TITLE: Is $a_n=\sum_{\alpha=0}^n\;\text{round}(\sin\alpha)$ bounded?
QUESTION [14 upvotes]: Is the sequence $(a_n)$ bounded if 
$$a_n=\sum_{\alpha=0}^n\;\text{round}(\sin\alpha)?$$
Edit: round$(x)$ simply means round $x$ to the nearest integer; the case where $x=\pm0.5$ may be ignored since $\pi$ is irrational.
Please see comments for our current progress.

REPLY [5 votes]: It is not an answer to your question (I believe the answer should be no), but rather to a related question, which is already nontrivial:
Is there a subsequence of $(a_n)_n$ which is bounded ?
The answer is yes.
It is a direct application of the Denjoy-Koksma inequality (see https://en.wikipedia.org/wiki/Denjoy%E2%80%93Koksma_inequality), which in this case states that, if we denote $f:[0,1[\to [0,1[, f(x)=\{x+\frac1{2\pi} \}$ and 
$$\phi(x)=round(sin(2\pi x)),$$
(which is clearly of bounded variation, namely $Var(\phi)=4$, and $\int_0^1 \phi=0$) then for any integer couple $(p,q)$ such that the following diophantine inequality holds
$$\left|\frac1{2\pi}-\frac{p}q\right|\leq \frac{1}{q^2},$$
then
$$|a_{q-1}|=\left| \sum_ {k=0}^{q-1} \phi(f^k (0)) \right|\leq Var(\phi)=4.$$
There exist infinitely many couples $p,q$ satisfying the above diophantine inequality. For example, the convergents of the continued fraction expansion of $1/2\pi$. Thus, there exists a bounded, 'explicit' subsequence of $(a_n)_n$.