One way to do it is to make a set of coupled recurrences. Imagine building your $2 \times 2 \times n$ prism sitting on the table extending away from you. Let $A(n)$ be the number of ways to build a prism. Now think you fill the cube nearest you, with priority to top left, top right, bottom left, bottom right in that order. Define a new variable for each configuration of cubes beyond a solid block, so $B(n)$ is the number of ways to make a prism of length $n$ plus a $1 \times 1 \times 2$ sticking horizontally away from you. You have $B(n)=A(n)$ because you can't get this shape except by starting with a complete prism and adding one block. There are two other ways you can add a block to a full prism, horizontally or vertically. Those are two more configurations. Now imagine adding a block to each. Eventually there won't be any more configurations and you have your coupled recurrences. Now you can eliminate variables to get down to some minimal number. There are two recurrences given in the formulas to the OEIS sequence. You can check your result against those.