What specifically doesn't sound right?

The claim is that every bb(n) is computable but I don’t think you can compute bb(6) without knowing which machines won’t halt. That doesn’t seem like a finite calculation?

But given the answer, I suppose you could write a program that just returns it. This seems to hinge on the definition of “computable.” It’s an integer, so that fits the definition of a computable number.

My mistake.

	▲	Kranar 7 days ago \| parent \| next [-]
		Yes exactly, imagine a function HH(n) that returns 0 if the Turing machine represented by the integer n halts, and 1 if it doesn't. Then HH the function itself is not computable, but the numbers 0 and 1, which are the only two outputs of HH are computable. Integers themselves are always computable, even if they are the output of functions that are themselves uncomputable.
	▲	tux3 7 days ago \| parent \| prev \| next [-]
		Yes. A valid question for a specific n would be whether you can prove the value of BB(n). If you don't care about provability, you can indeed just produce a number that happens to be the right one. So as you noticed, it only makes sense to talk about whether a function is computable, we can't meaningfully talk of computable numbers.
	▲	edanm 6 days ago \| parent \| prev [-]
		The main thing to make it clear is that BB(n) for a specific n isn't a function - it's a number. Just like Mult(10,4) isn't a function, it's a number (40). So a specific BB(n) is just a number and is computable.