9.1. How many natural numbers $n \in [20182019 ; 20192018]$ are there for which the number $\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right]$ is even? (Here, $[x]$ denotes the greatest integer not exceeding $x$.)
Answer: 4999.

Solution. Consider the sequences

$$
x_{n}=\left(\frac{1+\sqrt{5}}{2}\right)^{n}, \quad y_{n}=\left(\frac{1-\sqrt{5}}{2}\right)^{n}, \quad a_{n}=x_{n}+y_{n}
$$

Since $-1<\frac{1-\sqrt{5}}{2}<0$, we have $y_{n} \in(0 ; 1)$ for even $n$, and $y_{n} \in(-1 ; 0)$ for odd $n$. Then $\left[x_{n}\right]=\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}\right]$ for even $n$ equals $a_{n}-1$, and for odd $n$ it is simply $a_{n}$. Moreover, we have $a_{n+2}=a_{n+1}+a_{n}$ for all natural $n$. Since $a_{1}=1, a_{2}=3$, then $a_{3}=4$ and we get such a parity sequence for $a_{n}(n=1,2, \ldots)$: odd, odd, even, odd, odd, even, $\ldots$

Then $\left[x_{n}\right]$ is related to $a_{n}$ as follows:

| $n$ | $6 k$ | $6 k+1$ | $6 k+2$ | $6 k+3$ | $6 k+4$ | $6 k+5$ |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $a_{n}$ | even | odd | odd | even | odd | odd |
| $\left[x_{n}\right]$ | $a_{n}-1$ | $a_{n}$ | $a_{n}-1$ | $a_{n}$ | $a_{n}-1$ | $a_{n}$ |
| $\left[x_{n}\right]$ | odd | odd | even | even | even | odd |

Since $x_{n}=a_{n}-y_{n}$, for numbers $n$ of the form $6 k, 6 k+1, 6 k+5$, the number $\left[x_{n}\right]$ is odd, and for numbers $n$ of the form $6 k+2, 6 k+3, 6 k+4$ it is even. Since $20182019=6 k_{1}+5, 20192018=6 k_{2}+2$, among the numbers $n=20182019, \ldots, 20192018$ the desired number is $3 \cdot \frac{20192018-20182022}{6}+1=4999$.