Suppose $w:= a_1 a_2 \cdots a_n$ is a binary word, i.e. $a_i \in \{0,1\}$ for all $1\le i \le n$. Let $$\sigma(w) = (\text{# subwords } 00, \text{# subwords } 01, \text{# subwords } 10,\text{# subwords } 11)$$ For example, we have $\sigma(01101110101) = (0,4,3,3)$. Note that the entries of $\sigma(w)$ should always add up to $n-1$.
We say a binary word $w$ is $\sigma$-determined if $\sigma(v) = \sigma(w)$ implies $v=w$.
Let $f(n)$ denote the number of length $n$ $\sigma$-determined words. How can we calculate $f(n)$? Numerically, we seem to have $f(2n) = f(2n+1) = 4n+2$ for $n\ge 2$.