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\begin{center}
\vskip 1cm{\LARGE\bf On Divisibility of \\
\vskip .1in
Fibonomial Coefficients by $3$}
\vskip 1cm
\large
Diego Marques\\
Departamento de Matem\' atica\\
Universidade de Bras\' ilia\\
Bras\' ilia, DF \\
Brazil\\
\href{mailto:diego@mat.unb.br}{\tt diego@mat.unb.br} \\
\ \\
Pavel Trojovsk\' y\\
Department of Mathematics\\
University of Hradec Kr\'alov\'e\\
Faculty of Science \\
Rokitansk\'eho 62\\
Hradec Kr\'alov\'e, 500 03 \\
Czech Republic\\
\href{mailto:pavel.trojovsky@uhk.cz}{\tt pavel.trojovsky@uhk.cz}
\end{center}

\vskip .2 in
\begin{abstract}
Let $F_n$ be the $n$th Fibonacci number. For $1\le k\le
m-1$ let
\begin{equation}\label{def}
{m\brack k}_F= \frac{F_m F_{m-1}\cdots F_{m-k+1}}{F_1\cdots F_k}
\end{equation}
be the corresponding Fibonomial coefficient. In this paper, we present some divisibility properties of ${sn \brack n}_F$ by $3$, for some positive integers $n$ and $s$. In par\-ti\-cu\-lar, among other things, we prove that $3 \mid {3^{a+1} \brack 3^a}_F$ , for all $a\geq 1$.
\end{abstract}

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%%  SECTION 1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}\label{sec:1}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Let $(F_n)_{n\geq 0}$ be the Fibonacci sequence given by the recurrence relation $F_{n+2}=F_{n+1}+F_n$, with $F_0=0$ and $F_1=1$. These numbers are 
well-known for possessing amazing properties 
(consult \cite{FIB3} together with its very extensive annotated bibliography for additional references and history). 

In 1915 Fonten\'e published a one--page note \cite{FONT} suggesting a generalization of binomial coefficients, 
replacing the natural numbers by the terms of an arbitrary sequence $(A_n)$ of real or complex numbers.


Since 1964, there has been an accelerated interest in the \textit{Fibonomial coefficients} ${m\brack k}_F$, which correspond to the choice 
$A_n = F_n$, thus are defined, for $1\leq k\leq m$, in the following way
$$
{m\brack k}_F= \frac{F_m F_{m-1}\cdots F_{m-k+1}}{F_1\cdots F_k}.
$$

It is surprising that this quantity will always take integer values. This can be shown by an induction argument and the recursion formula
$$
{m\brack k}_F=F_{k+1}{m-1\brack k}_F+F_{m-k-1}{m-1\brack k-1}_F,
$$
which is a consequence of the formula $F_m=F_{k+1}F_{m-k}+F_kF_{m-k-1}$.

Several authors became
interested in the 
divisibility properties of binomial coefficients. Among several interesting results on this subject, we mention the following facts:

\begin{itemize}
\item An integer $n\geq 2$ is prime if and only if all the binomial
coefficients ${n\choose 1},\ldots,{n\choose n-1}$ are divisible by $n$.

\item A surprising result, proved by D. Singmaster \cite{almost}, is that any integer divides almost all binomial coefficients. More precisely, let $d$ be an integer and let $f(N)$ be the number of binomial coefficients ${n\choose k}$ divisible by $d$, with $n &lt; N$. Then
$$
\displaystyle\lim_{N\to \infty}\displaystyle\frac{f(N)}{N(N+1)/2}=1.
$$
Since there are $N(N+1)/2$ binomial coefficients ${n\choose k}$, with $n&lt;N$, 
the density of the set of binomial coefficients divisible by $d$ is $1$. 

\item Recently Zhi-Wei Sun \cite{zhi} proved,
for example, that for any positive integers $k$, $\ell$ and $n$ the following holds $$ \ell n+1 \mid k{kn+\ell n\choose \ell n}.$$

\end{itemize}


Other interesting results concerning divisibility properties of binomial coefficients can be found in \cite{FINE,GLAI}. For example the following holds: $3\mid {sn \choose n}$, for all $n\geq 1$ if and only if $3\mid s$.

In a very recent paper, the authors \cite{MT} proved, among other things, that $2\mid {2n\brack n}_F$ for all integers $n&gt;1$. However, the same is not valid when we replace $2$ by $3$,
as can be seen by the example $3\nmid {3\cdot 2\brack 2}_F=40$.

In this paper, we shall study similar problems for the Fibonomial
coefficients. Thus we shall deal with the divisibility of ${sn \brack
n}_F$ by $3$ for some positive integers $n$ and $s$.

Our first result gives a necessary and sufficient condition for that $3\mid {3n\brack n}_F$. 

\begin{theorem}\label{main}
We have $3\nmid {3n\brack n}_{F}$ if and only if $n=1$ or $n={2\cdot 3^{k}}$ for $k\geq 0$.
\end{theorem}

As we said before, we have $3\mid {sn \choose n}$ for all $n\geq 1$ if and only if $3\mid s$. Our next theorem gives a related result in the Fibonomial context.

\begin{theorem}\label{main2}
Let $s&gt;0$ be an integer. The number ${sn\brack n}_F$ is a multiple of $3$ for all $n\geq 1$ if and only if $s\equiv 0\pmod {12}$.
\end{theorem}

We organize this paper as follows. In Section \ref{aux}, we will recall some useful properties of the Fibonacci numbers such as a result concerning the $3$-adic order of $F_n$. Sections \ref{s1} and \ref{s2} are devoted to the proof of Theorems \ref{main} and \ref{main2}, respectively.

\section{Auxiliary results}\label{aux}

Before proceeding further, we recall some facts about the Fibonacci numbers for the convenience of the reader. 

\begin{lemma}\label{l1}
We have
\begin{itemize}
\item[{\rm(a)}] $F_n\mid F_m$ if and only if $n\mid m$.
\item[{\rm (b)}] If $m&gt;k&gt;1$, then
$$
{m\brack k}_F=\displaystyle\frac{F_m}{F_k}{m-1\brack k-1}_F.
$$
%\item[\mbox{(c)}] (d'Ocagne's identity) $(-1)^{n}F_{m-n}=F_mF_{n+1}-F_nF_{m+1}$.
%\item[\mbox{(d)}] $F_{p-(\frac{5}{p})}\equiv 0\pmod p$, for all primes $p$.

\end{itemize}
\end{lemma}

Item (a) can be proved by using the well-known Binet formula
\begin{center}
$F_n=\displaystyle\frac{\alpha^n-\beta^n}{\alpha-\beta}$, for $n\geq 0$,
\end{center}
where $\alpha=(1+\sqrt{5})/2$ and $\beta=(1-\sqrt{5})/2$. The proof of item (b) follows directly from definition (\ref{def}). We refer the reader to \cite{fib1,fib2,fibonacci,paulo} for more properties and additional bibliography.


The $p$-adic order (or valuation) of $r$, $\nu_p(r)$, is the exponent of the highest power of a prime $p$ which divides $r$. The $p$-adic order of Fibonacci numbers was completely characterized, see \cite{21,order,22,23}. For instance, from the main results of Lengyel \cite{order}, we extract the following result.

\begin{lemma}\label{l2}
For $n\geq 1$, we have
\begin{center}

\begin{equation*}
\nu_3(F_n)= 
\begin{cases} 
\nu_3(n)+1, &amp; \text{if }  n\equiv 0\pmod {4}; \\
0,  &amp; \text{if } n\not \equiv 0\pmod{4}.
\end{cases}
\end{equation*}



\end{center}
\end{lemma}
A proof of a more general result can be found in \cite[pp.\ 236--237
and Section 5]{order}.

\begin{lemma}\label{Est}
For any integer $k\geq 1$ and $p$ prime, we have
\begin{equation}\label{est}
\frac{k}{p-1}-\left\lfloor \frac{\log k}{\log p}\right\rfloor -1\leq \nu_p(k!)\leq \frac{k-1}{p-1},
\end{equation}
where, as usual, $\lfloor x\rfloor$ denotes the largest integer less than or equal to $x$.
\end{lemma}

\begin{proof}
Recall the well-known Legendre formula \cite{leg}:
\begin{equation}\label{a}
\nu_p(k!)=\frac{k-s_p(k)}{p-1},
\end{equation}
where $s_p(k)$ is the sum of digits of $k$ in base $p$. Since $k$ has $\lfloor \log k/\log p\rfloor +1$ digits in base $p$, and each digit is at most $p-1$, we get
\begin{equation}\label{b}
1\leq s_p(k)\leq (p-1)\left(\left\lfloor \frac{\log k}{\log p}\right\rfloor +1\right).
\end{equation}
Therefore, the inequality in (\ref{est}) follows from (\ref{a}) and (\ref{b}).
\end{proof}

Now we are ready to deal with the proofs of our theorems.

\section{Proof of Theorem \ref{main}}\label{s1}

In order to make our proof clearer, we shall split the statement of Theorem \ref{main} in four propositions.
\begin{proposition} $($The ``if"\ part$)$
For all integers $k\geq 0$, we have that $3\nmid {2\cdot 3^{k+1}\brack 2\cdot 3^k}_F$.
\end{proposition}

\begin{proof}
Using the definition of the Fibonomial coefficients, we have
\begin{center}
$\displaystyle{2\cdot 3^{k+1}\brack2\cdot 3^{k}}_{F}=\displaystyle\prod\limits_{i=1}^{2\cdot
3^{k}}\frac{F_{2\cdot 3^{k+1}-2\cdot 3^k+i}}{F_i}=\displaystyle\prod\limits_{i=1}^{2\cdot 3^{k}}\frac{F_{{4\cdot 3^{k}+i}%
}}{F_{i}}$
\end{center}
and hence 
\begin{center}
$\nu _{3}\left( \displaystyle{2\cdot 3^{k+1}\brack 2\cdot 3^{k}}_{F}\right) =\nu _{3}\left(
\displaystyle\prod\limits_{i=1}^{2\cdot 3^{k}}\frac{F_{{4\cdot 3^{k}+i}}}{F_{i}}%
\right) =\displaystyle\sum_{i=1}^{2\cdot 3^{k}}\left( \nu _{3}\left( F_{{4\cdot
3^{k}+i}}\right) -\nu _{3}\left( F_{i}\right) \right)$.
\end{center}

Thus, for proving that the assertion holds, it suffices to show that $\nu _{3}\left( F_{i}\right)=\nu _{3}\left( F_{{%
4\cdot 3^{k}+i}}\right)$ for all $i=1,2,\ldots
,2\cdot 3^{k}$. Since $4\cdot 3^{k}+i\equiv i\pmod 4$ and $3\mid F_n$ if and only if $4\mid n$ (Lemma \ref{l1} (a)), we need only to consider the case when $4\mid i$, that is, when $i=4t_i$, for some positive integer $t_i$. From this fact together with Lemma \ref{l2}, we obtain
\[
\nu_3(F_i)=\nu_3(F_{4t_i})=\nu_3(t_i)+1
\]
while
\[
\nu_3(F_{4\cdot 3^k+i})=\nu_3(F_{4(3^k+t_i)})=\nu_3(3^k+t_i)+1=\nu_3(t_i)+1,
\]
where in the last equality above, we used that $t_i&lt;3^k$ (since $4t_i=i\leq 2\cdot 3^k$) and the clear identity $\nu_p(a+b) = \min\{\nu_p(a),\nu_p(b)\}$, when $\nu_p(a)\neq \nu_p(b)$, where $p$ is any prime. This completes the proof.
\end{proof}

For the ``only if"\ part, we have
\begin{proposition}
For all integers $a\geq 2$ and $k\geq 1$, we have that $3\mid {2^a\cdot 3^{k+1}\brack 2^a\cdot 3^{k}}_F$.
\end{proposition}

\begin{proof}
By Lemma \ref{l1} (b), we can write
\[
{2^a\cdot 3^{k+1}\brack 2^a\cdot 3^{k}}_F=\displaystyle\frac{F_{2^a\cdot 3^{k+1}}}{F_{2^a\cdot 3^{k}}}{2^a\cdot 3^{k+1}-1\brack 2^a\cdot 3^{k}-1}_F
\]
and so it suffices to prove that $3\mid F_{2^a\cdot 3^{k+1}}/F_{2^a\cdot 3^{k}}$. Indeed, using Lemma \ref{l2} and the fact that $a\geq 2$ to get
\[
\nu_3\left(\displaystyle\frac{F_{2^a\cdot 3^{k+1}}}{F_{2^a\cdot 3^{k}}}\right)=\nu_3(F_{2^a\cdot 3^{k+1}})-\nu_3(F_{2^a\cdot 3^k})=\nu_3(2^a\cdot 3^{k+1})-\nu_3(2^a\cdot 3^k)=1.
\]
\end{proof}



\begin{proposition}\label{03}
For all integers $a\geq 1$, we have that $3\mid {3^{a+1}\brack 3^{a}}_F$.
\end{proposition}

\begin{proof}
Let us suppose, without loss of generality, that $a$ is even (the case of $a$ odd can be handled in much the same way). Since $3\mid {27\brack 9}_F$, we may assume that $a&gt;2$. By definition of the Fibonomial coefficient, we have
\[
\displaystyle{3^{a+1}\brack 3^a}_F=\displaystyle\frac{F_{3^{a+1}}\cdots F_{2\cdot 3^a+1}}{F_1\cdots F_{3^a}}.
\]

So we must to compare the $3$-adic order of the numerator and denominator of the previous fraction. Since $3\mid F_n$ if and only if $4\mid n$, we need only to consider the $3$-adic order of the $\lfloor 3^a/4\rfloor$ numbers $F_4,\ldots,F_{3^a-1}$, for the denominator, and $F_{2\cdot 3^a+2},\ldots,F_{3^{a+1}-3}$, for the numerator. So, in the first case, we use Lemma \ref{l2} to obtain
\begin{eqnarray}\label{eq1}
\mathcal{S}_1 &amp; := &amp; \nu_3(F_1\cdots F_{3^a})\nonumber \\
&amp; = &amp; \nu_3(F_4)+\nu_3(F_8)+\cdots + \nu_3(F_{3^a-1})\nonumber \\
&amp; = &amp; (\nu_3(4)+1)+(\nu_3(8)+1)+\cdots + (\nu_3(3^a-1)+1)\nonumber \\
&amp; = &amp; \nu_3(4)+\nu_3(8)+\cdots + \nu_3(3^a-1)+\left\lfloor \frac{3^a}{4}\right\rfloor.\\
\nonumber
\end{eqnarray}
We note that (\ref{eq1}) could be rewritten as
\begin{eqnarray}
\nu_3(F_1\cdots F_{3^a}) &amp; = &amp; \nu_3(12)+\nu_3(24)+\cdots + \nu_3\left(12\left\lfloor \frac{3^a-1}{12}\right\rfloor\right)+\left\lfloor \frac{3^a}{4}\right\rfloor \nonumber\\
&amp; = &amp; \nu_3\left(\left\lfloor \frac{3^a-1}{12}\right\rfloor !\right)+\left\lfloor \frac{3^a-1}{12}\right\rfloor+\left\lfloor \frac{3^a}{4}\right\rfloor.\nonumber \\
\nonumber
\end{eqnarray}
For the $3$-adic order of numerator, we proceed as before to get
\begin{eqnarray}\label{eq2}
\mathcal{S}_2 &amp;:=&amp; \nu_3(F_{3^{a+1}}\cdots F_{2\cdot 3^a+1})  =  \nu_3(F_{3^{a+1}-3})+\cdots + \nu_3(F_{2\cdot 3^a+2})\nonumber \\
&amp; = &amp; \nu_3(3^{a+1}-3)+\cdots + \nu_3(2\cdot 3^a+2)+\left\lfloor \frac{3^a}{4}\right\rfloor\nonumber \\
&amp; = &amp; \nu_3(3(3^a-1))+\cdots + \nu_3(3(3^a-(3^{a-1}-2)))+ \left\lfloor \frac{3^a}{4}\right\rfloor\nonumber \\
&amp; = &amp; \nu_3(3^a-1)+\cdots + \nu_3(3^a-(3^{a-1}-2))+\nonumber\\
&amp; + &amp; \frac{3^{a-1}+1}{4}+\left\lfloor \frac{3^a}{4}\right\rfloor. \\
\nonumber
\end{eqnarray}

Observe that there exist several common terms in sums (\ref{eq1}) and (\ref{eq2}), so combining them gives
\begin{eqnarray}
\mathcal{S}_2-\mathcal{S}_1 &amp; = &amp; \frac{3^{a-1}+1}{4} - (\nu_3(4)+\cdots + \nu_3(3^a-(3^{a-1}+3)))\nonumber \\
&amp; = &amp; \frac{3^{a-1}+1}{4}-(\nu_3(12)+\cdots + \nu_3\left(12\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor\right))\nonumber \\
&amp; = &amp; \frac{3^{a-1}+1}{4}-\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor-\nu_3\left(\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor !\right).\\
\nonumber
\end{eqnarray}
Hence, when $a$ is even, we have
\begin{equation}\label{M1}
\nu_3\left(\displaystyle{3^{a+1}\brack 3^a}_F\right)=\frac{3^{a-1}+1}{4}-\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor-\nu_3\left(\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor !\right).
\end{equation}
The fact that $\lfloor x \rfloor \leq x$ yields the following estimate
\begin{equation}\label{est1}
\nu_3\left(\displaystyle{3^{a+1}\brack 3^a}_F\right)\geq \frac{3^{a-1}+6}{12}-\nu_3\left(\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor !\right).
\end{equation}
By applying Lemma \ref{Est} to the $3$-adic order in the right-hand side of (\ref{est1}), we obtain
\begin{equation}\label{est2}
\nu_3\left(\left\lfloor \frac{2\cdot 3^{a-1}-3}{12}\right\rfloor !\right)\leq \frac{2\cdot 3^{a-1}-15}{24}.
\end{equation}
Now, we combine (\ref{est1}) and (\ref{est2}) to derive
\[
\nu_3\left(\displaystyle{3^{a+1}\brack 3^a}_F\right)\geq \frac{3^{a-1}+6}{12}-\frac{2\cdot 3^{a-1}-15}{24}=\frac{27}{24}&gt;0
\]
as desired. Since $27/24=1.125$, we actually proved that $\nu_3\left({3^{a+1}\brack 3^a}_F\right)\geq 2$, when $a&gt;2$ is even.
\end{proof}

For the sake of completeness, we remark that the related formula to (\ref{M1}), for $a$ odd is
\begin{equation}\label{MM}
\nu_3\left(\displaystyle{3^{a+1}\brack 3^a}_F\right)=\frac{3^{a-1}-1}{4}-\left\lfloor \frac{2\cdot 3^{a-1}-2}{12}\right\rfloor-\nu_3\left(\left\lfloor \frac{2\cdot 3^{a-1}-2}{12}\right\rfloor !\right).
\end{equation}

To finish the ``only if"\ case, all that remains is to prove the following.
\begin{proposition}
For all integers $k\geq 1$ and every prime $p&gt;3$, we have that $3\mid {3pk\brack pk}_F$.
\end{proposition}

\begin{proof}
To prove this assertion, we take the same approach as in the proof of Proposition \ref{03}. Instead of demonstrating the general case, which is notationally complicated, we restrict ourselves to a particular case that captures the exact essence of our idea. For that, we shall consider $p\equiv k\equiv 1\pmod {12}$. Although there are several cases to consider (48 cases depending on the residue of $p$ and $k$ modulo $12$), the proofs are very similar. 

First, we write
\[
{3pk\brack pk}_F=\displaystyle\frac{F_{3pk}\cdots F_{2pk-1}}{F_1\cdots F_{pk}}.
\]

We note that again, by Lemma \ref{l1} (a) (for $n=4$), we need to take care only of the following sequences of indexes: $4,8,\dots,pk-1$ and $2pk+2,\dots,3pk-3$ which correspond to indexes of the denominator and numerator respectively, having non-zero 3-adic valuation. Thus
\begin{eqnarray}\label{eq4}
\mathcal{M}_1:=\nu_3(F_1F_2\cdots F_{pk}) &amp; = &amp; \nu_3(F_4)+\nu_3(F_8)+\cdots + \nu_3(F_{pk-1})\nonumber \\
&amp; = &amp; (\nu_3(4)+1)+\cdots + (\nu_3(pk-1)+1)\nonumber \\
&amp; = &amp; \nu_3(4)+\cdots + \nu_3(pk-1)+\left\lfloor \frac{pk}{4}\right\rfloor\\
\nonumber
\end{eqnarray}
and 


\begin{eqnarray}\label{eq5}
\mathcal{M}_2&amp;:= &amp; \nu_3(F_{3pk}F_{3pk-1}\cdots F_{2pk+1})\nonumber\\
 &amp; = &amp; \nu_3(F_{3pk-3})+\nu_3(F_{3pk-7})+\cdots + \nu_3(F_{2pk+2})\nonumber \\
&amp; = &amp; \nu_3(3pk-3)+\cdots + \nu_3(3pk-(pk-2))+\left\lfloor \frac{pk}{4}\right\rfloor\nonumber \\
&amp; = &amp; \nu_3(3(pk-1))+\cdots + \nu_3\left(3\left(pk-\frac{pk-10}{3}\right)\right)+\left\lfloor \frac{pk}{4}\right\rfloor\nonumber \\
&amp; = &amp; \nu_3(pk-1)+\cdots + \nu_3\left(pk-\frac{pk-10}{3}\right)+ \frac{pk-1}{12}+\nonumber\\
&amp; + &amp; \left\lfloor \frac{pk}{4}\right\rfloor. \\
\nonumber
\end{eqnarray}


Observe that there exist several common terms in sums (\ref{eq4}) and (\ref{eq5}), thus combining them
\begin{eqnarray}
\mathcal{M}_2-\mathcal{M}_1 &amp; = &amp; \frac{pk-1}{12}-(\nu_3(4)+\cdots + \nu_3\left(\frac{2pk+2}{3}\right))\nonumber \\
&amp; = &amp; \frac{pk-1}{12}-(\nu_3(12)+\cdots + \nu_3\left(12\left\lfloor \frac{2pk+2}{36}\right\rfloor\right))\nonumber \\
&amp; = &amp; \frac{pk-1}{12}-\left\lfloor \frac{2pk+2}{36}\right\rfloor-\nu_3\left(\left\lfloor \frac{2pk+2}{36}\right\rfloor !\right).\\
\nonumber
\end{eqnarray}
Hence
\begin{eqnarray}
\nu_3\left( {3pk\brack pk}_F \right)
 &amp;= &amp; \frac{pk-1}{12}-\left\lfloor \frac{2pk+2}{36}\right\rfloor-\nu_3\left(\left\lfloor \frac{2pk+2}{36}\right\rfloor !\right)\nonumber\\
 &amp; \geq &amp; \frac{pk-5}{36}-\nu_3\left(\left\lfloor \frac{2pk+2}{36}\right\rfloor !\right)\label{A}\\
 &amp; \geq &amp; \frac{pk-5}{36}-\frac{pk-17}{36}\label{B}\\
 &amp; = &amp; \frac{1}{3}&gt;0,\nonumber\\
 \nonumber
\end{eqnarray}
where we used that $\lfloor x \rfloor \leq x$ (in (\ref{A})) and that $\nu_3(\lfloor (2pk+2)/36\rfloor !)\leq (pk-17)/36$, by Lemma \ref{Est} (in (\ref{B})). The proof is then complete.
\end{proof}

\section{Proof of Theorem \ref{main2}}\label{s2}

\begin{proof}
For the ``if"\ part, we write $s=12k$, then 
\begin{center}
$\displaystyle{sn\brack n}_{F}=\displaystyle{12kn\brack n}_{F}=\displaystyle\frac{F_{12kn}}{F_{n}}{12kn-1\brack n-1}_F$.
\end{center}%
Now, it suffices to prove that $3\mid F_{12kn}/F_{n}$. For that we use Lemma \ref{l2} to obtain
\[
\nu_3\left(\frac{F_{12kn}}{F_{n}}\right)=\nu_3(F_{12kn})-\nu_3(F_n)=\nu_3(kn)+2-\nu_3(F_n)
\]
and so
\begin{equation*}
\nu_3\left(\displaystyle\frac{F_{12kn}}{F_{n}}\right) = 
\begin{cases} 
2+\nu_3(kn),&amp; \mbox{if}\  4\nmid n;\\
1+\nu_3(k),  &amp; \mbox{if}\  4\mid n.
\end{cases}
\end{equation*}

Summarizing, we conclude that $\nu_3\left(F_{12kn}/F_{n}\right)\geq 1$ and this completes the proof of this case.

Let $k$ be an integer belonging to $\{1,\ldots,11\}$. Suppose that $s\equiv k%
\pmod {12}$, in order to prove the ``only if"\ part, it suffices to exhibit a
positive integer $N_k$ such that $3\nmid {sN_k\brack N_k}_F$. Of course, $%
N_k=1$ is an example of such number for $k=1,2,3,5,6,7,9,10,11$, because ${s\brack 1}%
_F=F_s$ is not a multiple of $3$, if $4\nmid s$. We claim that $N_4=N_8=4$ are also
examples. In fact, we have
\begin{center}
$\nu_3\left(\displaystyle{4s\brack 4}_F\right)=\nu_3\left(\displaystyle\frac{F_{4s}F_{4s-1}F_{4s-2}F_{4s-3}}{%
F_1F_2F_3F_4}\right)=\nu_3\left(\displaystyle\frac{F_{4s}}{3}\right)=\left(\nu_3(4s)+1\right) -1=0$, 
\end{center}
where we used that $3\nmid s$ when $s\equiv 4,8\pmod{12}$.
\end{proof}

\section{Conclusion}

In this paper, we study divisibility properties of the Fibonomial coefficients ${m\brack k}_F$ by $3$. Among other things, we give necessary and sufficient conditions for ${sn\brack n}_F$ being divisible by $3$, for some integers $s$ and $n$. Our method is effective and possibly can be used to work on divisibility by larger primes. However, it is important to get noticed that for each prime, this study brings a lot of particular technicalities.

\section{Acknowledgements}
The authors would like to express their gratitude to the anonymous
referee for his/her corrections and comments. The first author thanks
to DPP-UnB, FAP-DF, FEMAT and CNPq-Brazil for financial support. The
second author is supported by Specific research 2103 UHKCZ.

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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11B39.

\noindent \emph{Keywords: } 
Fibonacci number, Fibonomial coefficient, divisibility.

\bigskip
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\noindent (Concerned with sequences \seqnum{A000045} and \seqnum{A144712}.)

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\vspace*{+.1in}
\noindent
Received March 1 2012;
revised version received   June 11 2012.
Published in {\it Journal of Integer Sequences}, June 19 2012.

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\noindent
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