AI 中文总结
针对给定数 N，提出避免因式分解的随机化算法，计算满足条件的除数对 (r,f)，并给出总数公式及比值估计，同时研究 r>N 的情形。
AI 中文摘要
对于给定的数 $N$，我们考虑计算两个整数 $1\leq r,f < N$ 的问题，使得集合 $$\mathcal{X}(N,r,f) = \{(a+b)-(f+\frac{Nr+1}{f}): ab=Nr\}$$ 仅由正整数组成。计算该问题的一个解等价于找到一对 $(r,f)$ 满足 $l(Nr) < f \leq l(Nr+1)$，其中 $l(x)$ 是 $x$ 的不超过 $\sqrt{x}$ 的最大除数。这需要对 $Nr$ 和 $Nr+1$ 进行因式分解。我们提出一个简单的随机化算法，该算法避免因式分解即可计算对 $(r,f)$。我们给出了可能的对 $(r,f)$ 总数的精确公式，并借助经验数据估计比值 $$\frac{\phi(N)-2}{|\mathfrak{F}(N)|}$$ 大约为 $c*\log \log N$。这里，$\mathfrak{F}(N)$ 是所有可能对 $(r,f)$ 中出现的唯一 $r$ 的集合，$\phi(.)$ 是欧拉 totient 函数，$c$ 是一个常数，对于素数 $N$ 等于 2，对于合数 $N$ 则波动较大。作为独立的情况，我们研究相同的问题，即计算满足 $r>N$ 的 $(r,f)$。我们提出一个寻找这样的 $r$ 的过程，该过程需要在算术级数中找到最小素数。
英文摘要
For a given number $N$, we consider the problem of computing two integers $1\leq r,f < N$ such that the set $$\mathcal{X}(N,r,f) = \{(a+b)-(f+\frac{Nr+1}{f}): ab=Nr\}$$ consists only of positive integers. Computing a solution to the problem is equivalent to finding a pair $(r,f)$ satisfying $l(Nr) < f \leq l(Nr+1)$, where $l(x)$ is the largest divisor of $x$ bounded by $\sqrt{x}$. This requires factoring both $Nr$ and $Nr+1$. We present a simple randomized algorithm that - avoiding factoring - computes pairs $(r,f)$. We give an exact formula for the total number of possible pairs $(r,f)$, and with the aid of empirical data we estimate that the ratio $$\frac{ϕ(N)-2}{|\mathfrak{F}(N)|}$$ is approximately about $c*\log \log N$. Here, $\mathfrak{F}(N)$ is the set of unique $r$ appearing among all possible pairs $(r,f)$, $ϕ(.)$ is the Euler's Totient function, and $c$ is a constant equal to 2 for prime $N$ and oscillates much for composite $N$. As a separate and independent case, we study the same problem of computing $(r,f)$ with $r>N$. We present a procedure to find such an $r$, which requires finding the least prime in an arithmetic progression.
