Let $S=\{(a,b)|a=1,2,\dots,n,b=1,2,3\}$. A [i]rook tour[/i] of $S$ is a polygonal path made up of line segments connecting points $p_1,p_2,\dots,p_{3n}$ is sequence such that 

(i) $p_i\in S,$ 

(ii) $p_i$ and $p_{i+1}$ are a unit distance apart, for $1\le i<3n,$ 

(iii) for each $p\in S$ there is a unique $i$ such that $p_i=p.$ 

How many rook tours are there that begin at $(1,1)$ and end at $(n,1)?$

(The official statement includes a picture depicting an example of a rook tour for $n=5.$ This example consists of line segments with vertices at which there is a change of direction at the following points, in order: $(1,1),(2,1),(2,2),(1,2), (1,3),(3,3),(3,1),(4,1), (4,3),(5,3),(5,1).$)
1. **Define the problem and notation:**
   We are given a set \( S = \{(a,b) \mid a = 1, 2, \ldots, n, b = 1, 2, 3\} \). A *rook tour* of \( S \) is a sequence of points \( p_1, p_2, \ldots, p_{3n} \) such that:
   - \( p_i \in S \),
   - \( p_i \) and \( p_{i+1} \) are a unit distance apart for \( 1 \leq i < 3n \),
   - Each \( p \in S \) appears exactly once in the sequence.
   We need to count the number of rook tours that start at \( (1,1) \) and end at \( (n,1) \).

2. **Initial observations and base cases:**
   By examining small cases, we find:
   - For \( n = 2 \), there is 1 rook tour.
   - For \( n = 3 \), there are 2 rook tours.
   - For \( n = 4 \), there are 4 rook tours.
   - For \( n = 5 \), there are 8 rook tours.
   This suggests a pattern \( R(n) = 2^{n-2} \).

3. **Define primitive tours:**
   A primitive tour for a board of width \( n \) is defined as follows:
   - Choose \( m \) such that \( 0 \leq m \leq n-2 \).
   - Go right \( m \) spaces.
   - Go up one space.
   - Go left \( m \) spaces.
   - Go up one space.
   - Go right \( n-1 \) spaces.
   - Go down one space.
   - Go left \( n-m-2 \) spaces.
   - Go down one space.
   - Go right \( n-m-2 \) spaces to finish the tour.

4. **Recursive structure:**
   Every tour on a board of width \( n \) is either primitive or a chain of connected primitives for narrower boards connected by one rightward move between any two primitives.

5. **Set up the recursion:**
   - The base cases are \( R(2) = 1 \) and \( R(3) = 2 \).
   - For \( n \geq 4 \), the recursion is:
     \[
     R(n) = (n-1) + \sum_{j=2}^{n-2} (j-1) R(n-j)
     \]

6. **Induction proof:**
   - **Base Case:** For \( n = 2 \), \( R(2) = 1 = 2^0 \). For \( n = 3 \), \( R(3) = 2 = 2^1 \).
   - **Induction Hypothesis:** Assume \( R(n) = 2^{n-2} \) holds for \( 2 \leq n \leq k \).
   - **Induction Step:** For \( n = k+1 \),
     \[
     R(k+1) = (k+1-1) + \sum_{j=2}^{k-1} (j-1) R(k+1-j)
     \]
     Using the induction hypothesis \( R(k+1-j) = 2^{k-j-1} \),
     \[
     R(k+1) = k + \sum_{j=2}^{k-1} (j-1) 2^{k-j-1}
     \]
     Change the index of summation \( i = k-j \),
     \[
     R(k+1) = k + \sum_{i=1}^{k-2} (k-i-1) 2^{i-1}
     \]
     Simplify the sum,
     \[
     R(k+1) = k + \sum_{i=1}^{k-2} (k-i-1) 2^{i-1}
     \]
     Recognize the geometric series,
     \[
     R(k+1) = k + \sum_{i=1}^{k-2} 2^{i-1} (k-i-1)
     \]
     Simplify further,
     \[
     R(k+1) = k + 2^{k-2} - 1
     \]
     Therefore,
     \[
     R(k+1) = 2^{k-1}
     \]
     This completes the induction.

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