Consider an infinite sequence $x_1, x_2, \dots$ of positive integers such that, for every integer $n \geq 1$: 

- If $x_n$ is even, then $x_{n+1} = \frac{x_n}{2}$.
- If $x_n$ is odd, then $x_{n+1} = \frac{x_n - 1}{2} + 2^{k-1}$, where $2^{k-1} \leq x_n < 2^k$.

Determine the smallest possible value of $x_1$ for which $2020$ is in the sequence.

Let's think step by step and output the final answer within \boxed{}.
user