Let $n \geqslant 3$ be integer. Given convex $n-$polygon $\mathcal{P}$. A $3-$coloring of the vertices of $\mathcal{P}$ is called [i]nice[/i] such that every interior point of $\mathcal{P}$ is inside or on the bound of a triangle formed by polygon vertices with pairwise distinct colors. Determine the number of different nice colorings. 
([I]Two colorings are different as long as they differ at some vertices. [/i])
1. **Label the vertices and define the problem**:
   Let the vertices of the convex $n$-polygon $\mathcal{P}$ be labeled as $A_1, A_2, \ldots, A_n$ in clockwise order. We need to determine the number of different nice 3-colorings of these vertices such that every interior point of $\mathcal{P}$ is inside or on the boundary of a triangle formed by vertices with pairwise distinct colors.

2. **Characterize the nice coloring**:
   A nice coloring requires that every interior point of $\mathcal{P}$ is inside or on the boundary of a triangle formed by vertices with pairwise distinct colors. This implies that no two adjacent vertices can have the same color, and all three colors must be used.

3. **Reduction to smaller polygons**:
   If $A_i, A_{i+1}, A_{i+2}$ have pairwise distinct colors, we can reduce the problem to a smaller polygon by removing $A_{i+1}$. This motivates the following characterization:
   - If the coloring uses all three colors and no two adjacent vertices have the same color, the coloring is nice.
   - If two adjacent vertices have the same color, then the coloring is not nice.

4. **Induction and base cases**:
   We proceed by induction. If $A_i, A_{i+1}, A_{i+2}$ pairwise distinct does not exist, then the color of $A_i$ is the same as the color of $A_{i+2}$ for every $i$, which implies the polygon is 2-colorable. Hence, such a configuration must exist.

5. **Counting the number of nice colorings**:
   Let $f(n)$ be the number of ways to color the vertices such that no two adjacent vertices have the same color and all three colors are used. Let $g(n)$ be the number of ways to color the vertices such that no two adjacent vertices have the same color, and the first and last vertices are different colors.

6. **Recurrence relations**:
   - $f(n) + 2g(n) = 2^n$: Each vertex has 2 choices depending on the color of the previous vertex.
   - $f(n) = 2g(n-1)$: The first and last vertices must be different.
   - $g(n) = f(n-1) + g(n-1)$: The last vertex can be either of the two colors different from the previous vertex.

7. **Solving the recurrence relations**:
   - From $f(n) = 2g(n-1)$ and $g(n) = f(n-1) + g(n-1)$, we get $f(n) - g(n) = -(f(n-1) - g(n-1))$.
   - Using the initial conditions $f(2) = 2$ and $g(2) = 1$, we solve the recurrence relations to get:
     \[
     f(n) = \frac{2^n + 2(-1)^n}{3}
     \]
     \[
     g(n) = \frac{2^n - (-1)^n}{3}
     \]

8. **Final calculation**:
   Since there are three choices for the first color, the total number of ways to color such that neighboring colors are distinct is:
   \[
   3 \cdot \frac{2^n + 2(-1)^n}{3} = 2^n + 2(-1)^n
   \]
   We need to remove those colorings with at most 2 colors satisfying neighboring colorings are distinct. If $n$ is odd, this is 0. If $n$ is even, this is 6. So the final answer is:
   \[
   2^n + 2(-1)^n - 3(1 + (-1)^n)
   \]

The final answer is $\boxed{2^n - 3 - (-1)^n}$.