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.

Someone a day ago | parent | next [-]

> Given a constanct C, you can construct a constant sequence f(n) = C.

> Since f(n) is trivially computable

That only is true if C is computable.

Integers are what's relevant to BB, and all integers are computable.

	▲	Scarblac a day ago \| parent [-]
		So what if we set C = BB(10)?

Ah I see now, thank you.