Closeness of prime powers

Question

For any integer $n$, denote by $pp(n)$ the smallest prime power greater or equal to $n$.

Bound $pp(n)-n$ from above.

Is $pp(n)-n=O(\log n)$, for instance?

Answer 1

Notice that $pp(n) = P^2$, where $P$ denotes the smallest prime greater than $\sqrt n$. Let $\pi(P) =k$, hence $P \sim k\log k$ where $\pi$ is the prime counting function.Since we have $k> \sqrt n$, we have $$pp(n)=P^2 \sim k^2(\log k)^2 > \frac{1}{4}n(\log n)^2 $$

Hence $$ pp(n) -n = O(n(\log n)^2)$$

Notice that the error term $n(\log n)^2$ cannot be improved.