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by btilly
361 days ago
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No, the original was correct. Any computable sequence S(n) must be computed by a specific finite program of fixed length. Once n gets big enough, BB(n) will include the function S(2^n), and therefore will exceed that computable sequence. Given computable sequences may exceed BB(n) for a finite number of terms. But eventually BB(n) will outgrow them, and will never look back. |
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