Combinatorics and T-tetromino In how many ways can I fill a grid $4 \times N$ with only T-tetrominoes? 
my idea is that it's possible to fill it if and only if $N$ is a multiple of $4$, in that case the solution is:
\begin{equation}
\bigg \{
\begin{array}{rl}
2^{N/4} &  N \text{ is multiple of } 4,\\
0 & \text{otherwhise}.\\
\end{array}
\end{equation}
Is that correct? It seems too easy!
 A: At first we can show that if $N$ is not divisible by $4$ the answer is $0$. 
Let's consider $4$ types of cells identified by diagonals:
$$\begin{array}{|c|c|c|c|c}
\hline
1 & 2 & 3 & 4 & 1 & \\\hline
4 & 1 & 2 & 3 & 4 & \\\hline
3 & 4 & 1 & 2 & 3 & \\\hline
2 & 3 & 4 & 1 & 2 & \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
It is easy to see that any $4 \times N$ field has equal number of cell of each type. Each T-tetramino takes $2$ cells of one type (let's call it main type), $1$ cell of each of two other types, the $0$ cells of the remaning type. This means that filling has the same number of tetraminoes for each of $4$ main types. Therefore total number of cells is divisible by $16$ and then $N$ is divisible by $4$.
Let's call a filling solid if each inner vertical line (crossing $4$ lines) contains inner point of some tetramino. Each filling can be splitted into a set of solid fillings. If $f(N)$ is the number of all fillings of field $4 \times N$ and $g(N)$ is the number of solid fillings of the same field, then $$f(n) = \sum_{K = 1}^{N/4}g(4K)f(N - 4K).$$
Now let's compute $g(N)$. Upper left corner cell can be covered in two ways:
$$\begin{array}{|c|c|c|c}
\hline
1 & 1 & 1 & ~ & \\\hline
  & 1 &   &   & \\\hline
  &   &   &   & \\\hline
  &   &   &   & \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}\qquad
\begin{array}{|c|c|c|c}
\hline
1 &   &  ~ & \\\hline
1 & 1 &    & \\\hline
1 &   &    & \\\hline
  &   &    & \\\hline
\end{array}\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
That implies
$$\begin{array}{|c|c|c|c}
\hline
1 & 1 & 1 &   & \\\hline
2 & 1 &   &   & \\\hline
2 & 2 & 3 &   & \\\hline
2 & 3 & 3 & 3 & \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}\qquad
\begin{array}{|c|c|c|c}
\hline
1 & 3 & 3 & 3 & \\\hline
1 & 1 & 3 &   & \\\hline
1 & 2 &   &   & \\\hline
2 & 2 & 2 &   & \\\hline
\end{array}\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
These two cases are symmetrical reflection of each other, so we can consider any of them and multiply result by $2$. If $N = 4$ we can place one more tetramino to fininish filling:
$$\begin{array}{|c|c|c|c|}
\hline
1 & 1 & 1 & 4 \\\hline
2 & 1 & 4 & 4 \\\hline
2 & 2 & 3 & 4 \\\hline
2 & 3 & 3 & 3 \\\hline
\end{array}$$
If $N > 4$ the only way to continue is the following:
$$\begin{array}{|c|c|c|c}
\hline
1 & 1 & 1 & 4 & 7 & 7 & 7 &   \\\hline
2 & 1 & 4 & 4 & 4 & 7 &   &   \\\hline
2 & 2 & 3 & 5 & 5 & 5 & 6 &   \\\hline
2 & 3 & 3 & 3 & 5 & 6 & 6 & 6 \\\hline
\end{array}
\begin{array}{c}
\cdots\\\cdots\\\cdots
\end{array}$$
This is absolutely the same situation: for $N = 8$ we can finish filling and for $N > 8$ we have exactly $1$ way to continue. Therefore $g(N) = 2$.
$f(4) = 2$ and we have
$$f(N) = \sum_{K = 1}^{N / 4}g(4K)f(N - 4K) = 2\sum_{K = 1}^{N / 4}f(N - 4K)\\
 = 2f(N - 4) + 2\sum_{K = 1}^{N / 4 - 1}f(N - 4 - 4K) = 3f(N - 4).$$
Therefore we have $f(N) = [4 \mid N]2 \cdot 3^{N / 4 - 1}$ for $N \ge 1$ and $f(0) = 1$.
A: I think your answer is acceptable. 
I think you mean to have T-tetronimos made of 4 bricks as in the picture.
This way I can see just 2 solutions for N=4 and 2^2 solutions for N = 8.
There are no solutions for N%4 != 0.
See the image below:

