\;  G_{n}^{*}(x)= (2x+1) G_{n - 1}^{*}(x) + G_{n - 2}^{*}(x) \text{ for } n\ge 2.$$ After some calculations we can see that $\gcd(G_{m}^{*}(x), G_{n}^{'}(x))$ is one, two or $G_{\gcd(m,n)}^{*}(x)$. Using the same polynomials we can see also that $\gcd(G_{m}^{*}(x), G_{n}^{*}(x))$ is one, two or $G_{\gcd(m,n)}^{*}(x)$. If we do the same calculations for numerical sequences (Fibonacci and Lucas numbers), we can see that they have the same behaviour.

In this section we use the notation $E_{2}(n)$ to represent the *integer exponent base two* of a positive integer $n$ which is defined to be the largest integer $k$ such that $2^{k}\mid n$.

\[Dic1\] If $R(x)$, $S(x)$ and $T(x)$ are polynomial in $\mathbb{Z}[x]$, then $$\gcd(R(x),T(x))=\gcd(R(x),R(x)S(x)-T(x)).$$

\[propiedadDivision\] Let $\{ G_{n}^{*}(x)\}$ be a GFP sequence of the Lucas type. If $m \mid n$ and $E_2(n)=E_{2}(m)$, then $
\gcd(G_{n}^{*}(x),G_{m}^{*}(x))=G_{m}^{*}(x).
$

First of all we recall that $E_{2}(n)$ is the largest integer $k$ such that $2^{k}\mid n$. We suppose that $n=mq$ with $q\in\mathbb{N}$. Since $E_{2}(m)=E_{2}(n)=E_{2}(mq)$, we conclude that $q$ is odd. This, Lemma \[Dic1\] and Proposition $\ref{Dic2}$ part (2) imply that $$\begin{aligned}
\gcd(G_{n}^{*}(x),G_{m}^{*}(x))&=&\gcd(G_{qm}^{*}(x),G_{m}^{*}(x))\\
&=&\gcd (G_{m}^{*}(x)T(x)+(-1)^{(n-1)m+n}(g(x))^{(n-1)m} G_{m}^{*}(x),G_{m}^{*}(x))\\
&=&G_{m}^{*}(x).   \hspace{10.1cm}\end{aligned}$$ This proves the proposition.

\[divisity:property:firstype\] Let $G_{m}^{*}(x)$ be a GFP of Lucas type. If $m>0$ is not a power 