Let $A=\bigg\lbrace k \geq 1 \ \bigg| \ \lbrace k\sqrt{2} \rbrace \lt \frac{1}{2}\bigg\rbrace$ (where $\lbrace x \rbrace$ denotes the fractional part of $x$), $a_n=|A\cap [1,n]|$ and $d_n=\frac{a_n}{n}-\frac{1}{2}$. By the equidistribution theorem, we know that $d_n \to 0$ when $n\to\infty$. However, numerical evidence suggests the following conjectures :
(1) $0 \leq d_n \leq \frac{4}{n}$ for any $n\geq 1$.
(2) $d_n=0$ for infinitely many $n$.
(3) $d_n=\frac{4}{n}$ for infinitely many $n$.
The numerical evidence is as follows : for $n\leq 10^6$, (1) always holds, $d_n=0$ holds for 5102 values of $n$ and $d_n=\frac{4}{n}$ holds for 4374 values of $n$. Any ideas on how to prove or disprove any of (1)-(2)-(3) ?
Update(06/01/2019) : I expand here on fedja's comment. We have $a_n=f(n\sqrt{2})+\frac{1}{2}$ where $f$ is the $1$-periodic function satisfying $f(x)=\frac{1}{2}$ for $x\in [0,\frac{1}{2}]$ and $f(x)=-\frac{1}{2}$ for $x\in [\frac{1}{2},1)$.
It follows that $d_n=\frac{\sum_{k=1}^n f(k\sqrt{2})}{n}$. Now, since the Dirichlet conditions are satisfied for $f$ we have the Fourier expansion (for $x\not\in \frac{1}{2}+{\mathbb Z}$)
$$ f(x)=\frac{-2}{\pi}\sum_{q=0}^{\infty} \frac{1}{2q+1}sin((2q+1)2\pi x) \tag{4} $$
Taking $x=k\sqrt{2}$ in (4) above, summing on $k$ and using the well-known identity $\sum_{k=1}^n \sin(k\theta)=\frac{\sin(\frac{(n+1)\theta}{2})}{\sin(\frac{\theta}{2})}\sin(\frac{n\theta}{2})$, we obtain
$$ d_n=\frac{-2}{n\pi}\sum_{q=0}^{\infty} \frac{1}{(2q+1)}\frac{\sin(\frac{(n+1)}{2}2\pi(2q+1)\sqrt{2})}{\sin(\pi(2q+1)\sqrt{2})}\sin(n\pi(2q+1)\sqrt{2}) \tag{5} $$
But I do not see how to continue from here.