How many different ways are there to tile dominos of size 2x1 on a 3x2n grid?
I'm regards to this question Dominos ($2 \times 1$ on $2 \times n$ and on $3 \times 2n$)
I understand the logic behind the first part, but I don't understand how the function $g_{n}$ helps you, and why you would add it twice to the previous number of ways
Because there are two ways to cover the left-most 3 squares so that the remaining uncovered squares are in the configuration of a $3\times(2n-1)$ board with one corner removed. There is only one way to cover the left-most three squares so that what remains uncovered is a $3\times(2n-2)$ board. This gives $f_n = 2g_{n-1}+f_{n-1}$.
On the other hand, given a $3\times(2n+1)$ board with one corner removed, the other two on that edge can be either covered by one domino, and you are left with a $3\times 2n$ board, or you they are covered by two dominos, which leaves a straggler in the next row, which has to be covered to remove a corner from the remaining. So you have $g_{n}=f_n+g_{n-1}$