, 2020

Solution:
Let $a_{0} = \frac{1}{\sqrt{2}^{2020}}\left(t + \frac{1}{t}\right)$, with $t \geq 1$. (If $a_{0} < \frac{1}{\sqrt{2}^{2018}}$ then no real $t$ exists, but we ignore these values because $a_{0}$ is smaller.) Then, we can prove by induction that
$$a_{n} = \frac{1}{\sqrt{2}^{2020 \cdot 2^{n}}}\left(t^{2^{n}} + \frac{1}{t^{2^{n}}}\right) .$$
For this to be bounded, it is easy to see that we just need
$$\frac{t^{2^{n}}}{\sqrt{2}^{2020 \cdot 2^{n}}} = \left(\frac{t}{\sqrt{2}^{2020}}\right)^{2^{n}}$$
to be bounded, since the second term approaches $0$. We see that this is equivalent to $t \leq 2^{2020 / 2}$, which means
$$a_{0} \leq \frac{1}{\sqrt{2}^{2020}}\left(\sqrt{2}^{2020} + \left(\frac{1}{\sqrt{2}}\right)^{2020}\right) = 1 + \frac{1}{2^{2020}}$$