1. **Problem statement:** Find the 9th term $a_9$ of the recursive sequence defined by: $$a_1 = a_2 = 1, \quad a_n = a_{n - \lfloor n/2 \rfloor} + a_{n - \lfloor n/2 \rfloor},$$ and then find $a_{n+1}$. 2. **Understanding the recursion:** The formula simplifies to $$a_n = 2 \times a_{n - \lfloor n/2 \rfloor}$$ because the same term is added twice. 3. **Calculate terms step-by-step:** - $a_1 = 1$ - $a_2 = 1$ - $a_3 = 2 \times a_{3 - \lfloor 3/2 \rfloor} = 2 \times a_{3-1} = 2 \times a_2 = 2 \times 1 = 2$ - $a_4 = 2 \times a_{4 - \lfloor 4/2 \rfloor} = 2 \times a_{4-2} = 2 \times a_2 = 2 \times 1 = 2$ - $a_5 = 2 \times a_{5 - \lfloor 5/2 \rfloor} = 2 \times a_{5-2} = 2 \times a_3 = 2 \times 2 = 4$ - $a_6 = 2 \times a_{6 - \lfloor 6/2 \rfloor} = 2 \times a_{6-3} = 2 \times a_3 = 2 \times 2 = 4$ - $a_7 = 2 \times a_{7 - \lfloor 7/2 \rfloor} = 2 \times a_{7-3} = 2 \times a_4 = 2 \times 2 = 4$ - $a_8 = 2 \times a_{8 - \lfloor 8/2 \rfloor} = 2 \times a_{8-4} = 2 \times a_4 = 2 \times 2 = 4$ - $a_9 = 2 \times a_{9 - \lfloor 9/2 \rfloor} = 2 \times a_{9-4} = 2 \times a_5 = 2 \times 4 = 8$ 4. **Find $a_{10}$:** $$a_{10} = 2 \times a_{10 - \lfloor 10/2 \rfloor} = 2 \times a_{10-5} = 2 \times a_5 = 2 \times 4 = 8$$ 5. **Answer:** The 9th term $a_9$ is $8$ and the 10th term $a_{10}$ is also $8$. This matches the pattern of doubling the term at the index shifted by the floor of half the current index.