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Let $\mathbb{N}$ denote the set of non-negative integers. For $A\subseteq \mathbb{N}$ we let the (upper) density of $A$ be defined by $d^+(A) = \lim\sup_{n\to\infty} \frac {|A\cap \{0,\ldots, n\}|}{n+1}$.

Moreover, we define the "logarithm" of $A$ by $$\log(A) = \big\{\max\{0,\lfloor\log_2(a+1)\rfloor\}: a\in A \big\}.$$ Inductively, we set $\log^{(0)}(A) = A$ and $\log^{(n+1)}(A) = \log(\log^{(n)}(A))$ for all $n\in\mathbb{N}$.

For any $r\in\mathbb{R}\setminus \mathbb{Q}$, let $$\alpha_r(n)=\min\{|r-\frac{m}{n}|: m\in\mathbb{N}\}$$ be the best approximation of $r$ that can be obtained using $n$ as the denominator.

We inductively define $\text{appr}_r:\mathbb{N} \to \mathbb{N}$, the approximation sequence of $r$:

  • $\text{appr}_r(0) = 1$, and
  • $\text{appr}_r(n+1) = \min\{m\in\mathbb{N}: m \geq \text{appr}_r(n) \text{ and } \alpha_r(m) < \alpha_r(\text{appr}_r(n))\}$ for all $n\in \mathbb{N}$.

Let $\text{Approx(r)} = \{\text{appr}_r(n):n\in\mathbb{N}\}\subseteq \mathbb{N}$. The $\log$-class of $r\in\mathbb{R}\setminus\mathbb{Q}$ is set to $n$ if $n$ is the smallest integer such that $$d^+(\log^{(n)}(\text{Approx}(r)) > 0,$$ and we say the $\log$-class of $r$ is $\infty$ if $d^+(\log^{(k)}(\text{Approx}(r)) = 0$ for all $k\in\mathbb{N}$.

Questions. For which $n\in\mathbb{N}$ is there $r\in\mathbb{R}\setminus\mathbb{Q}$ with $\log$-class $n$? And is there $r\in\mathbb{R}\setminus\mathbb{Q}$ with $\log$-class $\infty$?

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    $\begingroup$ Choose your favorite integer-valued function $f$ such that $\lim_{n\to\infty} \log^n f(n)=\infty$. Then $\sum 1/2^{f(n)}$ will have infinite $\log$-class. $\endgroup$ Commented May 8, 2023 at 14:53
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    $\begingroup$ The reason why \text{appr} rather than \operatorname{appr} is incorrect LaTeX or MathJax usage can be seen in the behavior of things like \cos, \det, \log, \max, etc. In $\cos x$, coded as \cos x, you see more space to the right of $\cos$ than you see in $\cos(x),$ coded as \cos(x), and $2\cos(x),$ coded as 2\cos(x) does not look the same as $2\text{cos}(x),$ coded as 2\text{cos}(x). The thing that makes one font look far better to you than another is a large number of little details like this that most people don't notice. Donald Knuth had that in$\,\ldots\qquad$ $\endgroup$ Commented May 8, 2023 at 15:38
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    $\begingroup$ $\ldots\,$mind when he invented TeX. And he said so. $\qquad$ $\endgroup$ Commented May 8, 2023 at 15:38
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    $\begingroup$ Thanks @MichaelHardy for the TeX hints! $\endgroup$ Commented May 8, 2023 at 18:30

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