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 $w$?