The easiest way to construct the recurrence relation for this one was to realize that, in
promoting a legal ternary string from length n-1 to another legal string of length n, we can
append either of two characters to the end. Specifically, if a legal string of length n - 1 ends
in a 0, then we can append either a 1 or a 2 to the end to get a legal string of length n. This
leads to the simple formula: