# Irrational natural density set, intersected with odd polynomial

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.

• $\alpha$ 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? Oct 30, 2018 at 22:31
• 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 Oct 30, 2018 at 23:08

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 $$p(x) = 2 x$$.