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	\begin{center}
		\vskip 1cm{\LARGE\bf 
		The Golden Ratio, Factorials, and the Lambert $W$ Function\\
			\vskip 0.1in
			
		}
		\vskip 1cm
		\large
		Aleksandar Petojevi\'{c}, Marjana Gorjanac Ranitovi\'{c}, Dragan Rastovac,  
		and Milinko Mandi\'{c}\\
		University in Novi Sad\\ Faculty of Education in Sombor\\ 
		Podgori\v cka 4 \\
		25000 Sombor\\ Serbia\\
		\href{mailto:apetoje@uns.ac.rs}{\tt apetoje@uns.ac.rs} \\
		\href{mailto:ranitovicm@uns.ac.rs}{\tt ranitovicm@uns.ac.rs}\\
		\href{mailto:rastovacd@gmail.com}{\tt rastovacd@gmail.com}\\
		\href{mailto:milinmand@gmail.com}{\tt milinmand@gmail.com}\\[5mm]
    \end{center}
	
	\begin{abstract}
	We study the relationship of the integer sequence \seqnum{A214048}
	with the Lambert $W$ function and the left factorial numbers
	\seqnum{A003422}.
	\end{abstract}
	
	\section{Notation and Introduction}
	We use the following notation \cite{So}:
		\begin{itemize}
			\item $\mathbb{N}$: the set of positive integers.
			\item $\mathbb{N}_{0}$: the set of non-negative integers.
			\item $\mathbb{Z}_{0}^-$: the set of negative integers and zero.
			\item $\mathbb{R}$: the set of real numbers.
			\item $\mathbb{R}^+$: the set of real positive numbers.
			\item $\mathbb{C}$: the set of complex numbers.
			\item $L_m$: the $m$'th Lucas number \cite[\seqnum{A000032}]{IS}.
			\item $F_m$: the $m$'th Fibonacci number \cite[\seqnum{A000045}]{IS}.
			\item $\AM_m$: the $m$'th associated Mersenne number \cite[\seqnum{A001350}]{IS}. 
			\item $\varphi$: the golden ratio, $\varphi=\frac{1+\sqrt5}{2}$.
			\item $\lfloor \cdot \rfloor$: the floor function of a real number, which is the largest integer not exceeding that real number.
			\item $W(\cdot)$: the Lambert $W$ function \cite{Mez} is the solution to the equation $xe^{x} = z$ for  $z\in\mathbb C$.
			\item $\re (\cdot)$: the real part of the complex number.
			\item $\Gamma(\cdot)$: the gamma function, 
			defined by \cite[p.\ 1]{Sr}  $$\Gamma(z)=\int_{0}^{+\infty}e^{-t}t^{z-1}dt \quad (\re(z)&gt;0).$$
			\item $ \psi^{(m)}(\cdot)$: the polygamma function of order $m$, \cite[p.\ 22]{Sr}, defined by
			\begin{align*}
				\psi^{(m)}(z) &amp;=\frac{\partial^m }{\partial z^m}\psi(z)=\frac{\partial^{m+1}}{\partial z^{m+1}}\ln\bigl(\Gamma(z)\bigr),\\
				\psi^{(0)}(z) &amp;=\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}\qquad (m\in \mathbb{N}_{0}, \,z\in \mathbb C\backslash \mathbb Z_0^-).
			\end{align*}
			\item $\arg(\cdot)$: the argument of the complex number.
			\item $ \gamma(\cdot, \cdot)$: the incomplete gamma function, \cite[p.\ 11]{Sr}, defined by
			$$\gamma(s,a)=\int_{0}^{a}e^{-t}t^{s-1}dt \quad (\re(s)&gt;0,\, |\arg(a)|&lt;\pi).$$
		 \end{itemize}
	
	Next, let us define the function
	$$!z(x)=\int_{0}^{x}e^{-t}\frac{t^{z}-1}{t-1}dt\quad (x\in \mathbb R,\, z\in \mathbb C).$$
	On the basis of known properties of the Lambert $W$ function and the sequence \cite[\seqnum{A214048}]{IS}, we determine $x\in\mathbb R$ so that
	$$!n(x)&lt;\ !n&lt; \ !n(x)+1\quad (n\in\mathbb N).$$

	\section{Connection to the Lambert $W$ function}
	 As a multi-valued function, $W(z)$ has infinitely many complex branches and two real branches \cite[p.\ 2]{Lo}:
	\begin{align*}
		W_0 &amp;:[-1/e, \infty) \rightarrow [-1, \infty),\\
		W_{-1} &amp;:[-1/e, 0)  \rightarrow (-\infty, -1].
	\end{align*}
	In what follows, we consider only the real branch $W_{-1}(z)$ of the Lambert  $W$ function. The function $W_{-1}(z)$ strictly decreases on $(-1/e, 0)$ and $W_{-1}(-1/e) = -1$. We also have the following well-known inequality~\cite[p.\ 3]{Lo} for the function $W_{-1}(z)$:
	\begin{equation}\label{1.3}
		\frac{e\ln(-x)}{e-1}\leq W_{-1}(x)\leq \ln(-x)-\ln(-\ln(-x))\quad (x\in[-1/e, 0)).
	\end{equation}

	In proving one of the main results of the paper, the polygamma function of order $m$ plays a key role. The polygamma function satisfies the inequality
	\begin{equation}\label{1.4}
		\frac{(m-1)!}{x^m}+\frac{m!}{2x^{m+1}}\leq(-1)^{(m+1)}\psi^{(m)}(x)\leq\frac{(m-1)!}{x^m}+\frac{m!}{x^{m+1}}\quad (m\geq 1,\, x&gt;0).
	\end{equation}
	
	\subsection{The real sequence $(a_n)_{n\geq 4}$}
	We have $-\frac{1}{n-1}(4/7)^\frac{1}{n-1} \in \bigl(-\frac{1}{e}, 0\bigr)$, $(4 \leq n\in\mathbb N)$. Hence the equation $ xe^x=-\frac{1}{n-1}(4/7)^\frac{1}{n-1}$ has two real solutions, namely
	$ x=W_0\bigl(-\frac{1}{n-1}(4/7)^\frac{1}{n-1}\bigr)$ and $ x=W_{-1}\bigl(-\frac{1}{n-1}(4/7)^\frac{1}{n-1}\bigr)$.
	Let us define the sequence of real positive numbers $(a_n)_{n\geq 4}$ as follows:
	\begin{equation}\label{2.2}
		a_n=-(n-1)W_{-1}\biggl(-\frac{1}{n-1}(4/7)^{\frac{1}{n-1}}\biggr)\quad (a_n\in \mathbb R^+).
	\end{equation}
	\begin{remark}\label{re:0}
		Note that the sequence $a_n$ is strictly increasing. We have
		\[a_n^{-(n-1)}e^{a_n}=a_{n+1}^{-n}e^{a_{n+1}}\,\, \text{and}\,\, a_{n+1} = a_n+\ln \biggl( a_{n+1}\biggl(\frac{a_{n+1}}{a_{n}}\biggr)^{n-1} \biggr).\]
	\end{remark}
	
	\subsection{The golden sequence $(\alpha_n)_{n\geq 1}$}
	In 2012,  Kimberling \cite{IS} defined the {\it golden sequence}  $(\alpha_n)_{n\geq 1}$ \cite[\seqnum{A214048}]{IS} by
	\begin{equation}\label{2.0}
		\alpha_n=m \Longleftrightarrow \varphi^{m-1}\leq n!\leq\varphi^{m}\quad (m\in \mathbb N),
	\end{equation}
	or 
	$$\alpha_1=1,\quad \alpha_n=m \Longleftrightarrow L_{m-1}\leq n!&lt;\L_m\quad (1&lt;n\in\mathbb N).$$
	Definition \eqref{2.0} produces the following lemma:
	\begin{lemma}\label{l.1.0} The following inequalities hold:
		\[\alpha_{n-1}+\biggl\lfloor \frac{\ln n}{\ln \varphi }\biggr\rfloor \leq \alpha_n \leq\alpha_{n-1}+\left\lfloor\frac{\ln n}{\ln \varphi }\right\rfloor+1\quad (1&lt;n \in \mathbb N).\]
	\end{lemma}

	Consider the numbers $\lfloor\frac{\ln n}{\ln \varphi }\rfloor$ \cite[\seqnum{A371672}]{IS}. For this purpose, we consider the integer sequence $(d_m)_{m\geq 0}$ \cite[\seqnum{A181716}]{IS} (see also \cite[\seqnum{A098600}]{IS}) which  we define as follows:
	\[ d_m=d_{m-1}+d_{m-2}+(-1)^m,\quad d_0=0,\,d_1=1.\]
	The following equalities hold \cite{IS}:
	\[d_{m}=F_{m-2}+F_{m}+(-1)^m=\AM_{m-1}+1\quad (1&lt;m\in \mathbb N).\]
	\begin{lemma}\label{l.1.1} The number $d_{m}$ is the number of natural numbers $n$ for which the equality $$\biggl\lfloor\frac{\ln n}{\ln \varphi }\biggr\rfloor=m\quad (m\in  \mathbb N)$$
		holds.
	\end{lemma}
	\begin{proof} The proof presented here is due to J. Shallit. If we want to count the number of $n$ for which $\bigl\lfloor\frac{\ln n}{\ln \varphi }\bigr\rfloor=m$, this is the same as counting the number of integers $n$ such that $m \leq \frac{\ln n}{\ln \varphi } &lt; m+1$, which by rearrangement is the number of integers $n$ such that $\varphi^m \leq n &lt; \varphi^{m+1}$. However, we know from the Binet form of the Lucas numbers that
		\begin{displaymath} 
			\bigl\lfloor\varphi^m \bigr\rfloor=\begin{cases} L_m - 1, &amp; \text{if}  \  m \ \text{is\ even};\\
				L_m, &amp; \text{if} \  m \ \text{is\ odd},
			\end{cases}
		\end{displaymath}
		so now trivially we get $d_m = L_{m-1} + 1$ if $m$ is even and $d_m = L_{m-1} - 1$ if $m$ is odd, that is,
		\begin{equation}\label{2.0.00}
			d_m = L_{m-1} + (-1)^m\quad (m\in \mathbb N).
		\end{equation}
	\end{proof}


	{\tabcolsep 12pt \begin{table}[ht]\label{table3}
			\begin{center}
				\begin{tabular}{|c|rrrrrrrrrrrrr|}
					\hline
					$n$&amp;$1$&amp;$2$&amp;$3$&amp;$4$&amp;$5$&amp;$6$&amp;$7$&amp;$8$&amp;$9$&amp;$10$&amp;$11$&amp;$12$&amp;$13$ \cr
					\hline
					$\bigl\lfloor\frac{\ln n}{\ln \varphi }\bigr\rfloor$ &amp; 0 &amp; 1&amp; 2 &amp; 2 &amp;3 &amp; 3 &amp; 4&amp; 4&amp; 4&amp; 4&amp; 4&amp; 5&amp; 5\cr
					\hline
				\end{tabular}
			\end{center}
			\caption{The numbers $\bigl\lfloor\frac{\ln n}{\ln \varphi }\bigr\rfloor$ for $n=1,2,\ldots,13.$}
	\end{table}}

	\begin{remark}
		Equality~\eqref{2.0.00} was first stated without proof by G. C. Greubel in \cite[\seqnum{A181716}]{IS}. We do not use Lemma~\ref{l.1.0} and Lemma~\ref{l.1.1} when proving the new results.
	\end{remark}
	\begin{example} We have
		\begin{align*}
			\biggl\lfloor\frac{\ln n}{\ln \varphi }\biggr\rfloor &amp;=1\,\,\text{for}\,\,n=2 \qquad\qquad\quad\Rightarrow d_1=1,\\
			\biggl\lfloor\frac{\ln n}{\ln \varphi }\biggr\rfloor &amp;=2\,\,\text{for}\,\,n=3,4 \qquad\qquad\Rightarrow d_2=2,\\
			\biggl\lfloor\frac{\ln n}{\ln \varphi }\biggr\rfloor &amp;=3\,\, \text{for}\,\,n=5,6 \qquad\qquad \Rightarrow d_3=2,\\
			\biggl\lfloor\frac{\ln n}{\ln \varphi }\biggr\rfloor &amp;=4\,\, \text{for}\,\,n=7,8,9,10,11 \,\Rightarrow d_4=5.
		\end{align*}
	\end{example}
	
	\subsection{The main result}	
	We give our main result in the following theorem:
	\begin{theorem}\label{le:3.6}  The following inequality holds:
		\[\alpha_n &gt;a_n\quad (4&lt;n\in \mathbb N).\]
	\end{theorem}
	\begin{proof} 	
	The inequality is easily checked for  $4 \leq n \leq 78$. We prove
	that it holds for $n &gt; 78$.
	Since $\varphi^{\alpha_n}&gt;n!$, we have $\alpha_n &gt; \frac{\ln n!}{\ln \varphi}&gt;2\ln(n!)$. Inequality~\eqref{1.3} produces
	\[-(n-1)\frac{e}{e-1}\ln\biggl(\frac{1}{n-1}(4/7)^\frac{1}{n-1}\biggr)\geq a_n.\] 
	Therefore we have to prove $2\ln(n!)&gt;a_n$, or, equivalently
	\begin{equation}\label{2.19}
	 2\ln(n!)&gt;-(n-1)\frac{e}{e-1}\ln\biggl(\frac{1}{n-1}(4/7)^\frac{1}{n-1}\biggr).
	\end{equation}
	The last inequality holds for $n=79$.  Let us define the function
	\[\omega(x)=2\ln\bigl( \Gamma(x+1)\bigr)+(x-1)\frac{e}{e-1}\ln\biggl(\frac{1}{x-1}(4/7)^\frac{1}{x-1}\biggr).\]
	Therefore we need to prove that the first derivative of the function $\omega(x)$ is higher than $0$ for $79\leq x\in \mathbb R$. It is easy to check that
	\[\omega'(x)= 2\psi(x+1)+\frac{e}{e-1}\biggl(\ln\biggl(\frac{1}{x-1}(4/7)^\frac{1}{x-1}\biggr)-1+\frac{\ln7/4}{x-1}\biggr)&gt;0 \quad (x=38).\]
	Further, it is necessary to show that $\omega'(x)$ is an increasing function for $79\leq x\in \mathbb R$. We need to prove the following inequality:
	\[ \frac{\partial^2 }{\partial x^2 }\omega(x)= 2\psi^{(1)}(x+1)-\frac{e}{(e-1)(x-1)}&gt;0\quad (79\leq x\in \mathbb R).\]
	Hence using Inequality~\eqref{1.4} it is sufficient to show that
	\[2\biggl(\frac{1}{x+1}+\frac{1}{2(x+1)^2}\biggr)&gt;\frac{e}{(e-1)(x-1)}.\]
	Equivalently, it suffices to show that
	$$2x^2+x-3&gt;(x+1)^2\frac{16}{10}&gt;(x+1)^2\frac{e}{(e-1)}.$$
    It follows straightforwardly that 
	$$2x^2+x-3&gt;(x+1)^2\frac{16}{10}$$
	holds for  $x\geq 8$, which completes our proof.
	\end{proof} 
	
    \section{The left factorial}
	In 1971,  Kurepa \cite{Ku1} defined the function {\it left factorial} \cite[\seqnum{A003422}]{IS} for natural numbers $!n$ by
	\[!0=0, \quad !n=\sum_{k=0}^{n-1}k! \quad (n\in \mathbb N).\]
	In the aforementioned paper  \cite[p.\ 151]{Ku1} Kurepa extended left factorial function to the complex half-plane $\re(z)&gt;0$ as
	$$!z=\int_{0}^{+\infty}e^{-t}\frac{t^{z}-1}{t-1}dt.$$	
	We can also extend such function analytically to the whole complex plane \cite{Ku2} by
	$$!z=\ !(z+1)-\Gamma(z+1).$$ 
	\begin{remark}
		In \cite[p.\ 149]{Ku1}, Kurepa proposed the conjecture for the left factorial as follows:
		$$\text{If\ } 1&lt;n\in\mathbb N,\ \text{then}\ \gcd(!n,n!)=2.$$
		Over the past fifty years there have been many attempts to find a solution to Kurepaâ€™s
		conjecture. The problem remains open. For more details, see \cite{Pe}.
	\end{remark}
	\subsection{The left factorial and the Lambert $W$ function}
	Let us define the sequence of real positive numbers $(b_n)_{n\geq 4}$ as follows:
	$$b_n=\int_{0}^{a_n}e^{-x}\frac{x^{n}-1}{x-1}dx=\sum_{k=0}^{n-1}\gamma(k+1,a_n)\quad (b_n\in \mathbb{R}^+).
	$$
	The sequence $a_n$ is given by \eqref{2.2}.
	{\tabcolsep 12pt
		\begin{table}[ht]\label{table1}
			\begin{center}
				\begin{tabular}{|c|rrrrrr|}
					\hline
					$ n$&amp;$4$&amp;$5$&amp;$6$&amp;$7$&amp;$8$&amp;$9$ \cr
					\hline
					$a_n $ &amp; 5.87 &amp; 9.61 &amp; 13.62 &amp; 17.85 &amp; 22.29  &amp; 26.90\cr
					\hline
					$b_n$  &amp; 8.86 &amp;33.01 &amp; 153.08 &amp; 873.12 &amp; 5913.15 &amp; 46233.17\cr
					\hline
					$!n $ &amp; 10&amp; 34&amp; 154 &amp; 874 &amp; 5914  &amp; 46234\cr
					\hline
				\end{tabular}
			\end{center}
			\caption{The numbers $a_n$, $b_n$ and $!n$ for $n=4,5,6,7,8,9.$ }
	\end{table}}
    \begin{lemma}\label{le:3.5,0}  We have
	 \[2n-4&lt;\frac34 a_n\quad (4\leq n\in \mathbb{N}).\]
	\end{lemma}
    \begin{proof}
    Applying Remark~\ref{re:0}, we have
        \begin{align*}
        	a_{n+1} &amp;&gt;a_n+\ln a_{n+1}\qquad\quad\,\,\,\,  (4\leq n\in 
		\mathbb{N})\\
        	&amp;&gt;a_n+2.8&gt;a_n+\frac83\quad (\ln a_{6+1}&gt;2.88).
        \end{align*}
    Hence by induction on $n$ we obtain our inequality.	
    \end{proof}	 
    \begin{lemma}\label{le:3.5}  The following inequality holds:
     	\[\sum_{t=0}^{k}\frac{a_n^t}{t!}&lt;\frac{a_{n}^{k+1}}{(k+1)!}\quad (5\leq n\in \mathbb{N},\, n-2&gt;k\in \mathbb{N}_{0}).\]
     \end{lemma}
     \begin{proof} 
     	We prove the lemma by induction on $k$. For $k=0$ we have
     	$1&lt;a_n$, which is valid. Assume that the statement holds for $k-1&lt;n-3$, i.e.,
     	$$\sum_{t=0}^{k-1}\frac{a_n^t}{t!}&lt;\frac{a_{n}^{k}}{k!}.$$
     	Then
     	\[\sum_{t=0}^{k}\frac{a_n^t}{t!}=\frac{a_n^k}{k!}+\sum_{t=0}^{k-1}\frac{a_n^t}{t!}&lt;2\frac{a_{n}^{k}}{k!}.\]
     	Hence Lemma~\ref{le:3.5,0} produces
     	\[2\frac{a_{n}^{k}}{k!}&lt;\frac{a_{n}^{k+1}}{(k+1)!}\quad (k&lt;n-2),\]
     	which completes the proof.
     \end{proof}
     \begin{theorem}\label{thm:2.1} The following inequalities hold:
     \[b_n&lt;\ !n&lt;b_n+1\quad (5\leq n \in \mathbb N).\]
     \end{theorem}
	\begin{proof} 
	The claim holds for $5\leq n \leq 9$. Since 
	$$e^{-x}\frac{x^{n}-1}{x-1}&gt;0,$$
	the left side of the inequality holds.  Furthermore, we have
	\begin{equation}\label{4.1}
		\int_{0}^{a_n}e^{-x}\frac{x^{n}-1}{x-1}dx= \ !n - e^{-a_n}\sum_{t=0}^{n-1}\frac{!n-!t}{t!}a_n^t.
	\end{equation}
	To prove the right side of the inequality,
	we use Equation~\eqref{4.1}.  It suffices to show
	$$\sum_{t=0}^{n-1}\frac{!n-!t}{t!}a_n^t&lt;e^{a_n}.$$
	Firstly, by induction on $n$ we have the proof of the following inequality:
	\begin{equation}\label{3.2}
		2n+1&lt;\frac{3}{4}\,a_n\qquad (9\leq n\in \mathbb N).
	\end{equation}
	Next, we have
	\begin{align*}
		\sum_{t=0}^{n-1}\frac{!n-!t}{t!}a_n^t    &amp;=a_n^{n-1}+na_n^{n-2}+\sum_{t=0}^{n-3}\frac{!n-!t}{t!}a_n^t&lt;a_n^{n-1}+na_n^{n-2}+\ !n\sum_{t=0}^{n-3}\frac{a_n^t}{t!}\\ 
		&amp;&lt; a_n^{n-1}+na_n^{n-2}+\ !n\frac{a_n^{n-2}}{(n-2)!}\qquad\quad  (\text{Lemma~\ref{le:3.5}}\,\,\text{for}\,\, k=n-3)\\
		&amp;&lt; a_n^{n-1}+na_n^{n-2}+(n+1)a_n^{n-2}\qquad \biggl(\frac{!(n-2)}{(n-2)!}&lt;1\biggr)\\
		&amp;&lt; \frac74a_n^{n-1}.\qquad\qquad\qquad\qquad\qquad \quad\,  (\text{Equation~\eqref{3.2}}).
	\end{align*}
	Since one real solution to the equation 
	$$e^{x}-\frac74x^{n-1}=0$$
	is $a_n$, we have
	$$\sum_{t=0}^{n-1}\frac{!n-!t}{t!}a_n^t &lt;\frac74a_n^{n-1}= e^{a_n},$$
	which proves our theorem.
	\end{proof} 
	\begin{remark}
	If we replace the constant $\frac47$ with numbers $\frac{4+\frac1{31}}7$ or  $\frac{4-\frac1{31}}7$ in Definition~\eqref{2.2}, then Lemma~\ref{le:3.5} and Inequality~\eqref{3.2} hold for the sequence $(a_n)_{n\geq 4}$ defined in this way. We conclude that our constant $\frac47$ is not optimal. We do not deal with this issue in this paper.
	\end{remark}

   \subsection{The left factorial and the golden sequence}
   Let us define the sequence of real numbers $(\beta_n)_{n\geq 1}$ as follows:
   $$\beta_n=\int_{0}^{\alpha_n}e^{-x}\frac{x^{n}-1}{x-1}dx\quad (\beta_n\in \mathbb R).$$
   The sequence $\alpha_n$ is given by \eqref{2.0}.
   {\tabcolsep 12pt
   	\begin{table}[ht]\label{table2}
   		\begin{center}
   			\begin{tabular}{|c|rrrrrrrr|}
   				\hline
   				$ n$&amp;$1$&amp;$2$&amp;$3$&amp;$4$&amp;$5$&amp;$6$&amp;$7$&amp;$8$ \cr
   				\hline
   				$\alpha_n $ &amp; 1 &amp; 2&amp; 4 &amp; 7 &amp; 10  &amp; 14 &amp; 18&amp; 23\cr
   				\hline
   				$\beta_n$  &amp; 0.63 &amp;1.46&amp; 3.13 &amp; \,\,9.44 &amp; 33.23 &amp; 153.29 &amp; 873.21&amp; 5913.49 \cr
   				\hline
   				$!n $ &amp; 1 &amp; 2&amp; 4 &amp; 10 &amp; 34  &amp; 154 &amp; 874&amp; 5914\cr
   				\hline
   			\end{tabular}
   		\end{center}
   		\caption{The numbers $!n$, $\alpha_n$ and $\beta_n$ for $n=1,2,\ldots,8.$}
   \end{table}}
   \begin{theorem}\label{thm:2.2} The following inequalities hold:
   	\[\beta_n&lt;\ !n&lt; \beta_n+1\quad (n\in \mathbb N).\]
   \end{theorem}
   \begin{proof} 
   	Let us define the family of functions $f_n(x)$ as follows:
   	\[f_n(x)=x-(n-1)\ln x-\ln\frac{7}{4}\quad (0&lt;x\in \mathbb R,\, 3&lt;n\in \mathbb N).\]
   	Then
   	\[\frac{\partial }{\partial x }f_n(x)=f'_n(x)=1-\frac{n-1}{x},\quad\frac{\partial^2 }{\partial x^2 }f_n (x)=\frac{n-1}{x^2},\quad f'_n(n-1)=0,\quad f_n(n-1)&lt;0.\]
   	For the equation $f_n(x)=0$, we have the real solution $x=a_n$. On the interval $(0,+\infty)$ the function $f_n(x)$ has the minimum for $x = n-1$. The function $f_n(x)$ increases on the interval $(0,+\infty)$. 
   	For the sequence $(\alpha_n)_{n\geq 5}$ the analogs of Lemma~\ref{le:3.5} and Inequality~\eqref{3.2} are valid. Hence Theorem~\ref{le:3.6} produces
   	\[\beta_n&lt;\ !n&lt;\beta_n+1.\]
   	\end{proof}
   	\begin{remark}\label{re:1} If Lemma~\ref{le:3.5} and Inequality~\eqref{3.2} hold for the positive real sequence $(c_n)_{n\geq n_0}$, and if $c_n\geq a_n,\, (n\geq n_0\geq 5)$, then
   		\[\int_{0}^{c_n}e^{-x}\frac{x^{n}-1}{x-1}dx&lt;\ !n&lt;\int_{0}^{c_n}e^{-x}\frac{x^{n}-1}{x-1}dx+1 \quad (n\geq n_0).\]
   	\end{remark}

    \subsection{Generalization}
    Let us define two-dimensional real sequences $(A_{n,m})_{n\geq 5,m\geq 1}$ and $(B_{n,m})_{n \geq 5,m\geq 1 }$ as follows:
    \begin{align*}
    	A_{n,m} &amp;=-(n-1)W_{-1}\biggl(-\frac{1}{n-1}\left(\frac{4}{7m}\right)^{\frac{1}{n-1}}\biggr),\\ 
    	B_{n,m} &amp;=\int_{0}^{A_{n,m}}e^{-x}\frac{x^{n}-1}{x-1}dx.
    \end{align*}
    {\tabcolsep 12pt
    	\begin{table}[ht]\label{table4}
    		\begin{center}
    			\begin{tabular}{|c|rrrrr|}
    				\hline
    				$ n$&amp;$5$&amp;$6$&amp;$7$&amp;$8$&amp;$9$ \cr
    				\hline
    				$B_{n,1}$   &amp;33.01 &amp; 153.08 &amp; 873.12 &amp; 5913.15 &amp; 46233.17\cr
    				\hline
    				$B_{n,2}$   &amp;33.53 &amp; 153.56 &amp; 873.57 &amp; 5913.58 &amp; 46233.60\cr
    				\hline
    				$B_{n,3}$   &amp;33.70 &amp; 153.71 &amp; 873.72 &amp; 5913.73 &amp; 46233.70\cr
    				\hline
    				$B_{n,4}$   &amp;33.78 &amp; 153.79 &amp; 873.79 &amp; 5913.80 &amp; 46233.80\cr
    				\hline
    				$B_{n,5}$   &amp;33.82 &amp; 153.83 &amp; 873.83 &amp; 5913.84 &amp; 46233.84\cr
    				\hline
    				$!n $ &amp; 34&amp; 154 &amp; 874 &amp; 5914  &amp; 46234\cr
    				\hline
    			\end{tabular}
    		\end{center}
    		\caption{The numbers $B_{n,m}$ and $!n$ for $n=5,6,7,8,9$ and $m=1,2,3,4,5.$}
    \end{table}}
    \begin{theorem}\label{thm:2.3} The following inequalities hold:
    	\[ B_{n,m}&lt;\ !n&lt;B_{n,m}+\frac1m\quad (5\leq n\in \mathbb N,\ m\in \mathbb N).\]
    \end{theorem}
    \begin{proof} The Theorem~\ref{thm:2.3} follows from the application of the procedures used in the proof of Theorem~\ref{thm:2.1} and Remark~\ref{re:1}.
    \end{proof}
    \begin{corollary}\label{co:2.5} The following equalities hold:
    	\begin{align*}
    		!n &amp;=\lfloor \beta_n\rfloor +1\qquad\qquad\qquad\quad\,\,\,\, (n\in \mathbb N) \\ 
    		   &amp;=\lfloor b_n\rfloor + 1 = \biggl\lfloor B_{n,m}+\frac{1}{m} \biggr\rfloor \quad (4&lt;n\in \mathbb N).
    	\end{align*}
    \end{corollary}
    
     \section{Acknowledgment}
     We would like to thank the referee for his helpful suggestions.
    
   	
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		\Dj. Kurepa, On the left factorial function $!n$, {\it Math. Balkanica}  {\bf 1} (1971), 147--153. 
				
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		\Dj. Kurepa, Left factorial function in complex domain, {\it Math. Balkanica} {\bf 3} (1973), 297--307.
		
		\bibitem{Lo}
		L. L\'{o}czi,  Guaranteed and high precision evaluation of the Lambert $W$ function, {\it Appl. Math. Comput.} {\bf 433} (2022), Article 127406. 
		
		\bibitem{Mez}
		I. Mez\"{o}, {\it The Lambert $W$ Function, Its Generalisations and Applications},   Chapman and Hall/CRC, 2022.
				
		\bibitem{Pe}  A. Petojevi\'{c},  S. Gordi\'{c}, M.  Mandi\'{c}, and M. Gorjanac Ranitovi\'{c},   New equivalents of Kurepa's hypothesis for left factorial, {\it  Axioms} {\bf 12} (8) (2023), Article 785. 
		
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		N. J. A. Sloane et al., The On-Line Encyclopedia of Integer Sequences, 2024. Available at \url{https://oeis.org}. 
		
		\bibitem{So} S. T. Somu and D. V. K. Tran, On sums of practical numbers and polygonal numbers, \textit{J. Integer Sequences} \textbf{27} (2024), 
		\href{https://cs.uwaterloo.ca/journals/JIS/VOL27/Somu/somu5.pdf}{Article 24.5.1}.
		
		\bibitem{Sr}
		H. M. Srivastava and J. Choi, {\it Series Associated with the Zeta and Related Functions}, Kluwer Academic Publishers, 2001.
								
	\end{thebibliography}
	
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	\noindent 2020 {\it Mathematics Subject Classification}:
	Primary 11B83; Secondary 11B39.\\
	
	\noindent \emph{Keywords:}  golden ratio, Fibonacci number, Lucas number, associated Mersenne number, Lambert $W$ function, left factorial, Kurepa hypothesis.
	
	
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	\noindent (Concerned with sequences 
	\seqnum{A000032},
	\seqnum{A000045},
	\seqnum{A001350},
	\seqnum{A003422},
	\seqnum{A098600},
	\seqnum{A181716},
	\seqnum{A214048} and
	\seqnum{A371672}.)

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\vspace*{+.1in}
\noindent
Received April 4 2024;
revised versions received April 5 2024; May 4 2024; May 5 2024; May 7 2024; 
May 15 2024.
Published in {\it Journal of Integer Sequences}, June 3 2024.

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