Let $A = \{ [n \sqrt{2}] : 1 \leq n \leq N \}$ then can we estimate the size of the sumset $A+A$ ?
$$ A+A = \big\{ [m \sqrt{2}] + [n \sqrt{2}] : 1 \leq m \leq N, 1 \leq n \leq N \big\} $$
Here are two possibilities:
$$ |A| \leq | A + A| \leq \binom{|A|}{2} $$
Can we do any better for the size of $A+A$? I know that if $|A+A| = |A|$ then $A$ is the coset of a finite group. For example, do we have that
$$|A+A| < 2.5 |A| $$
for sufficiently large $N$ ?
Computer models suggest $\sigma(A) = |A+A|/|A| \approx 2 \sqrt{2}$ as $N \to \infty$ but also $\sigma(A) > 2\sqrt{2} - \epsilon$ for e.g. $\epsilon > 0.1$
For $N = 2500$ we have that $|A+A| \approx (2.827)|A|$. And maybe in general it's always close to twice the spacing. At least, for the type of sequence I have written down.
In fact, I think we've argued in the comments that $|A+A| < 3|A|$ and certainly that ratio is $>1$. That's could be sufficient for many argments.
I wonder if the answer to this problem changes if I replace integer part $[\cdot ]$ with round up, or nearest integer function.