Three, a deck of cards consists of 52 cards, among which, "Diamonds", "Clubs", "Hearts", and "Spades" each have 13 cards, numbered sequentially as $2,3, \cdots, 10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}$. Cards of the same suit and adjacent numbers are called "straight flush" cards, and $\mathrm{A}$ can be considered as 1, thus $\mathrm{A}$ and 2 are also considered as straight flush cards. Determine the number of ways to draw 13 cards from this deck such that each number appears and no straight flush cards are included.
（Supplied by Tao Pingsheng）
$$
\begin{array}{l}
\text { Let } n \geqslant 3 \text {, from } \\
A=\left(a_{1}, a_{2}, \cdots, a_{n}\right), B=\left(b_{1}, b_{2}, \cdots, b_{n}\right) \text {, } \\
C=\left(c_{1}, c_{2}, \cdots, c_{n}\right), D=\left(d_{1}, d_{2}, \cdots, d_{n}\right) \\
\end{array}
$$

Select $n$ terms from these four sequences, satisfying:
(1) Each index $1,2, \cdots, n$ appears;
(2) Any two terms with adjacent indices do not come from the same sequence (indices $n$ and 1 are considered adjacent).
Let the number of selection methods be $x_{n}$. We will determine the expression for $x_{n}$.
As shown in Figure 5, 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 circular disk with four colors such that no two adjacent sectors have the same color. It is easy to see that
$$
\begin{array}{l}
x_{1}=4, x_{2}=12, \\
x_{n}+x_{n-1}=4 \times 3^{n-1}(n \geqslant 3) .
\end{array}
$$

Rewrite equation (1) as
$$
\begin{array}{l}
(-1)^{n} x_{n}-(-1)^{n-1} x_{n-1}=-4 \times(-3)^{n-1} \text {. } \\
\text { Hence }(-1)^{n-1} x_{n-1}-(-1)^{n-2} x_{n-2}=-4 \times(-3)^{n-2} ; \\
\cdots \cdots \\
(-1)^{3} x_{3}-(-1)^{2} x_{2}=-4 \times(-3)^{2} ; \\
(-1)^{2} x_{2}=-4 \times(-3) .
\end{array}
$$

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