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\begin{center}
\vskip 1cm{\LARGE\bf 
The Inverse of Power Series and \\
\vskip .1in
the Partial Bell Polynomials 
}
\vskip 1cm
\large
Miloud Mihoubi\footnote{This research is supported by PNR project 8/u160/3172.} and Rachida Mahdid$^{1}$  \\
Faculty of Mathematics\\
University of Science and Technology Houari Boumediene (USTHB)\\
PB 32 \\
El Alia 16111 Algiers\\
Algeria\\
\href{mailto:miloudmihoubi@gmail.com}{\tt miloudmihoubi@gmail.com} \\
\href{mmihoubi@usthb.dz}{\tt mmihoubi@usthb.dz} \\
\href{mailto:rachida-mahdid@hotmail.fr}{\tt rachida-mahdid@hotmail.fr} \\
\end{center}

\vskip .2 in
\begin{abstract}
Using the Bell polynomials,
in this paper we give the explicit compositional inverses and/or
the reciprocals of some power series. We illustrate the obtained
results by some examples on Stirling numbers.
\end{abstract}

\section{Introduction}

The applications of the partial Bell polynomials have attracted the attention
of several authors.  Comtet \cite{4} studied these polynomials, and Riordan
\cite{10} used them in combinatorial analysis and Roman \cite{11} in
umbral calculus.  Recently, more applications of these polynomials
have appeared in different frameworks, including integration \cite{14},
inverse relations \cite{6}, congruences \cite{7}, Dyck paths \cite{15}
and Blissard problem \cite{3}, all of which
motivate us to apply these
polynomials to determine the explicit compositional inverses and/or
reciprocals of some power series.

Indeed, recall that the study of
the existence of the compositional inverses of power series is a
well-known result of complex analysis; see Forsyth \cite{20} and
Stanley \cite[Proposition 5.4.1]{24}.
Under some conditions on a function $f,$ to find
the compositional inverse $f^{ \langle -1 \rangle }$ of $f$ around
zero, three methods are used to compute the coefficients $y_{n} $ for which
the series $f^{ \langle -1 \rangle } (  t )  =\underset
{n\geq0}{\sum}y_{n}\frac{t^{n}}{n!}$ is the compositional inverse of the
series $f (  t )  =\underset{n\geq0}{\sum}x_{n}\frac{t^{n}}{n!}.$ The
first method
is based on the solution on $\textbf{\emph{y}}=(y_1,y_2,\ldots)$ 
in the equation%
\[
t=\underset{k\geq0}{\sum}\frac{y_{k}}{k!} \biggl(  \underset{j\geq0}{\sum}%
x_{j}\frac{t^{j}}{j!} \biggr)  ^{k}.
\]
If we set $x_{n}=y_{n}=0$ if $n\leq0,$ it was shown by Whittaker \cite{23}
that
\begin{equation*}
y_{1}=-\frac{1}{x_{1}},\ \ y_{n}=-\frac{ (  -1 )  ^{n-1}}%
{n!x_{1}^{2n-1}}\det \biggl(   (   (  1+i-j )  n+j-1 )
x_{2+i-j} \biggr)  _{1\leq i,j\leq n-1},\ \ n\geq2.%
\end{equation*}
This can be reduced to solve the equation%
\[
t=\underset{k\geq1}{\sum}\frac{y_{k}}{k!} \biggl(  \underset{j\geq1}{\sum}%
x_{j}\frac{t^{j}}{j!} \biggr)  ^{k}=\underset{n\geq1}{\sum}\frac{t^{n}}%
{n!}\underset{k=1}{\overset{n}{\sum}}y_{k}B_{n,k} (  x_{1},x_{2}%
,\ldots )  ,
\]
which is equivalent to solving the system%
\[
\underset{k=1}{\overset{n}{\sum}}y_{k}B_{n,k} ( x_{1},x_{2}%
,\ldots )  =\delta_{n-1},\text{ \ \ }n\geq1,
\]
where $\delta_{n}$ is the Kronicker's symbol, i.e. $\delta_{0}=1$ and $\delta
_{n}=0$ if $n\geq1,$ and the 
polynomials $B_{n,k} (  x_{1},x_{2},\ldots )$ are the (exponential)
partial Bell polynomials defined by their generating
function%
\begin{equation*}
\underset{n=k}{\overset{\infty}{\sum}}B_{n,k} (  x_{1},x_{2},\ldots )
\frac{t^{n}}{n!}=\frac{1}{k!} \biggl(  \underset{m=1}{\overset{\infty}{\sum}%
}x_{m}\frac{t^{m}}{m!} \biggr)  ^{k}.%
\end{equation*}
The second method is based on Lagrange's inversion formula \cite{21} for which we have%
\begin{equation*}
y_{n}=\frac{d^{n-1}}{d\varphi^{n-1}} \biggl(  \frac{\varphi}{f (  \varphi )  } \biggr)^{n}\biggl|_{\varphi=0}.%
\end{equation*}
The third method is based on the $n^{th}$ nested derivative of a function $g$
defined in \cite{22} by%
\[
\mathcal{D}^{0} [  g ]   (  \varphi)  :=1,\ \ \mathcal{D}%
^{n} [  g ]   (  \varphi )  :=\frac{d}{d\varphi} \biggl(  f (
\varphi )  \mathcal{D}^{n-1} [  f ]   ( \varphi )   \biggr)
,\ \ n\geq1,
\]
for which Dominici \cite{22} showed that if $f (  t )  =\int_{\alpha}%
^{t}\frac{1}{g (  \varphi)  }d\varphi,$ with $g (  \alpha )  \neq0,\pm
\infty,$ we have%
\begin{equation*}
f^{ \langle -1 \rangle } (  t )  =\alpha+g (  \alpha )
\underset{n\geq1}{\sum}\mathcal{D}^{n} [  g ]   (  \alpha )
\frac{t^{n}}{n!}.
\end{equation*}
For our contribution, we show that the partial Bell polynomials define two
families of power series for which we can obtain explicit 
compositional inverses.
We also give the reciprocals of power series connected to these families.
For the applications, we give some examples on
power series whose coefficients are related to Stirling and
$r$-Stirling numbers. Indeed, let $r,s,d$ be integers with $d\geq1$ and $ \textbf{\emph{x}}=(
x_{1},x_{2},\ldots )  $ be a sequence of real numbers with $x_{1}=1.$ For
$r&gt;\max (  s,-2s )  $ we consider%
\begin{equation}
H_{1} (  t )  =t \biggl(  1-\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !s}{rdn-s}\frac{\binom{ (  r+1 )  dn-s}{ (
d-1 )  n}}{\binom{ (  rd+1 )  n-s}{n}}B_{ (  rd+1 )
n-s,rdn-s} ( \textbf{\emph{x}} )  \frac{t^{dn}}{n!}\biggr )  ,\label{4}%
\end{equation}
and for $r&gt;\max (  -s, (  d+1 )  s )  $ we consider%
\begin{equation}
H_{2} (  t )  =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}\frac{B_{ (
r+1 )  n-s,\ rn-s} (\textbf{\emph{x}})  }{\binom{ (  r+1 )
n-s}{rn-s}}\frac{t^{dn}}{n!} \biggr)  .\label{5}%
\end{equation}
We give below the explicit compositional inverses $H_{1}^{ \langle
-1 \rangle }$ of $H_{1}$ (Theorem \ref{T1}) and $H_{2}^{ \langle
-1 \rangle }$ of $H_{2}$ (Theorem \ref{T2}). Also, we present the
explicit reciprocal power series $\frac{t}{H_{1} (  t )  }$ of
$\frac{H_{1} (  t )  }{t}.$\\
The mathematical tools used are based on the connection between the
partial Bell polynomials and the polynomials of binomial type.
For a given real number $\alpha $ and for any sequence of binomial type $(f_{n}(\varphi)),$
in what follows
we let $ (f_{n}(\varphi;\alpha) ) $ denote any sequence of binomial type such that%
\begin{equation}
f_{n}(\varphi;\alpha):=\frac{\varphi}{\alpha n+\varphi}f_{n} (  \alpha n+\varphi ), \label{3}%
\end{equation}
see Comtet \cite[pp.\ 133--175]{4}, Aigner \cite[pp.\ 99--116]{16} and Proposition 1 given in \cite{5}. 

For example, the sequences $(\varphi^{n})$ and $(\varphi(\alpha n+\varphi)^{n-1})$ are sequences of binomial type. 

Below, we use the following notation: 
$B_{n,k}( x_j )$ or $B_{n,k}( \textbf{\emph{x}} )$ with $\textbf{\emph{x}}=(  x_{1},x_{2},\ldots )$ for the partial Bell polynomial $B_{n,k} (  x_{1},x_{2},\ldots )$,\\ \ \\
$D_{z=0}^{k}f (  z )$ for $\frac{d^{k}f}{dz^{k}} (  0 )  , \ k\geq2,$  and
$D_{z=0}f (  z )$ for $\frac{df}{dz} (  0 ),$\\ \ \\
$\genfrac{[}{]}{0pt}{}{n}{k}$ and $\genfrac{[}{]}{0pt}{}{n}{k}_{r}$  for the  unsigned Stirling and $r$-Stirling numbers of the first kind, respectively, \\ \ \\
$\genfrac{\{}{\}}{0pt}{}{n}{k}$ and $\genfrac{\{}{\}}{0pt}{}{n}{k}_{r}$ for the Stirling and $r$-Stirling numbers of the second kind, respectively.

\section{The first family of power series and their inverses}

We give in this section the reciprocal and/or the compositional inverse of a
power series given from a large family of power series which have coefficients
can be expressed in terms of partial Bell polynomials.

\begin{proposition}
\label{P1}Let $ \textbf{x}=(x_{1},x_{2},\ldots )  $ be a sequence of real numbers with $x_{1}=1.$
Then, for $r,s$ integers such that $r&gt;\max (  s,-2s )  ,$
the reciprocal and compositional inverse of the power series%
\begin{align}
H (  t )  =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}\frac{B_{ (
r+1 )  n-s,rn-s} ( \textbf{x} )  }{\binom{ (  r+1 )
n-s}{n}}\frac{t^{n}}{n!} \biggr) \label{5a}
\end{align}
are given by%
\begin{align}
H^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(  1+\sum
_{n\geq1}\frac{s}{ (  r+s )  n+s}\frac{B_{ (  r+s+1 )
n+s, (  r+s )  n+s} (\textbf{x} )  }{\binom{ (
r+s+1 )  n+s}{n}}\frac{t^{n}}{n!} \biggr)  ,\label{6}\\
\frac{t}{H (  t )  }  &amp; =1+\sum_{n\geq1}\frac{s}{rn+s}\frac
{B_{ (  r+1 )  n+s,rn+s} ( \textbf{x} )  }{\binom{ (
r+1 )  n+s}{n}}\frac{t^{n}}{n!}.\label{7}%
\end{align}
\end{proposition}

\begin{proof}
We consider only the case $s\geq0$ (for $s&lt;0,$ we can proceed
similarly). Let $ ( f_{n} (  \varphi )  )  $ be a
sequence of binomial-type polynomials such that $f_{n} (  1 )
=x_{n+1}/(n+1)$ with $x_{2}\neq0$ to ensure that $Df_{1} (
\varphi )  \neq0.$ Then, by Proposition 1 given in \cite{5}, we get
\begin{equation}
f_{n} (  k )  =\binom{n+k}{k}^{-1}B_{n+k,k} (\textbf{\emph{x}} ). \label{a}%
\end{equation}
In \cite[Theorem 1]{6}, we proved that
\begin{align*}
H(t)  &amp; =t \biggl(  1-\sum_{n\geq1}\frac{s}{r n-s}f_{n}(r n-s)\frac{t^{n}}%
{n!} \biggr )  ,
\end{align*}
has inverse
\begin{align*}
H^{\langle-1\rangle}(t)  &amp; =t \biggl(  1+\sum_{n\geq1}\frac{s}{ (
r +s )  n+s}f_{n}( (  r +s )  n+s)\frac{t^{n}}{n!} \biggr),
\end{align*}
and because for any binomial-type sequence of polynomials $ (f_{n} ( \varphi )   )  $ we have%
\[
 \biggl(  1+\sum_{n\geq1}f_{n}(-s)\frac{t^{n}}{n!} \biggr)  ^{-1}=1+\sum_{n\geq
1}f_{n}(s)\frac{t^{n}}{n!},
\]
then, by replacing in the last identity $  f_{n} (  s )   $ by $f_{n} (  s;r ) $ defined by (\ref{3}) we get%
\[
\frac{t}{H (  t )  }=1+\sum_{n\geq1}\frac{s}{r n+s}f_{n}%
(r n+s)\frac{t^{n}}{n!}.%
\]
Then, replace $f_{n}(rn-s)$ and $f_{n}( (  r+s )  n+s)$ by their
expressions obtained from the identity (\ref{a}) by
\begin{equation*}
f_{n}(rn-s)=\frac{B_{ (  r+1 )  n-s,rn-s} ( \textbf{\emph{x}} )
}{\binom{ (  r+1 )  n-s}{n}}\text{ \ and \ }f_{n}( (  r+s )
n+s)=\frac{B_{ (  r+s+1 )  n+s, (  r+s )  n+s} (
\textbf{\emph{x}} )  }{\binom{ (  r+s+1 )  n+s}{n}}.%
\end{equation*}
The proposition remains true for the case $x_{2}=0$ by continuity.
\end{proof}

\begin{lemma}
\label{L1}Let $ \textbf{x}=(x_{1},x_{2},\ldots )  $ be a sequence of real numbers with
$x_{1}=1, $ $x_n=0$ if $d\nmid n-1,$ and $\textbf{y}=(y_1,y_2,\ldots)$ with
$ y_j=j!x_{d (  j-1 )  +1}/ (d (  j-1 )  +1 )  !.$
Then, we have%
\begin{align*}
\frac{B_{dn+k,k} ( \textbf{x} )}{(dn+k)!}   =\frac{B_{n+k,k} (\textbf{y} )}{(n+k )!}  \ \ and \ \
B_{dn+k+l,k} ( \textbf{x} ) =0\text{ if }1\leq l\leq d-1.
\end{align*}
\end{lemma}

\begin{proof}
Setting $\textbf{z}=(z_1,z_2,\ldots)$ with $ z_j=y_{j+1}/(j+1). $ \\
From the definition of partial Bell polynomials we have%
\[
\sum_{n\geq k}B_{n,k} ( \textbf{\emph{x}} )  \frac{t^{n}}{n!}
=\frac{t^{k}}{k!} \biggl(  1+\sum_{j\geq1}z_j\frac{ (  t^{d} )  ^{j}}{j!} \biggr)  ^{k}
=\frac{t^{k}}{k!}\underset{l=0}{\overset{k}{\sum}}\frac{ (  k )
_{l}}{l!} \biggl(  \sum_{j\geq1}z_j \frac{ (  t^{d} )  ^{j}}{j!} \biggr)  ^{l}
\]
i.e.
\[
\sum_{n\geq k}B_{n,k} ( \textbf{\emph{x}} )  \frac{t^{n}}{n!}
=\frac{t^{k}}{k!}\underset{l=0}{\overset{k}{\sum}} (  k )
_{l}   \sum_{n\geq l}B_{n,l} (  \textbf{\emph{z}} )  \frac{t^{dn}}{n!}
=\frac{t^{k}}{k!}\sum_{n\geq0}\frac{t^{dn}}{n!}\underset{l=0}{\overset
{\min (  n,k )  }{\sum}} (  k )  _{l}B_{n,l} (\textbf{\emph{z}} )  .
\]
Now, from the identity $ [  3l ]  $ given in Comtet \cite[pp. 136]%
{4}, we have%
\[
\underset{l=0}{\overset{\min (  n,k )  }{\sum}} (  k )
_{l}B_{n,l} ( \textbf{\emph{z}} )
=\binom{n+k}{k}^{-1}B_{n+k,k} (  \textbf{\emph{y}} )
\]
and the above expansion becomes%
\begin{align*}
\sum_{n\geq k}B_{n,k} ( \textbf{\emph{x}} )  \frac{t^{n}}{n!}  =\frac{1}%
{k!}\sum_{n\geq0}\frac{t^{dn+k}}{n!}\binom{n+k}{k}^{-1}B_{n+k,k} (\textbf{\emph{y}})
 =\sum_{n\geq0}\frac{ (  dn+k )  !}{ (  n+k )  !}%
B_{n+k,k} (  \textbf{\emph{y}})  \frac{t^{dn+k}}{ (  dn+k )  !}.
\end{align*}
This gives $B_{dn+k,k} ( \textbf{\emph{x}} )  =\frac{ (  dn+k )
!}{ (  n+k )  !}B_{n+k,k} (  \textbf{\emph{y}})  $ and $B_{dn+k+l,k}%
 ( \textbf{\emph{x}} )  =0$ if $1\leq l\leq d-1.$
\end{proof}

\begin{theorem}
\label{T1}Let $\textbf{y}= (  y_{1},y_{2}, \ldots )  $ be a sequence of real numbers with
$y_{1}=1 ,$  $r,s,d$ be integers such that $d\geq1,$  $r&gt;\max (
s,-2s )  $  and let
\begin{equation}
U_{n} (  r,s;d )  =\frac{ (   (  d-1 )  n )
!s}{rdn+s}\frac{\binom{ (  r+1 )  dn+s}{ (  d-1 )  n}%
}{\binom{ (  rd+1 )  n+s}{n}}B_{ (  rd+1 )  n+s,rdn+s}%
 (  \textbf{y} )  ,\text{ \ }n\geq1,  \ \ U_{0} (  r,s;d )
=1.\label{13}%
\end{equation}
Then for%
\begin{align}
H_{1} (  t )  =t \biggl(  1+\sum_{n\geq1}U_{n} (  r,-s;d )
\frac{t^{dn}}{n!} \biggr) \label{11}%
\end{align}
we have%
\begin{align}
H_{1}^{ \langle -1 \rangle } (  t )  &amp;=t \biggl(  1+\sum
_{n\geq1}U_{n} (  r+s,s;d )  \frac{t^{dn}}{n!} \biggr), \label{b} \\%
\frac{t}{H_{1} (  t )  }&amp;=1+\sum_{n\geq1}U_{n} (  r,s;d ) \frac{t^{dn}}{n!},\label{12}%
\end{align}
\end{theorem}

\begin{proof}
In Proposition \ref{P1}, choice $ \textbf{\emph{x}}=(x_{1},x_{2},\ldots )  $ such that  $x_n=0,$ if $d\nmid n-1$  and $ x_{dn+1}=(dn+1)!y_{n+1}/(n+1)!,$ after that, apply Lemma \ref{L1}.
\end{proof}

\begin{example}
For $ \textbf{y}=(  1!,2!,\ldots, (  q+1 )  !,0,\ldots )$ in Theorem
\ref{T1}, the identity%
\begin{equation}
B_{n,k} (  \textbf{y} )  =\frac{n!}{k!}\binom{k}{n-k}_{q}, \label{16}%
\end{equation}
see \cite{1}, gives%
\begin{align*}
H_{1} (  t )   &amp; =t \biggl(  1-s\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !}{rdn-s}\frac{\binom{ (  r+1 )  dn-s}{ (
d-1 )  n}}{\binom{ (  rd+1 )  n-s}{n}}\binom{rdn-s}{n}_{q}%
\frac{t^{dn}}{n!} \biggr)  ,\\
H_{1}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+s\sum_{n\geq1}\frac{ (   (  d-1 )  n )  !}{ (
r+s )  dn+s}\frac{\binom{ (  r+s+1 )  dn+s}{ (  d-1 )
n}}{\binom{ (   (  r+s )  d+1 )  n+s}{n}}\binom{ (
r+s )  dn+s}{n}_{q}\frac{t^{dn}}{n!} \biggr)  ,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+s\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !}{rdn+s}\binom{ (  r+1 )  dn+s}{ (
d-1 )  n}\binom{rdn+s}{n}_{q}t^{dn},
\end{align*}
where $\binom{k}{n}_{q}$ is the coefficients defined by $(1+\varphi+\varphi^{2}%
+\cdots+\varphi^{q})^{k}=\sum\limits_{n\geq0}\binom{k}{n}_{q}\varphi^{n}.$
\end{example}

\begin{example}
For $\textbf{y}=(  0!,1!,\ldots )$ in Theorem \ref{T2}, the identity $B_{n,k} (  \textbf{y} ) =\genfrac{[}{]}{0pt}{}{n}{k}$ gives
\begin{align*}
H_{1} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !s}{rdn-s}\frac{\binom{ (  r+1 )  dn-s}{ (
d-1 )  n}}{\binom{ (  rd+1 )  n-s}{rdn-s}}%
\genfrac{[}{]}{0pt}{}{ (  rd+1 )  n-s}{rdn-s} \frac{t^{dn}}{n!} \biggr)  ,\\
H_{1}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{ (   (  d-1 )  n )  !s}{ (
r+s )  dn+s}\frac{\binom{ (  r+s+1 )  dn+s}{ (  d-1 )
n}}{\binom{ (   (  r+s )  d+1 )  n+s}{ (  r+s )
dn+s}} \genfrac{[}{]}{0pt}{}{ (   (  r+s )  d+1 )  n+s}{ (r+s )  dn+s} \frac{t^{dn}}{n!} \biggr)  ,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\sum_{n\geq1}\frac{ (   (d-1 )  n )  !s}{rdn+s}\frac{\binom{ (  r+1 )  dn+s}{ (
d-1 )  n}}{\binom{ (  rd+1 )  n+s}{rdn+s}} \genfrac{[}{]}{0pt}{}{ (  rd+1 )  n+s}{rdn+s} \frac{t^{dn}}{n!}%
\end{align*}
\end{example}

\begin{example}
For $\textbf{y}=( 1,1,\ldots )$ in Theorem \ref{T2}, the identity $B_{n,k} ( \textbf{y} ) =\genfrac{\{}{\}}{0pt}{}{n}{k}$ gives
\begin{align*}
H_{1} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !s}{rdn-s}\frac{\binom{ (  r+1 )  dn-s}{ (
d-1 )  n}}{\binom{ (  rd+1 )  n-s}{rdn-s}} \genfrac{\{}{\}}{0pt}{}{ (  rd+1 )  n-s}{rdn-s}\frac{t^{dn}}{n!} \biggr),\\
H_{1}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{ (   (  d-1 )  n )  !s}{ (
r+s )  dn+s}\frac{\binom{ (  r+s+1 )  dn+s}{ (  d-1 )
n}}{\binom{ (   (  r+s )  d+1 )  n+s}{ (  r+s )
dn+s}} \genfrac{\{}{\}}{0pt}{}{ (   (  r+s )  d+1 )  n+s}{ (r+s )  dn+s} \frac{t^{dn}}{n!} \biggr)  ,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !s}{rdn+s}\frac{\binom{ (  r+1 )  dn+s}{ (
d-1 )  n}}{\binom{ (  rd+1 )  n+s}{rdn+s}} \genfrac{\{}{\}}{0pt}{}{ (  rd+1 )  n+s}{rdn+s} \frac{t^{dn}}{n!},
\end{align*}
\end{example}

\begin{proposition}
\label{P3}Let $r,s,d$ be integers with $s\neq0,\ d\geq1$ and $(f_{n}(\varphi))$ be binomial-type polynomials.
Then, for $r&gt;\max (  s,-2s )  $ we get%
\begin{align*}
H_{1} (  t )   &amp; =t \biggl(  1-\varphi \sum_{n\geq1} (   (
d-1 )  n )  !\binom{ (  r+1 )  dn-s}{ (  d-1 )
n}\frac{f_{n} (  \alpha n-\varphi )  }{\alpha n-\varphi}\frac{t^{dn}}{n!} \biggr)  ,\\
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\varphi \sum_{n\geq
1} (   (  d-1 )  n )  !\binom{ (  r+s+1 )
dn+s}{ (  d-1 )  n}\frac{f_{n} (   (  \alpha +sd )
n+\varphi )  }{ (  \alpha +sd )  n+\varphi }\frac{t^{dn}}{n!} \biggr)  ,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\varphi \sum_{n\geq1} (   (
d-1 )  n )  !\binom{ (  r+1 )  dn+s}{ (  d-1 )
n}\frac{f_{n} (  \alpha n+\varphi )  }{\alpha n+\varphi}\frac{t^{dn}}{n!}.
\end{align*}
\end{proposition}

\begin{proof}
For $H_{1},$ take $y_{n}=f_{n} (  \varphi;\alpha )  =\frac{\varphi}{\alpha n+\varphi}f_{n} (
\alpha n+\varphi)  $ in Theorem \ref{T1} and use Proposition 1 given in \cite{5}, after that, replace $\alpha$ by $\alpha-rd$ and $\varphi$ by $\frac{\varphi}{s}.$
\end{proof}

\begin{example}
For $f_{n}(\varphi)=\varphi^{n}$ in Proposition \ref{P3} we get%
\begin{align*}
H_{1} (  t )   &amp; =t \biggl(  1-\varphi \sum_{n\geq1} (   (
d-1 )  n )  !\binom{ (  r+1 )  dn-s}{ (  d-1 )
n} (  \alpha n-\varphi )  ^{n-1}\frac{t^{dn}}{n!} \biggr)  ,\\
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\varphi \sum_{n\geq
1} (   (  d-1 )  n )  !\binom{ (  r+s+1 )
dn+s}{ (  d-1 )  n} (   (  \alpha+sd )  n+\varphi )
^{n-1}\frac{t^{dn}}{n!} \biggr)  ,\\
\frac{t}{H_{1}(t)}  &amp; =1+\varphi \sum_{n\geq1} (   (  d-1 )  n )
!\binom{ (  r+1 )  dn+s}{ (  d-1 )  n} (  \alpha n+\varphi )
^{n-1}\frac{t^{dn}}{n!},
\end{align*}
and for $f_{n}(\varphi)=n!\binom{\varphi}{n}$ in Proposition \ref{P3} we get%
\begin{align*}
H_{1} (  t )   &amp; =t \biggl(  1-\varphi \sum_{n\geq1} (   (  d-1 )
n )  !\binom{ (  r+1 )  dn-s}{ (  d-1 )  n}\binom
{\alpha n-\varphi}{n}t^{dn} \biggr)  ,\\
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\varphi \sum_{n\geq1} (
 (  d-1 )  n )  !\binom{ (  r+s+1 )  dn+s}{ (
d-1 )  n}\binom{ (  \alpha +sd )  n+\varphi}{n}t^{dn} \biggr)  ,\\
\frac{t}{H_{1}(t)}  &amp; =1+\varphi \sum_{n\geq1} (   (  d-1 )  n )
!\binom{ (  r+1 )  dn+s}{ (  d-1 )  n}\binom{\alpha n+\varphi}{n}t^{dn}.
\end{align*}
\end{example}

The following theorem generalizes Theorem \ref{T1}.

\begin{theorem}
\label{T3}Let $ \textbf{x}=(x_{1},x_{2},\ldots )  $ be a sequence of real numbers with
$x_{1}=1 $ and $r,s,u,v,d$ be integers such that $d\geq1,$ $r&gt;\max (
s,-2s )  $ and $u&gt;\max (   \vert v \vert ,-sd-v )  .$
Then for%
\[
H_{1} (  t )  =t \biggl(  1-\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !v}{un-v}\binom{ (  u+1 )  n-v}{un-v}^{-1}\binom{ (  r+1 )  dn-s}{ (d-1 )  n}B_{ (  u+1 )
n-v,un-v} (  \textbf{x} )  \frac{t^{dn}}{n!} \biggr)
\]
we have%
\begin{align*}
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\sum_{n\geq1}%
\frac{ (   (  d-1 )  n )  !v}{ (  u+sd )  n+v}%
\frac{\binom{ (  r+s+1 )  dn+s}{ (  d-1 )  n}}%
{\binom{ (  u+sd+1 )  n+v}{ (  u+sd )  n+v}}B_{ (
u+sd+1 )  n+v, (  u+sd )  n+v} (  \textbf{x} )  \frac
{t^{dn}}{n!} \biggr)  ,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !v}{un+v}\binom{ (  u+1 )  n+v}{un+v}^{-1}\binom{ (  r+1 )  dn+s}{ (
d-1 )  n}B_{ (  u+1 )n+v,un+v} (\textbf{x} )  \frac{t^{dn}}{n!}.
\end{align*}
\end{theorem}

\begin{proof}
Let $ ( f_{n} (  \varphi ) )  $ be a sequence of of binomial
type of polynomials such that $f_{n} (  1 )  =x_{n+1}/(n+1)$
with $x_{2}\neq0$ to ensure that $Df_{1} (  \varphi )  \neq0.$ Then, $f_n(\varphi)$ satisfies (\ref{a}). Set $\alpha=u$ and $\varphi=v$ in Proposition \ref{P3}, after that, use the identity (\ref{a})
in the three power series of Proposition \ref{P3}, respectively, for $ k=un-v ,$ $k=(  u+sd )  n+v$ and $k= un+v.$ \\
The theorem remains true for the case $x_{2}=0$ by continuity.
\end{proof}

For $u=rd$ and $v=s$ in Theorem \ref{T3} we obtain Theorem \ref{T1}.

\begin{example}
Set $ \textbf{x}=(x_{1},x_{2},\ldots )  $ with $x_{n}=n!\binom{p+n-2}{p-1},\ p\geq1,$
in Theorem \ref{T3}. The identity
\cite[Example 13]{9}%
\begin{equation}
B_{n,k} ( \textbf{x} )  =\frac{n!}{k!}\binom{n+k (p-1 )  -1}{kp-1}\label{31}%
\end{equation}
implies for the power series%
\[
H_{1} (  t )  =t \biggl(  1-\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !v}{un-v}\binom{ (  r+1 )  dn-s}{ (
d-1 )  n}\binom{ (  up+1 )  n-vp-1}{upn-vp-1}t^{dn} \biggr)
\]
we have%
\begin{align*}
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\sum_{n\geq1}%
\frac{ (   (  d-1 )  n )  !v}{ (  u+sd )
n+v}\binom{ (  r+s+1 )  dn+s}{ (  d-1 )  n}\binom{ (
 (  u+sd )  p+1 )  n+vp-1}{ (  u+sd )  pn+vp-1}%
t^{dn} \biggr)  ,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\sum_{n\geq1}\frac{ (   (
d-1 )  n )  !v}{un+v}\binom{ (  r+1 )  dn+s}{ (
d-1 )  n}\binom{ (  up+1 )  n+vp-1}{upn+vp-1}t^{dn}.
\end{align*}
\end{example}

\begin{example}
Set $ \textbf{x}=(x_{1},x_{2},\ldots )  $ with $x_{n}=p\binom{j+p-1}{p-1}^{-1}
\genfrac{[}{]}{0pt}{}{p+q+j-1}{p+q}_{q}, \ p\geq1,\ q\geq0,$
in Theorem \ref{T3}. The identity \cite[Example 13]{9}%
\begin{equation}
B_{n,k} (  \textbf{x} )  =\binom{kp}{k}\binom{n+ (  p-1 )  k}{ (
p-1 )  k}^{-1} \genfrac{[}{]}{0pt}{}{n+ (  p+q-1 )  k}{ (  p+q )  k}_{kq}\label{14}%
\end{equation}
implies for the power series%
\[
H_{1} (  t )  =t \biggl(  1-\sum_{n\geq1}v\frac{ (   (
r+1 )  dn-s )  !}{ (   (  rd+1 )  n-s )  !}%
\frac{\binom{pA}{A} \genfrac{[}{]}{0pt}{}{n+ (  p+q )  A}{ (  p+q )  A}
_{Aq}}{A\binom{n+A}{A}\binom{n+pA}{ (  p-1 )  A}}\frac{t^{dn}}{n!} \biggr)  ,\ \ A=un-v,
\]
we have%
\begin{align*}
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\sum_{n\geq
1}v\frac{ (   (  r+s+1 )  dn+s )  !}{ (   (   (
r+s )  d+1 )  n+s )  !}\frac{\binom{pB}{B}%
\genfrac{[}{]}{0pt}{}{n+ (  p+q )  B}{ (  p+q )  B}%
_{Bq}}{B\binom{n+B}{B}\binom{n+pB}{ (  p-1 )  B}}\frac{t^{dn}}%
{n!} \biggr)  ,\text{ \ }B= (  u+sd )  n+v,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\sum_{n\geq1}v\frac{ (   (
r+1 )  dn+s )  !}{ (   (  rd+1 )  n+s )  }%
\frac{\binom{pC}{C} \genfrac{[}{]}{0pt}{}{n+ (  p+q )  C}{ (  p+q )  C}%
_{Cq}}{C\binom{n+C}{C}\binom{n+pC}{ (  p-1 )  C}}\frac{t^{dn}}%
{n!},\ \ C=un+v,
\end{align*}
\end{example}

\begin{example}
Set $ \textbf{x}=(x_{1},x_{2},\ldots )  $ with $x_{n}=p\binom{j+p-1}{p-1}^{-1} \genfrac{\{}{\}}{0pt}{}{p+q+j-1}{p+q}%
_{q},\ p\geq1,\ q\geq0,$ in Theorem \ref{T3}. The identity \cite[Example 13]{9}%
\begin{equation}
B_{n,k} (  \textbf{x})  =\binom{kp}{k}\binom{n+ (  p-1 )  k}{ (
p-1 )  k}^{-1} \genfrac{\{}{\}}{0pt}{}{n+ (  p+q-1 )  k}{ (  p+q )  k}%
_{kq}\label{15}%
\end{equation}
implies for the power series%
\[
H_{1} (  t )  =t \biggl(  1-\sum_{n\geq1}v\frac{ (   (
r+1 )  dn-s )  !}{ (   (  rd+1 )  n-s )  !}%
\frac{\binom{pA}{A} \genfrac{\{}{\}}{0pt}{}{n+ (  p+q )  A}{ (  p+q )  A}%
_{Aq}}{A\binom{n+A}{A}\binom{n+pA}{ (  p-1 )  A}}\frac{t^{dn}}%
{n!} \biggr)  ,\ \ A=un-v,
\]
we have%
\begin{align*}
H_{1}^{ (  -1 )  } (  t )   &amp; =t \biggl(  1+\sum_{n\geq
1}v\frac{ (   (  r+s+1 )  dn+s )  !}{ (   (   (
r+s )  d+1 )  n+s )  !}\frac{\binom{pB}{B}%
\genfrac{\{}{\}}{0pt}{}{n+ (  p+q )  B}{ (  p+q )  B}%
_{Bq}}{B\binom{n+B}{B}\binom{n+pB}{ (  p-1 )  B}}\frac{t^{dn}}%
{n!} \biggr)  ,\text{ \ }B= (  u+sd )  n+v,\\
\frac{t}{H_{1} (  t )  }  &amp; =1+\sum_{n\geq1}v\frac{ (   (
r+1 )  dn+s )  !}{ (   (  rd+1 )  n+s )  }%
\frac{\binom{pC}{C} \genfrac{\{}{\}}{0pt}{}{n+ (  p+q )  C}{ (  p+q )  C}%
_{Cq}}{C\binom{n+C}{C}\binom{n+pC}{ (  p-1 )  C}}\frac{t^{dn}}%
{n!},\ \ C=un+v,
\end{align*}
\end{example}

\section{The second family of power series and their inverses}

We give in this section the compositional inverse of a power series given from
a second family of power series which have coefficients can be expressed in
terms of partial Bell polynomials.

\begin{lemma}
\label{L2}Let $ \textbf{x}=(x_{0},x_{1},\ldots )  $ with $x_0=1$ and $ \textbf{y}=(y_{1},y_{2},\ldots )  $
be sequences of real numbers with $y_j=jx_{j-1}.$ Then, the compositional inverse of%
\begin{equation}
H (  t )  =\frac{t}{1+\sum_{n\geq1}x_{n}\frac{t^{n}}{n!}}\label{17}%
\end{equation}
is given by%
\begin{equation}
H^{ \langle -1 \rangle } (  t )  =t \biggl(  1+\sum_{n\geq
1}\frac{n!}{ (  2n+1 )  !}B_{2n+1,n+1} (  \textbf{y} )
t^{n} \biggr)  .\label{18}%
\end{equation}
\end{lemma}

\begin{proof}
Let $ ( f_{n} ( \varphi) )  $ be a sequence of polynomials such that $f_{n} (  1 )  =x_{n}$
and assume that $x_{1}\neq0$ to ensure that $Df_{1} (  \varphi )  \neq0.$ Then we get $f_{n} (  k )  =\binom{n+k}{k}^{-1}B_{n+k,k} ( \textbf{y} ).$
In \cite[Theorem 1]{6}, we proved that
\[
H (  t )  =\frac{t}{1+\sum_{n\geq1}f_{n} (  1 )  \frac
{t^{n}}{n!}}=\sum_{n\geq1}nf_{n-1} (  -1 )  \frac{t^{n}}%
{n!},\ \ H^{\langle-1\rangle}(t)=\sum_{n\geq1}f_{n-1}(n)\frac{t^{n}}{n!}.
\]
Then, from the last identity, $f_{n-1}(n)$ can be expressed by partial
Bell polynomials as%
\[
f_{n-1}(n)=\binom{2n-1}{n}^{-1}B_{2n-1,n} ( \textbf{y} )  .
\]
The lemma holds for the case $x_{1}=0$ by continuity.
\end{proof}

\begin{proposition}
\label{P2}Let $ \textbf{x}=(x_{1},x_{2},\ldots ),  $   $ \textbf{y}=(y_{1},y_{2},\ldots )  $ be sequences of real
 numbers with $x_{1}=1, y_j=jx_{j-1}$ and $d$ be a positive integer.
 Then, we have the pair of compositional inverse power series%
\begin{align}
H (  t )   &amp; =\frac{t}{1+\sum_{n\geq1}x_{n}\frac{t^{dn}}{n!}%
},\label{20}\\
H^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(  1+\sum
_{n\geq1}\frac{ (  dn )  !}{ (   (  d+1 )  n+1 )
!}B_{ (  d+1 )  n+1,dn+1} ( \textbf{y} )  t^{dn} \biggr)
.\label{21}%
\end{align}
\end{proposition}

\begin{proof}
For $d\geq2,$ choice in Lemma \ref{L2} $x_{n}=0$ if $d\nmid n.$
On using the notations of Lemma \ref{L1} we get $y_n=nx_{n-1}=0 $ if $d\nmid n-1$ and
\[
1+\sum_{n\geq1}x_{n}\frac{t^{dn}}{n!}  =1+\sum_{n\geq1}x_{dn}\frac{t^{dn}}{ (  dn )  !}.
\]
Let  $ \textbf{\emph{z}}=(z_{1},z_{2},\ldots )$ with $z_j=j!y_{d (  j-1 )  +1}/ (d (  j-1 )  +1 )  !.$\\
Then, on using Lemma \ref{L1} we obtain:\\
If $d\nmid n-1$ then $B_{2n+1,n+1} (  \textbf{\emph{y}} )=0$ and if $d\nmid n-1$ we get
\begin{align*}
\frac{(dn)!}{ (  2dn+1 )  !}B_{2dn+1,dn+1} (  \textbf{\emph{y}} ) =
\frac {(dn)!}{((d+1)n+1)!}B_{(d+1)n+1,dn+1} (  \textbf{\emph{z}} ).
\end{align*}
Therefore, the pair of the power series given in Lemma \ref{L2} can be written as
\[
H(t)=\frac{t}{1+\sum_{n\geq1}x_{dn}t^{dn}/ (  dn )  !}, \ \ H^{ \langle -1 \rangle } (  t )=1+\sum_{n\geq1}\frac {(dn)!}{((d+1)n+1)!}B_{(d+1)n+1,dn+1} (  \textbf{\emph{z}} )t^{dn}.
\]
To finish this proof, replace $n!x_{dn}/(dn)!$ by $x_n.$
\end{proof}

\begin{example}
For $ \textbf{x}=(\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots)$ in Proposition \ref{P2} we get%
\[
H (  t )  =\frac{t^{d+1}}{e^{t^{d}}-1},\text{ \ }H^{ \langle
-1 \rangle } (  t )  =t\sum_{n\geq0}\frac{1}{dn+1}\binom{ (  d+1 )  n+1}{dn+1}^{-1}
\genfrac{\{}{\}}{0pt}{}{ (  d+1 )  n+1}{dn+1}\frac{t^{dn}}{n!},
\]
for $ \textbf{x}=(1!, 2!, 3!, \ldots)$ in Proposition \ref{P2} we get%
\begin{equation}
H (  t )  =t (  1-t^{d} )  ,\ \ H^{ \langle
-1 \rangle } (  t )  =t\sum_{n\geq0}\frac{1}{dn+1}\binom{ (
d+1 )  n}{dn}t^{dn}\nonumber
\end{equation}
and for $ \textbf{x}=(\frac{1!}{2}, \frac{2!}{3}, \frac{3!}{4}, \ldots)$ in Proposition \ref{P2} we get%
\begin{equation}
H (  t )  =-\frac{t^{d+1}}{\ln (  1-t^{d} )  }%
,\ \ H^{ \langle -1 \rangle } (  t )  =t\sum_{n\geq0}%
\frac{1}{dn+1}\binom{ (  d+1 )  n+1}{dn+1}^{-1}\genfrac{[}{]}{0pt}{}{ (  d+1 )  n+1}{dn+1}%
\frac{t^{dn}}{n!}.\nonumber
\end{equation}
\end{example}

\begin{theorem}
\label{T2}Let $ \textbf{x}=(x_{1},x_{2},\ldots ) $ be a sequence of real numbers with
$x_{1}=1,$ $r,s,d$ be integers such that $r&gt;\max (  s,- (  d+1 )
s ) , $  $d\geq1$ and let
\begin{align}
V_{n} (  r,s )  =\frac{s}{rn+s}\frac{B_{ (  r+1 )
n+s,\ rn+s} ( \textbf{x} )  }{\binom{ (  r+1 )  n+s}{rn+s}%
},\text{ \ }n\geq1, \ \ V_{0} (  r,s )  =1.\label{24}%
\end{align}
Then, we have the pair of compositional inverses
\begin{align}
H_{2} (  t )  &amp;=t \biggl(  1+\sum_{n\geq1}V_{n} (  r,-s )
\frac{t^{dn}}{n!} \biggr) \label{22} \\%
H_{2}^{ \langle -1 \rangle } (  t )  &amp;=t \biggl(  1+\sum
_{n\geq1}V_{n} (  r+sd,s )  \frac{t^{dn}}{n!} \biggr)  ,\label{23}%
\end{align}
\end{theorem}

\begin{proof}
Let  $ (  f_{n} (  \varphi )   )  $ be a sequence of binomial type such
that $f_{n} (  1 )  =\frac{x_{n+1}}{n+1}.$ Necessarily $f_{n} (\varphi)$
satisfies (\ref{a})  . Then because
\[
\frac{ s }{rn+s} \binom{ (  r+1 )  n+ s }{rn+ s}^{-1}
B_{ (  r+1 )  n+s , rn+ s } ( \textbf{\emph{x}} )=f_{n} ( s;r )
\]
and
\[
H (  t )  =t \biggl(  1+\sum_{n\geq1}\frac{ s }{rn+ s }\frac{B_{(r+1 )n+ s
,\ rn+ s } (  \textbf{\emph{x}} )  }{\binom{ (r+1 )  n+ s }{rn+ s }}\frac{t^{dn}}{n!} \biggr)  ^{-1}
 = t \biggl(  1+\sum_{n\geq1}f_{n} (  s,r )\frac{t^{dn}}{n!} \biggr) ^{-1}
\]
we can state, on using Proposition \ref{P2} and Proposition 1 given \cite{5}, that
\[
H^{ \langle -1 \rangle } (t)  =t\biggl(1+\sum_{n\geq1}\frac{s}{(r+ s d )n+s }
f_{n}((r+ s d )n+s) \frac{t^{dn}}{n!} \biggr)
\]
and by the identity (\ref{a}) we obtain%
\[
H^{ \langle -1 \rangle }(t)=t \biggl(  1+\sum_{n\geq
1}\frac{ s }{( r+sd)n+s }\frac{B_{ (
r+ s d+1 )  n+ s ,\ (r+sd)n+s } (\textbf{\emph{x}} )  }{\binom{ (  r+ s d+1 )
n+ s }{ (  r+ s d )n+ s }}\frac{t^{dn}}{n!} \biggr).
\]
It suffices to remark that from Proposition \ref{P1} we have
\[
\biggl(1+s \sum_{n\geq1}\frac{B_{(r+1)n+s ,\ rn+s}(\textbf{\emph{x}})}{(rn+s)
\binom{ (r+1)n+ s }{rn+s}}\frac{t^{dn}}{n!} \biggr)^{-1}
=1-s \sum_{n\geq1}\frac{B_{(r+1)n-s ,\ rn-s} (\textbf{\emph{x}})}{ (
rn-s)  \binom{(r+1)n-s }{rn-s}}\frac{t^{dn}}{n!}.
\]
\end{proof}

\begin{example}
With $ \textbf{x}=(  1!,2!,\ldots, (  q+1 )  !,0,\ldots )$ in Theorem
\ref{T2}, the identity (\ref{16}) gives
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}\binom
{rn-s}{n}_{q}t^{dn} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{s}{ (  r+sd )  n+s}\binom{ (  r+sd )
n+s}{n}_{q}t^{dn} \biggr)  .
\end{align*}
With $ \textbf{x}=(  1!,2!,\ldots )$  in Theorem \ref{T2} we obtain
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}%
\binom{ (  r+1 )  n-s-1}{rn-s-1}t^{dn} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{s}{ (  r+sd )  n+s}\binom{ (  r+sd+1 )
n+s-1}{ (  r+sd )  n+s-1}\frac{t^{dn}}{n!} \biggr)  .
\end{align*}
With $ \textbf{x}=(  1,1,\ldots )$  in Theorem \ref{T2} we obtain%
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}\binom{ (  r+1 )  n-s}{rn-s}^{-1}
\genfrac{\{}{\}}{0pt}{}{ (  r+1 )  n-s}{rn-s}\frac{t^{dn}}{n!} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{s}{ (  r+sd )  n+s}\binom{ (  r+sd+1 )  n+s}{ (  r+sd )  n+s}^{-1}
\genfrac{\{}{\}}{0pt}{}{ (  r+sd+1 )  n+s}{ (  r+sd )  n+s}\frac{t^{dn}}{n!} \biggr)  .
\end{align*}
With $ \textbf{x}=(  0!,1!,\ldots )$  in Theorem \ref{T2} we obtain%
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}\binom{ (  r+1 )  n-s}{rn-s}^{-1}
\genfrac{[}{]}{0pt}{}{ (  r+1 )  n-s}{rn-s}\frac{t^{dn}}{n!} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{s}{ (  r+sd )  n+s}\binom{ (  r+sd+1 )  n+s}{ (  r+sd )  n+s}^{-1}
\genfrac{[}{]}{0pt}{}{ (  r+sd+1 )  n+s}{ (  r+sd )  n+s}\frac{t^{dn}%
}{n!} \biggr)  .
\end{align*}
Other examples can be derived by using the identities (\ref{31}), (\ref{14})
and (\ref{15}).
\end{example}

\begin{proposition}
\label{P5}Let $r,s,d$ be integers with $r&gt;\max (  s,- (
d+1 )  s )  ,$  $d\geq1$ and $ (  f_{n} (  \varphi )   )  $ be a sequence of 
binomial type.
Then, we have the pair of compositional inverses
\begin{align*}
H_{2} (t) &amp;= t \biggl(  1-\sum_{n\geq1}\frac{\varphi}{\alpha n-\varphi}f_{n} (\alpha n-\varphi )  \frac{t^{dn}}{n!} \biggr), \\
H_{2}^{ \langle -1 \rangle } (t) &amp;=t \biggl(  1+\sum_{n\geq1}\frac{\varphi}{ (  \alpha+d\varphi )  n+\varphi}f_{n} (   (  \alpha+d\varphi )n+\varphi)  \frac{t^{dn}}{n!} \biggr)  .
\end{align*}
\end{proposition}

\begin{proof}
Set $x_{n}=nf_{n-1} (  \varphi;\alpha )  $ in Theorem \ref{T2} to get
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{s\varphi}{ (
\alpha+r\varphi )  n-s\varphi}f_{n} (   (  \alpha+r\varphi )  n-s\varphi )  \frac{t^{dn}%
}{n!} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{s\varphi}{ ( \alpha+r\varphi+sd\varphi )  n+s\varphi}f_{n} (   (
\alpha+r\varphi+sd\varphi)  n+s\varphi)  \frac{t^{dn}}{n!} \biggr)
\end{align*}
and change $s\varphi$ by $\varphi$ $a$ and $\alpha-r\varphi$ and $s\varphi$ by $\varphi.$
\end{proof}

\begin{example}
With $f_{n}(\varphi)=\varphi^{n}$ in Proposition \ref{P5} we get%
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\varphi\sum_{n\geq1} (  \alpha n-\varphi )
^{n-1}\frac{t^{dn}}{n!} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+x\sum_{n\geq1} (   (  \alpha+d\varphi )  n+x )  ^{n-1}%
\frac{t^{dn}}{n!} \biggr)
\end{align*}
With $f_{n}(\varphi)=n!\binom{\varphi}{n}$ in Proposition \ref{P5} we get%
\begin{align*}
H_{2} (  t )   &amp; =t \biggl(  1-\sum_{n\geq1}\frac{\varphi}{\alpha n-\varphi}\binom
{\alpha n-\varphi}{n}t^{dn} \biggr)  ,\\
H_{2}^{ \langle -1 \rangle } (  t )   &amp; =t \biggl(
1+\sum_{n\geq1}\frac{\varphi}{ (  \alpha+d\varphi )  n+x}\binom{ (  \alpha+d\varphi )
n+\varphi}{n}t^{dn} \biggr)  .
\end{align*}
\end{example}

\section{Consequences and complementary results}

We give in this section some properties and complementary remarks on the compositional inverse by taking particular cases of Theorems (\ref{T1}) and (\ref{T2}).

\begin{corollary}
\label{C3}Let $ \textbf{x}=(x_{1},x_{2},\ldots ) $ be a sequence of real numbers and $r,s,d,f$ be
integers with $d\geq1$ and $f\geq2.$ Then, for $r&gt;\max (  s,-2s )  $
the inverse power series given by (\ref{11}) and (\ref{b}) hold for%
\begin{equation*}
U_{n} (  r,s;d )  =s\frac{ (   (  r+1 )  dn+s )
!}{rdn+s}\binom{ (  rd+1 )  n+s}{n}^{-1}\frac{B_{ (  rd+1 )
n+s- (  f-1 )   [  n/f ]  ,rdn+s} (  \textbf{x} )
}{ (   (  rd+1 )  n+s- (  f-1 )   [  n/f ] )  !},%
\end{equation*}
and for $r&gt;\max (  s,- (  d+1 )  s )  $ the inverse power
series given by (\ref{22}) and (\ref{23}) hold for%
\begin{equation*}
V_{n} (  r,s )  = (  rn+s-1 )  !n!s\frac{B_{ (r+1 )  n+s- (  f-1 )   [  n/f ]  ,\ rn+s}
 (\textbf{x} )  }{ (   (  r+1 )  n+s- (  f-1 )   [n/f ]   )  !},%
\end{equation*}
where $ [ \varphi ]  $ is the largest integer $\leq \varphi.$
\end{corollary}

\begin{proof}
Set $y_{1}=1,\ y_{2}=\cdots=y_{f}=0$ and $n=mf+\delta$ $ (  0\leq\delta\leq
f-1 )  $ in Theorem \ref{T1} to get%
\begin{align*}
\frac{B_{n+k,k} (  y_{j} )  }{\binom{n+k}{k}} &amp;  =\frac
{B_{mf+\delta+k,k} (  y_{j} )  }{\binom{mf+\delta+k}{k}}\\
&amp;  =\sum_{i=0}^{mf+\delta}\frac{k!}{ (  k-i )  !}B_{mf+\delta
,i} \biggl(  \frac{y_{j+1}}{j+1} \biggr) \\
&amp;=\frac{ (  mf+\delta )  !}{ (  m+\delta )  !}\sum
_{i=0}^{m+\delta}\frac{k!}{ (  k-i )  !}B_{m+\delta,i} \biggl(
\frac{j!y_{j+f}}{ (  j+f )  !} \biggr)\\
&amp;= \frac{ (  mf+\delta )  !}{ (  m+\delta )  !}%
\binom{m+\delta+k}{k}^{-1}B_{m+\delta+k,k} \biggl(  \frac{j!y_{j+f-1}}{ (j+f-1 )  !} \biggr) \\
 &amp;  =\frac{n!k!}{ (n+k- (  f-1 )   [  n/f ]   )  !}B_{n+k- (f-1 )   [  n/f ]  ,k}
 \biggl(  \frac{j!y_{j+f-1}}{ (j+f-1 )  !} \biggr)  .
\end{align*}
Then, for $x_{n}=n!y_{n+f-1}/ (  n+f-1 )  !$ we obtain%
\begin{align*}
U_{n} (  r,s;d )   &amp; =\frac{ (   (  d-1 )  n )!s}{rdn+s}
\binom{ (  rd+1 )  n+s}{n}^{-1}\binom{ (  r+1 )  dn+s}{ (  d-1 )  n}B_{ (  rd+1 )  n+s,rdn+s} (  y_{j} ) \\
&amp; =s\frac{ (   (  r+1 )  dn+s )  !}{rdn+s}\binom{ (
rd+1 )  n+s}{n}^{-1}\frac{B_{ (  rd+1 )  n+s- (  f-1 )
 [  n/f ]  ,rdn+s} (  x_{j} )  }{ (   (
rd+1 )  n+s- (  f-1 )   [  n/f ]   )  !},\\
V_{n} (  r,s )   &amp; =\frac{s}{rn+s}\frac{B_{ (  r+1 )
n+s,\ rn+s} (  y_{j} )  }{\binom{ (  r+1 )  n+s}{rn+s}%
}= (  rn+s-1 )  !n!s\frac{B_{ (  r+1 )  n+s- (
f-1 )   [  n/f ]  ,\ rn+s} (  x_{j} )  }{ (
 (  r+1 )  n+s- (  f-1 )   [  n/f ]   )  !}.
\end{align*}
\end{proof}

\begin{proposition}
\label{P4}Let $r,s$ be integers, $a,x$ be real numbers and $(f_{n}(\varphi))$ be a
binomial-type polynomials. Then, for $r&gt;\max (  s,-2s )  $ the
inverse power series given by (\ref{11}) and (\ref{b}) hold for%
\begin{equation*}
U_{n} (  r,s;d )  =\frac{ (   (  d-1 )  n )
!s}{rdn+s}\binom{ (  r+1 )  dn+s}{ (  d-1 )  n}%
D_{z=0}^{rdn+s} (  e^{z}f_{n} (   (  rdn+s )  \varphi+z;\alpha )
 )  ,%
\end{equation*}
and for $r&gt;\max (  s,- (  d+1 )  s )  $ the inverse power
series given by (\ref{22}) and (\ref{23}) hold for%
\begin{equation*}
V_{n} (  r,s )  =\frac{s}{rn+s}D_{z=0}^{rn+s} (  e^{z}%
f_{n} (   (  rn+s )  \varphi+z;\alpha )   )  .%
\end{equation*}
\end{proposition}

\begin{proof}
For $y_{n}=nD_{z=0} (  e^{z}f_{n-1} (  \varphi+z;\alpha )   )  $ in
Theorems \ref{T1} and \ref{T2} and use the identity given in \cite[Lemma 1]{8}
by $B_{n,k} (  jD_{z=0} (  e^{z}f_{j-1} (  \varphi+z;\alpha )   )
 )  =\binom{n}{k}D_{z=0}^{k} (  e^{z}f_{n-k} (  k\varphi+z;\alpha ) )  .$
\end{proof}

Theorems \ref{T1} and \ref{T2} remain true when one use a finite
product of power series as follows:

\begin{proposition}
Let $ \textbf{x}=( x_1,x_2, \ldots) $ be a sequence of real numbers with $x_{1}=1,$ $s_{1},\ldots,s_{m},$ $r,s,m$
be integers and $H_{r,s}$ be power series defined by%
\[
H_{r,s} (  t )  =t \biggl(  1-\sum_{n\geq1}\frac{s}{rn-s}%
\frac{B_{ (  r+1 )  n-s,rn-s} ( \textbf{x} )  }{\binom{ (
r+1 )  n-s}{n}}\frac{t^{n}}{n!} \biggr)  ,\ \ r&gt;\max (  s,0 )
:=s^{+}.%
\]
Then, we have%
\[
t\underset{i=1}{\overset{m}{{\displaystyle\prod}
}}\frac{H_{r,s_{i}} (  t )  }{t}=H_{r,s} (  t )  \text{
\ with \ }r&gt;\max (  s^{+},s_{1}^{+},\ldots,s_{m}^{+} )
,\ \ s=\overset{m}{\underset{i=1}{\sum}}s_{i}.%
\]
\end{proposition}

\begin{proof}
Let $ ( f_{n} (  \varphi )  )  $ be a sequence of
polynomials of binomial type such that $f_n(1)=x_{n+1}/(n+1)$  and
let $ (  f (  t;r )   )  ^{\varphi}$ be the exponential generating
function of the sequence of binomial type $ (  f_{n} (  t;r )
 )  .$ Assume that $x_2\neq 0.$ By Proposition 1 given in \cite{5} we get%
\[
\underset{i=1}{\overset{m}{{\displaystyle\prod}
}}\frac{H_{r,s_{i}} (  t )  }{t}=\underset{i=1}{\overset{m}{{\displaystyle\prod}
}} \biggl(  \sum_{n\geq0}f_{n} (  -s_{i};r )  \frac{t^{n}}{n!} \biggr)
=\underset{i=1}{\overset{m}{{\displaystyle\prod}
}} (  f (  t;r )   )  ^{-s_{i}}=\sum_{n\geq0}f_{n} (
-s;r )  \frac{t^{n}}{n!}=\sum_{n\geq0}f_{n} (  -s;r ) \frac{t^{n}}{n!},
\]
and this is exactly
\[
1-\sum_{n\geq1}\frac{s}{rn-s}%
\frac{B_{ (  r+1 )  n-s,rn-s} ( \textbf{\emph{x}})  }{\binom{ (
r+1 )  n-s}{n}}\frac{t^{n}}{n!}=\underset{i=1}{\overset{m}{{\displaystyle\prod}
}}\frac{H_{r,s_{i}} (  t )  }{t}.
\]
The proposition remains true for the case $x_{2}=0$ by continuity.
\end{proof}

\begin{corollary}
\label{C1}Let $ \textbf{y}=( y_1,y_2, \ldots) $ be a sequence of real numbers with $y_{1}=1,$
$r,s,d$ be integers with $d\geq1$ and let $U_{n} (  r,s;d ),$ $V_{n} (  r,s;d )   $
be given by Theorem \ref{T1} and Theorem \ref{T2}. Then, for $r&gt;\max (s,-2s )  $ we have%
\begin{align*}
\underset{k=1}{\overset{n}{\sum}}B_{n,k} (  U_{j} (  r,-s;d )
 )   (  dn+k )  _{k-1}  &amp; =U_{n} (  r+s,s;d )  ,\\
\underset{k=1}{\overset{n}{\sum}}B_{n,k} (  U_{j} (  r+s,s;d )
 )   (  dn+k )  _{k-1}  &amp; =U_{n} (  r,-s;d )  ,
\end{align*}
and for $r&gt;\max (  s,- (  1+d )  s )  $ we have%
\begin{align*}
\underset{k=1}{\overset{n}{\sum}}B_{n,k} (  V_{j} (  r,-s )
 )   (  dn+k )  _{k-1}  &amp; =V_{n} (  r+sd,s )  ,\\
\underset{k=1}{\overset{n}{\sum}}B_{n,k} (  V_{j} (  r+sd,s )
 )   (  dn+k )  _{k-1}  &amp; =V_{n} (  r,-s ) .
\end{align*}
\end{corollary}

\begin{proof}
From Comtet \cite[pp. 151]{4}, for $h (  t )  =t (
1+\underset{n\geq1}{\sum}a_{n}\frac{t^{dn}}{n!} )  ,$ we have
\[
h^{ \langle -1 \rangle } (  t )  =t \biggl(  1+\underset
{n\geq1}{\sum}b_{n}\frac{t^{dn}}{n!} \biggr) \ \ \text{ with } \ \ b_{n}=\underset
{k=1}{\overset{n}{\sum}} (  -1 )  ^{k} (  dn+k )
_{k-1}B_{n,k} (  a_{1},a_{2},\ldots )  .
\]
Then, it suffices to combine with Theorem \ref{T1} by taking $a_{n}%
=U_{n} (  r,-s;d )  $ and $b_{n}=U_{n} (  r+s,s;d )  $ and
combine with Theorem \ref{T2} by taking $a_{n}=V_{n} (  r,-s )  $ and
$b_{n}=V_{n} (  r+sd,s )  $.
\end{proof}

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\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B73; Secondary 30B10, 70H03, 05A10.

\noindent \emph{Keywords: } 
Partial Bell polynomials; binomial-type polynomials;
reciprocals and compositional inverses; Stirling numbers; binomial coefficients.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 28 2011;
revised version received March 15 2012.
Published in {\it Journal of Integer Sequences}, March 25 2012.

\bigskip
\hrule
\bigskip

\noindent
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