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\begin{center}
\vskip 1cm{\LARGE\bf
Nonlinear Inverse Relations for Bell \\
\vskip .02in
Polynomials via the Lagrange  \\
\vskip .1in
Inversion Formula
}
\vskip 1cm
Jin Wang\\
Department of Mathematics\\
Zhejiang Normal University\\
Jinhua 321004 \\
P.~R.~China \\
\href{mailto:jinwang2016@yahoo.com}{\tt jinwang2016@yahoo.com}\\
\end{center}


\vskip .2 in

\begin{abstract}
In this paper, by using  the classical Lagrange inversion formula, we establish a nonlinear inverse relation
that involves the Bell polynomials. As
  applications of  this  inverse relation, we not only find a short proof  of  another nonlinear inverse relation due to Birmajer et al.,
  but also deduce some allied combinatorial identities.
  Finally,  we  propose the general problem of finding nonlinear inverse relations  and  give  a  positive solution to it.
\end{abstract}



\section{Introduction}

Throughout this paper, we adopt the same notation of Henrici \cite{Henrici}.  For instance, we use $\mathbb{C}[[x]]$ to denote the ring of formal power series (in short, fps) over  the complex  number field $\mathbb{C}$ and for any $f(x)=\sum_{n\geq 0}a_nx^n\in \mathbb{C}[[x]]$, the coefficient functional
$$\boldsymbol\lbrack x^n\boldsymbol\rbrack f(x)=a_n, n\geq 0.$$
For  convenience,  define
\begin{align*}
 \mathcal{L}_0&=\bigg\{\sum_{n=0}^\infty a_nx^n| a_0\neq 0\bigg\},\\
  \mathcal{L}_1&=\bigg\{\sum_{n=0}^\infty a_nx^n| a_0=0, a_1\neq 0\bigg\}.
\end{align*}
Moreover, for $f(x),g(x)\in \mathbb{C}[[x]]$, $g(x)$ is said to be the composite inverse of $f(x)$ if $f(g(x))=g(f(x))=x$. As conventions, we denote the composite inverse $g(x)$ of $f(x)$ by  $f^{\langle-1\rangle}(x)$.

\begin{lemma}\label{inverxinzhi}
 Given $f(x)\in \mathbb{C}[[x]]$,
  $f(x)$ has the  composite inverse  if and only if $f(x)\in \mathcal{L}_1$.
   \end{lemma}
   We also need the concept of the ordinary Bell polynomials (referenced by  \seqnum{A263633} in the OEIS \cite{Sloane}).
\begin{definition} For integers $n\geq k\geq 0$ and
variables
 $(x_n)_{n\geq  1}$, the sums
 \begin{align}
 \sum_{\sigma_k(n)}\frac{k!}{i_1!i_2!\cdots i_{n-k+1}!}x_{1}^{i_1}x_{2}^{i_2}\cdots x_{n-k+1}^{i_{n-k+1}}\label{bellexpression}
\end{align}
are called  the \textit{ordinary Bell polynomials} in $x_1,x_2,\ldots,x_{n-k+1}$, where $\sigma_k(n)$ denotes the set of  partitions of $n$ with
 $k$ parts, namely, all nonnegative integers $i_1,i_2,\ldots,i_{n-k+1}$ subject to
\begin{align}
\left\{\begin{array}{l}
i_1+i_2+\cdots+i_{n-k+1}=k\\ i_1+2i_2+\cdots+(n-k+1)i_{n-k+1}=n.
\end{array}\right.
\end{align}
We let $B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$  denote (\ref{bellexpression}).
\end{definition}
As of today, the Bell polynomials have played  very important roles in analysis, combinatorics, and number theory.  It should be pointed out here that
 the above Bell polynomials are in agreement with the exponential Bell polynomials \cite[Def., p.\ 133]{comtet} with the specialization $x_n\to x_n/n!$ and multiplied by $n!/k!$.  The exponential Bell polynomials are referenced by  \seqnum{A111785} in the OEIS \cite{Sloane}.

 The ordinary generating function of   the Bell polynomials $B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})$ will be often used in our discussions.
\begin{lemma}\label{final}
For any fps
$f(x)=\sum_{n\geq 1} x_n x^n$,
 it holds that
 \begin{align}f^k(x)=\sum_{n=k}^\infty B_{n,k}(x_1,x_2,\ldots,x_{n-k+1})x^n.\label{bellnewone}\end{align}
 \end{lemma}
\begin{proof} According to the Cauchy product of the fps, for given $f(x)=\sum_{n\geq 1} x_n x^n$, let
 \begin{align*}f^k(x)=\sum_{n=k}^\infty A_{n,k}x^n.\end{align*}
Then
\begin{align*}A_{n,k}=\sum_{\stackrel{i_1+i_2+\cdots+i_k=n}{i_j\geq 1}}x_{i_1}x_{i_2}\cdots x_{i_k}.
\end{align*}
It is easy to see that  $$\max\{i_j\geq 1|i_1+i_2+\cdots+i_k=n\}=n-k+1.$$
Consequently, after a bit of series rearrangement, we have
\begin{align*}A_{n,k}&=\sum_{\stackrel{i_1+2i_2+\cdots+(n-k+1)i_{n-k+1}=n}{i_1+i_2+\cdots+i_{n-k+1}=k}}
\frac{k!}{i_1!i_2!\cdots i_{n-k+1}!}x_{1}^{i_1}x_{2}^{i_2}\cdots x_{n-k+1}^{i_{n-k+1}}\\
&=B_{n,k}(x_1,x_2,\ldots,x_{n-k+1}),
\end{align*}
as claimed.
\end{proof}

 As far as the Bell polynomials are concerned,  it is very natural to study  inverse relations lurking behind it. Hereafter, the word ``inverse"  means a pair of equivalent relations  expressing $(x_n)_{n\geq 1}$ in terms of  the Bell polynomials in variables $(y_n)_{n\geq 1}$  and vice versa. To the best of our knowledge, it is one of the most interesting problems first  posed and solved by Riordan  \cite[Chaps.\ 2 and 3]{riodan},
 and  also investigated  by  Chou  et al.\ \cite{laoshi} and   Mihoubi  \cite{mihoubi}.
 The reader may consult  Riordan \cite[Sect.\ 5.3]{riodan} for  further details and Mihoubi \cite{mihoubi} for many of such inverse relations.
  It is especially noteworthy  that in their paper \cite{birmajer},  via the establishment of many interesting identities for the Bell polynomials,
   Birmajer  et al.\  achieved   the following   somewhat unusual (essentially different from  \cite{laoshi,xrma-huang,riodan})   inverse relation.
\begin{theorem}[{\cite[Thm.\ 17]{birmajer}}]\label{Gosper} Let $B_{n,k}(x_1,x_2,\ldots,x_{n-k})$ denote the Bell polynomials as above. Then for any integers $a,b,m$ with $m\geq 1$, $a^2+b^2\neq 0$, and
 any sequence $(x_m)_{m\geq 1}$, the system of   nonlinear relations
\begin{align}
z_m(b)=\displaystyle\sum_{k=1}^m\frac{a m+bk}{k(a m+b)}{-a m-b\choose k-1}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1})\label{gosper-idi}
\end{align}
is equivalent to the system of nonlinear  relations
\begin{align}
x_m=\displaystyle\sum_{k=1}^m\frac{1}{k}{a m+bk\choose k-1}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b)).\label{gosper-idii}
\end{align}
 \end{theorem}
In the above theorem and what  follows, we use  the notation
${x\choose n}$ to denote the generalized binomial coefficients $(x)_n/n!$ and
$(x)_n$ to  the usual  falling factorial $x(x-1)\cdots(x-n+1).$

Motivated by  Birmajer et al.'s result,  the aim of the present paper is to establish the following nonlinear inverse relation. \footnote{
As anonymous reviewer pointed out that  Theorem  \ref{mainewthm} has  now been generalized to Cor.\ 2 of  \cite{birmajer2}, which appeared in
{\it Discrete  Math.\ } in January 2019. However,  as our argument of Theorem  \ref{mainewthm} shows,  Cor.\ 2 of  \cite{birmajer2} is still a direct consequence of the Lagrange inversion formula. See Section \ref{sect4}.}
\begin{theorem}\label{mainewthm} Let $B_{n,k}(x_1,x_2,\ldots,x_{n-k})$ denote the Bell polynomials as above. For any integers $m\geq 1$ and $a,b\in\mathbb{C}$, and
 any sequence $(x_m)_{m\geq 1}$, define
 \begin{align}
y_m(b)=\displaystyle\frac{1}{am+b}\sum_{k=1}^m{-am-b\choose k}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1})\label{uuu}.
\end{align}
Then
\begin{align}
x_m=\displaystyle\frac{1}{am+1}
\sum_{k=1}^m{-(am+1)/b\choose k}b^kB_{m,k}(y_1(b),y_2(b),\ldots,y_{m-k+1}(b)), \label{uuuvvv}
\end{align}
and vice versa.
\end{theorem}

Later as  we shall see, Theorem \ref{mainewthm} sheds a new light on the mystery of Theorem \ref{Gosper}.
Furthermore, by Theorem \ref{mainewthm}, we easily extend Theorem \ref{Gosper} to the following
\begin{corollary}\label{zxbellrela} With the same notation as in Theorem \ref{Gosper},  for any integers $m\geq j\geq 1$, the system of nonlinear  relations
\begin{align}
&B_{m,j}(z_1(b),z_2(b),\ldots,z_{m-j+1}(b))\label{gosper-idadded}\\
&=
\sum_{k=j}^m\frac{j(a m+bk)}{k(a m+bj)}{-a m-bj\choose k-j}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1})\nonumber
\end{align}
is equivalent to the system of nonlinear  relations
\begin{align}
&B_{m,j}(x_1,x_2,\ldots,x_{m-j+1})\label{gosper-idadded-dual}\\
&=
\sum_{k=j}^m\frac{j}{k}{a m+bk\choose k-j}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b)).\nonumber
\end{align}
\end{corollary}

Corollary \ref{zxbellrela} leads us to a short proof  of another combinatorial identity of \cite{birmajer}.

\begin{corollary}[{\cite[Thm.\ 15 and Rem.]{birmajer}}]\label{lizi} With all assumptions of Theorem \ref{Gosper}, for integers
$m\geq s\geq 1$ and arbitrary $\lambda\in \mathbb{C}$, it holds that
\begin{align}
&\sum_{k=s}^m\frac{1}{k}{\lambda\choose k-s}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1})\label{xxxx}\\
&=\sum_{k=s}^m\frac{1}{k}{\lambda+a m+bk\choose k-s}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b)).\nonumber
\end{align}
\end{corollary}

\section{Proof of Theorem \ref{mainewthm}}
Our  argument for Theorem \ref{mainewthm} is merely based on the  classical Lagrange inversion formula \cite[Thms.\ C and D, p.\ 150]{comtet}. This celebrated formula is now known to be a  basic but useful tool of finding of the composite inverse of fps. We refer the reader to \cite{Gessel,hof, spr2}  for further details.
\begin{lemma}[The Lagrange inversion formula]\label{lagrangeinversion}
 Let $\phi(x)\in \mathcal{L}_0$. Then for any fps $F(x)$, it always holds that \begin{align}
F(x)=\sum_{n=0}^\infty a_n\biggl(\frac{x}{\phi(x)}\biggr)^n,
\end{align}
where
\begin{align}
a_n=\frac{1}{n}\boldsymbol\lbrack x^{n-1}\boldsymbol\rbrack F'(x)\phi^n(x).
\end{align}
\end{lemma}

Next is the full proof of Theorem \ref{mainewthm}.

\begin{proof} We only need to show the theorem for integers $a,b\geq 1$, since both right-hand sides of \eqref{uuu} and \eqref{uuuvvv} are polynomials in $a,b$, the equalities remain true for $a, b\in \mathbb{C}$ provided that they hold for all integers $a,b\geq 1$.
Now  we proceed to establish \eqref{uuuvvv} from \eqref{uuu}.   As such, for integers $a,b\geq 1$,  we may consider
\begin{align}
f(x)=x\big(1+\sum_{m=1}^\infty x_m x^{am}\big).\label{ssss}
\end{align}
Then identity \eqref{uuu} amounts to
\begin{align}\label{123-123-123}
\big(f^{\langle-1\rangle}(x)\big)^b=x^b\big(1+b\sum_{m=1}^\infty y_m(b) x^{am}\big).
\end{align}
The proof for this goes as follows. At first, according to Lemma \ref{inverxinzhi},  it is reasonable to assume that there exists the expansion
\begin{align}\label{123-123}
\big(f^{\langle-1\rangle}(x)\big)^b=\sum_{n=1}^\infty \lambda_n x^{n}.
\end{align}
Further replacing $x$ with $f(x)$ in \eqref{123-123}, we obtain
\begin{align}\label{456-456}
x^b=\sum_{n=1}^\infty\lambda_n f^n(x).
\end{align}
Now we are  able to apply the Lagrange inversion formula to \eqref{456-456}. As a consequence, we arrive at
\begin{align*}
\lambda_n&=\frac{b}{n}\boldsymbol\lbrack x^{n-b}\boldsymbol\rbrack \biggl(\frac{f(x)}{x}\biggr)^{-n}\\
&=\frac{b}{n}\boldsymbol\lbrack x^{n-b}\boldsymbol\rbrack\biggl(1+\sum_{i=1}^\infty{-n\choose i} \biggl(\sum_{m=1}^\infty x_m x^{am}\biggr)^{i}\biggr)\\
&=\frac{b}{n}\boldsymbol\lbrack x^{n-b}\boldsymbol\rbrack\biggl(1+\sum_{i=1}^\infty {-n\choose i}\sum_{m=i}^\infty B_{m,i}(x_1,x_2,\ldots,x_{m-i+1})x^{am}\biggr).
\end{align*}
Relabeling the corresponding parameters,
we conclude that for $n\neq b \pmod a$, $\lambda_n=0$; for $n=b$, $\lambda_n=1$; and for $n=am+b$,
\begin{align*}
\lambda_n=\frac{b}{am+b}
\sum_{k=1}^m{-am-b\choose k}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1})=by_m(b).
\end{align*}
Having \eqref{123-123-123} been shown, by Lemma  \ref{inverxinzhi} again, we now assume that
\begin{align}\label{1233-1233-1233}
f(x)=\sum_{n=1}^\infty \mu_n x^{n}.
\end{align}
Clearly, to show \eqref{uuuvvv} we need
 to express $\mu_n$ in terms of $y_m(b)$. To do this,  we solve  \eqref{123-123-123} for $f^{\langle-1\rangle}(x)$ to obtain
\begin{align}\label{1233-1233-12334}
f^{\langle-1\rangle}(x)=x\big(1+b\sum_{m=1}^\infty y_m(b) x^{am}\big)^{1/b}.
\end{align}
Note that \eqref{1233-1233-1233}, after the replacement of $x$ with $f^{\langle-1\rangle}(x)$,  reduces to
\begin{align*}
x=\sum_{n=1}^\infty \mu_n (f^{\langle-1\rangle}(x))^{n}.
\end{align*}
Now by making use of the Lagrange inversion formula and substituting \eqref{1233-1233-12334} for $f^{\langle-1\rangle}(x)$,
 we easily compute
 \begin{align*}
\mu_n&=\frac{1}{n}\boldsymbol\lbrack x^{n-1}\boldsymbol\rbrack \biggl(\frac{f^{\langle-1\rangle}(x)}{x}\biggr)^{-n}\\
&=\frac{1}{n}\boldsymbol\lbrack x^{n-1}\boldsymbol\rbrack \biggl(1+b\sum_{m=1}^\infty y_m(b) x^{am}\biggr)^{-n/b}\\
&=\frac{1}{n}\boldsymbol\lbrack x^{n-1}\boldsymbol\rbrack \biggl(1+\sum_{i=1}^\infty{-n/b\choose i} \biggl(b\sum_{m=1}^\infty y_m(b)x^{am}\biggr)^{i}\biggr)\\
&=\frac{1}{n}\boldsymbol\lbrack x^{n-1}\boldsymbol\rbrack \biggl(1+\sum_{i=1}^\infty{-n/b\choose i}b^i\sum_{m=i}^\infty B_{m,i}(y_1(b),y_2(b),\ldots,y_{m-i+1}(b))x^{am}\biggr).
\end{align*}
Relabeling the corresponding parameters, we conclude that $\mu_n=0$ when $n\neq 1 \pmod a$ and $\mu_1=1$; for $n=am+1$,
\begin{align*}
\mu_n=\frac{1}{am+1}
\sum_{k=1}^m{-(am+1)/b\choose k}b^kB_{m,k}(y_1(b),y_2(b),\ldots,y_{m-k+1}(b)).
\end{align*}
Comparing \eqref{1233-1233-1233} with \eqref{ssss}, we thereby obtain
\begin{align*}
x_m=\frac{1}{am+1}
\sum_{k=1}^m{-(am+1)/b\choose k}b^kB_{m,k}(y_1(b),y_2(b),\ldots,y_{m-k+1}(b)).
\end{align*}
Hence \eqref{uuuvvv} is proved. It is quite clear that the above derivations are valid in reverse direction. The theorem is therefore proved.
\end{proof}

Some necessary  comments on Theorem \ref{mainewthm} are in order.
\begin{remark}\label{remark88}
  A simple comparison between Theorem \ref{Gosper}  and Theorem \ref{mainewthm} reminds us that the requirement
  that $a,b$ be integral and $a^2+b^2\neq 0$ can be removed by appealing to the fact that all $y_m(b)$ are polynomials in $a$ and $b$, whereas these conditions are very necessary to our argument. For instance, when $a=b=0$, Theorem \ref{mainewthm} reduces,  treated as the limiting case as a and b tend to zero, to a pair of inverse relations
  \begin{align}
  \left\{\begin{array}{ll}
&\displaystyle y_m(0)=\sum_{k=1}^{m}\frac{(-1)^{k}}{k}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1})\\
&\\
&\displaystyle x_m=\sum_{k=1}^{m}\frac{(-1)^k}{k!}B_{m,k}(y_1(0),y_2(0),\ldots,y_{m-k+1}(0)).
\end{array}
\right.
\end{align}
It is obviously equivalent, in terms of generating functions,  to the composite inverse relation between the logarithmic and the exponential fps
  \begin{align}
  \left\{\begin{array}{ll}
&\displaystyle -\sum_{m=1}^{\infty}y_m(0) x^m=\log\biggl(1+\sum_{m=1}^{\infty}x_m x^m\biggr)\\
&\\
&\displaystyle\sum_{m=1}^{\infty}x_m x^m=\exp\biggl(-\sum_{m=1}^{\infty}y_m(0)x^m\biggr)-1.
\end{array}
\right.
\end{align}
\end{remark}
\begin{remark} It is noteworthy that $x_m$'s in \eqref{uuu}  are independent of $b$, suggesting that for $y_i=y_i(1)$,
\begin{align}
x_m=\sum_{k=1}^m\frac{(-1)^{k}}{am+1}{am+k\choose k}B_{m,k}(y_1,y_2,\ldots,y_{m-k+1}).
\end{align}
\end{remark}
To illustrate Theorem \ref{mainewthm}, we consider the special case   $x_m=1/m!$ as an example.  In this case, by making use of the well-known identity for
 the Stirling numbers of the second kind
 \begin{align}
x^m=\sum_{k=0}^mS(m,k)(x)_k,\label{stirling}
\end{align}
we may set up  an interesting  combinatorial identity.
\begin{example} For arbitrary complex numbers $a, b$ and integers $m\geq 1$, we have
\begin{align}
\sum_{k=1}^m(-1)^{m-k}{m-1\choose k-1}\frac{(am+bk)^{m-1}}{
am+1+bk}=\frac{(m-1)! b^{m-1}}{\prod_{k=1}^m(am+1+bk)}.
\end{align}
\end{example}
\begin{proof} As indicated as above,  putting  $x_m=1/m!$ in  \eqref{uuu}  and using \eqref{stirling}, it is easily found that
$
y_m(b)=(-1)^m(am+b)^{m-1}/m!
$.
Observe that $b y_m(b)$'s are just the Abel polynomials \cite[(1c), p.\ 128]{comtet} in $b$, whose generating function turns out to be
\begin{align}
1+b \sum_{m=1}^\infty  y_m(b) (xe^{ax})^m=e^{-bx}.
\end{align}
Hence, from Lemma \ref{final} it follows that
\begin{align*}
&b^k\sum_{m=k}^\infty B_{m,k}(y_1(b),y_2(b),\ldots, y_{m-k+1}(b))(xe^{ax})^m\\
=&(e^{-bx}-1)^k=\sum_{i=0}^k(-1)^{k-i}{k\choose i}e^{-bix}.
\end{align*}
On applying the Lagrange inversion formula to this expansion, we immediately obtain
\begin{align*}
&b^k B_{m,k}(y_1(b),y_2(b),\ldots, y_{m-k+1}(b))\\
&=\frac{1}{m}\boldsymbol\lbrack x^{m-1}\boldsymbol\rbrack \biggl(-b\sum_{i=0}^k(-1)^{k-i}i{k\choose i}e^{-bix}\biggr)e^{-amx}\\
&=(-1)^m\frac{b}{m!}\sum_{i=0}^k(-1)^{k-i}i{k\choose i}(am+bi)^{m-1}.
\end{align*}
Next,  substituting  the last expression for $b^k B_{m,k}(y_1(b),y_2(b),\ldots, y_{m-k+1}(b))$ of  \eqref{uuuvvv}  and interchanging the order of two sums,  we at once obtain
\begin{align}
1=\sum_{i=1}^m\frac{(-1)^{m-i}}{(i-1)!}(am+bi)^{m-1}H(m,i),\label{key}
\end{align}
where
\begin{align*}
H(m,i)&:=\sum_{k=i}^m\frac{b^{1-k}}{(k-i)!}\prod_{j=1}^{k-1}(am+1+bj)\\
&=\frac{b^{1-m}}{(m-i)!}\prod_{j=1,j\neq i}^{m}(am+1+bj).
\end{align*}
This reduces \eqref{key} to the claimed identity.
\end{proof}

\section{Proof of Theorem \ref{Gosper}}

Analogous to the proof of Theorem \ref{mainewthm}, we are able to show 
the nonlinear inverse relation of
Birmajer et al.,
(i.e., Theorem \ref{Gosper}) in a shorter way. To make this point  clear, we need
\begin{lemma}\label{goodyl} Define
\begin{align}
\Gamma_m(j,k,a,b):=\sum_{i=0}^{k}(-1)^{k-i}{k\choose i}\frac{bk-i}{am+bk-i+j}\binom{am+bk-i+j}{j}.\label{deffunc-i}
\end{align}
Then, for $1\leq k\leq j\leq m$,
\begin{align}
\Gamma_m(j,k,a,b)=(-1)^k\frac{k(am+bj)}{j(am+bk+j-k)}{am+bk+j-k\choose j-k},\label{deffunc-ii}
\end{align}
and for $m\geq k\geq j+1$, $\Gamma_m(j,k,a,b)=0$.
\end{lemma}
\begin{proof}
 It suffices, for $1\leq j\leq k-1$, to reformulate the term
\begin{align*}
\frac{bk-i}{am+bk-i+j}{am+bk-i+j\choose j}:=\sum_{s=0}^{j}\eta_s i^s.
\end{align*}
Note that all $\eta_s$ are $i$-free. In this form, by using of the usual
forward-difference operator $\Delta: f(x)\to f(x+1)-f(x)$, it is clear that
\begin{align*}
\Gamma_m(j,k,a,b)=\sum_{s=0}^{j}\eta_s\sum_{i=0}^k(-1)^{k-i}{k\choose i} i^s=\sum_{s=0}^{j}\eta_s\Delta^k(x^s)\bigg|_{x=0}=0.
\end{align*}

 Next we proceed to show \eqref{deffunc-ii} for  $m\geq j\geq k\geq 1$.  For the same reason as indicated at beginning of the proof of Theorem \ref{mainewthm} or Remark \ref{remark88}, we only need to show this under $a \geq 1,b\geq 2$.
For this, we invoke  the classical   Hagen-Rothe formula  \cite{gould} as follows:
\begin{align}
\sum_{i=0}^n \frac{p}{ic+p}{ic+p\choose i} \frac{q}{(n-i)c+q}{(n-i)c+q\choose n-i}&=\frac{p+q}{nc+p+q}{nc+p+q\choose n}.\label{ha}
\end{align}
Evidently, by putting $ c=1,n=am+bk$ in  \eqref{ha}  and then letting $p=-k, q=j$,  we immediately  obtain
 \begin{align*}
&\sum_{i=0}^{am+bk} \frac{-k}{i-k}{i-k\choose i} \frac{j}{am+bk-i+j}{am+bk-i+j\choose am+bk-i}\\&=\frac{-k+j}{am+bk-k+j}{am+bk-k+j\choose am+bk}.
\end{align*}
Multiplying  both sides by $(am+j)/j$ and using the basic relation
\begin{align*}
{i-k \choose i}=\frac{(-1)^i(k-i)}{k}{k\choose i},
\end{align*}
then we obtain
  \begin{align}
&\sum_{i=0}^{k} (-1)^i{k\choose i} \frac{am+j}{am+bk-i+j}{am+bk-i+j\choose \nonumber am+bk-i}\\&=\frac{(-k+j)(am+j)}{j(am+bk-k+j)}{am+bk-k+j\choose am+bk}.\label{eq1}
\end{align}
On the other hand, we have
 \begin{align}
 \sum_{i=0}^k (-1)^i{k \choose i}{am+bk-i+j\choose am+bk-i}={am+bk-k+j\choose am+bk}.\label{eq2}
 \end{align}
This is in fact  the special case of $ c=1$ and $n=am+bk$ in another  Hagen-Rothe formula \cite{gould}
\begin{align}
\sum_{i=0}^n\frac{p}{ic+p}{ic+p \choose i} {(n-i)c+q\choose n-i}&={p+q+nc\choose n},\label{other}
\end{align}
 while $p=-k, q=j$ at the same time.
Upon subtracting \eqref{eq1} from  \eqref{eq2},  we obtain
\begin{align*}
&\sum_{i=0}^{k}(-1)^{i}{k\choose i}\frac{bk-i}{am+bk-i+j}{am+bk-i+j \choose am+bk-i}\\
&=\biggl(1-\frac{(-k+j)(am+j)}{j(am+bk-k+j)}\biggr){am+bk-k+j\choose am+bk}\\
&=\frac{j(bk-k)+(am+j)k}{j(am+bk-k+j)}{am+bk-k+j\choose am+bk}\\
&=\frac{k(am+bj)}{j(am+bk-k+j)}{am+bk-k+j\choose am+bk}.
\end{align*}
Hence, we have
\begin{align*}
&\sum_{i=0}^{k}(-1)^{i}{k \choose i}\frac{bk-i}{am+bk-i+j}{am+bk-i+j\choose am+bk-i}
\\&=\frac{k(am+bj)}{j(am+bk-k+j)}{am+bk-k+j\choose am+bk}.
\end{align*}
 This, after multiplying $(-1)^k$ on both sides,  yields  \eqref{deffunc-ii}.
\end{proof}

Now we are ready to show Birmajer et al.'s  nonlinear inverse relation,
i.e., Theorem \ref{Gosper}, via the use of Theorem \ref{mainewthm}. For
the same reason as indicated at beginning of the proof of Theorem
\ref{mainewthm} or Remark \ref{remark88}, we only need to prove the
theorem for integers $a\geq 1,b\geq 2$.

\begin{proof} First, we  show \eqref{gosper-idii} from \eqref{gosper-idi}. To this end, by taking  both \eqref{gosper-idi} and  \eqref{uuu} in Theorem \ref{mainewthm}  into account and noting that  $x_m$'s  are independent of $b$,  we easily verify that
\begin{align}
z_m(b)=(b-1)y_m(b-1)-by_m(b).\label{yzrelation}
\end{align}
By virtue of this relation, we can show a key fact that
 \begin{align}
B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b))=\sum_{i=0}^k(-1)^{k-i}{k \choose i}(bk-i)y_m(bk-i).\label{bellexpression-new}
\end{align}
The proof goes as follows. By the definition of the Bell polynomials, it holds
\begin{align*}
\biggl(\sum_{m=1}^\infty z_m(b)x^{am}\biggr)^k=\sum_{m=k}^\infty B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b))x^{am}.
\end{align*}
On the other hand, referencing  \eqref{123-123-123} and \eqref{yzrelation}, we may deduce
\begin{align*}
\biggl(\sum_{m=1}^\infty z_m(b)x^{am}\biggr)^k&=\biggl((b-1)\sum_{m=1}^\infty y_m(b-1)x^{am}-b\sum_{m=1}^\infty y_m(b)x^{am}\biggr)^k\\
&=\biggl((f^{\langle-1\rangle}(x)/x\big)^{b-1}-\big(f^{\langle-1\rangle}(x)/x\big)^{b}\biggr)^k\\
&=\sum_{i=0}^k(-1)^{k-i}{k \choose i}\big(f^{\langle-1\rangle}(x)/x\big)^{bk-i}\\
&=\sum_{i=0}^k(-1)^{k-i}{k\choose i}\biggl(1+(bk-i)\sum_{m=1}^\infty y_m(bk-i)x^{am}\biggr)\\
&=\delta_{k,0}+\sum_{m=1}^\infty\biggl( \sum_{i=0}^k(-1)^{k-i}{k \choose i}(bk-i)y_m(bk-i)\biggr)x^{am},\end{align*}
where $\delta_{k,0}$ denotes the usual Kronecker symbol. Upon equating the coefficients of $x^{am}$ on both sides of these expansions, we immediately obtain  \eqref{bellexpression-new}.

Now, with  \eqref{bellexpression-new} in hand,  we commence  evaluating the sum on the right side of   \eqref{gosper-idii}, namely,
\begin{align*}
\Omega&:=\sum_{k=1}^m\frac{1}{k}{am+bk\choose k-1}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b))\\ &=\sum_{k=1}^m\frac{1}{k}{am+bk\choose k-1}\sum_{i=0}^k(-1)^{k-i}{k\choose i}(bk-i)y_m(bk-i).
\end{align*}
Upon inserting the expression of $y_m(b)$ given by \eqref{uuu} into this identity and after some rearrangement,  it  reduces to
\begin{align*}
\Omega=\sum_{j=1}^m\chi_m(j,a,b)B_{m,j}(x_1,x_2,\ldots,x_{m-j+1}),
\end{align*}
where
\begin{align*}
\chi_m(j,a,b):=\sum_{k=1}^{m}\frac{(-1)^j}{k}{am+bk\choose k-1} \Gamma_m(j,k,a,b)
\end{align*}
with  $\Gamma_m(j,k,a,b)$  given by Lemma \ref{goodyl}. By virtue of the same lemma, we further  find
\begin{align*}
&\chi_m(j,a,b)\\
&=\biggl(
\sum_{k=1}^{j}+\sum_{k=j+1}^m\biggr)\frac{(-1)^j}{k}{am+bk\choose k-1}
\Gamma_m(j,k,a,b)\\
&=\frac{1}{j}\sum_{k=1}^{j}\frac{(-1)^{j-k}(am+bj)}{am+bk+j-k}{am+bk\choose k-1}
{am+bk+j-k\choose j-k}\\
&=\frac{1}{j}\sum_{k-1=0}^{j-1}\frac{am+bj}{am+bk}{am+bk\choose k-1}{-am-bk\choose j-k}
=\frac{am+bj}{j(am+b)}{0\choose j-1}=\delta_{j,1}.
\end{align*}
The penultimate equality is the special case    of   \eqref{other}, wherein  setting $p=am+b, q=-am-bj$  and $c=b$, after taking $(n,i)\to (j-1,k-1)$.
The preceding computation simplifies
\[\Omega=B_{m,1}(x_1,x_2,\ldots,x_{m})=x_m.\]
Hence,  \eqref{gosper-idii} is confirmed.

Conversely,  assume  \eqref{gosper-idii} is given. For simplicity, we let $w_m$ denote the right-hand side of  \eqref{gosper-idi}, namely,
\begin{align}
w_m:=\sum_{k=1}^m\frac{a m+bk}{k(a m+b)}{-a m-b\choose k-1}B_{m,k}(x_1,x_2,\ldots,x_{m-k+1}).\label{gosper-widi}
\end{align}
In the sequel, in order to show \eqref{gosper-idi}, we only need to show $z_m(b)=w_m$ for all $m\geq 1$.
To that end, performing as above, we first solve for $x_m$ in \eqref{gosper-widi} and find that
\begin{align}
x_m=\sum_{k=1}^m\frac{1}{k}{am+bk\choose k-1}B_{m,k}(w_1,w_2,\ldots,w_{m-k+1}).\label{gosper-widiv}
\end{align}
In the meantime,  \eqref{gosper-idii}  states that
\begin{align}
x_m=\sum_{k=1}^m\frac{1}{k}{am+bk\choose k-1}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b)).
\end{align}
Therefore,  we  have
\begin{align}
&B_{m,1}(z_m(b))+\sum_{k=2}^m\frac{1}{k}{am+bk\choose k-1}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b))\nonumber\\
&=B_{m,1}(w_m)+\sum_{k=2}^m\frac{1}{k}{am+bk\choose k-1}B_{m,k}(w_1,w_2,\ldots,w_{m-k+1}).\label{wwwmmm}
\end{align}
All that remains is  to  show $z_m(b)=w_m$ by induction on $m$ via the use of \eqref{wwwmmm}. When $m=1$, it is easy to see
that
\[x_1=B_{1,1}(z_1(b))=B_{1,1}(w_1),\]
so $z_1(b)=w_1$. Suppose further  $z_k(b)=w_k$ for all $k\leq m-1$. Then we need to prove that $z_m(b)=w_m$.
Since $m-k+1\leq m-1$ for $k\geq 2$, by the hypothesis, we have $z_i(b)=w_i$ for $1\leq i\leq m-k+1$, thereby
\[B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b))=B_{m,k}(w_1,w_2,\ldots,w_{m-k+1}).\]
Thus we obtain
$B_{m,1}(z_m(b))=B_{m,1}(w_m)$, i.e., $z_m(b)=w_m$.
Summing up, for all integers $m\geq 1, z_m(b)=w_m$, so \eqref{gosper-idi} follows from \eqref{gosper-widi}.  The theorem is proved.
\end{proof}

Next is a sketched proof of Corollary \ref{zxbellrela} by use of Theorem \ref{mainewthm}.

\begin{proof}
This is a direct consequence of substituting \eqref{uuu} of Theorem \ref{mainewthm}  into  \eqref{bellexpression-new}, and then applying Lemma \ref{goodyl} again to the obtained one. Identity \eqref{gosper-idadded-dual} follows from \eqref{gosper-idadded} by using once again the special case of the Hagen-Rothe formula \eqref{other}
\begin{align*}
\sum_{k=r}^{j}\frac{am+bj}{am+bk}{am+bk\choose k-r}{-am-bk\choose j-k}
=\delta_{j,r}.
\end{align*}
All other details are left to the interested reader.
\end{proof}

We end this section by  a short proof of Corollary \ref{lizi}.
\begin{proof} It suffices to compute,  by making use of  \eqref{gosper-idadded-dual}, in a straightforward manner that
\begin{align*}
&\sum_{j=s}^m\frac{1}{j}{\lambda\choose j-s}B_{m,j}(x_1,x_2,\ldots,x_{m-j+1})\\
&=
\sum_{k=s}^m\frac{1}{k}\biggl(\sum_{j=s}^k{\lambda\choose j-s}{a m+bk\choose k-j}\biggr)B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b))\\
&=\sum_{k=s}^m\frac{1}{k}{\lambda+a m+bk\choose k-s}B_{m,k}(z_1(b),z_2(b),\ldots,z_{m-k+1}(b)).
\end{align*}
Note that the inner summation on the second equality can be evaluated in closed form by the Vandermonde convolution formula \cite[(1)]{gould}.
\end{proof}

\section{Recent results on nonlinear inverse relations}\label{sect4}
 Thanks to the anonymous reviewer's suggestions, we are led to the latest  work \cite{birmajer2}. Indeed,  Birmajer et al.~\cite{birmajer2} found a general and more beautiful nonlinear inverse relation for the Bell polynomials. Their result  inspires us to consider the following research problem.

\begin{problem}
For any integers $m\geq 1$, let $p$, $(a_k)_{k=1}^m$, and $(q_k)_{k=1}^m$ be $2m+1$ complex numbers subject to  $$p\neq0,~\sum_{k=1}^m a_k=0,~\sum_{k=1}^m a_kq_k\neq 0.$$  Assume  further that
$F(x)=\sum_{n\geq 1}x_nx^n$ and $\phi(x)=1+\sum_{n\geq 1}y_nx^n$  satisfy
 \begin{align}
F(x/\phi^p(x))=\sum_{k=1}^m a_k\phi^{q_k}(x).\label{equationknown-gen}
\end{align}
Find any relationship between the sequences $(x_n)_{n\geq 1}$ and $(y_n)_{n\geq1}$.
\end{problem}

Using the Lagrange inversion formula as in the proof of Theorem \ref{mainewthm}, we can find a positive solution to this problem as follows.


\begin{theorem}\label{333-new}
With the same notation and assumptions as above,  the system of nonlinear  relations
\begin{align}
x_n=\sum_{k=1}^n \biggl(\sum_{i=1}^m \frac{a_iq_i}{np+q_i}{np+q_i\choose k}\biggr)B_{n,k}(y_1,y_2,\ldots,y_{n-k+1})\label{eqone}
\end{align}
is equivalent to the system of nonlinear  relations
\begin{align}
y_n=\sum_{k=1}^n\frac{\lambda_k(-1/p+n)}{1-pn}B_{n,k}(x_1,x_2,\ldots,x_{n-k+1}),\label{eqtwo-gen}
\end{align}
where $\lambda_n(s)$ is defined by
\begin{align}(n+1)\lambda_{n+1}(s)\sum_{k=1}^m a_kq_k&=-\biggl(ps+n\sum_{k=1}^m a_kq_k\lambda_1(-q_k/p)\biggr)\lambda_n(s) \\
&-\sum_{k=1}^m a_kq_k\sum_{j=1}^{n-1}\lambda_{n+1-j}(-q_k/p)j\lambda_{j}(s).
\nonumber\end{align}
\end{theorem}
 Evidently,  \eqref{eqone} can be obtained via the direct application of the Lagrange inversion formula to \eqref{equationknown-gen}, while \eqref{eqtwo-gen} can be derived by the very similar method as we did
  for  \eqref{uuuvvv} of Theorem \ref{mainewthm}. For instance, when  $m=2$, the nonlinear relation (39)/(40) reduces to
\begin{corollary}\label{333-old}Let $c=r-q\neq 0$. Then the system of nonlinear  relations
\begin{align}
x_n=\frac{1}{c}\sum_{k=1}^n\biggl(\frac{q}{q+np}{q+np\choose k}
-\frac{r}{r+np}{r+np\choose k}\biggr)B_{n,k}(y_1,y_2,\ldots,y_{n-k+1})
\end{align}
is equivalent to the system of nonlinear  relations
\begin{align}
y_n=-\sum_{k=1}^n\frac{1}{k!}
\prod_{j=1}^{k-1}\big(cj+np+k q-1\big)B_{n,k}(x_1,x_2,\ldots,x_{n-k+1}).\label{eqtwo}
\end{align}
\end{corollary}
We remark that under the parametric replacement $(p,q,r)\to (-p/d,-q/d,-r/d)$ and $(x_n,y_n)\to (-dx_n,dy_n)$, Corollary  \ref{333-old} is in agreement with  Cor.\ 2 of \cite{birmajer2}. A full proof of Theorem \ref{333-new}  will be 
given in a forthcoming paper.

\section{Acknowledgments}
The author is very indebted to the anonymous reviewer for  his/her
timely drawing our attention to the latest paper \cite{birmajer2} by
Birmajer et al.,  which greatly improved the exposition of this paper.
Thanks are also due to Professors X. R. Ma and R. Z. Wei for helpful
discussions and suggestions. This  work was supported by the National
Natural Science Foundation of China [Grant No.~11471237].

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\bibitem{gould}H. W. Gould, Some generalizations of Vandermonde's convolution,
\textit{ Amer.  Math.  Monthly} \textbf{63} (1956), 84--91.
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\end{thebibliography}


\bigskip
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\noindent 2010 {\it Mathematics Subject Classification}: Primary 05A10; Secondary 05A19.

\bigskip
\noindent \emph{Keywords:}
Bell polynomial, inverse relation, Lagrange inversion formula.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A111785},
\seqnum{A133932},
\seqnum{A134264},
\seqnum{A178867},
\seqnum{A187082},
and
\seqnum{A263633}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  July 9 2018;
revised versions received  October 17 2018; March 25 2019; March 31 2019.
Published in {\it Journal of Integer Sequences}, May 22 2019.

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