My Number is WAYYY larger [game]

[unname4798]

CARuler and rutabaga, you violated rule 1! (no TREE(x))
New round:
12

[rutabaga]

Ignoring unname's post (I explained why it was okay to post TREE there) I say:

f_{TREE(w)}(8974)

[get_Snacked]

Fine. I'm sorry, but it has to be done: TREE(3).

Ideas to progress: You can make some recursive version of TREE(n), you can whip out another function like SSCG or BB (not RAYO quite yet, please), or you could shove TREE(n) into some sort of thing involving FGH.

Please don't start another round just because I mentioned TREE. I just wanted to change things up a bit because right now it's just a battle of FGH and pure confusion. TREE is not remotely close to unbeatable, it's just unoriginal-- and so is the FGH.

i forgor what TREE's growth rate is i think it's something like e0
also, don't use BB. the busy beaver function is uncomputable.
and lastly, thank you for not starting a new round. i think this is the longest-lasting round so far, no? (as in most turns)
EDIT: the TREE function's growth rate is approximately θ(Ω^ω*ω,0), which, correct me if i'm wrong, is f_{ψ(Ω^(Ω^ω*ω))}(n).

CARuler and rutabaga, you violated rule 1! (no TREE(x))
New round:
12

erm actually rule 1 says "tree(x)" which is a completely different function from "TREE(x)"

[ababa11e]

Ignoring unname's post (I explained why it was okay to post TREE there) I say:

f_{TREE(w)}(8974)

New round. I SPECIFFICALLY said no tree.
CRAZY(n) = f_{ψ(Ω^^n)}(n)
CRAZY(3) >> TREE(3) >> CRAZY(2)

CRAZY(CRAZY(CRAZY...CRAZY(n)...))) iteterated is CRAZY^2[n]
CRAZY^k(n) = CRAZY^k-1(CRAZY^k-1(CRAZY^k-1(...CRAZY^k-1(n)))) iterated n times.

My number is CRAZY^10^100(10^100)

[confocaloid]

Ignoring unname's post (I explained why it was okay to post TREE there) I say:

f_{TREE(w)}(8974)

New round. I SPECIFFICALLY said no tree.
CRAZY(n) = f_{ψ(Ω^^n)}(n)
[...]

My number is LOOK(HOW(CLEVER(I(AM(18))))), where

• LOOK(n) = n!
• HOW(n) = n^^n
• CLEVER(n) = floor((CRAZY^10^n(10^n))^2 + pi^(AM(n))^2)
• I(n) = f_{TREE(w)}(n)
• AM(n) is the number of subsets of {1..n} (counting the empty set) such that all sums of pairs of distinct elements are pairwise distinct

[...]
With all due respect, that title only seems to apply to certain parts of the sandbox, not the forum as a whole.

[rutabaga]

My number is YES(YOU(ARE(VERY(CLEVER(18!!!!!!!!!!!!!!!!!!))))) where:

YES(n) = f{dunno_(LVO)+1}(n)
the ordinal dunno_n = the least fixed point that is a solution to dunno_n = LOOK(HOW(CLEVER(I(AM(LOOK(HOW(CLEVER(I(AM(...dunno_(n-1) times...(dunno_n)...)))))))))), where dunno_0 = w
LVO is the Large Veblen Ordinal
YOU(n) = f{number of w-glider collisions}(n)
ARE(n) = 4n^7 - 9n + 8
VERY(n) = BB_LVO(n); Rayo describes Oracle Turing Machines here: https://web.mit.edu/arayo/www/brink-duel.pdf
CLEVER(n) has already been defined as floor((CRAZY^10^n(10^n))^2 + pi^(AM(n))^2).

[unname4798]

The rules say "No tree(x)".

[CARuler]

The rules say "No tree(x)".

TREE_(TREE(3))(TREE(3))

[rutabaga]

The rules say "No tree(x)".

If every instance of the TREE function and BB function in my number were replaced with the successor function, it would still be larger than the previous number. Also, you've said this multiple times now; we hear you.

TREE_(TREE(3))(TREE(3))

That's much smaller, but at least you're proving a point. Thank you :)

[get_Snacked]

• AM(n) is the number of subsets of {1..n} (counting the empty set) such that all sums of pairs of distinct elements are pairwise distinct

is AM computable? just wanting to make sure, but even then, the owner of the game says "no TREE(x)".

New round. I SPECIFFICALLY said no tree.
CRAZY(n) = f_{ψ(Ω^^n)}(n)
CRAZY(3) >> TREE(3) >> CRAZY(2)

CRAZY(CRAZY(CRAZY...CRAZY(n)...))) iteterated is CRAZY^2[n]
CRAZY^k(n) = CRAZY^k-1(CRAZY^k-1(CRAZY^k-1(...CRAZY^k-1(n)))) iterated n times.

My number is CRAZY^10^100(10^100)

which is exactly f_{ψ(Ω_2)+10^100-1}(10^100).
in that case, I'll use the Buchholz Hydra:
+(0(2))(0(2)){3}, where the rules of +(...){n} are:
if +(...) is a limit ordinal, expand to +(...)[n]{n}
else, add 1 to n
this is H_{ψ(Ω_2)*2}(3) where H is the Hardy hierarchy, which is f_{ψ(Ω_2)*2}(3) as well because the Hardy hierarchy catches up at ψ(Ω).
EDIT: @rutabaga, we still are technically going on the same game because ababa11e's first number in the new round is bigger than the last one.

[eRroR_6o6]

this thread seems to be slowly devolving into insanity

[rutabaga]

No, this is EVOLVING into beautiful mayhem.

I'll use my previous number, which would be larger, but since it seems that it has been ignored, I'll replace every instance of any "invalid functions" (like BB and TREE) with the successor function. [Edit: AM is apparently computable so I won't remove that lol]

If this is unacceptable or if it's somehow smaller, let me know and I'll see if I can come up with something else.

My number is YES(YOU(ARE(VERY(CLEVER(18!!!!!!!!!!!!!!!!!!))))) where:

YES(n) = f{dunno_(LVO)+1}(n)
the ordinal dunno_n = the least fixed point that is a solution to dunno_n = LOOK(HOW(CLEVER(I(AM(LOOK(HOW(CLEVER(I(AM(...dunno_(n-1) times...(dunno_n)...)))))))))), where dunno_0 = w
LVO is the Large Veblen Ordinal
YOU(n) = f{number of w-glider collisions}(n)
ARE(n) = 4n^7 - 9n + 8
VERY(n) = BB_LVO(n); Rayo describes Oracle Turing Machines here: https://web.mit.edu/arayo/www/brink-duel.pdf
CLEVER(n) has already been defined as floor((CRAZY^10^n(10^n))^2 + pi^(AM(n))^2).

[confocaloid]

• AM(n) is the number of subsets of {1..n} (counting the empty set) such that all sums of pairs of distinct elements are pairwise distinct

is AM computable? just wanting to make sure, [...]

https://oeis.org/A196723

[get_Snacked]

YOU(n) = f{number of w-glider collisions}(n)

isn't this undefined?

VERY(n) = BB_LVO(n); Rayo describes Oracle Turing Machines here: https://web.mit.edu/arayo/www/brink-duel.pdf

also, isn't BB_m(n) uncomputable?
EDIT: oh, i see. BB is replaced with the successor function. in that case, that's still undefined--what is "S_LVO(n)"?

[eRroR_6o6]

YOU(n) = f{number of w-glider collisions}(n)

isn't this undefined?

I think it can be defined as an ordinal.

also, I suggest a small change: for a number to be accepted, you need to prove that it is larger than the previous.

[CARuler]

TREE(w^^^^^w)

[confocaloid]

[...] also, I suggest a small change: for a number to be accepted, you need to prove that it is larger than the previous.

If someone is not willing to make efforts contributing something nontrivial and interesting, not willing to search and read about and understand what is already known / what already exists, and so on, then they are unlikely to care about proving to someone else that some not entirely well defined number is larger than another not entirely well defined number.

this thread seems to be slowly devolving into insanity

No, this is EVOLVING into beautiful mayhem. [...]

I think this thread is devolving from a non-CA-related chatty thread into a non-CA-related chatty thread with increased likelihood of further violations of several different forum rules.

[eRroR_6o6]

TREE(w^^^^^w)

aint w infinite therefore tree w^^^^^w is infinite

[CARuler]

TREE(w^^^^^w)

aint w infinite therefore tree w^^^^^w is infinite

Yes

[eRroR_6o6]

you can't just have an ordinal in here (unless it's fgh) ordinals aren't really numbers

[rutabaga]

As long as I'm already breaking every single rule, I'll go with RAYO(FOOT(TREE(BB_dunno_LVO(SSCG(f{dunno_dunno_LVO}(LOOK(HOW(CLEVER(I(AM(YES(YOU(ARE(VERY(CLEVER(Utter Oblivion^^^^^^Meamealokkapoowa Oompa + CRAZY(G(Croutonillion!!!!!!!))))))))))))))))))

[CARuler]

As long as I'm already breaking every single rule, I'll go with RAYO(FOOT(TREE(BB_dunno_LVO(SSCG(f{dunno_dunno_LVO}(LOOK(HOW(CLEVER(I(AM(YES(YOU(ARE(VERY(CLEVER(Utter Oblivion^^^^^^Meamealokkapoowa Oompa + CRAZY(G(Croutonillion!!!!!!!))))))))))))))))))

Ordenials?

[unname4798]

Restart!
37

[rutabaga]

10^37

[confocaloid]

10^37

702^13