This market gets recommended to me way too often, so let me comment on it.

BB(8000) is independent of ZFC [Yedidia, Aaronson, 2016], and for any given currently-known large cardinal axiom H, it's reasonable to say that BB(10^5) is independent of ZFC+H (or use BB(10^100) if you want to be really conservative). So in what sense does the top-rated describe a single number? Does "1 if Con(ZFC), otherwise 0" descibe a number? I would say it does not: ZFC is widely believed to be consistent, but if so, we will never be able to prove it (without assuming a stronger theory), and so we won't be able to prove that this number is 1. In the same way I'd argue that BB(8000) doesn't describe a number.

The fifth-rated answer (using TREE) is tiny compared to anything involving Rayo's function, since you can prove in ZFC that TREE is finite, and hence define the tree function. Since you can also define composition and Graham's number, you should be easily able to write this answer with less than 9! symbols in ZFC. So Rayo(9!) should already be much larger than this answer.

The third answer (composition of a lot of Rayo) is clearly much larger than Rayo(9!), so I would say that answer wins, if the answers using BB are indeed disqualified.

@FlorisvanDoorn Actually, thinking about it more, I believe that my problem with the busy beaver function also applies to Rayo numbers. For some formula it is independent of ZFC whether it specifies a number. Looking over googology some people don't think this is an issue if you don't work with provability but with truth in a model, but then this depends on a model, which itself we cannot be proven to exist. And even if we could know the existence of such a model, the answer is still model-dependent...

@FlorisvanDoorn Although it will be impossible to prove any specific number is BB(8000), there does exist a program which can be specified in finite symbols that will produce it, by definition. Furthermore, it is very easy to produce lower bounds for BB(8000).

@FlorisvanDoorn Rayo() grows faster than BB(), so I don't think it makes sense to consider the former fair game and not the latter.

@IsaacKing You are right, I missed that in my first post. I would suggest that you disqualify both, because of the rule "If I have to ask you for clarification in order to know what number you're describing, that answer is invalid."

These answers depend on the answer of the question "In what model of set theory are you evaluating this number?" or perhaps "What axiom system do you use to prove that this is a concrete number (a numeral)". The answers involving Rayo and BB depend on the answer to these clarifications.

@A is right that any lower bound for the questions involving BB and Rayo are much larger than the value of the TREE number, but I think that is irrelevant since the precise value is ill-defined without clarification.

Alright, down to three answers that could be valid. I've been procrastinating determining which is actually the largest.

Let me convert them all into similar notation for ease of comparison. Define Comp(f, input, repetitions) to mean applying function f to input, then applying f to the result, and repeat until f has been applied this way repetition times. Comp(f, 2, 3) = f(f(f(2)))

Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number), then square the answer.

=Rayo(Comp(BB, G64, G64 + 2))^2

Apply TREE to 9, BIG-G times. Where TREE is Friedman's function in the context of Kruskal's tree theorem, and BIG-G is TREE applied to Graham's Number... then square the answer.

=Comp(TREE, 9, TREE(G64))^2

notation F[n] = composition of func F with itself n times. R=RayosFunc

Extending Graham's-number-like towers

Lyr(1)=R(9!), Lyr(n)=R[Lyr(n-1)](9!)

Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9)

Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9)

Row[Row(9)](9)

=Comp(Row, 9, Row(9) + 1), where:

Lyr(1) = Rayo(9!)

Lyr(n>1) = Comp(Rayo, 9!, Lyr(n-1) + 1)

Twr(1) = Lyr(9)

Twr(n>1) = Comp(Rayo, 9, Twr(n-1) + 1)

Row(1) = Twr(9)

Row(n>1) = Comp(Rayo, 9, Row(n-1) + 1)

(Is there a simpler way to define this that involves calling Comp() on itself? I feel like there might be, but I couldn't figure it out.)

@IsaacKing Heuristically feels like the last one is biggest, because it's calling Rayo on itself a Rayo-y number of times, whereas the first is calling Rayo once on BB called on itself a Graham-y number of times, and Rayo >> BB >> Graham.

@Tetraspace Agreed. It's also the only answer that tried to use its character limit efficiently; the others were extremely lazy.

I don't think it's conclusive though, because it starts by just calling Rayo on 9!, which may well be smaller than Graham's Number. I'd like to be a little more certain before resolving to it.

@IsaacKing The good news is that if a breakthrough in the philosophy of numbers shows a different answer you can now petition the admins to re-resolve it.

@YaakovSaxon I'm going to predict that this ends up being the winner, since it's the only answer that made any effort to use its character allocation well.

Let BBC(N) be BB composed with itself N times onto N, that is BB(BB(... BB(N)...)) with N layers of BB

I believe this is ambiguous due to an off-by-one issue. The first part of your definition, "Let BBC(N) be BB composed with itself N times onto N", defines the number of function compositions as N. But the second part of the definition, "BB(BB(... BB(N)...)) with N layers of BB", defines the number of function applications as N. So it's unclear whether BBC(3) = BB(BB(BB(3))) or BB(BB(BB(BB(3)))).

I'm therefore going to disqualify this answer, unless someone points out something I missed.

Apply TREE to 9, BIG-G times

This is potentially ambiguous. I assume it means "Apply TREE to 9. Take the result and apply TREE to it. Take the result of that and apply TREE to it. Repeat this process Big-G times", but it doesn't actually say that. What it says is to perform a function call several times, and get that many different answers, and then it says to "square the answer". I'm tempted to disqualify this for being too vague, but that also seems like I might be being too pedantic.

@IsaacKing Yeah I think that's too pedantic; it's completely clear what you mean.

Ok, there are 4 answers that cannot be trivially ruled out:

Rayo(Graham's Number passed to the busy beaver function, with the result of that computation passed to the busy beaver function - repeat a number of times equal to Graham's Number), then square the answer.

Apply TREE to 9, BIG-G times. Where TREE is Friedman's function in the context of Kruskal's tree theorem, and BIG-G is TREE applied to Graham's Number... then square the answer.

notation F[n] = composition of func F with itself n times. R=RayosFunc

Extending Graham's-number-like towers

Lyr(1)=R(9!), Lyr(n)=R[Lyr(n-1)](9!)

Twr(1)=Lyr(9), Twr(n)=Lyr[Twr(n-1)](9)

Row(1)=Twr(9), Row(n)=Twr[Row(n-1)](9)

Row[Row(9)](9)

BB=Busy Beaver Function. Let BBC(N) be BB composed with itself N times onto N, that is BB(BB(... BB(N)...)) with N layers of BB. Similarly let BBCC be BBC composed with itself N times onto N. Answer is BBCC(BBCC(BBCC(BBCC(Rayo's number))))

Figuring out which of these is the largest seems doable. I will attempt to do it over the course of the next few weeks.

@a Guaranteed to be smaller than @MattP's answer that squares your result.

@a Guaranteed to be smaller than @MattP's answer that squares this one.

@wavedash Invalid due to an infinite loop.

@Ibozz91 Not with that attitude!

@WilliamEhlhardt Invalid due to an infinite loop.

@jorge Invalid due to an infinite loop.

@MartinRandall Invalid

@JamesGrugett Trivially smaller than other answers.

@a Invalid

@IsaacKing Is it valid now that Jorge sold his shares?

@Yev No, since the other answer still exists and can be chosen.