We count the number $a(n)$ of valid binary strings, i.e. strings which do not contain $101$ by partitioning them according to their matching length with the initial parts of the bad string $101$.
\begin{align*}
a_n=a^{[\emptyset]}_n+a^{[1]}_n+a^{[01]}_n\tag{1}
\end{align*}
The number $a^{[\emptyset]}_n$ counts the valid strings of length $n$ which do not start with the rightmost character of the bad word $10\color{blue}{1}$, i.e. start with $0$.
The number $a^{[1]}_n$ counts the valid strings of length $n$ which do start with the rightmost character of the bad word $10\color{blue}{1}$, i.e. $\color{blue}{1}$.
The number $a^{[01]}_n$ counts the valid strings of length $n$ which do start with the two rightmost characters of the bad word $1\color{blue}{01}$, i.e. start with $\color{blue}{01}$.
We get a relationship between valid strings of length $n$ with those of length $n+1$ as follows:
If a word counted by $a^{[\emptyset]}_n$ is appended by $0$ from the left it contributes to $a^{[\emptyset]}_{n+1}$. If it is appended by $1$ from the left it contributes to $a^{[1]}_{n+1}$.
If a word counted by $a^{[1]}_n$ is appended by $0$ from the left it contributes to $a^{[01]}_{n+1}$. If it is appended by $1$ from the left it contributes to $a^{[1]}_{n+1}$.
If a word counted by $a^{[01]}_n$ is appended by $0$ from the left it contributes to $a^{[\emptyset]}_{n+1}$. Appending from the left by $1$ is not allowed since then we have an invalid string starting with $101$.
This relationship can be written as
\begin{align*}
a^{[\emptyset]}_{n+1}&=a^{[\emptyset]}_{n}+a^{[01]}_{n}\tag{2}\\
a^{[1]}_{n+1}&=a^{[\emptyset]}_{n}+a^{[1]}_n\tag{3}\\
a^{[01]}_{n+1}&=a^{[1]}_n\tag{4}
\end{align*}
We can now derive a recurrence relation from (1) - (4):
We obtain for $n\geq 3$:
\begin{align*}
\color{blue}{a_{n+1}}&=a^{[\emptyset]}_{n+1}+a^{[1]}_{n+1}+a^{[01]}_{n+1}\tag{$ \to (1)$}\\
&=\left(a^{[\emptyset]}_{n}+a^{[01]}_{n}\right)+\left(a^{[\emptyset]}_{n}+a^{[1]}_n\right)+\left(a^{[1]}_n\right)\tag{$\to (2),(3),(4)$}\\
&=2a_n-a^{[01]}_{n}\tag{$\to (1)$}\\
&=2a_n-a^{[1]}_{n-1}\tag{$\to (4)$}\\
&=2a_n-a_{n-2}+a^{[\emptyset]}_{n-2}+a^{[01]}_{n-2}\tag{$\to (1)$}\\
&=2a_n-a_{n-2}+a^{[\emptyset]}_{n-3}+a^{[01]}_{n-3}+a^{[1]}_{n-3}\tag{$\to (2),(4)$}\\
&\,\,\color{blue}{=2a_n-a_{n-2}+a_{n-3}}\tag{$\to (1)$}
\end{align*}
We finally derive manually the starting values
\begin{align*}
\color{blue}{a_0=1,a_1=2,a_2=4,a_3=7}
\end{align*}
and obtain a sequence $(a_n)_{n\geq 0}$ starting with
\begin{align*}
(a_n)_{n\geq 0}=(1,2,4,7,12,21,37,65,114,200,351,\ldots)
\end{align*}
Hint: We have $a_5=\color{blue}{21}$ valid strings of length $5$. The $2^5-21=11$ invalid strings are
\begin{align*}
\color{blue}{101}00\qquad0\color{blue}{101}0\qquad00\color{blue}{101}\\
\color{blue}{101}01\qquad0\color{blue}{101}1\qquad01\color{blue}{101}\\
\color{blue}{101}10\qquad1\color{blue}{101}0\qquad\color{lightgrey}{10101}\\
\color{blue}{101}11\qquad1\color{blue}{101}1\qquad11\color{blue}{101}
\end{align*}
