\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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} \hyphenation{com-ple-ting} \def\noi {\noindent} \def\Ker {{\rm Ker}} \def\Im {{\rm Im}} \def\A {\mathcal{A}} \def\B {\mathcal{B}} \def\O {\mathcal{O}} \def\P {\mathfrak{P}} \def\Q {\mathfrak{Q}} \def\N {\mathbb{N}} \def\Z {\mathbb{Z}} \def\Q {\mathbb{Q}} \def\R {\mathbb{R}} \def\RR {\mathcal{R}} \def\QQ {\overline{\Q}} \def\QQQ {\QQ} \def\C {\mathbb{C}} \def\cF {\mathcal{F}} \def\cR {\mathcal{R}} \def\e {\epsilon} \def\d {\delta} \def\g {\gamma} \def\fle {\longrightarrow } \def\d {\displaystyle} \vskip .2in %% 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 $n < N$. 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 $n1$. 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 $s>0$ 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 $m>k>1$, 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, & \text{if } n\equiv 0\pmod {4}; \\ 0, & \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 $t_i<3^k$ (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 $a>2$. 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 & := & \nu_3(F_1\cdots F_{3^a})\nonumber \\ & = & \nu_3(F_4)+\nu_3(F_8)+\cdots + \nu_3(F_{3^a-1})\nonumber \\ & = & (\nu_3(4)+1)+(\nu_3(8)+1)+\cdots + (\nu_3(3^a-1)+1)\nonumber \\ & = & \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}) & = & \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\\ & = & \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 &:=& \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 \\ & = & \nu_3(3^{a+1}-3)+\cdots + \nu_3(2\cdot 3^a+2)+\left\lfloor \frac{3^a}{4}\right\rfloor\nonumber \\ & = & \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 \\ & = & \nu_3(3^a-1)+\cdots + \nu_3(3^a-(3^{a-1}-2))+\nonumber\\ & + & \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 & = & \frac{3^{a-1}+1}{4} - (\nu_3(4)+\cdots + \nu_3(3^a-(3^{a-1}+3)))\nonumber \\ & = & \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 \\ & = & \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 \[ \nu_3\left(\displaystyle{3^{a+1}\brack 3^a}_F\right)\geq \frac{3^{a-1}+6}{12}-\frac{2\cdot 3^{a-1}-15}{24}=\frac{27}{24}>0 \] 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 $a>2$ 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 $p>3$, 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}) & = & \nu_3(F_4)+\nu_3(F_8)+\cdots + \nu_3(F_{pk-1})\nonumber \\ & = & (\nu_3(4)+1)+\cdots + (\nu_3(pk-1)+1)\nonumber \\ & = & \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&:= & \nu_3(F_{3pk}F_{3pk-1}\cdots F_{2pk+1})\nonumber\\ & = & \nu_3(F_{3pk-3})+\nu_3(F_{3pk-7})+\cdots + \nu_3(F_{2pk+2})\nonumber \\ & = & \nu_3(3pk-3)+\cdots + \nu_3(3pk-(pk-2))+\left\lfloor \frac{pk}{4}\right\rfloor\nonumber \\ & = & \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 \\ & = & \nu_3(pk-1)+\cdots + \nu_3\left(pk-\frac{pk-10}{3}\right)+ \frac{pk-1}{12}+\nonumber\\ & + & \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 & = & \frac{pk-1}{12}-(\nu_3(4)+\cdots + \nu_3\left(\frac{2pk+2}{3}\right))\nonumber \\ & = & \frac{pk-1}{12}-(\nu_3(12)+\cdots + \nu_3\left(12\left\lfloor \frac{2pk+2}{36}\right\rfloor\right))\nonumber \\ & = & \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) &= & \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\\ & \geq & \frac{pk-5}{36}-\nu_3\left(\left\lfloor \frac{2pk+2}{36}\right\rfloor !\right)\label{A}\\ & \geq & \frac{pk-5}{36}-\frac{pk-17}{36}\label{B}\\ & = & \frac{1}{3}>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),& \mbox{if}\ 4\nmid n;\\ 1+\nu_3(k), & \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. \begin{thebibliography}{9999} \bibitem{fib1} A. Benjamin and J. Quinn, The Fibonacci numbers---exposed more discretely. \textit{Math. Mag.} \textbf{76} (2003), 182--192. \bibitem{FINE} N. J. Fine, Binomial coefficients modulo a prime, {\it Amer. Math. Monthly} {\bf 54} (1947), 589--592. \bibitem{FONT} G. Fonten\'e, G\'en\'eralisation d'une formule connue, {\it Nouv. Ann. Math.} \textbf{4} (15) (1915), 112. \bibitem{GLAI} J. W. L. 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Vorobiev, \textit{Fibonacci Numbers}, Birkh\"auser, 2003. \bibitem{zhi} Zhi-Wei Sun, On divisibility concerning binomial coefficients, preprint, May 6 2010, available at \url{http://arxiv.org/abs/1005.1054}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 11B39. \noindent \emph{Keywords: } Fibonacci number, Fibonomial coefficient, divisibility. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000045} and \seqnum{A144712}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received March 1 2012; revised version received June 11 2012. Published in {\it Journal of Integer Sequences}, June 19 2012. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}