Given a constanct C, you can construct a constant sequence f(n) = C.
Since f(n) is trivially computable, we know that BB(n) grows faster than f(n). It follows that there must be some n such that BB(n) > C.
Since f(n) is trivially computable, we know that BB(n) grows faster than f(n). It follows that there must be some n such that BB(n) > C.
> Since f(n) is trivially computable
That only is true if C is computable.
However, the BB function itself is not computable. It makes sense to ask "for what inputs is the BB function computable?". When people talk about BB(n) not being computable, they are really talking about the computability of the BB function at n, not the computability of the integer that happens to be BB(n).