Final answer:
A recurrence relation for the number of ternary strings of length n without two consecutive 0s is a(n) = a(n-1) * 3 + a(n-2), with initial conditions a(0) = 1 and a(1) = 3.
Step-by-step explanation:
The question is asking for a recurrence relation for the number of ternary strings of length n that do not contain two consecutive 0s. Let's denote the number of such strings of length n as a(n). We can build a ternary string of length n by appending a digit to a string of length n-1 provided it does not create a string with two consecutive 0s.
There are two cases to consider:
If the string of length n-1 ends in a 0, the next digit can only be a 1 or 2. So, the number of such strings is equal to a(n-2) because we need to avoid double zeros.
If the string of length n-1 ends in a 1 or 2, we can append any of the three digits. So, the number of such strings is 3 times a(n-1).
Combining these cases gives the recurrence relation: a(n) = a(n-1) * 3 + a(n-2).
For the initial conditions, consider:
a(0) = 1, as there is one empty string that does not contain any consecutive zeros.
a(1) = 3, since with a string of length 1 we can have '0', '1', or '2' without violating our condition.