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\begin{center}
\vskip 1cm{\LARGE\bf 
Explicit Formulas for the $p$-adic  \\
\vskip .03in
Valuations of Fibonomial Coefficients
}
\vskip 1cm
\large
Phakhinkon Phunphayap and 
Prapanpong Pongsriiam\footnote{Prapanpong Pongsriiam receives financial support jointly from The Thailand Research Fund and Faculty of Science Silpakorn University, grant number RSA5980040. Prapanpong Pongsriiam is the 
corresponding author.} \\
Department of Mathematics \\
Faculty of Science\\
Silpakorn University\\
Nakhon Pathom 73000 \\
Thailand\\
\href{mailto:phakhinkon@gmail.com}{\tt phakhinkon@gmail.com} \\
\href{mailto:pongsriiam_p@silpakorn.edu}{\tt pongsriiam\_p@silpakorn.edu}, 
\href{mailto:prapanpong@gmail.com}{\tt prapanpong@gmail.com} \\
\end{center}


\newcommand{\lrf}[1]{\left\lfloor #1\right\rfloor}

\vskip .2 in

\begin{abstract}
We obtain explicit formulas for the $p$-adic valuations of Fibonomial coefficients which extend some results in the literature.
\end{abstract}

\section{Introduction}

The \textit{Fibonacci sequence} $(F_n)_{n\geq 1}$ is given by the recurrence relation $F_n = F_{n-1} + F_{n-2}$ for $n\geq 3$ with the initial values $F_1 = F_2 = 1$. For each $m\geq 1$ and $1\leq k \leq m$, the \textit{Fibonomial coefficients} ${m \choose k}_F$ are defined by
\begin{equation*}
{m \choose k}_F = \frac{F_1F_2F_3\cdots F_m}{(F_1F_2F_3\cdots F_k)(F_1F_2F_3\cdots F_{m-k})} = \frac{F_{m-k+1}F_{m-k+2}\cdots F_m}{F_1F_2F_3\cdots F_k},
\end{equation*}
where $F_n$ is the $n$th Fibonacci number. If $k=0$, we define ${m \choose k}_F=1$ and if $k&gt;m$, we define ${m \choose k}_F=0$. It is well known that ${m \choose k}_F$ is an integer for all positive integers $m$ and $k$. So it is natural to consider the divisibility properties and the $p$-adic valuation of ${m \choose k}_F$. As usual, $p$ always denotes a prime and the \textit{$p$-adic valuation (or $p$-adic order)} of a positive integer $n$, denoted by $\nu_p (n)$, is the exponent of $p$ in the prime factorization of $n$. In addition, the \textit{order (or the rank) of appearance} of $n$ in the Fibonacci sequence, denoted by $z(n)$, is the smallest positive integer $k$ such that $n\mid F_k$. The Fibonacci sequence and the triangle of Fibonomial coefficients are, respectively, \seqnum{A000045} and \seqnum{A010048} in OEIS \cite{S}. Also see
\seqnum{A055870} and \seqnum{A003267} for signed Fibonomial triangle and central Fibonomial coefficients, respectively.    

In 1989, Knuth and Wilf \cite{KW} gave a short description of the $p$-adic valuation of ${m\choose k}_C$ where $C$ is a regularly divisible sequence. However, this does not give explicit formulas for ${m\choose k}_F$. Then recently, there has been some interest in explicitly evaluating the $p$-adic valuation of Fibonomial coefficients of the form ${p^b \choose p^a}_F$. For example, Marques and Trojovsk\'y \cite{Mar2,Mar3} and Marques, Sellers, and Trojovsk\'y \cite{Mar4} deal with the case $b=a+1$, $a\geq 1$. Ballot \cite[Theorem 4.2]{ballot} extends the Kummer-like theorem of Knuth and Wilf \cite[Theorem 2]{KW}, which gives the $p$-adic valuation of Fibonomials, to all Lucasnomials, and, in particular, uses it to determine explicitly the $p$-adic valuation of Lucasnomials of the form ${p^b \choose p^a}_U$, for all nondegenerate fundamental Lucas sequences $U$ and all integers $b&gt;a\geq 0$, \cite[Theorem 7.1]{ballot}.

Note that in the formula given by Marques and Trojovsk\'y \cite[Theorem 1]{Mar3} for $U=F$ and $b=a+1$, only the case of $a$ even is actually explicitly computed. It appears, using the theorem of Ballot \cite[Theorem 7.1]{Apostol}, that their stated result for $a$ odd is correct only for primes $p$ for which $p^2$ does not divide $F_{z(p)}$, where $z(p)$ is the rank of appearance of $p$ in the Fibonacci sequence. Also see Examples \ref{rem16} and \ref{rem18} in this article.

Our purpose is to extend Ballot's theorem, Theorem 7.1, in the case $U=F$ and $b\geq a&gt;0$ and obtain explicit formulas for ${\ell_1 p^b \choose \ell_2 p^a}_F$, where $\ell_1$ and $\ell_2$ are arbitrary positive integers such that $\ell_1 p^b &gt; \ell_2 p^a$. This leads us to study the $p$-adic valuations of integers of the forms  
\begin{equation}\label{eq1.1}
\left\lfloor\frac{\ell p^a}{m}\right\rfloor! \quad\text{or}\quad \left\lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right\rfloor!,
\end{equation}
where $p\equiv \modd{\pm 1} {m}$. For instance, we obtain in Example \ref{example2} the following result: for positive integers $a,b,\ell$ with $b\geq a$, and a prime $p$ distinct from 2 and 5, if $p \equiv \modd{\pm 1} {5}$, then
	\begin{equation*}
	\nu_p\left( {\ell p^b \choose p^a}_F \right) =
		\begin{cases}
			b + \nu_p(F_{z(p)}) + \nu_p(\ell), &amp;\text{if $z(p) \mid \ell$;}\\
			0, &amp;\text{otherwise.}
		\end{cases}
	\end{equation*}
	Furthermore, if $p \equiv \modd{\pm 2} {5}$, then
	\begin{equation*}
	\nu_p\left( {\ell p^b \choose p^a}_F \right) =
	\begin{cases}
	0, &amp;\text{if $\ell \equiv \modd{1 - 2 \varepsilon} {z(p)}$;}\\
	b + \nu_p(F_{z(p)}) + \nu_p(\ell), &amp;\text{if $\ell \equiv 
	\modd{0} {z(p)}$};\\
	\frac{a}{2}, &amp;\text{if $\ell \not\equiv \modd{0,1 - 2 \varepsilon} 
	{z(p)}$ and $a$ is even;}\\
	\frac{a-1}{2} + \nu_p (F_{z(p)}), &amp;\text{if $\ell \not\equiv 
	\modd{0,1 - 2 \varepsilon} {z(p)}$ and $a$ is odd,}
	\end{cases}
	\end{equation*}
	where $\varepsilon = 1$ if $a$ and $b$ have different parity and $\varepsilon = 0$ otherwise. We also obtain the corresponding results for $p \in \{ 2,5 \}$ in Examples \ref{example1} and \ref{example3}. These extend all the main results in \cite{Mar2,Mar3,Mar4} and Ballot's theorem, Theorem 7.1, in the case $U=F$.
	
Recall that for each $x\in\mathbb{R}$, $\lfloor x\rfloor$ is the largest integer less than or equal to $x$, $\{ x\}$ is the \textit{fractional part} of $x$ given by $\{ x\}=x-\lfloor x\rfloor$, and $\lceil x\rceil$ is the smallest integer larger than or equal to $x$. In addition, we write $a\bmod{m}$ to denote the least nonnegative residue of $a$ modulo $m$. We also use the \textit{Iverson notation}: if $P$ is a mathematical statement, then
\begin{equation*}
[P] = 
\begin{cases}
	1, &amp; \text{if $P$ holds}; \\
	0, &amp; \text{otherwise}.
\end{cases}
\end{equation*}
For example, $[5\equiv \modd{-1} {4}]=0$ and $[3\equiv \modd{-1} {4}]=1$.

We organize this article as follows. In Section \ref{sec2}, we give
some preliminaries and useful results which are needed in the proof of
the main theorems. In Section \ref{sec3}, we give exact formulas for
the $p$-adic valuations of integers \eqref{eq1.1}. In Section
\ref{sec4}, we apply the results obtained in Section \ref{sec3} to
Fibonomial coefficients. Our most general theorem is Theorem
\ref{theorem4.3}. Finally, in Section \ref{sec5}, we give the $p$-adic
valuations of some specific sub-families of Fibonomial coefficients of
type (1), since generally, the more specific the family, the shortest
the formula becomes.

For more information related to Fibonacci numbers, we invite the readers to visit the second author's Researchgate account \cite{P9} which contains some freely downloadable versions of his publications \cite{KP,KP2,KP3,OP,OP2,P,P2,P3,P4,P5,P6,P7,P8}.


\section{Preliminaries and lemmas}\label{sec2}
 
Recall that for each odd prime $p$ and $a\in\mathbb{Z}$, the \textit{Legendre symbol} $(\frac{a}{p})$ is defined by
\begin{equation*}
\left (\frac{a}{p}\right )=
\begin{cases}
0, &amp;\text{if $p\mid a$};\\
1, &amp; \text{if $a$ is a quadratic residue of $p$};\\
-1, &amp; \text{if $a$ is a quadratic nonresidue of $p$.}
\end{cases}
\end{equation*}
Then we have the following result.

\begin{lemma}\label{lemma2.2}
Let $p\neq 5$ be a prime and let $m$ and $n$ be positive integers. Then the following statements hold.
\begin{itemize}
\item[\em{(i)}] If $p&gt;2$, then $F_{p-(\frac{5}{p})}\equiv \modd{0} {p}$.
\item[\em{(ii)}] $n\mid F_m$ if and only if $z(n)\mid m$.
\item[\em{(iii)}] $z(p)\mid p+1$ if and only if $p\equiv 2\text{ or }
\modd{-2} {5}$, and $z(p)\mid p-1$ otherwise.
\item[\em{(iv)}] $\gcd (z(p),p)=1$.
\end{itemize}
\end{lemma}

\begin{proof}
These are well known results. For example, (i) and (ii) can be found in \cite[p.\ 410]{koshy} and \cite{Wall}, respectively. Then (iii) follows from (i) and (ii). By (iii), $z(p)\mid p\pm 1$. Since $\gcd(p,p\pm 1)=1$, we obtain $\gcd(z(p),p)=1$. This proves (iv).
\end{proof}

Lengyel's result and Legendre's formula given in the following lemmas are important tools in evaluating the $p$-adic valuation of Fibonomial coefficients. We also refer the reader to \cite{Mar2,Mar3,Mar4,P} for other similar applications of Lengyel's result.

\begin{lemma}\label{lengyel}
{\rm (Lengyel \cite{Len})} For $n\geq 1$, we have
\begin{equation*}
\nu_2(F_n)=
\begin{cases}
0, &amp; \text{if $n\equiv \modd{1,2} {3}$};\\
1, &amp; \text{if $n\equiv \modd{3} {6}$};\\
\nu_2(n)+2, &amp; \text{if $n\equiv \modd{0} {6}$},
\end{cases}
\end{equation*}
\begin{equation*}
\text{$\nu_5(F_n)=\nu_5(n)$, and if $p$ is a prime  distinct from $2$ and $5$, then}
\end{equation*}
\begin{equation*}
\nu_p(F_n)=
\begin{cases}
\nu_p(n)+\nu_p(F_{z(p)}), &amp; \text{if $n\equiv \modd{0} {z(p)}$}:\\
0, &amp; \text{if $n\not\equiv \modd{0} {z(p)}$},
\end{cases}
\end{equation*}
\end{lemma}
\begin{lemma}
{\rm(Legendre's formula)}
Let $n$ be a positive integer and let $p$ be a prime. Then
\begin{equation*}
\nu_p(n!)=\sum_{k=1}^\infty \left\lfloor\frac{n}{p^k}\right\rfloor.
\end{equation*}
\end{lemma}

In the proof of the main results, we will deal with a lot of calculation involving the floor function. So it is useful to recall the following results.

\begin{lemma}\label{floorlemma}
For $n\in\mathbb{Z}$ and $x\in\mathbb{R}$, the following holds
\begin{itemize}
\item[\em{(i)}] $\lfloor n+x\rfloor = n+\lfloor x\rfloor$,
\item[\em{(ii)}] $\{n+x\}=\{x\}$,
\item[\em{(iii)}] $\lfloor x\rfloor + \lfloor -x \rfloor=
\begin{cases}
-1, &amp;\text{if $x\not\in\mathbb{Z}$;}\\
0, &amp;\text{if $x\in\mathbb{Z}$,}
\end{cases}$
\item[\em{(iv)}] $\{-x\} =
\begin{cases}
1-\{x\}, &amp;\text{if $x\not\in\mathbb{Z}$;}\\
0, &amp;\text{if $x\in\mathbb{Z}$,}
\end{cases}$
\item[\em{(v)}] $\lfloor x+y\rfloor =
\begin{cases}
\lfloor x\rfloor +\lfloor y\rfloor, &amp;\text{if $\{x\}+\{y\}&lt;1$;}\\
\lfloor x\rfloor +\lfloor y\rfloor+1, &amp;\text{if $\{x\}+\{y\}\geq 1$,}
\end{cases}$
\item[\em{(vi)}] $\left\lfloor\frac{\lfloor x\rfloor}{n}\right\rfloor = \left\lfloor\frac{x}{n}\right\rfloor$ for $n\geq 1$.
\end{itemize}
\end{lemma}

\begin{proof}
These are well-known results and can be proved easily. For more details, see in \cite[Exercise 13, p.\ 72]{Apostol} or in \cite[Chapter 3]{Concrete}. We also refer the reader to \cite{OP2} for a nice application of (v).
\end{proof}

The next lemma is used often in counting the number of positive integers $n\leq x$ lying in a residue class $a\bmod{q}$, see for instance in \cite[Proof of Lemma 2.6]{PV}.

\begin{lemma}\label{sumlemma}
For $x\in [1,\infty)$, $a$, $q \in \mathbb{Z}$ and $q\geq 1$, we have
\begin{equation}\label{eq2.1}
\sum_{\substack{1\leq n\leq x \\
	   n\, \equiv\, a\ (\text{\rm mod}\ q)}} 1 = \left\lfloor\frac{x-a}{q}\right\rfloor - \left\lfloor-\frac{a}{q}\right\rfloor.
\end{equation}
\end{lemma}
\begin{proof}
Replacing $a$ by $a+q$ and applying Lemma \ref{floorlemma}, we see that the value on the right-hand side of \eqref{eq2.1} is not changed. Obviously, the 
left-hand side is also invariant when we replace $a$ by $a+q$. So it is enough to consider only the case $1\leq a\leq q$. Since $n\equiv \modd{a} {q}$,
we write $n=a+kq$ where $k\geq 0$ and $a+kq\leq x$. So $k\leq \frac{x-a}{q}$. Therefore
\begin{equation*}
\sum_{\substack{1\leq n\leq x \\
	   n\,\equiv\, a\ (\text{mod}\ q)}} 1 = \sum_{0\leq k\leq \frac{x-a}{q}} 1 = \left\lfloor\frac{x-a}{q}\right\rfloor +1 = \left\lfloor\frac{x-a}{q}\right\rfloor - \left\lfloor-\frac{a}{q}\right\rfloor.
\end{equation*}
\end{proof}

It is convenient to use the Iverson notation and to denote the least nonnegative residue of $a$ modulo $m$ by $a\bmod{m}$. Therefore we will do so from this point on.

\begin{lemma}\label{lemma2.7}
Let $n$ and $k$ be integers, $m$ a positive integer, $r=n \bmod{m}$, and $s=k \bmod{m}$. Then
\begin{equation*}
	\left\lfloor\frac{n-k}{m}\right\rfloor = \left\lfloor\frac{n}{m}\right\rfloor - \left\lfloor\frac{k}{m}\right\rfloor -[r&lt;s].
\end{equation*}
\end{lemma}

\begin{proof}
By Lemma \ref{floorlemma}(i) and the fact that $0\leq r&lt; m$, we obtain
\begin{equation*}
\left\lfloor\frac{n}{m}\right\rfloor = \left\lfloor\frac{n-r}{m} + \frac{r}{m}\right\rfloor = \frac{n-r}{m} + \left\lfloor\frac{r}{m}\right\rfloor = \frac{n-r}{m}.
\end{equation*}
Similarly, $\left\lfloor\frac{k}{m}\right\rfloor = \frac{k-s}{m}$. Therefore $\left\lfloor\frac{n-k}{m}\right\rfloor$ is equal to
\begin{equation*}
\left\lfloor\frac{n-r}{m} - \frac{k-s}{m} + \frac{r-s}{m}\right\rfloor = \frac{n-r}{m} - \frac{k-s}{m}+\left\lfloor\frac{r-s}{m}\right\rfloor =
\begin{cases}
\left\lfloor\frac{n}{m}\right\rfloor - \left\lfloor\frac{k}{m}\right\rfloor, &amp;\text{if $r\geq s$};\\
\left\lfloor\frac{n}{m}\right\rfloor - \left\lfloor\frac{k}{m}\right\rfloor - 1, &amp;\text{if $r&lt;s$}.
\end{cases}
\end{equation*}
\end{proof}


\section{The $p$-adic valuation of integers in special forms}\label{sec3}
In this section, we calculate the $p$-adic valuation of $\left\lfloor \frac{\ell p^a}{m}\right\rfloor !$ and other integers in similar forms.
\begin{theorem}\label{theorem1}
Let $p$ be a prime and let $a\geq 0$, $\ell \geq 0$, and $m\geq 1$ be integers. Assume that $p\equiv \modd{\pm 1} {m}$ and let $\delta = 
[\ell \not\equiv \modd{0} {m}]$. Then
\begin{equation*}
\nu_p \left(\left\lfloor\frac{\ell p^a}{m}\right\rfloor !\right) =
\begin{cases}
\frac{\ell (p^a -1)}{m(p-1)}-a\left\{\frac{\ell}{m}\right\} + \nu_p\left(\left\lfloor\frac{\ell}{m}\right\rfloor!\right), &amp;\text{if $p\equiv 
\modd{1} {m}$;}\\[.1in]
\frac{\ell (p^a -1)}{m(p-1)} - \frac{a}{2}\delta + \nu_p\left(\left\lfloor\frac{\ell}{m}\right\rfloor!\right), &amp;\text{if $p\equiv 
\modd{-1} {m}$ and $a$ is even;}\\[.1in]
\frac{\ell (p^a -1)}{m(p-1)} - \frac{a-1}{2}\delta - \left\{\frac{\ell}{m}\right\} + \nu_p\left(\left\lfloor\frac{\ell}{m}\right\rfloor!\right), &amp;\text{if $p\equiv \modd{-1} {m}$ and $a$ is odd.}
\end{cases}
\end{equation*}
\end{theorem}

We remark that if $m=1$ or $2$, then the expressions in each case of this theorem are all equal.

\begin{proof}
The result is easily verified when $a=0$ or $\ell=0$. So we assume throughout that $a\geq 1$ and $\ell \geq 1$. We also use Lemmas \ref{floorlemma}(i), \ref{floorlemma}(vi), and \ref{sumlemma} repeatedly without reference. By Legendre's formula, we obtain
\begin{equation}\label{eq1}
\nu_p \left(\left\lfloor\frac{\ell p^a}{m}\right\rfloor !\right) = \sum_{j =1}^\infty \left\lfloor\frac{\ell p^a}{mp^j}\right\rfloor = \sum_{j=1}^a \left\lfloor\frac{\ell p^{a-j}}{m}\right\rfloor + \sum_{j=a+1}^\infty \left\lfloor\frac{\ell p^{a-j}}{m}\right\rfloor = \sum_{j=1}^a \left\lfloor\frac{\ell p^{a-j}}{m}\right\rfloor + \nu_p\left(\left\lfloor\frac{\ell}{m}\right\rfloor!\right).
\end{equation}
From \eqref{eq1}, it is immediate that for $m=1$, we obtain
\begin{equation*}
	\nu_p\left((\ell p^a)!\right) = \frac{\ell (p^a -1)}{p-1} + \nu_p (\ell !).
\end{equation*}
So we assume throughout that $m\geq 2$.

\medskip

\noindent{\bf Case 1.} $p\equiv \modd{1} {m}$. Then, for every $k\geq 0$, $p^k \equiv \modd{1} {m}$ and 
\begin{equation*}
\left\lfloor\frac{\ell p^k}{m}\right\rfloor=\left\lfloor\frac{\ell (p^k -1)}{m} +\frac{\ell}{m}\right\rfloor=\frac{\ell (p^k -1)}{m} + \left\lfloor\frac{\ell}{m}\right\rfloor.
\end{equation*}
Therefore the sum $\sum_{j=1}^a \left\lfloor\frac{\ell p^{a-j}}{m}\right\rfloor$ appearing in \eqref{eq1} is equal to
\begin{equation*}
\sum_{j=1}^a \left(\frac{\ell (p^{a-j} -1)}{m} + \left\lfloor\frac{\ell}{m}\right\rfloor\right)  = \left(\frac{\ell}{m}\sum_{j=1}^a p^{a-j}\right) - a\left(\frac{\ell}{m} - \left\lfloor \frac{\ell}{m}\right\rfloor\right) = \frac{\ell (p^a -1)}{m(p-1)} - a\left\{\frac{\ell}{m}\right\}.
\end{equation*}

\noindent{\bf Case 2.} $p\equiv \modd{-1} {m}$. Then for $k\geq 0$, we have $p^k \equiv \modd{1} {m}$ if $k$ is even, and $p^k \equiv 
\modd{-1} {m}$ if $k$ is odd. Therefore
\begin{align*}
\left\lfloor\frac{\ell p^k}{m}\right\rfloor=\left\lfloor\frac{\ell (p^k - 1)}{m}+\frac{\ell}{m}\right\rfloor=\frac{\ell (p^k -1)}{m} + \left\lfloor\frac{\ell }{m} \right\rfloor \quad\quad&amp;\text{if $k\geq 0$ and $k$ is even,}\\
\left\lfloor\frac{\ell p^k}{m}\right\rfloor=\left\lfloor\frac{\ell (p^k + 1)}{m}-\frac{\ell}{m}\right\rfloor = \frac{\ell (p^k +1)}{m} + \left\lfloor - \frac{\ell}{m}\right\rfloor \quad\quad&amp;\text{if $k\geq 0$ and $k$ is odd.}
\end{align*}
Therefore the sum $\sum_{j=1}^a \left\lfloor\frac{\ell p^{a-j}}{m}\right\rfloor$ appearing in \eqref{eq1} is equal to
\begin{align}\label{eq2}
&amp;\sum_{\substack{1\leq j\leq a \\
	   a-j\,\equiv \, 0\ (\text{mod}\ 2)}} \left(\frac{\ell (p^{a-j} -1)}{m} + \left\lfloor\frac{\ell }{m} \right\rfloor\right) +  \sum_{\substack{1\leq j\leq a \\
	   a-j\, \equiv\, 1\ (\text{mod}\ 2)}} \left(\frac{\ell (p^{a-j} +1)}{m} + \left\lfloor - \frac{\ell}{m}\right\rfloor\right) \nonumber\\
&amp;=\frac{\ell}{m}\sum_{1\leq j\leq a} p^{a-j} - \sum_{\substack{1\leq j\leq a \\
	   j\, \equiv a\, \ (\text{mod}\ 2)}} \left(\frac{\ell}{m} - \left\lfloor\frac{\ell}{m}\right\rfloor\right)  + \sum_{\substack{1\leq j\leq a \\
	   j\, \equiv\, a-1\ (\text{mod}\ 2)}} \left(\frac{\ell}{m} + \left\lfloor - \frac{\ell}{m} \right\rfloor\right) \nonumber\\
&amp;=\frac{\ell (p^a -1)}{m(p-1)} +\left\lfloor-\frac{a}{2}\right\rfloor \left(\frac{\ell}{m} - \left\lfloor\frac{\ell}{m}\right\rfloor\right) - \left\lfloor-\frac{a-1}{2}\right\rfloor\left(\frac{\ell}{m} + \left\lfloor - \frac{\ell}{m} \right\rfloor\right).
\end{align}
By Lemma \ref{floorlemma}(iii), we see that
\begin{equation*}
\left\lfloor\frac{\ell }{m} \right\rfloor + \left\lfloor - \frac{\ell}{m}\right\rfloor = -[\ell \not\equiv \modd{0} {m}]=-\delta.
\end{equation*}
Therefore if $a$ is even, then \eqref{eq2} is equal to
\begin{align*}
&amp;\frac{\ell (p^a -1)}{m(p-1)}-\frac{a}{2}\left(\frac{\ell}{m} - \left\lfloor\frac{\ell}{m}\right\rfloor\right) + \frac{a}{2}\left(\frac{\ell}{m} + \left\lfloor - \frac{\ell}{m} \right\rfloor\right) \\
&amp;= \frac{\ell (p^a -1)}{m(p-1)} + \frac{a}{2}\left( \left\lfloor\frac{\ell}{m}\right\rfloor + \left\lfloor - \frac{\ell}{m}\right\rfloor\right)=\frac{\ell (p^a -1)}{m(p-1)} - \frac{a}{2}\delta,
\end{align*}
and if $a$ is odd, then \eqref{eq2} is equal to
\begin{align*}
&amp;\frac{\ell (p^a -1)}{m(p-1)}-\frac{a+1}{2}\left(\frac{\ell}{m} - \left\lfloor\frac{\ell}{m}\right\rfloor\right) + \frac{a-1}{2}\left(\frac{\ell}{m} + \left\lfloor - \frac{\ell}{m} \right\rfloor\right) \\
&amp;= \frac{\ell (p^a -1)}{m(p-1)}+\frac{a-1}{2}\left(\left\lfloor\frac{\ell}{m}\right\rfloor + \left\lfloor - \frac{\ell}{m}\right\rfloor\right) + \left\lfloor\frac{\ell}{m}\right\rfloor - \frac{\ell}{m}\\
&amp;= \frac{\ell (p^a -1)}{m(p-1)} - \left(\frac{a-1}{2}\right)\delta - \left\{\frac{\ell}{m}\right\}.
\end{align*}
This completes the proof.
\end{proof}

We can combine every case in Theorem \ref{theorem1} into a single form as given in the next corollary.

\begin{corollary}\label{corollary1}
	Assume that $p$, $a$, $\ell$, $m$, and $\delta$ satisfy the same assumptions as in Theorem \ref{theorem1}. Then the $p$-adic valuation of $\left \lfloor\frac{\ell p^a}{m}\right \rfloor !$ is
	\begin{align}\label{eqcol1}
		&amp;\frac{\ell(p^a -1)}{m(p-1)} - \left \lfloor\frac{a}{2}\right \rfloor \delta - \left \{\frac{\ell}{m}\right \}[a\equiv \modd{1} {2}] \nonumber\\
		&amp;+ \delta\left \lfloor\frac{a}{2}\right \rfloor \left (1-2\left \{\frac{\ell}{m}\right \}\right )[p\equiv \modd{1} {m}] + \nu_p \left (\left \lfloor \frac{\ell}{m}\right \rfloor ! \right ).
	\end{align}
\end{corollary}

\begin{proof}
	This is merely a combination of each case from Theorem \ref{theorem1}. For example, when $p\equiv \modd{-1} {m}$, the right-hand side of \eqref{eqcol1} reduces to 
	\begin{align*}
		&amp;\frac{\ell(p^a -1)}{m(p-1)} - \left \lfloor\frac{a}{2}\right \rfloor \delta - \left \{\frac{\ell}{m}\right \}[a\equiv 
		\modd{1} {2}] + \nu_p\left ( \left \lfloor\frac{\ell}{m}\right \rfloor ! \right )\\
		&amp;=
		\begin{cases}
			\frac{\ell(p^a -1)}{m(p-1)} - \frac{a}{2}\delta + \nu_p\left ( \left \lfloor\frac{\ell}{m}\right \rfloor ! \right ), &amp;\text{if $a$ is even};\\[.1in]
			\frac{\ell(p^a -1)}{m(p-1)} - \left(\frac{a-1}{2}\right)\delta - \left\{ \frac{\ell}{m}\right\} + \nu_p\left ( \left \lfloor\frac{\ell}{m}\right \rfloor ! \right ), &amp;\text{if $a$ is odd},
		\end{cases}
	\end{align*}
which is the same as Theorem \ref{theorem1}. The other cases are similar. We leave the details to the reader.
\end{proof}

Next we deal with the $p$-adic valuation of an integer of the form $\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor  !$ where $a$, $b$, $\ell_1$, $\ell_2$, and $m$ are positive integers. It is natural to assume $\ell_1 p^b -\ell_2 p^a &gt;0$. In addition, if $a=b$, then the above expression is reduced to $\left \lfloor\frac{(\ell_1 -\ell_2)p^b}{m}\right \rfloor!$, which can be evaluated by using Theorem \ref{theorem1}. We consider the case $b\geq a$ in Theorem \ref{theorem2} and the other case in Theorem \ref{theorem3}.

\begin{theorem}\label{theorem2}
	Let $p$ be a prime, let $a$ be a nonnegative integer, and let $b$, $m$, $\ell_1$, $\ell_2$ be positive integers satisfying $b\geq a$ and $\ell_1 p^b -\ell_2 p^a &gt;0$. Assume that $p\equiv \modd{\pm 1} {m}$. Then the following statements hold.
	\begin{itemize}
		\item[\em{(i)}] If $p\equiv \modd{1} {m}$, then
		\begin{equation*}
			\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor !\right ) = \frac{(\ell_1 p^{b-a} -\ell_2)(p^a -1)}{m(p-1)} - a\left \{\frac{\ell_1 - \ell_2}{m}\right \} + \nu_p \left (\left \lfloor\frac{\ell_1 p^{b-a} - \ell_2}{m}\right \rfloor ! \right ).
		\end{equation*}
		\item[\em{(ii)}] If $p\equiv \modd{-1} {m}$ and $a\equiv 
		\modd{b} {2}$, then
		\begin{align*}
			\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor !\right ) &amp;= \frac{(\ell_1 p^{b-a} -\ell_2)(p^a -1)}{m(p-1)} - \left \{\frac{\ell_1 -\ell_2}{m}\right \}[a\equiv \modd{1} {2}]  \\
			&amp;\phantom{=}- \left \lfloor\frac{a}{2}\right \rfloor[\ell_1 \not\equiv \modd{\ell_2} {m}] + \nu_p \left (\left \lfloor\frac{\ell_1 p^{b-a} - \ell_2}{m}\right \rfloor ! \right ).
		\end{align*}
		\item[\em{(iii)}] If $p\equiv \modd{-1} {m}$ and $a\not\equiv 
		\modd{b} {2}$, then
		\begin{align*}
			\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor !\right ) &amp;= \frac{(\ell_1 p^{b-a} -\ell_2)(p^a -1)}{m(p-1)} - \left \{-\frac{\ell_1 +\ell_2}{m}\right \}[a\equiv \modd{1} {2}]  \\
			&amp;\phantom{=}- \left \lfloor\frac{a}{2}\right \rfloor[\ell_1 \not\equiv \modd{-\ell_2} {m}] + \nu_p \left (\left \lfloor\frac{\ell_1 p^{b-a} - \ell_2}{m}\right \rfloor ! \right ).
		\end{align*}
	\end{itemize}
	We remark that if $m=1$, the expressions in each case of this theorem are equal.
\end{theorem}

\begin{proof}
	The result is easily checked when $a=0$, and as discussed above, if $b=a$, then the result can be verified using Theorem \ref{theorem1}. So we assume throughout that $a\geq 1$ and $b&gt;a$. Similar to the proof of Theorem \ref{theorem1}, we use Lemmas \ref{floorlemma}(i), \ref{floorlemma}(vi), and \ref{sumlemma} repeatedly without reference. Then, as for \eqref{eq1}, we obtain 
	\begin{align}\label{thm2eq1}
		\nu_p \left (\left \lfloor\frac{\ell_1 p^b - \ell_2 p^a}{m}\right \rfloor!\right ) &amp;= \sum_{j=1}^{a} \left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor + \sum_{j=a+1}^{\infty} \left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor \nonumber\\
		&amp;=\sum_{j=1}^{a} \left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor + \nu_p \left (\left \lfloor\frac{\ell_1 p^{b-a} - \ell_2}{m}\right \rfloor ! \right ).
	\end{align}
	We see that when $m=1$, \eqref{thm2eq1} becomes
	\begin{equation*}
		\nu_p \left((\ell_1 p^b - \ell_2 p^a) !\right) = \frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{p-1} + \nu_p \left((\ell_1 p^{b-a} - \ell_2)!\right).
	\end{equation*}
	So assume throughout that $m\geq 2$. We begin with the proof of (i). Suppose that $p\equiv \modd{1} {m}$. For each $1\leq j\leq a$, we have 
	\begin{align*}
		\left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor &amp;= \left\lfloor\frac{\ell_1 p^{b-j}-\ell_1}{m} - \frac{\ell_2 p^{a-j} - \ell_2}{m} + \frac{\ell_1 -\ell_2}{m}\right\rfloor \\
		&amp;= \frac{\ell_1 p^{b-j}- \ell_2 p^{a-j}}{m} - \frac{\ell_1 -\ell_2}{m} + \left \lfloor \frac{\ell_1 -\ell_2 }{m}\right \rfloor.		
	\end{align*}
	Then the sum $\sum_{j=1}^{a} \left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor$ appearing in \eqref{thm2eq1} is equal to
	\begin{align*}
		&amp;\frac{\ell_1}{m}\sum_{1\leq j\leq a} p^{b-j} - \frac{\ell_2}{m}\sum_{1\leq j\leq a} p^{a-j} - a\left (\frac{\ell_1 - \ell_2}{m} - \left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor \right )\\
		&amp;= \frac{\ell_1}{m}\left (\frac{p^{b-a}(p^a -1)}{p-1}\right ) - \frac{\ell_2}{m}\left (\frac{p^a -1}{p-1}\right ) - a\left \{\frac{\ell_1 -\ell_2}{m}\right \} \\
		&amp;= \frac{(\ell_1 p^{b-a} -\ell_2)(p^a -1)}{m(p-1)} - a\left \{\frac{\ell_1 - \ell_2}{m}\right \}.
	\end{align*}
This proves (i). So from this point on, we assume that $p\equiv 
\modd{-1} {m}$. For each $1\leq j\leq a$, we have
	\begin{align*}
			\left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor &amp;= \left\lfloor\frac{\ell_1 p^{b-j} - (-1)^{b-j}\ell_1}{m} - \frac{\ell_2 p^{a-j} - (-1)^{a-j}\ell_2}{m} + \frac{(-1)^{b-j} \ell_1 - (-1)^{a-j}\ell_2}{m}\right\rfloor \\[.1in]
			&amp;= \frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m} - \frac{(-1)^{b-j} \ell_1 - (-1)^{a-j} \ell_2}{m} + \left \lfloor\frac{(-1)^{b-j}\ell_1 - (-1)^{a-j}\ell_2}{m}\right \rfloor\\
			&amp;= 
			\begin{cases}
				\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m} - \frac{(-1)^{b-j}(\ell_1 - \ell_2)}{m} + \left \lfloor\frac{(-1)^{b-j}(\ell_1 -\ell_2)}{m}\right \rfloor, &amp;\text{if $a\equiv \modd{b} {2}$};\\[.1in]
				\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m} - \frac{(-1)^{b-j}(\ell_1 + \ell_2)}{m} + \left \lfloor\frac{(-1)^{b-j}(\ell_1 +\ell_2)}{m}\right \rfloor, &amp;\text{if $a\not\equiv \modd{b} {2}$}.
			\end{cases}
	\end{align*}

\noindent{\bf Case 1.} $a\equiv \modd{b} {2}$. Then the sum $\sum_{j=1}^{a} \left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor$ appearing in \eqref{thm2eq1} is equal to
	\begin{align}\label{thm2eq2}
		&amp;\frac{\ell_1}{m}\sum_{1\leq j\leq a} p^{b-j} - \frac{\ell_2}{m}\sum_{1\leq j\leq a} p^{a-j} - \left (\frac{\ell_1 - \ell_2}{m} \right ) \sum_{1\leq j\leq a} (-1)^{b-j} + \sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 -\ell_2)}{m}\right \rfloor \nonumber\\
		&amp;= \frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{m(p-1)} - \left (\frac{\ell_1 - \ell_2}{m} \right ) \sum_{1\leq j\leq a} (-1)^{b-j} + \sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 -\ell_2)}{m}\right \rfloor.
	\end{align}
	Observe that 
	\begin{equation*}
	\sum_{1\leq j\leq a} (-1)^{b-j} = 
	\begin{cases}
		0, &amp;\text{if $a$ is even};\\
		1, &amp;\text{if $a$ is odd}.
	\end{cases}
	\end{equation*}
	So we have
	\begin{equation*}
		\left (\frac{\ell_1 - \ell_2}{m} \right ) \sum_{1\leq j\leq a} (-1)^{b-j} = \left (\frac{\ell_1 - \ell_2}{m}\right ) [a\equiv \modd{1} {2}].
	\end{equation*}
	It remains to calculate the last term in \eqref{thm2eq2}. If $\ell_1 \equiv \modd{\ell_2} {m}$, then we obtain by Lemma \ref{floorlemma}(iii) that 
	\begin{align*}
		\sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 -\ell_2)}{m}\right \rfloor &amp;= 
		\begin{cases}
			0, &amp;\text{if $a$ is even};\\
			\left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor, &amp;\text{if $a$ is odd};
		\end{cases}\\
&amp;= \left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}].
	\end{align*}
	Similarly, if $\ell_1 \not\equiv \modd{\ell_2} {m}$, then we obtain by Lemma \ref{floorlemma}(iii) that
	\begin{align*}
		\sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 -\ell_2)}{m}\right \rfloor &amp;= 
		\begin{cases}
			-\frac{a}{2}, &amp;\text{if $a$ is even};\\[.1in]
			\left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor - \frac{a-1}{2}, &amp;\text{if $a$ is odd};\\
		\end{cases}\\[.1in]
&amp;= \left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor [a\equiv 
\modd{1} {2}] - \left \lfloor\frac{a}{2}\right \rfloor.
	\end{align*}
	In any case,
$$\sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 -\ell_2)}{m}\right \rfloor = \left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}] - \left \lfloor\frac{a}{2}\right \rfloor[\ell_1 \not\equiv \modd{\ell_2} {m}].$$
Therefore \eqref{thm2eq2} is equal to
	\begin{align*}
		 &amp;\frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{m(p-1)} - \left (\frac{\ell_1 - \ell_2}{m}\right ) [a\equiv \modd{1} {2}] + \left \lfloor\frac{\ell_1 -\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}] \\
		 &amp; - \left \lfloor\frac{a}{2}\right \rfloor[\ell_1 \not\equiv
		 \modd{\ell_2} {m}]\\
		 &amp;=\frac{(\ell_1 p^{b-a} -\ell_2)(p^a -1)}{m(p-1)} - \left \{\frac{\ell_1 -\ell_2}{m}\right \}[a\equiv \modd{1} {2}] - \left \lfloor\frac{a}{2}\right \rfloor[\ell_1 \not\equiv \modd{\ell_2} {m}].
	\end{align*}
	This proves (ii). Next we prove (iii).

\medskip
	
\noindent{\bf Case 2.} $a\not\equiv \modd{b} {2}$. Similar to Case 1, the sum $\sum_{j=1}^{a} \left \lfloor\frac{\ell_1 p^{b-j} - \ell_2 p^{a-j}}{m}\right \rfloor$ appearing in \eqref{thm2eq1} is equal to
	\begin{align}\label{thm2eq3}
		&amp;\frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{m(p-1)} - \left (\frac{\ell_1 + \ell_2}{m} \right ) \sum_{1\leq j\leq a} (-1)^{b-j} + \sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 +\ell_2)}{m}\right \rfloor\nonumber\\
		&amp;= \frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{m(p-1)} + \left (\frac{\ell_1 + \ell_2}{m} \right )[a\equiv \modd{1} {2}] + \sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 +\ell_2)}{m}\right \rfloor.
	\end{align}
	If $\ell_1 \equiv \modd{-\ell_2} {m}$, then we obtain by Lemma \ref{floorlemma}(iii) that 
	\begin{align*}
		\sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 +\ell_2)}{m}\right \rfloor &amp;= 
		\begin{cases}
				0, &amp;\text{if $a$ is even};\\
				\left \lfloor-\frac{\ell_1 +\ell_2}{m}\right \rfloor, &amp;\text{if $a$ is odd};
		\end{cases}\\
		&amp;= \left \lfloor-\frac{\ell_1 +\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}].
	\end{align*}
	Similarly, if $\ell_1 \not\equiv \modd{-\ell_2} {m}$, then we obtain by Lemma \ref{floorlemma}(iii) that 
	\begin{align*}
		\sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 +\ell_2)}{m}\right \rfloor &amp;= 
		\begin{cases}
			-\frac{a}{2}, &amp;\text{if $a$ is even};\\
			\left \lfloor-\frac{\ell_1 +\ell_2}{m}\right \rfloor - \frac{a-1}{2}, &amp;\text{if $a$ is odd};
		\end{cases}\\
		&amp;= \left \lfloor-\frac{\ell_1 +\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}] - \left \lfloor\frac{a}{2}\right \rfloor.
	\end{align*}
	In any case, $\sum_{1\leq j\leq a} \left \lfloor\frac{(-1)^{b-j}(\ell_1 +\ell_2)}{m}\right \rfloor = \left \lfloor-\frac{\ell_1 +\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}] - \left \lfloor\frac{a}{2}\right \rfloor [\ell_1 \not\equiv \modd{-\ell_2} {m}]$. Therefore \eqref{thm2eq3} is equal to
	\begin{align*}
		&amp;\frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{m(p-1)} + \left (\frac{\ell_1 + \ell_2}{m} \right )[a\equiv \modd{1} {2}] + \left \lfloor-\frac{\ell_1 +\ell_2}{m}\right \rfloor [a\equiv \modd{1} {2}]\\
		&amp; - \left \lfloor\frac{a}{2}\right \rfloor [\ell_1 \not\equiv 
		\modd{-\ell_2} {m}]\\
		&amp;=\frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{m(p-1)} - \left \{-\frac{\ell_1 +\ell_2}{m}\right \}[a\equiv \modd{1} {2}] - \left \lfloor\frac{a}{2}\right \rfloor [\ell_1 \not\equiv \modd{-\ell_2} {m}].
	\end{align*}
	This completes the proof.
\end{proof}

Next we replace the assumption $b\geq a$ in Theorem \ref{theorem2} by $b&lt;a$. The calculation follows from the same idea so we skip the details of the proof. Although we do not use it in this article, it may be useful for future reference. So we record it in the next theorem.

\begin{theorem}\label{theorem3}
	Let $p$ be a prime, let $b$ be a nonnegative integer, and let $a$, $m$, $\ell_1$, $\ell_2$ be positive integers satisfying $b&lt;a$ and $\ell_1 p^b -\ell_2 p^a &gt;0$. Assume that $p\equiv \modd{\pm 1} {m}$. Then the following statements hold.
	\begin{itemize}
		\item[\em{(i)}] If $p\equiv \modd{1} {m}$, then
		\begin{equation*}
			\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor !\right ) = \frac{(\ell_1  -\ell_2 p^{a-b})(p^b -1)}{m(p-1)} - b\left \{\frac{\ell_1 - \ell_2}{m}\right \} + \nu_p \left (\left \lfloor\frac{\ell_1  - \ell_2 p^{a-b}}{m}\right \rfloor ! \right ).
		\end{equation*}\\
		\item[\em{(ii)}] If $p\equiv \modd{-1} {m}$ and $a\equiv 
		\modd{b} {2}$, then
		\begin{align*}
			\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor !\right ) &amp;= \frac{(\ell_1  -\ell_2 p^{a-b})(p^b -1)}{m(p-1)} - \left \{\frac{\ell_1 -\ell_2}{m}\right \}[b\equiv \modd{1} {2}]  \\
			&amp;\phantom{=}- \left \lfloor\frac{b}{2}\right \rfloor[\ell_1 \not\equiv \modd{\ell_2} {m}] + \nu_p \left (\left \lfloor\frac{\ell_1  - \ell_2 p^{a-b}}{m}\right \rfloor ! \right ).
		\end{align*}\\
		\item[\em{(iii)}] If $p\equiv \modd{-1} {m}$ and
		$a\not\equiv \modd{b} {2}$, then
		\begin{align*}
			\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor !\right ) &amp;= \frac{(\ell_1  -\ell_2 p^{a-b})(p^b -1)}{m(p-1)} - \left \{\frac{\ell_1 +\ell_2}{m}\right \}[b\equiv \modd{1} {2}]  \\
			&amp;\phantom{=}- \left \lfloor\frac{b}{2}\right \rfloor[\ell_1 \not\equiv \modd{\ell_2} {m}] + \nu_p \left (\left \lfloor\frac{\ell_1  - \ell_2 p^{a-b}}{m}\right \rfloor ! \right ).
		\end{align*}
	\end{itemize}
\end{theorem}

\begin{proof}
	We begin by writing $\nu_p \left (\left \lfloor\frac{\ell_1 p^b -\ell_2 p^a}{m}\right \rfloor!\right ) $ as 
	\begin{equation*}
		\sum_{j=1}^{b} \left \lfloor\frac{\ell_1 p^{b-j}- \ell_2 p^{a-j}}{m}\right \rfloor + \sum_{j=b+1}^{\infty} \left \lfloor\frac{\ell_1 p^{b-j}- \ell_2 p^{a-j}}{m}\right \rfloor.
	\end{equation*}
	The second sum above is $\nu_p \left (\left \lfloor\frac{\ell_1  - \ell_2 p^{a-b}}{m}\right \rfloor ! \right )$. The first sum can be evaluated in the same way as in Theorem \ref{theorem2}. We leave the details to the reader.
\end{proof}

When we put more restrictions on the range of $\ell_1$ and $\ell_2$, the expression $\nu_p \left (\left \lfloor\frac{\ell_1 p^{b-a} - \ell_2}{m}\right \rfloor ! \right )$ appearing in Theorems \ref{theorem2} and \ref{theorem3} can be evaluated further. Nevertheless, since we do not need it in our application, we do not give them here. In the future, we plan to put it in the second author's Researchgate account. So the interested reader can find it there.


\section{The $p$-adic valuations of Fibonomial coefficients}\label{sec4}

Recall that the \textit{binomial coefficients} ${m\choose k}$ is defined by
\begin{equation*}
{m\choose k}=
\begin{cases}
\frac{m!}{k!(m-k)!}, &amp;\text{if $0\leq k\leq m$;}\\
0, &amp;\text{if $k&lt;0$ or $k&gt;m$.}
\end{cases}
\end{equation*}
A classical result of Kummer states that for $0\leq k\leq m$, $\nu_p\left( {m \choose k} \right)$ is equal to the number of carries when we add $k$ and $m-k$ in base $p$. From this, it is not difficult to show that for all primes $p$ and positive integers $k$, $b$, $a$ with $b\geq a$, we have
\begin{equation*}
\nu_p\left({p^b \choose p^a}\right) = b-a, \quad\text{or more generally,}\quad \nu_p\left({p^a \choose k}\right) = a-\nu_p (k).
\end{equation*}
Knuth and Wilf \cite{KW} also obtain the result analogous to that of Kummer for a C-nomial coefficient. However, our purpose is to obtain $\nu_p\left( {m \choose k}_F \right)$ is an explicit form. So we first express $\nu_p\left( {m \choose k}_F \right)$ in terms of the $p$-adic valuation of some binomial coefficients in Theorem \ref{theorem5}. Then we write it in a form which is easy to use in Corollary \ref{corollary5}. Then we apply it to obtain the $p$-adic valuation of Fibonomial coefficients of the form ${\ell_1 p^b \choose \ell_2 p^a}_F$.

\begin{theorem}\label{theorem5}
Let $0\leq k\leq m$ be integers. Then the following statements hold.
\begin{itemize}
\item[\em{(i)}] Let $m'=\left\lfloor\frac{m}{6}\right\rfloor$, $k'=\left\lfloor\frac{k}{6}\right\rfloor$, and let $r=m \bmod 6$ and $s=k\bmod 6$ be the least nonnegative residues of $m$ and $k$ modulo $6$, respectively. Then
\begin{align*}
\nu_2\left({m \choose k}_F\right) &amp;= \nu_2\left({m' \choose k'}\right) + \left\lfloor\frac{r+3}{6}\right\rfloor - \left\lfloor\frac{r-s+3}{6}\right\rfloor - \left\lfloor\frac{s+3}{6}\right\rfloor-3\left\lfloor\frac{r-s}{6}\right\rfloor\\
&amp; + [r&lt;s]\nu_2\left (\left \lfloor\frac{m-k+6}{6}\right \rfloor\right ).
\end{align*}
\item[\em{(ii)}] $\nu_5\left({m \choose k}_F\right) = \nu_5\left({m \choose k}\right)$.
\item[\em{(iii)}] Suppose that $p$ is a prime, $p\neq 2$, and $p\neq 5$. Let $m'=\left\lfloor\frac{m}{z(p)}\right\rfloor$, $k'=\left\lfloor\frac{k}{z(p)}\right\rfloor$, and let $r=m\bmod z(p)$, and $s=k\bmod z(p)$ be the least nonnegative residues of $m$ and $k$ modulo $z(p)$, respectively. Then
\begin{align*}
\nu_p\left({m \choose k}_F\right) &amp;=  \nu_p \left({m' \choose k'}\right) + [r&lt;s]\left (\nu_p\left(\left\lfloor\frac{m-k+z(p)}{z(p)}\right\rfloor\right) +\nu_p(F_{z(p)})\right ).
\end{align*}
\end{itemize}
\end{theorem}

\begin{proof}
We will use Lemmas \ref{floorlemma}(i) and \ref{sumlemma} repeatedly without reference. In addition, it is useful to recall that for every $a,b \in \mathbb{N}$, $\nu_p (ab) = \nu_p(a) + \nu_p (b)$ and  if $b \mid a$, then $\nu_p (\frac{a}{b})= \nu_p (a)- \nu_p (b)$. Since the formulas to prove clearly hold when $k=0$ or $m$, we assume $m\geq 2$ and $1\leq k &lt; m$. 
By Lemma \ref{lengyel}, we obtain, for every $\ell \geq 1$,
\begin{align}\label{eq8}
\nu_2(F_1 F_2 F_3\cdots F_\ell) &amp;= \sum_{\substack{1\leq n\leq \ell\\
n\, \equiv\, \text{3 (mod 6)}}} \nu_2(F_n) + \sum_{\substack{1\leq n\leq \ell\\
n\, \equiv\, \text{0 (mod 6)}}} \nu_2(F_n)\nonumber\\
&amp;= \sum_{\substack{1\leq n\leq \ell\\
n\, \equiv\, \text{3 (mod 6)}}} 1 + \sum_{\substack{1\leq n\leq \ell\\
n\, \equiv\, \text{0 (mod 6)}}} (\nu_2(n)+2)\nonumber\\
&amp;=\left\lfloor\frac{\ell +3}{6}\right\rfloor + 2\left\lfloor\frac{\ell}{6}\right\rfloor + \sum_{1\leq j\leq \frac{\ell}{6}} \nu_2 (6j)\nonumber\\
&amp;=\left\lfloor\frac{\ell +3}{6}\right\rfloor + 3\left\lfloor\frac{\ell}{6}\right\rfloor + \sum_{1\leq j\leq \frac{\ell}{6}} \nu_2 (j)\nonumber\\
&amp;=\left\lfloor\frac{\ell +3}{6}\right\rfloor +3\left\lfloor\frac{\ell}{6}\right\rfloor+\nu_2\left(\left\lfloor\frac{\ell}{6}\right\rfloor !\right).
\end{align}
Then we obtain from the definition of ${m \choose k}_F$ and from \eqref{eq8} that
\begin{align}\label{eq9}
\nu_2\left({m \choose k}_F\right) &amp;= \nu_2(F_1 F_2 \cdots F_m) - \nu_2(F_1 F_2 \cdots F_{m-k}) - \nu_2(F_1 F_2\cdots F_k)\nonumber\\
&amp;= \left(\left\lfloor\frac{m+3}{6}\right\rfloor - \left\lfloor\frac{m-k+3}{6}\right\rfloor - \left\lfloor\frac{k+3}{6}\right\rfloor\right) + 3\left(\left\lfloor\frac{m}{6}\right\rfloor - \left\lfloor\frac{m-k}{6}\right\rfloor -\left\lfloor\frac{k}{6}\right\rfloor\right)\nonumber\\
&amp;\phantom{=} +\nu_2\left(\left\lfloor\frac{m}{6}\right\rfloor !\right) -\nu_2\left(\left\lfloor\frac{m-k}{6}\right\rfloor !\right) - \nu_2\left(\left\lfloor\frac{k}{6}\right\rfloor !\right).
\end{align}
The expression in the first parenthesis in \eqref{eq9} is equal to
\begin{align*}
&amp;\left\lfloor\frac{m-r}{6}+\frac{r+3}{6}\right\rfloor - \left\lfloor\frac{(m-r)-(k-s)}{6} + \frac{r-s+3}{6}\right\rfloor - \left\lfloor \frac{k-s}{6}+\frac{s+3}{6}\right\rfloor \nonumber\\
&amp;= \frac{m-r}{6} + \left\lfloor\frac{r+3}{6}\right\rfloor - \frac{(m-r)-(k-s)}{6} - \left\lfloor\frac{r-s+3}{6}\right\rfloor - \frac{k-s}{6} - \left\lfloor\frac{s+3}{6}\right\rfloor \nonumber\\
&amp;=\left\lfloor\frac{r+3}{6}\right\rfloor - \left\lfloor\frac{r-s+3}{6}\right\rfloor - \left\lfloor\frac{s+3}{6}\right\rfloor.
\end{align*}
Similarly, the expression in the second parenthesis is
\begin{equation*}
3\left(\left\lfloor\frac{r}{6}\right\rfloor - \left\lfloor\frac{r-s}{6}\right\rfloor - \left\lfloor\frac{s}{6}\right\rfloor\right) = -3\left\lfloor\frac{r-s}{6}\right\rfloor.
\end{equation*}
Therefore \eqref{eq9} becomes
\begin{equation}\label{eq10}
\nu_2\left({m \choose k}_F\right) = \left\lfloor\frac{r+3}{6}\right\rfloor - \left\lfloor\frac{r-s+3}{6}\right\rfloor - \left\lfloor\frac{s+3}{6}\right\rfloor-3\left\lfloor\frac{r-s}{6}\right\rfloor + \nu_2\left(\frac{\lfloor x+y\rfloor !}{\lfloor x\rfloor !\lfloor y\rfloor !}\right)
\end{equation}
where $x=\frac{m-k}{6}$ and $y=\frac{k}{6}$. By Lemma \ref{floorlemma}(v), 
we see that
\begin{align*}
\frac{\lfloor x+y\rfloor !}{\lfloor x\rfloor !\lfloor y\rfloor !} &amp;= 
\begin{cases}
{\lfloor x+y\rfloor \choose \lfloor y\rfloor}, &amp;\text{if $\{x\}+\{y\} &lt;1$;}\\[.1in]
{\lfloor x+y\rfloor \choose \lfloor y\rfloor}(\lfloor x\rfloor +1), &amp;\text{if $\{x\}+\{y\} \geq1$;}
\end{cases}\\
&amp;= 
\begin{cases}
{m' \choose k'}, &amp;\text{if $\{x\}+\{y\} &lt;1$;}\\[.1in]
{m' \choose k'}\left(\left\lfloor\frac{m-k+6}{6}\right\rfloor\right), &amp;\text{if $\{x\}+\{y\} \geq1$.}
\end{cases}
\end{align*}
By Lemma \ref{floorlemma}(ii), we obtain
\begin{equation*}
\{x\}=\left\{\frac{(m-r)-(k-s)}{6} + \frac{r-s}{6}\right\} = \left\{\frac{r-s}{6}\right\} \text{ and } \{y\}=\left\{\frac{k-s}{6} + \frac{s}{6}\right\} = \frac{s}{6}.
\end{equation*}
If $r\geq s$, then $\{x\}+\{y\}=\left\{\frac{r-s}{6}\right\} + \frac{s}{6} = \frac{r-s}{6} + \frac{s}{6} = \frac{r}{6} &lt;1$. If $r&lt;s$, then we obtain by Lemma \ref{floorlemma}(iv) that $\{x\}+\{y\}= \left\{-\frac{s-r}{6}\right\} + \frac{s}{6} = 1-\frac{s-r}{6} + \frac{s}{6} = 1 + \frac{r}{6} \geq 1$. Therefore 
\begin{equation}\label{eq11}
\frac{\lfloor x+y\rfloor !}{\lfloor x\rfloor !\lfloor y\rfloor !} =
\begin{cases}
{m' \choose k'}, &amp;\text{if $r\geq s$;}\\[.1in]
{m' \choose k'}\left(\left\lfloor\frac{m-k+6}{6}\right\rfloor\right), &amp;\text{if $r&lt;s$.}
\end{cases}
\end{equation}
Substituting \eqref{eq11} in \eqref{eq10}, we obtain part (i) of this theorem. The calculation in parts (ii) and (iii) are similar, so we give fewer details than given in part (i). By Lemma \ref{lengyel}, for every $\ell \geq 1$, we have 
\begin{equation*}
\nu_5(F_1 F_2\cdots F_\ell) = \sum_{1\leq n\leq \ell} \nu_5 (F_n) = \sum_{1\leq n \leq \ell} \nu_5(n) = \nu_5(\ell !),
\end{equation*}
which implies
\begin{equation*}
\nu_5\left({m \choose k}_F\right) = \nu_5(m!) - \nu_5(k!) - \nu_5((m-k)!) = \nu_5\left({m \choose k}\right).
\end{equation*}
For (iii), we apply Lemmas \ref{lengyel} and \ref{lemma2.2}(iv) to obtain 
\begin{align*}
\nu_p(F_1 F_2 \cdots F_\ell) &amp;= \sum_{\substack{1\leq n\leq \ell\\
										n\, \equiv\, 0\ (\text{mod}\ z(p))}} \nu_p (F_n) = \sum_{\substack{1\leq n\leq \ell\\
										n\, \equiv\, 0\ (\text{mod}\ z(p))}} (\nu_p(n) + \nu_p(F_{z(p)}))\\
&amp;= \sum_{1\leq k\leq \frac{\ell}{z(p)}} \nu_p(kz(p)) + \left\lfloor\frac{\ell}{z(p)}\right\rfloor\nu_p(F_{z(p)}) \\
&amp;= \nu_p\left(\left\lfloor\frac{\ell}{z(p)}\right\rfloor !\right) + \left\lfloor\frac{\ell}{z(p)}\right\rfloor\nu_p(F_{z(p)}).
\end{align*}
As in part (i), the above implies that
\begin{equation}\label{eq13}
\nu_p\left({m \choose k}_F\right) = \nu_p\left(\frac{\lfloor x+y\rfloor !}{\lfloor x\rfloor !\lfloor y\rfloor !}\right) + (\lfloor x+y \rfloor - \lfloor x\rfloor - \lfloor y\rfloor )\nu_p(F_{z(p)}),
\end{equation}
where $x=\frac{m-k}{z(p)}$ and $y=\frac{k}{z(p)}$. In addition, if $r \geq s$, then $\{x\}+\{y\}&lt;1$ and if $r&lt;s$, then $\{x\}+\{y\}\geq 1$. Therefore \eqref{eq13} can be simplified to the desired result. This completes the proof.
\end{proof}

By Theorem \ref{theorem5}(ii), we see that the 5-adic valuations of Fibonomial and binomial coefficients are the same. So we focus our investigation only on the $p$-adic valuations of Fibonomial coefficients when $p\neq 5$. Calculating $r$ and $s$ in Theorem \ref{theorem5}(i) in every case and writing Theorem \ref{theorem5}(iii) in another form, we obtain the following corollary.

\begin{corollary}\label{corollary5}
Let $m$, $k$, $r$, and $s$ be as in Theorem \ref{theorem5}. Let 
\begin{equation*}
A_2 = \nu_2\left(\left\lfloor \frac{m}{6} \right\rfloor!\right) - \nu_2\left(\left\lfloor \frac{k}{6} \right\rfloor!\right) -\nu_2\left(\left\lfloor \frac{m-k}{6} \right\rfloor!\right),
\end{equation*}
and for each prime $p\neq 2,5$, let $A_p = \nu_p\left(\left\lfloor \frac{m}{z(p)} \right\rfloor!\right) - \nu_p\left(\left\lfloor \frac{k}{z(p)} \right\rfloor!\right) -\nu_p\left(\left\lfloor \frac{m-k}{z(p)} \right\rfloor!\right)$. Then the following statements hold.
\begin{itemize}
\item[\em{(i)}] $\displaystyle \nu_2\left({m \choose k}_F\right) = 
\begin{cases}
A_2, &amp;\text{if $r\geq s$ and $(r,s)\neq (3,1),(3,2),(4,2)$;}\\
A_2 + 1, &amp;\text{if $(r,s) = (3,1),(3,2),(4,2)$;}\\
A_2+3, &amp;\text{if $r&lt;s$ and $(r,s)\neq (0,3),(1,3),(2,3)$,}\\
&amp;(1,4),(2,4),(2,5);\\
A_2 + 2, &amp;\text{if $(r,s)= (0,3),(1,3),(2,3),(1,4),(2,4)$,}\\
&amp; (2,5).
\end{cases}$
\item[\em{(ii)}] For $p\neq 2,5$, we have 
\begin{equation*}
	\nu_p\left({m \choose k}_F \right) = 
	\begin{cases}
		A_p, &amp;\text{if $r\geq s$;}\\
		A_p + \nu_p(F_{z(p)}), &amp;\text{if $r&lt;s$.}
	\end{cases}
\end{equation*}
\end{itemize}
\end{corollary}

\begin{proof}
For (i), we have $0\leq r\leq 5$ and $0\leq s\leq 5$, so we can directly consider every case and reduce Theorem \ref{theorem5}(i) to the result in this corollary. In addition, (ii) follows directly from  \eqref{eq13}.
\end{proof}

In a series of papers (see \cite{Mar3} and references therein), Marques and Trojovsk\'y obtain a formula for $\nu_p\left({p^b \choose p^a}_F\right)$ only when $b=a+1$. Then Ballot \cite{ballot} extends it to any case $b&gt;a$. Corollary \ref{corollary5} enables us to compute $\nu_p\left({\ell_1 p^b \choose \ell_2 p^a}_F\right)$. We illustrate this in the next theorem.

\begin{theorem}\label{theorem4.3}
	Let $a$, $b$, $\ell_1$, and $\ell_2$ be positive integers and $b\geq a$. Let $p\neq 5$ be a prime. Assume that $\ell_1 p^b &gt;\ell_2 p^a$ and let $m_p = \left\lfloor\frac{\ell_1 p^{b-a}}{z(p)}\right\rfloor$ and $k_p =\left\lfloor\frac{\ell_2}{z(p)}\right\rfloor$. Then the following statements hold.
	\begin{itemize}
		\item[\em{(i)}] If $a\equiv \modd{b} {2}$, then $\nu_2\left( {\ell_1 2^b \choose \ell_2 2^a}_F \right)$ is equal to
		\begin{equation*}
			\begin{cases}
				\nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{\ell_2} {3}$ or $\ell_2 \equiv \modd{0} {3}$;}\\
				a + 2 + \nu_2 \left( m_2	- 	k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
				\modd{0} {3}$ and $\ell_2 \not\equiv 
				\modd{0} {3}$;}\\
				\left\lceil \frac{a}{2} \right\rceil + 1 + \nu_2 \left( m_2	- k_2 \right) + \nu_2 \left( { m_2 \choose  k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{1} {3}$ and $\ell_2 \equiv \modd{2} {3}$;}\\
				\left\lceil\frac{a+1}{2}\right\rceil + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{2} {3}$ and $\ell_2 \equiv \modd{1} {3}$,}
			\end{cases}
		\end{equation*}
		and if $a\not\equiv \modd{b} {2}$, then $\nu_2\left( {\ell_1 2^b \choose \ell_2 2^a}_F \right)$ is equal to
		\begin{equation*}
			\begin{cases}
				\nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{-\ell_2} {3}$ or $\ell_2 \equiv \modd{0} {3}$;}\\
				a + 2 + \nu_2 \left( m_2	- k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
				\modd{0} {3}$ and $\ell_2 \not\equiv 
				\modd{0} {3}$;}\\
				\left\lceil \frac{a+1}{2} \right\rceil + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
				\modd{1} {3}$ and $\ell_2 \equiv 
				\modd{1} {3}$;}\\
				\left\lceil\frac{a}{2}\right\rceil + 1 + \nu_2 \left( m_2	- k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{2} {3}$ and $\ell_2 \equiv \modd{2} {3}$.}
			\end{cases}
		\end{equation*}
		\item[\em{(ii)}] Let $p\neq 5$ be an odd prime and let $r=\ell_1 p^b \mod{z(p)}$ and $s=\ell_2 p^a \mod{z(p)}$. If $p\equiv \pm 
		\modd{1} {5}$, then
		\begin{equation*}
			\nu_p\left( {\ell_1 p^b \choose \ell_2 p^a}_F \right) = [r&lt;s]\left(a + \nu_p \left( m_p	- k_p \right) + \nu_p(F_{z(p)}) \right) + \nu_p \left( {m_p \choose k_p} \right),
		\end{equation*} 
		and if $p\equiv \modd{\pm 2} {5}$, then $\nu_p\left( {\ell_1 p^b \choose \ell_2 p^a}_F \right)$ is equal to
		\begin{equation*}
			\begin{cases}
				\nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r=s$ or $\ell_2 \equiv \modd{0} {z(p)}$;}\\
				a + \nu_p (F_{z(p)}) + \nu_p \left( m_p	- 	k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 
\equiv \modd{0} {z(p)}$ and}\\
				&amp; \ell_2 \not\equiv \modd{0} {z(p)};\\
				\frac{a}{2} + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&gt; s$, $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$,}\\
				&amp;\text{and $a$ is even;}\\ 
				\frac{a}{2} + \nu_p (F_{z(p)}) + \nu_p \left( m_p	- 	k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&lt;s$, $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$,}\\
				&amp;\text{and $a$ is even;}\\
				\frac{a+1}{2} + \nu_p \left( m_p	- k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&gt;s$, $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$,}\\
				&amp;\text{and $a$ is odd;}\\
				\frac{a-1}{2} + \nu_p(F_{z(p)}) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&lt;s$, $\ell_1,\ell_2 \not\equiv 
				\modd{0} {z(p)}$,}\\
				&amp;\text{and $a$ is odd.}
			\end{cases}
		\end{equation*}	
	\end{itemize} 
\end{theorem}

\begin{remark}\label{remark4.4}
 In the proof of this theorem, we also show that the condition $r=s$ in Theorem \ref{theorem4.3}(ii) is equivalent to 
$\ell_1 \equiv \modd{\ell_2 - 2\ell_2 [a\not\equiv  \modd{b} {2}]} {z(p)}$.
It seems more natural to write $r=s$ in the statement of the theorem,
but it is more convenient in the proof
to use the condition 
$\ell_1 \equiv \modd{\ell_2 - 2\ell_2 [a\not\equiv 
\modd{b} {2}]} {z(p)}$.
\end{remark}

\begin{proof}[Proof of Theorem \ref{theorem4.3}]
	We apply Corollary \ref{corollary5} to calculate $\nu_2\left( {\ell_1 2^b \choose \ell_2 2^a}_F \right)$ with $m=\ell_1 2^b$, $k= \ell_2 2^a$, $r = \ell_1 2^b \mod{6}$, and $s=\ell_2 2^a \mod 6$. For convenience, we also let $r' = \ell_1 \mod{3}$, and $s' = \ell_2 \mod{3}$. Therefore $A_2$ given in Corollary \ref{corollary5} is
	\begin{equation}\label{eq1cor4.3}
		A_2 = \nu_2\left(\left\lfloor \frac{\ell_1 2^{b-1}}{3} \right\rfloor!\right) - \nu_2 \left(\left\lfloor \frac{\ell_2 2^{a-1}}{3} \right\rfloor!\right) - \nu_2 \left(\left\lfloor \frac{\ell_1 2^{b-1} - \ell_2 2^{a-1}}{3} \right\rfloor!\right).
	\end{equation}
	By Corollary \ref{corollary1}, the first term on the right-hand side of \eqref{eq1cor4.3} is equal to
	\begin{align}\label{eq1thm4.3}
		&amp;\frac{\ell_1 (2^{b-1}-1)}{3} - \left\lfloor \frac{b-1}{2} \right\rfloor [\ell_1 \not\equiv \modd{0} {3}] - \left\{\frac{\ell_1}{3}\right\}[b\equiv \modd{0} {2}] + \nu_2 \left( \left\lfloor \frac{\ell_1}{3} \right\rfloor ! \right)\nonumber\\
		&amp;= \frac{\ell_1 (2^{b-1}-1)}{3} - \left\lfloor \frac{b-1}{2} \right\rfloor [r' \neq 0] - \frac{r'}{3}[b\equiv \modd{0} {2}] + \nu_2 \left( \left\lfloor \frac{\ell_1}{3} \right\rfloor ! \right).
	\end{align}
	Similarly, the second term is 
	\begin{equation}\label{eq2thm4.3}
		\frac{\ell_2 (2^{a-1}-1)}{3} - \left\lfloor \frac{a-1}{2} \right\rfloor [s' \neq 0] - \frac{s'}{3}[a\equiv \modd{0} {2}] + \nu_2 \left( \left\lfloor \frac{\ell_2}{3} \right\rfloor ! \right).
	\end{equation}
	To evaluate the third term on the right-hand side of \eqref{eq1cor4.3}, we divide the proof into two cases according to the parity of $a$ and $b$.

\medskip
	
\noindent{\bf Case 1.} $a\equiv \modd{b} {2}$.
Observe that $\ell_1 \equiv \modd{\ell_2} {3}$ if
and only if $r'=s'$. In addition, $\left\{ \frac{\ell_1 - \ell_2}{3} \right\} = \left\{\frac{r'-s'}{3} \right\}$ and $\left \lfloor\frac{r'-s'}{3}\right \rfloor = -[r'&lt;s'] $. Then by Theorem \ref{theorem2}, the third term on the right-hand side of \eqref{eq1cor4.3} is equal to 
	\begin{align}\label{eq3thm4.3}
		&amp;\frac{(\ell_1 2^{b-a} - \ell_2 )(2^{a-1} -1)}{3} - \left\{\frac{r'-s'}{3}\right\}[a \equiv \modd{0} {2}] - \left\lfloor\frac{a-1}{2}\right\rfloor[r' \neq s'] + \nu_2 \left( \left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor ! \right)\nonumber\\
		&amp;= \frac{(\ell_1 2^{b-a} - \ell_2 )(2^{a-1} -1)}{3} - \left(\frac{r'-s'}{3} + [r'&lt;s']\right)[a \equiv \modd{0} {2}] - \left\lfloor\frac{a-1}{2}\right\rfloor[r' \neq s'] \nonumber\\
		&amp;\phantom{=} + \nu_2 \left( \left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor ! \right).
	\end{align}
	Recall that $m_2 = \left \lfloor\frac{\ell_1 2^{b-a}}{3}\right \rfloor$ and $k_2 =\left \lfloor\frac{\ell_2}{3}\right \rfloor$. Since $b-a$ is even, $2^{b-a} \equiv \modd{1} {3}$ and we obtain by Lemma \ref{lemma2.7} that
	\begin{equation*}
		\left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor = m_2 - k_2 - [r'&lt;s'].
	\end{equation*}
	Therefore $\nu_2 \left( \left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor ! \right)$ is equal to
	\begin{equation*}
		\nu_2(m_2 !) -[r'&lt;s']\nu_2(m_2 -k_2) - \nu_2 \left(\frac{m_2 !}{(m_2 - k_2)!} \right).
	\end{equation*}
	By Corollary \ref{corollary1}, $\nu_2 (m_2 !)$ is equal to
	\begin{equation*}
		\frac{\ell_1 (2^{b-a} -1)}{3} - \frac{b-a}{2}[r'\neq 0] + \nu_2 \left(\left\lfloor\frac{\ell_1}{3}\right\rfloor !\right).
	\end{equation*}
	We substitute the value of $\nu_2 \left( \left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor ! \right)$ in \eqref{eq3thm4.3} and then substitute \eqref{eq1thm4.3}, \eqref{eq2thm4.3}, and \eqref{eq3thm4.3} in \eqref{eq1cor4.3} to obtain $A_2$. We see that there are some cancellations. For instance,
	\begin{equation*}
		\frac{r'}{3}\Big([a\equiv \modd{0} {2}] - [b\equiv 
		\modd{0} {2}]\Big)=0. 
	\end{equation*} 
	and 
	\begin{equation*}
		\nu_2\left( \frac{m_2 !}{(m_2 -k_2)!} \right) - \nu_2 \left( \left \lfloor\frac{\ell_2}{3}\right \rfloor ! \right) = \nu_2\left({m_2 \choose k_2}\right)
	\end{equation*}
	Then we obtain
	\begin{align}\label{eq2cor4.3}
		A_2 &amp;= - \left\lfloor \frac{b-1}{2} \right\rfloor [r' \neq 0] + \left\lfloor\frac{a-1}{2}\right\rfloor [s'\neq 0] +[r'&lt;s'][a\equiv 
		\modd{0} {2}]  \nonumber\\
		&amp;\phantom{=} + \left\lfloor\frac{a-1}{2}\right\rfloor[r'\neq s'] + \frac{b-a}{2}[r'\neq 0] + [r'&lt;s']\nu_2 (m_2-k_2) + \nu_2\left({m_2 \choose k_2}\right).
	\end{align}
	Next we divide the calculation of $A_2$ into 4 cases:
	\begin{itemize}
		\item Case 1.1. $\ell_1 \equiv 
		\modd{\ell_2} {3}$ or $\ell_2\equiv \modd{0} {3}$,
		\item Case 1.2. $\ell_1 \equiv 
		\modd{0} {3}$ and $\ell_2\not\equiv 
		\modd{0} {3}$,
		\item Case 1.3. $\ell_1 \equiv 
		\modd{1} {3}$ and $\ell_2\equiv 
		\modd{2} {3}$,
		\item Case 1.4. $\ell_1 \equiv 
		\modd{2} {3}$ and $\ell_2\equiv 
		\modd{1} {3}$.
	\end{itemize}
	Since the calculation in each case is similar, we only show the details in Case 1.1 and Case 1.2. So assume that $\ell_1 \equiv 
	\modd{\ell_2} {3}$. Then $r'=s'$ and \eqref{eq2cor4.3} becomes
	\begin{equation*}
		A_2 = -\left \lfloor\frac{b-1}{2}\right \rfloor[r'\neq 0] + \left \lfloor\frac{a-1}{2}\right \rfloor[r'\neq 0] + \frac{b-a}{2}[r'\neq 0] + \nu_2\left({m_2 \choose k_2} \right)
	\end{equation*}
	Since $-\left \lfloor\frac{b-1}{2}\right \rfloor + \left \lfloor\frac{a-1}{2}\right \rfloor + \frac{b-a}{2}=0$, we see that $A_2 = \nu_2\left({m_2 \choose k_2} \right)$. Next if $\ell_2 \equiv 
	\modd{0} {3}$, then $s'=0$ and the same calculation leads to $A_2 = \nu_2\left({m_2 \choose k_2} \right)$. Next assume that $\ell_1 \equiv 
	\modd{0} {3}$ and $\ell_2\not\equiv 
	\modd{0} {3}$. Then $r'=0$, $s'\neq 0$, and \eqref{eq2cor4.3} becomes
	\begin{equation*}
		A_2 = \left \lfloor\frac{a-1}{2}\right \rfloor + [a\equiv 
		\modd{0} {2}] + \left \lfloor\frac{a-1}{2}\right \rfloor + \nu_2(m_2 - k_2) + \nu_2 \left( {m_2 \choose k_2} \right).
	\end{equation*}
	Observing that the sum of the first three terms above is equal to $a-1$, we obtain $A_2 = a - 1 + \nu_2(m_2 - k_2) + \nu_2 \left( {m_2 \choose k_2} \right)$. The other cases are similar. Therefore $A_2$ is 
	\begin{equation*} 
		\begin{cases}
			\nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{\ell_2} {3}$ or $\ell_2 \equiv 
			\modd{0} {3}$;}\\	
			a - 1 + \nu_2 \left( m_2 - k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
			\modd{0} {3}$ and $\ell_2 \not\equiv 
			\modd{0} {3}$;}\\
			\left\lfloor \frac{a}{2} \right\rfloor  + \nu_2 \left( m_2 - k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
\modd{1} {3}$ and $\ell_2 \equiv \modd{2} {3}$;}\\
			\left\lfloor\frac{a-1}{2}\right\rfloor + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
			\modd{2} {3}$ and $\ell_2 \equiv 
			\modd{1} {3}$.}
		\end{cases}
	\end{equation*}
	Recall that $r = \ell_1 2^{b} \mod{6}$ and $s = \ell_2 2^{a} \mod{6}$. Therefore
	\begin{equation*}
		r = 
		\begin{cases}
			0, &amp;\text{if $\ell_1 \equiv 
			\modd{0} {3}$;}\\
			2, &amp;\text{if $b$ is even and $\ell_1 \equiv 
			\modd{2} {3}$ or if $b$ is odd and $\ell_1 \equiv 
			\modd{1} {3}$;}\\
			4 &amp;\text{if $b$ is even and $\ell_1 \equiv 
			\modd{1} {3}$ or if $b$ is odd and $\ell_1 \equiv 
			\modd{2} {3}$,}
		\end{cases}
	\end{equation*}
	and
	\begin{equation*}
		s = 
		\begin{cases}
			0, &amp;\text{if $\ell_2 \equiv 
			\modd{0} {3}$;}\\
			2, &amp;\text{if $a$ is even and $\ell_2 \equiv 
			\modd{2} {3}$ or if $a$ is odd and $\ell_2 \equiv 
			\modd{1} {3}$;}\\
			4 &amp;\text{if $a$ is even and $\ell_2 \equiv 
			\modd{1} {3}$ or if $a$ is odd and $\ell_2 \equiv 
			\modd{2} {3}$.}
		\end{cases}
	\end{equation*}
	To obtain the formula for $\nu_2\left( {\ell_1 2^b \choose \ell_2 2^a}_F \right)$, we divide the calculation into 4 cases: Case 1.1 to Case 1.4 as before. Then we consider the values of $r$ and $s$ in each case, and substitute $A_2$ in Corollary \ref{corollary5}. This leads to the desired result. Since the calculation in each case is similar, we only give the details in Case 1.3. In this case, $A_2 = \left\lfloor\frac{a}{2}\right\rfloor + \nu_2 (m_2 - k_2) + \nu_2\left({m_2 \choose k_2}\right)$, $(r,s)=(2,4)$ if $a$ and $b$ are odd, and $(r,s)=(4,2)$ if $a$ and $b$ are even. By Corollary \ref{corollary5}, we obtain
	\begin{align*}
		\nu_2\left( {\ell_1 2^b \choose \ell_2 2^a}_F \right) &amp;= 
		\begin{cases}
			A_2 + 2, &amp;\text{if $a$ and $b$ are odd;}\\
			A_2 + 1, &amp;\text{if $a$ and $b$ are even,}
		\end{cases}\\
		&amp;= \left\lceil\frac{a}{2}\right\rceil + 1 +\nu_2(m_2 - k_2) + \nu_2 \left({m_2 \choose k_2}\right), 
	\end{align*}
	as required. The other cases are similar.

\medskip
	
	\noindent{\bf Case 2.} $a\not\equiv \modd{b} {2}$. The calculation in this case is similar to Case 1, so we omit some details. By Theorem \ref{theorem2}, the third term on the right-hand side of \eqref{eq1cor4.3} is equal to 
	\begin{align}
		&amp;\frac{(\ell_1 2^{b-a} - \ell_2)(2^{a-1}-1)}{3} - \left\{-\frac{r'+s'}{3}\right\}[a\equiv \modd{0} {2}] - \left\lfloor \frac{a-1}{2} \right\rfloor [\ell_1 \not \equiv \modd{-\ell_2} {3}] \nonumber\\
		&amp;+ \nu_2\left(\left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor !\right).
	\end{align}
	Since $b-a$ is odd, $\ell_1 2^{b-a} \equiv \modd{-r'} {3}$ and we obtain by Lemma \ref{lemma2.7} that
	\begin{equation*}
	\left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor = m_2 - k_2 - B
	\end{equation*}
	where $B = [(r',s') \in \{(0,1),(0,2),(2,2)\}]$. Similar to Case 1, $\nu_2 \left( \left\lfloor \frac{\ell_1 2^{b-a} - \ell_2}{3} \right\rfloor ! \right)$ is
	\begin{equation*}
		\nu_2 \left( m_2 ! \right) - B\nu_2 \left( m_2 - k_2 \right) - \nu_2 \left( \frac{m_2 !}{(m_2 - k_2)!} \right).
	\end{equation*}
	Then we evaluate $\nu_2 (m_2 !)$ by Corollary \ref{corollary1}, and substitute all of these in \eqref{eq1cor4.3} to obtain that $A_2$ is equal to
	\begin{align*}
		&amp;\left(\frac{b-a-1}{2} - \left\lfloor \frac{b-1}{2} \right\rfloor \right) [r' \neq 0] - \frac{r'}{3}[b\equiv \modd{0} {2}] + \left\lfloor \frac{a-1}{2} \right\rfloor [s' \neq 0]   \\
		&amp;\phantom{=} + \left(\frac{s'}{3} + \left\{-\frac{r'+s'}{3}\right\}\right)[a\equiv \modd{0} {2}] + \left\lfloor \frac{a-1}{2} \right\rfloor [\ell_1 \not \equiv \modd{-\ell_2} {3}] + \frac{r'}{3} \\
		&amp;\phantom{=} + B\nu_2 \left( m_2 - k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right). 
	\end{align*}
	Then we divide the calculation into 4 cases and obtain that $A_2$ is
	\begin{equation*}
		\begin{cases}
		\nu_2 \left( {m_2  \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{-\ell_2} {3}$ or $\ell_2 \equiv \modd{0} {3}$;}\\
		a-1 + \nu_2 \left( m_2 - k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{0} {3}$ and
		$\ell_2 \not\equiv \modd{0} {3}$;}\\
		\left\lfloor \frac{a-1}{2} \right \rfloor + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv \modd{1} {3}$ and
		$\ell_2 \equiv \modd{1} {3}$;}\\
		\left\lfloor \frac{a}{2} \right \rfloor + \nu_2 \left( m_2 - k_2 \right) + \nu_2 \left( {m_2 \choose k_2} \right), &amp;\text{if $\ell_1 \equiv 
		\modd{2} {3}$ and $\ell_2 \equiv \modd{2} {3}$.}
		\end{cases}
	\end{equation*}
	We illustrate the calculation of $A_2$ above only for the case $\ell_2 \equiv \modd{0} {3}$ since the other cases are similar. So suppose $\ell_2 \equiv \modd{0} {3}$. So $s'=0$. If $r'=0$, then it is easy to see that $A_2$ is equal to $\nu_2\left({m_2 \choose k_2}\right)$. So assume that $r'\neq 0$. Then $A_2$ is equal to $x + y + \nu_2\left({m_2 \choose k_2}\right)$, where
	\begin{equation*}
		x = \frac{b-a-1}{2} - \left \lfloor\frac{b-1}{2}\right \rfloor + \left \lfloor\frac{a-1}{2}\right \rfloor = 
		\begin{cases}
			0, &amp;\text{if $a$ is odd;}\\
			-1, &amp;\text{if $a$ is even,}
		\end{cases}
	\end{equation*}
	\begin{equation*}
		y = \frac{-r'}{3}[b\equiv \modd{0} {2}] + \left\{-\frac{r'}{3} \right\} [a\equiv \modd{0} {2}] + \frac{r'}{3} = 
		\begin{cases}
			0, &amp;\text{if $a$ is odd;}\\
			1, &amp;\text{if $a$ is even.}
		\end{cases}
	\end{equation*}
	Therefore $A_2 = \nu_2\left({m_2 \choose k_2}\right)$, as required. As in Case 1, we divide the calculation of $\nu_2 \left({\ell_1 2^b \choose \ell_2 2^a}_F \right)$ into 4 cases according to the value of $A_2$, which leads to the desired result. This proves (i).
	
	For (ii), we apply Corollary \ref{corollary5} with $m=\ell_1 p^b$ and $k= \ell_2 p^a$. For convenience, we let $r' = \ell_1 \mod{z(p)}$ and $s' = \ell_2 \mod{z(p)}$. The calculation of this part is similar to that of part (i), so we omit some details. We have
	\begin{equation}\label{eq3cor4.3}
		A_p = \nu_p\left(\left\lfloor \frac{\ell_1 p^{b}}{z(p)} \right\rfloor!\right) - \nu_p \left(\left\lfloor \frac{\ell_2 p^{a}}{z(p)} \right\rfloor!\right) - \nu_p \left(\left\lfloor \frac{\ell_1 p^{b} - \ell_2 p^{a}}{z(p)} \right\rfloor!\right).
	\end{equation}

\noindent{\bf Case 1.} $p \equiv \pm \modd{1} {5}$. Then by Lemma \ref{lemma2.2}(iii), $p\equiv \modd{1} {z(p)}$. By Corollary \ref{corollary1}, the first term on the right-hand side of \eqref{eq3cor4.3} is equal to
	\begin{align*}
		&amp;\frac{\ell_1 (p^b - 1)}{z(p)(p-1)} + \nu_p \left(\left \lfloor\frac{\ell_1}{z(p)}\right \rfloor ! \right) - \left\lfloor\frac{b}{2}\right\rfloor [r'\neq 0] - \frac{r'}{z(p)}[b\equiv \modd{1} {2}] + \left\lfloor\frac{b}{2}\right\rfloor [r'\neq 0]\left(1 - \frac{2r'}{z(p)}\right) \\
		&amp;= \frac{\ell_1(p^b -1)}{z(p)(p-1)} - \frac{br'}{z(p)} + \nu_p \left ( \left \lfloor \frac{\ell_1}{z(p)} \right \rfloor ! \right ),
	\end{align*}
	and similarly, the second term is 
	\begin{equation*}
		\frac{\ell_2(p^a -1)}{z(p)(p-1)} - \frac{as'}{z(p)} + \nu_p \left ( \left \lfloor \frac{\ell_2}{z(p)} \right \rfloor ! \right ).
	\end{equation*}
	By Theorem \ref{theorem2}, the third term is
	\begin{equation*}
		\frac{(\ell_1 p^{b-a} - \ell_2)(p^a -1)}{z(p)(p-1)} - a\left ( \frac{r' - s'}{z(p)} +[r'&lt;s'] \right ) + \nu_p \left ( \left \lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right \rfloor ! \right ).
	\end{equation*}
	Since $p \equiv \modd{1} {z(p)}$, we obtain by Lemma \ref{lemma2.7} that 
	\begin{equation*}
		\left\lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right\rfloor = m_p - k_p - [r'&lt;s'].
	\end{equation*}
	Therefore $\nu_p \left ( \left \lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right \rfloor ! \right )$ is equal to
	\begin{equation*}
		\nu_p \left( m_p ! \right) - [r'&lt;s']\nu_p 	\left( m_p - k_p \right) - \nu_p \left(\frac{m_p !}{(m_p - k_p)!}\right).
	\end{equation*}
	As usual, the first term above can be evaluated by Corollary \ref{corollary1} and is equal to
	\begin{equation*}
		\frac{\ell_1 (p^{b-a} -1)}{z(p)(p-1)} - (b-a)\frac{r'}{z(p)} + \nu_p \left(\left\lfloor\frac{\ell_1}{z(p)}\right\rfloor !\right).
	\end{equation*}
	We substitute all of these in \eqref{eq3cor4.3} to obtain
	\begin{equation*}
		A_p = [r'&lt;s']\left(a + \nu_p \left( 
		m_p - k_p \right) \right) + \nu_p \left( {m_p \choose k_p} \right).
	\end{equation*}
	Since $p\equiv \modd{1} {z(p)}$, $r=r'$ and $s=s'$. Substituting $A_p$ and applying Corollary \ref{corollary5}, we obtain the desired result.

\medskip
	
	\noindent{\bf Case 2.} $p\equiv \pm \modd{2} {5}$. Then by Lemma \ref{lemma2.2}(iii), $p\equiv \modd{-1} {z(p)}$. By Corollary \ref{corollary1}, the first term on the right-hand side of \eqref{eq3cor4.3} is equal to
	\begin{equation*}
		\frac{\ell_1(p^b -1)}{z(p)(p-1)} - \left\lfloor \frac{b}{2}\right\rfloor[r'\neq 0] - \frac{r'}{z(p)}[b\equiv \modd{1} {2}] + \nu_p \left ( \left \lfloor \frac{\ell_1}{z(p)} \right \rfloor ! \right ).
	\end{equation*}
	Similarly, the second term is 
	\begin{equation*}
		\frac{\ell_2(p^a -1)}{z(p)(p-1)} - \left \lfloor \frac{a}{2} \right \rfloor[s'\neq 0] - \frac{s'}{z(p)}[a\equiv \modd{1} {2}] + \nu_p \left ( \left \lfloor \frac{\ell_2}{z(p)} \right \rfloor ! \right ).
	\end{equation*}
	For the third term, we divide the proof into two cases according to the parity of $a$ and $b$.

\medskip
	
	\noindent{\bf Case 2.1.} $a\equiv \modd{b} {2}$. Then by Theorem \ref{theorem2}, the third term on the right-hand side of \eqref{eq3cor4.3} is equal to 
	\begin{align*}
		&amp;\frac{(\ell_1 p^{b-a} - \ell_2 )(p^{a} -1)}{z(p)(p-1)} - \left(\frac{r'-s'}{z(p)} + [r'&lt;s']\right)[a \equiv \modd{1} {2}] - \left\lfloor\frac{a}{2}\right\rfloor[r' \neq s']\\
		&amp; + \nu_p \left( \left\lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right\rfloor ! \right).
	\end{align*}
	As in Case 1, we apply Lemma \ref{lemma2.7} to write
	\begin{equation*}
		\left\lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right\rfloor = m_p - k_p - [r'&lt;s'],
	\end{equation*}
	and then use Corollary \ref{corollary1} to show that $\nu_p \left ( \left \lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right \rfloor ! \right )$ is equal to
	\begin{equation*}
		\frac{\ell_1 (p^{b-a} -1)}{z(p)(p-1)} - \frac{b-a}{2}[r'\neq 0] + \nu_p \left(\left\lfloor \frac{\ell_1}{z(p)}\right\rfloor !\right) - [r'&lt;s']\nu_p \left( m_p - k_p \right) - \nu_p\left(\frac{m_p !}{(m_p - k_p)!}\right).
	\end{equation*}
	Substituting all of these in \eqref{eq3cor4.3}, we see that $A_p$ is equal to
	\begin{align*}
		&amp;-\left\lfloor \frac{b}{2} \right\rfloor[r'\neq 0] + \left\lfloor\frac{a}{2}\right\rfloor[s'\neq 0] + [r'&lt;s'][b \equiv \modd{1} {2}] + \left\lfloor \frac{a}{2}\right\rfloor[r'\neq s'] +\frac{b-a}{2}[r'\neq 0] \\
		&amp;+ [r'&lt;s']\nu_p \left( m_p - k_p \right) + \nu_p \left( {m_p \choose k_p} \right)\\
		&amp;=
		\begin{cases}
			\nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv \modd{\ell_2} {z(p)}$ or $\ell_2 \equiv \modd{0} {z(p)}$;}\\
			a + \nu_p \left( m_p - k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv \modd{0} {z(p)}$ and $\ell_2 \not\equiv \modd{0} {z(p)}$;}\\
			\left\lfloor\frac{a}{2}\right\rfloor + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$ and  $r'&gt; s'$;}\\
			\left\lceil\frac{a}{2}\right\rceil + \nu_p \left( m_p - k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$ and  $r'&lt;s'$.}
		\end{cases}
	\end{align*}
	Recall that $r = \ell_1 p^b \mod{z(p)}$ and $s = \ell_2 p^a \mod{z(p)}$. If $a$ and $b$ are even, then $p^b \equiv p^a \equiv \modd{1} {z(p)}$, $r=r'$, and $s=s'$, and we can obtain $\nu_p\left( {\ell_1 p^b \choose \ell_2 p^a}_F \right)$ by substituting $A_p$ in Corollary \ref{corollary5}. Suppose $a$ and $b$ are odd. Then $r\equiv \modd{-r'} {z(p)}$ and $s \equiv \modd{-s'} {z(p)}$ and thus when $r$ and $s$ are both nonzero or are both zero, we have
	\begin{equation*}
	r\geq s \quad\text{if and only if}\quad r'\leq s'.
	\end{equation*}
	Similar to the above, we can obtain $\nu_p\left( {\ell_1 p^b \choose \ell_2 p^a}_F \right)$ by the substitution of $A_p$ in Corollary \ref{corollary5}. We see that $\nu_p \left( {\ell_1 p^b \choose \ell_2 p^a} \right)$ is equal to
	\begin{equation*}
		\begin{cases}
			\nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv \modd{\ell_2} {z(p)}$ or $\ell_2 \equiv \modd{0} {z(p)}$;}\\
			a + \nu_p (F_{z(p)}) + \nu_p \left( m_p	- k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv 
			\modd{0} {z(p)}$ and $\ell_2 \not\equiv 
			\modd{0} {z(p)}$;}\\
			\frac{a}{2} + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&gt; s$, $\ell_1,\ell_2 \not\equiv 
			\modd{0} {z(p)}$, and $a$ is even;}\\ 
			\frac{a}{2} + \nu_p (F_{z(p)}) + \nu_p \left( m_p - k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&lt;s$, $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$, and $a$ is even;}\\
			\frac{a+1}{2} + \nu_p \left( m_p - k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&gt;s$, $\ell_1,\ell_2 \not\equiv 
			\modd{0} {z(p)}$, and $a$ is odd;}\\
			\frac{a-1}{2} + \nu_p(F_{z(p)}) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&lt;s$, $\ell_1,\ell_2 \not\equiv 
			\modd{0} {z(p)}$, and $a$ is odd.}
		\end{cases}
	\end{equation*}
	Since $a\equiv \modd{b} {2}$, we see that $p^a \equiv 
	\modd{p^b} {z(p)}$ and therefore
	\begin{equation}\label{eq4thm4.3}
		\ell_1 \equiv \modd{\ell_2} {z(p)} \Leftrightarrow r=s
	\end{equation}
	So the condition $\ell_1 \equiv \modd{\ell_2} {z(p)}$ can be replaced by $r=s$.

\medskip
	
	\noindent{\bf Case 2.2.} $a\not\equiv \modd{b} {2}$. The calculation in this case is similar to that given before. So we skip some details. By Theorem \ref{theorem2}, the third term on the right-hand side of \eqref{eq3cor4.3} is equal to 
	\begin{equation*}
		\frac{(\ell_1 p^{b-a} - \ell_2 )(p^{a} -1)}{z(p)(p-1)} - [a \equiv \modd{1} {2}]B_1 - \left\lfloor\frac{a}{2}\right\rfloor[\ell_1 \not\equiv
\modd{-\ell_2} {z(p)}] + \nu_p \left( \left\lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right\rfloor ! \right),
	\end{equation*}
	where $B_1 = \left\{ - \frac{r'+s'}{z(p)} \right\} = -\frac{r'+s'}{z(p)} + [r'+s'&gt;0] + [r'+s'&gt;z(p)]$. Since $p\equiv \modd{-1} {z(p)}$, we obtain by Lemma \ref{lemma2.7} and a straightforward verification that
	\begin{equation*}
		\left\lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right\rfloor = m_p - k_p - \varepsilon,
	\end{equation*}
	where $\varepsilon = [-r' \bmod{z(p)} &lt; s'] = [r'=0 \text{ and } s'\neq 0] + [r'+s'&gt;z(p)]$. Then by Corollary \ref{corollary1}, $\nu_p \left ( \left \lfloor \frac{\ell_1 p^{b-a} - \ell_2}{z(p)} \right \rfloor ! \right )$ is equal to
	\begin{equation*}
		\frac{\ell_1 (p^{b-a} -1)}{z(p)(p-1)} - \left\lfloor\frac{b-a}{2}\right\rfloor[r'\neq 0] - \frac{r'}{z(p)} + \nu_p \left(\left\lfloor\frac{\ell_1}{z(p)}\right\rfloor !\right) - B_2 - \nu_p\left( \frac{m_p !}{(m_p - k_p)!} \right),
	\end{equation*}
	where $B_2 = \varepsilon\nu_p \left( m_p - k_p  \right)$. 
Since $a\not\equiv \modd{b} {2}$,
$[b\equiv \modd{1} {2}] = 1 - 
[a\equiv \modd{1} {2}]$ and $\left\lfloor\frac{b-a}{2}\right\rfloor + \left\lfloor\frac{a}{2}\right\rfloor - \left\lfloor\frac{b}{2}\right\rfloor + 
[a\equiv \modd{1} {2}]$. We substitute all of these in \eqref{eq3cor4.3} to obtain that $A_p$ is equal to
	\begin{align*}
		&amp;\left(\left\lfloor\frac{b-a}{2}\right\rfloor - \left\lfloor\frac{b}{2}\right\rfloor \right)[r'\neq 0] + \left\lfloor\frac{a}{2}\right\rfloor\Big([s'\neq 0] + [\ell_1 \not\equiv \modd{-\ell_2} {z(p)}]\Big) \\
		&amp; + \left([r'+s'&gt;0] + [r'+s'&gt;z(p)]\right)
[a\equiv \modd{1} {2}] + B_2 + \nu_p\left({m_p \choose k_p}\right)\\
		&amp;=
		\begin{cases}
			\nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv \modd{-\ell_2} {z(p)}$ or $\ell_2 \equiv 
			\modd{0} {z(p)}$;}\\
			a + \nu_p \left( m_p	- k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv 
\modd{0} {z(p)}$ and $\ell_2 \not\equiv \modd{0} {z(p)}$;}\\
			\left\lfloor\frac{a}{2}\right\rfloor + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1,\ell_2 \not\equiv 
\modd{0} {z(p)}$ and  $r'+s'&lt;z(p)$;}\\
			\left\lceil\frac{a}{2}\right\rceil + \nu_p \left( m_p	- k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1,\ell_2 \not\equiv \modd{0} {z(p)}$ and  $r'+s'&gt;z(p)$.}
		\end{cases}
	\end{align*}
	Recall that $r = \ell_1 p^b \mod{z(p)}$ and $s = \ell_2 p^a \mod{z(p)}$. Suppose that $a$ is odd and $b$ is even. Then $r=r'$ and $s\equiv 
\modd{-s'} {z(p)}$. Moreover, if $s'\neq 0$, then $s=z(p)-s'$ and thus 
	\begin{equation*}
		r&lt;s \Leftrightarrow r'+s'&lt;z(p) \quad\text{and}\quad r&gt;s \Leftrightarrow r'+s'&gt;z(p).
	\end{equation*}
	Similarly, if $a$ is even and $b$ is odd, then $r\equiv 
	\modd{-r'} {z(p)}$ and $s=s'$, and for $r'\neq 0$, we have
	\begin{equation*}
		r&lt;s \Leftrightarrow r'+s'&gt;z(p) \quad\text{and}\quad r&gt;s \Leftrightarrow r'+s'&lt;z(p).
	\end{equation*}
	From the above observation and the substitution of $A_p$ in Corollary \ref{corollary5}, we see that $\nu_p\left( {\ell_1 p^b \choose \ell_2 p^a}_F \right)$ is equal to
	\begin{equation*}
		\begin{cases}
			\nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv \modd{-\ell_2} {z(p)}$ or $\ell_2 \equiv 
\modd{0} {z(p)}$;}\\
			a + \nu_p (F_{z(p)}) + \nu_p \left( m_p	- 	k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $\ell_1 \equiv 
\modd{0} {z(p)}$ and $\ell_2 \not\equiv 
\modd{0} {z(p)}$;}\\
			\frac{a+1}{2} + \nu_p \left( m_p	- k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&gt;s$, $\ell_1,\ell_2 \not\equiv 
\modd{0} {z(p)}$, and $a$ is odd;}\\
			\frac{a-1}{2} + \nu_p(F_{z(p)}) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&lt;s$, $\ell_1,\ell_2 \not\equiv 
\modd{0} {z(p)}$, and $a$ is odd;	}\\
			\frac{a}{2} + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&gt; s$, $\ell_1,\ell_2 \not\equiv 
\modd{0} {z(p)}$, and $a$ is even;}\\ 
			\frac{a}{2} + \nu_p (F_{z(p)}) + \nu_p \left( m_p - k_p \right) + \nu_p \left( {m_p \choose k_p} \right), &amp;\text{if $r&lt;s$, $\ell_1,\ell_2 \not\equiv 
\modd{0} {z(p)}$, and $a$ is even.}\\
		\end{cases}
	\end{equation*}	
	Since $a\not \equiv 
\modd{b} {2}$, we see that $p^a \equiv 
\modd{-p^b} {z(p)}$ and therefore 
	\begin{equation*}
		\ell_1 \equiv 
\modd{-\ell_2} {z(p)} \Leftrightarrow r=s.
	\end{equation*}
	Combining this with \eqref{eq4thm4.3}, we conclude that
	\begin{equation*}
		\ell_1 \equiv \modd{\ell_2 - 2\ell_2 [a\not\equiv 
\modd{b} {2}]} {z(p)} \Leftrightarrow r=s.
	\end{equation*}
	This completes the proof.
\end{proof}

\section{Examples}\label{sec5}
In this last section, we give several examples to show applications of our main results. We also recall from Remark \ref{remark4.4} that the condition $r=s$ in Theorem \ref{theorem4.3}(ii) can be replaced by $\ell_1 \equiv
\modd{\ell_2 - 2\ell_2 [a\not\equiv \modd{b} {2}]} {z(p)}$. In the calculation given in this section, we will use this observation without further reference. 
\begin{example}\label{example1}
	Let $a$, $b$, and $\ell$ be positive integers and $b\geq a$. We assert that for $\ell \not\equiv \modd{0} {3}$, we have
	\begin{equation}\label{eq1exam1}
		\nu_2\left({\ell \cdot 2^b \choose 2^a}_F\right) = \left\lceil\frac{a+1}{2}\right\rceil\left( \varepsilon_1 \varepsilon_2 + \varepsilon_1' \varepsilon_2' \right),
	\end{equation}
	where $\varepsilon_1 = [\ell \equiv \modd{2} {3}]$,
$\varepsilon_2 = [a\equiv \modd{b} {2}]$,
$\varepsilon_1' = [\ell \equiv \modd{1} {3}]$,
and $\varepsilon_2' = [a\not\equiv \modd{b} {2}]$. In addition, if $\ell 
\equiv \modd{0} {3}$, then
	\begin{equation}\label{eq2exam1}
		\nu_2 \left({\ell \cdot 2^b \choose 2^a}_F\right) = b + 2 + \nu_2(\ell).
	\end{equation}
\end{example}
\begin{proof}
	We apply Theorem \ref{theorem4.3} to verify our assertion. Here $m_2 = \left\lfloor\frac{\ell\cdot 2^{b-a}}{3}\right\rfloor$ and $k_2 = \left\lfloor\frac{1}{3}\right\rfloor = 0$. So we immediately obtain the following: 
if $a\equiv \modd{b} {2}$, then
	\begin{equation*}
		\nu_2 \left( {\ell\cdot 2^b \choose 2^a}_F \right) = 
		\begin{cases}
			0, &amp;\text{if $\ell \equiv 
\modd{1} {3}$;}\\
			a + 2 + \nu_2 (m_2), &amp;\text{if $\ell \equiv 
\modd{0} {3}$;}\\
			\left\lceil \frac{a+1}{2} \right\rceil, &amp;\text{if $\ell \equiv \modd{2} {3}$,}
		\end{cases}
	\end{equation*}
	and if $a\not\equiv \modd{b} {2}$, then
	\begin{equation*}
		\nu_2 \left( {\ell\cdot 2^b \choose 2^a}_F \right) = 
		\begin{cases}
			0, &amp;\text{if $\ell \equiv 
\modd{1} {3}$;}\\
			a + 2 + \nu_2 (m_2), &amp;\text{if $\ell \equiv 
\modd{0} {3}$;}\\
			\left\lceil \frac{a+1}{2} \right\rceil, &amp;\text{if $\ell \equiv \modd{1} {3}$.}
		\end{cases}
	\end{equation*}
	This proves \eqref{eq1exam1}. If $\ell \equiv 
\modd{0} {3}$, then $m_2 = \frac{\ell}{3}\cdot 2^{b-a}$ and $\nu_2 (m_2)$ is equal to
	\begin{equation*}
		\nu_2 (m_2) = \nu_2 \left(\ell\right) + \nu_2 (2^{b-a}) - \nu_2 (3) = b-a + \nu_2 \left(\ell\right),
	\end{equation*}
	which implies \eqref{eq2exam1}.
\end{proof}
\begin{example}\label{rem16}
Substituting $\ell = 1$ in Example \ref{example1}, we see that
\begin{align}\label{eq1remark1}
	\nu_2\left({2^b \choose 2^a}_F\right) &amp;= \left\lceil \frac{a+1}{2}\right\rceil [a\not\equiv \modd{b} {2}]\nonumber\\
	&amp;=
	\begin{cases}
		0, &amp;\text{if $a\equiv \modd{b} {2}$;}\\
		\left\lceil\frac{a+1}{2}\right\rceil, &amp;\text{if $a\not\equiv 
\modd{b} {2}$.}
	\end{cases}
\end{align}
Our example also implies that \eqref{eq1remark1} still holds for the 2-adic valuations of ${2^{b+2c} \choose 2^a}_F$, ${7\cdot 2^{b} \choose 2^a}_F$, ${5\cdot 2^{b+1} \choose 2^a}_F$, ${13\cdot 2^{b} \choose 2^a}_F$, etc.
\end{example}
\begin{example}\label{example2}
	Let $a$, $b$, and $\ell$ be positive integers, $b\geq a$, and $p$ a prime distinct from 2 and 5. If $p \equiv \modd{\pm 1} {5}$, then
	\begin{equation*}
		\nu_p\left( {\ell p^b \choose p^a}_F \right) = \left(b + \nu_p(F_{z(p)}) + \nu_p(\ell) \right)[\ell \equiv \modd{0} {z(p)}],
	\end{equation*}
	and if $p \equiv \modd{\pm 2} {5}$, then
	\begin{equation*}
		\nu_p\left( {\ell p^b \choose p^a}_F \right) =
		\begin{cases}
			0, &amp;\text{if $\ell \equiv \modd{1 - 2 \varepsilon} {z(p)}$;}\\
			b + \nu_p(F_{z(p)}) + \nu_p(\ell), &amp;\text{if $\ell 
\equiv \modd{0} {z(p)$};}\\
			\frac{a}{2}, &amp;\text{if $\ell \not\equiv 
\modd{0,1 - 2 \varepsilon} {z(p)}$ and $a$ is even;}\\
			\frac{a-1}{2} + \nu_p (F_{z(p)}), &amp;\text{if $\ell 
\not\equiv \modd{0,1 - 2 \varepsilon} {z(p)}$ and $a$ is odd,}
		\end{cases}
	\end{equation*}
	where $\varepsilon = [a\not\equiv \modd{b} {2}]$.
\end{example}
\begin{proof}
	Similar to Example \ref{example1}, we verify this by applying Theorem \ref{theorem4.3}. Here $m_p = \left\lfloor\frac{\ell p^{b-a}}{z(p)}\right\rfloor$, $k_p = \left\lfloor\frac{1}{z(p)}\right\rfloor = 0$, $r=\ell p^b \bmod{z(p)}$, and $s = p^a \bmod{z(p)}$. 
We first assume that $p \equiv \modd{\pm 1} {5}$.
Then by Lemma \ref{lemma2.2}, we have
$p\equiv \modd{1} {z(p)}$.
Therefore $s=1$, $r\equiv \modd{\ell} {z(p)}$, and 
	\begin{equation*}
		\nu_p\left( {\ell p^b \choose p^a}_F \right) = \left(a + \nu_p (m_p) + \nu_p(F_{z(p)}) \right)[\ell \equiv \modd{0} {z(p)}].
	\end{equation*}
	Similarly, if $p \equiv \modd{\pm 2} {5}$ and $a\equiv 
\modd{b} {2}$, then we obtain by Lemma \ref{lemma2.2} and Theorem \ref{theorem4.3} that
	\begin{equation*}
		\nu_p\left( {\ell p^b \choose p^a}_F \right) =
		\begin{cases}
			0, &amp;\text{if $\ell \equiv \modd{1} {z(p)}$;}\\
			a + \nu_p(m_p) + \nu_p(F_{z(p)}),
&amp;\text{if $\ell \equiv \modd{0} {z(p)$};}\\
\frac{a}{2}, &amp;\text{if $\ell \not\equiv \mod{0, 1} {z(p)}$ and $a$ is even;}\\
			\frac{a-1}{2} + \nu_p (F_{z(p)}), &amp;\text{if $\ell \not\equiv \modd{0,1} {z(p)}$ and $a$ is odd.}
		\end{cases}
	\end{equation*}
	In addition, if $p\equiv \modd{\pm 2} {5}$ and $a\not\equiv 
\modd{b} {2}$, then
	\begin{equation*}
		\nu_p\left( {\ell p^b \choose p^a}_F \right) =
		\begin{cases}
0, &amp;\text{if $\ell \equiv 
\modd{-1} {z(p)}$;}\\
a + \nu_p(m_p) + \nu_p(F_{z(p)}),
&amp;\text{if $\ell \equiv \modd{0} {z(p)$};}\\
\frac{a}{2}, &amp;\text{if $\ell \not\equiv \modd{0, -1} {z(p)}$ and
$a$ is even;}\\
\frac{a-1}{2} + \nu_p (F_{z(p)}),
&amp;\text{if $\ell \not\equiv \modd{0,-1} {z(p)}$ and $a$ is odd.}
		\end{cases}
	\end{equation*}
	It remains to calculate $\nu_p(m_p)$ when $\ell \equiv 
\modd{0} {z(p)}$. In this case, we have
	\begin{equation*}
		\nu_p(m_p) = \nu_p \left(\frac{\ell p^{b-a}}{z(p)}\right) = \nu_p (\ell) + \nu_p (p^{b-a}) - \nu_p (z(p)) = b-a + \nu_p (\ell).
	\end{equation*}
	This implies the desired result.
\end{proof}
\begin{example}\label{rem18}
Substituting $\ell = 1$ in Example \ref{example2}, we see that for $p\neq 2, 5$, we have
\begin{equation}\label{eq1remark2}
	\nu_p\left( {p^b \choose p^a}_F \right) =
	\begin{cases}
		0, &amp;\text{if $p\equiv 
\modd{\pm 1} {5}$ or $a\equiv \modd{b} {2}$;}\\
		\frac{a}{2}, &amp;\text{if $p\equiv 
\modd{\pm 2} {5}$, $a\not\equiv \modd{b} {2}$, and $a$ is even;}\\
		\frac{a-1}{2} + \nu_p (F_{z(p)}),
&amp;\text{if $p\equiv \modd{\pm 2} {5}$, $a\not\equiv 
\modd{b} {2}$, and $a$ is odd.}
	\end{cases}
\end{equation}
Our example also implies that \eqref{eq1remark2} still holds for the $p$-adic valuations of ${p^{b+2c} \choose p^a}_F$ and ${(z(p)+1)\cdot p^{b} \choose p^a}_F$. Similarly, for $p\neq 2,5$, we have
\begin{equation}\label{eq2remark2}
	\nu_p\left( {2p^b \choose p^a}_F \right) =
	\begin{cases}
		0, &amp;\text{if $p\equiv \modd{\pm 1} {5}$;}\\
		\frac{a}{2}, &amp;\text{if $p\equiv 
\modd{\pm 2} {5}$ and $a$ is even;}\\
		\frac{a-1}{2} + \nu_p (F_{z(p)}),
&amp;\text{if $p\equiv \modd{\pm 2} {5}$ and $a$ is odd.}
	\end{cases}
\end{equation}
In addition, \eqref{eq2remark2} also holds when ${2p^b \choose p^a}_F$ is replaced by ${\ell p^b \choose p^a}_F$ for
$\ell \not\equiv \modd{0,\pm 1} {z(p)}$ and
$p\neq 2,5$. Furthermore, replacing ${2p^b \choose p^a}_F$ by ${(z(p)-1) p^b \choose p^a}_F$, the formula becomes
\begin{equation*}
\begin{cases}
0, &amp;\text{if $p\equiv \modd{\pm 1} {5}$ or $a\not \equiv \modd{b} {2}$;}\\
\frac{a}{2}, &amp;\text{if $p\equiv \modd{\pm 2} {5}$,
$a\equiv \modd{b} {2}$, and $a$ is even;}\\
\frac{a-1}{2} + \nu_p (F_{z(p)}),
&amp;\text{if $p\equiv \modd{\pm 2} {5}$,
$a\equiv \modd{b} {2}$, and $a$ is odd.}
\end{cases}
\end{equation*}
\end{example}
\begin{example}\label{example3}
	We know that the $5$-adic valuations of Fibonomial coefficients are the same as those of binomial coefficients. For example, by Theorem \ref{theorem5}(ii) and Kummer's theorem, we obtain
	\begin{equation*}
		\nu_5\left( {\ell \cdot 5^b \choose 5^a}_F \right) =\nu_5\left( {\ell \cdot 5^b \choose 5^a} \right) = b-a + \nu_5(\ell),
	\end{equation*}
	for every $a, b, \ell \in \mathbb{N}$ with $b\geq a$. Similarly, $\nu_5\left( {5^b \choose \ell \cdot 5^a}_F \right) = b-a - \nu_5(\ell)$ for every $a, b, \ell \in \mathbb{N}$ such that $5^b &gt;\ell \cdot 5^a$.
\end{example}
\begin{example}\label{example4}
	Let $a$, $b$, and $\ell$ be positive integers and $2^b &gt; \ell\cdot 2^a$. Let $m_2 = \left\lfloor \frac{2^{b-a}}{3}\right\rfloor$ and $k_2 = \left\lfloor\frac{\ell}{3}\right\rfloor $. Then 
	\begin{equation}\label{eq1exam4}
		\nu_2\left( {2^b \choose \ell\cdot 2^a}_F \right) = \nu_2\left({m_2 \choose k_2}\right) + \left(\left\lceil \frac{a+2}{2}\right\rceil + \nu_2(m_2 - k_2)\right)\varepsilon_1 \varepsilon_2 + \left\lceil \frac{a+1}{2}\right\rceil\varepsilon_3 \varepsilon_4,
	\end{equation}
	where $\varepsilon_1 = [a\equiv 
\modd{b} {2}]$,
$\varepsilon_2 = [\ell \equiv \modd{2} {3}]$,
$\varepsilon_3 = [a\not\equiv \modd{b} {2}]$,
and $\varepsilon_4 = [\ell \equiv \modd{1} {3}]$. 
\end{example}
\begin{proof}
Similar to Example \ref{example1}, this follows from the application of Theorem \ref{theorem4.3}. So we leave the details to the reader.

\end{proof}
\begin{example}
	Let $k\geq 2$. We observe that
	\begin{equation*}
		\left\lfloor\frac{2^k}{3}\right\rfloor =
		\begin{cases}
			\frac{2^k - 1}{3}, &amp;\text{if $k$ is even;}\\
			\frac{2(2^{k-1} -1)}{3}, &amp;\text{if $k$ is odd,}
		\end{cases}
	\end{equation*}
	which implies, 
	\begin{equation}\label{eq2exam4}
\nu_2 \left( \left\lfloor\frac{2^k}{3}\right\rfloor\right) = 
[k\equiv \modd{1} {2}].
	\end{equation}
	By a similar reason, we also see that for $k\geq 3$, 
	\begin{equation}\label{eq3exam4}
	\nu_2\left(\left\lfloor \frac{2^k}{3}\right\rfloor -1 \right) = 2[k\equiv \modd{0} {2}].
	\end{equation}
	From \eqref{eq1exam4}, \eqref{eq2exam4}, and \eqref{eq3exam4}, we obtain the following results:
	\begin{itemize}
		\item[(i)] if $b-a \geq 2$, then $\nu_2 \left( {2^b \choose 3\cdot 2^a}_F \right) = [a\not\equiv \modd{b} {2}]$,
		\item[(ii)] if $b-a\geq 3$, then $\nu_2 \left( {2^b \choose 5\cdot 2^a}_F \right)$ is equal to
		\begin{align*}
			 &amp;[a\not\equiv \modd{b} {2}] + \left( \left\lceil \frac{a+2}{2} \right\rceil +2[a\equiv \modd{b} {2}] \right)[a\equiv \modd{b} {2}]\\
			&amp;= 1 + \left\lceil \frac{a+4}{2}\right\rceil[a\equiv 
\modd{b} {2}],
		\end{align*}
		\item[(iii)] if $b-a \geq 3$, then $\nu_2 \left( {2^b \choose 6\cdot 2^a}_F \right) = [a\equiv \modd{b} {2}]$,
		\item[(iv)] if $b-a \geq 4$, then $\nu_2 \left( {2^b \choose 7\cdot 2^a}_F \right) = [a\equiv \modd{b} {2}] + \left\lceil \frac{a+1}{2}\right\rceil [a\not\equiv \modd{b} {2}]$.
	\end{itemize}
\end{example}

\begin{example}
	Let $p\neq 5$ be an odd prime and let $a$, $b$, and $\ell$ be positive integers, $p^b &gt; \ell p^a$, $m_p = \left\lfloor \frac{p^{b-a}}{z(p)}\right\rfloor$, and $k_p = \left\lfloor\frac{\ell}{z(p)}\right\rfloor $. Then the following statements hold.
	\begin{itemize}
		\item[(i)] If $p\equiv \modd{\pm 1} {5}$, then 
		\begin{equation*}
			\nu_p\left( {p^b \choose \ell p^a}_F \right) = \left(a + \nu_p \left( m_p	-
k_p \right) + \nu_p(F_{z(p)}) \right)[\ell \not\equiv \modd{0,1} {z(p)}] +
\nu_p \left( {m_p \choose k_p} \right),
		\end{equation*} 
		\item[(ii)] If $p\equiv 
\modd{\pm 2} {5}$,
then $\nu_p\left( {p^b \choose \ell p^a}_F \right)$ is equal to
		\begin{equation}\label{eq1exam6}
		\nu_p \left({m_p \choose k_p}\right) + \varepsilon_1 \varepsilon_2 \varepsilon_5 \left(\left\lceil \frac{a}{2}\right\rceil + \nu_p (m_p - k_p) + \varepsilon_3 \nu_p(F_z(p)) \right) + \varepsilon_1 \varepsilon_4 (1-\varepsilon_5) \left(\left\lfloor \frac{a}{2}\right\rfloor + \varepsilon_3 \nu_p(F_z(p)) \right)\\
		\end{equation}
		where $\varepsilon_1 = [\ell \not\equiv 
\modd{0} {z(p)}]$,
$\varepsilon_2 = [\ell \not\equiv \modd{1} {z(p)}]$,
$\varepsilon_3 = [b \equiv \modd{0} {2}]$,
$\varepsilon_4 = [\ell \not\equiv \modd{-1} {z(p)}]$,
and $\varepsilon_5 = [a\equiv \modd{b} {2}]$.
	\end{itemize}
\end{example}
\begin{proof}
Similar to Example \ref{example2}, this follows from the application of Lemma \ref{lemma2.2} and Theorem \ref{theorem4.3}. Since (i) is easily verified, we only give the proof of (ii). The calculation is done in two cases. 
If $p\equiv \modd{\pm 2} {5}$ and 
$a\equiv \modd{b} {2}$, then $\nu_p\left({p^b \choose \ell p^a}_F\right)$ is equal to
\begin{align*}
	 	&amp;\begin{cases}
		\nu_p \left({m_p \choose k_p}\right),
		&amp;\text{if $\ell \equiv \modd{0,1} {z(p)}$;}\\
		\frac{a}{2} + \nu_p (F_{z(p)}) + \nu_p (m_p - k_p) + \nu_p \left({m_p \choose k_p}\right), &amp;\text{if $\ell \not\equiv 
\modd{0,1} {z(p)}$ and $a$ is even;}\\
		\frac{a+1}{2} + \nu_p (m_p - k_p) + \nu_p \left({m_p \choose k_p}\right), &amp;\text{if $\ell \not\equiv 
\modd{0,1} {z(p)}$ and $a$ is odd,}
	\end{cases}\\
	&amp;= \nu_p \left({m_p \choose k_p}\right) + \varepsilon_1 \varepsilon_2 \left(\left\lceil \frac{a}{2}\right\rceil + \nu_p (m_p - k_p) + \varepsilon_3 \nu_p(F_{z(p)}) \right),
\end{align*}
where $\varepsilon_1 = [\ell \not\equiv 
\modd{0} {z(p)}]$,
$\varepsilon_2 = [\ell \not\equiv \modd{1} {z(p)}]$,
and $\varepsilon_3 = [b \equiv \modd{0} {2}]$. 
If $p\equiv \modd{\pm 2} {5}$ and $a\not\equiv \modd{b} {2}$, then
\begin{align*}
	\nu_p\left({p^b \choose \ell p^a}_F\right) &amp;=
	\begin{cases}
		\nu_p \left({m_p \choose k_p}\right),
&amp;\text{if $\ell \equiv \modd{0,-1} {z(p)}$;}\\
\frac{a}{2} + \nu_p \left({m_p \choose k_p}\right),
&amp;\text{if $\ell \not\equiv \modd{0,-1} {z(p)}$ and $a$ is even;}\\
		\frac{a-1}{2} +\nu_p(F_{z(p)}) + \nu_p \left({m_p \choose k_p}\right), &amp;\text{if $\ell \not\equiv \modd{0,-1} {z(p)}$ and $a$ is odd,}
	\end{cases}\\
	&amp;= \nu_p \left({m_p \choose k_p}\right) + \varepsilon_1 \varepsilon_4 \left(\left\lfloor \frac{a}{2}\right\rfloor + \varepsilon_3 \nu_p(F_{z(p)}) \right),
\end{align*}
where $\varepsilon_1,\varepsilon_2,\varepsilon_3$ are as above and $\varepsilon_4 = [\ell \not\equiv \modd{-1} {z(p)}]$. 
Let $\varepsilon_5 = [a\equiv \modd{b} {2}]$.
Then both cases can be combined to obtain (ii).
\end{proof}
\begin{example}
Let $k\geq 2$. We observe that $z(7) = 8$ and
\begin{equation*}
\left\lfloor\frac{7^k}{8}\right\rfloor =
\begin{cases}
\frac{7^k - 1}{8}, &amp;\text{if $k$ is even;}\\
\frac{7(7^{k-1} - 1)}{8}, &amp;\text{if $k$ is odd.}
\end{cases}
\end{equation*}
Therefore
\begin{equation}\label{eq2exam6}
	\nu_7 \left(\left\lfloor\frac{7^k}{8}\right\rfloor\right) = 
[k\equiv \modd{1} {2}] \quad\text{and}\quad
\nu_7 \left(\left\lfloor\frac{7^k}{8}\right\rfloor - 1\right) = 0.
\end{equation}
From \eqref{eq1exam6} and \eqref{eq2exam6}, we obtain the following results:
\begin{itemize}
	\item[(i)] if $b-a \geq 2$, then $\nu_7 \left( {7^b \choose 8\cdot 7^a}_F \right) = [a\not\equiv \modd{b} {2}]$,
	\item[(ii)] if $b-a \geq 2$, then $\nu_7 \left( {7^b \choose 9\cdot 7^a}_F \right) = \left(\left\lfloor\frac{a +2}{2}\right\rfloor + [b\equiv 
\modd{0} {2}] \right)[a\not\equiv \modd{b} {2}] $,
	\item[(iii)] if $b-a\geq 2$, then $\nu_7 \left( {7^b \choose 15\cdot 7^a}_F \right)$ is equal to
	\begin{equation*}
	[a\not\equiv \modd{b} {2}] + \left( \left\lceil \frac{a}{2} \right\rceil +[b\equiv \modd{0} {2}] \right)[a\equiv \modd{b} {2}].
	\end{equation*}
\end{itemize}
\end{example}
To keep this article not too lengthy, we plan to give more applications of our main results in the next article.

\section{Acknowledgments}

We are very grateful to the anonymous referee for his/her careful reading, kind words of praise, and many valuable suggestions which greatly improve the presentation of this article. Phakhinkon Phunphayap receives a scholarship from Science Achievement Scholarship of Thailand(SAST). Prapanpong Pongsriiam receives financial support jointly from The Thailand Research Fund and Faculty of Science Silpakorn University, grant number RSA5980040.
 

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\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11B65.

\noindent \emph{Keywords: }
Fibonacci number, binomial coefficient, Fibonomial coefficient, $p$-adic valuation, $p$-adic order, divisibility.

\bigskip
\hrule
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\noindent (Concerned with sequences 
\seqnum{A000045},
\seqnum{A003267},
\seqnum{A010048}, and
\seqnum{A055870}.)

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\vspace*{+.1in}
\noindent
Received October 1 2017;
revised version received March 7 2018. 
Published in {\it Journal of Integer Sequences}, March 8 2018.

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