Number of ternary sequences ${0,1,2}$ of length n without two consecutive even numbers.

Question

(I edited the question and erased my last try, cause my understanding of it, was poor)

any help would be appreciated.

Answer 1

As @Element118 says above, the recurrence will be $a_n=a_{n-1}+2a_{n-2}$ with initial conditions $a_0=1$ and $a_1=3$.

It is worth going through the effort to find the closed form as well, as this is where the difficulty usually comes into play.

We find our characteristic polynomial is $x^2-x-2=0=(x-2)(x+1)$

Our equation should then look something like $a_n=c_1(2)^n+c_2(-1)^n$ for some appropriate numbers $c_1$ and $c_2$.

Using our initial conditions, we have the following system of equations:

$\begin{cases}1=c_1+c_2\\3=2c_1-c_2\end{cases}$

Solving gives $c_1=\frac{4}{3}$ and $c_2=-\frac{1}{3}$ for a final answer of

$$a_n = \frac{4}{3}(2)^n-\frac{1}{3}(-1)^n$$

Answer 2

Your previous method of solving this still stands. If $a_n$ is the number of sequences of length $n$, we can break it down into cases:

If you start with $0$ or $2$, the next number has to be $1$, this case has $2a_{n-2}$ ways.

Otherwise, you start with $1$, this case has $a_{n-1}$ ways.

The base cases would be $a_0=1$ and $a_1=3$.