TITLE: What is the subword complexity function of this infinite word?
QUESTION [8 upvotes]: Let $w_{0}$ denote the finite word $01$ in the free monoid $\{ 0, 1 \}^{\ast}$, and for $i \in \mathbb{N}$ define $w_{i}$ as the word obtained by adjoining the first $\left\lfloor \frac{\ell(w_{i-1})}{2}  \right\rfloor$ entries in $w_{i-1}$ to the right of $w_{i-1}$. We thus have that:
\begin{align*}
 w_{0}   &  =   01   \\
 w_{1}   &  =   010   \\
 w_{2}   &  =   0100   \\
 w_{3}   &  =   010001   \\
 w_{4}   &  =   010001010   \\
      & \text{etc.}
\end{align*}
Let $$ w = 0100010100100010001010001010010001010010000100010100100010001010100 \ldots$$ 
 denote the infinite binary word obtained in the limit, with respect to the sequence $(w_{i} : i \in \mathbb{N}_{0})$.
Since the construction of this infinite word is very simple and natural, it is surprising that the integer sequence $$(0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, \ldots)$$ given by the consecutive entries in $w$ is not currently in the On-Line Encyclopedia of Integer Sequences (OEIS).
Recall that the subword complexity function $\sigma_{v} = \sigma : \mathbb{N} \to \mathbb{N}$ of an infinite word $v$ is the function on $\mathbb{N}$ that maps $n \in \mathbb{N}$ to the number of distinct factors of $v$ of length $n$. Given the simple definition of the binary word $w$, it is natural to ask: what is $\sigma_{w}$? It is not obvious to me how to find a closed-form evaluation of the sequence $$(\sigma_{w}(n) )_{n \in \mathbb{N}} = (2, 3, 5, 8, 12, \ldots),$$ since proving a statement of the form $\sigma_{w}(n) = m$ for fixed $n \in \mathbb{N}$ (where $m \in \mathbb{N}$) appears to be nontrivial in general. However, for certain 'small' values of $n \in \mathbb{N}$, the evaluation of $\sigma_{w}(n)$ is relatively trivial. For example, using induction, it is easily seen that $\sigma_{w}(2)=3$.
It is also natural to ask: What is the abelian complexity function of $w$?

REPLY [2 votes]: Not a full answer, but probably a useful reference.
You mention that the integer sequence $$(0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, \ldots)$$ given by the consecutive entries in $w$ is not currently in OEIS. However, the sequence of positions of $1$ in this sequence, namely:
$$
1, 5, 7, 10, 14, 18, 20, 24, 26, 29, 33, 35, \dotsm
$$
appears as A020942, First column of 3rd-order Zeckendorf array, in OEIS. 
Quoting OEIS

Any number $n$ has unique representation as a sum of terms from $\{1,
 2, 3, 4, 6, 9, 13, 19, ...\}^{*}$ such that no two terms are adjacent
  or pen-adjacent; e.g., $7=6+1$. Sequence gives all $n$ where that
  representation involves $1$.

$^{(*)}\scriptsize\text{ These terms obey the recurrence equation $a(n) = a(n-1) + a(n-3)$.}$
Two references are given
[1] Larry Ericksen and Peter G. Anderson, Patterns in differences between rows in k-Zeckendorf arrays, The Fibonacci Quarterly 50, February 2012.
[2] C. Kimberling, The Zeckendorf array equals the Wythoff array, Fibonacci Quarterly 33 (1995) 3-8.
Reference [1] gives interesting connections with the sequence of words $w_k$ on a $k$-letter alphabet defined as follows: 
\begin{align}
w_0 &= a_0,\\
w_1 &= a_0a_1,\\
&\ \vdots\\
w_{k-1} &= a_0a_1 \dotsm a_{k-1}
\end{align}
and $w_i = w_{i-1}w_{i-k}$ for $i \geqslant k$. This makes this article a reasonable approach towards the solution of your problem.