Now, suppose we knew the Nth Busy Beaver number, which we’ll call BB(N). Then we could decide whether any Turing machine with N rules halts on a blank tape. We’d just have to run the machine: if it halts, fine; but if it doesn’t halt within BB(N) steps, then we know it never will halt, since BB(N) is the maximum number of steps it could make before halting. Similarly, if you knew that all mortals died before age 200, then if Sally lived to be 200, you could conclude that Sally was immortal. So no Turing machine can list the Busy Beaver numbers—for if it could, it could solve the Halting Problem, which we already know is impossible.

But here’s a curious fact. Suppose we could name a number greater than the Nth Busy Beaver number BB(N). Call this number D for dam, since like a beaver dam, it’s a roof for the Busy Beaver below. With D in hand, computing BB(N) itself becomes easy: we just need to simulate all the Turing machines with N rules. The ones that haven’t halted within D steps—the ones that bash through the dam’s roof—never will halt. So we can list exactly which machines halt, and among these, the maximum number of steps that any machine takes before it halts is BB(N).

Conclusion? The sequence of Busy Beaver numbers, BB(1), BB(2), and so on, grows faster than any computable sequence. Faster than exponentials, stacked exponentials, the Ackermann sequence, you name it. Because if a Turing machine could compute a sequence that grows faster than Busy Beaver, then it could use that sequence to obtain the D‘s—the beaver dams. And with those D’s, it could list the Busy Beaver numbers, which (sound familiar?) we already know is impossible. The Busy Beaver sequence is non-computable, solely because it grows stupendously fast—too fast for any computer to keep up with it, even in principle.

https://www.scottaaronson.com/writings/bignumbers.html

EDIT: For fun I converted it to Rust and expected to see it spew a few million numbers into my terminal, but no, this equivalent loop actually terminates after 15 steps, which is fascinating given that the Turing machine takes 47 million steps:

    let mut x = 0;
    loop {
        x = match x % 3 {
            0 => 5 * x + 18,
            1 => 5 * x + 22,
            _ => break
        } / 3;
    }

In the old days we used to just chop wood, and burn it to keep warm. Then sit down and watch the sportsball game on TV to waste time.

[0] https://web.archive.org/web/20251027173129/https://benbrubak...

The sequence is (truly/fairly) random in its distribution of mods 1/2.

Even fair coins flipped infinitely would - on occasion - have arbitrary long results of heads or tails.

So the question becomes, is the anti-hydra sequence 'sufficiently' random?