Q:

Bounded sequence with $a_n=|a_{n+1}-a_{n+2}|$

Suppose $a_0,a_1>0$ are distinct, and $a_n=|a_{n+1}-a_{n+2}|$ for all $n\geq 0$. Is it possible that the sequence is bounded? From my experiment, the sequence seem to always be unbounded.
We have $\pm a_n=a_{n+1}-a_{n+2}$, so $a_{n+2}=a_{n+1}\pm a_n$. For each $n$ we can choose plus or minus. What can we do to show that the magnitude of the sequence must grow?

A:

Although it can be shown
$$
(*)\ \limsup_{n\to\infty} a_n = +\infty,
$$
it is not necessarily the case that
$$
\liminf_{n\to\infty} a_n = +\infty.
$$
Example: $1,2,3,5,2,7,9,2,11,13,2,\ldots$.
I now show $(*)$ above.
We know: $a_0,a_1>0$. $a_0 \ne a_1$. $a_n = |a_{n+2} - a_{n+1}|$.
Clearly, $a_n\ge 0$ and for all $n$, either (1) $a_{n+2} = a_{n+1} + a_n$, or (2) $a_{n+1} > a_n$ and $a_{n+2} = a_{n+1} - a_n$.
Fix an $\epsilon>0$ such that $a_0, a_1, \mbox{ and } |a_1-a_0|>\epsilon$.
Observe that $a_n, a_{n+1}, \mbox{ and } |a_{n+1} - a_n|>\epsilon$ for all $n$. To see this, use induction. It's true for $a_0$ and $a_1$ by assumption. Suppose it's true for $a_n$ and $a_{n+1}$. We know
$$
a_{n+2} \ge |a_{n+1} - a_n| > \epsilon
$$
and
$$
|a_{n+2} - a_{n+1}| = a_n > \epsilon.
$$
Thus, the induction step is proved.
Finally, observe that for all $n$, either $a_{n+1} - a_n > \epsilon$ or $a_{n+2} - a_n > \epsilon$. If $a_{n+1} > a_n$, then $a_{n+1} - a_n > \epsilon$ by the lemma shown above. If $a_{n+1} < a_n$, then $a_{n+2} = a_n + a_{n+1}$, and so $a_{n+2} - a_n = a_{n+1} > \epsilon$, again by the lemma above.
Since for all $n$, $a_{n+2} - a_n > \epsilon$ or $a_{n+1} - a_n > \epsilon$, it follows $a_n$ is arbitrarily large for some $n$.

