## Task 1.

Let $a=0.365$. The sequence $a_{1}, a_{2}, \ldots, a_{2020}$ is defined by the formulas

$$
a_{1}=a, \quad a_{n+1}=a^{a_{n}} \quad \text { for } \quad n=1, \ldots, 2019
$$

Arrange the numbers $a_{1}, a_{2}, \ldots, a_{2020}$ from the smallest to the largest.
## Solution.

Since $0a^{y}$.

If we apply this rule for $x=a$ and $y=1$, we conclude $a^{a}>a$, i.e., $a_{2}>a_{1}$.

Furthermore, due to $1>a_{2}>a_{1}$, it follows that $a_{1}<a_{3}<a_{2}$.

If we continue to conclude in the same manner, it follows that

$$
a_{1}<a_{3}<\cdots<a_{2 k-1}<a_{2 k-2}<\cdots<a_{2}<1
$$

and

$$
a_{1}<a_{3}<\cdots<a_{2 k-1}<a_{2 k}<\cdots<a_{2}<1
$$

for every $k=2,3, \ldots, 1010$. For $k=1010$, the last line gives the desired order.