Set up a system of recurrences. You have four cases, call even 0, odd 1; denote the number of sequences of length $n$ and parities $x y$ by $s_{x y}^{(n)}$. Considering how you can make up the strings (add 0, add 1, add 2 or 3) you have:
\begin{align}
s_{00}^{(n + 1)}
&= 2 s_{00}^{(n)} + s_{01}^{(n)} + s_{10}^{(n)} \\
s_{01}^{(n + 1)}
&= s_{00}^{(n)} + 2 s_{01}^{(n)} + s_{11}^{(n)}\\
s_{10}^{(n + 1)}
&= s_{00}^{(n)} + 2 s_{10}^{(n)} + s_{11}^{(n)} \\
s_{11}^{(n + 1)}
&= s_{01}^{(n)} + s_{10}^{(n)} + 2 s_{11}^{(n)}
\end{align}
You are interested only in $s_{00}^{(n)}$.
You also have $s_{00}^{(0)} = 1$, and $s_{01}^{(0)} = s_{10}^{(0)} = s_{11}^{(0)} = 0$.
Define the generating functions
$$
S_{xy}(z) = \sum_{n \ge 0} s_{xy}^{(n)} z^n
$$
Multiply the recurrences by $z^n$, add over $n \ge 0$ and recognize e.g.:
$$
\sum_{n \ge 0} s_{xy}^{(n + 1)} z^n
= \frac{S_{xy}(z) - s_{xy}^{(0)}}{z}
$$
to get a system of equations in the functions $S_{xy}(z)$. Solving this gives:
$$
S_{00}(z)
= \frac{1}{4}
+ \frac{1}{2} \cdot \frac{1}{1 - 2 z}
+ \frac{1}{4} \cdot \frac{1}{1 - 4 z}
$$
This is just geometric series, except for the constant term, which appears only when $n = 0$; use Iverson's convention to represent it:
$$
s_{00}^{(n)}
= \frac{1}{4} \cdot [n = 0] + 2^{n - 1} + 4^{n - 1}
$$
This matches hand count:
\begin{align}
0 & 1 & \\
1 & 2 & 2, 3 \\
2 & 6 & 00, 11, 22, 23, 32, 33 \\
3 & 20 & 002, 003, 020, 030,
112, 113, 121, 131, \\
& & 200, 211, 222, 223, 232, 233,
300, 311, 322, 323, 332, 333
\end{align}