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\begin{center}
\vskip 1cm{\LARGE\bf GCD Property of the Generalized Star of \\
\vskip .11in
David in the Generalized Hosoya Triangle}
\vskip 1cm
\large
Rigoberto Fl\'orez\\
Department of Mathematics and Computer Science\\
The Citadel\\
Charleston, SC 29409\\
USA \\
\href{mailto:rigo.florez@citadel.edu}{\tt rigo.florez@citadel.edu} \\
\ \\
Robinson A. Higuita\\
Departamento de Matem\'aticas\\
Universidad de Antioquia\\
Medell\'in\\
Colombia\\
\href{mailto:robinharra@yahoo.es}{\tt robinharra@yahoo.es}\\
\ \\
Leandro Junes\\
Department of Mathematics, Computer Science and Information Systems\\
California University of Pennsylvania\\
California, PA  15419\\
USA \\
\href{mailto:junes@calu.edu}{\tt junes@calu.edu}
\end{center}

\vskip .2in

\begin{abstract}
The \emph{generalized Hosoya triangle}  is an arrangement of
numbers in which each entry is a product of two generalized Fibonacci
numbers. We prove the GCD property for the star  of David of length two. We give necessary and sufficient conditions such that the star  of David of length three satisfies the GCD property. We propose some open questions and a conjecture for the star  of David of length bigger than or equal to four. We also study GCD properties and modularity properties of generalized Fibonacci numbers.
\end{abstract}

\section{Introduction}
The $n$th \emph{generalized Fibonacci number} is a linear combination of the elements $F_{n-2}$ and $F_{n-1}$ with integer coefficients. That is, if we take two integers $a$ and $b$, then the first and second generalized Fibonacci numbers are $a$ and $b$, respectively, and the $n$th generalized Fibonacci number has the form $aF_{n-2}+bF_{n-1}$, where $n>2$. If $a=b=1$, then  we obtain the $n$th regular Fibonacci number.

The \emph{generalized Hosoya triangle}  is a triangular arrangement of
numbers that resembles the Pascal triangle. Each entry is the  product of two generalized Fibonacci
numbers instead of a binomial coefficient. In particular, the regular Hosoya triangle \cite{hosoya, koshy} consists of a triangular array of numbers where each entry is the product of two Fibonacci numbers. Hosoya \cite{hosoya} published this triangle in 1976.

Several authors have studied greatest common divisor (GCD) properties of some geometric configurations in the Pascal triangle and the regular Hosoya triangle (see, for example, \cite{florezjunes, gould, hillman, hoggatt, korntved, long}). We study GCD properties of a special configuration of points in the generalized Hosoya triangle called the \emph{generalized star  of David (GSD)}. We form a GSD by the vertices of two triangles from a regular hexagon in a generalized Hosoya triangle. We say that a star  of David has length $l$ if a side of the regular hexagon has length $l$.

Many authors have studied properties of the star
of David of length two and/or its generalizations (see, for example
\cite{florezjunes, gould, hillman, hoggatt, korntved, long}). With
the exception of the work done by Fl\'orez and Junes \cite{florezjunes} all these papers consider the star  of David in the Pascal triangle.

Hoggatt and Hansell \cite{hoggatt} proved that the product of all points in a star  of David of length two in the Pascal triangle forms a perfect square. Hillman and Hogart \cite{hillman} proved that the GCD of each triangle of the star  of David of length two in the Pascal triangle gives the same number. These
two properties are called the \emph{product} and \emph{GCD properties}, respectively, of the star  of David. Fl\'{o}rez and Junes
\cite{florezjunes} proved that both properties are true for any star  of David of length two in the regular Hosoya triangle.

We prove the GCD property of the star  of David of length two in the generalized Hosoya triangle. We give necessary and sufficient conditions such that the star  of David of length three in the generalized Hosoya triangle satisfies the GCD property. These conditions depend on $L_{2}$--divisibility, where $L_{2}$ is the second Lucas number. To prove these results, we introduce new properties of generalized Fibonacci numbers. In particular, we provide an analog to the identity $\gcd(F_{n},F_{m})=F_{\gcd(n,m)}$ for generalized Fibonacci numbers and show that if a generalized Fibonacci number $G_{n}$ is divisible by $F_{w}$, then the GCD of $G_{n}$ and  $G_{n+kw}$ is $F_{w}$ for any non--zero integer $k$.

We propose a conjecture that gives necessary and sufficient conditions to determine whether the star  of David of length $l \geq 4$ has the GCD property.  These conditions depend on $L_{l-1}$--divisibility, where $L_{l-1}$ is the $(l-1)$th Lucas number.

\section{Preliminaries and  Examples}\label{preliminaries}
In this section we give some examples and introduce notation and
definitions that we are going to use throughout the paper. Some of
them are well known, but we prefer to restate them here to
avoid ambiguities.
\subsection{The Generalized Hosoya Triangle}
We denote by $ \lbrace G_{n}(a,b) \rbrace _{n \in \mathbb{N}}$ the
\emph{generalized Fibonacci sequence} with integers $a$ and $b$. That
is,
\[ G_{1}(a,b)=a,  G_{2}(a,b)=b \text{ and  } G_{n}(a,b)=G_{n-1}(a,b)+G_{n-2}(a,b) \text{ for all } n \in \mathbb{N} \setminus \{1,2\}. \]
It is clear that $G_{n}(1,1)=F_{n}$. When there
is no ambiguity with $a$ and $b$, we denote the $n$th term of the generalized
Fibonacci sequence by $G_{n}$ instead of $G_{n}(a,b)$. The first eight terms of the generalized Fibonacci sequence with integers $a$ and $b$ are
\[ a, b, a + b, a + 2 b, 2 a + 3 b, 3 a + 5 b, 5 a + 8 b, \text{ and } 8 a + 13 b.\]

Notice that every element in this sequence is a linear combination of the integers $a$ and $b$ with Fibonacci coefficients. In general, we have that $G_{n} = a\,F_{n-2} + b\,F_{n-1}$ for all $n \in \mathbb{N}$ (see, for example, \cite[Thm.~7.1, p.~109]{koshy}.)

The \emph{generalized
Hosoya sequence} $\left\{H_{a,b}(r,k)\right\}_{r \ge k\ge 1}$ is defined by the double recursion
\[ H_{a,b}(r,k)= H_{a,b}(r-1,k)+H_{a,b}(r-2,k) \]
\text{ and }
\[ H_{a,b}(r,k)= H_{a,b}(r-1,k-1)+H_{a,b}(r-2,k-2)\]
%
$\text { where  }  r>2 \text { and } 1\le k \le r, \text{ with initial conditions }$
\[H_{a,b}(1,1)=a^2; \quad H_{a,b}(2,1)=ab; \quad H_{a,b}(2,2)= ab; \quad H_{a,b}(3,2)=b^2.\]
It is easy to see that if we let $a=b=1$ in the generalized
Hosoya sequence, then we obtain the regular Hosoya sequence $\left\{H(r,k)\right\}_{r \ge k\ge 1}$ as Koshy defines it \cite[pp.~187--188]{koshy}. It is known that
\[H(r,k)= F_{k} \, F_{r-k+1}\]
for all natural numbers $r, k$ such that $k \le r$, see \cite[p.~188]{koshy}. This and Proposition \ref{prop:GenHos} show that our definition of $\left\{H_{a,b}(r,k)\right\}_{r \ge k\ge 1}$ is the ``right" generalization for $\left\{H(r,k)\right\}_{r\ge k\ge 1}$.
\begin{proposition}\label{prop:GenHos}If $r$ and $k$ are natural numbers such that $ k \le r$, then
     \[H_{a,b}(r,k)= G_{k} \, G_{r-k+1},\]
for all integers $a, b \in \mathbb{Z}$.
\end{proposition}

\begin{proof} We prove first that $H_{a,b}(r,1)= a \, G_{r}$ for all $r \in \mathbb{N}$.

Since $H_{a,b}(r,1)= H_{a,b}(r-1,1)+H_{a,b}(r-2,1)$
for all $r \geq 3$, where $H_{a,b}(1,1)=a^2$ and $H_{a,b}(2,1)=ab$, a straightforward argument by strong induction on $r$ shows that
\[ H_{a,b}(r,1)= a \, G_{r} \text{ for all } r \in \mathbb{N} . \tag{a}\]
Using a similar argument, we can prove that
\[ H_{a,b}(r,2)= b \, G_{r-1} \text{ for all } r \in \mathbb{N} \setminus \{ 1 \}. \tag{b}\]

Now, for any $k \in \mathbb{N}$ let $P(k)$ be the statement: $H_{a,b}(r,k)= G_{k} \, G_{r-k+1}$ for all natural numbers $r$ such that $r \geq k$. We want to prove by strong induction that $P(k)$ is true for all $k \in \mathbb{N}$.

Statements (a) and (b) show that $P(1)$ and $P(2)$ are true. This proves the basis step. We assume now that $P(1), \ldots, P(k)$ are true for a fixed natural number $k \ge 2$ and prove that $P(k+1)$ is true. Let $r$ be a natural number such that $r \geq k+1$. Therefore, $r-1 \geq k$ and  $r-2 \geq k-1$. Since $P(k-1)$ and $P(k)$ are true, we can write
   \[H_{a,b}(r-2,k-1)= G_{k-1} \, G_{r-k} \text{ \quad and \quad } H_{a,b}(r-1,k)= G_{k} \, G_{r-k}.\]
This and the recurrence $H_{a,b}(r,k+1)= H_{a,b}(r-1,k)+H_{a,b}(r-2,k-1)$ show that
\[H_{a,b}(r,k+1)= G_{k+1} \, G_{r-k}.\]
Thus, $P(k+1)$ is true. This proves the proposition. \end{proof}


If there is no ambiguity with the integers $a$ and $b$ we write $H(r,k)$ instead of $H_{a,b}(r,k)$. The generalized Hosoya sequence gives rise to the \emph{generalized Hosoya triangle}
where the entry in position $k$, taken from left to right, of the
$r$th row is equal to $H(r,k)$ (see Table \ref{tabla1}).
\begin{table} [!h] \small
\begin{center} \addtolength{\tabcolsep}{-3pt} \scalebox{.9}{
\begin{tabular}{ccccccccccc}
&&&&&                                                    	 $H(1,1)$                                                          				              &&&&&\\
&&&&                                          $H(2,1)$      	    &&       $H(2,2)$                                              		   	                 &&&&\\
&&&                               $H(3,1)$         &&       $H(3,2)$     &&         $H(3,3)$                                   				&&&\\
&&                      $H(4,1)$       &&       $H(4,2)$      &&       $H(4,3)$      &&       $H(4,4)$                         	            	   &&\\
&             $H(5,1)$     &&       $H(5,2)$     &&        $H(5,3)$     &&       $H(5,4)$     &&        $H(5,5)$               	              &\\
     $H(6,1)$    &&     $H(6,2)$    &&        $H(6,3)$      &&       $H(6,4)$      &&       $H(6,5)$     &&       $H(6,6)$                 \\
\end{tabular}}
\end{center}
\caption{Generalized Hosoya Triangle.} \label{tabla1}
\end{table}

If $P$ is a point in a generalized Hosoya triangle, then it is clear that there are two unique positive integers $r$ and $k$ such that $r \geq k$ with $P=H(r,k)$. We call the pair $(r,k)$ the \emph{rectangular coordinates} of the point $P$.

We now give a more convenient system of coordinates for points in the generalized Hosoya triangle. Proposition \ref{prop:GenHos} shows that every entry of the generalized Hosoya triangle is the product of two generalized Fibonacci numbers. In particular, if we use  Proposition \ref{prop:GenHos} for all entries of Table \ref{tabla1}, we obtain Table \ref{tablaWithGs}.

\begin{table} [!h] \small
\begin{center} \addtolength{\tabcolsep}{-1pt} \scalebox{.9}{
\begin{tabular}{ccccccccccc}
&&&&&                                                    	 $G_1 \, G_1$                                                          				              &&&&&\\
&&&&                                          $G_1\, G_2$      	    &&       $G_2 \,G_1$                                              		   	                 &&&&\\
&&&                               $G_1\,G_3$         &&       $G_2\,G_2$     &&         $G_3\,G_1$                                   				&&&\\
&&                      $G_1\,G_4$       &&       $G_2\,G_3$      &&       $G_3\,G_2$      &&       $G_4\,G_1$                         	            	   &&\\
&             $G_1\,G_5$     &&       $G_2\,G_4$     &&        $G_3\,G_3$     &&       $G_4\,G_2$     &&        $G_5\,G_1$               	              &\\
     $G_1\,G_6$    &&     $G_2\,G_5$    &&        $G_3\,G_4$      &&       $G_4\,G_3$      &&       $G_5\,G_2$     &&       $G_6\,G_1$                 \\
\end{tabular}}
\end{center}
\caption{Generalized Hosoya Triangle.} \label{tablaWithGs}
\end{table}

Notice that any diagonal of Table \ref{tablaWithGs} is the collection of all generalized Fibonacci numbers multiplied by a particular $G_{n}$. More precisely, an $n$th \emph{diagonal} in the generalized Hosoya triangle is the collection
of all generalized Fibonacci numbers  multiplied by $G_n$. We distinguish between \emph{slash diagonals} and \emph{backslash diagonals}, with the obvious meaning. We write $\SD(G_{n})$ and $\BD(G_{m})$ to mean
the slash diagonal and backslash diagonal, respectively (see Figure \ref{slashBackslashDiag}). We formally define these two diagonals as
%
\[\SD(G_{n})=\{ H(n+i-1,n) \}_{i=1}^{\infty} =\{  G_{n}\,G_{i} | i \in \mathbb{N}\},  \]
%
and
\[\BD(G_{m})=\{ H(m+i-1,i) \}_{i=1}^{\infty} = \{ G_{i}\,  G_{m} | i \in \mathbb{N} \}. \]
%


\begin{figure}[!ht]
\begin{center}
\includegraphics[scale=3.9]{slash_backslash.eps}
\end{center}
\caption{The slash diagonal $\SD(G_3)$ and backslash diagonal  $\BD (G_4)$.} \label{slashBackslashDiag}
\end{figure}
We can associate an ordered pair of natural numbers to every element of a generalized
Hosoya triangle. If $P$ is a point in a generalized Hosoya triangle, then
there are two generalized Fibonacci numbers $G_{m}$ and  $G_{n}$ such that $P
\in \BD(G_{m}) \cap \SD(G_{n})$ (Figure \ref{slashBackslashDiag} depicts this fact for $m=4$ and $n=3$.) Thus, $P=G_{m} \, G_{n}$. Therefore, the point $P$ corresponds to the pair $(m,n)$. It is clear that this correspondence is a bijection between points of a generalized Hosoya triangle and ordered pairs of natural numbers. The pair $(m,n)$ is called the \emph{diagonal coordinate} of $P$. Thus, we can view Proposition \ref{prop:GenHos} as a way to change from rectangular coordinates to diagonal coordinates and vice--versa.

Throughout this paper we use only diagonal coordinates to refer to any point in a generalized Hosoya triangle.

We now give some examples of generalized Hosoya triangles. We can construct different triangles by fixing values for the integers $a$ and $b$. For the first example we fix $a=b=1$ and obtain the following numerical sequence.
\[G_{1}= 1, \quad G_{2}=1,  \quad G_{3}=2,  \quad G_{4}=3,  \quad G_{5}=5,  \quad G_{6}=8,   \ldots \]


\noindent Substituting these values in Table \ref{tablaWithGs} we obtain the regular Hosoya triangle (see \cite{hosoya,koshy} and Table \ref{tabla:regHos}).
\begin{table} [!h] \small
\begin{center} \addtolength{\tabcolsep}{4pt} \scalebox{.9}{
\begin{tabular}{ccccccccccc}
&&&&&                                                    	 $1$                                                          				              &&&&&\\
&&&&                                          $1$      	    &&       $1$                                              		   	                 &&&&\\
&&&                               $2$         &&       $1$     &&         $2$                                   				&&&\\
&&                      $3$       &&       $2$      &&       $2$      &&       $3$                         	            	   &&\\
&             $5$     &&       $3$     &&        $4$     &&       $3$     &&        $5$               	              &\\
     $8$    &&     $5$    &&        $6$      &&       $6$      &&       $5$     &&       $8$                 \\
\end{tabular}}
\end{center}
\caption{Regular Hosoya Triangle.} \label{tabla:regHos}
\end{table}

We now fix $a=7$ and $b=2$ and obtain the following numerical sequence.
\[G_{1}= 7, \quad G_{2}=2,  \quad G_{3}=9,  \quad G_{4}=11,  \quad G_{5}=20,  \quad G_{6}=31,   \ldots \]
\noindent Substituting these values in Table \ref{tablaWithGs} we obtain a Hosoya triangle with $a=7$ and $b=2$ (see Table \ref{tabla2} part  (a)). Similarly, if we fix  $a=5$ and  $b=8$ we obtain another numerical sequence.
\[G_{1}= 5, \quad G_{2}=8,  \quad G_{3}=13,  \quad G_{4}=21,  \quad G_{5}=34,  \quad G_{6}=55,  \ldots \]

The Hosoya triangle generated by this new sequence is depicted in Table \ref{tabla2} part  (b).
\begin{table} [!ht]
\small
\begin{center} \addtolength{\tabcolsep}{-1pt} \scalebox{.82}{
\begin{tabular}{cccccccccccc cccccccccccc  ccccccccccccc }
&&&&&&                     49                                      &&&&&&     		&&&&&&                         25                                         &&&&&& \\
&&&&&                  14   && 14                                 &&&&&   		&&&&&                   40   &&   40                                     &&&&& \\
&&&&              63   &&   4  && 63                              &&&&    		&&&&                65   &&   64    &&  65                               &&&&\\
&&&            77  &&  18  && 18  && 77                         &&&               &&&          105   &&   104   &&   104  && 105                      &&&\\
&&        140  &&  22  && 81&& 22   && 140                   &&			&&      170    &&   168   &&  169   &&  168  && 170               && \\
&  217  &&  40  && 99  && 99  && 40  && 217                  &			&     275  &&  272  &&   273   &&  273  &&  272  && 275        & \\
&          &&        &&      & (a) &       &&      &&         &&                                 	     &&        &&       &(b)&        &&       &&          &&       \\
\end{tabular}}
\end{center}
\caption{Generalized Hosoya triangles.} \label{tabla2}
\end{table}

\subsection{Generalized Star of David}
Let $E$ be a regular hexagon formed with points lying in a generalized Hosoya triangle. We say that $E$ has length $l$ if a side of the
hexagon contains $l$ points from the generalized Hosoya triangle. We denote the corner points of $E$ by $a_1, a_2, a_3$ and $b_1, b_2, b_3$.  The \emph{star  of David} of length $l$ is a
configuration formed by the six corner points of $E$. That is, the star  of David is a configuration of six points in the Hosoya triangle formed by two triangles with
vertices $a_1, a_2, a_3$ and $b_1, b_2, b_3$ of $E$. Figure \ref{figure1} part (a) represents a star  of David of length two. The continuous lines in Figure
\ref{figure1} parts (b) and (c) show stars of David of length three and four, respectively.


\begin{figure} [ht!]
\begin{center}
\includegraphics[scale=1.8]{StartDavid1.eps}
\end{center}
\caption{Stars of David.} \label{figure1}
\end{figure}


We first consider some examples of the generalized star  of David of length two as in Figure \ref{figure1} part (a). We can obtain a complete characterization of its vertices $a_1, a_2, a_3$ and $b_1, b_2, b_3$, by knowing the location of one. For instance, if $(m, n)$ are the diagonal coordinates of $a_{2}$, then we have
\begin{center}
\begin{tabular}{llll}
$a_{1}=G_{m+1} \, G_{n-2},$ \quad &$a_{2}=G_{m} \, G_{n}$ & \text{ and } \quad& $a_{3}=G_{m+2} \, G_{n-1},$ \\
$b_{1}=G_{m} \, G_{n-1},$  &$b_{2}=G_{m+2} \, G_{n-2}$ & \text{ and }&  $b_{3}=G_{m+1} \, G_{n}.$
\end{tabular}
\end{center}
We need to give values to $a, b, m$ and $n$ to consider some particular examples. For instance, if we take $a=b=1$ (the regular Hosoya triangle) and  $m= 5$, $n= 4$,  then we obtain
\[a_{1}=8, \quad a_{2}= 15, \quad   a_{3}=26,  \quad b_{1}=10, \quad b_{2}=13, \quad b_{3}=24.\]

Similarly, if we take $a=7$, $b=2$ (generalized Hosoya triangle in Table \ref{tabla2} part (a)), $m= 3$ and $n= 3$, then we obtain
\[a_{1}=77, \quad a_{2}=81,  \quad a_{3}=40,  \quad b_{1}=18, \quad b_{2}=140, \quad b_{3}=99.\]

If we now take  $a=5$, $b=8$ (generalized Hosoya triangle in Table \ref{tabla2} part (b)),  $m= 2$ and $n= 4$, then we have
\[a_{1}=104, \quad a_{2}=168,  \quad a_{3}= 273,  \quad b_{1}=104, \quad b_{2}=168, \quad b_{3}= 273.\]
Notice that in all three cases $\gcd(a_1, a_2,a_3)=\gcd(b_1,b_2,b_3)=1$. 
Fl{\'o}rez
and Junes proved in \cite[Thm.~3.2]{florezjunes} that this is always true for any star  of David of length two in the regular Hosoya triangle. We prove a more general result. Namely, we show that  $\gcd(a_1, a_2,a_3)=\gcd(b_1,b_2,b_3)= (\gcd(a,b))^2$ for any star  of David of length two in any generalized Hosoya triangle (see Theorem \ref{n=2}).


We now consider some examples of the generalized star  of David of length three as in Figure \ref{figure1} part (b). We can obtain a complete characterization of its vertices $a_1, a_2, a_3$ and $b_1, b_2, b_3$ by knowing the location of one. For instance, if $(m, n)$ are the diagonal coordinates of $a_{2}$, then we have
\begin{center}
\begin{tabular}{llll}
$a_{1}=G_{m+2} \, G_{n-4},$\quad &$a_{2}=G_{m} \, G_{n}$& \text{ and } \quad& $a_{3}=G_{m+4} \, G_{n-2},$\\
 $b_{1}=G_{m} \, G_{n-2},$&  $b_{2}=G_{m+4} \, G_{n-4}$&  \text{ and } &  $b_{3}=G_{m+2} \, G_{n}.$
\end{tabular}
\end{center}
We also need to give values to $a, b, m$ and $n$ to consider some particular examples. If we take $a=b=1$ (the regular Hosoya triangle) and $m= 3$,  $n= 5$, then we get
\[a_{1}=5, \quad a_{2}=10,  \quad a_{3}= 26,  \quad b_{1}=4, \quad b_{2}=13, \quad b_{3}=25.\]

Therefore, $\gcd(a_1, a_2,a_3)=\gcd(b_1,b_2,b_3)=1$. However, if we take $m= 8$ and  $n= 16$, then
\[a_{1}=7920, \quad a_{2}=20727,  \quad a_{3}= 54288,  \quad b_{1}=7917, \quad b_{2}=20736, \quad b_{3}=54285,\]
$\gcd(a_1, a_2,a_3)=9$ and $\gcd(b_1,b_2,b_3)=3$. In this case $\gcd(a_1, a_2,a_3) \neq \gcd(b_1,b_2,b_3)$. We give in this paper a complete characterization of all stars of David of length three in the generalized Hosoya triangle for which  $\gcd(a_1, a_2,a_3) = \gcd(b_1,b_2,b_3)$ (see Theorem \ref{n=3equality}).

We can construct similar examples for the star of David of length four. In general, if $(m, n)$ are the diagonal coordinates of $a_{2}$, then the vertices of the star  of David of length $l$ in the generalized Hosoya triangle are given by

\begin{center}
\begin{tabular}{llll}
$a_{1}=G_{m+(l-1)} \, G_{n-2(l-1)},$\quad &$a_{2}=G_{m} \, G_{n}$& \text{ and } \quad& $a_{3}=G_{m+2(l-1)} \, G_{n-(l-1)},$\\
 $b_{1}=G_{m} \, G_{n-(l-1)},$&  $b_{2}=G_{m+2(l-1)} \, G_{n-2(l-1)}$&  \text{ and } &  $b_{3}=G_{m+(l-1)} \, G_{n},$
\end{tabular}
\end{center}
for positive integers $m$ and $n$ such that $n > 2(l-1)$. We require that $n > 2(l-1)$ so that $a_{1},  a_{3},  b_{1}$ and $b_{2}$ correspond to ``real" points in the generalized Hosoya triangle.

We finish this subsection with a definition. We say that a star  of David of length $l$ has the \emph{GCD property} if $\gcd(a_1, a_2,a_3) = \gcd(b_1,b_2,b_3)$.

\subsection{Auxiliary Results}
The \emph{multiplication property of the GCD} says that if $a$ and
$b$ are co--prime, then $\gcd(ab,c)=\gcd(a,c)\gcd(b,c)$. We use this
property to prove Lemma \ref{lemmaparateoreman=2}, which plays a special
role in the proof of the main results, Theorems \ref{n=2} and \ref{n=3equality}. In addition, we also use the following proposition.

\begin{proposition} [{\cite[Proposition 2.2]{florezjunes}}] \label{gcd} If  $\gcd(a,c)=\gcd(b,d)=1$, then $\gcd(a b,c
d)=\gcd(a,d) \gcd(b,c)$, where $a, b, c, d\in \mathbb{Z}.$
\end{proposition}


\section{GCD Properties of Generalized Fibonacci Numbers}\label{PropGenFib}
In this section we study GCD properties of generalized Fibonacci numbers as well as some of its modularity properties. In particular, we provide an analog to the identity $\gcd(F_{n},F_{m})=F_{\gcd(n,m)}$ for generalized Fibonacci numbers. That is,  $\gcd(G_{n},G_{m})$ is always a divisor of $\gcd(a,b)\; F_{\left | m-n \right |}$ with equality when $  \left | m-n \right | \in \{1,  2\}$. We also prove that if a generalized Fibonacci number $G_{n}$ is divisible by $F_{w}$, then the GCD of $G_{n}$ and  $G_{n+kw}$ is $F_{w}$ for any non--zero integer $k$.

We use these results to prove the GCD property for the star  of David of length two and three in the generalized Hosoya triangle.
We write $a\mid b $ to mean that $a$ divides $b$.

\begin{lemma}\label{gcdproperty}
If $\gcd(a,b)=d$, $a'=a/d$ and $b'=b/d$, then $G_{n}(a,b)=dG_{n}(a',b')$ for $n \in \mathbb{N}$.
\end{lemma} \begin{proof} We know that $G_n(a,b)=aF_{n-2}+bF_{n-1}$ for all $n \in \mathbb{N}$. Thus,
\[G_{n}(a,b)=da'F_{n-2}+db'F_{n-1}=d(a'F_{n-2}+b'F_{n-1})=dG_{n}(a',b').  \]
This proves the Lemma.
\end{proof}
Recall that we use $G_{m}$ for $G_{m}(a,b)$ for any integer $m$ if there is no ambiguity. Similarly, we use $G'_{m}$ for $G_{m}(a',b')$  if $d=\gcd(a,b)$, $a'=a/d$ and  $b'=b/d$.

\begin{lemma}\label{lepropo54} If $a, b, n$ and $w$ are integers, then
\[G_{n+w}(a,b)=(aF_{w-1}+bF_w)F_{n-2}+(aF_{w}+bF_{w+1})F_{n-1}=G_{n}\left(G_{w+1},G_{w+2}\right).\]
\end{lemma}
\begin{proof} It is well known that  $F_{n+m}=F_{n-1}F_m+F_nF_{m+1}$, see \cite[Ex.~47, p.~89]{koshy}. Thus,
\begin{eqnarray*}
G_{n+w}(a,b)
			&=&aF_{w+n-2}+bF_{n-1+w}\\
			&=&a(F_{w-1}F_{n-2}+F_wF_{n-1})+b(F_{n-2}F_{w}+F_{n-1}F_{w+1})\\
			&=&(aF_{w-1}+bF_w)F_{n-2}+(aF_w+bF_{w+1})F_{n-1}\\
			&=&G_{w+1}F_{n-2}+G_{w+2}F_{n-1}\\
			&=&G_{n}\left(G_{w+1},G_{w+2}\right).
\end{eqnarray*}
This proves the lemma.
\end{proof}

\begin{theorem}\label{thm:GcdOfTwoFibonacci} Let  $d=\gcd(a,b)$. If  $n$ and $w$ are natural numbers,  then  $ \gcd(G_{n},G_{n+w}) \mid d\,F_{w}$. Moreover, if $w=1, 2$, then $ \gcd(G_{n},G_{n+w}) =d$.
\end{theorem}
\begin{proof}The Lemma \ref{gcdproperty} implies that $\gcd(G_{n},G_{n+w})= d \gcd(G'_{n},G'_{n+w})$.
Thus, it is enough to prove the theorem for  $ \gcd(G'_{n},G'_{n+w})$.

We first prove that $\gcd(G'_n,G'_{n+1})=1$. We use induction on $n$. If $n=1$, then
 \[\gcd(G'_1,G'_2)=\gcd(a',b')=1.\]
This proves the basis step. We now suppose that $\gcd(G'_{k},G'_{k+1}) =1$
for a fixed natural number $k$ and prove $\gcd(G'_{k+1},G'_{k+2}) =1$.

Let $r$ be the natural number such that $\gcd(G'_{k+1},G'_{k+2})=r$. Thus,
\begin{equation}  \label{(1)}
		r \mid G'_{k+1}   \quad \text{ and } \quad r \mid G'_{k+2}.
\end{equation}
Therefore, $r \mid (G'_{k+2}-G'_{k+1})$. Thus, $r \mid G'_{k}$. This and (\ref{(1)}) imply that $r \mid  \gcd(G'_{k},G'_{k+1})$. Therefore, the
inductive hypothesis implies that $r=1$. This proves that
\begin{equation}  \label{extra}
		\gcd(G'_n,G'_{n+1})=1   \quad \text{ for every }  n \in \mathbb{N}.
\end{equation}

We now prove that $\gcd(G'_n,G'_{n+w})$ divides $F_{w}$. Let $d'$ be a divisor of  $\gcd(G'_n,G'_{n+w})$. Therefore,
\begin{eqnarray} \label{extra1}
d'\mid G'_n\quad \text{  and  } \quad  d'\mid G'_{n+w}.
\end{eqnarray}
So,  $d'$ divides any linear combination of  $G'_n $ and $G'_{n+w}$. In particular,
\begin{eqnarray} \label{equ3}
d' \mid (G'_{n+w}-F_{w-1}G'_n).
\end{eqnarray}
We prove that  $d'\mid F_wG'_{n+1}$. Since $G'_{k}=G_{k}(a',b')$ for any positive integer $k$, from Lemma \ref{lepropo54} we have
\[G'_{n+w}=(a'F_{w-1}+b'F_w)F_{n-2}+(a'F_w+b'F_{w+1})F_{n-1}.\]
Therefore,
\[ G'_{n+w}-F_{w-1}G'_n=\left[(a'F_{w-1}+b'F_w)F_{n-2}+(a'F_w+b'F_{w+1})F_{n-1}\right]-F_{w-1}(a'F_{n-2}+b'F_{n-1}). \]
That is,
\begin{eqnarray*}
G'_{n+w}-F_{w-1}G'_n &=&b'F_w F_{n-2}+(a'F_w+b'(F_{w+1}-F_{w-1}))F_{n-1}\\
					    &=&b'F_w F_{n-2}+(a'F_w+b'F_{w})F_{n-1}\\
					    &=&F_w(a'F_{n-1}+b'(F_{n-2}+F_{n-1}))\\
					    &=&F_{w}(a'F_{n-1}+b'F_{n})\\
					    &=&F_wG'_{n+1}.
\end{eqnarray*}
This and (\ref{equ3}) imply that $d' \mid F_wG'_{n+1}$. From (\ref{extra}) and  (\ref{extra1}) we have that $\gcd(d',G'_{n+1})=1$. This and $d' \mid F_wG'_{n+1}$ imply that  $d'\mid F_w$. We have shown that any divisor of $\gcd(G'_n,G'_{n+w})$ is a divisor of $F_{w}$. This proves that $\gcd(G'_n,G'_{n+w})$ divides $F_{w}$.

We now prove the second claim. Equality (\ref{extra}) proves the second claim for $w=1$. If $w=2$, then $\gcd(G'_n,G'_{n+2})$ divides $F_{2}$. Therefore, $\gcd(G'_n,G'_{n+2})=1=\gcd(a',b')$. This proves the theorem.
\end{proof}


\begin{corollary}\label{generalizationlemma2} Let $m,n,s$ and $t$ be positive integers.
If $ |m-n| \in \{1, 2\} $ and $ |s-t| \in \{1, 2\} $, then
 \[\gcd(G_{m}\,G_{s}, G_{n}\,G_{t})= \gcd(G_{m}, G_{t})\, \gcd(G_{s}, G_{n}).\]
\end{corollary}

\begin{proof} Let $d=\gcd(a,b)$, $a'=a/d$ and $b'=b/d$.
Since $ |m-n| \in \{1, 2\} $ and $ |s-t| \in \{1, 2\} $, Theorem \ref{thm:GcdOfTwoFibonacci} shows that
$\gcd(G'_m, G'_{n})=1$  and $\gcd(G'_s, G'_{t})=1$. This and Proposition \ref{gcd} imply that
\[
 \gcd(G'_m G'_{s}, G'_{n} G'_{t})=\gcd(G'_m , G'_{t} ) \gcd(G'_{s},  G'_{n} ).
\]
Multiplying both sides of this equality by $d^2$ we obtain
\begin{equation}\label{formula-a}
 d^2\gcd(G'_m G'_{s}, G'_{n} G'_{t})=d\gcd(G'_m , G'_{t} ) d \gcd(G'_{s},  G'_{n} ).
\end{equation}
Using Lemma \ref{gcdproperty}  one can easily see that
\[ \gcd(G_{m}, G_{t})=d\, \gcd(G'_{m}, G'_{t}), \]
 \[ \gcd(G_{s}, G_{n})=d\, \gcd(G'_{s}, G'_{n}) , \text{ and } \]
 \[\gcd(G_{m} \, G_{s}, G_{n}\,G_{t})= d^2 \gcd(G'_{m} G'_{s}, G'_{n} G'_{t}).\]
Substituting these three equalities in (\ref{formula-a}) we obtain that
\[
 \gcd(G_m G_{s}, G_{n} G_{t})=\gcd(G_m , G_{t} ) \gcd(G_{s},  G_{n} ).
\]
This proves the corollary. \end{proof}

\begin{theorem}  \label{lemma:gcds}  Let  $a,b \in \mathbb{Z}$ and  $n,w \in \mathbb{N}$ such that $\gcd(a,b)=1$. Then
\[ G_{n}\equiv 0 \pmod{ F_{w}} \text{ if and only if } \gcd(G_{n},G_{n-w})=\gcd(G_{n},G_{n+w})=F_{w}.\]
\end{theorem}

\begin{proof}The proof of necessity is straightforward.

We prove sufficiency. We suppose that $G_{n}\equiv 0 \pmod{F_{w}}$ and prove that $\gcd(G_{n},G_{n-w})=\gcd(G_{n},G_{n+w}) = F_{w}$.
Since $G_{n}\equiv 0 \pmod{ F_w}$, there is a $k\in\mathbb{Z}$ such that $G_{n}=kF_w$.  From Lemma  \ref{lepropo54} we have that  for every $m \in\mathbb{Z}$
\begin{eqnarray*}
G_{m+w}&=&(aF_{w-1}+bF_w)F_{m-2}+(aF_w+bF_{w+1})F_{m-1} \\
		     &=&(aF_{w-1}+bF_w)F_{m-2}+(aF_w+b(F_{w-1}+F_w))F_{m-1}\\
		     &=&(aF_{m-2}+bF_{m-1})F_{w-1}+F_w(bF_{m-2}+aF_{m-1}+bF_{m-1}).
\end{eqnarray*}
Thus,
\begin{equation}\label{formula1}
G_{m+w}=G_{m}F_{w-1}+F_w(bF_{m-2}+aF_{m-1}+bF_{m-1}).
\end{equation}
Taking $n$ instead of $m$ in (\ref{formula1}) and using that $G_{n}=kF_{w}$ we obtain
\begin{eqnarray*}
G_{n+w}&=&kF_wF_{w-1}+F_w(bF_{n-2}+aF_{n-1}+bF_{n-1})\\
		    &=&F_w(kF_{w-1}+(bF_{n-2}+aF_{n-1}+bF_{n-1})).
\end{eqnarray*}
Therefore, $G_{n+w}\equiv 0 \pmod{ F_{w}} $. Thus,  $F_{w} \mid \gcd(G_{n+w},G_{n})$. This and Theorem \ref{thm:GcdOfTwoFibonacci} imply that  $F_w=\gcd(G_{n+w},G_{n})$.
Now, taking $n-w$ instead of $m$ in (\ref{formula1}) we obtain
\[ G_{n} = G_{n-w}F_{w-1}+F_w(bF_{n-w-2}+aF_{n-w-1}+bF_{n-w-1}). \]
Since $G_{n}=kF_w$, we have that
\begin{eqnarray*}
G_{n-w}F_{w-1}&=&kF_w-F_w(bF_{n-w-2}+aF_{n-w-1}+bF_{n-w-1})\\
                                 &=&F_w(k-(bF_{n-w-2}+aF_{n-w-1}+bF_{n-w-1})).
\end{eqnarray*}
This and $\gcd(F_w,F_{w-1})=1$ imply that $G_{n-w} \equiv 0 \pmod{ F_w }$. So,
$F_w|\gcd(G_{n-w},G_{n})$. This and Theorem \ref{thm:GcdOfTwoFibonacci} imply that
$F_w=\gcd(G_{n-w},G_{n})$, proving sufficiency.  \end{proof}


\begin{lemma} \label{lemmaparateoreman=2} Let $\alpha, \beta, \delta, \gamma, \rho, \phi \in \mathbb{N}$ and $a,b \in \mathbb{Z}$ such that $\gcd(a,b)=1$ and let
$D=\gcd\left(G_{\alpha}\,G_{\beta},G_{\delta}\,G_{\gamma},G_{\rho}\,G_{\phi} \right)$.

\begin{enumerate}
 \item  If $\alpha =\delta+1, \rho = \delta+2, \; \beta =\gamma-2, \;  \text{ and } \phi = \gamma-1$, then
 $D=1$.

 \item If $ \; \delta=\alpha+1 , \; \rho=\alpha+2, \; \beta=\gamma-1, \text{ and } \phi = \gamma-2$, then
 $D=1$.

\item  If $\alpha = \delta+2, \; \rho=\delta+4, \;  \beta =\gamma-4, \;  \text{  and } \phi =\gamma-2$, then
 $ D= \gcd\left(G_{\beta},G_{\rho},G_{\delta}G_{\gamma}\right)$.

\item If $ \delta=\alpha+2, \; \rho=\alpha+4, \; \beta= \gamma-2, \;  \text{ and } \phi =\gamma-4$, then
$D=\gcd(G_{\alpha},G_{\gamma},G_{ \rho}G_{\phi})$.
\end{enumerate}
\end{lemma}
\begin{proof}It is known that
 \begin{eqnarray}
    \gcd\left(G_{\alpha}\,G_{\beta},G_{\delta}\,G_{\gamma},G_{\rho}\,G_{\phi} \right) &=& \gcd\left(\gcd(G_{\alpha}\,G_{\beta},G_{\delta}\,G_{\gamma}),G_{\rho}\,G_{\phi} \right) \label{31} \\
    					    &=& \gcd\left(\gcd(G_{\alpha}\,G_{\beta},G_{\rho}\,G_{\phi} ), G_{\delta}\,G_{\gamma}\right).   \label{32}
  \end{eqnarray}

We prove part (1). The proof of part (2) is similar and we omit it. Since $\left | \alpha - \delta \right | = 1$  and $\left | \beta - \gamma \right | = 2$, Corollary \ref{generalizationlemma2} and (\ref{31}) imply that
\begin{equation}\label{formula3-lemma36}
D= \gcd\left(\gcd(G_{\alpha},G_{\gamma}) \gcd(G_{\delta},G_{\beta}),G_{\rho}G_{\phi}\right).
\end{equation}
From Theorem  \ref{thm:GcdOfTwoFibonacci}, $\left | \alpha - \rho \right | = 1$ and $\left | \beta - \phi \right | = 1$, we have
$\gcd(G_{\rho},G_{\alpha})=\gcd(G_{\phi },G_{\beta})=1.$ Thus,
\[\gcd\left(\gcd(G_{\gamma},G_{\alpha}),G_{\rho}\right)=\gcd\left(\gcd(G_{\delta},G_{\beta}),G_{\phi}\right)=1.\]
 This, Proposition \ref{gcd} and (\ref{formula3-lemma36}) imply that
    \[D=\gcd\left(\gcd(G_{\alpha},G_{\gamma}),G_{\phi}\right) \; \gcd\left(\gcd(G_{\delta},G_{\beta}),G_{\rho} \right).\]
Using Theorem  \ref{thm:GcdOfTwoFibonacci}, $\left | \gamma - \phi \right | = 1$ and $\left | \delta - \rho \right | = 2$, we have that  $D=1$. This proves part (1).

We now prove part (3). The proof of part (4) is similar and we omit it. Theorem  \ref{thm:GcdOfTwoFibonacci}, $\left | \alpha - \delta \right | = 2$  and $\left | \gamma - \phi \right | = 2$ imply that $\gcd(G_{\alpha},G_{\delta})=1$ and $\gcd(G_{\gamma},G_{\phi})=1$. Therefore, $\gcd\left(\gcd(G_{\alpha},G_{\phi}),G_{\delta}\right)=1$ and $\gcd\left(\gcd(G_{\alpha},G_{\phi}),G_{\gamma}\right)=1$. Thus,
\begin{equation}\label{formula4-lemma36}
\gcd\left(\gcd(G_{\alpha},G_{\phi}),G_{\delta}G_{\gamma}\right)=1.
\end{equation}
Since $\left | \alpha - \rho \right | = 2$  and $\left | \beta - \phi \right | = 2$,
Corollary \ref{generalizationlemma2} and (\ref{32}) imply that
\begin{equation}\label{formula5-lemma36}
   D = \gcd(\gcd(G_{\alpha},G_{\phi}) \gcd(G_{\beta},G_{\rho}),G_{\delta}G_{\gamma}).
    \end{equation}
Theorem  \ref{thm:GcdOfTwoFibonacci}, $\left | \alpha - \rho \right | = 2$  and $\left | \beta - \phi \right | = 2$ imply that   $\gcd(G_{\alpha},G_{\rho})=\gcd(G_{\beta},G_{\phi})=1$. Therefore, $\gcd(\gcd(G_{\alpha},G_{\phi}),\gcd(G_{\beta},G_{\rho}))=1$. This and the multiplication property of the GCD imply that
   \[ \gcd\left(\gcd(G_{\alpha},G_{\phi}) \gcd(G_{\beta},G_{\rho}),G_{\delta}G_{\gamma}\right) = \gcd\left(\gcd(G_{\alpha},G_{\phi}),G_{\delta}G_{\gamma}\right) \; \gcd\left(\gcd(G_{\beta},G_{\rho}) ,G_{\delta}G_{\gamma}\right). \]
This, (\ref{formula4-lemma36}) and (\ref{formula5-lemma36}) prove that $D= \gcd\left(G_{\beta},G_{\rho},G_{\delta}G_{ \gamma} \right)$. This proves the lemma. \end{proof}
%
\section{GCD Property for the Star of David of Length Two and Three}\label{Main}
In this section we prove the GCD property for the star  of David of length two and give necessary and sufficient conditions for the star  of David of length three to have the GCD property. We also prove that $9$ divides the coordinates of the vertices $a_{2}$ or $b_{2}$ in Figure \ref{figure1} part (b) if and only if the GCD of each triangle gives distinct numbers.

We remind the reader that if $d=\gcd(a,b)$, $a'=a/d$ and  $b'=b/d$, then we use $G_{m}$ for $G_{m}(a,b)$ and $G'_{m}$ for $G_{m}(a',b')$ for any integer $m$.

\begin{theorem} \label{n=2}Let $d=\gcd(a,b)$. If $a_{1},a_{2},a_{3}$ and  $b_{1},b_{2},b_{3}$
are the vertices of the star  of David of length two,
 then $\gcd(a_{1},a_{2},a_{3})=\gcd(b_{1},b_{2},b_{3})=d^{2}.$
\end{theorem}

\begin{proof} Let $a'=a/d$, $b' =b/d$, and $(m,n)$ be the diagonal coordinates of $a_{2}$. We prove first that $\gcd(a_{1},a_{2},a_{3})=d^{2}$. It is known that $a_{1}=G_{m+1}\,G_{n-2}$,  $a_{2}=G_{m}\,G_{n}$ and $a_{3}=G_{m+2}\,G_{n-1}$. This and Lemma \ref{gcdproperty} imply that
       \[ \gcd(a_{1},a_{2},a_{3})=  d^{2}\gcd\left(G'_{m+1}\,G'_{n-2},G'_{m}\,G'_{n},G'_{m+2}\,G'_{n-1}\right). \]
 This and Lemma \ref{lemmaparateoreman=2} part (1) show that $\gcd(a_{1},a_{2},a_{3})= d^2$.

The proof of  $\gcd(b_{1},b_{2},b_{3})= d^{2}$ is analogous to the proof of $\gcd(a_{1},a_{2},a_{3})= d^2$. Indeed, the proof follows from Lemma \ref{gcdproperty} and Lemma \ref{lemmaparateoreman=2} part (2). This proves the theorem. \end{proof}

 \begin{lemma} \label{lemma:simplify} Let $d=\gcd(a,b)$ and $a' =a/d$, $b'=b/d$.
 If $a_{1},a_{2},a_{3}$ and  $b_{1},b_{2},b_{3}$ are the vertices of the star  of David of
 length three where $(m, n)$ are the diagonal coordinates of $a_{2}$, then
    \begin{enumerate}
         \item $\gcd(a_{1},a_{2},a_{3})=d^{2} \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$.
         \item $\gcd(b_{1},b_{2},b_{3})=d^{2} \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})$.
    \end{enumerate}
  \end{lemma}
\begin{proof}Since $(m,n)$ are the diagonal coordinates of $a_{2}$,
    \begin{center}
     \begin{tabular}{llll}
   $a_{1}=G_{m+2} \, G_{n-4},$\quad &$a_{2}=G_{m} \, G_{n}$& \text{ and } \quad& $a_{3}=G_{m+4} \, G_{n-2},$\\
   $b_{1}=G_{m} \, G_{n-2},$&  $b_{2}=G_{m+4} \, G_{n-4}$&  \text{ and } &  $b_{3}=G_{m+2} \, G_{n}.$
  \end{tabular}
\end{center}
This and Lemma \ref{gcdproperty} imply that
       \[ \gcd(a_{1},a_{2},a_{3})=  d^{2}\gcd\left(G'_{m+2} \, G'_{n-4},G'_{m}\,G'_{n},G'_{m+4} \, G'_{n-2}\right). \]
This and Lemma \ref{lemmaparateoreman=2} part (3) show that $\gcd(a_{1},a_{2},a_{3})=d^{2} \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$. This proves part (1). The proof of part (2) is similar and we omit it. \end{proof}

 \begin{theorem} \label{n=3}Let $d=\gcd(a,b)$ and $a' =a/d$, $b'=b/d$.
 If $a_{1},a_{2},a_{3}$ and  $b_{1},b_{2},b_{3}$ are the vertices of the star  of David of
 length three where $(m, n)$ are the diagonal coordinates of $a_{2}$, then
   \begin{enumerate}
       \item $G'_{m} \equiv 0 \pmod{ 9}$ and $G'_{n} \equiv 0 \pmod{ 9}$ if and only if $\gcd(a_{1},a_{2},a_{3}) < \gcd(b_{1},b_{2},b_{3})$.
       \item $G'_{m+4} \equiv 0 \pmod{ 9}$ and $G'_{n-4} \equiv 0 \pmod{ 9}$ if and only if $\gcd(a_{1},a_{2},a_{3}) > \gcd(b_{1},b_{2},b_{3})$.
   \end{enumerate}
\end{theorem}
 \begin{proof} We prove part (1). For sufficiency, we assume that $G'_{m} \equiv 0 \pmod{ 9}$ and $G'_{n} \equiv 0 \pmod{ 9}$ and prove that $\gcd(a_{1},a_{2},a_{3})=3\,d^{2}$ and $\gcd(b_{1},b_{2},b_{3})=9\,d^{2}$.

 Since $G'_{m} \equiv 0 \pmod{ 9}$ and $G'_{n} \equiv 0 \pmod{ 9}$, it is clear that $G'_{m} \equiv 0 \pmod{ 3}$ and $G'_{n} \equiv 0 \pmod{ 3}$.
 This and Theorem \ref{lemma:gcds} imply that
 \begin{equation}\label{equation1_theorem43}
 \gcd(G'_{m},G'_{m + 4})=3 \quad  \text{ and } \quad \gcd(G'_{n},G'_{n - 4})=3.
\end{equation}
Therefore, $3 \mid \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$. We now show that  $3$ is the only prime divisor of $\gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$. If $p$ is a prime number and $p \mid \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$, then one can easily see that $p \mid \gcd(G'_{n-4},G'_{n})$  or $p \mid \gcd(G'_{m+4},G'_{4})$. This and (\ref{equation1_theorem43}) imply that $p=3$. Thus, $\gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})=3^{k}$ for some integer $ k \geq 1$. We show that $k=1$. If $k >1$, then $9 \mid G'_{n-4}$ and $9 \mid G'_{m+4}$. Since $G'_{m} \equiv 0 \pmod{ 9}$ and $G'_{n} \equiv 0 \pmod{ 9}$, we can conclude that $9 \mid \gcd(G'_{m},G'_{m + 4})$ and $9 \mid \gcd(G'_{n},G'_{n - 4})$. This  contradicts (\ref{equation1_theorem43}). Thus,
 \begin{equation}\label{equation2_theorem43}
\gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})=3.
\end{equation}

We now prove that  $\gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})=9$. From $G'_{m} \equiv 0 \pmod{ 9}$, $G'_{n} \equiv 0 \pmod{ 9}$ and (\ref{equation1_theorem43}), we can write that $9 \mid \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})$. We show that  3 is the only prime divisor of $\gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})$. If $p$ is a prime and $p \mid \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})$, then one can easily see that $p \mid \gcd(G'_{n-4},G'_{n})$  or $p \mid \gcd(G'_{m+4},G'_{4})$. This and (\ref{equation1_theorem43}) imply that $p=3$. Thus, $\gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})=3^{l}$ for some integer $l \geq 2$. We show that $l=2$. If $l>2$, then  $3^3 \mid G'_{n-4}G'_{m+4}$. Therefore, $9 \mid G'_{n-4}$ or $9 \mid G'_{m+4}$. Since $G'_{m} \equiv 0 \pmod{ 9}$ and $G'_{n} \equiv 0 \pmod{ 9}$, we can conclude that $9 \mid \gcd(G'_{m},G'_{m + 4})$ or $9 \mid \gcd(G'_{n},G'_{n - 4})$. This  contradicts (\ref{equation1_theorem43}). Thus,
                   \[ \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})=9.  \]
This equality, Lemma \ref{lemma:simplify} and  (\ref{equation2_theorem43}) show that $\gcd(a_{1},a_{2},a_{3})=3\,d^{2}$ and $\gcd(b_{1},b_{2},b_{3})=9\,d^{2}$, proving sufficiency of part (1).

Conversely, we assume that $\gcd(a_{1},a_{2},a_{3}) < \gcd(b_{1},b_{2},b_{3})$. This and Lemma \ref{lemma:simplify} imply that
\begin{equation}\label{equation3_theorem43}
       \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n}) < \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4}).
\end{equation}
Therefore, $\gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4}) > 1$. If $p$ is a prime such that $p \mid \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4})$, then it is easy to see that $p \mid \gcd(G'_{m},G'_{m+4})$ or $p \mid \gcd(G'_{n},G'_{n-4})$. This and Theorem \ref{thm:GcdOfTwoFibonacci} imply that $p=3$. Thus,
\begin{equation}\label{equation4_theorem43}
   \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4}) = 3^{t} \text{ for some } t \geq 1.
\end{equation}
Therefore, $3 \mid G'_{m} $ and $3 \mid G'_{n}$. This and Theorem \ref{lemma:gcds} imply that $\gcd(G'_{m},G'_{m + 4})=3$ and $\gcd(G'_{n},G'_{n - 4})=3$. Thus, $3 \mid \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$. In particular, we have that $3 \leq \gcd(G'_{n-4},G'_{m+4},G'_{m}G'_{n})$. This, (\ref{equation3_theorem43}) and (\ref{equation4_theorem43}) imply that
   \[  \gcd(G'_{m},G'_{n},G'_{n-4}G'_{m+4}) = 3^{t} \text{ for some } t \geq 2. \]
Thus, $G'_{m} \equiv 0 \pmod{ 9}$ and $G'_{n} \equiv 0 \pmod{ 9}$. This proves necessity of part (1).

The proof of part (2) is analogous to the proof of part (1). Indeed, in the proof of part (1) we need to interchange the roles of $a_{i}$ and $b_{i}$ for $i=1, 2, 3$, instead of $m$ use $m+4$, instead of $n$ use $n-4$, and also use Theorem \ref{lemma:gcds}. This completes the proof.      \end{proof}

\begin{theorem} \label{n=3equality}Let $d=\gcd(a,b)$ and $a' =a/d$, $b'=b/d$.
 If $a_{1},a_{2},a_{3}$ and  $b_{1},b_{2},b_{3}$ are the vertices of the star  of David of
 length three where $(m, n)$ are the diagonal coordinates of $a_{2}$, then
\[ \gcd(G'_{m+i}, G'_{n-i}, L^{2}_{2}) \neq L^{2}_{2} \text{ for } i=0,4 \text{ if and only if } \gcd(a_{1},a_{2},a_{3}) = \gcd(b_{1},b_{2},b_{3}).\]
\end{theorem} \begin{proof}  It is a straightforward application of Theorem \ref{n=3}. \end{proof}

\section{Open Questions and a Conjecture}
Theorem \ref{n=2} shows that any star  of David of length two has the GCD property. Theorem \ref{n=3equality} proves that the star  of David of length three has the GCD property provided that the integers $a, b, m$ and $n$ satisfy some $L^{2}_{2}-$divisibility conditions. It is natural to ask the following questions.

\begin{table} [!h]
\begin{center}
\begin{tabular}{ccc} \hline
    $l$ & $Q$ & Relationship between Q \\
    &  & and Lucas number \\ \hline \hline \noalign{\smallskip}
    3  & $3^{2}$ & $L^{2}_{2}$ \\ \noalign{\smallskip}
    4 &  $4^{2}$ & $L^{2}_{3}$ \\ \noalign{\smallskip}
    5 &  $7^{2}$ & $L^{2}_{4}$ \\ \noalign{\smallskip}
    6 &  $11^{2}$ & $L^{2}_{5}$  \\ \noalign{\smallskip}
    7 &  $18$ & $L_{6}$ \\  \noalign{\smallskip}
    8 &  $29^{2}$ & $L^{2}_{7}$ \\  \noalign{\smallskip}
    9 &  $47^{2}$ & $L^{2}_{8}$ \\  \noalign{\smallskip}
    10 &  $76^{2}$ & $L^{2}_{9}$ \\  \noalign{\smallskip}
    11 &  $123^{2}$ & $L^{2}_{10}$ \\  \noalign{\smallskip}
    12 &  $119^{2}$ & $L^{2}_{11}$ \\  \noalign{\smallskip}
    13 &  $322$ & $L_{12}$ \\  \noalign{\smallskip}  \hline

\end{tabular}
\end{center}
\caption{Values of $Q$ for some $l$.} \label{conjecture}
\end{table}


\begin{description}
\item[Open Question 1.] What conditions should we impose on $a, b, m$ and $n$ so that the star  of David of length $l \geq 2$ satisfies the GCD property?
\item[Open Question 2.] If a star  of David of length $l \geq 2$ satisfies the GCD property, then what is the common value of $\gcd(a_1, a_2,a_3)$ and $\gcd(b_1,b_2,b_3)$?
\end{description}
Theorem \ref{n=2} answers open questions 1 and 2 for $l=2$. Theorem \ref{n=3equality} answers open question 1  for $l=3$. We propose the following conjecture to answer open question 1 for $l \geq 4$.

\begin{conjecture} Let $d=\gcd(a,b)$ and $a' =a/d$, $b'=b/d$.
 If $a_{1},a_{2},a_{3}$ and  $b_{1},b_{2},b_{3}$ are the vertices of the star  of David of
 length $l$ where $(m, n)$ are the diagonal coordinates of $a_{2}$, then
\[ \gcd(G'_{m+i}, G'_{n-i},Q) \neq Q \text{ for } i=0,\ldots, 2(l-1) \text{ if and only if } \gcd(a_{1},a_{2},a_{3}) = \gcd(b_{1},b_{2},b_{3}),\]
where $Q$ depends only on $l$.
\end{conjecture}
Table \ref{conjecture} gives empirical values of $Q$ for some $l  \geq 4$. Notice that the value of $Q$ for $l=3$ is given by Theorem \ref{n=3equality}. We were not able to find the values of $Q$ for $l \geq 14$. We believe that the values of $Q$ for $l=19$ and $l=25$ follow the same pattern as $l=7$ and $l=13$. These numerical results, for $l  \geq 4$ were found using  \emph{Mathematica}\textsuperscript{\textregistered}.


\section{Acknowledgement}
The first author was partially supported by The Citadel Foundation.
The third author was supported by the Small Grant from the FPDC at California University of Pennsylvania.

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\bibitem{florezjunes} R.~Fl\'orez and L.~Junes,
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\emph{Fibonacci Quart.} \textbf{50} (2012) 163--174.                      			

\bibitem{gould} H. W.~Gould,                                  				
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coefficients,
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\bibitem{hillman} A. P.~Hillman and V. E.~Hoggatt, Jr,
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\emph{Fibonacci Quart.} \textbf{10} (1972) 565--568, 598.

\bibitem{hoggatt} V. E.~Hoggatt Jr. and W.~Hansell,
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\emph{Fibonacci Quart.} \textbf{9} (1971) 120, 133.

\bibitem{hosoya} H.~Hosoya,
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\emph{Fibonacci Quart.} \textbf{14} (1976) 173--178.

\bibitem{korntved} E.~Korntved,
Extensions to the GCD star of David theorem,
\emph{Fibonacci Quart.} \textbf{10} (1994) 160--166.

\bibitem{koshy} T.~Koshy,
\emph{Fibonacci and Lucas numbers with applications},
John Wiley, New York, 2001.

\bibitem{long} C. T.~Long and S.~Ando,
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\end{thebibliography}


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 20D60.

\noindent \emph{Keywords: }
Hosoya triangle, generalized Fibonacci numbers, star of David,
GCD properties, triangular arrangements.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000032} and \seqnum{A000045}.)

\bigskip
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\vspace*{+.1in}
\noindent
Received August 22 2013;
revised versions received  December 11 2013; January 28 2014.
Published in {\it Journal of Integer Sequences}, February 16 2014.

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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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