The number of binary strings of length $n$ without the sequence (1;0;1) in it. First i know that a similat question had been answered here: The number of binary strings of length $n$ with no three consecutive ones But i am really a numb on this subject so i want to be sure to understand it correctly.
But my question is a little bit different as i am looking for the number of binary word of length $n$ without a different sequence, more precisally without the sequence $(1;0;1)$ in it.
Let writte $s_{n}$ all the correct sequence of length $n$.
So $s_{n}=s_{n-1}0 + s_{n-1, valid}1$ that means the total number of valid sequences of length $n$ equal the total number of valid sequences length $n-1$ that end with a zero + the total number of valid sequences length $n-1$ that end with a 1.
No matter what is the valid sequence of length $n-1$ if we add a zero at the end our sequence is still valid. The difficulty is concerning: $s_{n-1, valid}1$. Indeed if $s_{n-1}$ is a valid sequence of length $n-1$ that end with $10$ if we add a $1$ at the end to it we get the forbidden sequence.
So let focus on the total number of valid sequence of length $n-1$ that end with a $1$
$s_{n-1, valid}1 = s_{n-2}11 + s_{n-2, valid}01$
Now we encounter the same difficulty as describe above for $s_{n-2, valid}01$.
But similarly we can writte:$s_{n-2,valid}01=s_{n-3}001$
That means that the only valid sequences of length $n-2$ that after we add 01 to them are still valid sequences of numbers are the one who finish with a zero.
So we can conclude that:
$s_{n}=s_{n-1}0+s_{n-1, valid}1=s_{n-1}0+s_{n-2}11+s_{n-2, valid}01=s_{n-1}0+s_{n-2}11+s_{n-3}001$
So: $s_{n}=s_{n-1}+s_{n-2}+s_{n-3}$
With as initial condition: $s_1=2, s_2=4$ and $s_3=7=2^3-1$
Am i correct?

Hint: First I'd like to point at the somewhat problematic part of OPs conclusion. Let's have a look at
\begin{align*}
s_{n}&=s_{n-1}0+s_{n-1, \text{valid}}1\\
&=s_{n-1}0+\color{blue}{s_{n-2}11}+s_{n-2, \text{valid}}01\\
&=s_{n-1}0+s_{n-2}11+s_{n-3}001\\
\\
&s_{n}\stackrel{?}{=}s_{n-1}+s_{n-2}+s_{n-3}
\end{align*}

The pitfall here is the blue marked part $\color{blue}{s_{n-2}11}$. We are not allowed to take any valid sequence of length $n-2$ followed by $11$, since there are valid sequences ending in $10$ which combined with $11$ give the invalid sequence $\color{blue}{101}1$.

Note: Besides the nicely given answers here you might also find this answer helpful which essentially follows your approach (and which makes this post regrettably a duplicate).
Generating function approach
Nevertheless I'd like to add another approach based upon generating functions, from which the recurrence relation can be easily derived. It is the so-called Goulden-Jackson Cluster Method.

We consider binary words and the set $B=\{101\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $A(z)$ with the coefficient of $z^n$ being  the number of wanted words of length $n$ and derive from it the currence relation.

According to the referred paper (p.7) the generating function $A(z)$  is
\begin{align*}
A(z)=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\tag{1}
\end{align*}
with $d=2$, the size of the alphabet and $\mathcal{C}$ the weight-numerator of bad words with
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[101])
\end{align*}
We calculate according to the paper
\begin{align*}
\text{weight}(\mathcal{C})=\text{weight}(\mathcal{C}[101])&=-z^3-z^2\text{weight}(\mathcal{C}[101])\tag{2}\\
\end{align*}
where $z^3$ marks the length $3$ of the bad word $101$ and the term $z^2$ respects overlapping of bad words as in $101\color{blue}{01}$.

It  follows from (1) and (2):
\begin{align*}
\color{blue}{A(z)}&=\frac{1}{1-dz-\text{weight}(\mathcal{C})}\\
&=\frac{1}{1-2z+\frac{z^3}{1+z^2}}\\
&\,\,\color{blue}{=\frac{1+z^2}{1-2z+z^2-z^3}}\tag{3}\\
\end{align*}

We recall if a generating function has a representation as rational function of the form
\begin{align*}
A(z)=\sum_{n=0}^\infty a_n z^n=\frac{P(z)}{Q(z)}
\end{align*}
with $P(z), Q(z)$ polynomials, $\deg Q=q>\deg P$ and
\begin{align*}
Q(z)=1+\alpha_1 z+\alpha_2 z^2+\cdots + \alpha_q z^q
\end{align*} then the coefficients $a_n$ follow the recurrence relation
\begin{align*}
a_{n+q}+\alpha_1 a_{n+q-1}+\alpha_2 a_{n+q-2}+\cdots +\alpha_q a_{n}=0\qquad\qquad n\geq 0
\end{align*}
See for instance theorem 4.1.1 in Enumerative Combinatorics, Vol. I by R. P. Stanley.

Thanks to this theorem we can derive the recurrence relation from (3) as
\begin{align*}
a_{n+3}-2a_{n+2}+a_{n+1}-a_{n}=0\qquad\qquad n\geq 0
\end{align*}
resp. by shifting the indices we get
\begin{align*}
\color{blue}{a_n}&\color{blue}{=2a_{n-1}-a_{n-2}+a_{n-3}\qquad\qquad n\geq 3}\tag{4}\\
\color{blue}{a_0}&\color{blue}{=1,a_1=2,a_2=4,a_3=7}
\end{align*}
in accordance with already given answers. The initial conditions of the recurrence relation follow easily.

Two recurrence relations: Besides (4) another recurrence relation
\begin{align*}
\color{blue}{s_n=s_{n-1}+s_{n-2}+s_{n-4}\qquad\qquad n\geq 4}\tag{5}
\end{align*}
is stated by @Mateo. In terms of generating functions we observe that expanding in (3) $A(z)$ with $1+z$ results in
\begin{align*}
A(z)&=\frac{1+z^2}{1-2z+z^2-z^3}=\\
&=\frac{\left(1+z^2\right)(1+z)}{\left(1-2z+z^2-z^3\right)(1+z)}\\
&=\frac{1+z+z^2+z^3}{\color{blue}{1-z-z^2-z^4}}\tag{6}
\end{align*}
and the denominator in (6) provides us with the recurrence relation (5).
