How can I show that this sequence is aperiodic and is not even eventually-periodic.

Question

Let $\alpha$ be an irrational real number and let $f:\mathbf{N} \to \mathbf{N}$ be defined as $$ f(n) = \lceil \alpha (n+1) \rceil - \lceil \alpha n \rceil. $$ Here, as usual, $\lceil \cdot \rceil$ is the "ceiling" operator.

I want to show that $f$ is aperiodic and, more generally, it isn't even "eventually periodic".

Recall that a sequence is eventually periodic if there is a positive period $P\in \mathbf{N}$ and number $M\in \mathbf{N}$ such that for all $m\ge M$ and all $l\in \mathbf{N}$ we have $$ f(m +l * P) = f(m). $$

I have tried a variety of things, but I end up getting stuck in the weeds.

Note that this is not true if $\alpha$ is rational, as in that case $f$ will have a period equal to the denominator of $\alpha$.

Any ideas are appreciated. I want to say something like this is discussed in Knuth's TAOCP books, but I don't have access to them right now.

Answer 1

For simplicity, assume $\alpha>0$. Write $\alpha=k+\theta$ with $k=\lfloor\alpha\rfloor\in\mathbb Z_{\ge0}$ and $\theta\in (0,1)$. Since $\alpha$ is irrational, $\theta$ is irrational as well. For every $n\in\mathbb N$ the number $\alpha n$ is not an integer and hence $\lceil \alpha n\rceil=\lfloor\alpha n\rfloor+1$.

This gives $$\begin{align*} f(n) &= \lceil\alpha(n+1)\rceil-\lceil\alpha n\rceil\\ &= \bigl(\lfloor\alpha(n+1)\rfloor+1\bigr)-\bigl(\lfloor\alpha n\rfloor+1\bigr)\\ &= \lfloor\alpha(n+1)\rfloor-\lfloor\alpha n\rfloor\\ &= k+\bigl(\lfloor\theta(n+1)\rfloor-\lfloor\theta n\rfloor\bigr) \end{align*}$$ since $\alpha=k+\theta$.

Define $g(n)=\lfloor\theta(n+1)\rfloor-\lfloor\theta n\rfloor$. Then $g(n)\in \{0,1\}$ and $$\sum_{n=1}^N g(n)=\lfloor\theta(N+1)\rfloor -\lfloor\theta\rfloor=\lfloor\theta(N+1)\rfloor$$ as $0<\theta<1$. This implies $$\frac{1}{N}\sum_{n=1}^N g(n)=\frac{\lfloor\theta(N+1)\rfloor}{N}\xrightarrow[N\to\infty]{}\theta$$ and hence the density of $1$'s in the sequence $\{g(n)\}$ exists and equals $\theta$.

But if $g$ were eventually periodic with periodic $P$, then this average would converge to a rational of the form $r/P +E$ where $r$ is the number of $1$'s per period and $E$ is the (finite) number of $1$'s that came from counting $1$'s in the interval where $g$ is not periodic. This is a contradiction hence completing the proof.