Consider a recursive sequence. Find all values of x for which this sequence is bounded

Question

Consider a recursive sequence $a_{n+1} = a_n + a_{n-1}$ for all $n \geq 2$ with $a_{1} = 1$ and $a_2 = x$. Find all values of $x$ for which this sequence is bounded.

Answer 1

Note that (similarly to the Fibonacci sequence) all such sequences must satisfy $$a_n=\alpha \cdot a^n+\beta \cdot b^n$$ where $a,b$ are the roots of $x^2-x-1=0$ i.e. $a=\frac{1 +\sqrt{5}}{2},b=\frac{1 -\sqrt{5}}{2}$. Since $|b|<1$ but $|a|>1$ it is necessary (and sufficient) for the sequence to be bounded to satisfy $\alpha=0$. So $a_n=\beta b^n$ i.e. $\beta=\frac{1}{b}$ (by the condition $a_1=1$) and so $x=a_2=b=\frac{1-\sqrt{5}}{2}$ is the unique solution.