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.

翻译：暂无翻译