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\begin{center}
\vskip 1cm
{\LARGE\bf A Family of Riordan Group Automorphisms} 
\vskip 1cm 
\large
\^Angela Mestre and Jos\'e Agapito \\
Centro de An\'alise Funcional, Estruturas Lineares e Aplica\c{c}\~oes\\
Grupo de Estruturas Alg\'ebricas, Lineares e Combinat\'orias\\
Departamento de Matem\'atica\\
Faculdade de Ci\^encias, Universidade de Lisboa\\
1749-016 Lisboa\\
Portugal\\
\href{mailto:amestre@fc.ul.pt}{\tt amestre@fc.ul.pt} \\
\href{mailto:jaruiz@fc.ul.pt}{\tt jaruiz@fc.ul.pt} \\
\end{center}

\begin{abstract}
In 2006, Bacher introduced a family of Riordan group automorphisms 
parametrized by three complex numbers. Bacher's family is a subgroup of 
the group of automorphisms of the Riordan group and so is the subfamily 
parametrized only by two real numbers. Here, we study some of the algebraic 
properties of this subfamily and use the elements to point out isomorphisms 
between Riordan subgroups. In this context, we prove that the set of Riordan 
arrays whose row sum sequence is a sequence of partial sums, forms a Riordan 
subgroup.
Moreover, we show that the well-known recursive matrices may be constructed 
from sequences of images of a Riordan array under automorphisms. Our 
construction also discloses a correspondence between the recursive matrices and 
a pair of well-defined Riordan arrays.
\end{abstract}

\section{Introduction}
\label{sec:intro}
Let 
$f(\te)=\sum_{k=0}^\infty f_k\te^k$ 
be a formal power series whose coefficient 
$f_k=[\te^k] f(\te)$ 
is over a field, say, the field of complex numbers 
$\C$. 
Recall that 
$[\te^k]$ 
is the usual coefficient extraction operator. Often, 
$f(\te)$ 
is called the generating function of the sequence 
$(f_k)_{k\ge0}$. 
Let 
$\dd(\te)$ and $\h(\te)$ 
be two generating functions and consider the infinite matrix 
$\DD=(\DD_{k,n})_{k\geq n\geq1}$ 
whose generic entry is given by 
\begin{equation}
\label{eq:entries}
\DD_{k,n}=[\te^{k-1}]\dd(\te)\h(\te)^{n-1} .
\end{equation}
If 
$h_0=0$, 
then 
$\DD$ 
is an infinite lower triangular matrix, known as a 
{\em Riordan array},  
usually represented as the pair 
$\DD=(\dd(\te),\h(\te))$. 
If, in addition, 
$d_0\neq 0$ 
and 
$h_1\neq 0$, 
then 
$\DD$ 
is said to be a 
{\em proper} 
Riordan array. The matrix multiplication rule for Riordan arrays in terms of 
their generating functions is 
\begin{equation}
\label{eq:multR}
(\dd_{1}(\te),\h_{1}(\te)){}
(\dd_{2}(\te),\h_{2}(\te))=(\dd_{1}(\te)\dd_{2}(\h_{1}(\te)),\h_{2}(\h_{1}
(\te))) .
\end{equation}
Also, if 
$f(\te)$ 
and 
$g(\te)$ 
are the generating functions of the sequences 
$(f_k)_{k\ge0}$ 
and 
$(g_k)_{k\ge0}$, 
then the following identity is often referred to as the 
{\em Fundamental Theorem of Riordan Arrays} (FTRA)
\begin{equation}
\label{eq:FTRA}
g(\te)=(\dd(\te),\h(\te)){} f(\te)=\dd(\te)f(\h(\te)) .
\end{equation}
The first instances where Riordan arrays were considered in the literature are 
probably Jabotinsky's work on Bell-type arrays 
\cite{ja1,ja2,ja3}, 
Shapiro's Catalan triangle 
\cite{sha:catalan}, 
and Rogers' 
{\em renewal arrays} 
\cite{rogers}. The 
formal definition of (proper) Riordan 
arrays was introduced years later by Shapiro, Getu,  Woan, and Woodson 
\cite{ShGeWoWo:Rgroup}. 
The extra conditions, namely,
$d_0\neq 0$ 
and 
$h_1\neq 0$ 
allow for the set of proper Riordan arrays to form a group. Here, the Riordan 
group, denoted by 
$\Rio$, 
is the set 
$\Rio = 
\bigl\{\big(\dd(\te),\h(\te)\big)\,|\,\dd_{0}=1,\h_{0}=0,\h_{1}=1\bigr\}$,
where the group multiplication is given by formula 
\eqref{eq:multR}. 
The group identity is 
$\id {\Rio}=(1,\te)$
and the inverse of any element 
$\DD=(\dd(\te),\h(\te))$ 
is 
$\DD^{-1}=(1/\dd(\bar{\h}(\te)),\bar{\h}(\te))$, 
where 
$\bar{\h}(\te)$ 
denotes the compositional inverse of the generating function 
$\h(t)$, 
which satisfies 
$\h(\bar{\h}(t))=\bar{\h}(\h(t))=t$. 
Throughtout the paper, unless otherwise 
stated, by {\em Riordan array} we actually mean {\em Riordan group element}.

This paper aims at giving further insight into the Riordan group 
$\Rio$ 
through some of its automorphisms. A Riordan group  automorphism 
$\phi$ 
is a 1-1 mapping from $\Rio$ onto itself such that 
$\phi(\DD_1\DD_2)=\phi(\DD_1)\phi(\DD_2)$ 
for all
$\DD_1,\DD_2$ in $\Rio$.  
We denote the
set of Riordan group automorphisms by 
$\aut \Rio$. 
Here, we consider a family of Riordan group automorphisms, denoted by
$\prs$, 
which depend 
on two real numbers, 
$r$ 
and 
$s$ 
with 
$s\neq0$. 
The definition is 
$\prs(\dd(\te),\h(\te))=((\h(\te)/\te)^r\dd(\te)^s,\h(\te))$.
They are a special case of Bacher's automorphisms on three complex numbers 
introduced in the context of the Lie algebra structure of the Riordan group 
\cite{bacher}.

The paper is organized as follows. Let 
$\R$ 
denote the real numbers and 
$\R^*=\Rzero$.
Section 
\ref{sec:family} 
shows that the set 
$\family$, 
has a very rich algebraic structure.  We give the involutions, the 
commutator subgroup of 
$\family$,
and show that the set 
$\{\p{r}{\pm1}\}_{r\in\R}$ 
is a normal subgroup. Moreover, let 
$\aaut\Rio$ 
denote the group of automorphisms and anti-automorphisms of  
the Riordan group
$\Rio$. 
Let 
$\ars$
denote the anti-automorphism induced by 
$\prs$.
We  give the multiplication and inverse rules for 
the subgroup of 
$\aaut\Rio$
consisting of 
$\afamily$.
We end up this section by posing two questions on the general characterization 
of the automorphisms of 
$\Rio$.

Section 
\ref{sec:subgroups} 
determines the images of most of the well-known 
Riordan subgroups under the automorphisms 
$\prs$. 
This points out an infinity 
of isomorphisms between Riordan subgroups. We use the row sum 
generating function 
\cite{bijections,HeSha:sum} 
to show that the image of the 
stochastic subgroup under the automorphism 
$\p 1 1$ 
is the subgroup of Riordan arrays whose row sum sequence is the partial sum of 
the coefficients of a generating function 
$\h(\te)$. 
We also give a generic closed form for the row-sum sequence of the arrays in 
the Riordan subgroup obtained by applying 
$\p 1 n$ 
to the stochastic subgroup, 
for any
$n\geq1$.

Furthermore, Section 
\ref{sec:minusInfinite} 
shows that for a given a Riordan 
array
$\DD$ 
({\em base array}), 
for very large  
$m$, 
the Riordan array 
$\p{-m}1(\DD)$ 
is an approximation to a bi-infinite triangle which coincides with the 
recursive matrix of 
Luz\'on, Merlini, Mor\'on, and Sprugnoli 
\cite{complementary} 
associated with 
$\DD$. 
The fact that the automorphism setting does not involve Laurent series is an 
advantage over the recursive matrices. Moreover, we show that both the east and 
west side of the bi-infinite triangles can be described by well-defined Riordan 
arrays. 
In this context, a bi-infinite triangle turns out to be equivalent to a pair of 
Riordan arrays. Finally, we observe that the
$[-m]$-complementary of a Riordan array 
$\DD$ 
introduced 
in the former paper 
(see also \cite{sprugnoli,CheonJin,LMMP-MS}),
is the image of the inverse matrix 
$\DD^{-1}$ 
under a Bacher automorphism with appropriate parameters.

Throughtout the paper, the integer sequences which are well-known are referred 
to by their number from the 
On-Line Encyclopedia of Integer Sequences 
\cite{oeis}.
\section{A family of automorphisms of $\Rio$}
\label{sec:family}
We study some of the algebraic structure of an infinite subgroup of 
$\aut \Rio$ 
parametrized by two real numbers. We obtain the involutions, the 
commutator subgroup, and the normal subgroups of the considered subgroup 
of
$\aut \Rio$. 
We also raise 
the question of the general characterization of the automorphisms of the 
Riordan group.

We consider the following transformation:
\begin{eqnarray*}
\nonumber
\Phi :\,\R\times\RzeroP &amp;\to    &amp;\aut\Rio\\
\label{eq:Phi}
(r,s)                   &amp;\mapsto&amp;\prs ,
\end{eqnarray*}
where for any 
$r\in\R$ 
and 
$s\in \R^*$
the mappings 
$\prs$ 
are defined by:
\begin{eqnarray}
\nonumber
\prs:\Rio         &amp;\to    &amp;\Rio\\
\label{eq:prs}
(\dd(\te),\h(\te))&amp;\mapsto&amp;
\Biggl(
\biggl(\frac{\h(\te)}{\te}\biggr)^r\dd(\te)^s,\h(\te)
\Biggr) .
\end{eqnarray}
Note that, 
$\p{0}{1}=\id{\aut\Rio}$. 
It is easy to verify that the mappings 
$\prs$ 
are automorphisms of the Riordan group.
Moreover, 
$\prs=\varphi_{s,r,0}$, 
where for any  
$\lambda,\mu\in\C$ 
and 
$\kappa\in \CzeroP$, 
with 
$\CzeroP=\Czero$,  
the automorphisms 
$\varphi_{\lambda,\kappa,\mu}$ 
are defined by Bacher 
\cite{bacher} 
as follows:
\begin{eqnarray}
\nonumber
\varphi_{\lambda,\kappa,\mu}:\Rio&amp;\to    &amp;\Rio\\
\label{eq:bacher}
\DD=(\dd(\te),\h(\te))           &amp;\mapsto&amp;
\left(
\biggl(\frac{\h(\te)}{\te}\biggr)^\lambda
\dd(\te)^\kappa\h'(\te)^\mu,\h(\te)
\right) ,
\end{eqnarray}
where 
$\h'(\te)$ 
denotes the derivative of 
$\h(\te)$. 
Note that here the order of the parameters  
$\lambda$ 
and 
$\kappa$  
is reversed in 
$\varphi_{\lambda,\kappa,\mu}$ 
in comparison with 
\cite{bacher}.

The automorphism 
$\p 1 1$ 
(as well as its iterations) has recently been studied in the 
literature. Actually,  
$\p1 1$  
is the diagonal translation operator used by 
Luz\'on, Merlini, Mor\'on, and Sprugnoli 
\cite{complementary} 
(see also \cite{bacher}). 
The automorphism 
$\p{1}{1}$  
is also the mapping 
$T:\Rio\to\Rio$ 
of He 
\cite{he:char}, 
defined by 
$T(\DD)=U\DD U^T$, 
where 
$(U_{i,j})_{i,j\ge1}=(\delta_{i+1,j})_{i,j\geq1}$. 

For any real numbers 
$r$ 
and 
$s\neq0$,
the array
$\prs(\dd(\te),\h(\te))$ 
has the following semidirect product factorization:
\begin{equation}
\label{eq:factorization}
\prs(\dd(\te),\h(\te))=
\bigl(
\dd(\te)^s,\te
\bigr)
\Biggl(
\biggl(\frac{\h(\te)}{\te}\biggr)^r,\h(\te)
\Biggr) .
\end{equation}
Clearly, the  set 
$\Phi(\R\times\RzeroP)=\family$
forms a group under composition.
\begin{proposition}
$$
\Phi(\R\times\RzeroP) \leq \Aut(\Rio) .
$$
\end{proposition}
\begin{proof}
Any mapping 
$\prs$ 
is invertible for all 
$r\in \R$ 
and 
$s\in \RzeroP$, 
the inverse being given by
\begin{equation}
\label{eq:prs inv}
\prs^{-1}=\p {-r/s} {1/s} .
\end{equation}
The set is  clearly closed under composition:
\begin{equation}
\label{eq:prs circ}
\p{r'}{s'}\circ\prs=\p{r'+rs'}{ss'} .
\end{equation}
\end{proof}
Now, let 
$\Z$
and 
$\Q$ 
denote the integer and rational numbers, respectively. Let 
$\Z^*=\Zzero$ 
and 
$\Q^*=\Qzero$. 
Observe that whereas 
$\Phi(\Q\times\Q^*)$ 
remains a subgroup of 
$\aut\Rio$, 
the same does not hold for 
$\Phi(\Z\times\Z^*)$, 
except for 
$\Phi(\Z\times\{\pm1\})$. 
Next, we explore some of the algebraic properties of the sets 
$\Phi(\Z\times\{\pm1\})$.
\begin{proposition}  
Any element in 
$\Phi(\R\times\{-1\})$
is an involution.
\end{proposition}
\begin{proof} 
By equation 
\eqref{eq:prs circ}, 
we have
$
\p r {-1}^2 = 
\p r {-1}\circ\p r {-1}=\p 0 1
$.  
\end{proof}
Moreover, it follows from equation  
\eqref{eq:prs circ} 
that, for all 
integers 
$n\ge0$, 
the 
$n$th 
iterate of 
$\prs$ 
is 
\begin{equation*}
\label{eq:prs iterate}
\prs^{n}=\p{r(1+s+s^2+\cdots+s^{n-1})}{s^n} .
\end{equation*}
Hence, there are no elements in 
$\Phi(\R\times\R^*)$ 
of order higher than 2, which implies that the set 
$\Phi(\R\times\{-1\})$ 
contains all the involutions of 
$\Phi(\R\times\R^*)$.
\begin{proposition}
\label{prop:commutator} 
The set 
$\Phi(\R\times\{1\})$ 
is the commutator subgroup of 
$\Phi(\R\times\R^*)$.
\end{proposition}
\begin{proof}
It is clear that 
$\Phi(\R\times\{1\})$
is a subgroup of 
$\Phi(\R\times\R^*)$. 
Also, by equations 
\eqref{eq:prs inv} 
and 
\eqref{eq:prs circ}, 
we have
\begin{eqnarray*}
\nonumber
[\p{r'}{s'},\prs]
&amp;=&amp;
\p{r'}{s'}^{-1}\circ\prs^{-1}\circ\p{r'}{s'}\circ\prs\\
\nonumber
&amp;=&amp;
\p{-r'/s'}{1/s'}\circ\p{-r/s}{1/s}\circ\p{r'}{ s'}\circ\prs\\
\nonumber
&amp;=&amp;
\p{-r'/{s'}-r/{ss'}}{1/{ss'}}\circ\p{r'+rs'}{ss'}\\
\label{eq:commutator}
&amp;=&amp;
\p{(r'-r+rs'-r's)/{ss'}}{1} .
\end{eqnarray*}
Therefore, the commutator of any two elements of the group 
$\Phi(\R\times\R^*)$ 
is in 
$\Phi(\R\times\{1\})$
and so is any finite product of commutators of 
$\Phi(\R\times\R^*)$.
\end{proof}
\begin{proposition}
The set 
$\Phi(\R\times\{-1,1\})$ 
is a normal subgroup of 
$\Phi(\R\times\R^*)$.
\end{proposition}
\begin{proof}
It is straightforward to check that 
$\Phi(\R\times\{-1,1\})$ 
is a subgroup of 
$\Phi(\R\times\R^*)$. 
The normality property follows from Proposition 
\ref{prop:commutator} 
and from equations 
\eqref{eq:prs inv} 
and 
\eqref{eq:prs circ}:
$$
\prs\circ\p{r'}{-1}\circ\prs^{-1}
=
\p{r+r's}{-s}\circ\p{-r/{s}}{1/{s}}
=
\p{2r+r's}{-1} ,
$$
$$
\prs\circ\p{r'}{1}\circ\prs^{-1}
=
\p{r+r's}{s}\circ\p{-r/{s}}{1/{s}} 
=
\p{r's}{1} .
$$
\end{proof}

We define as well a family of anti-automorphisms of 
$\Rio$ 
induced by 
$\Phi$ as follows:
\begin{eqnarray*}
\nonumber
\Psi :\,\R\times\RzeroP &amp;\to    &amp;\aaut \Rio\\
\label{eq:Psi}
(r,s)                   &amp;\mapsto&amp;\ars ,
\end{eqnarray*}
where for any 
$r\in\R$ 
and 
$s\in \RzeroP$,
the mapping 
$\ars$ 
is defined by:
\begin{eqnarray*}
\nonumber
\ars:\Rio&amp;\to    &amp;\Rio\\
\label{eq:psi general}
\DD      &amp;\mapsto&amp;\prs(\DD^{-1})
=\Biggl(
\biggl(\frac{\bar{\h}(\te)}{\te}\biggr)^r
\frac{1}{\dd(\bar{\h}(\te))^s},\bar{\h}(\te)
\Biggr) .
\end{eqnarray*}
It is easy to see that 
$\Phi(\R\times\RzeroP)\cup \Psi(\R\times\RzeroP)\leq\aaut\Rio$.
Moreover, the automorphisms 
$\prs$ 
commute with the mapping that sends a 
Riordan group element to its inverse.
\begin{lemma}
Let
$\Inv:\Rio\to\Rio ;\DD\mapsto \DD^{-1}$
be the mapping that sends a Riordan group element 
$\DD$ 
to its inverse 
$\DD^{-1}$. 
Then, we have
$$
\prs\circ\Inv=\Inv\circ\,\prs .
$$
\end{lemma}
The following four equations are immediately derived from the lemma above:
\begin{eqnarray*}
\label{eq:psi inv}
\ars^{-1}            &amp;=&amp;\at{-r/s}{1/s} ,\\
\label{eq:psi circ}
\at {r'}{s'}\circ\ars&amp;=&amp;\p{r'+rs'}{ss'} ,\\
\label{eq:phi circ psi}
\p{r'}{s'}\circ\ars  &amp;=&amp;\at{r'+rs'}{ss'} ,\\
\label{eq:psi circ phi}
\ars\circ\p{r'}{s'}  &amp;=&amp;\at{r+r's}{ss'} .
\end{eqnarray*}

Now, let
$\DD_1=(\dd_1(\te),\h_1(\te))$ 
and 
$\DD_2=(\dd_2(\te),\h_2(\te))$ 
denote two Riordan matrices. One can partition the set of Riordan arrays into 
equivalence classes by setting  
$\DD_1\sim\DD_2$ 
if and only if 
$\h_1(\te)=\h_2(\te)$. 
Clearly, the family 
$\family$ 
(as well as Bacher's larger family) 
preserves these equivalence 
classes, something which does not happen, for instance, with 
the inner automorphisms. Recall that for the Riordan group, the 
inner automorphism
$\phi_{\mathfrak{I}}$ 
associated with the element 
$\mathfrak{I}=(g(\te),f(\te))$
is
\begin{eqnarray*}
\nonumber
\phi_{\mathfrak{I}}
&amp;=&amp;
(g(\te),f(\te)){}(\dd(\te),\h(\te)){}(g(\te),f(\te))^{-1}\\
\label{eq:inner}
&amp;=&amp;
\left(
g(\te)\dd(f(\te))\frac{1}{\g(\bar{f}(\h(f(\te))))},\bar{f}(\h(f(\te)))
\right) .
\end{eqnarray*} 
 We did not find any automorphism, other than the inner automorphisms 
(or a composition with them), that modifies the second element of the pair 
$(\dd(\te),\h(\te))$. 
Thus, we pose the following two questions:
\begin{question}
\label{quiz:h}
Is  there any automorphism in 
$\Aut(\Rio)$ 
up to composition with inner automorphisms, that sends a Riordan array 
$(\dd_1(\te), \h_1(\te))$ 
to 
$(\dd_2(\te), \h_2(\te))$ 
with
$\h_1(\te)\neq\h_2(\te)$?
\end{question}
Due to Bacher's definition of the Riordan group as the semidirect product of 
the group of formal power series with constant term 1 by the group of formal 
power series under substitution 
\cite{bacher}. 
We believe that the results of 
Muckenhoupt 
\cite{benjamin} 
and Johnson 
\cite{johnson} 
are likely to be 
relevant to answer the question above in the negative. Moreover, a more general 
question in this context is the following:
\begin{question}
\label{quiz:char}
Which is the general characterization of the group of automorphisms of the 
Riordan group? That is, in terms of the generating functions 
$\dd(\te)$ 
and 
$\h(\te)$ 
such that 
$\dd_{0}=1,\h_{0}=0$, 
and 
$\h_{1}=1$, 
which properties must a mapping 
$(\dd_1(\te),\h_1(\te))\mapsto(\dd_2(\te), \h_2(\te))$ 
satisfy in order to be a Riordan group automorphism?
\end{question}
\section{Isomorphisms between Riordan subgroups via $\prs$}
\label{sec:subgroups}
Since the mappings 
$\prs$ 
are automorphisms of the Riordan group 
$\Rio$, 
for any subgroup 
$\sub\leq\Rio$, 
the sets 
$\prs(\sub)$ 
are also subgroups of 
$\Rio$. 
In this section, we characterize the images under 
$\prs$ 
of several well-known subgroups of the Riordan group. For brevity, we shall 
write  
$\sub_{r,s}=\prs(\sub)$. 
Observe that the groups 
$\sub_{r,s}$ 
and 
$\sub_{r',s'}$ 
are isomorphic via 
$\p{r'-rs'/{s}}{s'/{s}}$ 
for all pairs 
$(r,s)$ 
and 
$(r',s')$. 
\subsection{
The $c$-Appell subgroup 
$
\Big\{
\big(\dd(\te),c\te\big)
\,\big|\,c\neq0
\Big\}
$
}
\begin{proposition}
Let 
$c\text{-}\Appell_{r,s}=\Big\{(c^r\dd^s(\te),c\te)\,\big|\,c\neq0\Big\}$. 
Then, 
for all 
$r\in\R$ 
and 
$s\in\RzeroP$:
$$
c\text{-}\Appell_{r,s}\leq\Rio .
$$
\end{proposition}
The usual Appell subgroup 
$\Appell$ 
is obtained for 
$c=1$
and 
$s=1$; 
that is 
$1\text{-}\Appell_{r,s}=\Appell_{s}$.
\subsection{
The Lagrange subgroup 
$
\Big\{
\big(1,\h(\te)\big)
\Big\}
$
}
\begin{proposition}
Let 
$
\Lagrange_{r}=\Big\{\big(\bigl(\h(\te)/\te
\big)^r ,
\h(\te)\bigr)\Big\}
$. 
Then, for all 
$r\in\R$:
$$
\Lagrange_{r}\leq\Rio .
$$
\end{proposition} 
Note that when 
$r\neq0$ 
the group 
$\Lagrange_{1/r}$ 
coincides with 
the $r$-power-Bell subgroup of Jean-Louis and Nkwanta 
\cite{jean-louis}, 
whose arrays are of the form 
$((\h(\te)/\te)^{1/r},\h(\te))$. 
When 
$r=1$ 
we recover the well-known isomorphism between the Lagrange and the 
Bell subgroups given  by 
Jean-Louis and Nkwanta 
\cite{jean-louis} 
and by He 
\cite{he:char}. 
Furthermore, observe that by equation 
\eqref{eq:factorization}, 
the group 
$\familyRio$ 
can be written as the 
semidirect product of the groups 
$\Appell_{s}$ 
and 
$\Lagrange_r$.
\subsection{The $c$-Bell subgroup 
$
\Big\{
\big(\h(\te)/\te,c\,\h(\te)\big)
\,\big|\,c\neq0
\Big\}
$
}
\label{sec:bell}
\begin{proposition}
Let 
$
c\text{-}\Bell_{r,s}=
\Big\{
\Bigl(c^r\bigl(\h(\te)/\te\bigr)^{r+s},c\,\h(\te)\Bigr)\,\big|\,c\neq0
\Big\}
$.
Then, for all 
$r\in\R$ 
and 
$s\in\RzeroP$:
$$
c\text{-}\Bell_{r,s}\leq\Rio .
$$
\end{proposition}
Here, the  Bell subgroup for 
$c=1$ 
is simply denoted by 
$\Bell$. 
We have 
$\Bell_{r,s}=\Lagrange_{r+s}$. 
\begin{example}
\label{ex:catalan}
\normalfont
Recall that 
$c(\te)=(1-\sqrt{1-4\te})/2\te$ 
is the generating function of the Catalan numbers 
$C_n=1/(n+1){2n\choose n}$, 
$n\geq0$ 
\seqnum{A000108}. 
The function 
$c(\te)$ 
satisfies the functional equation 
$c(\te)=1+\te c(\te)^2$. 
There are several well-known Bell matrices in the literature generated by 
$c(\te)$ 
and accordingly referred to as Catalan triangles, namely,
\begin{itemize}
\item 
Aigner's array 
$C=(c(\te),\te c(\te))$ 
of the ballot numbers
\cite{aigner:CatLike2,aigner};
\item 
Shapiro's Catalan triangle 
$B=(c(\te)^2,\te c(\te)^2)$ 
\cite{sha:catalan};
\item 
Radoux's triangle of numbers 
$R=(c(\te),\te c(\te)^2)$ 
\cite{radoux}.
\end{itemize}
It is easy to see that  
$\prs(B)=R$ 
when 
$r+s=1/2$.
Also, 
$\prs(R)=B$ 
when 
$2r+s=2$. 
We thus have a system of two linear equations with two unknowns 
$r$ 
and 
$s$. 
The solution gives  the involution 
$\p{3/2}{-1}$ 
such that 
$\p{3/2}{-1}(B)=R$ 
and 
$\p{3/2}{-1}(R)=B$. 

Moreover, recall that Shapiro's triangle satisfies the following factorization 
property 
$B=CP$, 
where 
$P$ 
is the classical Pascal triangle 
(see, e.g., \cite{agapito2}). 
Obviously, 
$\prs(B)=\prs(C)\prs(P)$. 
A similar 
factorization holds for Radoux's triangle. Indeed,  
for every pair 
$(r,s)$ 
such that 
$r+s=1/2$ 
it follows that 
$R=\prs(C)\prs(P)$. 
For the involution 
$\p{3/2}{-1}$, 
we have 
$R=(c(\te)^{1/2},\te c(\te))(p(\te)^{1/2},\te p(\te))$ 
and 
$\prs(R)=\p{r+3s/2}{-s}(C)\p{r+3s/2}{-1}(P)$.

Furthermore, the authors, Petrullo, and Torres 
\cite{AMPT3} 
showed that the matrices above are particular cases of a wide family of Catalan 
triangles given by 
\begin{equation*}
C^{a,b}(q,z)=\big(c(q,z;\te)^a,\,\te\,c(q,z;\te)^b\big),
\quad 
a,b\geq1 ,
\end{equation*}
where
\begin{equation*}
\label{def:c(q,t;z)}
c(q,z;t)=\frac{1-zt-\sqrt{(1-zt)^2-4qt}}{2qt}
\quad
\text{with}
\quad
q,z\geq0 
.
\end{equation*}
The family 
$C^{a,b}(q,z)$ 
includes:
\begin{itemize}
\item 
He's parametric Catalan triangles 
$C^{1,1}(q,z-q)$ 
\cite{he};
\item 
Yang's generalized Catalan triangles  
$C^{a,b}(q,0)$ 
\cite{yang}.
\end{itemize}
It is easy to check that 
$\prs(C^{a,b}(q,z))=C^{br+as,b}(q,z)$. 
Also, note that the action of the automorphisms 
$\prs$ 
on the generalized Catalan triangles of He yields  
$\prs(C^{1,1}(q,z-q))=C^{r+s,1}(q,z-q)$. 
Hence, the image under 
$\prs$ 
of any He's parametric Catalan triangle is not a He's parametric Catalan 
triangle unless 
$r+s=1$. 
\end{example}
\subsection{The checkerboard subgroup 
$
\Big\{
\big(\dd(\te),\h(\te)\big) 
\,\big|\, 
\dd(\te)=\dd(-\te)\,, 
\h(\te)=-\h(-\te)
\Big\}
$
}
\begin{proposition}
Let 
$
\Checkboard_{r,s}=
\Big\{
\left(
\bigl(\h(\te)/\te\bigr)^r\dd(\te)^s,\h(\te)
\right)\,\big|
\dd(\te)=\dd(-\te)\,, 
\h(\te)=-\h(-\te)
\Big\}
$. 
Then, for all 
$r\in\R$ 
and 
$s\in\RzeroP$:
$$
\Checkboard_{r,s}\leq\Rio .
$$
\end{proposition}
Jean-Louis and Nkwanta 
\cite{jean-louis} 
proved that the checkerboard subgroup is the centralizer of the element 
$(1,-\te)$. 
We note that the checkerboard subgroup 
$\Checkboard$ 
is also the centralizer of 
$\prs(1,-\te)$ 
for every 
pair 
$(r,s)$.
\subsection{
The derivative subgroup 
$
\Big\{
\big(\h'(\te),\h(\te)\big)
\Big\}
$
}
\begin{proposition}
Let
$
\Derivative_{r,s}=
\Big\{
\left(\bigl(\h(\te)/\te\bigr)^r\h'(\te)^s,\h(\te)\right)
\Big\}
$. 
Then, for
all
$r\in\R$ 
and 
$s\in\RzeroP$:
$$
\Derivative_{r,s}\leq\Rio .
$$
\end{proposition}
The
derivative subgroup is isomorphic to the Lagrange subgroup via 
Bacher's automorphism 
$\varphi_{0,1,-1}: (\h'(t),\h(\te))\mapsto (1,\h(\te))$. 
The analogous result for the subgroups 
$\Derivative_{r,s}$ 
and 
$\Lagrange_{r}$ 
follows via 
$
\varphi_{0,1,-s}: 
\left(
\bigl(\h(\te)/\te\bigr)^r\h'(\te)^s,\h(\te)
\right)
\mapsto 
\left(
\bigl(\h(\te)/\te\bigr)^r,\h(\te)
\right)
$. 
Note that the subgroup 
$\Derivative_{r,s}$
was previously introduced in 
\cite{complementary}.
\subsection{
The hitting time subgroup 
$
\Big\{
\big(\te\,\h'(\te)/\h(\te),\h(\te)\big)
\Big\}
$
}
\begin{proposition}
Let 
$
\Hittingtime_{r,s}=
\Big\{
\left(
\bigl(\h(\te)/\te\bigr)^{r-s}\h'(\te)^s,\h(\te)
\right)
\Big\}
$. 
Then, for 
all 
$r\in\R$ 
and 
$s\in\RzeroP$:
$$
\Hittingtime_{r,s}\leq\Rio .
$$
\end{proposition}
The traditional hitting time subgroup corresponds to the case 
$r=-1$ 
and 
$s=1$. 
As a  direct consequence of the above definition, we have 
$\Hittingtime_{r,s}=\Derivative_{r-s,s}$. 
Also,  the groups 
$\Hittingtime_{r,s}$ 
and 
$\Lagrange_{r-s}$ 
are isomorphic via Bacher's automorphism
$\varphi_{1,1,-s}$.
\subsection{
The Jean-Louis--Nkwanta subgroup 
$
\Big\{
\big(\f(\te)/\f(\h(\te)),\h(\te)\big)
\Big\}
$
}
Recall that the stochastic Riordan subgroup is the stabilizer of the 
column vector 
$(1,1,1,\ldots)$. 
More generally, let 
$f(t)=f_0+f_1t+f_2t^2+\cdots$ 
be the generating function of the sequence 
$(f_0,f_1,f_2,\ldots)$. 
By using the fundamental theorem of Riordan arrays, Jean-Louis and Nkwanta 
\cite{jean-louis} 
showed that the set of Riordan matrices of the form 
$(\f(\te)/\f(\h(\te)),\h(\te))$ 
is the stabilizer of the column vector 
$(f_0,f_1,f_2,\ldots)$. 
In particular, the elements of the stochastic Riordan subgroup are of 
the form 
$\big((1-\h(\te))/(1-t),\h(\te)\big)$. 
\begin{proposition}
Let 
$
\Jean_{r,s}=
\Big\{
\left(
\bigl(\h(\te)/\te\bigr)^{r}\bigl(f(\te)/f(\h(\te))\bigr)^s,\h(\te)
\right)
\Big\}
$. 
Then, for all 
$r\in\R$ 
and 
$s\in\RzeroP$:
$$
\Jean_{r,s}\leq\Rio .
$$
\end{proposition} 
\subsubsection{The stochastic subgroup}
\normalfont
Let  
$\SSs{h} = \p 1 1 \big((1-\h(\te))/({1-\te}),\h(\te)\big)$. 
Note that the row sum sequence of 
$\SSs{h}$, 
here denoted 
$\rowsum {} {\SSs h}$, 
is the partial sum of the coefficients of the generating function 
$\h(\te)$. 
That is, 
$\rowsum{n}{\SSs h}=\sum_{i=1}^n \h_i$. 
This follows from observing that the 
generating function for the row sum of a Riordan array is given by 
$\dd(\te)/(1-\h(\te))$ 
\cite{bijections,HeSha:sum}. 
Therefore, the Riordan subgroup
$\Stoch_{1,1}=\{\SSs h\}$
is the set of Riordan arrays whose row sum sequence is the 
partial sum of the coefficients of the generating function 
$\h(\te)$.
\begin{example}
\label{ex:stoch1}
\normalfont
Let 
$h(t)/t=p(t),c(t),c^2(t)$, 
where 
$p(t)=1/(1-\te)$ 
and 
$c(t)$ 
the generating function of the Catalan numbers. The elements of the group 
$\Stoch_{1,1}$ 
related to these functions are:
$$
\SSs {\te p}
=
\Big(\frac{p(t)-tp^2(t)}{1-t},tp(t)\Big)
=
\left[ 
\begin{array}{rrrrrrr}
1     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
0     &amp; 2    &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
-2    &amp; 2    &amp; 3    &amp; 1    &amp; 0    &amp; 0    &amp;\cdots\\ 
-5    &amp; 0    &amp; 5    &amp; 4    &amp; 1    &amp; 0    &amp;\cdots\\ 
-9    &amp; -5   &amp; 5    &amp; 9    &amp;  5   &amp; 1    &amp;\cdots\\ 
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots
\end{array}
\right] ,
$$
$$
\SSs {\te c}
=
\Big(\frac{c(t)-tc^2(t)}{1-t},tc(t)\Big)
=
\left[ 
\begin{array}{lllllll}
1     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 2    &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 4    &amp; 3    &amp; 1    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 9    &amp; 8    &amp; 4    &amp; 1    &amp; 0    &amp;\cdots\\ 
1     &amp; 23   &amp; 22   &amp; 13   &amp; 5    &amp; 1    &amp;\cdots\\
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots
\end{array}
\right] ,
$$
$$
\SSs {\te {c^2}}
=
\Big(\frac{c^2(t)-tc^4(t)}{1-t},tc^2(t)\Big)
=
\left[
\begin{array}{rrrrrrr}
1     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
2     &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
3     &amp; 4    &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
3     &amp; 12   &amp; 6    &amp; 1    &amp; 0    &amp; 0    &amp;\cdots\\ 
-3    &amp; 33   &amp; 25   &amp; 8    &amp; 1    &amp; 0    &amp;\cdots\\ 
-36   &amp; 88   &amp; 91   &amp; 42   &amp; 10   &amp; 1    &amp;\cdots\\ 
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots 
\end{array}
\right] .
$$
Their row sum sequence numbers are  
\seqnum{A000027}, 
\seqnum{A014137}, 
and
\seqnum{A014138}. 
Obviously, we have 
$\SSs {\te c}\SSs{\te p}=\SSs {\te c^2}$.
\end{example}
The above result fixes 
$s=1$. 
Restricting 
$s$ 
to integers greater than 1 gives the next result, which is a closed form for 
the row-sum sequences of the the arrays in the image of the Stochastic subgroup 
under the mapping 
$\p 1 n$.
\begin{proposition}
\label{prop:stoch gen} 
Let 
$
\SSsm{h}{n} = \p 1 n 
\big(
(1-\h(\te))/({1-\te}),\h(\te)
\big)
$, 
where 
$h(t)=h_1t+h_2t^2+h_3t^3+\cdots$ 
with 
$h_1\neq 0$ 
is any power series over the complex numbers. Then, for all 
$n\geq1$
the Riordan subgroup 
$\Stoch_{1,n}=\{\SSsm {h}{n}\}$
is the set of Riordan arrays whose row sum sequence is given by:
\begin{eqnarray*}
\rowsum {} {\SSsm{h}{n}} 
&amp;=&amp;
\sum_{j=0}^{\infty}
\sum_{i=0}^j
\sum_{k=0}^{n-1}
\sum_{i_1+i_2+\cdots+i_{k+1}=i}
h_{i_1+1}\ldots h_{i_{k+1}+1}
\times\\
&amp; &amp;
(-1)^{k}{n+j-i-1\choose n-1} 
{n-1\choose k}
t^{j+k} ,
\end{eqnarray*}
where the third sum runs over all sequences of non-negative 
integers 
$(i_1,\ldots,i_{k+1})$ 
satisfying  
$i_1+i_2+\cdots+i_{k+1}=i$. 
\end{proposition}
\begin{proof}
The proof is a straightforward computation of the product of the power series 
$h(1-h(t))^{n-1}/t$ 
and 
$1/(1-t)^n=\sum_{i=0}^\infty{n+i-1\choose n-1}t^i$.
\end{proof}
\begin{example}
\label{ex:stoch2}
\normalfont
We consider again 
$h(t)/t=p(t),c(t),c^2(t)$. 
The row sum sequences associated 
with  the   elements 
$\SSsm{\te p}{2}$, 
$\SSsm {\te c}{2}$,  
and 
$\SSsm {\te{c^2}}{2}$ 
of the group 
$\Stoch_{1,2}$ 
are:
\begin{eqnarray*}
\big(
\rowsum{}{\SSsm{tp}{2}}
\big)
&amp;=&amp; 
1, 2, 2, 0, -5, -14, -28, -48, -75, -110,-154, -208, -273,-350, -440,  \ldots\\
&amp; &amp; 
(\seqnum{A005586} \text{ from the fourth term on})  
,\\
\big(
\rowsum{}{\SSsm{tc}{2}}
\big)
&amp;=&amp;
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,15, 16, 17, 18, 19, 20, 21,\ldots 
(\seqnum{A000027}) 
,\\
\big(
\rowsum{}{\SSsm{tc^2}{2}}
\big)
&amp;=&amp;
1, 3, 6, 9, 6, -30, -209, -960, -3921, -15280, -58293,-220170, -827787, \ldots .
\end{eqnarray*}
\end{example}
\section{Bi-infinite triangles defined via automorphisms}
\label{sec:minusInfinite}
This section uses automorphisms to introduce bi-infinite triangles. 
These bi-infinite triangles coincide
with the recursive matrices of Luz\'on, Merlini, Mor\'on, and Sprugnoli 
\cite{complementary}. 
Recall that a recursive matrix 
$(d_{k',n'})_{k'\geq n'\in\mathbb{Z}}=\chi(d(t),h(t))$ 
is defined by
\begin{equation*}
d_{k',n'}=
[t^{k'}]d(t)h(t)^{n'}\quad \forall 
k',n'\in\Z, \quad k'\geq n' ,
\end{equation*}
where 
$d(t)$ 
and 
$h(t)$ 
are formal power series as in the definition of Riordan arrays. Also, 
$d(t)h(t)^{n'}=\sum_{i={n'}}^\infty l_it^i$ 
is a formal Laurent series 
(which yields a formal power series if $n'\geq0$).

Let $m$ be a non-negative integer. Let $\DD$ be any Riordan array.
We examine the matrix 
$\p {-m} 1 (\DD)$ 
for large 
$m$.   
Though, for simplicity, we fix the second parameter of the automorphism equal 
to 1, our results  generalize straightforwardly for 
$\p {-m} s(\DD)$, 
with 
$s\in \RzeroP$, 
by replacing 
$d(t)$ 
by 
$d(t)^s$.

For any  non-negative integer 
$m$, 
the submatrix of  
$\p {-m} 1(\DD)$ 
that results from deleting the first 
$m$ 
columns and the first 
$m$ 
rows is the base Riordan array 
$\DD$. 
Moreover, any array 
$\p{-m}1(\DD)$ 
may be obtained from 
$\p {-(m+1)}1(\DD)$ 
by removing the first row and the first column. That is to say that the array 
sequence 
$(\p{-m}1(\DD))_{m\geq1}$ 
is an ascending chain of matrices, in the north west direction. Thus, every 
array in the sequence
$(\p{-m}1(\DD))_{m\geq1}$
is a submatrix of a bi-infinite triangle, denoted by
$\p{-\infty} 1(\DD)$, 
constructed as follows.  
Take the matrix 
$\DD$
and place on the right of 
$\DD$,
the first column of each of the matrices 
$\p{-m}1(\DD)$ 
(from right to left). 
Then,  add infinite rows of zeros on the top of 
each column 
(including the columns of $\DD$) 
to yield bi-infinite columns of a 
bi-infinite triangle. Then, for any pair of integers 
$(k',n')$
the entries of 
$\p{-\infty} 1(\DD)$
are given by
\begin{displaymath}
\big(\p{-\infty} 1(\DD)\big)_{k'\geq n'} = 
\begin{cases}
	\DD_{k',n'},                             &amp; \text{if $n'\geq1$;} \\
	\big(\p{n'-1}{1}(\DD)\big)_{k'-n'+1,1}, &amp; \text{otherwise,}
	\end{cases}
\end{displaymath}
where
$
\big(\p{n'-1}{1}(\DD)\big)_{k'-n'+1,1}=
\coef{t^{k'-1}}d(t)h(t)^{n'-1}
$.
Therefore, our construction of the bi-infinite triangle 
$\p{-\infty}1(\DD)$ 
leads to the definition of the recursive array 
$\chi(d(t),h(t))$ 
of Luz\'on, Merlini, Mor\'on, and Sprugnoli 
\cite{complementary},
up to a column phase shift of one unit. Observe that the phase shift is due to 
the fact that in Equation 
\eqref{eq:entries} 
the rows and columns, indexed by 
$k$ 
and 
$n$, 
respectively, are numbered from 1 on,
while in 
\cite{complementary} 
they go from 0 on.

The array 
$\p{-\infty}1(\DD)$
can be visualized as follows:
\begin{center}
$
\p {-\infty} {1}(\DD)
=
\left[
\begin{array}{llllll}
\mathfrak{L}_\DD&amp; \mathfrak{R}_\DD\\
\end{array}
\right] ,
$
\end{center}
where 
$
\mathfrak{R}_\DD
=
\left[
\begin{array}{llllll}
&amp; \DD
\end{array}
\right]^T$
and the matrix 
$\mathfrak{L}_\DD$ 
is a lower-triangular array which grows to the west region of 
$\p{-\infty}1(\DD)$, 
while the matrix 
$\DD$ 
grows to the south east region.

Although the bi-infinite triangle 
$\p{-\infty}1(\DD)$ 
is not a Riordan array, this can be approximated by a Riordan array
$\p{-m} 1 (\DD)$ 
when 
$m$ 
is large. For practical calculations, one can thus work with a Riordan 
array 
$\p{-m} 1(\DD)$
after choosing 
$m$ 
large enough to fit the purpose. Indeed, the matching between the columns of 
both bi-infinite triangles can also be seen from the large 
$m$ 
approximation of 
$\p{-\infty}1(\DD)$. 
We have
$$
{\p {-m} {1}(\DD)}_{k,n}=[\te^{k-m-1}]d(t)\h(\te)^{n-m-1},
\quad 
\forall 
k\geq n \geq1 .
$$ 
Let 
$k'=k-m-1$ 
and 
$n'=n-m-1$. 
Then, when  
$m$ 
is large, say 
$m$
approaches infinity, both 
$k'$ 
and 
$n'$ 
run through all integers in 
$\mathbb{Z}$. 
Hence, the array 
$\p {-\infty} 1(\DD)$
has the same columns as the recursive matrix 
$\chi(d(t),h(t))$. 
However, since the image of a Riordan array under an automorphism is a Riordan 
array, the expansion of the columns of 
$\p{-m}1(\DD)$ 
never involves formal Laurent series, only formal power series whose 
coefficient sequences coincide with the coefficient sequences of the formal 
Laurent series in the corresponding recursive matrices. Note that in this case, 
there is a phase shift of 
$m+1$ 
units between the index of a column in 
$\p{-m}1(\DD)$
and the index of the same column in 
$\chi(d(t),h(t))$. 
A recursive matrix 
$\chi(\DD)$ 
can thus be studied through a Riordan array 
$\p {-m} 1(\DD)$ 
after choosing an appropriate value for 
$m$. 

Furthermore, the matrix  
$\mathfrak{L}_\DD$ 
(read from right to left)  
mirrors a well-defined Riordan array.
\begin{definition}
\label{def:lD}
Let 
$\DD$ 
be any Riordan array. Consider the matrix sequence 
$(\p {-m}1(\DD))_{m\ge1}$. 
We define the lower triangular matrix  
$\mathfrak{l}_{\DD}$ 
as one whose 
$m$th 
column is the first column of the matrix 
$\p {-m} 1(\DD)$ 
for each 
$m\geq1$, 
with 
$m$ 
rows of zeros added to the top of the column. Hence, the entries of 
$\mathfrak{l}_{\DD}$ 
read as follows:
$$
({\mathfrak{l}_{\DD}})_{k,n}
=
[\te^{k-2n}]\dd(\te)\h(\te)^{-n}, 
\quad 
k\ge\ene\ge1 .
$$
Thus, by definition, 
$\mathfrak{l}_{\DD}$  
is the following Riordan array:
\begin{equation}
\label{eq:lD}
\mathfrak{l}_{\DD}
=
\biggl(\frac{t}{h(t)}d(t), \frac{t^2}{\h(t)}\biggr) .
\end{equation}
\end{definition}
Note that a similar idea to that of defining the matrices 
$\p {-\infty} 1(\DD)$ 
and
$\mathfrak{l}_{\DD}$
from a sequence of images of the Riordan array 
$\DD$ 
under some automorphisms, was recently given by the authors for the 
construction of some square matrices related to Riordan arrays
\cite{MeAga:JIS1}.
\begin{remark}
\label{rem:pair}
Let 
$\DD$ 
be any Riordan array. The bi-infinite triangle 
$\p{-\infty}1(\DD)$ 
corresponds to the following pair of Riordan arrays:
$$
(\mathfrak{l}_\DD,\DD)=
\left(
\biggl(\frac{t}{h(t)}d(t),\frac{t^2}{\h(t)}\biggr),
\bigl(d(t),h(t)\bigr)
\right) 
,
$$
in the sense of yielding exactly the same columns.
\end{remark}
Now, recall briefly that for
$m\in\mathbb{Z}$,
the  $[m]$-complementary
array $\chi(\DD)^{[m]}$ of a recursive matrix
$\chi(\DD)$, 
is defined by
$\chi(\DD)^{[m]}_{k,n}=\chi(\DD)_{m-n,m-k}$
\cite{complementary}. 
Next, for fixed 
$m\ge1$, 
we introduce the following  lower triangular array with 
$m$ 
columns 
$\mathfrak{L}_\DD^{(m)}$:
\begin{equation*}
({\mathfrak{L}_\DD^{(m)}})_{k,n}=[\te^{k-m-1}]d(t)\h(\te)^{n-m-1}, 
\quad
1\leq n\leq k, m . 
\end{equation*}
That is,
$$
({\mathfrak{L}_\DD^{(m)}})_{k,n}
=
{\p{-\infty}{1}(\DD)}_{k-m,n-m}
=
\chi(\DD)^{[-m]}_{-n+1,-k+1} .
$$
We also remark that the
$[m]$-complementary array  of a Riordan array 
$\DD$  
reads as
$$
\DD^{[m]}=
\biggl(
\left(\frac{\te}{\bar{\h}(\te)}\right)^{m+1}
\dd(\bar{\h}(\te))\bar{\h}'(\te),\bar{\h}(\te)
\biggr)
$$
yields 
$\varphi_{-(m+1),-1,1}(\DD^{-1})$, 
where Bacher's automorphism 
$\varphi_{-(m+1),-1,1}$ 
is defined by formula 
\eqref{eq:bacher}. 

Additionally, the first 
$l\geq1$ 
rows of the matrix 
$\mathfrak{L}_\DD^{(m)}$ 
yield a finite matrix, say, 
$\mathfrak{L}_\DD^{(l,m)}$. 
Moreover, consider the submatrix, say 
$\DD^{(l',m')}$, 
of the base array 
$\DD$ 
consisting of the first 
$l'\ge1$ 
rows and 
$m'\geq1$ 
columns of 
$\DD$. 
In this context, for fixed 
$q\ge0$ 
we consider the following finite 
$(2q+1)\times(2q+1)$  
and 
$(2q+2)\times (2q+2)$ 
matrices:
\begin{equation*}
A_q=
\left[
\begin{array}{llllll}
\mathfrak{L}_\DD^{(2q+1,q)} &amp; \mathfrak{R}_\DD^{(q+1,q+1)}\\
\end{array}
\right],
\quad
\text{where}
\quad 
\mathfrak{R}_\DD^{(q+1,q+1)}=
\left[
\begin{array}{llllll}
\updownarrow q 
\text{ rows of zeros}\\ 
\DD^{(q+1,q+1)}
\end{array}
\right] ,
\end{equation*}
\begin{equation*}
B_q=
\left[
\begin{array}{llllll}
\mathfrak{L}_\DD^{(2q+2,q)}&amp; \mathfrak{R}_\DD^{(q+2,q+2)}\\
\end{array}
\right] ,
\quad
\text{where}
\quad 
\mathfrak{R}_\DD^{(q+2,q+2)}=
\left[
\begin{array}{llllll}
\updownarrow q 
\text{ rows of zeros}\\ 
\DD^{(q+2,q+2)}
\end{array}
\right] .
\end{equation*}
Note that the west side matrices do not occur in 
$A_0$ 
and 
$B_0$.
We observe that 
$(A_q)_{q\geq0}$ 
and 
$(B_q)_{q\geq0}$ 
are sequences of finite matrices such that 
$A_q=\gamma_q(\DD)$, 
$B_q=\delta_q(\DD)$, 
where 
$(\gamma_q)_{q\ge0}$ 
and 
$(\delta_q)_{q\ge0}$ 
are sequences of projections introduced by 
Luz\'on, Merlini, Mor\'on, Prieto-Martinez, and Sprugnoli 
\cite{LMMP-MS}.
\begin{example}
\label{ex:pascalBarnabei}
\normalfont
Consider Pascal's array 
$P=(p(t),tp(t))$,
where 
$p(t)=1/(1-t)$, 
sequence 
\seqnum{A007318}:
\begin{center}
\begin{equation}
\label{eq:pascal}
P=
\left[
\begin{array}{rrrrrrrr}
1     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 2    &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 3    &amp; 3    &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 4    &amp; 6    &amp; 4    &amp; 1    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 5    &amp; 10   &amp; 10   &amp; 5    &amp; 1    &amp; 0    &amp;\cdots\\ 
1     &amp; 6    &amp; 15   &amp; 20   &amp; 15   &amp; 6    &amp; 1    &amp;\cdots\\ 
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots  
\end{array}
\right] .
\end{equation}
\end{center}
Hence, 
$P_{k,n}={k-1\choose n-1}$, 
$n\geq k\geq1$. 
Let 
$m\geq1$.
We now examine the matrix 
$\p {-m} 1 (P)=((1-t)^{m-1},t/(1-t))$ 
for large 
$m$.  
For instance, we have for 
$m=10$:
\begin{center}
$
\p{-10} 1 (P)=
\left[
\begin{array}{rrrrrrrrrrrrrrrr}
1     &amp; 0    &amp; 0    &amp; 0    &amp;0     &amp;0     &amp;0     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
-9    &amp; 1    &amp; 0    &amp; 0    &amp;0     &amp;0     &amp;0     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\
36    &amp; -8   &amp; 1    &amp; 0    &amp;0     &amp;0     &amp;0     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\
-84   &amp; 28   &amp; -7   &amp; 1    &amp;0     &amp;0     &amp;0     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
126   &amp; -56  &amp; 21   &amp;-6    &amp;1     &amp;0     &amp;0     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\
-126  &amp; 70   &amp; -35  &amp; 15   &amp;-5    &amp;1     &amp;0     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\
84    &amp; -56  &amp; 35   &amp; -20  &amp; 10   &amp;-4    &amp;1     &amp;0     &amp; 0    &amp; 0    &amp; 0    &amp;  
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
-36   &amp; 28   &amp; -21  &amp; 15   &amp; -10  &amp;6     &amp;-3    &amp;1     &amp; 0    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
9     &amp; -8   &amp; 7    &amp; -6   &amp; 5    &amp; -4   &amp;3     &amp;-2    &amp; 1    &amp; 0    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\
-1    &amp; 1    &amp; -1   &amp; 1    &amp; -1   &amp;1     &amp;-1    &amp;1     &amp;-1    &amp; 1    &amp; 0    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp; 
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp;
0     &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp;
2     &amp;1     &amp; 0    &amp; 0    &amp;\cdots\\ 
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp; 
3     &amp; 3    &amp; 1    &amp; 0    &amp;\cdots\\ 
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp;
4     &amp; 6    &amp; 4    &amp; 1    &amp;\cdots\\
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots\\
\end{array}
\right] .
$
\end{center} 
We observe that the matrix 
$\mathfrak{L}_P^{(10)}$ 
is zero below the equator. Note that is not true in general for the matrix 
$\mathfrak{L}_\DD^{(m)}$, 
with 
$m\ge1$ 
associated with an arbitrary Riordan array 
$\DD$. 
Indeed, this happens for all 
$m\ge1$ 
with the Pascal array  by chance, for only the first two coefficients of  
$p^{-1}(t)$ 
are different from zero.

When 
$m$ 
is large, in the north west region of the matrix 
$\p {-m} 1(P)$, 
the transpose of the reflection of the matrix  
$\mathfrak{L}_P$ 
(read from right to left) 
over the equator yields the matrix 
$P^{-1}=(1/(1+t),t/(1+t))$, 
the inverse of Pascal's array.
The columns of 
$\mathfrak{L}_P$ 
(read from right to left) 
also coincide with the sequence 
\seqnum{A110555} 
when seen as an upper right triangle. Besides, the lower triangular
matrix 
$\mathfrak{l}_P$ 
of formula 
\eqref{eq:lD} 
is 
\begin{center}
$
\mathfrak{l}_P=
\left[
\begin{array}{rrrrrrrr}
1     &amp; 0    &amp; 0    &amp;0     &amp;0     &amp;0     &amp; 0    &amp;\cdots\\ 
0     &amp; 1    &amp; 0    &amp;0     &amp;0     &amp;0     &amp; 0    &amp;\cdots\\
0     &amp;-1    &amp; 1    &amp;0     &amp;0     &amp;0     &amp; 0    &amp;\cdots\\
0     &amp; 0    &amp; -2   &amp;1     &amp;0     &amp;0     &amp; 0    &amp;\cdots\\
0     &amp; 0    &amp; 1    &amp;-3    &amp;1     &amp;0     &amp; 0    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp;3     &amp;-4    &amp;1     &amp; 0    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp;-1    &amp;6     &amp;-5    &amp; 1    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp;-4    &amp;10    &amp;-6    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp;-10   &amp;15    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;5     &amp;-20   &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;-1    &amp; 15   &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;-6    &amp;\cdots\\
0     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 1    &amp;\cdots\\
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots\\
\end{array}
\right] .
$
\end{center}
Then, 
$(\mathfrak{l}_P)_{k,n}=(-1)^{k-n}{n-1\choose k-n}$. 
The diagonals of  
$\mathfrak{l}_P$ 
yield  the columns in the north east region of the bi-infinite matrix of 
Barnabei 
\cite[p. 1132]{barnabei}. 
Note that a similar correspondence also holds for the matrix of p. 1139 of the 
same paper.
\end{example}
\begin{example}
We compute the array 
$\mathfrak{l}_T(x,y)$ 
of formula 
\eqref{eq:lD}, where 
$T(x,y)$ 
is a Pascal-like array of polynomials in the variables 
$x$ and $y$, 
introduced by Barry 
\cite{barry411}:
\begin{eqnarray*}
T(x,y)
&amp;=&amp;
\biggl(
\frac{1}{1-yt},\frac{yt(1+xt)}{1-yt}
\biggr)\\
&amp;=&amp;
\left[
\begin{array}{cccccc}
1     &amp; 0          &amp; 0                 &amp;0         &amp;0     &amp;\cdots\\ 
y     &amp; y          &amp; 0                 &amp;0         &amp;0     &amp;\cdots\\ 
y^2   &amp;xy+2y^2     &amp; y^2               &amp;0         &amp;0     &amp;\cdots\\ 
y^3   &amp;2xy^2+3y^3  &amp; 2xy^2+3y^3        &amp;y^3       &amp;0     &amp;\cdots\\
y^4   &amp; 3xy^3+4y^4 &amp; x^2y^2+6xy^3+6y^4 &amp;3xy^3+4y^4&amp;y^4   &amp;\cdots\\
\vdots&amp;\vdots      &amp;\vdots             &amp;\vdots    &amp;\vdots&amp;\ddots\\
\end{array}
\right] .
\end{eqnarray*}
The general expression for the entries  is:
$$
T_{k,n}(x,y)
= 
\sum_{j=0}^{n-1} 
{n-1\choose j}
{k-1-j\choose n-1}x^j y^{k-1-j} ,
\quad 
k,n\geq1 .
$$
Then, the matrix 
$\mathfrak{l}_T$ 
is
\begin{eqnarray*}
\mathfrak{l}_T(x,y)
&amp;=&amp;
\biggl(
\frac{1}{y(1+xt)},\frac{t(1-yt)}{y(1+xt)}
\biggr)\\
&amp;=&amp;
\left[
\begin{array}{rrrrrrr}
\frac{1}{y}         &amp;0                      &amp;0                       &amp;  
0                   &amp;0                      &amp;\cdots\\
-\frac{x}{y}        &amp;\frac{1}{y^2}          &amp;0                       &amp; 
0                   &amp;0                      &amp;\cdots\\ 
\frac{x^2}{y}       &amp;-\frac{2x+y}{y^2}      &amp;\frac{1}{y^3}           &amp; 
0                   &amp;0                      &amp;\cdots\\ 
-\frac{x^3}{y}      &amp;\frac{3x^2+2xy}{y^2}   &amp; -\frac{3x+2y}{y^3}     &amp; 
\frac{1}{y^4}       &amp;0                      &amp;\cdots\\
\frac{x^4}{y}       &amp;-\frac{4x^3+3x^2y}{y^2}&amp;\frac{6x^2+6xy+y^2}{y^3}&amp; 
-\frac{4x+3y}{y^4}  &amp;\frac{1}{y^5}          &amp;\cdots\\
\vdots              &amp;\vdots                 &amp;\vdots                  &amp;
\vdots              &amp;\vdots                 &amp;\ddots\\
\end{array}
\right] .
\end{eqnarray*}
Hence, the complete central coefficients sequence introduced by Barry 
\cite{barry411} 
of the matrix 
$T(x,y)$ 
is the sequence:  
$$
T_{k,\lceil \frac{k}{2}\rceil}(x,y)
= 
\sum_{j=0}^{\lceil{\frac{k}{2}}\rceil-1} 
{\lceil\frac{k}{2}\rceil-1\choose j}
{k-1-j\choose \lceil\frac{k}{2}\rceil-1} 
x^j y^{k-1-j} ,
\quad 
k,n\geq1 .
$$
Note that 
Barry  considers 
$\lfloor k /2 \rfloor$ 
instead of 
$\lceil k /2 \rceil$ 
for in 
\cite{barry411} 
$k,n\geq0$, 
while here 
$k,n\geq1$.
The general term of the Riordan array 
$\mathfrak{l}_T(x,y)$  
is given by:
$$
(\mathfrak{l}_T)_{k,n}(x,y)
=
\frac{(-1)^{k+1}}{y^n}
\sum_{j=0}^{n-1} 
{n-1\choose j}
{k-1-j-1\choose n-1}
x^{k-j-n} 
y^{j} ,
\quad 
k,n\geq1 .
$$
Therefore, the complete central coefficients sequence of Barry 
of the matrix 
$\mathfrak{l}_T(x,y)$ 
is the sequence:
$$
(\mathfrak{l}_T)_{k,\lceil\frac{k}{2}\rceil}(x,y)
=
\frac{(-1)^{k+1}}{y^{\lceil\frac{k}{2}\rceil}} 
\sum_ { j=0 } ^ {\lceil\frac{k}{2}\rceil-1} 
{\lceil\frac{k}{2}\rceil-1\choose j}
{k-1-j-1\choose\lceil\frac{k}{2}\rceil-1}
x^{k-j-\lceil\frac{k}{2}\rceil}y^{j} ,
\quad
k,n\geq1 .
$$
\end{example}
\begin{example}
\label{ex:fibcat}
We compute the Riordan array 
$\mathfrak{l}_{FC}$, 
where 
\begin{eqnarray*}
FC
&amp;=&amp;
\biggl(
\frac{1}{1-t-t^2},tc(t)
\biggr)\\
&amp;=&amp;
\biggl(
\frac{1}{1-t-t^2},\frac{1-\sqrt{1-4t}}{2}
\biggr)\\
&amp;=&amp;\left[
\begin{array}{rrrrrrr}
1     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
1     &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
2     &amp; 2    &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
3     &amp; 5    &amp; 3    &amp; 1    &amp; 0    &amp; 0    &amp;\cdots\\ 
5     &amp; 12   &amp; 9    &amp; 4    &amp; 1    &amp; 0    &amp;\cdots\\ 
8     &amp; 31   &amp;  26  &amp; 14   &amp; 5    &amp; 1    &amp;\cdots\\
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots  
\end{array}
\right]
\end{eqnarray*}
is the Fibonacci-Catalan triangle whose first column yields the Fibonacci 
numbers 
\seqnum{A000045} 
(without the first element which is zero). 
Moreover, the second column and the row sum sequence number both yield the 
convolution of the Fibonacci and the Catalan numbers  
\seqnum{A090826}. 
The sequence numbers for the first three descending diagonals are 
\seqnum{A000012}, 
\seqnum{A000027}, 
and 
\seqnum{A000096}. 
The next three descending diagonals yield:
\begin{eqnarray*}
FC_{3+n,n}&amp;=&amp;\frac{n^3+9n^2+20n-12}{3!} , 
\quad 
n\geq1 ,\\
FC_{4+n,n}&amp;=&amp;\frac{n^4+18n^3+107n^2+162n-168}{4!} , 
\quad 
n\geq1 ,\\
FC_{5+n,n}&amp;=&amp;\frac{n^5+30n^4+335n^3+1530n^2+1824n-2760}{5!} , 
\quad 
n\geq1 .
\end{eqnarray*}
The array 
$\mathfrak{l}_{FC}$ 
of formula 
\eqref{eq:lD} 
is
\begin{eqnarray*}
\mathfrak{l}_{FC}
&amp;=&amp;
\biggl(\frac{2t}{(1-t-t^2)(1-\sqrt{1-4t})}, 
\frac{2t^2}{1-\sqrt{1-4t}}
\biggr)\\
&amp;=&amp;
\left[
\begin{array}{rrrrrrr}
1     &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
0     &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
0     &amp; -1   &amp; 1    &amp; 0    &amp; 0    &amp; 0    &amp;\cdots\\ 
-2    &amp; -1   &amp; -2   &amp; 1    &amp; 0    &amp; 0    &amp;\cdots\\ 
-7    &amp; -4   &amp;  -1  &amp; -3   &amp; 1    &amp; 0    &amp;\cdots\\ 
-23   &amp; -10  &amp; -4   &amp; 0    &amp; -4   &amp; 1    &amp;\cdots\\
\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\vdots&amp;\ddots  
\end{array}
\right] .
\end{eqnarray*}
The coefficients of the generating function 
$2t/(1-\sqrt{1-4t})$ 
yield the sequence 
\seqnum{A115140}.  
Then, the first column of 
$\mathfrak{l}_{FC}$ 
is the convolution of 
\seqnum{A115140} 
and 
\seqnum{A000045} 
(after removing the first element of the latter).
The first descending diagonal is of course  
\seqnum{A000012}. 
The second is 
\seqnum{A001478}. 
The third is 
\seqnum{A000096} 
from the fourth element on. 
We now give recursion formulas for the fourth and fifth descending 
diagonals from the sixth and seventh element on, respectively. The following 
holds:
\begin{eqnarray*}
(\mathfrak{l}_{FC})_{3+n,n}
&amp;=&amp;
-\frac{n^3-12n^2+41n-18}{3!} , 
\quad 
n\geq6 ,\\
(\mathfrak{l}_{FC})_{4+n,n}
&amp;=&amp;
\frac{n^4-22n^3+167n^2-434n+120}{4!} , 
\quad 
n\geq7 .
\end{eqnarray*}
Alternatively, we have the following recursion formulas:
\begin{eqnarray*}
(\mathfrak{l}_{FC})_{9,6}
&amp;=&amp;
-2 ,\quad(\mathfrak{l}_{FC})_{10,7}=-4 ,\\
(\mathfrak{l}_{FC})_{3+n,n}
&amp;=&amp;
-\bigl(
(\mathfrak{l}_{ FC})_{2+n,n-2}-2(\mathfrak{l}_{FC})_{3+n,n-1}+n-5
\bigr) , 
\quad 
n\geq8 ,\\
(\mathfrak{l}_{FC})_{11,7}
&amp;=&amp;
5,
\quad
(\mathfrak{l}_{FC})_{12,8}=7 ,
\quad
(\mathfrak{l}_{FC})_{13,9}=11 ,\\
(\mathfrak{l}_{FC})_{4+n,n}
&amp;=&amp;
3(\mathfrak{l}_{FC})_{4+n,n-1}-3
(\mathfrak{l}_{ FC})_{3+n,n-2}+
(\mathfrak{l}_{FC})_{2+n,n-3}+n-7 , 
\quad 
n\geq10 .\\
\end{eqnarray*}
\end{example}
\section{Acknowledgments}
We would like to thank the anonymous referee for his valuable 
suggestions. This research was made within the activities of the 
Group for Linear, Algebraic and Combinatorial Structures 
of the 
Center for Functional Analysis, Linear Structures and Applications 
(University of Lisbon, Portugal).
The first and the second authors were supported by the research fellowships 
SFRH/BPD/100567/2014 
and 
SFRH/BPD/111317/2015, 
respectively, provided by the 
Portuguese Foundation for Science and Technology (FCT). 
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 05B20; Secondary 18A30, 20E18, 20H20.

\noindent \emph{Keywords: } Riordan group,
Riordan subgroup,
automorphism,
isomorphism,
bi-infinite triangle.

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\noindent 
(Concerned with sequences
\seqnum{A000012},
\seqnum{A000027},
\seqnum{A000045},
\seqnum{A000096},
\seqnum{A000108},
\seqnum{A001478},
\seqnum{A005586},
\seqnum{A007318},
\seqnum{A014137},
\seqnum{A014138},
\seqnum{A090826},
\seqnum{A110555}, and
\seqnum{A115140}.)

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\vspace*{+.1in}
\noindent
Received July 10 2019;
revised versions received  December 23 2019; December 24 2019.
Published in {\it Journal of Integer Sequences}, December 26 2019.

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\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
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