How many ways are there to cover a $2\times 16$ rectangle with $2\times 2,$ $2\times 3$ and $2\times 4$ rectangles? 
How many ways are there to cover a $2\times 16$ rectangle with $2\times 2,$ $2\times 3$ and $2\times 4$ rectangles?

I already dealt with a similar problem, which is how many ways are there to cut a $1\times 8$ rectangle into $1\times 1$ and $1\times 2$ rectangles. The answer to this problem can be calculated using the number of ways to divide $1\times k$ strip for $3\le k\le8$ sequentially, to arrive at $34$. However, this problem is completely different. How can I solve this problem?
 A: 
If you have a $2 \times n$ rectangle then call $a_n$ the number of ways to cover that rectangle with $2\times p$ ($p\in \{2,3,4\}$) rectangles. 
If you start with $2\times 2$ then you will face the problem $a_{n-2}$, but if you start with $2\times 3$ you will find the problem $a_{n-3}$ and finally if you start with $2\times 4$ you will face $a_{n-4}$ then you have
$$a_{n}=a_{n-2}+a_{n-3}+a_{n-4}$$
and
$$a_2=1\\
a_3=1\\
a_4=2$$
and we also need define $a_1=0$.
now you can go forward and find $a_{16}$.
A: We are dealing with a $2 \times 16$ grid. The first column of two cells can be filled by a $2 \times 2$, a $2 \times 3$ or a $2 \times 4$ grid. Let us call $f(n)$ the number of ways to fill a $2 \times n$ grid. We then have:
$$f(n) = f(n-2) + f(n-3) + f(n-4), n \geq 4$$
We also know that:
$$f(0) = 1$$
$$f(1) = 0$$
$$f(2) = 1$$
$$f(3) = 1$$
A closed form is not straightforward, but we find:
$$f(4) = 2$$
$$f(5) = 2$$
$$f(6) = 4$$
$$f(7) = 5$$
$$f(8) = 8$$
$$f(9) = 11$$
$$f(10) = 17$$
$$f(11) = 24$$
$$f(12) = 36$$
$$f(13) = 52$$
$$f(14) = 77$$
$$f(15) = 112$$
$$f(16) = 165$$
