Five, (15 points) 1 and 0 alternate to form the following sequence of numbers:
$$
101,10101,1010101,101010101, \cdots
$$

Please answer, how many prime numbers are there in this sequence? And please prove your conclusion.
Obviously, 101 is a prime number.
Below is the proof that $N=\underbrace{101010 \cdots 01}_{k \uparrow 1}(k \geqslant 3)$ are all composite numbers (with $k-1$ zeros in between).
$$
\begin{aligned}
11 N= & 11 \times \underbrace{10101 \cdots 01}_{k \uparrow 1} \\
& =\underbrace{1111 \cdots 11}_{2 k \uparrow 1}=\underbrace{11 \cdots 1}_{k \uparrow 1} \times\left(10^{k}+1\right) .
\end{aligned}
$$
(1) When $k$ is an odd number not less than 3, according to the divisibility rule for 11, 11 does not divide $\underbrace{11 \cdots 11}_{k \uparrow 1}$, so, $111\left(10^{k}+1\right)$, i.e.,
$$
\frac{10^{k}+1}{11}=M_{1}>1 \text {. }
$$

Thus, $N=\underbrace{11 \cdots 11}_{k \uparrow 1} \times\left(\frac{10^{k}+1}{11}\right)=\underbrace{11 \cdots 11}_{k \uparrow 1} \times M_{1}$. Therefore, $N$ is a composite number.
(2) When $k$ is an even number not less than 3, it is easy to see that $11 \mid \underbrace{11 \cdots 11}_{k \uparrow 1}$.

i.e., $\frac{11 \cdots 11}{11}=M_{2}>1$.
Thus, $N=\frac{\overbrace{11 \cdots 11}^{k \uparrow 1}}{11} \times\left(10^{k}+1\right)=M_{2}\left(10^{k}+1\right)$.
Therefore, $N$ is a composite number.
In summary, when $k \geqslant 3$, $N=\underbrace{10101 \cdots 01}_{k \uparrow 1}$ must be a composite number.
Therefore, in the sequence $101,10101,1010101,101010101, \cdots \cdots$, only 101 is a prime number.
(Provided by Zhou Chunluo, Capital Normal University)