If $A$ is a forbidden-divisor set with $|A \cap [1,x]| = o(\sqrt{x})$ and $B = \{b_1 < b_2 < \cdots\}$ the sifted set, must $x^{-1} \sum_{b_i < x} (b_{i+1} - b_i)^2$ converge to a finite limit?

Let $A\subseteq \mathbb{N}$ be a set such that $\lvert A\cap [1,x]\rvert=o(x^{1/2})$. Let $B=\{ n\geq 1 : a\nmid n\textrm{ for all }a\in A\}$. If $B=\{b_1 < b_2 < \cdots\}$ then is it true that $$\lim_{x \to \infty} \frac{1}{x}\sum_{b_i < x}(b_{i+1}-b_i)^2$$ exists (and is finite)?

For example, when $A=\{p^2: p\textrm{ prime}\}$ then $B$ is the set of squarefree numbers, and the existence of this limit was proved by Erdős.

See also [208].