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[alemagno12]

You just have to create new notations, bro.
Here's one:
n[0] = f_Γ0(n)
n[0][0][0]...[0][0][0] = f_Γ0(n[0][0][0]...[0][0])
n[000...000] = n[000...00][000...00][000...00]...[000...00][000...00][000...00] with n [000...00]'s
n[000...000][0][0][0]...[0][0][0] = f_Γ0(n[000...000][0][0][0]...[0][0])
n[000...000][000...00][000...00][000...00]...[000...00][000...00][000...00] = n[000...000][000...00][000...00][000...00]...[000...00][000...00][000...0][000...0][000...0]...[000...0][000...0][000...0] with n [000...0]'s
Number = 100[000...000] with 100 0's

[alemagno12]

You just have to create new notations, bro.
Here's one:
n[0] = f_Γ0(n)
n[0][0][0]...[0][0][0] = f_Γ0(n[0][0][0]...[0][0])
n[000...000] = n[000...00][000...00][000...00]...[000...00][000...00][000...00] with n [000...00]'s
n[000...000][0][0][0]...[0][0][0] = f_Γ0(n[000...000][0][0][0]...[0][0])
n[000...000][000...00][000...00][000...00]...[000...00][000...00][000...00] = n[000...000][000...00][000...00][000...00]...[000...00][000...00][000...0][000...0][000...0]...[000...0][000...0][000...0] with n [000...0]'s
Number = 100[000...000] with 100 0's

Exactly. It's just being redundant at this point.

Also, f_Σ3. I win.

Define f_Σ(n)

[alemagno12]

Is that uncomputable?

[alemagno12]

Is that uncomputable?

Sigma in this case is not. However, it does collapse into other ordinals, which gives it the ability to be computed relative to the problem. Even the earlier example with gamma- gamma is an ordinal infinity that, in the case of fast-growing hierarchies, collapses into others that eventually result in a value being computed. In other words, it doesn't need to be.

EDIT: Actually, that is a bit unfair. Let me do a different equation:

f(ω₁^ck,G), where ω₁^ck is the supremum of computable ordinals.

Epic Fail.
f(ω₁^ck,G) is uncomputable.
So it breaks the rules.

[alemagno12]

Is that uncomputable?

Sigma in this case is not. However, it does collapse into other ordinals, which gives it the ability to be computed relative to the problem. Even the earlier example with gamma- gamma is an ordinal infinity that, in the case of fast-growing hierarchies, collapses into others that eventually result in a value being computed. In other words, it doesn't need to be.

EDIT: Actually, that is a bit unfair. Let me do a different equation:

f(ω₁^ck,G), where ω₁^ck is the supremum of computable ordinals.

Epic Fail.
f(ω₁^ck,G) is uncomputable.
So it breaks the rules.

f(ψ_(εΩ+1),G), where the ordinal is the Bachmann–Howard ordinal.

Ok then.
f_ψ(ψI(0))(G), where G is Graham's Number.

[alemagno12]

Prove it. Also, i changed my entry.

[FallenAngel]

A(G,G), where A is the Ackermann function, and G is Graham's Number.
This might not be larger than what you have, so I've developed a one-up to my own number.

A(A(A(A(A(A(...(G,G)...), where the number of A() is equal to A(A(A(A(...(G,G)...), in which the number of A() is Graham's Number.

[alemagno12]

Not even even even even EVEN EVEN close to my number.

[FallenAngel]

Hmmm...

4HIKJHHGFTEWTFTGBY in base A(G,G) might be larger.
Probably not, but it is worth a shot.

[alemagno12]

nope

[FallenAngel]

I'm willing to dispute your statements that A(G,G) is smaller than your number.
Given that a fairly small set of values for the Ackermann Function can and will give large values, say A(4,1) (which equals 65533, or (2^2^2^2)-3, or 2 tetrated to 3 minus three), A(G,G) can also be written as A(G-1, A(G, G-1)), which can also be written as A(G-2, A(G-1, A(G, G-1)-1)). This holds no real significance, I just found it interesting.
A less confusing way of representing the equation would be 2(↑^(G-2)) (G+3)-3. I'd make that more legible, but I honestly don't know how to do that thing to make the math functions as seen on Wikipedia/Mediawiki.
Because of how Knuth's up-arrow notation works, there would be G-2 up-arrows - since one arrow takes the first number to the power of the second, two tetrates the first to the second, and so on and so fort, that would lead to 2 being taken to the hyper-G of (G-3), with 3 subtracted from that.
That is a very, very large number. It dwarfs Graham's Number by a long shot. If tetration (or hyper-4, its other name) can develop large numbers, with pentation (hyper-5) creating even larger numbers, hyper-G will easily create obnoxiously large numbers, even if the base is infinitesimally larger than 1. Since A(G,G) is sometimes called the "xkcd number", I shall call hyper-G the xkcd hyperoperation.

EDIT: Miscalculated how up-arrows relate to hyperoperators.

[alemagno12]

Ugh.
A(G,G) is aproximately f_ω(G).
And your other number is aproximately f_ω+1(f_ω+1(G)).
Neither of them are close to my number.

[MagmaMcFry]

I'm willing to dispute your statements that A(G,G) is smaller than your number.

Apparently you have no idea how large large numbers can actually be.

First of all, A(G) = A(G,G) = f_ω(G), and A^G(G) = f_ωⁿ(G) = f_ω+1(G), and your number is A^AG(G)(G) < A^AG(G)(A^G(G)) = f_ω+1(f_ω+1(G)) = f_ω+1²(G) < f_ω+1^G(G) = f_ω+2(G). Note how incredibly unexact those inequalities are. And since ω+2 is so very very much smaller than ψ(ψ_I(0)), your number is so incredibly much smaller than f_ψ(ψI(0))(G) you can't even imagine how hard it would be to even come up with this shit in the first place.

EDIT: Suddenly I know so much about ordinals. Thanks, thread.

EDIT2: Aaaaaaactually, after reading through the definition of fast-growing hierarchies again, my previous statements are all not really true. Since alemagno's number is dependent on the choice of the fast-growing hierarchy (which is defined by the choice of fundamental sequences of limit ordinals) and since both ψ(ψ_I(0)) and ω are limit ordinals, fψ(ψ_I(0))(G) could very well be smaller than f_ω+1(G), namely if you choose the first G elements of ω's and ψ(ψ_I(0))'s limit sequences to both be the first G integers, then A^AG(G)(G) > f_ω+1(G) > f_ω(G) = f_ω[n](G) = f_G(G) = f_{ψ(ψI(0))[n]}(G) = f_ψ(ψI(0))(G). So alemagno, you need to specify your fundamental sequences first before making such claims, because your number was not completely defined.

[FallenAngel]

I know how large numbers can get - I just wanted to pry some info out of the situation.
Going by the numbers that have been derived from the xkcd function, I have a new, larger number.
f_ω+1(f_ω+1(G)), my large ...thing made of nested Ackermann functions with the root of it all being A(G,G), is not close to your number, or f_ψ(ψI(0))(G).
Now, to take this information. Since f_ω+1(f_ω+1(G)) is approximately A(A(A(A(A(A(...(G,G)...), where the number of A() is equal to A(A(A(A(...(G,G)...), in which the number of A() is Graham's Number,  the two can technically be interchangeable - we're at large enough numbers where most differences between similar numbers can be ignored safely - A(G,G)-1 can be written as A(G,G) to no ill effect.
From this, my new number is f_ω+1(f_ω+1(f_ω+1(f_ω+1(f_ω+1(...(G)...), in which function the f_ω+1(...) are nested G taken to the xkcd hyperoperation taken to the xkcd hyperoperation times. Plus 1. Because one is my seventh-favorite number, and 7 is considered "lucky".

[MagmaMcFry]

In that case your number is f_ω+1^{G↑G-2G↑G-2..↑G-2G}(G) = f_ω+1^G↑G-1(G+1)(G) < f_ω+1^G↑G-1(G+1)(G↑^G-1(G+1)) = f_ω+2(G↑^G-1(G+1)) = f_ω+2((G+1)↑^G2) < f_ω+2((2↑^GG)↑^G2) < f_ω+2(2↑^G(2↑^G2)) = f_ω+2(2↑^G+13) < f_ω+2(2↑^G+1(G+2)) < f_ω+2(f_G+2(G+2)) = f_ω+2(f_ω(G+2)) < f_ω+2(f_ω(f_ω(G))) < f_ω+3(G). At least for the standard (Wainer) definition of f_ω. Try harder, bro.