Apologise in advance if this problem isn't research-level (I'm quite certain it isn't). It's just I found it quite intriguing because it turned out to be much more subtle than it appeared at my first glance. Also, this problem didn't get an answer so far (nor much attention) from MSE, so I decided to post it here in hopes I can get some valuable insights. Cheers!

Here goes the original post from MSE.

Let $b=\dfrac{\sqrt{5}-1}2$ and $a_n:=(-1)^{[nb]\bmod 2}$ where $[\cdot]$ denotes the floor function. Then is $\sum a_n$ bounded?

As far as I know $nb\bmod 2$ is equidistributed in $[0,2]$ and 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 $\sum a_n\in o(n)$. By testing on computer softwares, it seems to be $O(\log n)$ (but of course any such "test" is worthless). But I don't know what else I can do.

Also, is there anything special about $b$ besides being an irrational real number (or, being the golden ratio)? What if I replace it by, say, $\pi$?

PS: A comment suggested a probable connection with the 1D random walk, but I really can't see a possible way to formulate such a connection beyond the intuitive level.

continued fractions, saying he proved it isnot boundedsome time ago but apparently have forgot and lost all the details. $\endgroup$ – Vim Apr 11 '17 at 15:585more comments