21.146 **A deck of cards consists of 52 cards, with 13 cards of each suit: "Diamonds," "Clubs," "Hearts," and "Spades." The cards are numbered $2, 3, \cdots, 10, J, Q, K, A$, where cards of the same suit and consecutive numbers are called "straight flush cards," 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 rank appears and no "straight flush cards" are included.**
Parse the problem as follows: Let $n \geqslant 3$, from
$$
\begin{array}{l}
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), B=\left(b_{1}, b_{2}, \cdots, b_{n}\right), \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right), D=\left(d_{1}, d_{2}, \cdots, d_{n}\right)
\end{array}
$$

select $n$ terms from these four sequences, satisfying:
(1) Each index $1,2, \cdots, n$ appears;
(2) Any two adjacent terms do not come from the same sequence (index $n$ and 1 are considered adjacent), and 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 sequentially as $1,2, \cdots, n$, and color the sequences $A, B, C, D$ with four different colors. For any selection scheme, 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 circular 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
$$
(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-4 \cdot(-3)^{n-1} \text {, }
$$

Therefore,
$$
\begin{array}{c}
(-1)^{n-1} x_{n-1}-(-1)^{n-2} x_{n-2}=-4 \cdot(-3)^{n-2}, \\
\cdots \cdots \\
(-1)^{3} x_{3}-(-1)^{2} x_{2}=-4 \cdot(-3)^{2}, \\
(-1)^{2} x_{2}=-4 \cdot(-3) .
\end{array}
$$

Adding these, we get $(-1)^{n} x_{n}=(-3)^{n}+3$, thus $x_{n}=3^{n}+3 \cdot(-1)^{n}(n \geqslant 2)$.
Therefore, $x_{13}=3^{13}-3$ is the number of ways to select the cards.