
1. Let $\left(a_{n}\right)_{n=1}^{\infty}$ be an infinite sequence such that for all positive integers $n$ we have

$$
a_{n+1}=\frac{a_{n}^{2}}{a_{n}^{2}-4 a_{n}+6}
$$

a) Find all values $a_{1}$ for which the sequence is constant.

b) Let $a_{1}=5$. Find $\left\lfloor a_{2018}\right\rfloor$.

(Vojtech Bálint)


Solution. a) Assume $\left(a_{n}\right)_{n=1}^{\infty}$ is constant. Then $a_{2}=a_{1}$, hence

$$
a_{1}=\frac{a_{1}^{2}}{a_{1}^{2}-4 a_{1}+6}
$$

which rewrites as $a_{1}\left(a_{1}-2\right)\left(a_{1}-3\right)=0$. It follows that $a_{1} \in\{0,2,3\}$. On the other hand, once $a_{2}=a_{1}$, the sequence is clearly constant (formally e.g. by mathematical induction), thus any such $a_{1}$ indeed works.

b) Let $a_{1}=5$. Using the recurrence, we obtain $a_{2} \approx 2.27, a_{3} \approx 2.49, a_{4} \approx 2.77$, and so on. This leads to the following conjecture: For any $n \geqslant 2$ we have $20, \\
& a_{n+1}-2=\frac{a_{n}^{2}}{a_{n}^{2}-4 a_{n}+6}-2=\frac{\left(6-a_{n}\right)\left(a_{n}-2\right)}{\left(a_{n}-2\right)^{2}+2}>0 .
\end{aligned}
$$

This proves $2<a_{n+1}<3$, hence the proof of the conjecture is complete. In particular, $2<a_{2018}<3$ and the answer is $\left\lfloor a_{2018}\right\rfloor=2$.
