Lucas or Fibonacci type, then

(1) $\gcd(d(x), G_{2n+1}(x))=G_1(x)$ for every positive integer $n$,

(2) If the GFP is of Lucas type, then $\gcd(d(x), G_{2n}^{*}(x))= 1$ and

    if the GFP is of Fibonacci type, then $\gcd(d(x), G_{2n}^{\prime}(x))= d(x)$

(3) $\gcd(g(x), G_n(x))=\gcd(g(x), G_{1}(x))=1,$ for every positive integer $n$.

We prove part (1) by induction, the proof of part (2) is similar and we omit it.

Let $\{G_n\}$ a GFP. Let $S(n)$ be the statement $$\rho=\gcd(d(x), G_{2n+1}(x))\text{  for  }n\ge 1.$$ To prove $S(1)$ we suppose that $\gcd(d(x), G_{3}(x))=r$. Thus, $r$ divides any linear combination of $d(x)$ and $G_3(x)$. Therefore, $r$ divides $G_3(x)-d(x) G_2(x)$. This and given that $G_3(x)= d(x) G_{2}(x) + g(x) G_{1}(x)$, imply that $r\mid g(x)G_{1}(x)$. So, $r\mid \gcd(d(x),g(x)G_{1}(x))$. Since $\gcd(d(x), g(x))=1$, we have that $r\mid \rho$. It is easy to see that $\rho \mid r$. Thus, $r=\gcd(d(x),G_1(x))$. This proves $S(1)$.

We suppose that $S(n)$ is true for $n=k-1$. That is, suppose that $\gcd(d(x), G_{2k-1}(x)) = \rho$. To prove $S(k)$ we suppose that $\gcd(d(x), G_{2k+1}(x))=r'$. Thus, $r'$ divides any linear combination of $d(x)$ and $G_{2k+1}(x)$. Therefore, $r'$ divides $G_{2k+1}(x)-d(x) G_{2k}(x)$. This and given that $G_{2k+1}(x)= d(x) G_{2k}(x) + g(x) G_{2k-1}(x)$, imply that $r'\mid g(x)G_{2k-1}(x)$. Therefore, $r'\mid \gcd(d(x),g(x)G_{2k-1}(x))$. Since $\gcd(d(x), g(x))=1$, we have 