Since Graham's number is computable, it's an expected result even. The Busy Beaver sequence grows faster than any computable sequence.
No, it's not an obvious result. There is a BB(N) > Graham's Number, but in no way N=10 is obvious.
I'm not sure what you mean, Graham's number is trivially computable since it is a natural number (finite).<p>Graham's number is not a sequence, so it doesn't matter that the BB sequence grows faster than any computable sequence.
Given a constanct C, you can construct a constant sequence f(n) = C.<p>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.
Would we expect that Graham's number is less than 47,176,870, since BB(5) = 47,176,870?
BB(5) is also computable in the intuitive sense, because we proved its value: 47,176,870. In this sense BB(5) is just as computable as Graham's number. However there is likely a Busy Beaver number >5 that is intuitively uncomputable, such that there is no non-cheating algorithm which computes it. A cheating algorithm would be one that requires the knowledge of the result in advance to write the algorithm, in which case we wouldn't be able to come up with the algorithm in the first place, and we wouldn't know that the result is correct.<p>There is also a proof that the function BB(n) (with the free argument n) is uncomputable. However, this alone doesn't mean that e.g. BB(6) (a function with no free parameter) is or isn't intuitively computable. Though if you allow cheating, which is the case for the conventional academic definition of "computability", every particular constant function (that is, without a free parameter) would be considered "computable". Though that definition doesn't mean it is intuitively computable, because if you cheat, you aren't really computing something something, you are just reporting something without having computed it.
I feel like I keep failing to wrap my mind around computability in the same way over and over.<p>Would the following statement be correct?<p>"There is a turing machine that cannot be proven whether it will terminate or not. For any n we can construct a proof that it will run longer than n. But we cannot construct a proof that for any n it will run longer than n. Due to this particular turing machine, BB(k) cannot be determined for k large enough to admit that turing machine."
I don't follow the logic.<p>Every BB(n) is trivially computable, since any integer is computable. So for every BB(n), there is a computable sequence that grows faster than the BB numbers through n (annoyingly, given the word "grows", this can just be the constant sequence of BB(n) + 1).<p>So for any given computable number, the fact that it's computable gives no intrinsic reason to believe it's smaller than BB(n) for any particular n.
No, high enough busy beaver numbers are not computable. They can be defined but there's no way to know what they are.<p>Even for very small n you can create Turing machines where it's incredibly hard to prove that they halt: for n=25 you can create a Turing machine that stops only if the Goldbach conjecture is false.<p>Though that isn't enough to prove it's not computable, the point stands that if you could compute BB(n) for arbitrary n, you could run a Turing machine for BB(n)+1 steps and if it hasn't stopped yet then you know that the input corresponds to a machine that never halts. So it's impossible to write a program that computes BB(n) for arbitrary n, because doing so is equivalent to the halting problem.<p>And worse, there are some Turing machines where "proving the Turing machine stops" means "proving that a given set of axioms is consistent", where Gödel's incompleteness theorem says that it's impossible to build such a proof. So, that is how there is some n after which the busy beaver number is defined, but is not computable at all.
What I mean is that any given integer is trivially computable.<p>The busy beaver sequence is not computable (I am aware that somewhere around BB(700), you've encoded the consistency of ZFC). You cannot necessarily compute that BB(n) is a given number. But you can compute larger numbers. There is no such thing as integer that is too large to be computable.<p>I did slightly fuck up--the English language "every BB(n) is computable" is a bit ambiguous. What I should say is: "for all n, BB(n) is trivially computable, since it is a single integer".
> for all n, BB(n) is trivially computable, since it is a single integer".<p>It is not computable, because for high enough n you can't prove that you got the right integer.<p>And also you can prove that if a sequence a_n is always larger than BB(n) for some n, that sequence a_n is also not computable.
You can always exhaustively run all Turing machines of a given size, at each step comparing the state & tape to all previous steps. If they ever reach the same state with the same tape as a previous step, they're in an infinite loop and never halt. If not, they eventually reach the halt state. BB(n) is computable. But BB is not computable, because there's no general algorithm requiring finite storage and time which computes the results of BB(n) for all n. You have to brute-force it for every positive integer n, which takes infinite time (and space) and is therefore not computable.
Is BB(n) "trivially computable"?