AI 中文总结
本文研究在固定循环运算顺序下从2出发构造1到N所有整数的最小步数，证明N-1步对N≥9足够，并给出N≤33的显式序列。
AI 中文摘要
从数字 $2$ 开始。每一步，将两个已经生成的数组合起来，但运算必须按固定的循环顺序 $+,\times,-,÷$ 使用。我们问：生成从 $1$ 到 $N$ 的每个整数所需的最少步数是多少。由于 $2$ 已经是目标数之一，且每一步最多生成一个新数，因此至少需要 $N-1$ 步。我们证明对于所有 $N\ge 9$，$N-1$ 步也足够。传统的归纳法无法奏效，因为在一个完整区间 $\{1,\dots,P\}$ 之后紧接着的除法会产生已经生成的数。相反，我们将完整区间 $\{1,\dots,P\}$ 一次性扩展到 $\{1,\dots,3P\}$，计数表明这是可行的最小乘法扩展 $P\to kP$，然后调整最后几步以到达所有其他 $N$。对于 $9\le N\le 33$，我们给出通过计算机搜索找到的显式序列。
英文摘要
Start with the number $2$. At each move, combine two numbers already made, but the operations must be used in the fixed repeating order $+,\times,-,÷$. We ask for the fewest moves needed to make every integer from $1$ to $N$. Since $2$ is already one of the numbers we want and each move makes at most one new number, at least $N-1$ moves are needed. We show that $N-1$ moves are also enough for every $N\ge 9$. Conventional induction cannot work, because a division that comes right after a completed interval $\{1,\dots,P\}$ produces numbers already made. Instead we extend a completed interval $\{1,\dots,P\}$ to $\{1,\dots,3P\}$ all at once, which counting shows is the smallest multiplicative extension $P\to kP$ that can work, and then adjust the last few moves to reach every other $N$. For $9\le N\le 33$ we give explicit sequences, found by computer search.
Comments25 pages, 2 figures, 9 tables. Verification code: https://github.com/poscle/AMSD-Number-Generator
