Find all bounded infinite sequences of positive integers $(a_n)$ with $a_n = (a_{n-1}+a_{n-2})/(\gcd(a_{n-1},a_{n-2}))$ for $n\ge 2$. 
Find all bounded infinite sequences of positive integers $(a_n)$ with $a_n = (a_{n-1}+a_{n-2})/(\gcd(a_{n-1},a_{n-2}))$ for $n\ge 2$.

I think the only possible sequence is the constant sequence with each term equal to $2$. Let $g_n = \gcd(a_{n},a_{n+1})$. One can show that $(g_n)$ is eventually constant by proving that it is nonincreasing, since it is bounded below by zero and a convergent sequence of integers is eventually constant (the sequence $g_n$ converges by the Monotone Convergence theorem).  If $d$ divides $a_{n+1}$ and $a_{n+2}$, then it divides $g_n a_{n+2} - a_{n+1} = a_n,$ so $g_{n+1} | g_n$, proving the claim.  Clearly one cannot have $g=1$, since then for all $n$ sufficiently large, $a_n = a_{n-1}+a_{n-2}$ so $(a_n)$ has a strictly increasing, unbounded subsequence. One should also be able to eliminate the case $g\ge 3$. Now I think once one proves $g=2$, one can use backwards induction to prove that the sequence $(a_n)$ must be as claimed, though I'm not sure about the details.

Here's some additional justification of the third case of the given answer by acreativename:

For the third case, I think they're assuming $A_{n} = 2$ for $n\ge k-2$ (even if not, the proof should still work if you instead assume this). We have that for all $n\ge k-2, A_n = 2.$ So in particular, $a_n$ and $a_{n+1}$ cannot both be divisible by $4$ for any $n\ge k-2.$ We have $a_{k+1} = \dfrac{a_{k-1}+a_{k}}2.$ So $a_k$ is even. If $a_k\equiv 0\mod 4,$ then $a_{k-1} + a_k\equiv 0\mod 4$ (as $a_{k+1}$ is even) implies $a_{k-1}\equiv 0\mod 4,$ contradicting $\gcd(a_k, a_{k-1}) = 2.$ Hence $a_k\equiv 2\mod 4,$ and so $a_{k-1}\equiv 2\mod 4.$ But from this point on, I don't know why $a_{k-1}=a_{k-2} = 2$ must hold. Once we show that $a_{k-1} = a_{k-2} = 2,$ we can then conclude the result by backwards induction ($a_{k-j-1} =a_{k-j} = 2\Rightarrow  a_{k-j-2}=2$).

$\textbf{Lemma:}$ when $j \in \mathbb{N}$ we have that $\text{gcd}(a_{j+1},a_{j+2})$ divides $\text{gcd}(a_{j},a_{j+1})$
$\textbf{Proof:} $ Put $\text{gcd}(a_{j},a_{j+1}) = A_{j}$ and write $a_{j} = A_{j}v_{j}, a_{j+1} = A_{j+1}v_{j+1}$ where $\text{gcd}(v_{j}, v_{j+1}) = 1$. We have $a_{j+2} = v_{j}+v_{j+1}$ and $\text{gcd}(a_{j+1}, a_{j+2}) = \text{gcd}(A_{j}v_{j+1}, v_{j}+v_{j+1}) = \text{gcd}(A_{j}, v_{j}+v_{j+1})$ as $\text{gcd}(v_{j}, v_{j+1}) = 1$.
Thus by the lemma the sequence $A_{j} := \text{gcd}(a_{j},a_{j+1})$ is a non-increasing sequence with $A_{j+1}|A_{j}$.
We now have three cases to consider;
$\bullet$ If $\lim_{j \rightarrow \infty} A_{j} = 1$ then there exists an integer $k$ so that $\text{gcd}(a_{k}, a_{k+1}) = 1$ and $a_{k}, a_{k+1} \geq 1$;  whence by the above lemma we will have $a_{n} = a_{n-1} + a_{n-2}$ for $n \geq k+2$ and therefore will have $a_{n} \rightarrow \infty$.
$\bullet$ If $\lim_{j \rightarrow \infty} A_{j} \geq 3$ then $a_{n} \leq \frac{a_{n-1}+a_{n-2}}{3}$ for some $n \geq k$ and will therefore have $a_{n} \rightarrow 0$ (trivial exercise) which is impossible.
$\bullet$ If $\lim_{j \rightarrow \infty} A_{j} = 2$ then we have $a_{n} = \frac{a_{n-1}+a_{n-2}}{2}$ for some $n \geq k$. The only way $a_{k-2},a_{k-1},...$ are positive integers under this condition is when $a_{k-1}=a_{k-2} = 2$.
Thus such bounded sequences must eventually collapse to $2$. Now note that if $w \in \mathbb{N}$ and $\frac{w+2}{\text{gcd}(w,2)} = 2$ then $w$ is forced to be $2$. Hence we must have $a_{1} = a_{2} = 2$.
