Suppose that a sequence $t_{0}, t_{1}, t_{2}, ...$ is defined by a formula $t_{n} = An^{2} +Bn +c$ for all integers $n \geq 0$. Here $A, B$ and $C$ are real constants with $A \neq 0$. Determine values of $A, B$ and $C$ which give the greatest possible number of successive terms of the Fibonacci sequence.[i] The Fibonacci sequence is defined by[/i] $F_{0} = 0, F_{1} = 1$ [i]and[/i] $F_{m} = F_{m-1} + F_{m-2}$ [i]for[/i] $m \geq 2$.
To determine the values of \(A\), \(B\), and \(C\) that give the greatest possible number of successive terms of the Fibonacci sequence, we start by assuming that the sequence \(t_n = An^2 + Bn + C\) matches the Fibonacci sequence for some consecutive terms.

1. **Assume the sequence matches the Fibonacci sequence for 5 consecutive terms:**
   \[
   t_n, t_{n+1}, t_{n+2}, t_{n+3}, t_{n+4}
   \]
   Given the Fibonacci property:
   \[
   t_{n+2} = t_{n+1} + t_n
   \]
   \[
   t_{n+3} = t_{n+2} + t_{n+1}
   \]
   \[
   t_{n+4} = t_{n+3} + t_{n+2}
   \]

2. **Express the terms using the quadratic formula:**
   \[
   t_n = An^2 + Bn + C
   \]
   \[
   t_{n+1} = A(n+1)^2 + B(n+1) + C = An^2 + (2A + B)n + (A + B + C)
   \]
   \[
   t_{n+2} = A(n+2)^2 + B(n+2) + C = An^2 + (4A + B)n + (4A + 2B + C)
   \]
   \[
   t_{n+3} = A(n+3)^2 + B(n+3) + C = An^2 + (6A + B)n + (9A + 3B + C)
   \]
   \[
   t_{n+4} = A(n+4)^2 + B(n+4) + C = An^2 + (8A + B)n + (16A + 4B + C)
   \]

3. **Set up the equations based on the Fibonacci property:**
   \[
   t_{n+2} = t_{n+1} + t_n
   \]
   \[
   An^2 + (4A + B)n + (4A + 2B + C) = [An^2 + (2A + B)n + (A + B + C)] + [An^2 + Bn + C]
   \]
   Simplifying:
   \[
   An^2 + (4A + B)n + (4A + 2B + C) = 2An^2 + (3A + 2B)n + (2A + 2B + 2C)
   \]
   Equating coefficients:
   \[
   A = 2A \implies A = 0 \quad (\text{contradiction since } A \neq 0)
   \]

4. **Assume the sequence matches the Fibonacci sequence for 4 consecutive terms:**
   \[
   t_n, t_{n+1}, t_{n+2}, t_{n+3}
   \]
   Using the same approach:
   \[
   t_{n+2} = t_{n+1} + t_n
   \]
   \[
   t_{n+3} = t_{n+2} + t_{n+1}
   \]

5. **Set up the equations:**
   \[
   An^2 + (4A + B)n + (4A + 2B + C) = [An^2 + (2A + B)n + (A + B + C)] + [An^2 + Bn + C]
   \]
   Simplifying:
   \[
   An^2 + (4A + B)n + (4A + 2B + C) = 2An^2 + (3A + 2B)n + (2A + 2B + 2C)
   \]
   Equating coefficients:
   \[
   A = 2A \implies A = 0 \quad (\text{contradiction since } A \neq 0)
   \]

6. **Assume the sequence matches the Fibonacci sequence for 3 consecutive terms:**
   \[
   t_n, t_{n+1}, t_{n+2}
   \]
   Using the same approach:
   \[
   t_{n+2} = t_{n+1} + t_n
   \]

7. **Set up the equations:**
   \[
   An^2 + (4A + B)n + (4A + 2B + C) = [An^2 + (2A + B)n + (A + B + C)] + [An^2 + Bn + C]
   \]
   Simplifying:
   \[
   An^2 + (4A + B)n + (4A + 2B + C) = 2An^2 + (3A + 2B)n + (2A + 2B + 2C)
   \]
   Equating coefficients:
   \[
   A = 2A \implies A = 0 \quad (\text{contradiction since } A \neq 0)
   \]

8. **Assume the sequence matches the Fibonacci sequence for 2 consecutive terms:**
   \[
   t_n, t_{n+1}
   \]
   Using the same approach:
   \[
   t_{n+1} = t_n + t_{n-1}
   \]

9. **Set up the equations:**
   \[
   An^2 + (2A + B)n + (A + B + C) = [An^2 + Bn + C] + [A(n-1)^2 + B(n-1) + C]
   \]
   Simplifying:
   \[
   An^2 + (2A + B)n + (A + B + C) = An^2 + Bn + C + A(n^2 - 2n + 1) + B(n-1) + C
   \]
   Equating coefficients:
   \[
   A = A \quad (\text{true})
   \]
   \[
   2A + B = B \implies 2A = 0 \quad (\text{contradiction since } A \neq 0)
   \]

10. **Assume the sequence matches the Fibonacci sequence for 1 term:**
    \[
    t_n
    \]
    This is trivially true for any \(A\), \(B\), and \(C\).

Thus, the maximum number of consecutive Fibonacci terms that can be matched by the sequence \(t_n = An^2 + Bn + C\) is 4. The values of \(A\), \(B\), and \(C\) that achieve this are:
\[
\boxed{(A, B, C) = \left(\frac{1}{2}, -\frac{1}{2}, 2\right) \text{ or } \left(\frac{1}{2}, \frac{1}{2}, 2\right)}
\]