Acknowledgements}

In this paper we point out some  relations between many sequences and
we think that a lot of interesting formulas will also rise, considering
the role played by Dickson and Chebyshev polynomials, their zeros and
the summation identities  involved. The author  sincerely thanks L.
Edson Jeffery for all his interesting work and for his invitation to
try a proof of his conjectures.



\begin{thebibliography}{9}
\bibitem{oeis} N. J. A. Sloane, \emph{The On--Line Encyclopedia of Integer Sequences}, \url{http://oeis.org} .
\bibitem{LMT} R. Lidl, G. L. Mullen, and G. Turnwald, \emph{Dickson Polynomials},  John Wiley and Sons, 1993.
\bibitem{Riv} T. J. Rivlin,  \emph{The Chebyshev Polynomials},  John Wiley and Sons, 1974.
\bibitem{MH} J. C. Mason and D. C. Handscomb, \emph{Chebyshev Polynomials},  Chapman and Hall, 2002.
\bibitem{GR} I. S. Gradshteyn and  I. M. Ryzhik, \emph{Tables of Integrals, Series and Products},  Academic Press, 2007.

\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11C08; Secondary 11B83.

\noindent \emph{Keywords:} 
Dickson polynomial, Chebyshev polynomial.

\bigskip
\hrule
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\noindent (Concerned with sequences 
\seqnum{A000027},
\seqnum{A005408},
\seqnum{A016921},
\seqnum{A114646},
\seqnum{A114698},
\seqnum{A114968},
\seqnum{A185095},
\seqnum{A186740},
\seqnum{A198632}, and
\seqnum{A198636}.)


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\vspace*{+.1in}
\noindent
Received  December 7 2013;
revised version received  January 25 2014.
Published in {\it Journal of Integer Sequences}, February 16 2014.
Corollary 3 and Proposition 4 corrected, May 21 2015.

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