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by btilly
3691 days ago
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It is unfortunately not that easy. Consider the following function, if n is even then f(n) = BB(n/2), else f(n) = BB(2n). Then f(n) grows faster than any computable function, but still if n is odd, then f(n+1) < f(n). However you have encapsulated the reason why I would be confident of this result. :-) |
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