Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[alemagno12]

n^^m = n^n^n^...^n^n^n with n exponentiated n times
n^^^m = n^^n^^n^^...^^n^^n^^n with n tetrated n times
n^^^...^^^m = n^^^...^^n^^^...^^n^^^...^^ ... ^^^...^^n^^^...^^n^^^...^^n with m levels
n^(x)^m = n^^^...^^^m with x arrows
Omniverse = 999^(999)^999

[alemagno12]

Grahamverse = G(G(G(...(999)...))) with 999 levels, where G(n) is defined as follows:
G(0) = 4
G(n+1) = 3^(G(n))^3

[FallenAngel]

Graham's Number octated to the Graham's Number.
Basically Graham's Number septrated Graham's Number times to the Graham's Number.
Think tetration, but four levels deeper. Let me work out the math, it'll take a bit.

It's g_64 heptrated by g_64 g_64 times, which would be g_64 hextated by g_64 to the power of g_64 g_64 times. I think.
That'd be g_64 pentated by g_64 tetrated by g_64 g_64 times, which'd be... and I've gone cross-eyed

[alemagno12]

Not even close to my number.

[FallenAngel]

x/x^3, when x equals 10 to the power of a negative googolplexian taken to the googolplexianth power.
In math notation, x = 10^-((10^10^10^100)^(10^10^10^100)) in the function x/x^3.
As the function x/x^3 increases dramatically in y value as x becomes smaller but not negative, this will be a painfully large number.

[Putnam]

Uh... we allowed to use numbers that aren't known, but are known to be really really big?

[FallenAngel]

In Base 2, 4 is unnaturally large.
Not 100 4, 4 4.
As in the number usually only possible in Base 5 and above.
Just like C is a large number in Base 10, even though it only can properly exist in Base 13 and above.

[Putnam]

Uh... we allowed to use numbers that aren't known, but are known to be really really big?

I suspect that the number I just posted is likely larger. By a lot. But probably!

S(Ω(3->3->100))?

[FallenAngel]

In Base 2, 4 is unnaturally large.
Not 100 4, 4 4.
As in the number usually only possible in Base 5 and above.
Just like C is a large number in Base 10, even though it only can properly exist in Base 13 and above.

That's not how bases work. 4 doesn't exist in base 2. Well, it does= 0100. If you have a 4 somewhere in there, you failed to convert bases- it isn't unnaturally large, it's incorrect.

It was a joke, sort of like "On a scale of 1 to 10, that's a 15".

[Putnam]

It's known that it's unable to be known because it's bigger than anything that can be known, ergo... it's a game ender, heh.

[alemagno12]

Damn, let me finish this:
Using Ispil's Omega Function:
Ω₀(n) = Ω(n)
Ω_m+1(n) = Ω_m(Ω_m(Ω_m(...(n)...))) with n levels
Number is Ω_G(G), where G is Graham's Number.

[MagmaMcFry]

[wrong]

[iceball3]

Σ(10↑↑10) = n
Uh, wait, does being unprovably finite make it finite or does it not count?

[alemagno12]

I think Ispil meaned f_Γ0(G) (see http://en.wikipedia.org/wiki/Fast-growing_hierarchy )

[MagmaMcFry]

Σ(10↑↑10) = n
Uh, wait, does being unprovably finite make it finite or does it not count?

Well, since it's defined as the largest of a nonempty finite set of finite numbers, it too is finite, and I just proved it, so it's also provably finite. But it definitely isn't the largest, because Putnam mentioned S(Ω(3->3->100)) earlier, and since S is monotonous and Ω(3->3->100) > 10↑↑10, S(Ω(3->3->100)) > S(10↑↑10) >= Σ(10↑↑10).

I think Ispil meaned f_Γ0(G) (see http://en.wikipedia.org/wiki/Fast-growing_hierarchy )

Yes, but since Γ₀ isn't finite, neither is f_Γ0(G).
EDIT: No, wait a minute, f is a natural function. Huh. My mistake. But that is indeed a huge-ass number then. Respect.