3 A deck of cards consists of 52 cards, with 13 cards of each of the four suits: "Diamonds," "Clubs," "Hearts," and "Spades." The cards are numbered $2, 3, \cdots, 10, J, Q, K, A$, where the same suit and adjacent numbered cards are called "sequential cards of the same suit," and $A$ can be considered as 1 (i.e., $A$ can be used as 1). Determine the number of ways to draw 13 cards from this deck such that each number appears and no "sequential cards of the same suit" are included. (Provided by Tao Pingsheng)
The problem can be generalized as follows: Let $n \geqslant 3$, and consider the four sequences
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), B=\left(b_{1}, b_{2}, \cdots, b_{n}\right), \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right), D=\left(d_{1}, d_{2}, \cdots, d_{n}\right)
\end{array}
$$

Select $n$ terms from these sequences such that:
(i) Each index from $1, 2, \cdots, n$ appears exactly once;
(ii) No two adjacent terms (with indices $n$ and 1 considered adjacent) come from the same sequence. The number of such selection methods is denoted as $x_{n}$.
Now, determine the expression for $x_{n}$:
Divide a circular disk into $n$ sectors, numbered $1, 2, \cdots, n$ in sequence, and color the sequences $A, B, C, D$ with four different colors. For any selection method, if the term with index $i$ is from a certain colored sequence, then the $i$-th sector is colored with that color.

Thus, $x_{n}$ becomes the number of ways to color the $n$ sectors of the disk with four colors such that no two adjacent sectors have the same color. It is easy to see that:
$$
\begin{array}{c}
x_{1}=4, x_{2}=12, \\
x_{n}+x_{n-1}=4 \cdot 3^{n-1}(n \geqslant 3),
\end{array}
$$

Rewrite (1) as $(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-4 \cdot(-3)^{n-1}$.
Thus, $(-1)^{n-1} x_{n-1}-(-1)^{n-2} x_{n-2}=-4 \cdot(-3)^{n-2}$;
$$
\begin{array}{c}
(-1)^{3} x_{3}-(-1)^{2} x_{2}=-4 \cdot(-3)^{2} ; \\
(-1)^{2} x_{2}=-4 \cdot(-3) .
\end{array}
$$

Adding these equations, we get $(-1)^{n} x_{n}=(-3)^{n}+3$, so $x_{n}=3^{n}+3 \cdot(-1)^{n}(n \geqslant 2)$.
Therefore, $x_{13}=3^{13}-3$. This is the number of selection methods.