If the first digit is 4, the remaining $n-1$ digits form a valid sequence: $a_{n-1}$
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- 2^8 = 256
- Now count the number of such numbers that do **not** contain two consecutive 3s. Let $a_n$ be the number of valid sequences of length $n$ with no two consecutive 3s, where each digit is 3 or 4.
- We define a recurrence:
- If the first digit is 3, the next must be 4, and the rest $n-2$ digits form a valid sequence: $a_{n-2}$
- a_1 = 2 \quad (\text{3 or 4}),\quad a_2 = 3 \quad (\text{34, 43, 44})
- Compute up to $a_8$:
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