XXVI OBM

The answer is yes. For instance, consider $x_0=-1$, $x_1=-n$ and $x_2=-n^2$, $n$ sufficiently large. All $x_i$'s are polynomials in $n$ and, for all large $n$ its sign is equal to the sign of the coefficient of the term of greatest degree. Let $a(x_n)$ such term. Then $$\begin{array}{rlrlrlrl}a(x_0)& =-1,& a(x_1)& =-n,& a(x_2)& =-n^2,& a(x_3)& =-n^3,\\ a(x_4)& =n^3,& a(x_5)& =n^4,& a(x_6)& =-n^5,& a(x_7)& =n^7\end{array}$$ Let's prove by induction that, for $k\ge 5$, $a(x_k)=a(x_{k-2})a(x_{k-3})$, which is equivalent to prove that the degree of $a(x_{k-1})$ is less than the degree of $a(x_{k-2}{x}_{k-3})$. This is true for $k=5,6,7$. Suppose that it is true for all $k$ less than $m$. So the degree of $a(x_{m-1})=a(x_{m-3})a(x_{m-4})$ is less than the degree of $a(x_{m-2})a(x_{m-3})$ because by the induction hypothesis the degree of $a(x_l)$ is less than the degree of $a(x_{l-1})$ and we're done. So the sign of $x_k$ is equal to the sign of $x_{k-2}{x}_{k-3}$ for $k\ge 5$. Thus the signs of the sequence $x_k$ are $$-,-,-,-,+,+,-,+,-,-,-,+,+,-,+,+,-,+,-,-,-,+,+,-,+,-,\dots$$ which are composed of several cycles of the form $-,-,-,+,+,-,+$. Since 4 out of the 7 terms in each cycle are negative, the result follows.