Four. (50 points) Let $n \in \mathbf{N}_{+}, f(n)$ be the number of all integer sequences $\left\{a_{k} \mid k=0,1, \cdots, n\right\}$ that satisfy the following conditions:
$$
\begin{array}{l}
\text { (1) } a_{0}=0, a_{n}=2 n, \text { and } \\
1 \leqslant a_{k+1}-a_{k} \leqslant 3(k=0,1, \cdots, n-1) ;
\end{array}
$$
(2) There do not exist $i, j(0 \leqslant i<j \leqslant n)$ such that
$$
a_{j}-a_{i}=n \text {. }
$$

Find the value of $3 f(16)-2 f(15)+f(10)$.
Divide a circle of length $2 \cdot n$ into $2n$ equal parts, and label the points sequentially as $0,1, \cdots, 2n$. Then color the points labeled $a_{i} (i=0,1, \cdots, n-1)$ black, and the other $n$ points white. The sequence given in the problem corresponds one-to-one with the following coloring method:
(1) The point labeled 0 is black, and the black points divide the circle into $n$ arcs, each of length 1, 2, or 3;
(2) There are no two black points that are diametrically opposite, i.e., black points and white points are paired, forming diametrically opposite points.

Clearly, there cannot be three consecutive black points. Otherwise, let $A, B, C$ be three consecutive black points. Then their diametrically opposite points $A', B', C'$ would be three consecutive white points, but the arc containing these three white points would be longer than 3, which is a contradiction.
Thus, the coloring method that satisfies (1) and (2) is to color the point labeled 0 black, and color the points $1 \sim n-1$ black or white such that no three consecutive points are the same color, and then color the points $n \sim 2n-1$ accordingly (point $i$ is black $\Leftrightarrow$ point $n+i$ is white).

First, color the points of a circular arc of length $k$ such that the endpoints are black and no three consecutive points are the same color.
Let the number of such coloring methods be $g(k)$. It is easy to see that:
$$
\begin{array}{l}
g(1)=g(2)=1, \\
g(3)=2^{2}-1=3, \\
g(4)=2^{3}-4=4 .
\end{array}
$$

For $k \geqslant 5$, consider the last segment of the circular arc with black endpoints.
If its length is 3, then the number of corresponding coloring methods is $g(k-3)$;
If its length is 2, then the number of corresponding coloring methods is $g(k-2)$;
If its length is 1, then the adjacent arc length is 2 or 3, and the number of coloring methods is $g(k-3)+g(k-4)$.
Thus, $g(k)=g(k-2)+2 g(k-3)-g(k-4)$.
Next, find the number of coloring methods $f(n)$ that satisfy (1) and (2).
If point $n-1$ is black, then the number of coloring methods is $g(n-1)$.
If point $n-1$ is white and point $n$ is white, then points $n-2$ and $n+1$ are black, and point 1 is white. If point 2 is black, then the number of coloring methods is $g(n-4)$; if point 2 is white, then point 3 is black, and the number of coloring methods is $g(n-5)$. Therefore,
$$
f(n)=g(n-1)+g(n-4)+g(n-5) .
$$

Calculating each term, we get
$$
\begin{array}{l}
g(5)=6, g(6)=11, g(7)=17, \\
g(8)=27, g(9)=45, g(10)=72, \\
g(11)=116, g(12)=189, g(13)=305, \\
g(14)=493, g(15)=799 .
\end{array}
$$

From equation (1), we get
$$
\begin{array}{l}
f(16)=g(15)+g(12)+g(11) \\
=799+189+116=1104, \\
f(15)=g(14)+g(11)+g(10) \\
=493+116+72=681, \\
f(10)=g(9)+g(6)+g(5) \\
=45+11+6=62 .
\end{array}
$$

Thus, $3 f(16)-2 f(15)+f(10)=2012$.