<s>[INST] In how many ways can you split a string of length n such that every substring has length at least m? Suppose you have a string of length 7 (abcdefg) and you want to split this string in substrings of length at least 2.
The full enumation of the possibilities is the following:
ab/cd/efg
ab/cde/fg
abc/de/fg
abc/defg
abcd/efg
abcde/fg
abcdefg

And perhaps some others.
In general, if we have a string of length n and we want to split the string in substrings of length at least m, in how many ways can we achieve this?
 [/INST] We can model the situation with generating functions. In order to do so we consider binary strings consisting of $0s$ and $1s$. Let $$0^\star=\{\varepsilon,0,00,000,\ldots\}$$ denote all strings containing only $0$s with length $\geq0$. The empty string is denoted with $\varepsilon$. The corresponding generating function is $$x^0+x^1+x^2+\ldots=\frac{1}{1-x},$$ with the exponent of $w^n$ marking the length $n$ of a string of $0s$ and the coefficient of $x^n$ marking the number of strings of length $n$.

We encode substrings of length at least $m(\geq 2)$ as strings starting with $1$ followed by at least $m-1$ zeros. So, each substring has the form
  \begin{align*}
(1)(0^{m-1})(0^\star)
\end{align*}
  with generating function
  \begin{align*}
\frac{x^m}{1-x}
\end{align*}

$$ $$

Since each string consists of one or more substrings, the strings can be encoded as
  \begin{align*}
(10^{m-1}0^\star)(10^{m-1}0^\star)^\star
\end{align*}
  with generating function
  \begin{align*}
\frac{\frac{x^m}{1-x}}{1-\frac{x^m}{1-x}}=\frac{x^m}{1-x-x^m}\qquad\qquad m\geq 2
\end{align*}

$$ $$
OPs example with $m=2$ results in a generating function for the Fibonacci Numbers
\begin{align*}
\frac{x^2}{1-x-x^2}=x^2+x^3+2x^4+3x^5+5x^6+8x^7+\mathcal{O}(x^8)
\end{align*}
So, the coefficient of $x^7$ is equal to $8$ corresponding to eight different strings of length $7$ with substrings of length at least $2$. These strings are 
\begin{align*}
&1010100\quad\quad(2,2,3)\\
&1010010\quad\quad(2,3,2)\\
&1001010\quad\quad(3,2,2)\\
&1010000\quad\quad(2,5)\\
&1001000\quad\quad(3,4)\\
&1000100\quad\quad(4,3)\\
&1000010\quad\quad(5,2)\\
&1000000\quad\quad(7)
\end{align*}
</s>