Start with the number . At each move, combine two numbers already made, but the operations must be used in the fixed repeating order . We ask for the fewest moves needed to make every integer from to . Since is already one of the numbers we want and each move makes at most one new number, at least moves are needed. We show that moves are also enough for every . Conventional induction cannot work, because a division that comes right after a completed interval produces numbers already made. Instead we extend a completed interval to all at once, which counting shows is the smallest multiplicative extension that can work, and then adjust the last few moves to reach every other . For we give explicit sequences, found by computer search.

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生成AI智能校验的中文版本