How to analyze $(-1)^{\left \lfloor n\theta \right \rfloor}$ (in which $\theta$ is an irrational number)?

Let $$\theta = \frac{\sqrt{5}-1}{2}$$. Define $$a_{n}=(-1)^{\left \lfloor n\theta \right \rfloor}$$. Please judge whether $$S_{n}=\sum_{k=1}^{n} a_{k}$$ is unbounded.

I tried to relate $$\left\lfloor n\theta \right\rfloor$$ to $$\left\lfloor\theta^n\right\rfloor$$, because $$\theta^n$$ can be written in the form "$$x_{n}\theta + y_{n}$$" in which $$x_{n}$$ and $$y_{n}$$ are related to Fibonacci sequence and $$\left \lfloor \theta^n \right \rfloor$$ is $$0$$.

But in this why I can only analyze some of the $$a_{n}$$.

Any ideas for help?

• Maybe you could use the Zeckendorf factorization of $n$: en.m.wikipedia.org/wiki/Zeckendorf%27s_theorem – Jacob Nov 7 '18 at 23:14
• It seems that $max(S_n, n \le 10^k)\ge k$ – Next Nov 8 '18 at 1:56
• @Jacob Thanks, but it's hard to analyze the precise equation of a certain number being represented as the sum of Fibonacci numbers. – Zero Nov 9 '18 at 0:34
• @Next How do you come to this conclusion? Could you give me some more-detailed hints please? Thanks – Zero Nov 9 '18 at 0:36

This is a very interesting problem.

Throughout this proof, let $$\rho(x,z)=\sum_{n=1}^{x}(-1)^{\left\lfloor\frac{n}{\varphi}+z\right\rfloor},$$ where $$x \in \mathbb{N}$$ and $$z\in \mathbb{R}$$, and let $$\rho(x)=\rho(x,0)=S_x$$. We may begin by noticing that $$\theta^{-1}=\varphi=\frac{1+\sqrt{5}}{2}.$$ We will use properties of the golden ratio throughout this proof.

Theorem 1. Let $$E(x) = x-[x]$$, where $$[x]$$ is the nearest integer function. If $$c for all $$n\leq x$$, then $$\rho(x)=\rho(x,c). \tag{1}$$

Proof. Since $$c, we have that $$\left\lfloor\frac{n}{\varphi}\right\rfloor=\left\lfloor\frac{n}{\varphi}+c\right\rfloor$$ for all $$n\leq x.$$ Thus, $$\rho(x)=\rho(x,c).$$

Theorem 2. $$$$\left\vert E\left(\frac{F_n}{\varphi}\right)\right\vert<\left\vert E\left(\frac{k}{\varphi}\right)\right\vert \quad \text{for all} \quad k where $$F_n=\frac{\varphi^n-(-\varphi)^{-n}}{\sqrt{5}}$$ is the $$n$$th Fibonacci number.

Proof. From this post, we have $$\vert F_{n-1}\varphi-F_n\vert<|q\varphi-p| \quad\text{for all} \quad p,q\in \mathbb{N} \quad\text{such that}\quad q since the Fibonacci numbers are the convergents of $$\varphi$$. Dividing by $$\varphi$$, we have $$\left\vert F_{n-1}-\frac{F_n}{\varphi}\right\vert<\left\vert q-\frac{p}{\varphi}\right\vert\quad\text{for all} \quad p,q\in \mathbb{N} \quad\text{such that}\quad q Also, $$\left[\frac{F_n}{\varphi}\right] = F_{n-1}$$, so $$\left[\frac{p}{\varphi}\right] for $$p. Letting $$q=\left[\frac{p}{\varphi}\right]$$, we arrive at $$(2)$$.

Theorem 3. For all $$k, $$\rho(F_n+k)=\rho(F_n)+(-1)^{F_{n-1}}\rho(k). \tag{3}$$

Proof. We write $$\rho(F_n+k)$$ as \begin{align}\rho(F_n+k)=\rho(F_n)+\rho\left(k,\frac{F_n}{\varphi}\right)=\rho(F_n)+\sum_{n=1}^k(-1)^{\left\lfloor\frac{n}{\varphi}+F_{n-1}+E\left(\frac{F_n}{\varphi}\right)\right\rfloor}\\ =\rho(F_n)+(-1)^{F_{n-1}}\rho\left(k,E\left(\frac{F_n}{\varphi}\right)\right).\end{align} Applying $$(1)$$ and $$(2)$$, we have, $$\rho\left(k,E\left(\frac{F_n}{\varphi}\right)\right)=\rho(k),$$ yielding $$(3).$$

Theorem 4. $$\rho(F_n)$$ is periodic with period $$6$$. Specifically, $$\rho(F_n,0)=\begin{cases}1 & n \equiv 1,2 \pmod{6} \\ 0 & n \equiv 0 \pmod{3} \\ -1 & n\equiv 4,5\pmod{6}\end{cases}\tag{4}$$

Proof. We shall induct on $$n$$. For our base case, we choose all $$n\leq 6$$: $$\rho(1)=1$$, $$\rho(2)=0$$, $$\rho(3)=-1$$, $$\rho(5)=-1$$, $$\rho(8)=0$$ (note that $$F_1=F_2=1$$). Now assume $$\rho(F_n)$$ obeys $$(4)$$ up to $$n=6k$$. Using $$(3)$$ and the fact that $$2\mid F_n$$ iff $$3\mid n$$, we have $$\begin{split}\rho(F_{6k+1})=\rho(F_{6k})+(-1)^{F_{6k-1}}\rho(F_{6k-1})=0+1=1 \\ \rho(F_{6k+2})=\rho(F_{6k+1})+(-1)^{F_{6k}}\rho(F_{6k})=1+0=1 \\ \rho(F_{6k+3})=\rho(F_{6k+2})+(-1)^{F_{6k+1}}\rho(F_{6k+1})=1-1=0\\ \rho(F_{6k+4})=\rho(F_{6k+3})+(-1)^{F_{6k+2}}\rho(F_{6k+2})=0-1=-1\\ \rho(F_{6k+5})=\rho(F_{6k+4})+(-1)^{F_{6k+3}}\rho(F_{6k+3})=-1+0=-1\\ \rho(F_{6k+6})=\rho(F_{6k+5})+(-1)^{F_{6k+4}}\rho(F_{6k+4})=-1+1=0.\end{split}$$ This proves the induction step, and thus, $$(4)$$.

Theorem 5. $$\rho(x)$$ is unbounded. Specifically, $$\rho:\mathbb{N}\rightarrow\mathbb{Z}$$ is surjective.

Proof. Let $$(n_i)_{i=1}^m$$ be a finite, sequence of positive integers congruent to $$1$$ modulo $$6$$ such that $$F_{n_i}>\sum_{k=i+1}^mF_{n_k}$$ for all $$i$$. Then, by using $$(3)$$ iteratively, \begin{align}\rho\left(\sum_{i=1}^mF_{n_i}\right)=\rho(F_{n_1})+\rho\left(\sum_{i=2}^mF_{n_i}\right)=\rho(F_{n_1})+\rho(F_{n_2})+\rho\left(\sum_{i=3}^mF_{n_i}\right)=\cdots\\ =\sum_{i=1}^m\rho(F_{n_i})=m.\end{align} Alternatively, let $$(n_i)_{i=1}^m$$ be a finite, sequence of positive integers congruent to $$4$$ modulo $$6$$ such that $$F_{n_i}>\sum_{k=i+1}^mF_{n_k}$$ for all $$i$$. Then, by using $$(3)$$ iteratively, \begin{align}\rho\left(\sum_{i=1}^mF_{n_i}\right)=\rho(F_{n_1})+\rho\left(\sum_{i=2}^mF_{n_i}\right)=\rho(F_{n_1})+\rho(F_{n_2})+\rho\left(\sum_{i=3}^mF_{n_i}\right)=\cdots\\ =\sum_{i=1}^m\rho(F_{n_i})=-m.\end{align} Since we also know $$\rho$$ takes on the value of $$0$$ at some numbers (e.g. $$\rho(2)=0$$), $$\rho$$ (i.e. $$S_n$$) is not only unbounded, but surjective.