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\begin{center}
\vskip 1cm{\LARGE\bf Solution of the Differential Equation $y^{(k)}=e^{ay}$, \\
\vskip .02in
Special Values of Bell Polynomials, and\\
\vskip .1in
$(k,a)$-Autonomous Coefficients}
\vskip 1cm
\Large
Ronald Orozco L\'opez\\
Departamento de Matem\'aticas \\
Universidad de los Andes\\
Bogot\'a 111711\\
Colombia \\
\href{mailto:rj.orozco@uniandes.edu.co}{\tt rj.orozco@uniandes.edu.co} \\
\end{center}

\vskip .2in

\begin{abstract}
We give special values of Bell polynomials by using the power series
solution of the equation $y^{(k)}=e^{ay}$. In addition, we define
complete and partial exponential autonomous functions, exponential
autonomous polynomials, autonomous polynomials, and $(k,a)$-autonomous
coefficients. Finally, we demonstrate the relationship between various numbers
counting combinatorial objects and binomial coefficients, Stirling
numbers of the second kind, and autonomous coefficients.
\end{abstract}



\section{Introduction}
It is a known fact that Bell polynomials are closely related to derivatives of the composition of functions. For example, Fa\`a di Bruno \cite{Faa}, Foissy \cite{Foissy}, and Riordan \cite{Riordan_j} showed that Bell polynomials are a very useful tool in mathematics to represent the $n$-th derivative of the composition of functions. Also, Bernardini and Ricci \cite{Bernardini}, Yildiz et al.\ \cite{Yildiz}, Cayley \cite{Caley}, and Wang \cite{Wang} showed the relationship between Bell polynomials and differential equations. On the other hand, Orozco \cite{Orozco} studied the convergence of the analytic solution of the autonomous differential equation $y^{(k)}=f(y)$ by using the formula of Fa\`a di Bruno. We can then look at differential equations as a source for investigating special values of Bell polynomials.

In this paper, we focus on finding special values of Bell polynomials when the vector field $f(x)$ of the autonomous differential equation $y^{(k)}=f(y)$ is the exponential function. We do not consider the convergence of the solutions, but we will show that well-known numbers such as reduced tangent numbers, Bernoulli numbers, Euler zigzag numbers, Blasius numbers, among others, can be constructed using Bell polynomials. In general, a special class of numbers, which have not yet been studied, are constructed using Bell polynomials. On the other hand, a new family of numbers called $(k,a)$-autonomous coefficients is obtained for each value of $k$. Four conjectures about these numbers are established.

This paper is divided as follows. We begin with a summary of results on complete and partial Bell polynomials, which are used to demonstrate the main results presented here. Next, we introduce the complete and partial exponential autonomous functions, the recurrence relations of these are constructed using Bell polynomials, and some recurrence relations of solutions of various initial value problems are given. In the fourth section, the $(k,a)$-autonomous coefficients are introduced. From these numbers we can obtain the triangular numbers, the 8-sequence numbers of $[1,n]$ with 2 contiguous pairs, among others. We finish this work by studying the cases $k=2,3,4$ for the autonomous differential equation $y^{(k)}=e^{ay}$. 

\section{Bell exponential polynomials}

The following basic results can be found at Comtet \cite[pp.\ 135--136]{Comtent}, and Riordan \cite[pp.\ 35--36, 49]{Riordan_b}. Exponential Bell polynomials are used to encode information on the ways in which a set can be partitioned, hence they are a very useful tool in combinatorial analysis. Bell polynomials are obtained from the derivatives of composite functions and are given by the formula of Fa\`a Di Bruno \cite{Faa}. Bell \cite{Bell}, Gould \cite{Gould_Q}, Mihoubi \cite{Mihoubi} and Qi et al.\  \cite{Qi_1,Qi_2}, provided important results on these polynomials. We start with the definition of partial Bell polynomials.
\begin{definition}
The exponential partial Bell polynomials are the polynomials
\begin{equation*}
    B_{n,k}(x_{1},x_{n},\ldots,x_{n-k+1})
\end{equation*}
in the variables $x_{1},x_{2},\ldots$ defined by the series expansion
\begin{equation}\label{eqn_egf}
\exp\left(u\sum_{j=1}^{\infty}x_{j}\frac{t^{j}}{j!}\right)=1+\sum_{n=1}^{\infty}\frac{t^{n}}{n!}\sum_{k=1}^{n}u^{k}B_{n,k}(x_{1},x_{2},\ldots,x_{n-k+1}).
\end{equation}
\end{definition}
The following result gives the explicit way to calculate the partial Bell polynomials.
\begin{theorem}
The partial or incomplete exponential Bell polynomials are given by
\begin{equation*}
B_{n,k}(x_{1},\ldots,x_{n-k+1})=\sum\frac{n!}{c_{1}!c_{2}!\cdots c_{n-k+1}!}\left(\frac{x_{1}}{1!}\right)^{c_{1}}\cdots\left(\frac{x_{n-k+1}}{(n-k+1)!}\right)^{c_{n-k+1}}
\end{equation*}
where the summation takes place over all integers $c_{1},c_{2},\ldots,c_{n-k+1}\geq0$, such that
\begin{align*}
c_{1}+2c_{2}+\cdots+(n-k+1)c_{n-k+1}&=n,\\
c_{1}+c_{2}+\cdots+c_{n-k+1}&=k.
\end{align*}
\end{theorem}

The following are special cases of partial Bell polynomials and will be very useful for proving results in this paper:
\begin{align*}
B_{n,1}(x_{1},\ldots,x_{n})&=x_{n},\\
B_{n,2}(x_{1},\ldots,x_{n-1})&=\frac{1}{2}\sum_{k=1}^{n-1}\binom{n}{k}x_{k}x_{n-k},\\
B_{n,n-a}(x_{1},\ldots,x_{a+1})&=\sum_{j=a+1}^{2a}\frac{j!}{a!}\binom{n}{j}x_{1}^{n-j}B_{a,j-a}\left(\frac{x_{2}}{2},\cdots,\frac{x_{2(a+1)-j}}{2(a+1)-j}\right),\nonumber\\
&\ 1\leq a< n,
\end{align*}
\begin{align}
B_{n,n}(x_{1})&=x_{1}^{n},\label{eqn_n_n}\\
B_{n,n-1}(x_{1},x_{2})&=\binom{n}{2}x_{1}^{n-2}x_{2},\label{eqn_n_n-1}\\
B_{n,n-2}(x_{1},x_{2},x_{3})&=\binom{n}{3}x_{1}^{n-3}x_{3}+3\binom{n}{4}x_{1}^{n-4}x_{2}^{2},\label{eqn_n_n-2}\\
B_{n,n-3}(x_{1},x_{2},x_{3},x_{4})&=\binom{n}{4}x_{1}^{n-4}x_{4}+10\binom{n}{5}x_{1}^{n-5}x_{2}x_{3}+15\binom{n}{6}x_{1}^{n-6}x_{2}^{3},\label{eqn_n_n-3}\\
B_{n,n-4}(x_{1},x_{2},x_{3},x_{4},x_{5})&=\binom{n}{5}x_{1}^{n-5}x_{5}+5\binom{n}{6}x_{1}^{n-6}(3x_{2}x_{4}+2x_{3}^{2})\nonumber\\
&+105\binom{n}{7}x_{1}^{n-7}x_{2}^{2}x_{3}+105\binom{n}{8}x_{1}^{n-8}x_{2}^{4}.\label{eqn_n,n-4}
\end{align}

Some values of partial Bell polynomials are
\begin{align*}
    B_{n,k}(0!,1!,\ldots,(n-k)!)&=\stirlingfk{n}{k}\ \ (\text{Unsigned Stirling number of the first kind}),\\
    B_{n,k}(1!,\ldots,(n-k)!)&=\binom{n-1}{k-1}\frac{n!}{k!}\ \ (\text{Lah number}),\\
    B_{n,k}(1,1,\ldots,1)&=\stirlingsk{n}{k}\ \ (\text{Stirling number of the second kind}),\\
    B_{n,k}(1,2,\ldots,n-k+1)&=\binom{n}{k}k^{n-k}\ \ (\text{Idempotent number}).
\end{align*}
Then we can see the beautiful relationship that exists between Bell polynomials and numbers like the above.

On the other hand, the partial Bell polynomials can be efficiently computed by means of the recurrence relation
\begin{equation}\label{eqn_recu_bep}
    B_{n,k}(x_{1},\ldots,x_{n-k+1})=\sum_{i=1}^{n-k+1}\binom{n-1}{i-1}x_{i}B_{n-i,k-1}(x_{1},\ldots,x_{n-i-k+2}).
\end{equation}

The definition of complete Bell polynomials is as follows.
\begin{definition}
The sum 
\begin{equation*}
B_{n}(x_{1},x_{2},\ldots,x_{n})=\sum_{k=1}^{n}B_{n,k}(x_{1},x_{2},\ldots,x_{n-k+1})
\end{equation*}
is called the
$n$-th complete exponential Bell polynomial 
with exponential generating function given by
to make $u=1$ in Eq.~(\ref{eqn_egf})
\begin{equation*}
\exp\left(\sum_{m=1}^{\infty}x_{m}\frac{t^{m}}{m!}\right)=\sum_{n=0}^{\infty}B_{n}(x_{1},x_{2},\ldots,x_{n})\frac{t^{n}}{n!}
\end{equation*}
and $B_{0}=1$.
\end{definition}
Some complete Bell polynomials are as follows:
\begin{align*}
B_{1}(x_{1})&=x_{1},\\
B_{2}(x_{1},x_{2})&=x_{1}^{2}+x_{2},\\
B_{3}(x_{1},x_{2},x_{3})&=x_{1}^{3}+3x_{1}x_{2}+x_{3},\\
B_{4}(x_{1},x_{2},x_{3},x_{4})&=x_{1}^{4}+6x_{1}^{2}x_{2}+4x_{1}x_{3}+3x_{2}^{2}+x_{4},\\
B_{5}(x_{1},x_{2},x_{3},x_{4},x_{5})&=x_{1}^{5}+10x_{1}^{3}x_{2}+15x_{1}x_{2}^{2}+10x_{1}^{2}x_{3}+10x_{2}x_{3}+5x_{1}x_{4}+x_{5}.
\end{align*}

\begin{theorem}
The complete Bell polynomials $B_{n}$ satisfy the identity
\begin{equation}\label{eqn_recu_bec}
    B_{n+1}(x_{1},\ldots,x_{n+1})=\sum_{i=0}^{n}\binom{n}{i}B_{n-i}(x_{1},\ldots,x_{n-i})x_{i+1}.
\end{equation}
\end{theorem}

From this it follows that
$    B_{2n+1}(0,x_{2},0,\ldots,0,x_{2n+1})=0$
for all $n\geq0$.

Another useful identity that Bell polynomials fulfill is as follows
\begin{equation}\label{eqn_bell_signo}
    B_{n}(-x_{1},x_{2},-x_{3},\ldots,(-1)^{n-1}x_{n})=(-1)^{n}B_{n}(x_{1},x_{2},x_{3},\ldots,x_{n}).
\end{equation}


\section{Exponential autonomous functions}

We study the solution of the equation 
\begin{equation}\label{eqn_exp_k}
y^{(k)}=e^{ay}    
\end{equation}
for $a\in\mathbb{C}$. Setting $y=u/a$ we obtain the equivalent equation 
\begin{equation}\label{eqn_equi}
u^{(k)}=ae^{u}.   
\end{equation}
Then without loss of generality we focus on Eq.~(\ref{eqn_equi}). Now 
by taking the derivative of Eq.~(\ref{eqn_equi}), we obtain another equation equivalent to Eq.~(\ref{eqn_exp_k}), namely
\begin{equation*}
    u^{(k+1)}=ae^{u}u^{\prime}=u^{(k)}u^{\prime}.
\end{equation*}
Let $(x_{1},x_{2},\ldots,x_{k})$ be the initial value problem $y(0)=x_{1}$, $y^{\prime}(0)=x_{2},\ldots,y^{(k-1)}(0)=x_{k}$. 
In this section we give the general solution and solutions with initial values
$(x,0,0,\ldots,0)$, $(x_{1}+k\ln c,cx_{2},\ldots,c^{k-1}x_{k})$, and $(x_{1},-x_{2},\ldots,x_{2k-1},-x_{2k})$ of Eq.~(\ref{eqn_equi}). We define the complete and partial exponential autonomous functions and the exponential autonomous polynomials, which are the coefficients of the power series solution of Eq.~(\ref{eqn_equi}). Moreover, we find special values of these functions by using Bell polynomials. We begin with the following definition.
\begin{definition}
Take $a\in\C$. Suppose $\x=(x_{1},\ldots,x_{k})$. Let $f_{n}(\x,a)$ 
denote the \textit{$n$-th complete exponential autonomous function of order $k$}, $k\geq1$, recursively defined as
\begin{align}
    f_{0}(\x,a)&=x_{1},\nonumber\\
    f_{1}(\x,a)&=x_{2},\nonumber\\
    &\quad \vdots\label{eqn_def_fac}\\
    f_{k-1}(\x,a)&=x_{k},\nonumber\\
    f_{k}(\x,a)&=ae^{x_{1}},\nonumber\\
    f_{n+k}(\x,a)&=ae^{x_{1}}B_{n}(f_{1}(\x,a),\ldots,f_{n}(\x,a)),\ \ n\geq1, \nonumber
\end{align}
where $B_{n}(y_{1},\ldots,y_{n})$ are the complete Bell polynomials. When $x_{1}=0$, we define the \textit{$n$-th exponential autonomous polynomial} as $q_{n}(x_{2},\ldots,x_{k})=f_{n}(0,x_{2},\ldots,x_{k})$, 
for $n\geq1$. 
\end{definition}
When $a=1$ in the above definition, we write $f_{n}(\x)=f_{n}(\x,1)$. In this section we restrict ourselves to exponential autonomous functions. Exponential autonomous polynomials are discussed in the next section.

The following are complete exponential autonomous functions for $k=1,2,3,4$. They will be very useful in the next section.
When $k=1$, $f_{n}(x,a)=(n-1)!a^{n}e^{nx}$.
When $k=2$, we have
\begin{align*}
f_{0}(x,y,a)&=x,\\
f_{1}(x,y,a)&=y,\\
f_{2}(x,y,a)&=ae^{x},\\
f_{3}(x,y,a)&=aye^{x},\\
f_{4}(x,y,a)&=ae^{x}(ae^{x}+y^{2}),\\
f_{5}(x,y,a)&=ae^{x}(4aye^{x}+y^{3}),\\
f_{6}(x,y,a)&=ae^{x}(4a^{2}e^{2x}+11ay^{2}e^{x}+y^{4}).
\end{align*}
When $k=3$, we have
\begin{align*}
f_{0}(x,y,z,a)&=x,\\
f_{1}(x,y,z,a)&=y,\\
f_{2}(x,y,z,a)&=z,\\
f_{3}(x,y,z,a)&=ae^{x},\\
f_{4}(x,y,z,a)&=aye^{x},\\
f_{5}(x,y,z,a)&=ae^{x}(z+y^{2}),\\
f_{6}(x,y,z,a)&=ae^{x}(ae^{x}+3yz+y^{3}),\\
f_{7}(x,y,z,a)&=ae^{x}(5ye^{x}+3z^{2}+6y^{2}z+y^{4}).
\end{align*}
And finally, when $k=4$, we have
\begin{align*}
f_{0}(x,y,z,w,a)&=x,\\
f_{1}(x,y,z,w,a)&=y,\\
f_{2}(x,y,z,w,a)&=z,\\
f_{3}(x,y,z,w,a)&=w,\\
f_{4}(x,y,z,w,a)&=ae^{x},\\
f_{5}(x,y,z,w,a)&=aye^{x},\\
f_{6}(x,y,z,w,a)&=ae^{x}(z+y^{2}),\\
f_{7}(x,y,z,w,a)&=ae^{x}(w+3yz+y^{3}),\\
f_{8}(x,y,z,w,a)&=ae^{x}(ae^{x}+3z^{2}+4yw+6y^{2}z+y^{4}).
\end{align*}

In the following result, we show that the exponential generating function
of the complete exponential autonomous functions is the solution of
Eq.~(\ref{eqn_equi}).
\begin{theorem}\label{theo_sol}
Let $\textbf{x}=(x_{1},\ldots,x_{k})$. The series 
\begin{equation*}
    E_{k}(t,\textbf{x},a)=\sum_{n=0}^{\infty}f_{n}(\textbf{x},a)\frac{t^{n}}{n!}
\end{equation*}
is the solution of the differential Eq.~(\ref{eqn_equi}).
\end{theorem}
\begin{proof}
Taking the $k$'th derivative of the series $E_{k}(t,\textbf{x},a)$ with respect
to $t$, using the definition of the autonomous functions $f_{n}(\textbf{x},a)$,
and Eq.~(\ref{eqn_egf}), we get
\begin{align*}
    \frac{\partial^{k}E_{k}(t,\textbf{x},a)}{\partial t^{k}}&=\sum_{n=0}^{\infty}f_{n+k}(\textbf{x},a)\frac{t^{n}}{n!}\\
    &=ae^{x_{1}}+\sum_{n=1}^{\infty}ae^{x_{1}}B_{n}(f_{1}(\x,a),\ldots,f_{n}(\x,a))\frac{t^{n}}{n!}\\
    &=e^{ax_{1}}\left(1+\sum_{n=1}^{\infty}B_{n}(f_{1}(\x,a),\ldots,f_{n}(\x,a))\frac{t^{n}}{n!}\right)\\
    &=ae^{x_{1}}e^{E_{k}(t,\textbf{x},a)-x_{1}}\\
    &=ae^{E_{k}(t,\textbf{x},a)}.
\end{align*}
\end{proof}

Now we define the partial exponential autonomous functions.
\begin{definition}
Let $g_{n,i}(\x,a)$ 
denote the \textit{partial exponential autonomous functions} by
\begin{equation*}
    g_{n,i}(\x,a)=B_{n,i}(f_{1}(\x,a),\ldots,f_{n-i+1}(\x,a))
\end{equation*}
with $g_{0,0}(\x,a)=1$, $g_{n,0}(\x,a)=0$, for $n\geq1$, and $g_{0,i}(\x,a)=0$ , for $i\geq1$. Then
\begin{equation*}
    f_{n+k}(\x,a)=ae^{x_{1}}\sum_{i=1}^{n}g_{n,i}(\x,a).
\end{equation*}
\end{definition}

In the following result we establish recurrence relations for the functions $f_{n}(\x,a)$ and $g_{n,i}(\x,a)$. Many important results of this paper are
proved using this theorem.
\begin{theorem}\label{theo_recu}
The autonomous functions $f_{n}(\textbf{x},a)$ and $g_{n,i}(\x,a)$ fulfill the recurrence relations
\begin{equation}\label{eqn_recu_fac}
    f_{n+k+1}(\textbf{x},a)=\sum_{i=0}^{n}\binom{n}{i}f_{n-i+k}(\textbf{x},a)f_{i+1}(\textbf{x},a)
\end{equation}
and
\begin{equation}\label{eqn_recu_fap}
    g_{n,i}(\textbf{x},a)=\sum_{j=1}^{n-i+1}\binom{n-1}{j-1}f_{j}(\textbf{x},a)g_{n-j,i-1}(\textbf{x},a).
\end{equation}
\end{theorem}
\begin{proof}
Setting $y_{j}=f_{j}(\textbf{x},a)$ in Eq.~(\ref{eqn_recu_bep}) and Eq.~(\ref{eqn_recu_bec}) and multiplying these by $ae^{x_{1}}$, we obtain the desired result.
\end{proof}
Now we study the behavior of the functions $f_{n}(\x,a)$ evaluated at $\x=(x_{1},0,0,\ldots,0)$. From previous result we can construct the first important sequence arising from the differential equation (\ref{eqn_equi}).
\begin{theorem}\label{theo_val_f}
The functions $f_{n}(\x,a)$ take the following values at $\x=(x_{1},0,\ldots,0)$
\begin{enumerate}
    \item $f_{kn+1}(x_{1},0,\ldots,0,a)=f_{kn+2}(x_{1},0,\ldots,0,a)=\cdots=f_{kn+k-1}(x_{1},0,\ldots,0,a)=0$, $n\geq0$;
    \item $f_{kn}(x_{1},0,\ldots,0,a)=A_{n}^{(k)}(a)e^{nx_{1}}$, $n\geq1$,
\end{enumerate}
where $A^{(k)}_{1}(a)=1$ and
\begin{equation}\label{eqn_recu_A}
    A^{(k)}_{n+2}(a)=\sum_{i=0}^{n}\binom{kn+k-1}{ki+k-1}A^{(k)}_{n-i+1}(a)A^{(k)}_{i+1}(a),\ \ 
\end{equation}
$n\geq0$, $k\geq1$.
\end{theorem}
\begin{proof}
Let $\mathbf{x}=(x_{1},0,\ldots,0)$. Clearly, $f_{1}(\mathbf{x},a)=0$, $f_{k+1}(\mathbf{x},a)=B_{1}(f_{1}(\mathbf{x},a))=f_{1}(\mathbf{x},a)=0$.  Now suppose it is true that $f_{ki+1}(\mathbf{x},a)=0$ for $2\leq i\leq n-1$. 
By Theorem \ref{theo_recu} we have
\begin{align*}
f_{kn+1}(\mathbf{x},a)&=f_{k(n-1)+k+1}(\mathbf{x},a)\\
&=\sum_{i=0}^{k(n-1)}\binom{k(n-1)}{i}f_{kn-i}(\mathbf{x},a)f_{i+1}(\mathbf{x},a).
\end{align*}
Since the product $f_{kn-i}(\mathbf{x},a)f_{i+1}(\mathbf{x},a)$ contains
the functions $f_{kj+1}(\mathbf{x},a)$, 
then for all $n$ we get that $f_{kn+1}(\mathbf{x},a)=0$.
The same conclusion holds 
for $f_{kn+j}(\mathbf{x},a)$, $j=2,\ldots,k-1$. 

Now we prove $2$. We know that $f_{k}(\mathbf{x},a)=ae^{x_{1}}$, $f_{2k}(\mathbf{x},a)=a^{2}e^{2x_{1}}$ and suppose that $f_{kn}(\mathbf{x},a)=A_{n}^{(k)}(a)e^{nx_{1}}$. Then
\begin{align*}
    f_{kn+k}(\mathbf{x},a)&=f_{(kn-1)+k+1}(\mathbf{x},a)\\
    &=\sum_{i=0}^{kn-1}\binom{kn-1}{i}f_{kn-1-i+k}(\mathbf{x},a)f_{i+1}(\mathbf{x},a)\\
    &=\sum_{i=0}^{n-1}\binom{kn-1}{ki+1}f_{k(n-i)}(\mathbf{x},a)f_{ki+k}(\mathbf{x},a)\\
    &=\sum_{i=0}^{n-1}\binom{kn-1}{ki+1}A_{n-i}^{(k)}(a)e^{(n-i)x_{1}}A_{i+1}^{(k)}(a)e^{(i+1)x_{1}}\\
    &=e^{(n+1)x_{1}}\sum_{i=0}^{n-1}\binom{kn-1}{ki+1}A_{n-i}^{(k)}(a)A_{i+1}^{(k)}(a)\\
    &=e^{(n+1)x_{1}}A_{n+1}^{(k)}(a).
\end{align*}
\end{proof}

It is easy to show that $A_{n}^{(1)}(a)=(n-1)!a^{n}$ when $k=1$. We use 
Eq.~(\ref{eqn_recu_A}) to prove this result. Suppose the result is true that $A_{i}^{(1)}(a)=(i-1)!a^{i}$ for $i$ ranging between $1$ and $n+1$. We have
\begin{align*}
A^{(1)}_{n+2}(a)&=\sum_{i=0}^{n}\binom{n}{i}A^{(1)}_{n-i+1}(a)A^{(1)}_{i+1}(a)\\
&=\sum_{i=0}^{n}\binom{n}{i}(n-i)!a^{n-i+1}i!a^{i+1}\\
&=a^{n+2}\sum_{i=0}^{n}\binom{n}{i}(n-i)!i!\\
&=a^{n+2}n!(n+1)=a^{n+2}(n+1)!.
\end{align*}
We can extend the above result to all $k\geq1$.
\begin{proposition}
For all $k\geq1$ we have
\begin{equation*}
    A_{n}^{(k)}(a)=a^{n}A_{n}^{(k)}(1).
\end{equation*}
\end{proposition}
\begin{proof}
Suppose by induction that $A_{i}^{(k)}(a)=a^{i}A_{i}^{(k)}(1)$ for all $i$ ranging between $1$ and $n+1$, then use the same steps as in the previous proof.
\end{proof}

From the above proposition it follows that 
\begin{equation*}
    E_{k}(t,(x,0,\ldots,0),a)=E_{k}(at,(x,0,\ldots,0),1).
\end{equation*}
Then without loss of generality it is sufficient to study the solution of Eq.~(\ref{eqn_equi}) with initial conditions $y(0)=x$,
$y^{\prime}(0)=y^{\prime}(0)=\cdots=y^{(k-1)}(0)=0$,
and $a=1$ to generate the sequence $A_{n}^{(k)}(1)$.

The following corollary of Theorem \ref{theo_val_f} shows us that the numbers $A_{n}^{(k)}(a)$ can be constructed using Bell polynomials.
\begin{corollary}\label{cor_val_f}
The numbers $A_{n}^{(k)}(a)$ satisfy the recurrence relation
\begin{equation*}
    A_{n}^{(k)}(a)=B_{n-k}(\overbrace{0,\ldots,0}^{k-1},A_{1}^{(k)}(a),\ldots,\overbrace{0,\ldots,0}^{k-1},A_{n-k}^{(k)}(a)),\ \ n\geq1.
\end{equation*}
\end{corollary}

In the following theorem we calculate some special values of the functions
$g_{n.i}(\x,a)$.
\begin{theorem}
Let $\x=(x_{1},0,\ldots,0)$. For all $a\in\R$ we have
\begin{enumerate}
    \item $g_{n,i}(\x,a)=0$, if $k\nmid n$.
    \item $g_{lk,1}(\x,a)=A_{l}^{(k)}(a)e^{lx_{1}}$.
    \item $g_{lk,2}(\x,a)=e^{lx_{1}}\sum_{j=1}^{l}\binom{kl-1}{kj-1}A_{j}^{(k)}(a)A_{l-j}^{(k)}(a)$.
    \item $g_{n,n}(\x,a)=g_{n,n-1}(\x,a)=g_{n,n-2}(\x,a)=g_{n,n-3}(\x,a)=g_{n,n-4}(\x,a)=0$, $k>1$.
\end{enumerate}
\end{theorem}
\begin{proof}
Suppose $k\nmid n$ and $g_{n-j,i-1}(\x,a)=0$ for all $k$ such that $k\nmid j$. Using Theorem \ref{theo_recu} and \ref{theo_val_f} we have that $f_{j}(\x,a)=0$. This proves 1. 

To prove 2 we have
\begin{align*}
    g_{lk,1}(\x,a)&=\sum_{j=1}^{lk}\binom{lk-1}{j-1}f_{j}(\x,a)g_{lk-j,0}(\x,a)\\
    &=\binom{lk-1}{lk-1}f_{lk}(\x,a)g_{0,0}(\x,a)\\
    &=A_{l}^{(k)}(a)e^{lx_{1}}.
\end{align*}

On the other hand, 
\begin{align*}
    g_{lk,2}(\x,a)&=\sum_{j=1}^{lk-1}\binom{lk-1}{j-1}f_{j}(\x,a)g_{lk-j,1}(\x,a)\\
    &=\sum_{j=1}^{l}\binom{lk-1}{jk-1}f_{kj}(\x,a)g_{lk-kj,1}(\x,a)\\
    &=\sum_{j=1}^{l}\binom{lk-1}{jk-1}A_{j}^{(k)}(a)e^{jx_{1}}A_{l-j}^{(k)}(a)e^{(l-j)x_{1}}\\
    &=e^{lx_{1}}\sum_{j=1}^{l}\binom{lk-1}{jk-1}A_{j}^{(k)}(a)A_{l-j}^{(k)}(a).
\end{align*}
Then this proves 3.

To prove 4 we use Eqs.~(\ref{eqn_n_n})--(\ref{eqn_n,n-4}).
\end{proof}

We conclude this section with the following properties of the exponential autonomous functions.
\begin{theorem}\label{theo_contra}
For all $n\geq1$, $k\geq1$ and for all $a,c\in\C$ is fulfilled
\begin{equation*}
    f_{n}(x_{1}+k\ln c,cx_{2},\ldots,c^{k-1}x_{k},a)=c^{n}f_{n}(x_{1},x_{2},\ldots,x_{k},a).
\end{equation*}
\end{theorem}
\begin{proof}
Let $\y=(x_{1}+k\ln c,cx_{2},\ldots,c^{k-1}x_{k})$ and $\x=(x_{1},x_{2},\ldots,x_{k})$. Suppose that the result is true for $i\leq n$. Then
\begin{align*}
f_{n+1}(\y,a)&=f_{(n+1-k)+k}(\y,a)\\
&=ac^{k}e^{x_{1}}B_{n+1-k}(f_{1}(\y,a),\ldots,f_{n+1-k}(\y,a))\\
&=ac^{k}e^{x_{1}}B_{n+1-k}(cf_{1}(\x,a),\ldots,c^{n+1-k}f_{n+1-k}(\x,a))\\
&=ac^{k}c^{n+1-k}e^{x_{1}}B_{n}(f_{1}(\x,a),\ldots,f_{n}(\x,a))\\
&=c^{n+1}f_{n+1}(\x,a).
\end{align*}
\end{proof}

The following is the corollary to Theorem \ref{theo_contra} that allows us to calculate the solutions of Eq.~(\ref{eqn_equi}) when the initial values are $(x_{1}+k\ln c,cx_{2},\ldots,c^{k-1}x_{k})$.
\begin{corollary}
\begin{equation*}
    E_{k}(t,(x_{1}+k\ln c,cx_{2},\ldots,c^{k-1}x_{k}),a)=k\ln c+E_{k}(ct,(x_{1},x_{2},\ldots,x_{k}),a).
\end{equation*}
\end{corollary}
\begin{proof}
Let $\y=(x_{1}+k\ln c,cx_{2},\ldots,c^{k-1}x_{k})$ and $\x=(x_{1},x_{2},\ldots,x_{k})$. From the above theorem and the definition of the function $E_{k}(t,\x,a)$ we have
\begin{align*}
E_{k}(t,\y,a)&=x_{1}+k\ln c+\sum_{n=1}^{\infty}f_{n}(\y,a)\frac{t^{n}}{n!}\\
&=x_{1}+k\ln c+\sum_{n=1}^{\infty}c^{n}f_{n}(\x,a)\frac{t^{n}}{n!}\\
&=x_{1}+k\ln c+\sum_{n=1}^{\infty}f_{n}(\x,a)\frac{(ct)^{n}}{n!}\\
&=k\ln c+E_{k}(ct,\x,a).
\end{align*}
\end{proof}

Finally, we compute $f_{n}(\x,a)$ when $\x=(x_{1},-x_{2},\ldots,x_{2k-1},-x_{2k})$.
\begin{theorem}\label{theo_alt}
For all $n\geq0$ and for all exponential autonomous functions of order $2k$ 
we have
\begin{equation*}
    f_{n}((x_{1},-x_{2},\ldots,x_{2k-1},-x_{2k}),a)=(-1)^{n}f_{n}((x_{1},x_{2},\ldots,x_{2k-1},x_{2k}),a).
\end{equation*}
\end{theorem}
\begin{proof}
Let $\y=(x_{1},-x_{2},\ldots,x_{2k-1},-x_{2k})$, and let $\x=(x_{1},x_{2},\ldots,x_{2k-1},x_{2k})$. Suppose the result is true for all values less than or equal to $n$. Then
\begin{align*}
f_{n+1}(\y,a)&=f_{(n+1-2k)+2k}(\y,a)\\
&=ae^{x_{1}}B_{n+1-2k}(f_{1}(\y,a),f_{2}(\y,a),\ldots,f_{n+1-2k}(\y,a))\\
&=ae^{x_{1}}B_{n+1-2k}(-f_{1}(\x,a),f_{2}(\x,a),\ldots,(-1)^{n+1-2k}f_{n+1-2k}(\x,a))\\
&=(-1)^{n+1-2k}ae^{x_{1}}B_{n+1-2k}(f_{1}(\x,a),f_{2}(\x,a),\ldots,f_{n+1-2k}(\x,a))\\
&=(-1)^{n+1}f_{n}(\x,a).
\end{align*}
\end{proof}

Finally, we have the corollary to Theorem \ref{theo_alt}.
\begin{corollary}\label{cor_sig_alt}
\begin{equation*}
E_{2k}(-t,(x_{1},-x_{2},\ldots,x_{2k-1},-x_{2k}),a)=E_{2k}(t,(x_{1},x_{2},\ldots,x_{2k-1},x_{2k}),a).
\end{equation*}
\end{corollary}
\begin{proof}
From the above theorem and the definition of the function $E_{2k}(t,x,a)$, we have
\begin{align*}
E_{2k}(-t,\y,a)&=\sum_{n=0}^{\infty}f_{n}(\y,a)\frac{(-t)^{n}}{n!}\\
&=\sum_{n=0}^{\infty}(-1)^{n}(-1)^{n}f_{n}(\x,a)\frac{t^{n}}{n!}\\
&=\sum_{n=0}^{\infty}f_{n}(\x,a)\frac{t^{n}}{n!}\\
&=E_{2k}(t,\x,a),
\end{align*}
where $\y=(x_{1},-x_{2},\ldots,x_{2k-1},-x_{2k})$ and $\x=(x_{1},x_{2},\ldots,x_{2k-1},x_{2k})$.
\end{proof}

\section{\texorpdfstring{$(k,a)$}{Lg}-autonomous coefficients}

When $k=1$ we obtain the equation $y^{\prime}=ae^{y}$, which is the easiest to solve for all $a\in\R$. Using the method of separation of variables,
we obtain the solution
\begin{equation*}
    y(t)=-\ln(e^{-x}-at)
\end{equation*}
with initial condition $y(0)=x$. On the other hand, by Theorem \ref{theo_sol} the solution in power series becomes
\begin{align*}
E_{1}(t,x,a)&=x+\sum_{n=1}^{\infty}A_{n}^{(1)}(a)\frac{t^{n}}{n!}\\
&=x+\sum_{n=1}^{\infty}(n-1)!a^{n}e^{nx}\frac{t^{n}}{n!}\\
&=x+\sum_{n=1}^{\infty}\frac{(ae^{x})^{n}}{n}=x-\ln(1-ae^{x}t).
\end{align*}
Now we can use the results of the previous section to prove some results
that are already known.
By the definition of complete exponential autonomous functions
\begin{align*}
n!a^{n+1}e^{a(n+1)x}&=ae^{x}\sum_{i=1}^{n}B_{n,i}(0!a^{1}e^{x},1!a^{2}e^{2x},\ldots,(n-i)!a^{n-i+1}e^{(n-i+1)x})\\
&=ae^{x}a^{n}e^{nx}\sum_{i=1}^{n}B_{n,i}(0!,1!,\ldots,(n-i)!)\\
&=e^{a(n+1)x}a^{n+1}\sum_{i=1}^{n}B_{n,i}(0!,1!,\ldots,(n-i)!)
=e^{a(n+1)x}a^{n+1}\sum_{i=1}^{n}\stirlingfk{n}{i}
\end{align*}
from which we get a result relating factorials and unsigned Stirling number of the first kind:
\begin{equation*}
    n!=\sum_{i=1}^{n}\stirlingfk{n}{i}.
\end{equation*}
Furthermore $g_{n,i}(\x,1)=\stirlingfk{n}{i}$ and by Eq.~(\ref{eqn_recu_fap}) we obtain the following finite-sum identity
\begin{align*}
    \stirlingfk{n+1}{i+1}&=\sum_{j=1}^{n-i+1}\binom{n}{j-1}(j-1)!\stirlingfk{n+1-j}{i}\\
    &=\sum_{j=n}^{i}\frac{n!}{j!}\stirlingfk{j}{i}\\
    &=\sum_{j=i}^{n}\frac{n!}{j!}\stirlingfk{j}{i}\\
    &=\sum_{j=0}^{n}\frac{n!}{j!}\stirlingfk{j}{i}.
\end{align*}
On the other hand, from Eq.~(\ref{eqn_recu_fac}) we obtain the trivial result
\begin{align*}
(n+1)!&=\sum_{i=0}^{n}\binom{n}{i}(n-i)!i!\\
&=\sum_{i=0}^{n}n!.
\end{align*}
The Stirling numbers of the first kind originally arose algebraically from the expansion of the falling factorial
\begin{equation*}
    (x)_{n}=x(x-1)(x-2)\cdots(x-n+1)
\end{equation*}
and in polynomial form is as follows
\begin{equation*}
    (x)_{n}=\sum_{i=0}^{n}(-1)^{n-k}\stirlingfk{n}{i}x^{i}.
\end{equation*}
Analogously, we want to define and study the coefficients of the expansion of the autonomous exponential polynomials $q_{n}(\x,a)$ with $\x=(0,x,x,\ldots,x)$. First we calculate the degree of $q_{n}(\x,a)$.

\begin{proposition}
Let $\x=(0,x,x,\ldots,x)$. Then the degree $\gr$ of $q_{n}(\x,a)$ is
\begin{equation*}
\gr(q_{n}(\x,a))=n-k,\ \ \ n\geq k.
\end{equation*}
\end{proposition}
\begin{proof}
By definition
\begin{align*}
q_{n+k}(\x,a)&=\sum_{i=1}^{n-1}a^{i}g_{n,i}(\x,a)+a^{n}g_{n,n}(\x,a)\\
&=\sum_{i=1}^{n-1}a^{i}g_{n,i}(\x,a)+a^{n}x_{1}^{n}.
\end{align*}
As $\gr(g_{n,i}(\x,a))\leq i$, then $\gr(q_{n+k}(\x,a))=n$.
\end{proof}

We now define the autonomous polynomials and autonomous coefficients.
\begin{definition}
For all $n \geq k$,
let $A_{n}^{(k)}(x,a)=q_{n}(0,x,\ldots,x,a)$ 
denote the \textit{autonomous polynomials} of degree $n-k$.
\end{definition}

Using Eq.~(\ref{eqn_def_fac}) we note that
\begin{equation}\label{eqn_pol_auto}
    A_{n+k}^{(k)}(x,a)=aB_{n}(A_{1}^{(k)}(x,a),\ldots,A_{n}^{(k)}(x,a))
\end{equation}
for all $n\geq1$. 

\begin{definition}
We define the \textit{$(k,a)$-autonomous coefficients}, 
denoted by $\cfauto{n}{i}_{(k,a)}$, as the coefficients of the autonomous polynomials $A_{n+k}^{(k)}(x,a)$, i.e.,
\begin{equation*}
    A_{n+k}^{(k)}(x,a)=\sum_{i=0}^{n}\cfauto{n}{i}_{(k,a)}x^{i}.
\end{equation*}
\end{definition}

Now we give some values of the $(k,a)$-autonomous coefficients.
\begin{theorem}\label{theo_val}
Some values of the coefficients $\cfauto{n}{i}_{(k,a)}$ are
\begin{align}
\cfauto{n}{0}_{(k,a)}&=
\begin{cases}
0,& \text{ if $k\nmid n$;}\\
a^{n/k}A_{n/k}^{(k)}(1),& \text{ if $k\vert n$}, 
\end{cases}\label{eqn_sti_n_0}\\
\cfauto{0}{i}_{(k,a)}&=0, \text{ if $i\geq1$},\label{eqn_sti_0_i}\\
\cfauto{n}{n-l}_{(k,a)}&=a\stirlingsk{n}{n-l},\ k> l+1,\ 0\leq l<n.\label{eqn_sti_n_n-l}
\end{align}
\end{theorem}
\begin{proof}
Eq.~(\ref{eqn_sti_n_0}) follows from Theorem \ref{theo_val_f}.  By definition, $B_{0,i}=0$ for $i\geq1$. Then Eq.~(\ref{eqn_sti_0_i}) is true. Finally, if $k>l+1$,
\begin{align*}
    B_{n,n-l}(A_{1}^{(k)}(x,a),\ldots,A_{l+1}^{(k)}(x,a))&=B_{n,n-l}(x,\ldots,x)\\
    &=\stirlingsk{n}{n-l}x^{n-l}
\end{align*}
and from this Eq.~(\ref{eqn_sti_n_n-l}) follows.
\end{proof}
We now show the relationship between the $(k,1)$-autonomous coefficients and the binomial coefficients.
\begin{theorem}
\begin{align*}
    \cfauto{n+1}{0}_{(k,1)}&=\binom{n}{k-1}\cfauto{n+1-k}{0}_{(k,1)}\nonumber\\
    &+\sum_{h=k+1}^{n}\binom{n}{h}\cfauto{n-h}{0}_{(k,1)}\cfauto{h+1-k}{0}_{(k,1)},
\end{align*}
for $1\leq i\leq n-k+3$
\begin{align*}
    \cfauto{n+1}{i}_{(k,1)}&=\sum_{h=0}^{k-2}\binom{n}{h}\cfauto{n-h}{i-1}_{(k,1)}\nonumber\\
    &+\binom{n}{k-1}\cfauto{n+1-k}{i}_{(k,1)}+\binom{n}{k}\cfauto{n-k}{i-1}_{(k,1)}\nonumber\\
    &+\sum_{h=k+1}^{n}\binom{n}{h}\sum_{j+l=i}\cfauto{n-h}{j}_{(k,1)}\cfauto{h+1-k}{l}_{(k,1)}
\end{align*}
and for $n-k+4\leq i\leq n+1$
\begin{align*}
    \cfauto{n+1}{i}_{(k,1)} &=\sum_{h=0}^{n-i+1}\binom{n}{h}\cfauto{n-h}{i-1}_{(k,1)}\nonumber\\
    &+\binom{n}{k-1}\cfauto{n+1-k}{i}_{(k,1)}+\binom{n}{k}\cfauto{n-k}{i-1}_{(k,1)}\nonumber\\
    &+\sum_{h=k+1}^{n}\binom{n}{h}\sum_{j+l=i}\cfauto{n-h}{j}_{(k,1)}\cfauto{h+1-k}{l}_{(k,1)}.
\end{align*}
\end{theorem}
\begin{proof}
As 
\begin{align*}
    A_{n+1+k}^{(k)}(x,1)&=\sum_{i=0}^{n}\binom{n}{i} A_{n-i+k}^{(k)}(x,1)A_{i+1}^{(k)}(x,1)\\
    &=\sum_{i=0}^{k-2}\binom{n}{i}A_{n-i+k}^{(k)}(x,1)x+\binom{n}{k-1}A_{n+1}^{(k)}(x,1)\\
    &+\binom{n}{k}A_{n}^{(k)}(x,1)x+\sum_{i=k+1}^{n}\binom{n}{i} A_{n-i+k}^{(k)}(x,1)A_{i+1}^{(k)}(x,1),
\end{align*}
then
\begin{align*}
    \sum_{i=0}^{n+1}\cfauto{n+1}{i}_{(k,1)}x^{i}&=\sum_{i=0}^{k-2}\binom{n}{i}\sum_{j=0}^{n-i}\cfauto{n-i}{j}_{(k,1)}x^{j+1}\\
    &+\binom{n}{k-1}\sum_{j=0}^{n+1-k}\cfauto{n+1-k}{j}_{(k,1)}x^{j}
    +\binom{n}{k}\sum_{j=0}^{n-k}\cfauto{n-k}{j}_{(k,1)}x^{j+1}\\
    &+\sum_{i=k+1}^{n}\binom{n}{i}\left(\sum_{j=0}^{n-i}\cfauto{n-i}{j}_{(k,1)}x^{j}\sum_{l=0}^{i+1-k}\cfauto{i+1-k}{l}_{(k,1)}x^{l}\right).
\end{align*}
We multiply the two autonomous polynomials within the last sum to get
\begin{align*}
\sum_{i=0}^{n+1}\cfauto{n+1}{i}_{(k,1)}x^{i}
    &=\sum_{i=0}^{k-2}\binom{n}{i}\sum_{j=0}^{n-i}\cfauto{n-i}{j}_{(k,1)}x^{j+1}\\
    &+\binom{n}{k-1}\sum_{j=0}^{n+1-k}\cfauto{n+1-k}{j}_{(k,1)}x^{j}
    +\binom{n}{k}\sum_{j=0}^{n-k}\cfauto{n-k}{j}_{(k,1)}x^{j+1}\\
    &+\sum_{i=k+1}^{n}\binom{n}{i}\sum_{h=0}^{n+1-k}\left(\sum_{j+l=h}\cfauto{n-i}{j}_{(k,1)}\cfauto{i+1-k}{l}_{(k,1)}\right)x^{h}.
\end{align*}
Then by rearranging the first and fourth sums we obtain
\begin{align*}
    \sum_{i=0}^{n+1}\cfauto{n+1}{i}_{(k,1)}x^{i}
    &=\sum_{i=0}^{n-k+3}\left(\sum_{h=0}^{k-2}\binom{n}{h}\cfauto{n-h}{i-1}_{(k,1)}\right)x^{i}\\
    &+\sum_{i=n-k+4}^{n+1}\left(\sum_{h=0}^{n+1-i}\binom{n}{h}\cfauto{n-h}{i-1}_{(k,1)}\right)x^{i}\\
    &+\binom{n}{k-1}\sum_{i=0}^{n+1-k}\cfauto{n+1-k}{i}_{(k,1)}x^{i}\\
    &+\binom{n}{k}\sum_{i=1}^{n-k+1}\cfauto{n-k}{i-1}_{(k,1)}x^{i}\\
    &+\sum_{i=0}^{n+1-k}\left(\sum_{h=k+1}^{n}\binom{n}{h}\sum_{j+l=i}\cfauto{n-h}{j}_{(k,1)}\cfauto{h+1-k}{l}_{(k,1)}\right)x^{i}.
\end{align*}
For a suitable value of $i$ the desired results are attained.
\end{proof}

Finally, we show without proof the relationship between the $(k,1)$-autonomous coefficients and the Stirling numbers of the second kind. 
\begin{conjecture}
Suppose that $A_{1}^{(k)}(1,1)=\cdots=A_{k}^{(k)}(1,1)=1$. Then
\begin{equation*}
    B_{n}(A_{1}^{(k)}(1,1),\ldots,A_{n}^{(k)}(1,1))=\sum_{i=1}^{n}\stirlingsk{n}{i}A_{i}^{(k)}(1,1),\ \ n\geq1.
\end{equation*}
Then,
\begin{equation}\label{eqn_conj1_1}
    A_{n+k}^{(k)}(1,1)=\sum_{i=1}^{n}\stirlingsk{n}{i}A_{i}^{(k)}(1,1),\ \ n\geq1
\end{equation}
and
\begin{equation*}
    \sum_{i=0}^{n}\cfauto{n}{i}_{(k,1)}=\sum_{j=1}^{k}\stirlingsk{n}{j}+\sum_{j=k+1}^{n}\stirlingsk{n}{j}\sum_{i=0}^{j-k}\cfauto{n}{i}_{(k,1)}.
\end{equation*}
\end{conjecture}
Eq.~(\ref{eqn_conj1_1}) corresponds to the number of shifts left $k-1$ places under Stirling transform.

\section{Sequences related to Eq.~(\ref{eqn_equi})}

We conclude this paper by showing sequences related to Eq.~(\ref{eqn_equi}) for values of $k=2,3,4$. Especially, we show that the numbers known as reduced tangent numbers, Bernoulli numbers, Euler zigzag numbers, Eulerian numbers, Blasius numbers, triangular numbers, number of shifts left 3 places under Stirling transform, and number of 8-sequences of $[1,n]$ with 2 contiguous pairs can be constructed using Bell polynomials, Stirling numbers of the second kind, binomial coefficient and autonomous coefficients. 

\subsection{Case \texorpdfstring{$k=2$}{Lg}}

The first case to be studied is
\begin{equation}\label{eqn_k=2}
y^{\prime\prime}=ae^{y}.
\end{equation}

Eq.~(\ref{eqn_k=2}) is equivalent to the equation
$y^{(3)}=y^{\prime\prime}y^{\prime}$, whose solution is
\begin{equation*}
    y^{\prime}=\sqrt{2}\sqrt{c_{1}}\tan\left(\frac{1}{2}\sqrt{2}\sqrt{c_{1}}t+\frac{1}{2}\sqrt{2}\sqrt{c_{1}}c_{2}\right)
\end{equation*}
and therefore
\begin{equation*}
    y=\ln\left(\sec^{2}\left(\frac{1}{2}\sqrt{2}\sqrt{c_{1}}t+\frac{1}{2}\sqrt{2}\sqrt{c_{1}}c_{2}\right)\right)+c_{3}
\end{equation*}
where $c_{1},c_{2}$ and $c_{3}$ are constants in $\C$. Since we want $y(0)=x$, $y^{\prime}(0)=y$ and $y^{\prime\prime}(0)=ae^{x}$, then
\begin{align*}
\ln\left(\sec^{2}\left(\frac{1}{2}\sqrt{2}\sqrt{c_{1}}c_{2}\right)\right)+c_{3}&=x,\\
\sqrt{2}\sqrt{c_{1}}\tan\left(\frac{1}{2}\sqrt{2}\sqrt{c_{1}}c_{2}\right)&=y,\\
c_{1}\sec^{2}\left(\frac{1}{2}\sqrt{2}\sqrt{c_{1}}c_{2}\right)&=ae^{x}.
\end{align*}
Hence
\begin{align*}
c_{1}&=ae^{x}-\frac{y^{2}}{2},\\
\frac{1}{2}\sqrt{2}\sqrt{c_{1}}c_{2}&=\arctan\left(\frac{y}{\sqrt{2ae^{x}-y^{2}}}\right),\\
c_{3}&=x-\ln\left(1+\frac{y^{2}}{2ae^{x}-y^{2}}\right).
\end{align*}
Thus, the function
\begin{align}\label{eqn_sol_k=2}
    E_{2}(t,(x,y),a)&=x+\ln\sec^{2}\left(\frac{\sqrt{2ae^{x}-y^{2}}t}{2}+\arctan\left(\frac{y}{\sqrt{2ae^{x}-y^{2}}}\right)\right)\nonumber\\
    &-\ln\left(1+\frac{y^{2}}{2ae^{x}-y^{2}}\right)
\end{align}
is the solution of Eq.~(\ref{eqn_k=2}) with initial values
$y(0)=x$ and $y^{\prime}(0)=y$.

The following is a list of particular solutions of Eq.~(\ref{eqn_k=2}) 
that can be obtained from Eq.~(\ref{eqn_sol_k=2}):
\begin{align}
E_{2}(t,(x,0),a)&=x+\ln\left(\sec^{2}\left(\frac{\sqrt{a}e^{x/2}t}{\sqrt{2}}\right)\right),\ \ a>0,\label{eqn_sol_x_0_p}\\
E_{2}(t,(x,0),-a)&=x+\ln\left(\sech^{2}\left(\frac{\sqrt{a}e^{x/2}t}{\sqrt{2}}\right)\right),\ \ a>0,\nonumber\\
E_{2}(t,(0,y),a)&=\ln\sec^{2}\left(\frac{\sqrt{2a-y^{2}}t}{2}+\arctan\left(\frac{y}{\sqrt{2a-y^{2}}}\right)\right)\nonumber\\
    &-\ln\left(1+\frac{y^{2}}{2a-y^{2}}\right),\ \ a>0,\label{eqn_sol_0_y_p}\\
E_{2}(t,(0,y),-a)&=\ln\sech^{2}\left(\frac{\sqrt{2a+y^{2}}t}{2}+\arctanh\left(\frac{y}{\sqrt{2a+y^{2}}}\right)\right)\nonumber\\
    &-\ln\left(1-\frac{y^{2}}{2a+y^{2}}\right),\ \ a>0.\nonumber
\end{align}

We now show the relationship between reduced tangent numbers (\seqnum{A002105} in \cite{sloane}) and Bell polynomials and binomial coefficients.
\begin{theorem}\label{theo_tan_bell}
Let 
\begin{equation*}
(T_{n})_{n\geq1}=(1,1,4,34,496,\ldots)    
\end{equation*}
denote the sequence of reduced tangent numbers. Then
\begin{enumerate}
\item $A_{n}^{(2)}(a)=a^{n}T_{n}$.
\item $T_{n}=B_{n}(0,T_{1},\ldots,0,T_{n-1})$, $n\geq2$.
\item $(-1)^{n}T_{n}=B_{n}(0,-T_{1},\ldots,0,(-1)^{n-1}T_{n-1})$, $n\geq2$.
\item $T_{n+2}=\sum_{i=0}^{n}\binom{2n+1}{2i+1}T_{n-i+2}T_{i+1}$, $n\geq0$.
\end{enumerate}
\end{theorem}
\begin{proof}
Another way to write Eq.~(\ref{eqn_sol_x_0_p}) is
\begin{equation*}
    E_{2}(t,(x,0),a)=x+\sqrt{2}\int_{0}^{\sqrt{a}e^{x/2}t}\tan\left(\frac{u}{\sqrt{2}}\right)du.
\end{equation*}
Then
\begin{align*}
    E_{2}(t,(x,0),a)&=x+\int_{0}^{\sqrt{a}e^{x/2}t}\sum_{n=1}^{\infty}T_{n}\frac{u^{2n-1}}{(2n-1)!}\\
    &=x+\sum_{n=1}^{\infty}T_{n}\frac{(\sqrt{a}e^{x/2}t)^{2n}}{(2n)!}.
\end{align*}
By comparison, $f_{2n}(x,0,a)=a^{n}T_{n}e^{nx}$. Thus parts
1, 2, and 3 follow.

The recurrence relation 4 follows from Eq.~(\ref{eqn_recu_A}).
\end{proof}

In general, Eq.~(\ref{eqn_sol_x_0_p}) is the generating function of the sequence
\begin{equation*}
    (a^{n}T_{n})_{n\geq1}=(a,a^{2},4a^{3},34a^{4},496a^{5},\ldots).
\end{equation*}

On the other hand, it is known that $T_{n}=\frac{2^{n}(2^{2n}-1)\vert b_{2n}\vert}{n}$, where the
$b_{2n}$ are the Bernoulli numbers (\seqnum{A000367}, \seqnum{A002445} in \cite{sloane}). Then Theorem \ref{theo_tan_bell} provides a relation between Bell polynomials and Bernoulli numbers, that is

\begin{equation*}
    \frac{2^{n}(2^{2n}-1)\vert b_{2n}\vert}{n}=B_{n}\left(0,6\vert b_{2}\vert,0,30\vert b_{4}\vert,\ldots,0,\frac{2^{n-1}(2^{2n-2}-1)\vert b_{2n-2}\vert}{n-1}\right).
\end{equation*}

We now show the relationship between Euler zigzag numbers (\seqnum{A000111} in \cite{sloane}) and Bell polynomials, binomial coefficients, and Stirling numbers of the second kind.
\begin{theorem}
Let
\begin{equation*}
(A_{n})_{n\geq0}=(1,1,1,2,5,16,61,272,\ldots)
\end{equation*}
denote the sequence of Euler zigzag numbers. Then
\begin{enumerate}
\item $A_{n+1}=B_{n}(A_{0},\ldots,A_{n-1})$, $n\geq1$.
\item $(-1)^{n}A_{n+1}=B_{n}(-A_{0},A_{1},\ldots,(-1)^{n-1}A_{n-1})$, $n\geq1$.
\item $A_{n+2}=\sum_{i=0}^{n}\binom{n}{i}A_{n-i+1}A_{i}$, $n\geq0$.
\item $A_{n+2}=\sum_{i=1}^{n}\stirlingsk{n}{i}A_{i}$, $n\geq1$.
\item $A_{2n+2}=T_{n}+\sum_{i=1}^{n}\cfauto{2n}{2i}_{(2,1)}$.
\item $A_{2n+3}=\sum_{i=1}^{n}\cfauto{2n+1}{2i+1}_{(2,1)}$.
\end{enumerate}
\end{theorem}
\begin{proof}
From Eq.~(\ref{eqn_sol_0_y_p}) we get
\begin{align*}
    E_{2}(t,(0,1),1)&=\ln\left(\sec^{2}\left(\frac{t}{2}+\frac{\pi}{4}\right)\right)-\ln(2)\\
    &=\ln(\sec^{2}(t)+\sec(t)\tan(t)).
\end{align*}
Then
\begin{align*}
E_{2}(t,(0,1),1)&=\int_{0}^{t}(\sec(u)+\tan(u))du\\
&=\int_{0}^{t}\left(1+\sum_{n=1}^{\infty}A_{n}\frac{u^{n}}{n!}\right)du\\
&=t+\sum_{n=1}^{\infty}A_{n}\int_{0}^{t}\frac{u^{n}}{n!}du\\
&=t+\sum_{n=1}^{\infty}A_{n}\frac{t^{n+1}}{(n+1)!}\\
&=\sum_{n=1}^{\infty}A_{n-1}\frac{t^{n}}{n!}.
\end{align*}
We apply Eq.~(\ref{eqn_def_fac}) to obtain 1. 

By Corollary \ref{cor_sig_alt}, it follows that
\begin{equation*}
    E_{2}(t,(0,-1),1)=E(-t,(0,1),1)=\ln(\sec^{2}(t)-\sec(t)\tan(t)).
\end{equation*}
From Eq.~(\ref{eqn_bell_signo}) we get 2.

Formula 3 follows from Eq.~(\ref{eqn_recu_fac}).

From Eq.~(\ref{eqn_conj1_1}) we get 4. 

Identities 5 and 6 follow because the Euler zigzag numbers are obtained when $x=1$ in $A_{n}^{(2)}(x,1)$. 
\end{proof}

When $k=2$, the exponential autonomous polynomials and the autonomous polynomials match. Some autonomous polynomials of Eq.~(\ref{eqn_k=2}) are as follows:
\begin{align*}
q_{1}(y,a)&=y,\\
q_{2}(y,a)&=a,\\
q_{3}(y,a)&=ay,\\
q_{4}(y,a)&=a(a+y^{2}),\\
q_{5}(y,a)&=a(4ay+y^{3}),\\
q_{6}(y,a)&=a(4a^{2}+11ay^{2}+y^{4}),\\
q_{7}(y,a)&=a(34a^{2}y+26ay^{3}+y^{5}),\\
q_{8}(y,a)&=a(34a^{3}+180a^{2}y^{2}+57ay^{4}+y^{6}).
\end{align*}
From the above we obtain the first $(2,a)$-autonomous coefficients
\begin{table}[H]
    \centering
    \begin{tabular}{l|c|c|c|c|c|c|c}
         \backslashbox{$n$}{$i$}& 0&1 & 2& 3& 4& 5&6  \\\hline
         0& $a$&  &  &  & & &\\
         1& 0&  $a$&  &  &  & &\\
         2& $a^2$&  0&  $a$&  &  & &\\
         3& 0&  $4a^2$&  0&  $a$&  & &\\
         4& $4a^3$&  0&  $11a^2$&  0& $a$ & &\\
         5& 0&  $34a^{3}$&  0&  $26a^{2}$& 0 &$a$ &\\
         6& $34a^{4}$&  0&  $180a^{3}$&  0& $57a^{2}$ &0 & $a$
    \end{tabular}
    \caption{$(2,a)$-autonomous coefficients.}
    
\end{table}


\begin{theorem}
Some values of $(2,a)$-autonomous coefficients are
\begin{equation}
\cfauto{2n}{2i+1}_{(2,a)}=\cfauto{2n+1}{2i}_{(2,a)}=0\label{eqn_par_impar}
\end{equation}
for all $i$.
\end{theorem}
\begin{proof}
Eq.~(\ref{eqn_par_impar}) follows from Theorem \ref{theo_val_f}. 
\end{proof}

\begin{conjecture}
\begin{equation*}
\cfauto{n}{n-2}_{(2,a)}=a^{2}(2^{n}-n-1).    
\end{equation*}
The sequence
\begin{align*}
    2^{n}-n-1&=(0, 0, 1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369,\nonumber\\
    &32752, 65519, 131054, 262125, 524268, 1048555, 2097130,\ldots)
\end{align*}
is known as the Eulerian numbers (\seqnum{A000295} in \cite{sloane}).
\end{conjecture}


\subsection{Case \texorpdfstring{$k=3$}{Lg}}
When $k=3$, we obtain the equation
\begin{equation}\label{eqn_k_3}
    y^{(3)}=ae^{y}.
\end{equation}
Solving Eq.~(\ref{eqn_k_3}) with initial conditions $(0,0,x)$ and $a=-1$, we get the solution of Blasius equation
\begin{equation*}
    u^{(3)}+u^{\prime\prime}u=0.
\end{equation*}
The Blasius equation \cite{Blasius} describes the velocity profile of the fluid in the boundary layer which forms when fluid flows along with a flat plate. Using Theorem \ref{theo_val_f} and Corollary \ref{cor_val_f}, we obtain the following result on Blasius numbers (\seqnum{A018893} in \cite{sloane}).
\begin{theorem}
Let
\begin{equation*}
    (\mathrm{b}_{n})_{n\geq1}=(1,1,11,375,27.897,\ldots)
\end{equation*}
denote the sequence of Blasius numbers. Then
\begin{enumerate}
\item $\mathrm{b}_{n}=B_{n}(0,0,\mathrm{b}_{1},\ldots,0,0,\mathrm{b}_{n-1})$, $n\geq2$.
\item $\mathrm{b}_{n+2}=\sum_{i=0}^{n}\binom{3n+2}{3i+2}\mathrm{b}_{n-i+1}\mathrm{b}_{i+1}$, $n\geq0$.
\end{enumerate}
\end{theorem}

On the other hand, the autonomous polynomials for Eq.~(\ref{eqn_k_3}) are
\begin{align*}
A_{3}^{(3)}(x,a)&=a,\\
A_{4}^{(3)}(x,a)&=ax,\\
A_{5}^{(3)}(x,a)&=a(x+x^{2}),\\
A_{6}^{(3)}(x,a)&=a(a+3x^{2}+x^{3}),\\
A_{7}^{(3)}(x,a)&=a(5ax+3x^{2}+6x^{3}+x^{4}),\\
A_{8}^{(3)}(x,a)&=a(11ax+16ax^{2}+15x^{3}+10x^{4}+x^{5}),\\
A_{9}^{(3)}(x,a)&=a(11a^{2}+84ax^{2}+(42a+15)x^{3}+45x^{4}+15x^{5}+x^{6}),\\
A_{10}^{(3)}(x,a)&=a(117a^{2}x+129ax^{2}+384ax^{3}+(99a+105)x^{4}+105x^{5}+21x^{6}+x^{7})
\end{align*}
and from here we obtain the following table with the first $(3,a)$-autonomous coefficients
\begin{table}[H]
    \centering
    \begin{tabular}{l|c|c|c|c|c|c|c|c}
         \backslashbox{$n$}{$i$}& 0&1 & 2& 3& 4 &5 &6&7\\\hline
         0& $a$&  &  &  & & &\\
         1& 0&  $a$&  &  & & &\\
         2& 0&  $a$&  $a$&  & & &\\
         3& $a^2$&  0&  $3a$&  $a$&  & &\\
         4& 0&  $5a^2$&  $3a$&  $6a$& $a$ & &\\
         5& 0& $11a^2$&  $16a^2$&  $15a$&  $10a$& $a$ &\\
         6& $11a^{3}$& 0&  $84a^2$&  $42a^2+15a$&  $45a$& $15a$ &$a$\\
         7& 0&$117a^{3}$& $129a^{2}$&  $384a^2$&  $99a^2+105a$&  $105a$& $21a$ &$a$
    \end{tabular}
    \caption{$(3,a)$-autonomous coefficients.}
    
\end{table}

\begin{theorem}
Some values of $(3,a)$-autonomous coefficients are
\begin{align*}
    \cfauto{n}{n}_{(3,a)}&=a,\\
    \cfauto{n}{n-1}_{(3,a)}&=a\binom{n}{2}.
\end{align*}
\end{theorem}
\begin{proof}
The results follow from Theorem \ref{theo_val} with $l=0,1$ and by keeping in mind that $\stirlingsk{n}{n-1}=\binom{n}{2}$.
\end{proof}

\begin{conjecture}
\begin{equation*}
\cfauto{n}{n-2}_{(3,a)}=a\binom{\binom{n}{2}}{2}.    
\end{equation*}
The numbers $\cfauto{n}{n-2}_{(3,1)}$ are the triangular numbers
\begin{equation*}
    (0,0,3,15,45,105,210, 378, 630, 990, 1485,\ldots)
\end{equation*}
(\seqnum{A050534} in \cite{sloane}).
\end{conjecture}

Finally, by Eqs.~(\ref{eqn_bell_signo}), (\ref{eqn_pol_auto}), (\ref{eqn_conj1_1}) we have the following theorem.

\begin{theorem}
Let
\begin{equation*}
    (\mathrm{e}_{n})_{n\geq1}=(A_{n}^{(3)}(1,1))_{n\geq1}=(1,1,1,1,2,5,15,53,213,\ldots)
\end{equation*}
denote the number of shifts 3 places left under exponentiation (\seqnum{A007548} in \cite{sloane}). Then
\begin{enumerate}
\item $\mathrm{e}_{n+3}=B_{n}(\mathrm{e}_{1},\ldots,\mathrm{e}_{n})$, $n\geq1$.
\item $(-1)^{n}\mathrm{e}_{n+3}=B_{n}(-\mathrm{e}_{1},\mathrm{e}_{2},\ldots,(-1)^{n-1}\mathrm{e}_{n})$, $n\geq1$.
\item $\mathrm{e}_{n+3}=\sum_{i=1}^{n}\stirlingsk{n}{i}\mathrm{e}_{i}$, $n\geq1$.
\item $\mathrm{d}_{3n}=\mathrm{b}_{n}+\sum_{i=1}^{3n}\cfauto{3n}{i}_{(3,1)}$.
\item $\mathrm{d}_{3n+j}=\sum_{i=1}^{3n+j}\cfauto{3n+j}{i}_{(3,1)}$, $j=1,2$.
\end{enumerate}
\end{theorem}


\subsection{Case \texorpdfstring{$k=4$}{Lg}}

The equation to be studied is
\begin{equation}\label{eqn_k_4}
    y^{(4)}=ae^{y}.
\end{equation}
This equation is not commonly studied in the literature. Here we show the relation of this equation with the number of shifts left 3 places under Stirling transform, and also the relation with the numbers $A_{n}^{(4)}(1)$.
A list of exponential autonomous polynomials of Eq.~(\ref{eqn_k_4}) is as follows:
\begin{align*}
q_{1}(y,z,w,a)&=y,\\
q_{2}(y,z,w,a)&=z,\\
q_{3}(y,z,w,a)&=w,\\
q_{4}(y,z,w,a)&=a,\\
q_{5}(y,z,w,a)&=ay,\\
q_{6}(y,z,w,a)&=a(z+y^{2}),\\
q_{7}(y,z,w,a)&=a(w+3yz+y^{3}),\\
q_{8}(y,z,w,a)&=a(a+3z^{2}+4yw+6y^{2}z+y^{4}).
\end{align*}

From Eq.~(\ref{eqn_recu_A}) we calculate the first few
numbers $A_{n}^{(4)}(1)$,
\begin{align*}
q_{4}(0,0,0,1)&=A_{1}^{(4)}(1)=1,\\
q_{8}(0,0,0,1)&=A_{2}^{(4)}(1)=\binom{3}{3}A_{1}^{(4)}A_{1}^{(4)}=1,\\
q_{12}(0,0,0,1)&=A_{3}^{(4)}(1)=\binom{7}{3}A_{2}^{(4)}A_{1}^{(4)}+\binom{7}{7}A_{1}^{(4)}A_{2}^{(4)}=35,\\
q_{16}(0,0,0,1)&=A_{4}^{(4)}(1)=\binom{11}{3}A_{3}^{(4)}A_{1}^{(4)}+\binom{11}{7}A_{2}^{(4)}A_{2}^{(4)}+\binom{11}{11}A_{1}^{(4)}A_{3}^{(4)}=6140.
\end{align*}
Following Theorem \ref{theo_val_f}, Corollary \ref{cor_val_f} and Eq.~(\ref{eqn_bell_signo}) we have the following recurrence relations for the numbers $A_{n}^{(4)}(1)$.

\begin{theorem}
Let $(\mathrm{c}_{n})_{n\geq1}=(A_{n}^{(4)}(1))_{n\geq1}=(1,1,35,6140,\ldots)$. Then
\begin{enumerate}
\item $\mathrm{c}_{n}=B_{n}(0,0,0,\mathrm{c}_{1},\ldots,0,0,0,\mathrm{c}_{n-1})$, $n\geq2$.
\item $\mathrm{c}_{n+2}=\sum_{i=0}^{n}\binom{4n+3}{4i+3}\mathrm{c}_{n-i+1}\mathrm{c}_{i+1}$, $n\geq0$.
\item $(-1)^{n}c_{n}=B_{n}(0,0,0,-\mathrm{c}_{1},\ldots,0,0,0,(-1)^{n}\mathrm{c}_{n-1})$.
\end{enumerate}
\end{theorem}

The autonomous polynomials associated with Eq.~(\ref{eqn_k_4}) are
\begin{align*}
A_{1}^{(4)}(x,a)&=A_{2}^{(4)}(x,a)=A_{3}^{(4)}(x,a)=x,\\
A_{4}^{(4)}(x,a)&=a,\\
A_{5}^{(4)}(x,a)&=ax,\\
A_{6}^{(4)}(x,a)&=a(x+x^{2}),\\
A_{7}^{(4)}(x,a)&=a(x+3x^2+x^{3}),\\
A_{8}^{(4)}(x,a)&=a(a+7x^2+6x^{3}+x^{4}),\\
A_{9}^{(4)}(x,a)&=a(6ax+10x^2+25x^{3}+10x^{4}+x^5),\\
A_{10}^{(4)}(x,a)&=a(16ax+32ax^2+75x^{3}+65x^{4}+15x^5+x^6),\\
A_{11}^{(4)}(x,a)&=a(36ax+136ax^2+(64a+175)x^3+315x^{4}+140x^{5}+21x^{6}+x^7).
\end{align*}

We now derive recurrence relations of the numbers $A_{n}^{(4)}(1,1)$ using Eqs.~(\ref{eqn_bell_signo}), (\ref{eqn_pol_auto}), (\ref{eqn_conj1_1}).
\begin{theorem}
Let 
\begin{equation*}
    (\mathrm{d}_{n})_{n\geq1}=(A_{n}^{(4)}(1,1))_{n\geq1}=(1,1,1,1,1,2,5,15,53, 222, 1115, 6698,\ldots)
\end{equation*}
denote the number of shifts left 3 places under Stirling transform (\seqnum{A336020} in \cite{sloane}). Then
\begin{enumerate}
\item $\mathrm{d}_{n+4}=B_{n}(\mathrm{d}_{1},\ldots,\mathrm{d}_{n})$, $n\geq1$.
\item $(-1)^{n}\mathrm{d}_{n+4}=B_{n}(-\mathrm{d}_{1},\mathrm{d}_{2},\ldots,(-1)^{n-1}\mathrm{d}_{n})$, $n\geq1$.
\item $\mathrm{d}_{n+4}=\sum_{i=1}^{n}\stirlingsk{n}{i}\mathrm{d}_{i}$, $n\geq1$.
\item $\mathrm{d}_{4n}=\mathrm{c}_{n}+\sum_{i=1}^{4n}\cfauto{4n}{i}_{(4,1)}$.
\item $\mathrm{d}_{4n+j}=\sum_{i=1}^{4n+j}\cfauto{4n+j}{i}_{(4,1)}$, $j=1,2,3$.
\end{enumerate}
\end{theorem}

The following is a table of the first $(4,a)$-autonomous coefficients:
\begin{table}[H]
    \centering
    \begin{tabular}{l|c|c|c|c|c|c|c|c}
         \backslashbox{$n$}{$i$}& 0&1 & 2& 3& 4&5 &6 & 7  \\\hline
         0& $a$&  &  &  & & & \\
         1& 0&  $a$&  &  &  & &\\
         2& 0&  $a$&  $a$&  &  & &\\
         3& 0&  $a$&  $3a$&  $a$&  & & \\
         4& $a^2$&  0&  $7a$&  $6a$& $a$& & \\
         5& 0&  $6a^2$&  $10a$&  $25a$& $10a$& $a$& \\
         6& 0&$16a^2$&  $32a^2$&  $75a$&  $65a$& $15a$& $a$  \\
         7& 0&$36a^2$&  $136a^2$&  $64a+175$&  $315a$& $140a$& $21a$& $a$\\
    \end{tabular}
    \caption{$(4,a)$-autonomous coefficients.}
    
\end{table}

\begin{theorem}
Some values of $(4,a)$-autonomous coefficients are
\begin{align}
    \cfauto{n}{n}_{(4,a)}&=a\label{eqn_sti_4_n_n},\\
    \cfauto{n}{n-1}_{(4,a)}&=a\binom{n}{2}\label{eqn_sti_4_n_n-1},\\
    \cfauto{n}{n-2}_{(4,a)}&=a\stirlingsk{n+2}{n}.\label{eqn_sti_4_n_n-2}
\end{align}
\end{theorem}
\begin{proof}
Eqs.~(\ref{eqn_sti_4_n_n}), (\ref{eqn_sti_4_n_n-1}), (\ref{eqn_sti_4_n_n-2}) arise from Theorem \ref{theo_val} with $l=0,1,2$.
\end{proof}

\begin{conjecture}
\begin{equation*}
    \cfauto{n}{n-3}_{(4,a)}=\frac{5a}{2}(n-1)\binom{n}{5}, \ n\geq5.
\end{equation*}
The sequence
\begin{align*}
    \cfauto{n}{n-3}_{(4,1)}&=(10, 75, 315, 980, 2520, 5670, 11550, 21780, 38610, 65065, \\
    &105105,163800, 247520, 364140, 523260, 736440, 1017450,\\
    &1382535, 1850695,2443980, 3187800, 4111250, 5247450,\ldots)
\end{align*}
counts the number of 8-sequences of $[1,n]$ with 2 contiguous pairs, (\seqnum{A027778} in \cite{sloane}).
\end{conjecture}

\section{Acknowledgment}
The author is thankful to the anonymous referee for helpful comments
and suggestions that helped to improve the paper.

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\bigskip
\hrule
\bigskip

\noindent 2020 {\it Mathematics Subject Classification}:
Primary 34A34; Secondary 11B37, 11B68, 11B73, 11B83. 

\noindent \emph{Keywords: } Bell polynomial, autonomous function, autonomous polynomial, autonomous coefficient, Stirling number.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000111},
\seqnum{A000295},
\seqnum{A000367},
\seqnum{A002105},
\seqnum{A002445},
\seqnum{A007548},
\seqnum{A018893},
\seqnum{A027778},
\seqnum{A050534}, and
\seqnum{A336020}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received April 3 2021;
revised versions received  April 8 2021; July 26 2021; August 1 2021;
August 6 2021.
Published in {\it Journal of Integer Sequences}, August 17 2021.

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

                                                                                

