Provide a formula for number of binary sequences of length n that have no consecutive 0's.

Let f(n) be the number of n-digit sequences with no 2 consecutive 0's. There are 2 1-digit sequences with no consecutive 0's. They are 0 and 1. Therefore, f(1)=2. There are 3 2-digit sequences with no consecutive 0's. They are 01, 10, and 11. Therefore, f(2)=3. Take any n>3. If an n-digit sequence starts with 1, then any n-1-digit sequence with no consecutive 0's can be the last n-1 digits. If an n-digit sequence starts with 0, then the next digit must be 1 to avoid 2 consecutive 0's. Then, any n-2-digit sequence with no consecutive 0's can be the last n-2 digits. Therefore, f(n)=f(n-1)+f(n-2). Then, f(n)=Fibonacci(n+2).

Edited on October 6, 2018, 9:09 am

Posted by Math Man on 2018-10-06 09:03:20