Let $n\ge 2$ be a positive integer. Find the number of functions $f:\{1,2,\ldots ,n\}\rightarrow\{1,2,3,4,5 \}$ which have the following property: $|f(k+1)-f(k)|\ge 3$, for any $k=1,2,\ldots n-1$.

[i]Vasile Pop[/i]
To solve the problem, we need to find the number of functions \( f: \{1, 2, \ldots, n\} \rightarrow \{1, 2, 3, 4, 5\} \) such that \( |f(k+1) - f(k)| \geq 3 \) for any \( k = 1, 2, \ldots, n-1 \).

1. **Define Sets**:
   Let \( A_n, B_n, C_n, D_n \) be the sets of \( n \)-tuples ending with \( 1, 2, 4, 5 \) respectively that satisfy the given condition. Notice that the number \( 3 \) cannot appear in the sequence because the difference between \( 3 \) and any other number in the set \( \{1, 2, 4, 5\} \) is less than 3.

2. **Symmetry Argument**:
   We observe that:
   \[
   |A_n| = |D_n| \quad \text{and} \quad |B_n| = |C_n|
   \]
   This is because to each \( (a_1, \dots, a_n) \in A_n \) corresponds an \( n \)-tuple \( \left(6 - a_1, \dots, 6 - a_n\right) \in D_n \). Similarly, for \( |B_n| \) and \( |C_n| \).

3. **Recursive Relations**:
   We can establish the following recursive relations:
   \[
   |B_n| = |C_n| = |A_{n-1}|
   \]
   This is because if we cancel the last number from \( C_n \), we obtain an \( (n-1) \)-tuple ending with \( 1 \).

   Also,
   \[
   |A_n| = |C_{n-1}| + |D_{n-1}| = |A_{n-2}| + |A_{n-1}|
   \]
   This follows from the previous observation and the fact that \( |A_n| = F_{n+1} \), where \( F_n \) is the Fibonacci number.

4. **Total Number of Functions**:
   Let \( x_n = |A_n| + |B_n| + |C_n| + |D_n| \). Then,
   \[
   x_n = 2|A_n| + 2|B_n| = 2|A_n| + 2|A_{n-1}|
   \]
   Using the recursive relation for \( |A_n| \):
   \[
   |A_n| = |A_{n-1}| + |A_{n-2}|
   \]
   We can derive:
   \[
   x_n = 4|A_{n-1}| + 2|A_{n-2}}
   \]
   Simplifying further, we get:
   \[
   x_n = x_{n-1} + x_{n-2}
   \]
   with initial conditions \( x_2 = 6 \) and \( x_3 = 10 \).

5. **Initial Conditions**:
   - For \( n = 2 \), the valid sequences are \( (1, 4), (1, 5), (2, 5), (4, 1), (5, 1), (5, 2) \), so \( x_2 = 6 \).
   - For \( n = 3 \), the valid sequences are \( (1, 4, 1), (1, 4, 2), (1, 5, 1), (1, 5, 2), (2, 5, 1), (2, 5, 2), (4, 1, 4), (4, 1, 5), (5, 1, 4), (5, 1, 5) \), so \( x_3 = 10 \).

6. **General Solution**:
   Using the recurrence relation \( x_n = x_{n-1} + x_{n-2} \) with the initial conditions, we can find \( x_n \) for any \( n \geq 2 \).

The final answer is \( \boxed{ x_n = x_{n-1} + x_{n-2} } \) with initial conditions \( x_2 = 6 \) and \( x_3 = 10 \).