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}