Just replying to say you’re right! Thanks!

Thanks for the good attitude!

Ah, I didn't realize 'computable' had a specific meaning in this domain -- showing the limits of my experience a bit. After looking at the wikipedia page and rereading your sketch, I think it is true that

> it eventually needs to grow faster than any computable operator we define

I'm not sure if this has implications about how quickly BB() grows compared to the "operator strength" ladder though? It's familiar/convenient to refer to tetration, pentation, etc as f() for convenience, but this is just notational -- not related to computability. When comparing BB(n) and tetration(n), pentation(n), etc(n), I don't think there is really anything that can be said 'easily' about which values are larger. BB(n) will be larger than anything that can be computed in n steps. But pentation(n) is not necessarily computable in n steps, it may take much more. We may find pentation(n) > BB(n) for all n greater than some threshold. I might be misunderstanding a logical step here though that connects them?

Sure, but the example I'm providing is just meant to illustrate that BB(n) is not greater than arbitrary f(n). I'm not trying to provide the biggest number, I'm trying to illustrate that the sketched-out proof is incorrect. If you want me to provide a bigger number, I suppose another easy example is to define f(n) = BB(BB(n)).

Edit: Oh sorry, I see I misread the direction of your greater than signs. Leaving comments as-is though, hopefully that results in the least confusion