Given a $3 \times 3$ symmetric real matrix $A$, we define $f(A)$ as a $3 \times 3$ matrix with the same eigenvectors of $A$ such that if $A$ has eigenvalues $a$, $b$, $c$, then $f(A)$ has eigenvalues $b+c$, $c+a$, $a+b$ (in that order). We define a sequence of symmetric real $3\times3$ matrices $A_0, A_1, A_2, \ldots$ such that $A_{n+1} = f(A_n)$ for $n \geq 0$. If the matrix $A_0$ has no zero entries, determine the maximum number of indices $j \geq 0$ for which the matrix $A_j$ has any null entries.
1. **Spectral Decomposition of \( A \)**:
   Given a symmetric real matrix \( A \), we can use spectral decomposition to write:
   \[
   A = Q \begin{pmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{pmatrix} Q^T = Q D Q^T
   \]
   where \( Q \) is an orthogonal matrix and \( D \) is a diagonal matrix with eigenvalues \( a, b, c \).

2. **Definition of \( f(A) \)**:
   The function \( f(A) \) is defined such that if \( A \) has eigenvalues \( a, b, c \), then \( f(A) \) has eigenvalues \( b+c, c+a, a+b \). Using the spectral decomposition, we can write:
   \[
   f(A) = Q \begin{pmatrix} b+c & 0 & 0 \\ 0 & c+a & 0 \\ 0 & 0 & a+b \end{pmatrix} Q^T
   \]
   Notice that:
   \[
   f(A) = Q (\text{Tr}(A)I - D) Q^T = \text{Tr}(A)I - A
   \]
   where \(\text{Tr}(A) = a + b + c\).

3. **Inductive Formula for \( f^n(A) \)**:
   We can establish the following formula for \( f^n(A) \) by induction:
   \[
   f^n(A) = \begin{cases} 
   \frac{2^n - 1}{3} \text{Tr}(A) I + A & \text{if } n \equiv 0 \pmod{2} \\
   \frac{2^n + 1}{3} \text{Tr}(A) I - A & \text{if } n \equiv 1 \pmod{2}
   \end{cases}
   \]

4. **Non-zero Off-diagonal Entries**:
   The off-diagonal entries of \( f^n(A) \) are clearly non-zero because they are derived from the orthogonal transformation of the diagonal matrix.

5. **Diagonal Entries and Null Entries**:
   Assume \(\text{Tr}(A) \neq 0\). Let the diagonal entries of \( A \) be \( a, b, c \). Suppose there are three even \( n \) such that \( f^n(A) \) has a null entry. This implies:
   \[
   x(a+b+c) + a = 0
   \]
   \[
   y(a+b+c) + b = 0
   \]
   \[
   z(a+b+c) + c = 0
   \]
   The only solution to these equations is \( a = b = c = 0 \), which contradicts the assumption that \( A_0 \) has no zero entries.

6. **Maximum Number of Indices**:
   Thus, there can be at most 2 even \( n \) such that \( f^n(A) \) has a null entry. Similarly, there can be at most 2 odd \( n \) such that \( f^n(A) \) has a null entry. If there are 3 overall indices, then 2 must be even and 1 must be odd, or vice versa. However, this leads to a contradiction as shown by the linear equations.

7. **Example**:
   Consider:
   \[
   A_0 = \begin{pmatrix} 1 & \pi & \pi \\ \pi & 5 & \pi \\ \pi & \pi & -7 \end{pmatrix}
   \]
   Using the formula, it can be shown that \( A_2 \) and \( A_4 \) both have a null entry.

The final answer is \(\boxed{2}\)