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Let $f(n)$ be the number of subsets $S\subseteq \{1,2,\ldots,2n\}$ such that $|S|=n$ and $a$ does not divide $b$ whenever $a,b \in S$ are distinct. Can we evaluate $f(n)$, at least asimptotically?

The question is related to this other one, where I had a more complicated (and unuseful) solution. In particular, is it true that $f(n)=o(n)$? Moreover, is it true that $f(n)$ is definitively bigger than $(\ln n)^k$ for any constant $k$?

Edit: the first conjecture has been proved to be false (see the answer of Robert below). Then, can we say that $\ln f(n)=O(n)$?

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Paolo Leonetti
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    See OEIS sequence A174094 (https://oeis.org/A174094) which has $f(n)$ for $n = 1..46$. The logarithmic scatterplot makes it plausible that $f(n) \sim a \exp(b n)$ for some $a,b$ with $b > 0$, but very implausible that $f(n) = o(n)$. – Robert Israel Aug 26 '15 at 00:18
  • See my blogpost regarding this question: https://anuragbishnoi.wordpress.com/2017/11/02/on-a-famous-pigeonhole-problem/. As per my knowledge, asymptotically the best lower bound on $f(n)$ is $2^{n/3}$ while the best upper bound is about $2^{0.78n}$ (see the comments). It would be really interesting for me to close this gap. – Anurag Nov 06 '17 at 11:47

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Here's one way to get a lower bound that grows exponentially. Start with the subset $S = \{n+1, \ldots, 2n\}$. Take an arbitrary subset $J \subseteq \{j \in S:\; j \text{ even},\; j > 4n/3\}$ and replace each $j \in J$ by $j/2$. This gives $2^{1 + \lfloor (n-1)/3 \rfloor}$ solutions.

EDIT: For the second question: of course $f(n)$ is at most the number of all subsets of $\{1, \ldots, 2n\}$, which is $2^{2n}$.

Robert Israel
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