Consider the following sequence `:a _(1) = a _(2) =1, a _(i) = 1 +` minimum ` ( a _(i-1) , a _(i-2))` for `i gt 2.` Then `a _( 2006)=`

Answer

To solve the problem step by step, we will analyze the sequence defined by the recurrence relation:

1. **Initial Conditions**:
- We are given that \( a_1 = 1 \) and \( a_2 = 1 \).

2. **Recurrence Relation**:
- For \( i > 2 \), the sequence is defined as:
\[ a_i = 1 + \min(a_{i-1}, a_{i-2}) \]

3. **Calculate the Next Terms**:
- Let's compute the next few terms of the sequence to identify a pattern:
- \( a_3 = 1 + \min(a_2, a_1) = 1 + \min(1, 1) = 1 + 1 = 2 \)
- \( a_4 = 1 + \min(a_3, a_2) = 1 + \min(2, 1) = 1 + 1 = 2 \)
- \( a_5 = 1 + \min(a_4, a_3) = 1 + \min(2, 2) = 1 + 2 = 3 \)
- \( a_6 = 1 + \min(a_5, a_4) = 1 + \min(3, 2) = 1 + 2 = 3 \)
- \( a_7 = 1 + \min(a_6, a_5) = 1 + \min(3, 3) = 1 + 3 = 4 \)
- \( a_8 = 1 + \min(a_7, a_6) = 1 + \min(4, 3) = 1 + 3 = 4 \)

4. **Identifying the Pattern**:
- From the computed values, we can observe the following pattern:
- The sequence appears to repeat each integer twice:
- \( 1, 1, 2, 2, 3, 3, 4, 4, \ldots \)
- This suggests that for every integer \( n \), \( a_{2n-1} = a_{2n} = n \).

5. **Finding \( a_{2006} \)**:
- Since \( 2006 \) is an even index, we can express it as:
\[ a_{2006} = a_{2 \times 1003} = 1003 \]

6. **Final Answer**:
- Therefore, the value of \( a_{2006} \) is:
\[ \boxed{1003} \]