44. Let $E$ be the set of all even numbers. For any real number $x$, define $d(x, E)$ as the distance from $x$ to the nearest even number, such as $d(1.2, E) = 2 - 1.2 = 0.8$, $d(0.7, E) = 0.7$, etc. Given the sequence $\{x_n\}$ satisfying $x_{n+1} = d(2x_n, E)$, $x_1 = \frac{1}{p}$, where $p$ is an odd prime. Find the period $T$ of the sequence $\{x_n\}$ (in this problem, the period $T$ of the sequence is the smallest positive integer such that there exists $N$ for which $n > N$ implies $x_{n+T} = x_n$).
Let $\mathbb{R}$ and $\mathbb{Z}$ denote the sets of all real numbers and all integers, respectively. Any real number $x$ can be expressed as $x = 2k \pm b$, where $k \in \mathbb{Z}$ and $0 \leq b \leq 1$. By the problem statement, we should have $d(x, E) = b$.

First, we prove the three properties of $d(x, E)$:

(1) $\forall y \in \mathbb{R}, k \in \mathbb{Z}, d(2k \pm y, E) = d(y, E)$.
Indeed, let $y = 2L \pm b, L \in \mathbb{Z}, 0 \leq b \leq 1$, then
the left side $= d(2k \pm 2L \mp b, E) = d(2(k \pm L) \mp b, E) = b$,
the right side $= d(2L \pm b, E) = b$.
$\therefore$ (1) holds.

(2) $\forall x \in \mathbb{R}, d(2d(x, E), E) = d(2x, E)$.
Let $x = 2k \pm b, k, b$ have the same meaning as above.
The left side $= d(2b, E)$.
The right side $= d(2(2k \pm b), E) = d(4k \pm 2b, E)$
$= d(2b, E)$ (by (1)),
$\therefore$ (2) holds.

(3) $d\left(2^{-1} d(2x, E), E\right) = d\left(2^{-1} x, E\right)$.
Let $x = 2k \pm b, k, b$ have the same meaning as above.
The left side $= d\left(2^{n-1} d(2(2k \pm b), E), E\right)$
$= d\left(2^{n-1} d(2b, E), E\right)$ (by (1))
If $0 \leq b \leq \frac{1}{2}$, then
the above expression $= d\left(2^{n-1} \cdot 2b, E\right) = d\left(2^n b, E\right)$.
If $\frac{1}{2} < b \leq 1$, then $0 \leq 2 - 2b < 1$.
$$
\begin{aligned}
\text{the above expression} & = d\left(2^{n-1} d(2 - (2 - 2b), E), E\right) \\
& = d\left(2^{n-1} (2 - 2b), E\right) = d\left(2^n - 2^n b, E\right) \\
& = d\left(2^n b, E\right) \text{ (by (1)), } \\
\text{the right side} & = d\left(2^n (2k \pm b), E\right) = d\left(2^{n+1} k \pm 2^n b, E\right) \\
& = d\left(2^n b, E\right).
\end{aligned}
$$
$\therefore$ (3) holds.

Now, we prove the original problem.
$$
\begin{aligned}
x_n & = d\left(2x_{n-1}, E\right) = d\left(2d\left(2x_{n-2}, E\right), E\right) \\
& = d\left(4x_{n-2}, E\right) \text{ (by (2))} \\
& = d\left(2^n d\left(2x_{n-3}, E\right), E\right) = d\left(2^3 x_{n-3}, E\right) \\
& = \cdots = d\left(2^{n-1} x_1, E\right) \\
& = d\left(\frac{2^{n-1}}{p}, E\right).
\end{aligned}
$$

Since $p$ is an odd prime, by Fermat's Little Theorem, we know $2^{p-1} \equiv 1 \pmod{p}$, so for sufficiently large $n$, $2^{n+p-2} \equiv 2^{n-1} \pmod{p}$.

Thus, $\frac{2^{n+p-2}}{p} = \frac{2^{n-1}}{p} + 2k, k \in \mathbb{N}$ (when $n$ is large, $2^{n+p-2} - 2^{n-1}$ is even).

From the above, we can derive $x_{n+p-1} = x_n$. Therefore, the period $T = p - 1$ (T is not necessarily the smallest period).
Note: The editor made significant modifications to the solution.
(Xie Qidao, Wang Liqing, Fudan University Mathematics Department 93rd Grade, 200433)