The 59th question: Given any positive integer $a$, define an integer sequence $x_{1}, x_{2}, x_{3}, \ldots$, such that $x_{1}=a, x_{n+1}=2 x_{n}+1$, and let $y_{n}=2^{x_{n}}-1$. Determine the largest possible integer $\mathrm{k}$ such that for some positive integer $a, y_{1}, y_{2}, \ldots, y_{k}$ are all prime numbers.
The 59th question:
Answer: The $\mathrm{k}_{\max }=2$ satisfies the conditions of the problem. The proof is given below.
We first prove a lemma: If $y=2^{x}-1$ is a prime number, then $\mathrm{x}$ is also a prime number.
Indeed, if $x=1$, then $y=1$, which is not a prime number. If $x$ is a composite number, take any prime factor $p$ of $x$, then $p \mid x$, and $12 x_{2}+1=x_{3}$, thus $y_{2}$ is a composite number, which contradicts the fact that $y_{1} 、 y_{2} 、 y_{3}$ are all prime numbers.
In summary, $\mathrm{k}_{\text {max }}=2$.
Note: Since $\left(\frac{2}{23}\right)=\left(\frac{5^{2}}{23}\right)=1$, according to Euler's criterion, we know: $23 \mid\left(2^{11}-1\right)$.