Ways of solving recurrence relation for $a(15)$

Question

I have this recurrence relation:

$$ R(1)=1, RE(1)=0, EE(1)=0$$

$$a(n)=R(n) + RE(n)$$

$$R(n)=EE(n-1)+RE(n-1),$$$$ RE(n)=R(n-1),$$$$ EE(n)=RE(n-1) $$

How do I get $a(15)$? What kind of method do I use?

Answer 1

Substitute $EE(n)$ from the last equation:

$$R(n)=RE(n-2)+RE(n-1).$$

Then substitute $RE(n)$:

$$R(n)=R(n-3)+RE(n-2) \ \ \ (1)$$

$$a(n)=R(n) + R(n-1) \ \ \ (2)$$

Now solve the recurrent equation ($1$) and then ($2$).

Answer 2

$$R(n)=EE(n-1)+RE(n-1)=RE(n-2)+RE(n-1)=R(n-3)+R(n-2) \,.$$ $$RE(n)=R(n-1)=EE(n-2)+RE(n-2)=RE(n-3)+RE(n-2) \,.$$

Adding them you get

$$a(n)=a(n-3)+a(n-2) \,.$$

Solve it!