Let recursive sequence be defined by the formula $$ s_{j+1}=as_j-s_{j-1}, $$ where $a>1$ is some integer number. Is it true that $s_0<0$, $s_1<0$ implies $s_j<0$ for $j \geq 0$?
Edit: No, its wrong. Under what conditions on $s_0$ and $s_1$ property $s_j<0$ holds?
First assume $a > 2$, the characteristic equation of the recurrence equation $$s_{j+1} = a s_{j} - s_{j-1}$$
is given by $$\lambda^2 - a \lambda + 1 = 0$$ It has roots at $\mu$ and $\mu^{-1}$ where $\displaystyle \mu = \frac{a + \sqrt{a^2-4}}{2} > 1$. The general solution of the recurrence equation have the form
$$s_j = A \mu^j + B \mu^{-j}$$
Solving $A$ and $B$ from $s_0$ and $s_1$, we get:
$$s_j = \frac{\mu s_1 - s_0}{\mu^2 - 1}\mu^j + \frac{\mu^2 s_0 - \mu s_1}{\mu^2 - 1} \mu^{-j}$$
Since $|\mu| > 1 > |\mu^{-1}|$, $s_j$ will be dominated by the term $A \mu^j$ for large $j$. If $A > 0$, then $s_j$ will be positive for sufficient large $j$. This means $A \le 0$ is an necessary condition for all $s_j < 0$. This condition is also sufficient. If $A \le 0$, then for $j > 1$, we have
$$s_j = A \mu^j + B \mu^{-j} = A \mu ( \mu^{j-1} - \mu^{1-j} ) + \mu^{1-j} (A \mu + B \mu^{-1})\\ = A \mu ( \mu^{j-1} - \mu^{1-j} ) + \mu^{1-j} s_1 \le \mu^{1-j} s_1 < 0$$
Conclusion, the necessary and sufficient condition for all $s_j < 0$ is $A \le 0 \iff \mu s_1 - s_0 \le 0$.
For $a = 2$, the general solution has the form
$$s_j = A j + B$$
It is easy to see the necessary and sufficient condition is again $A \le 0 \iff s_1 - s_0 \le 0$.