Let $A$ be a set of integers with irrational natural density. That is, suppose that

$\lim_{n\to\infty}\frac{\#(A\cap [-n,n))}{2n}$

exists and is irrational. Denote this value by $\alpha$. Now let $p$ be an odd polynomial with integer coefficients, and define the set $S_{p}(k) = \{p(i) : i \in [-k, k)\cap\mathbb{Z}\}$. Is is the case that any such $p$ gives

$\lim_{k\to\infty}\frac{\#(A\cap S_{p}(k))}{2k} = \alpha$?

A note on the necessity of these conditions: If $\alpha$ is rational, this fails. Take $\alpha = 1/2$, $A = 2\mathbb{Z}$, and $p(x) = 2x$. If $p$ is even, I suspect (but cannot prove) that this also fails. To see this, fix $\alpha$ irrational and construct $A$ in such a way that slightly more of the "mass" of $\alpha$ falls on $A\cap[0,\infty)$ than on $A\cap(-\infty,0)$. Then take $p(x) = x^2$.

I suspect that ergodic theory may be helpful here, but I have not found a good way to think about it yet! Any ideas are welcome.

  • $\begingroup$ $\alpha$ has not been defined. Is it supposed to be the natural density of $A$? And what does $A=\dots0101010101\dots$ supposed to mean? How do we interpret that as a set of integers? $\endgroup$ Oct 30, 2018 at 22:31
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    $\begingroup$ Ah, I have been freely interchanging between sets of integers and binary strings previous to this post. $\dots01010101\dots$ is supposed to mean $2\mathbb{Z}$. Moreover, $\alpha$ is the value of the limit described in the previous line. I've updated the post to reflect this $\endgroup$ Oct 30, 2018 at 23:08

1 Answer 1


A slight modification of your example shows the answer is no. Let $A$ consist of the even numbers together with a set of odd numbers with irrational density, and take $p(x) = 2 x$.

  • $\begingroup$ This is great, thank you for the insight!! $\endgroup$ Oct 30, 2018 at 23:10

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