Here's another approach. Use the graph as in my previous answer, but don't bother with the starting state $S$, we can start at $A$ (we start with no zeros and then can have a zero or a one so it's OK). The graph looks like this:

Let's define $a_n, b_n, c_n, d_n, e_n$ to be the number of walks on the graph that start from $A$ and after $n$ steps end up in $A, B, C, D, E$. What we want to calculate is the total number of approved strings:
$$x_n = a_n+b_n+c_n+d_n.$$
We notice these facts:
- $a_n = 1$
- $b_n = b_{n-2}+b_{n-3}$, and $b_0=0, b_1=b_2 = 1$
- $c_n = c_{n-1}+c_{n-2}-c_{n-4}$, and $c_0=c_1=0, c_2=c_3=1$
- $d_n = c_{n-1}$, and $d_0=0$
The first is obvious, since only the string $00\dots0$ can end up in $A$.
Also, the last since we can only arrive to $D$ from $C$.
Let's prove the other two inductively on $n$. First notice a helpful equation from the fact that we can arrive to $C$ from $B$ or from $D$:
- $\pmb{c_{m}} = b_{m-1} + d_{m-1} = \pmb{b_{m}+b_{m-1}-1}$, since $b_{m} = d_{m-1}+1$ (remember $a_m = 1$)
Let's now do the induction step for $b_{n+1}$:
$$b_{n+1} = a_{n}+ d_{n} = 1 + c_{n-1} = b_{n-1}+b_{n-2}$$
Then for $c_{n+1}$. I found it easier to start with what we want, that is:
$$c_n+c_{n-1}-c_{n-3} \\
= b_n+b_{n-1}-1 + b_{n-1}+b_{n-2}-1 - (b_{n-3}+b_{n-4}-1)\\
= b_n+b_{n-1}-1 + b_{n-1}+b_{n-2}-1 - b_{n-1} +1 \\
= b_n-1+ b_{n-1}+b_{n-2}\\
= b_n-1+ b_{n+1}\\
= c_{n+1}.
$$
The base cases can be checked easily (for example with the matrix approach), here are how the sequences begin
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1...
0, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28...
0, 0, 1, 1, 2, 3, 4, 6, 8, 11, 15, 20, 27, 36, 48...
0, 0, 0, 1, 1, 2, 3, 4, 6, 8, 11, 15, 20, 27, 36...
Then finally we also find a recursion for the sum of these, it is actually the same recurrence as for $c$ and $d$
$$x_n = x_{n-1}+x_{n-2}-x_{n-4} \text{ and } x_0=1, x_1=2, x_3=3$$
Let's prove this with induction and also here it's easier to start with the right hand side. The $c$ and $d$ terms gather nicely, $a$ is just the $1$ that survives and noticing that $b_{n}-b_{n-3}=b_{n-2}$ and then that $b_{n-2}+b_{n-3}=b_{n}$ proves the result.