This is a Fibonacci-type recurrence:
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- Solution: First, total number of 6-digit numbers with digits only 3 or 7: each digit has 2 choices, so $ 2^6 = 64 $. But 6-digit numbers cannot start with 0, but since digits are only 3 or 7, all 64 are valid.
- Now, subtract those that do **not** contain two consecutive 3s (i.e., no "33" in the number).
- Let $ a_n $ be the number of $ n $-digit sequences of 3 and 7 with no two consecutive 3s.
- Let $ a_n = b_n + c_n $, where:
- $ b_n $: sequences ending in 3
- $ c_n $: sequences ending in 7
