Let $\ x\in\mathbb{R}\setminus(\mathbb{Z}\ \cup [-1,1]).\ $ Does the set $\{\lfloor{x^k}\rfloor:k\in\mathbb{N}\}$ have roughly as many evens as odds? I couldn't fit the precise question in the title, but I tried. What I mean precisely is:

Let $\ x\in\mathbb{R}\setminus(\ \mathbb{Z}\ \cup [-1,1]\ ).\ $
Furthermore, for each $\ n\in\mathbb{N},\ $ define $\ A_n = \{\
 \lfloor{x^k}\rfloor:\ k\in\mathbb{N}\ \text{and}\ k\leq n\ \}\ $ where $\ \lfloor{\cdot}\rfloor\ $ is the floor function.
Proposition:
$$ \lim_{n\to\infty} \frac{\text{amount of odd integers in}\
 A_n}{\text{amount of integers in}\ A_n} = \frac{1}{2}\quad $$

It is simply the apparently chaotic nature of the sets that makes me think this. And I know nothing of Chaos theory, so it's possible I'm out of my depth here. But I don't know if this has anything to do with Chaos theory, so I don't think I'm out of place for asking this question.
Is this known to be true for some numbers $\ x\ $ and known to be false for others?
Is it even known whether or not the limit always converges?
Perhaps this can even be proven via a probabilistic approach? The arising sets seem to be chaotic in nature, so perhaps something to do with Chaos/Ergodic theory? Or perhaps there is a much more elementary approach? I've no idea...
Edit: The responses thus far make me think that for radical numbers $\ x,\ $ the limit will $\ \neq\frac{1}{2}.\ $ And some algebraic numbers may give rise to patterns. So transcendental numbers are "more interesting/more chaotic/random." But I am not sure of any this, and I know little about the properties of transcendental numbers.
 A: Questions involving floors of powers of a given number always invite solutions involving Pisot–Vijayaraghavan numbers. I will show how to make all the numbers in $A$ odd. Let $x_1=x$ be a quadratic irrational and $x_2$ be its algebraic conjugate, and $\sigma_1=x_1+x_2$ and $\sigma_2=x_1x_2$. We have that
$$x_1^{n+1}+x_2^{n+1}=\sigma_1(x_1^n+x_2^n)-\sigma_2(x_1^{n-1}+x_2^{n-1}).$$
So if we think of this as a recursive sequence $k_n=x_1^n+x_2^n$, $k_1=\sigma_1$, $k_2=\sigma_1^2-2\sigma_2$ we can make it always even by choosing $\sigma_1$ even, and then if $0<x_2<1$ we have that $x_1^n=k_n-x_2^n$, that is $x_1^n$ is an even number minus some small positive quantity, so its floor is odd. With a bit of trial and error that can be achieved with $\sigma_1=6$, $\sigma_2=3$, so $x_1=3+\sqrt 6$ and $x_2=3-\sqrt 6$.
A: 
Is this known to be true for some numbers $x$ and known to be false
for others?

For $x = \varphi^2 = \frac{3 + \sqrt{5}}{2}$, the integers $\lfloor x^n \rfloor$ (with $n > 0$) are precisely the even-indexed Lucas numbers minus one.  These repeat in the pattern even, even, odd, so the limit is $\frac{1}{3}$.  If $x = \varphi = \frac{1 + \sqrt{5}}{2}$ instead, then $\lfloor x^n \rfloor$ alternates between $L_n$ and $L_n - 1$, so the repeating pattern is odd, even, even, even, odd, odd; and the limit is $\frac{1}{2}$ in this case.
I will leave the harder questions (whether the limit always exists, and whether it equals $\frac{1}{2}$ for $100\%$ of choices of $x$) to others.
