Let $n$ be a positive integer and $\mathcal{P}_n$ be the set of integer polynomials of the form $a_0+a_1x+\ldots +a_nx^n$ where $|a_i|\le 2$ for $i=0,1,\ldots ,n$. Find, for each positive integer $k$, the number of elements of the set $A_n(k)=\{f(k)|f\in \mathcal{P}_n \}$.

[i]Marian Andronache[/i]
1. **Case \( k = 1 \)**:
   - Consider the polynomial \( f(x) = a_0 + a_1 x + \ldots + a_n x^n \) where \( |a_i| \leq 2 \) for \( i = 0, 1, \ldots, n \).
   - When \( k = 1 \), the polynomial simplifies to \( f(1) = a_0 + a_1 + \ldots + a_n \).
   - Each coefficient \( a_i \) can take any integer value from \(-2\) to \(2\), so the sum \( a_0 + a_1 + \ldots + a_n \) ranges from \(-2(n+1)\) to \(2(n+1)\).
   - Therefore, the set \( A_n(1) \) contains all integers from \(-2(n+1)\) to \(2(n+1)\), inclusive.
   - The number of elements in \( A_n(1) \) is \( 4(n+1) + 1 = 4n + 5 \).

2. **Case \( k = 2 \)**:
   - Consider the polynomial \( f(x) = a_0 + a_1 x + \ldots + a_n x^n \) where \( |a_i| \leq 2 \).
   - When \( k = 2 \), the polynomial becomes \( f(2) = a_0 + a_1 \cdot 2 + a_2 \cdot 2^2 + \ldots + a_n \cdot 2^n \).
   - Using only \( 0 \) and \( 1 \) for \( a_i \), we can generate all integers from \( 0 \) to \( 2^{n+1} - 1 \).
   - By including negative coefficients, we can generate all integers from \( -2(2^{n+1} - 1) \) to \( 2(2^{n+1} - 1) \).
   - Therefore, the set \( A_n(2) \) contains all integers from \( -2(2^{n+1} - 1) \) to \( 2(2^{n+1} - 1) \), inclusive.
   - The number of elements in \( A_n(2) \) is \( 1 + 2 \cdot (2^{n+1} - 1) = 2^{n+2} - 1 \).

3. **Case \( k = 3 \)**:
   - Consider the polynomial \( f(x) = a_0 + a_1 x + \ldots + a_n x^n \) where \( |a_i| \leq 2 \).
   - When \( k = 3 \), the polynomial becomes \( f(3) = a_0 + a_1 \cdot 3 + a_2 \cdot 3^2 + \ldots + a_n \cdot 3^n \).
   - Using only \( 0 \) and \( 1 \) for \( a_i \), we can generate all integers from \( 0 \) to \( 3^{n+1} - 1 \).
   - By including negative coefficients, we can generate all integers from \( -2(3^{n+1} - 1) \) to \( 2(3^{n+1} - 1) \).
   - Therefore, the set \( A_n(3) \) contains all integers from \( -2(3^{n+1} - 1) \) to \( 2(3^{n+1} - 1) \), inclusive.
   - The number of elements in \( A_n(3) \) is \( 1 + 2 \cdot (3^{n+1} - 1) = 2 \cdot 3^{n+1} - 1 \).

4. **Case \( k = 4 \)**:
   - Consider the polynomial \( f(x) = a_0 + a_1 x + \ldots + a_n x^n \) where \( |a_i| \leq 2 \).
   - When \( k = 4 \), the polynomial becomes \( f(4) = a_0 + a_1 \cdot 4 + a_2 \cdot 4^2 + \ldots + a_n \cdot 4^n \).
   - Using only \( 0 \) and \( 1 \) for \( a_i \), we can generate all integers from \( 0 \) to \( 4^{n+1} - 1 \).
   - By including negative coefficients, we can generate all integers from \( -2(4^{n+1} - 1) \) to \( 2(4^{n+1} - 1) \).
   - Therefore, the set \( A_n(4) \) contains all integers from \( -2(4^{n+1} - 1) \) to \( 2(4^{n+1} - 1) \), inclusive.
   - The number of elements in \( A_n(4) \) is \( 1 + 2 \cdot (4^{n+1} - 1) = 2 \cdot 4^{n+1} - 1 \).

5. **Case \( k > 4 \)**:
   - Consider the polynomial \( f(x) = a_0 + a_1 x + \ldots + a_n x^n \) where \( |a_i| \leq 2 \).
   - For \( k > 4 \), distinct polynomials in \( \mathcal{P}_n \) have distinct values at \( k \).
   - If \( a_0 + a_1 k + \ldots + a_n k^n = b_0 + b_1 k + \ldots + b_n k^n \), then \( (a_0 - b_0) + (a_1 - b_1) k + \ldots + (a_n - b_n) k^n = 0 \).
   - Taking modulo \( k \), we conclude \( a_0 = b_0 \). Canceling a factor of \( k \) and repeating the process, we find \( a_i = b_i \) for all \( i \).
   - Therefore, \( |A_n(k)| = |\mathcal{P}_n| = 5^{n+1} \).

The final answer is \( \boxed{ 5^{n+1} } \).