It's easy to prove by induction that $f(n)$ is equal to $1$ iff $n$ has an odd number of ones in the binary expansion and $0$ else. a) Since $2^{2020}-1$ has $2019$ ones, we have $f(2^{2020}-1)=1$. b) Suppose that $f$ is periodic. Let $p$ be a period. Let $n$ be a sufficiently large power of $2$. Then the number of ones in the expansion of $n+p$ will just be the some of those in $n$ and those in $p$. So for $f(n+p)=f(n)$ to be true we must have that $p$ contains an even number of ones. But on the other hand, choose $k$ such that $2^k \le p<2^{k+1}$. Then $2^k+p$ will have the same number of ones as $p$. So if again $n$ is a large power of $2$, then $f(n+2^k)=f(n+2^k+p)$ shows that $p$ contains an odd number of ones. Contradiction!