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](https://math.stackexchange.com/q/2228549/191404) from MSE.




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> 
> 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, $\pi$?


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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.



Is the following series consisting of equally distributed $\pm 1$ bounded?
The sequence $\sum a\_n$ is unbounded.


This is a consequence of a general result from Kesten,  

[On a conjecture of Erdös and Szüsz related to uniform distribution mod 1](https://eudml.org/doc/204796), Acta Arithmetica (1966). The proof is not very long but quite computational, using properties of continued fractions.


**Theorem**
*Let $\xi \in [0,1]$, $0\leq a < b \leq 1$, denote by $N(M, \xi, a,b)$ the number of integers less than $M$ such that $a \leq \{k\xi\} \leq b$. Then
$N(M,\xi, a,b) - M(b-a)$ is unbounded if $b-a$ is rational and $\xi$ is irrational.*


You are interested in the $1/2$ discrepancy of an irrational rotation of the circle. Note that from the [Denjoy-Koksma inequality](https://en.wikipedia.org/wiki/Denjoy%E2%80%93Koksma_inequality),
there is a sequence $q\_n$ such that
$\sum\_{k= 0}^{q\_n-1} a\_k$ is bounded by $2$. The sequence exists for any irrational rotation of the circle and is given by the denominators of the partial fraction decomposition of the rotation number. For the golden ratio, these numbers are given by the Fibonacci numbers.


The golden ratio is a bit special because its decomposition in continued fraction is just $[1,1,1...]$ so it is badly approximated by the rational numbers. So the computation of Kesten should be simpler in that case. 


EDIT: the theorem is also mentioned shortly in the book of Kuipers and Niederreiter, "uniform distribution of sequences". Some additional references can be found in the notes of chapter 2, section 3 p128.


