A sequence $\{n_{k}\}$ is lacunary if $\forall$ k $\in\mathbb{N}$, $\frac{n_{k+1}}{n_{k}}$ $\geq\lambda\gneq$ 1.

And natural upper density of a set S $\subset\mathbb{N}$ is defined as limsup$_{n\rightarrow\infty}$ $\frac{| S \cap \{ 1, 2, ... , n \}|}{n}$.

Only examples of lacunary sequences that I could think of were basically geometric sequences; however, they will have upper density zero. I was wondering if there exists lacunary sequences with positive upper density.

  • $\begingroup$ You can have lacunary sequences much sparser than geometric sequences. $n_k = 10^{k!}$, is somewhat well known. Perhaps you mean "[O]nly examples of lacunary sequences that might have positive upper density that I could think of were basically geometric sequences; ..." $\endgroup$ – Eric Towers May 20 at 21:33
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    $\begingroup$ (1) Title says density, body says upper density. (2) Why the weird symbol $\gneq$, how is that different from $\gt$? (3) In your definition of natural upper density, is $A=S$? $\endgroup$ – bof May 20 at 22:31

Lacunary sequences cannot have positive upper density. To see this, note that if $\lambda > 1$ is such that $n_{k+1} / n_{k} > \lambda$ for all $k$., then there are at most $\log(n) / \log(\lambda)$ elements in the sequence between $1$ and $n$.

Consider the sequence in log space. Then, note that if there are more than $\log(n) / \log(\lambda)$ elements in ${1,...,n}$, then by the pigeonhole principle, there must be a $k$ such that $\log(n_{k+1}) - \log(n_k) < \log(\lambda)$ which implies that $ n_{k+1} > \lambda n_{k}$. A contradiction.


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