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Does there exist an integer $k \ge 1$ such that ${\sf bd}^\ast(X) > 0$ whenever $X \subseteq \mathbf N^+$ and $\pi_X(n) \gg \frac{n}{\log^{[k]}(n)}$ as $n \to \infty$? Here, ${\sf bd}^\ast$ is the upper Banach (or uniform) density on $\mathbf N^+$, viz. the function $$\mathcal P(\mathbf N^+) \to \mathbf R: X \mapsto \lim_{n \to \infty} \frac{\max_{m \ge 1} |X \cap [m+1,m+n]|}{n},$$ and $\pi_X$ denotes, for each $X \subseteq \mathbf N^+$, the function $\mathbf N^+ \to \mathbf N: n \mapsto |X \cap [1,n]|$ (that is, the counting function of $X$), while $\log^{[k]}$ means, for each $k \in \mathbf N$, the $k$-th iterate of the natural logarithm.

I don't have a precise motivation for this, I'd just like to understand what is known about questions along this line of thought.

For what it is worth, let me note that $k = 1$ can't work, since the set of primes has zero density uniformly with respect to the actual choice of a subadditive and translational invariant function $\mu^\ast: \mathcal P(\mathbf N^+) \to \mathbf R$ such that $\mu^\ast(\mathbf N^+) = 1$ and $\mu^\ast(q \cdot X) = \frac{1}{q} \mu^\ast(X)$ for all $X \subseteq \mathbf N^+$ and $q \in \mathbf N^+$ (the upper Banach density on $\mathbf N^+$ belongs to this class).

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    $\begingroup$ Perhaps I am not understanding the question, but isn't the set $$A=\{n\lfloor\log^{[k]}(n)\rfloor : \ n\in\mathbb{N}\}$$ a counterexample for any $k$? $\endgroup$ Jun 19, 2015 at 16:30
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    $\begingroup$ Yes, and I was thinking of cancelling the post after having discussed the question with Alain Plagne. Should I? I see now the idea I had in mind can't work in any case. $\endgroup$ Jun 19, 2015 at 16:31

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