我想我知道为什么科拉茨会这样做。
这个术语通常被称为“狂躁的自大妄想”，所以请帮我一下：
3n走过的同余类 +1 每一步都不对齐 /2 从未改变这个事实<p>积累互质性是阻止n重复的“记忆”。这在2^k下是不变的，使其能够在混乱中向前推进。耗尽模3的同余类迫使n成为2的幂；游戏结束。<p>我让人工智能证明这些事情，它做到了。我想。<p>那么唯一的问题就是1.5n是否能足够快地增长残差向量，以超越耗尽的步行。我也问了人工智能，得到了一个一页的p词。我甚至不想打出来。<p>我可以用甜言蜜语让人工智能同意几乎任何事情，所以我被困住了。这是我知道的唯一一个始终有深思熟虑对话的论坛，而我无法想出解决办法，而且我还有真正的工作要做。<p>这里有数学家吗？p词在评论中。
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The normal term for this is "manic delusion of grandeur", so please help me out:
3n walks congruence classes +1 steps each one out of alignment /2 never changes that fact<p>Accumulating co-primeness is the "memory" that stops n from repeating. This is invariant under 2^k, which allows it to make forward progress through the chaos. Exhausting congruence classes mod 3 forces n to a power of 2; game over.<p>I asked AI to prove those things, and it did. I assume.<p>The only question then would be if 1.5n can grow the residue vector quickly enough to outrun the exhaustive walk. I asked AI that too, and got back a one page p-word. I'm not even going to type it.<p>I can sweet-talk AI into agreeing with damned near anything, so I'm stuck. This is the only forum I know with consistently thoughtful conversation, and I can't think my way out of this one, and I have real work to do.<p>Is there a mathematician in the house? The p-word is in a comment.