A [i]$n$-trønder walk[/i] is a walk starting at $(0, 0)$, ending at $(2n, 0)$ with no self intersection and not leaving the first quadrant, where every step is one of the vectors $(1, 1)$, $(1, -1)$ or $(-1, 1)$. Find the number of $n$-trønder walks.
1. **Define the problem and constraints:**
   - We need to find the number of $n$-trønder walks starting at $(0,0)$ and ending at $(2n,0)$.
   - The walk must not self-intersect and must stay within the first quadrant.
   - Each step is one of the vectors $(1,1)$, $(1,-1)$, or $(-1,1)$.

2. **Define the "depth" of a point $(a,b)$:**
   - The depth is given by $\frac{a+b}{2}$.
   - For the walk to stay within the first quadrant and have integer coordinates, the depth must be an integer.

3. **Analyze the effect of each step on the depth:**
   - The step $(1,1)$ increases the depth by 1.
   - The steps $(1,-1)$ and $(-1,1)$ keep the depth constant.

4. **Consider the points at each depth level:**
   - For any integer $k$ with $0 \leq k < n$, consider when we first move from depth $k$ to depth $k+1$.
   - At depth $k$, the possible points are $(0,2k), (1,2k-1), \ldots, (i,2k-i), \ldots, (2k,0)$.
   - There are exactly $2k+1$ such points.

5. **Count the valid $(1,1)$ steps:**
   - Each of the $2k+1$ points at depth $k$ can yield a valid $(1,1)$ step to depth $k+1$.
   - Once a point is chosen, the remaining steps that keep the depth constant can be filled in exactly one way.

6. **Calculate the total number of ways:**
   - The number of ways to choose the points and steps is given by the product of the number of choices at each depth level.
   - This product is $(2n-1)(2n-3)(2n-5)\ldots(1)$, which is the double factorial $(2n-1)!!$.

Therefore, the number of $n$-trønder walks is $(2n-1)!!$.

The final answer is $\boxed{(2n-1)!!}$