Comment by btilly

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.