Is the set of powerful numbers piecewise syndetic? Recall that a subset $A \subset \mathbb Z_+$ of positive integers syndetic if there exists a $d>0$ such that every positive integer has distance at most $d$ to an element of $A$. It is called piecewise syndetic if it is the intersection of a syndetic set with a subset of $[0,\infty)$ containing arbitrarily long intervals.
Let us call $n \in \mathbb Z_+$ powerful if for every prime $p$, the multiplicity $\nu_p(n) \neq 1$. (Or, equivalently: $n$ can be expressed as a product of a square and a cube.)
Here is the question: is the set of powerful numbers piecewise syndetic?
 A: The answer is no. A set $S$ to be piecewise syndetic iff there is an integer $d$ such that there exist intervals $I$ of arbitrary length such that distances between elements of $S\cap I$ are bounded by $d$. In particular, $|S\cap I|\geq\frac{1}{d}|I|$. I will show no such $d$ exists.
For any prime $p$, the fraction of those which are either not divisible by $p$ or divisible by $p^2$ is equal to $1-\frac{p-1}{p^2}$. Further, these conditions are independent - formally, if we consider take $P=p_1\dots p_k$, the product of first $k$ primes, then from the Chinese remainder theorem, in any interval $I$ of length $P^2$ the fraction of integers $n$ in $I$ such that $v_{p_i}(n)\neq 1$ for all $i$ is equal to
$$C:=\prod_{i=1}^k\left(1-\frac{p-1}{p^2}\right).$$
This gives an upper bound of $CP^2$ for the number of powerful numbers an interval of length $P^2$. Now, as $k$ tends to infinity, then $C$ tends to zero (this essentially follows from the fact sum of reciprocals of primes diverges), in particular for large enough $k$ it becomes smaller than $\frac{1}{d}$, so no interval $I$ of length $P^2$ contains more than $\frac{1}{d}|I|$ powerful numbers.
