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\begin{center}
\vskip 1cm{\LARGE\bf 
A Note About  Invariant Polynomial \\
\vskip .04in 
Transformations of Integer Sequences
}
\vskip 1cm
\large
Leonid Bedratyuk\\
Department of Applied Mathematics \\
Khmelnitskiy National University \\
Instituts'ka 11 \\
Khmelnitskiy, 29016 \\
Ukraine \\
\href{mailto:leonid.uk@gmail.com}{\tt leonid.uk@gmail.com} \\
\end{center}

\vskip .2 in

\newcommand{\wnt}[1]{{\bf #1}}

\begin{abstract} 
We present an algorithm to find invariant polynomial
transformations of integer sequences, using an approach based on
classical invariant theory.
\end{abstract} 

\section{Introduction}

Let  $\mathcal{A}=(a_n)_{n \geq 0}$ be an integer sequence. A sequence
$\textbf{F}(\mathcal{A})=\bigl(b_n=f_n(a_0,a_1,\ldots,a_m)\bigr)_{n
  \geq 0},$ where  $f_n \in  \mathbb{Z}[x_0,x_1,\ldots,x_m]$ and $m \geq n,$
% is  a polynomial of finitely many   variables  %$\mathbb{Z}[\mathcal{A}]:=\mathbb{Z}[a_0,a_1,\ldots ]$
 is called a  polynomial transformation of  the sequence $\mathcal{A}.$ In the sequel, only the polynomial transformations are considered. The composition  $\textbf{F} \circ \textbf{G}:= \textbf{F}(\textbf{G}(\mathcal{A}))$ of the two transformations $\textbf{F}$ and $\textbf{G}$ can be defined  in a natural way. A   transformation  $\textbf{G}$ is called the \textit{inverse} transformation of $\textbf{F}$, and it is denoted by $\textbf{F}^{-1},$  if for every sequence $\mathcal{A}$ we  have $\textbf{F}(\textbf{G}(\mathcal{A}))=\mathcal{A}.$
A transformation  $\textbf{F}$ is called  $\textbf{G}$-\textit{invariant}   if for every sequence $\mathcal{A}$ we  have $\textbf{F}(\textbf{G}(\mathcal{A}))=\textbf{F}(\mathcal{A})$.
         
 
For  instance, it is well known (see Layman {\cite{Lay} or Spivey and
  Steil \cite{SpS}}) that the Hankel  transformation  $\textbf{H}$
is $\textbf{B}_\mu$-invariant.    Here, for $\mu \in
\mathbb{Q}$,  $$\textbf{B}_\mu(\mathcal{A})=\left ( b_n=\sum_{i=0}^n
  {n \choose i} a_i \mu^{n-i} \right)_{n \geq 0},$$ denotes the
$\mu$-binomial transformation  and  $\textbf{H}(\mathcal{A})=(h_n)_{n
  \geq 0},$ where   $h_n$ is the determinant of the Hankel matrix for  the elements  $a_0,a_1,\ldots, a_{2n}$:
 $$
h_n=\begin{vmatrix} a_0 &amp;  a_1 &amp; a_2 &amp;\cdots  &amp; a_n  \\
a_1 &amp;  a_2 &amp; a_3 &amp;\cdots  &amp; a_{n+1}  \\
 \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
 a_{n-1} &amp;  a_{n} &amp; a_{n-1} &amp;\cdots  &amp; a_{2n-1}\\
 a_n &amp;  a_{n+1} &amp; a_{n+2} &amp;\cdots  &amp; a_{2n}  
\end{vmatrix}.
$$ 
This determinant arises in  classical invariant theory as the
\textit{catalecticant }of a binary form, see Grace and
Young\,{\cite[p.232]{GrY}}. It was first introduced by
Sylvester\,{\cite{Sylv}}. Also, the transformation $\textbf{B}_\mu$
appears in Hilbert's book  \cite[p.\,25]{Hilb}.  We can  prove that
the Hankel transformation is $\textbf{B}_\mu$-invariant via classical
invariant theory, as follows. Write $\partial_i$ for the partial derivative
$\partial/\partial a_i$ and let $\mathcal{D}$ be the differential operator:
$$
\mathcal{D}=a_0 \partial_1+2 a_1 \partial_2+3 a_2 \partial_3+\cdots +2n a_{2n-1} \partial_{2n}.
$$
Define
$$
h'_n=\begin{vmatrix} b_0 &amp;  b_1 &amp; b_2 &amp;\cdots  &amp; b_n  \\
b_1 &amp;  b_2 &amp; b_3 &amp;\cdots  &amp; b_{n+1}  \\
 \vdots &amp; \vdots &amp; \vdots &amp; \ddots &amp; \vdots \\
 b_{n-1} &amp;  b_{n} &amp; b_{n-1} &amp;\cdots  &amp; b_{2n-1}\\
 b_n &amp;  b_{n+1} &amp; b_{n+2} &amp;\cdots  &amp; b_{2n}  
\end{vmatrix}.
$$ 
Then, as in Lie  \cite{Lie}, we have 
$$
h'_n=h_n+\mathcal{D}(h_n)\mu+\mathcal{D}^2(h_n)\frac{\mu^2}{2!}+\cdots +
\mathcal{D}^i(h_n)\frac{\mu^i}{i!}+\cdots.
$$
By applying the determinant derivative rule we get, after some calculation,  that  $\mathcal{D}(h_n) = 0$ for all $n.$ 
Therefore  $h'_n=h_n.$ This condition is exactly equivalent to  the
$\textbf{B}_\mu$-invariance of  the Hankel transformation.

   This example motivates us to consider the following two general problems: 

\bigskip
   
\noindent \textbf{Problem 1.} For  a fixed transformation $\textbf{F}$, find all $\textbf{F}$-invariant transformations. 

\bigskip
  
\noindent  \textbf{Problem 2.} For   a fixed transformation  $\textbf{F}$, find a transformation  $\textbf{G}$   such that   $\textbf{F}$ is a $\textbf{G}$-invariant transformation.

\bigskip
  
The aim of this paper is to develop an effective method to solve
these two problems for some special kinds of  transformations. The
method is inspired by results in classical invariant theory and the
theory of locally nilpotent derivations. 

In the next section we introduce exponential transformations and then
prove  that for such transformations  Problem 1 is always solvable.

In Section~\ref{two} we also give a short introduction to the theory of locally
nilpotent derivations and offer algorithms to solve  Problems 1 and 2
for special classes of transformations.

In Section~\ref{three}  we  give  another   proof that  the Hankel
transformation is  $\textbf{B}_\mu$-invariant  and  introduce several
new  $\textbf{B}_\mu$-invariant transformations.  All of these transformations come from classical invariant theory. Further, we  describe all $\textbf{B}_\mu$-invariant polynomial transformations in terms  of derivations. 


In Section~\ref{four}  we give some examples to illustrate the theory. 
%================================================================





%==================================================================
\section{Derivations and automorphisms}
\label{two}



{Let $R = \mathbb{Z}[x_0, x_1, \ldots]$ be the polynomial ring in
countably many variables and let $\varphi: R \rightarrow R$ be a ring
homomorphism.} Such a map $\varphi$ is uniquely determined by the
sequence of polynomials  $( \varphi(x_n) \mid \,  n=0,1,\ldots ).$ To any polynomial   map $\varphi$ and an  integer sequence $( a_n)_{n \geq 0}$ we  assign the  transformation $( \varphi(a_n))_{n \geq 0}.$   A  polynomial map $\varphi$ is said to be a \textit{polynomial automorphism} if there is a polynomial map $\psi$ such that $\varphi(\psi(x_n))=x_n$ for all $n.$ 
%For $f\in \mathbb{Z}[x_0,x_1,\ldots,x_m]$ put 
%$$
%\varphi f:= f(\varphi(x_0),\varphi(x_1),\ldots, \varphi(x_m)).
%$$

Denote by   $\mathbb{Z}[x_0,x_1,\ldots,x_m]^{\varphi}$ the \textit{algebra of   $\varphi$-invariants}: 
$$
\mathbb{Z}[x_0,x_1,\ldots,x_m]^{\varphi}:=\left\{ f \in  \mathbb{Z}[x_0,x_1,\ldots,x_m] \mid f(\varphi(x_0),\varphi(x_1),\ldots, \varphi(x_m))=f(x_0,x_1,\ldots, x_m) \right \}.
$$

The following theorem will be our   main computing tool in finding  invariant  polynomial transformations.
\begin{theorem}
Let $\varphi$ be a polynomial map  and let   $\text{\textbf{F}}(\mathcal{A})=(\varphi(a_n))_{n \geq 0}$ be the corresponding integer transformation. Then the transformation   
$$\textbf{G}(\mathcal{A})= \left(g_n(a_0,a_1,\ldots,a_m) \right )_{n \geq 0},$$ is $\textbf{F}$-invariant if and only  if $g_n(x_0,a_1,\ldots,x_m) \in \mathbb{Z}[x_0,x_1,\ldots,x_m]^{\varphi}.$
\label{t1}
\end{theorem}
The proof follows immediately  from the definitions above.

  
In general, the  problem of finding   the algebras of
$\varphi$-invariants  is  difficult. But we will show that when
$\varphi$  is an exponential automorphism this  problem can be reduced
to the calculation of the kernel of a derivation. 

A {\it derivation} of the algebra $\mathbb{Z}[x_0,x_1,\ldots,x_n]$ is a linear  map  $D$ satisfying the Leibniz rule: 
$$
D(f_1 \, f_2)=D(f_1) f_2+f_1 D(f_2), \text{  for all }  f_1, f_2 \in \mathbb{Z}[x_0,x_1,\ldots,x_n].
$$
A derivation $D$  is called {\it locally nilpotent} if for every $f \in \mathbb{Z}[x_0,x_1,\ldots,x_n]$ there is an $r \in \mathbb{N}$ such that $D^r(f)=0.$ 
The subalgebra 
$$
\ker D:=\left \{ f \in \mathbb{Z}[x_0,x_1,\ldots,x_n]|  D(f)=0 \right \},
$$
is called the {\it kernel} of the derivation $D.$

Any derivation   $D$ is completely determined by the elements $D(x_i).$ A  derivation   $D$  is called  \textit{linear} if  $D(x_i)$ is a linear form. A  linear locally nilpotent derivation is called a \textit{Weitzenb\"ock derivation}.
The  Weitzenb\"ock derivation defined by  $\mathcal{D}(x_0)=0, \mathcal{D}(x_i)=i x_{i-1}$ is called the \textit{basic Weitzenb\"ock derivation}. 
%The kernel  of basic  Weitzenb\"ock derivation is well known.
 There exists an isomorphism between the kernel $\ker \mathcal{D}$
 and the algebra of covariants of a binary form,  a major object of
 research in classical invariant theory during the 19th century.  Here
 are a few examples of covariants: the discriminant, the resultant,
 the Jacobian, the Hessian, the catalectiant and the transvectant. The
 following theorem gives a  description of the algebra $\ker
 \mathcal{D}$. 
 
   
\begin{theorem} \label{kerbasic} The kernel of the basic Weitzenb\"ock derivation $\mathcal{D}$ of $\mathbb{Q}[x_0,x_1,\ldots,x_n]$ is a finitely generated algebra and 
$$
\ker \mathcal{D}=\mathbb{Q}[z_2,z_3,\ldots,z_n][x_0,x_0^{-1}] \cap \mathbb{Q}[x_0,x_1,\ldots,x_n],
$$
where  
$$
z_k=\sum_{i=0}^{k-2} (-1)^i {k \choose i} x_{k-i} x_1^i x_0^{k-i-1}+(k-1) (-1)^{k+1} x_1^k. 
$$
\label{t2}
\end{theorem}
Theorem~\ref{kerbasic} is a classical result due to Cayley, see Glenn
\cite[p.\ 164]{Gle}. One can also find more modern proofs in Nowicki
\cite{Now} and van den Essen \cite{Essen}. 

How are we to find the kernel of an arbitrary linear locally nilpotent derivation $D$?  Let us consider  the vector space (over $\mathbb{Q}$)  $X_n=\langle x_0,x_1,\ldots,x_n \rangle.$  Suppose that there exists an  isomorphism  $\Psi:X_n   \to X_n  $  such that  $\Psi \mathcal{D}=  D \Psi.$ This implies   that  $\ker D =\Psi \left( \ker \mathcal{D} \right), $ i.e.,
$$
\ker D=\mathbb{Q}[\Psi(z_2),\Psi(z_3),\ldots,\Psi(z_n)][\Psi(x_0),\Psi(x_0)^{-1}] \cap \mathbb{Q}[x_0,x_1,\ldots,x_n].
$$
 Such an  isomorphism $\Psi$ is called a \textit{$(\mathcal{D},D)$-intertwining} isomorphism. 
Therefore, to describe the kernel of an arbitrary Weitzenb\"ok derivation $D$ it is enough to know  the explicit  form of any   \textit{$(\mathcal{D},D)$-intertwining} isomorphism. 

An automorphism  $\varphi$ is called  \textit{exponential} if there exists a  locally nilpotent derivation $D$ such that 
$$
\varphi=\exp(D)=D^0+D+\frac{1}{2!}D^2+\cdots.
$$
For instance,    any automorphism of the
form $$\varphi(x_n)=x_n+f(x_0,x_1,\ldots,x_{n-1}), f \in
\mathbb{Z}[x_0,x_1,\ldots,x_{n-1}],$$ is exponential, see  Drensky and
Yu \cite{Dren}. Nowicki \cite[Proposition 6.1.4]{Now} shows that for any  exponential automorphism $\varphi=\exp(D)$, 
\begin{equation} \label{Now614}
    \mathbb{Q}[x_0,x_1,\ldots,x_n]^{\varphi}=\ker D. \end{equation} 
  
We introduce an analogue of these notions for the integer polynomial
  transformations. 

  \begin{definition} The transformation   $D({\bf F}(\mathcal{A})):=(D(f_n(x_0,x_1,\ldots,x_m))|_{(a_0,a_1,\ldots,a_m)})_{n \geq 0},$ 
is called the $D$-de\-ri\-va\-ti\-ve of the polynomial trans\-for\-mation ${\bf F}=(f_n(a_0,a_1,\ldots,a_m) )_{n \geq 0},$ $f \in \mathbb{Z}[x_0,\ldots,x_m].$
\end{definition}


\begin{definition}
A  transformation \text{ ${\bf F}$}  is called exponential if there exists  a locally nilpotent  derivation $D$ such that 
$$
{\bf F}(\mathcal{A})=\exp D (\mathcal{A}).
$$
\end{definition}
We can now  rewrite  Theorem~\ref{t1}  for an exponential transformation.
\begin{theorem}
Suppose  a transformation $\textbf{F}$ is exponential and ${\bf
  F}(\mathcal{A})=\exp D (\mathcal{A})$ for some locally nilpotent
derivation $D.$  Then a polynomial transformation  $\textbf{G}$ is
${\bf F}$-invariant if and only if   $D(\textbf{G}(\mathcal{A}))={\bf
  0},$ where ${\bf 0}$ stands for the zero sequence $(0,0,0,\ldots ).$
\label{t3}
\end{theorem}

\begin{proof}
Suppose that  the transformation $\textbf{G}$ is ${\bf F}$-invariant. Then by
Theorem~\ref{t1} we  have 
$$\textbf{G}(\mathcal{A})= \left(g_n(a_0,a_1,\ldots,a_m) \right )_{n
  \geq 0},$$ where $g_n(x_0,x_1,\ldots,x_m) \in
\mathbb{Z}[x_0,x_1,\ldots,x_m]^{\varphi},$  for $\varphi=\exp D.$
Since the automorphism $\varphi$ is exponential, we have that
$D(g_n(x_0,x_1,\ldots,x_m))=0$ by (\ref{Now614}). Thus $D(\textbf{G}(\mathcal{A}))=0.$

Suppose now  that the  transformation $\textbf{G}$  has  the form $$\textbf{G}(\mathcal{A})= \left(g_n(a_0,a_1,\ldots,a_m) \right )_{n \geq 0},$$ and $D(\textbf{G}(\mathcal{A}))=0.$  This implies  that 
$D(g_n(a_0,a_1,\ldots,a_m))=0$ for all $n.$ Then  we have that the polynomial $g_n(x_0,x_1,\ldots,x_m)$ belongs to $ \mathbb{Q}[x_0,x_1,\ldots,x_n]^{\varphi}$  where $\varphi=\exp D.$ By Theorem~\ref{t1}  the transformation $\textbf{G}$ is  ${\bf F}$-invariant.
\end{proof}


The Weitzenb\"ok derivations are related to   some special transformations by the following theorem:
\begin{theorem} \label{expthm}
Given an integer sequence $(\alpha_n)_{n \geq 0}$, the transformation  $F(\mathcal{A}){=}
\left (a_n+\sum\limits_{i=0}^{n-1} \alpha_i a_i  \right )_{n \geq 0}
$
is exponential and    $F(\mathcal{A})=\exp D (\mathcal{A}),$
where the derivation  $D$ is a  Weitzenb\"ok derivation defined by   
$$
D(f) =\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i} E^i(f),
$$
and   $E(x_k)=\sum\limits_{i=0}^{k-1} \alpha_k x_k$. 
\label{t4}
\end{theorem}
The proof follows from van den Essen \cite[Proposition 2.1.3]{Essen}. 

Theorem~\ref{expthm} yields an  algorithm to solve   Problem 1 in the  case when the transformation $\textbf{F}(\mathcal{A})=(b_n)_{n \geq 0}$ has  the special form 
$$
b_n=a_n+\sum\limits_{i=0}^{n-1} \alpha_i a_i, \alpha_i \in \mathbb{Z}.
$$
In this case, for the corresponding polynomial automorphism
$\varphi(x_n)=x_n+\sum\limits_{i=0}^{n-1} \alpha_i x_i $, we find the
explicit form of the Weitzenb\"ok derivation   $D$ such that
$\varphi=\exp(D)$ ({as predicted in } Theorem~\ref{t3}). After that we
find any $(\mathcal{D},D)$-intertwining automorphism $\Psi$ and obtain
that $\ker D =\Psi \left( \ker \mathcal{D} \right). $ Then an
arbitrary sequence of kernel elements defines an
$\textbf{F}$-invariant   transformation  (by Theorem~\ref{t1}). 


To solve Problem 2 for a transformation
$\textbf{F}(\mathcal{A})=(b_n=f_n(a_0,a_1,\ldots,a_m))_{n \geq 0}$ we
find a locally nilpotent derivation $D$ of
$\mathbb{Z}[x_0,x_1,\ldots,x_m]$ such that $f_n(x_0,x_1,\ldots,x_m)
\in \ker D.$ This can be done by the method of undetermined
coefficients. We define the automorphism $\varphi=\exp D$ and the
transformation $\textbf{G}(\mathcal{A})=(b_n=\varphi(a_n))_{n \geq
  0}.$ Then the transformation $\textbf{F}$ is $\textbf{G}$-invariant
by Theorem~\ref{t1}.
%
% WNT: I'm not sure why this produces all such transformations
% \textbf{G} though it certainly produces some of them. Perhaps some
% more explanation is possible?
%


%==========================================================


\section{The $\mu$-binomial transformations}
\label{three}
 
We use the techniques developed in Section~\ref{two}  to get  another  proof of   the following well known result.
%==========================================================
 
\begin{theorem}[\cite{Lay,SpS}]
The Hankel transformation $\textbf{H}$ is  $\textbf{B}_\mu$-invariant.
\label{t5}
\end{theorem}

\begin{proof} We follow  the algorithm from Section~\ref{t2}. The  automorphism $\varphi_\mu$ corresponding to $\textbf{B}_\mu$ has the form: $$\varphi_\mu(x_n)=\sum\limits_{i=0}^n {n \choose i} x_i \mu^{n-i}.$$
For the basic Weitzenb\"ok derivation $\mathcal{D}$  we have 
%$$\begin{array}{lllll}
%\exp (\mu \mathcal{D})(x_n) &amp; =&amp; \sum_{i \geq 0} \frac{1}{i!}
%\left(\mu\mathcal{D}\right)^i(x_n)&amp; =&amp; \sum_{i=0}^{n}
%\frac{n(n{-}1)\ldots (n{-}(i{-}1))}{i!}\, \mu^i x_{n-i}\\&amp; =&amp;
%\sum_{i=0}^{n} {n \choose i} \mu^i x_{n-i}&amp; =&amp;\sum\limits_{i=0}^n {n \choose i} x_i \mu^{n-i}.
%\end{array}$$
\begin{gather*}
\exp (\mu \mathcal{D})(x_n)=\sum_{i \geq 0} \frac{1}{i!} \left(\mu\mathcal{D}\right)^i(x_n)=\sum_{i=0}^{n} \frac{n(n{-}1)\cdots (n{-}(i{-}1))}{i!}\, \mu^i x_{n-i}=\\=\sum_{i=0}^{n} {n \choose i} \mu^i x_{n-i}=\sum\limits_{i=0}^n {n \choose i} x_i \mu^{n-i}.
\end{gather*}
Thus $\varphi_\mu=\exp (\mu \mathcal{D}).$ It follows that the
transformation $\textbf{B}_\mu$  is exponential, i.e.,
$\textbf{B}_\mu=\exp (\mu \mathcal{D}) \mathcal{A}.$  Since the
catalectiant belongs to the kernel of  the derivation $\mathcal{D}$ we  have   that $\mathcal{D}(\textbf{H}(\mathcal{A}))=\textbf{0}$.  Then by 
Theorem~\ref{t3}  we obtain  that  the transformation $\textbf{H}$ is  $\textbf{B}_\mu$-invariant. 

 The map $\exp (\mu \mathcal{D}):\mathbb{Q}[x_0,x_1,\ldots,x_n] \to
 \mathbb{Q}[x_0,x_1,\ldots,x_n]$ is a ring homomorphism, see van den
 Essen \cite[Proposition 1.2.24]{Essen}. It follows that
 $\varphi_{\mu_1+\mu_2}=\varphi_{\mu_1} \circ
 \varphi_{\mu_2}$. Therefore $\varphi_{\mu} \circ \varphi_{-\mu}$  is
 the identity map and  $\textbf{B}_\mu^{-1}=\textbf{B}_{-\mu}$. It
 follows  immediately that the inverse transformation
 $\textbf{B}_\mu^{-1}$ is also an $\textbf{H}$-invariant transformation.\end{proof}

See French \cite{French} for a proof that all
$\textbf{B}_\mu$-invariant transformations form a group.  The identity
$\varphi_{\mu_1+\mu_2}=\varphi_{\mu_1} \circ \varphi_{\mu_2}$ implies
that the group $(\mathbb{Z},+)$ is a subgroup of those groups.
 
 
 The following theorem gives a  solution to Problem 1   for  the $\mu$-binomial transformation.
 \begin{theorem} A   transformation  $\textbf{F}$ is   $\textbf{B}_\mu$-invariant  if and only if   ${\bf \mathcal{D}}\left(\textbf{F}(\mathcal{A})\right)={\bf 0}.$ 
 \label{t6}
 \end{theorem}
 The proof follows from Theorem~\ref{t3}.
 
 The next result follows from Bedratyuk \cite[Theorem 3.2]{B_App}. 
 
 \begin{theorem}
 Let $\textbf{F}$  be an arbitrary  $\textbf{B}_\mu$-invariant transformation. Then $\textbf{F}(\textbf{1})={\bf 0}$, where \\ {${\textbf{1}=(1,1,1,\ldots,1,\ldots).}$}
 \label{t7}
 \end{theorem}
 
 
 Below we describe  some Hankel-type transformations which arise in classical invariant theory. Note that all of these  transformations  are  $\textbf{B}_\mu$-invariant   and $\textbf{B}_\mu^{-1}$-invariant.

\subsection{Cayley transformation}

Put ${\rm \textbf{CAYLEY}}(\mathcal{A})=( b_{n+2})_{n \geq 0},$ where 
$$
b_{n}=\sum_{i=0}^{n-2} (-1)^i {n \choose i} a_{n-i} a_1^i a_0^{n-k-1}+(n-1) (-1)^{n+1} a_1^n.
$$
The definition of this transformation is inspired by Theorem~\ref{t2}.
\subsection{Transvectant transformation}
Let  $\mathcal{A}=(a_n)_{n \geq 0}$ and $\mathcal{C}=(c_n)_{n \geq 0}$ be two sequences. The  trans\-for\-ma\-tion  ${\rm \textbf{TR}}(\mathcal{A},\mathcal{C})=(b_n)_{n \geq 0},$ where
$$
b_n=\sum_{i=0}^n (-1)^i {n \choose i} a_i c_{n-i},
$$
is called the transvectant transformation.
We have  $${\rm \textbf{Tr}}(\textbf{B}_\mu(\mathcal{A}),\textbf{B}_\mu(\mathcal{C}))={\rm \textbf{Tr}}(\mathcal{A},\mathcal{C}).$$

In the case  $\mathcal{C}=\mathcal{A}$ we  get 
$$
b_n=\sum_{i=0}^n (-1)^i {n \choose i} a_i a_{n-i}.
$$

\subsection{Resultant   transformation}
Let  $\mathcal{A}=(a_n)_{n \geq 0}$ and $\mathcal{C}=(c_n)_{n \geq 0}$ be two sequences. The transformation  ${\rm \textbf{RES}}(\mathcal{A},\mathcal{C})=(b_n)_{n \geq 0}$,  where $b_n$
is  the leading coefficient  of the resultant of the polynomials 
$$
P_n(\mathcal{A})=\sum_{i=0}^{n} {n \choose i} a_i X^{n-i}, P_n(\mathcal{C})=\sum_{i=0}^{n} {n \choose i} c_i X^{n-i},
$$
is called the \textit{resultant transformation}.
\subsection{Discriminant   transformation}

The transformation  ${\rm \textbf{DISCR} }(\mathcal{A})=(b_n)_{n \geq 0}$,  where $b_n$
is  the discriminant of the polynomial 
$$
P_{n+2}(\mathcal{A})=\frac{1}{(n+2)^{n+2}}\sum_{i=0}^{n+2}a_i {n+2 \choose i} X^{n+2-i},
$$
is called the \textit{discriminant transformation}.

\bigskip

\noindent
\textbf{Problem 3.} What is   the explicit form of the $(\mathcal{D},D)$-intertwining isomorphism $\Psi(\textbf{F})$ for $$\textbf{F} \in \{\textbf{CAYLEY, H, RES, DISCRIM, TR} \}?$$

\section{Examples}
\label{four}

 
\subsection{ Transformation  ${\rm PSUM}(\mathcal{A})=(b_n=a_0+a_1+\cdots+a_n )_{n \geq 0}$} The corresponding locally nilpotent derivation  (see Theorem~\ref{t3}) has the form 
$$
D(x_n)=\sum_{i=1}^{\infty} \frac{(-1)^{i+1}}{i} E^i(x_n).
$$
We have 
\begin{align*}
&amp;E(x_0)=0,E(x_n)=x_0+x_1+x_2+\cdots+x_{n-1},\\
&amp;E^2(x_n)=\sum_{i=0}^{n-1}E(x_i)=\sum_{i=0}^{n-1}\sum_{j=0}^{i-1}x_j=\sum_{i=0}^{n-2} (n-1-i) x_i.
\end{align*}
By induction we obtain $E^i(x_n)=\displaystyle \sum\limits_{k=0}^{n-i} \displaystyle {n-i-1\choose i-1} x_k.$
Then \begin{gather*}
D(x_n)=\sum_{i=1}^{n} \frac{(-1)^{i+1}}{i} \sum_{k=0}^{n-i} \displaystyle {n-i-1\choose i-1} x_k=\sum_{k=0}^{n-1} \left( \sum_{i=0}^{n-1-k} \frac{(-1)^{i}}{i+1} { {n{-}1{-}k} \choose i}\right)x_k=\sum_{k=0}^{n-1} \frac{x_k}{n-k}.
\end{gather*}
Let us find  a $(\mathcal{D},D)$-intertwining transformation $\Psi$. We   show that   
$$
\Psi(\mathcal{A})=\left \{\Psi(x_n)=\sum_{k=0}^n (-1)^{n+k} k! \left\{\begin{matrix} n \\ k \end{matrix} \right\} x_k,  \Psi(x_0)=x_0 \right \},
$$
where $\left\{\begin{matrix} n \\ k \end{matrix} \right\}$ is  the
Stirling number of the second kind, is such a transformation. In fact,
\begin{gather*}
D\left(\Psi(x_n)\right)=D\left(\sum_{k=0}^n (-1)^{n+k} k! \left\{\begin{matrix} n \\ k \end{matrix} \right\} x_k\right)=\sum_{k=0}^n (-1)^{n+k} k! \left\{\begin{matrix} n \\ k \end{matrix} \right\} \sum_{i=0}^{k-1} \frac{x_i}{k-i}\\
=\sum_{i=0}^{n-1} \sum_{j=i+1}^n (-1)^{n+j} \left\{\begin{matrix} n \\ j \end{matrix}\right\} \frac{j!}{j-i}\,\, x_i=n \sum_{i=0}^{n-1} (-1)^{n-1+i} \left\{\begin{matrix} n-1 \\ i \end{matrix}\right\}i!\, x_i =\Psi(\mathcal{D}(x_n)).
\end{gather*}
Therefore, we  may now construct a {\rm  PSUM}-invariant
transformation using our  known $\textbf{B}_\mu $-invariant transformations and this $(\mathcal{D},D)$-intertwining transformation $\Psi$. For instance,    the transformation 
\begin{gather*}
\Psi(\textbf{H}(\mathcal{A}))=\{a_{{0}},-{a_{{1}}}^{2}-a_{{1}}a_{{0}}+2\,a_{{2}}a_{{0}},-4\,a_{{1}}a_{{2}}a_{{0}}+24\,a_{{1}}a_{{2}}a_{{3}}+24\,a_{{0}}a_{{1}}a_{{3}}+48\,a
_{{0}}a_{{2}}a_{{4}}-8\,{a_{{2}}}^{3}-\\-8\,a_{{0}}{a_{{2}}}^{2}-12\,a_{{
1}}{a_{{2}}}^{2}-36\,a_{{0}}{a_{{3}}}^{2}-4\,{a_{{1}}}^{2}a_{{2}}-24\,
{a_{{1}}}^{2}a_{{4}}+24\,{a_{{1}}}^{2}a_{{3}}-24\,a_{{0}}a_{{1}}a_{{4}
}, \ldots  \},
\end{gather*}
is {\rm  PSUM}-invariant.


\subsection{The Transformation ${\rm SUM}(\mathcal{A})=(b_n=a_n+a_{n-1} )_{n \geq 0}$}

We  have $\varphi(x_n)=x_n+x_{n-1},$  $E(x_n)=\varphi(x_n)-x_n=x_{n-1}$, and 
$$
D(x_n)=\sum_{i \geq 1} \frac{(-1)^{i+1}}{i} E^i(x_n)=\sum_{i=1}^n \frac{(-1)^{i+1}}{i} \, x_{n-i}.
$$ 
  Let  
$$
\Psi(x_0)=x_0, \Psi(x_n)=c_{n,1}x_1+c_{n,2} x_2 +\cdots+c_{n,n} x_n.
$$
The $(\mathcal{D},D)$-intertwining map satisfies the conditions   $D
(\Psi(x_n))=\Psi\left(\mathcal{D}(x_n)\right)$, $n=0,1,2,\ldots n.$
After a routine calculation we get that $c_{n,i}=i!
\left\{\begin{matrix} n \\ i \end{matrix} \right \}$ and a
$(\mathcal{D},D)$-intertwining map is given by 
$$
\Psi(x_n)=\sum_{i=1}^n i! \left\{\begin{matrix} n \\ i \end{matrix} \right\} x_i.
$$
Thus, the transformation 
$$
\Psi(\textbf{H}(\mathcal{A}))=\begin{vmatrix} \Psi(a_0) &amp;  \Psi(a_1) &amp; \Psi(a_2) &amp;\cdots  &amp; \Psi(a_n)  \\
\Psi(a_1) &amp;  \Psi(a_2) &amp; \Psi(a_3) &amp;\cdots  &amp; \Psi(a_{n+1})  \\
\hdotsfor{5}\\
 \Psi(a_{n-1}) &amp;  \Psi(a_{n}) &amp; \Psi(a_{n-1}) &amp;\cdots  &amp; \Psi(a_{2n-1})\\
 \Psi(a_n) &amp;  \Psi(a_{n+1}) &amp; \Psi(a_{n+2}) &amp;\cdots  &amp; \Psi(a_{2n})  
\end{vmatrix}.
$$
 is ${\rm SUM}$-invariant.

\subsection{ Transformation  ${\rm DIFF}(\mathcal{A}) =(b_n=a_n-a_{n-1} )_{n \geq 0}$} The corresponding automorphism has the form  $\varphi(x_n)=x_n-x_{n-1}.$ This  implies that $E(x_n)=-x_{n-1}$ and  $E^i(x_n)=(-1)^{i} x_{n-i}.$
Then the derivation $D$ is defined by 
$$
D(x_n)=\sum_{i \geq 1} \frac{(-1)^{i+1}}{i} E^i(x_n)=\sum_{i=1}^n \frac{(-1)^{i+1}}{i} \, (-1)^i x_{n-i}=-\sum_{i=1}^{n}\frac{x_{n-i}}{i}.
$$ 
A $( \mathcal{D},D)$-intertwining map  has  the form
$$
\Psi(x_n)=\sum_{i=1}^n (-1)^i i! \left\{\begin{matrix} n \\ i \end{matrix} \right\} x_i.
$$


\subsection{The  transformation $\textbf{F}=( b_n=\sum\limits_{i=0}^{2n} (-1)^i a_i a_{2n\!{-}i})_{n \geq 0}$.}

Let us try to solve   Problem 2  for this transformation.  To do so, we  need to find a suitable locally nilpotent derivation that satisfies the conditions   
$$D\left(\sum\limits_{i=0}^{2n} (-1)^i x_i x_{2n\!{-}i}\right)=0, n=0,1,\ldots .$$ 


Let us  consider the locally nilpotent derivation $D$ with
$D(x_i)=x_{i-1}.$ Now $$D\left(\sum\limits_{i=0}^{2n} (-1)^i x_i
  x_{2n\!{-}i}\right)=0,$$ as is shown in Bedratyuk \cite{A-Cas}. Let us calculate the exponential automorphism $\varphi=\exp D.$  We  have 
$$
\varphi(x_n)=D^0(x_n)+D(x_n)+\frac{1}{2!} D^2(x_n) +\cdots=x_n+x_{n-1}+\frac{1}{2!} \, x_{n-2}+\frac{1}{n!}\, x_0.
$$
Define a rational transformation   by $\textbf{G}(\mathcal{A}):=( \varphi(x_n) )_{n \geq 0}$. Then $\textbf{F}(\textbf{G}(\mathcal{A}))=\textbf{F}(\mathcal{A}).$  
 
 
 
  
\section{Acknowledgments}
 
  The author would  like to thank
the referees for many valuable suggestions that improved the  paper.
 
 
 

 
\begin{thebibliography}{15}
\bibitem{Lay} J. Layman, 
The Hankel transform and some of its properties,
\textit{J. Integer Seq.}, \textbf{4} (2001), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL4/LAYMAN/hankel.html}{Article 01.1.5}.

\bibitem{SpS} M.  Spivey and  L. Steil, The $k$-binomial transforms and
the Hankel transform, \textit{J. Integer Seq.}, \textbf{9} (2006), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL9/Spivey/spivey7.html}{Article
06.1.1}.

\bibitem{Gle} O. Glenn,  \textit{Treatise on Theory of Invariants},
Boston, 1915.

\bibitem{GrY} J. Grace and J. A. Young, \textit{The  Algebra  of
Invariants}, Cambrige Univ. Press, 1903.

\bibitem{Sylv}  J. J. Sylvester,   On the principles of the calculus of
forms, {\it Cambridge and Dublin Math. J.}, \textbf{7}
(1852), 52--97.

\bibitem{Lie} S. Lie, \textit{Theorie der Transformationsgruppen.
Erster Abschnitt}, Teubner, 1888.

\bibitem{Hilb}
D. Hilbert,  \textit{Theory of Algebraic Invariants}, Cambridge
University Press, 1993.

\bibitem{Now}  A. Nowicki, \textit{Polynomial Derivations and their
Rings of Constants}, UMK,  Torun, 1994.

\bibitem{Essen}
A. van den Essen,  \textit{Polynomial Automorphisms and the Jacobian
Conjecture}, Birkh\"auser, 2000.

\bibitem{Dren} V. Drensky  and  J.-T. Yu, Exponential automorphisms of
polynomial algebras, \textit{Comm. Algebra} {\bf 26} (1998), 2977--2985.

\bibitem{French}  C. French, 
Transformations preserving the Hankel transform,
\textit{J. Integer Seq.}, \textbf{10} (2007), 
\href{https://cs.uwaterloo.ca/journals/JIS/VOL10/French/french13.html}{Article 07.7.3}.

\bibitem{B_App} L. Bedratyuk, Semi-invariants of  binary forms and
identities for Bernoulli, Euler and Hermite polynomials,  \textit{Acta
Arith.}, \textbf{151} (2012), 361--376.

\bibitem{A-Cas} L. Bedratyuk, Kernels of derivations of polynomial
rings and Casimir elements, \textit{Ukr. Mat. Zh.}, \textbf{62} (2010),
435--452.

\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B75; Secondary 13A50.

\noindent \emph{Keywords: } 
invariant, polynomial transformation, Hankel determinant.

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\hrule
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\vspace*{+.1in}
\noindent
Received  March 3 2012;
revised versions received June 26 2012; August 6 2012;
August 23 2012.
Published in {\it Journal of Integer Sequences}, September 8 2012.

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