1. For the sequence $\left(a_{n}\right)_{n=1}^{\infty}$, we know that for all natural numbers $n$,

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

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

b) Let $a_{1}=5$. Determine the greatest integer not exceeding $a_{2018}$. (Vojtech Bálint)
SOLUTION. a) Suppose the sequence $\left(a_{n}\right)_{n=1}^{\infty}$ is constant. Then it must hold that $a_{2}=a_{1}$, which we can write using the relation from the problem as

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

We can easily rearrange this equation to $a_{1}\left(a_{1}-2\right)\left(a_{1}-3\right)=0$. From this, we get that $a_{1} \in\{0,2,3\}$. It is clear that for these values of $a_{1}$, the sequence $\left(a_{n}\right)_{n=1}^{\infty}$ is indeed constant and all its terms are equal to $a_{1}$. Formally, we could prove this by mathematical induction.

b) Let $a_{1}=5$. We will compute a few more terms of the sequence $\left(a_{n}\right)$. We get $a_{2} \approx 2.27, a_{3} \approx 2.49, a_{4} \approx 2.77$, etc. From this, we can infer that for all $n \geqq 2$ it holds that $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 both inequalities $2<a_{n+1}<3$, so the proof by mathematical induction is complete. It follows that $2<a_{2018}<3$, from which it follows that the greatest integer not exceeding $a_{2018}$ is 2.

GUIDING AND SUPPLEMENTARY PROBLEMS:

N1. For the sequence $\left(b_{n}\right)_{n=1}^{\infty}$, we know that for all natural numbers $n$ it holds that $b_{n+1}=b_{n}^{2}-2$. Find all values of $b_{1}$ for which all terms of $\left(b_{n}\right)_{n=1}^{\infty}$ are equal to $b_{1}$. [From the relation for $n=1$ we get $b_{1}=b_{1}^{2}-2$, from which $b_{1} \in\{-1,2\}$. We then verify that these values satisfy the condition.]

N2. For the sequence $\left(b_{n}\right)_{n=1}^{\infty}$, we know that $b_{1}=1$ and that for all natural numbers $n$ it holds that $b_{n+1}=3 b_{n} /\left(b_{n}+1\right)$. Prove that all terms of the sequence are in the interval $\langle 1,2)$. [Verify by mathematical induction that for all natural $n$ the inequalities $1 \leqq b_{n}<2$ hold.]

D1. For the sequence $\left(b_{n}\right)_{n=1}^{\infty}$, it is known that for all natural numbers $n$ it holds that $b_{n+1}=$ $=2 b_{n}^{2} /\left(b_{n}^{2}-3\right)$. Find all values of $b_{1}$ such that the sequence $b_{2}, b_{3}, b_{4}, \ldots$ is constant. [Derive that $b_{n+1}=b_{n}$ if and only if $b_{n} \in\{-1,0,3\}$, so $b_{2}$ must be equal to one of these values. We then compute the corresponding values of $b_{1}$. The result is $b_{1} \in\{-3,-1,0,1,3\}$.]

D2. Derive the explicit expression for the sequence $\left(b_{n}\right)_{n=1}^{\infty}$ from problem N2. [Rearrange the given relation to $3 \cdot 1 / b_{n+1}=1+1 / b_{n}$. The sequence $c_{n}=1 / b_{n}$ thus satisfies the equation $3 c_{n+1}=c_{n}+1$. By substituting $c_{n}=d_{n}+1 / 2$, we eliminate the constant term and get $3 d_{n+1}=d_{n}$. The sequence $d_{n}$ is thus geometric, so we can determine its explicit form. By reverse substitution, we can find the expression for $b_{n}$. The result is $b_{n}=2 \cdot 3^{n-1} /\left(3^{n-1}+1\right)$.]

D3. The most famous recursively defined sequence is the Fibonacci sequence. It is given by the relations $f_{1}=1, f_{2}=1$ and the recursive relation $f_{n}=f_{n-1}+f_{n-2}$ for every $n \geqq 3$. Prove that $\sum_{i=1}^{n} f_{i}=f_{n+2}-1$ and $\sum_{i=1}^{n} f_{i}^{2}=f_{n} \cdot f_{n+1}$. [Proceed by mathematical induction on $n$.]