Problem 1400

First of all, we note that the sequence is well defined: for each $n \in \mathbb{N}$, $n > 1$, the value of $f(n)$ is perfectly determined from the values of $f(r)$ with $r < n$.
We consider the sequence $g(n)$ given by $g(n) = 0$ if the binary expression of $n$ has an even number of ones; and $g(n) = 1$ if the binary expression of $n$ has an odd number of ones. It is obvious that $g$ satisfies all the conditions that define $f$, hence $f(n) = g(n)$ for every $n \in \mathbb{N}$.
Thus, since $2^{2020} - 1$ is written in binary with 2020 ones,
$$f(2^{2020} - 1) = 0$$
Finally, let us see that the sequence is not periodic. Suppose that it were, from a value $n_0$ onward and with period $t$. We take the integer, $r$, such that $2^r \le t < 2^{r+1}$. We observe that the binary expression of $2^r + t$ is obtained by replacing the first 1 (from the left) of the expression of $t$ with the two digits 10, so that $f(2^r + t) = f(t)$. We take an integer $k > r + 1$ such that $2^k > n_0$. Now, we have:
If $f(t) = 1$, then $f(2^k + t) = 0$ and therefore, $f(2^k + t) \ne f(2^k) = 1$.

If $f(t) = 0$, then $f(2^k + 2^r + t) = 1 + f(2^r + t) = 1 + f(t) = 1$ and therefore, $f(2^k + 2^r + t) \neq f(2^k + 2^r) = 0$.