How many series of length $n$ combined from the numbers $0,1,3,4$ , that don't contain the sequences $04,40,13,31$ or $0X0, 1X1, 3X3, 4X4$ are there? How many series of length $n$ combined from the numbers ${0,1,3,4}$ , that don't contain the sequences $04,40,13,31$ or $0X0, 1X1, 3X3, 4X4$ (where X can be any number from $0,1,3,4$) are there?
Let $f(n)$ be the number of the good series of length $n$.
If I start with $1$, then I have $f(n-1)$ options for the rest of the series without limitations, but I have to decrease the series of length $(n-2)$ starting with 3 (so I won't have $13$ in my series) and I have to decrease 3 times the series of length $(n-3)$ starting with 1 (so I won't have $1X1$ in my series).
Overall $f(n)$ is 4 times the process I just did because of symmetry depending on the first number in the series, i.e. $f(n)=4(f(n-1)-f(n-2)-3f(n-3))$.
I know I have a mistake when I'm decreasing the series of length $(n-2)$ becuse I'm decreasing too much.
Would appreciate any help in correcting that mistake.
Thanks:)

The restrictions can be put in a little simpler way by this small (not at all necessary) trick: modify the sequence at even times ($n=2,4,6\cdots)$ by remapping the values $0 \leftrightarrow 4$, $1 \leftrightarrow 3$... The prohibited subsequences then become $\cdots *XX*\cdots $ and $\cdots *X*X*\cdots$  . That is, we must have $x[n]\ne x[n-1]$ and $x[n]\ne x[x-2]$. Because the first restriction implies $x[n-1]\ne x[x-2]$, then the sequence construction follows a simple pattern: in each time we have $2$ alternatives, except for time $n=1$ when we have $4$, and time $n=2$ when we have $3$.
Hence, no need of recursions. The total number of sequences is $f(n) = 4 \times 3 \times 2 \times 2 \cdots 2$, or
$$ f(n)=\begin{cases}
4 & n=1\\
4 \times3 \times2^{n-2}=3\times2^n & n>1
\end{cases}
$$
