The smallest periods of the prefixes of the Fibonacci word

I'll start with some definitions to simplify the rest of the message.

Let's denote $f_0 = b, \ f_1 = a, \ f_n = f_{n-1} \cdot f_{n-2}$, where $\cdot$ stands for concatenation of two words. We call $f_n$ the $n$-th Fibonacci word. Let's also denote $f := \lim_{n \to \infty} f_n$ (the Fibonacci word). Let $s_{k}$ be the subword of $f$ consisting of the first $k$ letters of it. Let $w(k)$ be the $k$-th letter of a word $w$. Let $\mathrm{per}(k)$ stand for the smallest positive number such that
$$
\forall i \in \{ 1, \ldots, k - \mathrm{per}(k) \}. \ s_{k}(i) = s_{k}(i + \mathrm{per}(k))
$$
In other words, $\mathrm{per}(k)$ is the smallest period in the word $s_{k}$.

During one of my lectures, my lecturer mentioned an interesting property of the Fibonacci words. If we look at the sequence $(\mathrm{per}(k))_{k \in \mathbb{Z}_{+}}$, it looks like this
$$
(1, 2, 2, 3, 3, 3, 5, 5, 5, 5, 5, \ldots)
$$
So, one $1$, two $2$s, three $3$s, five $5$s, and so on. It is the non-decreasing sequence consisting of Fibonacci numbers where the number $F_n$ appears exactly $F_n$ times.

I've been trying to prove it today, but with little success. I'll be glad for any kind of help – a proof, some hints, remarks, pretty much anything. I'll list what I've managed to show below:
The smallest period of $f_k$ is $F_{k-1}$

I proved it inductively: the basis is obvious – we do it by hand. Then we make the induction step: let's look at $f_n$ and let $p$ be the smallest period in it. It's obvious $F_{n-1}$ is a period in that word. We also know that anything smaller than $F_{n-2}$ cannot be a period because then it would be a period in $f_{n-1}$ too and we'd get a contradiction. Thus, $p \in [F_{n-2}, F_{n-1}]$.

To show $p$ cannot be equal to $F_{n-2}$ it's sufficient to compare the last letters of $f_{n-1}$ and $f_{n-2}$ – they're different, so $p \neq F_{n-2}$. Thus $p \in (F_{n-2}, F_{n-1}]$.

Let's assume $p \in (F_{n-2}, F_{n-1})$. Note that $f_n = f_{n-3}f_{n-4}f_{n-3}f_{n-2}$. We'll fo

What we need to prove are

• the smallest period for $s_1$ is 1, which is trivially true.

• the smallest period for $s_i$ is $F_{k-1}$ for all $F_k-1\le i\le F_{k+1}-2$, where $k\ge3$.

Let $k\ge3$.

You have proved that the smallest period of $f_k=s_{F_k}$ is $F_{k-1}$.

• The smallest period of any string cannot be smaller than that of a prefix of itself.

• $f_{k+1}=f_{k-1}f_{k-2}f_{k-1}$. Since the suffix $f_{k-2}f_{k-1}$ is the same as the prefix $f_{k-1}f_{k-2}$ except the last two characters, $|f_{k-1}|$ is a period of $f_{k+1}$ without last two characters. That is, $F_{k-1}$ is a period of $s_{F_{k+1}-2}$. Hence for any $i\le F_{k+1}-2$, the smallest period of $s_i$ is at most $F_{k-1}$.

So, the smallest period for $s_i$ is $F_{k-1}$ for all $F_k\le i\le F_{k+1}-2$, where $k\ge3$.

Looking closely, we can see that you have also proved the smallest period of $s_{F_k-1}$ is $F_{k-1}$. $\quad\Box$

StackExchange Computer Science Q#154729, answer score: 2