Number of numbers in $n$th difference sequence

Question

Suppose that $r$ is an irrational number with fractional part between $1/3$ and $2/3$. Let $D_n$ be the number of distinct $n$th differences of the sequence $(\lfloor{kr}\rfloor)$. It appears that

$$D_n=(2,3,3,5,4,7,5,9,6,11,7,13,8,\ldots),$$

essentially A029579. Can someone verify that what appears here is actually true?

Example: for $r=(1+\sqrt{5})/2$, we find

$$(\lfloor{kr}\rfloor)=(1,3,4,6,8,9,11,12,14,16,17,\ldots) = A000201 = \text{ lower Wythoff sequence,}$$

\begin{align*} \Delta^1 =&(2,1,2,2,1,2,1,2,2,1,2,2,1,\ldots), & D_1=2, \\ \Delta^2 =&(1,-1,1,0,-1,1,-1,1,0,-1,1,\ldots), & D_2=3, \\ \Delta^3 =&(-2,2,-1,-1,2,-2,2,-1,-1,2,\ldots), & D_2=3. \end{align*}

Answer 1

It's easy to see that $D_n\leq \texttt{A029579}(n)$. Indeed, $\Delta^1$ is a Sturmian word, which is known to have exactly $n+1$ factors of length $n$.

Now, $\Delta^n$ is formed by values of the $(n-1)$-th difference operator on the factors of $\Delta^1$ of length $n$, i.e., $$\Delta^n = \left(\sum_{i=0}^{n-1} \binom{n-1}{i} (-1)^{n-1-i} \Delta^1_{k+i}\quad \big|\quad k=1,2,\dots\right).$$

For even $n$, we immediately have $D_n\leq n+1 = \texttt{A029579}(n)$.

For odd $n$, we additionally notice (I did not verify this carefully) that (i) the reverse of a Sturmian factor is a factor itself, (ii) values of the operator on a factor and its reverse are the same, and (iii) there are exactly two symmetric factors. Hence, here we have $$D_n\leq \frac{n+1-2}2 + 2 = \frac{n+3}2 = \texttt{A029579}(n).$$

It remains to prove that, besides the aforementioned cases, the operator values on factors of length $n$ are distinct.