(Link to cross-post to MathOverflow)
Let $b=\frac{\sqrt{5}-1}2$ and $a_n:=(-1)^{[nb]}$ where $[\cdot]$ denotes the floor function. Are the partial sums $A_N=\sum\limits_{n=0}^N a_n$ bounded?
The sequence $nb\bmod 2$ is equidistributed in $[0,2]$, so, as $n$ gets large, there will be about half of indices in $\{1,2,\cdots,n\}$ that correspond to $1$ and the other half to $-1$. This doesn't help much though because what it says is basically $A_N\in o(N)$.
By testing on computer softwares, it seems that $A_N\in O(\log N)$.
Also, is there anything special about $b$ besides being an irrational real number? What if one replaces it by, say, $\pi$?