In other words, given a sequence $(s_n)$, how can we tell if there exist irrationals $u>1$ and $v>1$ such that

$$s_n = \lfloor un\rfloor + \lfloor vn\rfloor$$

for every positive integer $n$?

A few thoughts: Graham and Lin (*Math. Mag.* 51 (1978) 174-176) give a test for $(s_n)$ to be a single Beatty sequence $(\lfloor un\rfloor)$ (which they call the spectrum of $u$). Perhaps someone knows a reference for a test for sums of two or more Beatty sequences? A special case would be a test for a given sequence $(s_n)$ to be the sum of two *complementary* Beatty sequences (i.e., $1/u + 1/v = 1$).

In response to comments, the part of the question that says "for every positive integer n" indicates that the intended sequence is infinite. It seems to me that the question, as stated above, is okay. If it's undecidable - well, that's of interest.

In any case, while it may be difficult to give a test that actually finds $u$ and $v$ when such numbers exist, there are some simple tests for deciding that $(s_n)$ is *not* a sum of two Beatty sequences:

(1) $\lim_{n\to\infty}s(n)/n$ must exist;

(2) if $u$ and $v$ exist, then $u$ + $v$ = $\lim_{n\to\infty}s(n)/n$;

(3) $\lfloor(u+v)n\rfloor \in \{[un]+[vn],[un]+[vn]+1\}$ for every $n$;

(4) if $(s_n)$ is a sum of two Beatty sequences, then the difference sequence of $(s_n)$ consists of at most three terms; and if there are three, then they are consecutive integers.

It's easy to see how each of those generalizes to give a "negative test"; that is, a way to see that a given $(s_n)$ is not a sum of any prescribed number of Beatty sequences. I hope that someone can find more "negative tests", or even better, a "positive test", perhaps similar to Graham and Lin's result.