In the net drawn below, in how many ways can one reach the point $3n+1$ starting from the point $1$ so that the labels of the points on the way increase?

[asy]
import graph; size(12cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.3,xmax=12.32,ymin=-10.66,ymax=6.3; draw((1,2)--(xmax,0*xmax+2)); draw((1,0)--(xmax,0*xmax+0)); draw((0,1)--(1,2)); draw((1,0)--(0,1)); draw((1,2)--(3,0)); draw((1,0)--(3,2)); draw((3,2)--(5,0)); draw((3,0)--(5,2)); draw((5,2)--(7,0)); draw((5,0)--(7,2)); draw((7,2)--(9,0)); draw((7,0)--(9,2)); 
dot((1,0),linewidth(1pt)+ds); label("2",(0.96,-0.5),NE*lsf); dot((0,1),linewidth(1pt)+ds); label("1",(-0.42,0.9),NE*lsf); dot((1,2),linewidth(1pt)+ds); label("3",(0.98,2.2),NE*lsf); dot((2,1),linewidth(1pt)+ds); label("4",(1.92,1.32),NE*lsf); dot((3,2),linewidth(1pt)+ds); label("6",(2.94,2.2),NE*lsf); dot((4,1),linewidth(1pt)+ds); label("7",(3.94,1.32),NE*lsf); dot((6,1),linewidth(1pt)+ds); label("10",(5.84,1.32),NE*lsf); dot((3,0),linewidth(1pt)+ds); label("5",(2.98,-0.46),NE*lsf); dot((5,2),linewidth(1pt)+ds); label("9",(4.92,2.24),NE*lsf); dot((5,0),linewidth(1pt)+ds); label("8",(4.94,-0.42),NE*lsf); dot((8,1),linewidth(1pt)+ds); label("13",(7.88,1.34),NE*lsf); dot((7,2),linewidth(1pt)+ds); label("12",(6.8,2.26),NE*lsf); dot((7,0),linewidth(1pt)+ds); label("11",(6.88,-0.38),NE*lsf); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); 
[/asy]
To solve this problem, we need to determine the number of ways to reach the point labeled \(3n+1\) starting from the point labeled \(1\) such that the labels of the points on the way increase. We will use a recursive approach to find the number of ways to reach each point.

1. **Define the recursive relation:**
   Let \(T_k\) be the number of ways to reach the point labeled \(k\). We need to establish a recursive relation for \(T_k\).

2. **Identify the base cases:**
   - \(T_1 = 1\) because there is only one way to start at point 1.
   - \(T_2 = 1\) because there is only one way to move from point 1 to point 2.
   - \(T_3 = 1\) because there is only one way to move from point 1 to point 3.

3. **Establish the recursive relations:**
   From the problem, we have the following relations:
   \[
   T_{3n+1} = T_{3n} + T_{3n-1}
   \]
   \[
   T_{3n} = T_{3n-3} + T_{3n-2}
   \]
   \[
   T_{3n-1} = T_{3n-4} + T_{3n-2}
   \]

4. **Simplify the recursive relations:**
   Subtract the third equation from the second:
   \[
   T_{3n} - T_{3n-1} = (T_{3n-3} + T_{3n-2}) - (T_{3n-4} + T_{3n-2})
   \]
   \[
   T_{3n} - T_{3n-1} = T_{3n-3} - T_{3n-4}
   \]
   This implies that \(T_{3n} - T_{3n-1}\) is constant.

5. **Determine the constant:**
   Since \(T_3 = 1\) and \(T_2 = 1\), we get:
   \[
   T_{3n} = T_{3n-1}
   \]
   for all \(n\).

6. **Find the general form:**
   Using the first recursive relation:
   \[
   T_{3n+1} = 2T_{3n}
   \]
   and substituting \(T_{3n} = T_{3n-1}\), we get:
   \[
   T_{3n+1} = 2T_{3n}
   \]
   Hence,
   \[
   T_{3n} = 3T_{3n-3}
   \]
   Therefore,
   \[
   T_{3n} = 3^{n-1}
   \]
   and
   \[
   T_{3n+1} = 2T_{3n} = 2 \cdot 3^{n-1}
   \]

Conclusion:
\[
T_{3n+1} = 2 \cdot 3^{n-1}
\]

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