Your number is, in fact, not bigger!

For all your silly time-killing forum games.

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Daggoth
Posts: 52
Joined: Wed Aug 05, 2009 2:37 am UTC

Re: Your number is, in fact, not bigger!

Postby Daggoth » Sat Jan 31, 2015 5:53 pm UTC

odf already has a gamma zero and above entry in the thread i believe. Plus wardafts notation seems still far from its limit

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Vytron
Posts: 429
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Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

Re: Your number is, in fact, not bigger!

Postby Vytron » Sat Jan 31, 2015 9:26 pm UTC

GoogologyMaster wrote:Your number is dwarfed by mine, I got past Γ0.


This is my personal battle with Daggot as I already defeated your number on this post (The notations I'm using against Daggot are stronger than the ones I'm using against you, because I'm building them more carefully, so I'll reach that will less symbols around here, and the other notation was rushed out, but it suffices.)

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Vytron
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Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Feb 01, 2015 9:35 am UTC

Anyway, so a clear problem with my notation is lack of clarity. This is because I keep switching what needs to advance to reach the next ordinal, and I think fixing this would make the progression neater.

So, there we go again:

n[0] = n+n+...+n+n n+s.
n[0],0 = n[0] n[0] n[0]...n[0] n []s computing the rightmost one to get a result you put in the previous.

Example:

2[0],0 = n[0] 2[0] = 2+2+2[0] = 6[0] = 6+6+6+6+6+6+6 = 42

n[0],0,0 = n[0],0 n[0],0...n[0],0 n[0],0 n []s, same rules.

Example:

2[0],0,0 = n[0],0 2[0],0 = 42[0],0 = n[0] n[0]...41 times...n[0] n[0] 42[0]
n[0] n[0]...40 times...n[0] n[0] 1806[0]
n[0] n[0]...39 times...n[0] n[0] 3263442[0]
n[0] n[0]...38 times...n[0] n[0] 10650056950806[0]

n[0],0,0,0 = n[0],0,0 n[0],0,0...n[0],0,0 n[0],0,0 n []s

n[0],1 = n[0],0,0,...n ,s...,0,0
n[0],1,0 = n[0],1 n[0],1 ... n[0],1
n[0],1,0,0 = n[0],1,0 n[0],1,0 ... n[0],1,0
n[0],1,1 = n[0],1,0,0,...n ,s...,0,0
n[0],0,1 = n[0],1,1,1,...n ,s...,1,1
n[0],0,0,1 = n[0],0,1,0,1,...n ,s...,0,1
n[0],0,0,0,1 = n[0],0,0,1,0,0,1,...n ,s...,0,0,1
n[0],2 = n[0],0,0,...n ,s...,0,0,1
n[0],3 = n[0],0,0,...n ,s...,0,0,1,2
n[0],[0] = n[0],n > f_ε_0(n)

Clarity is gained doing it like this, because, nY of above can be matched with any ordinal, which means:

n[0],X

Can be built, switching X for Y, for any ordinal up to ε_0, allowing:

n[0],[0],[0] = f_ε_0*2(n)
n[0],[0],[0],[0] = f_ε_0*3(n)
n[0],[0],[0],[0],[0] = f_ε_0*4(n)

n,0,[0] = n[0],[0],...,[0],[0] = f_ε_0*ω(n)

And now, X can happen here, so:

n,X,[0]

n[0-0] = n,n,[0] = f_ε_0^2(n)
n[0-0],0 = f_ε_0^2+1(n)
n[0-0],[0] = f_ε_0^2+ε_0(n)
n[0-0],[0-0] = f_ε_0^2*2(n)
n,0,[0-0] = f_ε_0^2*ω(n)
n[0],[0-0] = f_ε_0^3(n)
n[0],[0],[0-0] = f_ε_0^4(n)
n,0,[0],[0-0] = f_ε_0^ω(n)
n,1,[0],[0-0] = f_ε_0^ω^ω(n)
n,2,[0],[0-0] = f_ε_0^ω^ω^ω(n)
n[0-0,0] = f_ε_0^ε_0(n)
n[0-0,0],[0-0,0] = f_ε_0^ε_0*2(n)
n,0,[0-0,0] = f_ε_0^ε_0*ω(n)
n[0],[0-0,0] = f_ε_0^ε_0^2(n)
n[0],[0],[0-0,0] = f_ε_0^ε_0^3(n)
n[0],[0],[0],[0-0,0] = f_ε_0^ε_0^4(n)
n,0,[0],[0-0,0] = f_ε_0^ε_0^ω(n)
n[0-0],[0-0,0] = f_ε_0^ε_0^ε_0(n)
n[0-0],[0-0,0],[0-0],[0-0,0] = f_ε_0^ε_0^ε_0*2(n)
n,0,[0-0],[0-0,0] = f_ε_0^ε_0^ε_0*ω(n)
n[0],[0-0],[0-0,0] = f_ε_0^ε_0^ε_0^2(n)
n,0,[0],[0-0],[0-0,0] = f_ε_0^ε_0^ε_0^ω(n)
n[0-0,0,0] = f_ε_0^ε_0^ε_0^ε_0(n)
n[0-1] = n[0-0,0,...,0,0] = f_ε_1

n[0-X]

n[0-[0],[0]] = f_ε_ε_0
n[0-[0-[0],[0]]] = f_ε_ε_ε_0
n[0-[0-[0-[0],[0]]]] = f_ε_ε_ε_0

n[0,0-0] = n[0-[0-[0-...[0],[0]]]] = f_ζ_0
n[0,0-0],0 = f_ζ_0+1

n[0-[0,0-0],0] = f_ε_ζ_0+1
n[0-[0-[0,0-0],0]] = f_ε_ε_ζ_0+1
n[0-[0-[0-[0,0-0],0]] = f_ε_ε_ζ_0+1

n[0,0-0,0] = n[0-[0-[0-...[0,0-0],0]] = f_ζ_1

n[0,0-X]

n[0,0-[0,0-0]] = ζ_ζ_0
n[0,0-[0,0-[0,0-0]]] = ζ_ζ_ζ_0

n[0,0,0-0] = n[0,0-[0,0-...[0,0-0]]] = φ(3,0)
n[0,0,0,0-0] = n[0,0,0-[0,0,0-...[0,0,0-0]]] = φ(4,0)
n,[1-0] = φ(ω,0)

n,[X-0]

n,[[1-0]-0] = φ(φ(ω,0),0)
n,[[[1-0]-0]-0] = φ(φ(φ(ω,0),0),0)

n[0-0-0] = n,[[[1-0]...-0]-0] = φ(Γ_0,0)
n[0-0-0],0 = Γ_0+1
n[0-0-0],[0-0-0] = Γ_0*2
n,0,[0-0-0] = Γ_0*ω
n,[0-0-0,0] = n,[[[1-0]...-0]-0],[0-0-0] = Γ_0^2
n,[0-0-[0-0-0]] = Γ_0^Γ_0
n,[0-0,0-0] = n,[0-0-[0-0-[0-0-...[0-0-0],0]]] = φ(1,Γ_0+1)
n,[0-[1-0]-0] = φ(φ(ω,0),Γ_0+1)
n,[0-[[1-0]-0]-0] = φ(φ(φ(ω,0),0),Γ_0+1)
n,[0-[0-0-0]-0] = φ(Γ_0,1)
n,[0-[0-[0-0-0]-0]-0] = φ(Γ_0,2)
n,[0-[0-[0-[0-0-0]-0]-0]-0] = φ(Γ_0,3)
n[0,0-0-0] = φ(Γ_0,ω)

n[X-0-0]

n[[0-0-0]-0-0] = φ(Γ_0,φ(Γ_0,0))
n[[[0-0-0]-0-0]-0-0] = φ(Γ_0,φ(Γ_0,φ(Γ_0,0)))
n[[[[0-0-0]-0-0]-0-0]-0-0] = φ(Γ_0,φ(Γ_0,φ(Γ_0,φ(Γ_0,0))))

n[0-0-0-0] = n[[[[0-0-0]...-0-0]-0-0]-0-0] = Γ_1

My entry (now properly defeating GoogologyMaster)

4[0-0-0-0]

GoogologyMaster
Posts: 5
Joined: Sat Jan 10, 2015 2:49 pm UTC

Re: Your number is, in fact, not bigger!

Postby GoogologyMaster » Sun Feb 01, 2015 2:07 pm UTC

Okay, so I found 3 [3 [3] 3] 3 may become ambiguous, since there are two different ways I can represent it using my notation, namely 3 [3 [3] 3] 3 or 3 [(3 [3] 3)] 3, which is a lot smaller.
In order to avoid this, I decided to use {} for the outermost pair only. This allows me to distinguish between 3 {3 [3] 3} 3 and 3 {3 {3} 3} 3 (the first one is larger).

Daggoth
Posts: 52
Joined: Wed Aug 05, 2009 2:37 am UTC

Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 01, 2015 3:26 pm UTC

That is alot clearer vytron
just a tiny complaint. The n+s part was not fully clear to me till i saw you working out your examples. Perhaps phrasing it as n plus signs could be clearer
also, shouldnt 2+2+2 be 8 in your notation, since the leftmost two gets updated to a 4

Daggoth
Posts: 52
Joined: Wed Aug 05, 2009 2:37 am UTC

Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 01, 2015 3:40 pm UTC

what uve been interpreting as n*(n+1) is actually n*(2^n) so as you see i dont always call your function small :mrgreen:

GoogologyMaster
Posts: 5
Joined: Sat Jan 10, 2015 2:49 pm UTC

Re: Your number is, in fact, not bigger!

Postby GoogologyMaster » Sun Feb 01, 2015 7:57 pm UTC

Anyway, for Vytron, the limit of
φ(Γ_0,φ(Γ_0,0))
φ(Γ_0,φ(Γ_0,φ(Γ_0,0)))
φ(Γ_0,φ(Γ_0,φ(Γ_0,φ(Γ_0,0))))
is φ(Γ0+1, 0), not Γ1.
Γ1 is the limit of
φ(Γ_0,1)
φ(φ(Γ_0,1),0)
φ(φ(φ(Γ_0,1),0),0)

Right now, I'm at φ(Γ0, 1).
The {1 [1 \ 2 ¬ 2] 3} separator is defined as follows:
a {1 [1 \ 2 ¬ 2] 3} b = a {1 [1 [1 [1 [... ... [1 [1 [1 [1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] ... ...] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} a

So now I have the {1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} separator at φ(Γ0, 1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, 1)
{2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, 1) + 1
{1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, 1) * 2
{1 \ 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(1, φ(Γ0, 1)+1)
{1 \ 1 \ 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(2, φ(Γ0, 1)+1)
{1 \ 1 \ 1 \ 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(3, φ(Γ0, 1)+1)
{1 [2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(ω, φ(Γ0, 1)+1)
{1 [3 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(ω2, φ(Γ0, 1)+1)
{1 [1,2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(ωω, φ(Γ0, 1)+1)
{1 [1 [2] 2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(ωω^ω, φ(Γ0, 1)+1)
{1 [1 [1 \ 2] 2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(ε0, φ(Γ0, 1)+1)
{1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(ω,0), φ(Γ0, 1)+1)
{1 [1 [1 [1 [1 \ 2] 2 ¬ 2] 2] 2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(ε0,0), φ(Γ0, 1)+1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(Γ0, 2)
{2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(Γ0, 2) + 1
{1 \ 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(1, φ(Γ0, 2)+1)
{1 [2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(ω, φ(Γ0, 2)+1)
{1 [1,2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(ωω, φ(Γ0, 2)+1)
{1 [1 [1 \ 2] 2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(ε0, φ(Γ0, 2)+1)
{1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2} = φ(φ(ω,0), φ(Γ0, 2)+1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 4 [1 \ 2 ¬ 2] 2} = φ(Γ0, 3)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 5 [1 \ 2 ¬ 2] 2} = φ(Γ0, 4)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1,2 [1 \ 2 ¬ 2] 2} = φ(Γ0, ω)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2,2 [1 \ 2 ¬ 2] 2} = φ(Γ0, ω+1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1,3 [1 \ 2 ¬ 2] 2} = φ(Γ0, ω*2)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1,1,2 [1 \ 2 ¬ 2] 2} = φ(Γ0, ω2)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, ωω)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 \ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, ε0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 \ 1 \ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, ζ0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(ω,0))
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, Γ0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(Γ0, 1))
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(Γ0, 2))
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1,2 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(Γ0, ω))
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 \ 2] 2 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(Γ0, ε0))
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(Γ0, Γ0))

{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+1, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 \ 2 [1 \ 2 ¬ 2] 2} = φ(Γ0, φ(Γ0+1, 0)+1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 3 [1 \ 2 ¬ 2] 2} = φ(Γ0+1, 1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 1,2 [1 \ 2 ¬ 2] 2} = φ(Γ0+1, ω)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 1 [1 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+1, Γ0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 2 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+1, φ(Γ0+1, 0))
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 1 \ 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+2, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 1 \ 1 \ 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+3, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [3 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ02, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 \ 1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+φ(ω,0), 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 [1 \ 2] 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0+φ(ε0,0), 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*2, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*2 + 1, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*2 + ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*2 + ε0, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*3, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*4, 0)
{1 [2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*ω, 0)
{1 [3 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ02, 0)
{1 [4 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ03, 0)
{1 [1,2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0ω, 0)
{1 [1 [2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0ω^ω, 0)
{1 [1 [1 \ 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00, 0)
{1 [1 [1 [2 ¬ 2] 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*φ(ω,0), 0)
{1 [1 [1 [1,2 ¬ 2] 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*φ(ω^ω,0), 0)
{1 [1 [1 [1 [1 \ 2] 2 ¬ 2] 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*φ(ε0,0), 0)
{1 [1 [1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0*φ(φ(ω,0),0), 0)

{1 [1 [1 [1 \ 2 ¬ 2] 2] 3 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^2, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 4 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^3, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^ω^ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^φ(ω,0), 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [1 [1 [1 \ 2] 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^φ(ε0,0), 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ0^φ(φ(ω,0),0), 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 1 [1 [1 \ 2 ¬ 2] 2] 1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00^2, 0)
{1 [1 [2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ00^ω, 0)
{1 [1 [1 [1 \ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ000, 0)
{1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(Γ000, 0)
{1 [1 [1 \ 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(1, Γ0+1), 0)
{1 [1 [1 \ 3 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(1, Γ0+2), 0)
{1 [1 [1 \ 1 \ 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(2, Γ0+1), 0)
{1 [1 [1 \ 1 \ 1 \ 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(3, Γ0+1), 0)
{1 [1 [1 [2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(ω, Γ0+1), 0)
{1 [1 [1 [1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(ε0, Γ0+1), 0)
{1 [1 [1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(φ(ω,0), Γ0+1), 0)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(Γ0, 1), 0)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(Γ0, 2), 0)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 \ 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(Γ0+1, 0), 0)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(Γ02, 0), 0)
{1 [1 [1 [2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(Γ0ω, 0), 0)
{1 [1 [1 [1 \ 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(φ(1, Γ0+1), 0), 0)
{1 [1 [1 [1 \ 1 \ 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(φ(2, Γ0+1), 0), 0)
{1 [1 [1 [1 [2 ¬ 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(φ(ω, Γ0+1), 0), 0)
{1 [1 [1 [1 [1 [1 \ 2] 2 ¬ 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(φ(ε0, Γ0+1), 0), 0)
{1 [1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2} = φ(φ(φ(Γ0, 1), 0), 0)
...
{1 [1 [1 [1 [... ... [1 [1 [1 [1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] ... ...] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2}
= {1 [1 \ 2 ¬ 2] 3} = Γ1
My number:
3 {1 [1 \ 2 ¬ 2] 3} 3

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Sun Feb 01, 2015 9:19 pm UTC

The think limit of the single-colon-per-bracket notation is actually Γ0.

Anyway, the obvious next step is a second colon

[n|0[W:X:Y]Z] for non-empty W,X reduces to [n+1|n[W:dec n X:Y][dec n W:X:Y]Z]
[n|0[W:X:Y]Z] for empty W,X, reduces to [n+1|n[||f(n,0:[)||0:0:dec n Y||f(n,]:dec n Y)||]Z]
[n|0[W:X:Y]Z] for empty X reduces to [n+1|n[||f(n,0:[)||dec n W:0:Y||f(n,]:dec n Y)||]Z]
[n|0[W:X:Y][W':X':Y']Z] for empty W, non-empty X,W',X' reduces to:
[n|n[||f(n,n[)||n:X:Y||f(n,]:X:Y][W':X':Y']Z)||][dec n W':X':Y']Z]
For empty Y, simply discard the last colon.

So [2|0[ : :0[]]] has growth rate Γ0. It's fairly difficult for me to express something like φ(Γ0,1), but if we can hit Γ0, Γ1 shouldn't be far off, so I shall enter [2|1[0[]: :0[]] ] which is like f_{Γ1+1}(2).

EDIT: Almost overshot my target, had to reduce some things.
Last edited by WarDaft on Thu Feb 05, 2015 5:35 pm UTC, edited 1 time in total.
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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 02, 2015 9:33 pm UTC

Daggoth wrote:also, shouldnt 2+2+2 be 8 in your notation, since the leftmost two gets updated to a 4


I don't use the rules at the base.

But it's a good idea so I'll use it!

n[0] = n +n ... +n +n n plus signs

Where you keep updating n to the highest number reached.

2[0] = n +n +2 = n +2 +2 = 4 +4 = 8
3[0] = n +n +n +3 = n +n +3 +3 = n +6 +6 = 12 +12 = 24

GoogologyMaster wrote:Anyway, for Vytron, the limit of
φ(Γ_0,φ(Γ_0,0))
φ(Γ_0,φ(Γ_0,φ(Γ_0,0)))
φ(Γ_0,φ(Γ_0,φ(Γ_0,φ(Γ_0,0))))
is φ(Γ0+1, 0), not Γ1.
Γ1 is the limit of
φ(Γ_0,1)
φ(φ(Γ_0,1),0)
φ(φ(φ(Γ_0,1),0),0)


Thanks! Usually the corrections are of the kind "you get nowhere near Γ_0 because of <reasons>", this actually is easy to fix.

Spoiler that contains entire notation:
Spoiler:
n[0] = n +n ... +n +n n plus signs

Where you keep updating n to the highest number reached.

Example:
2[0] = n +n +2 = n +2 +2 = 4 +4 = 8
3[0] = n +n +n +3 = n +n +3 +3 = n +6 +6 = 12 +12 = 24

n[0],0 = n[0] n[0] n[0]...n[0] n []s computing the rightmost one to get a result you put in the previous.

Example:

2[0],0 = n[0] 2[0] = 2+2+2[0] = 8[0] =
n +n +n +n +n +n +n +8 +8 =
n +n +n +n +n +n +16 +16 =
n +n +n +n +n +32 +32 =
n +n +n +n +64 +64 =
n +n +n +128 +128 =
n +n +256 +256 =
n +512 +512 =
1024 +1024 =
2048

n[0],0,0 = n[0],0 n[0],0...n[0],0 n[0],0 n []s, same rules.

Example:

2[0],0,0 = n[0],0 2[0],0 = n[0],0 2048[0],0 = n[0],0 n[0] n[0]...2047 times...n[0] n[0] 2048[0]

2048[0] = n +n ...2048 times...+n +2048

n[0],0,0,0 = n[0],0,0 n[0],0,0...n[0],0,0 n[0],0,0 n []s

n[0],1 = n[0],0,0,...n ,s...,0,0
n[0],1,0 = n[0],1 n[0],1 ... n[0],1
n[0],1,0,0 = n[0],1,0 n[0],1,0 ... n[0],1,0
n[0],1,1 = n[0],1,0,0,...n ,s...,0,0
n[0],0,1 = n[0],1,1,1,...n ,s...,1,1
n[0],0,0,1 = n[0],0,1,0,1,...n ,s...,0,1
n[0],0,0,0,1 = n[0],0,0,1,0,0,1,...n ,s...,0,0,1
n[0],2 = n[0],0,0,...n ,s...,0,0,1
n[0],3 = n[0],0,0,...n ,s...,0,0,1,2
n[0],[0] = n[0],n > f_ε_0(n)

Clarity is gained doing it like this, because, nY of above can be matched with any ordinal, which means:

n[0],X

Can be built, switching X for Y, for any ordinal up to ε_0, allowing:

n[0],[0],[0] = f_ε_0*2(n)
n[0],[0],[0],[0] = f_ε_0*3(n)
n[0],[0],[0],[0],[0] = f_ε_0*4(n)

n,0,[0] = n[0],[0],...,[0],[0] = f_ε_0*ω(n)

And now, X can happen here, so:

n,X,[0]

n[0-0] = n,n,[0] = f_ε_0^2(n)
n[0-0],0 = f_ε_0^2+1(n)
n[0-0],[0] = f_ε_0^2+ε_0(n)
n[0-0],[0-0] = f_ε_0^2*2(n)
n,0,[0-0] = f_ε_0^2*ω(n)
n[0],[0-0] = f_ε_0^3(n)
n[0],[0],[0-0] = f_ε_0^4(n)
n,0,[0],[0-0] = f_ε_0^ω(n)
n,1,[0],[0-0] = f_ε_0^ω^ω(n)
n,2,[0],[0-0] = f_ε_0^ω^ω^ω(n)
n[0-0,0] = f_ε_0^ε_0(n)
n[0-0,0],[0-0,0] = f_ε_0^ε_0*2(n)
n,0,[0-0,0] = f_ε_0^ε_0*ω(n)
n[0],[0-0,0] = f_ε_0^ε_0^2(n)
n[0],[0],[0-0,0] = f_ε_0^ε_0^3(n)
n[0],[0],[0],[0-0,0] = f_ε_0^ε_0^4(n)
n,0,[0],[0-0,0] = f_ε_0^ε_0^ω(n)
n[0-0],[0-0,0] = f_ε_0^ε_0^ε_0(n)
n[0-0],[0-0,0],[0-0],[0-0,0] = f_ε_0^ε_0^ε_0*2(n)
n,0,[0-0],[0-0,0] = f_ε_0^ε_0^ε_0*ω(n)
n[0],[0-0],[0-0,0] = f_ε_0^ε_0^ε_0^2(n)
n,0,[0],[0-0],[0-0,0] = f_ε_0^ε_0^ε_0^ω(n)
n[0-0,0,0] = f_ε_0^ε_0^ε_0^ε_0(n)
n[0-1] = n[0-0,0,...,0,0] = f_ε_1

n[0-X]

n[0-[0],[0]] = f_ε_ε_0
n[0-[0-[0],[0]]] = f_ε_ε_ε_0
n[0-[0-[0-[0],[0]]]] = f_ε_ε_ε_0

n[0,0-0] = n[0-[0-[0-...[0],[0]]]] = f_ζ_0
n[0,0-0],0 = f_ζ_0+1

n[0-[0,0-0],0] = f_ε_ζ_0+1
n[0-[0-[0,0-0],0]] = f_ε_ε_ζ_0+1
n[0-[0-[0-[0,0-0],0]] = f_ε_ε_ζ_0+1

n[0,0-0,0] = n[0-[0-[0-...[0,0-0],0]] = f_ζ_1

n[0,0-X]

n[0,0-[0,0-0]] = ζ_ζ_0
n[0,0-[0,0-[0,0-0]]] = ζ_ζ_ζ_0

n[0,0,0-0] = n[0,0-[0,0-...[0,0-0]]] = φ(3,0)
n[0,0,0,0-0] = n[0,0,0-[0,0,0-...[0,0,0-0]]] = φ(4,0)
n,[1-0] = φ(ω,0)

n,[X-0]

n,[[1-0]-0] = φ(φ(ω,0),0)
n,[[[1-0]-0]-0] = φ(φ(φ(ω,0),0),0)

n[0-0-0] = n,[[[1-0]...-0]-0] = φ(Γ_0,0)
n[0-0-0],0 = Γ_0+1
n[0-0-0],[0-0-0] = Γ_0*2
n,0,[0-0-0] = Γ_0*ω
n[0-0-0,0] = n,[[[1-0]...-0]-0],[0-0-0] = Γ_0^2
n[0-0-[0-0-0]] = Γ_0^Γ_0
n[0-0,0-0] = n,[0-0-[0-0-[0-0-...[0-0-0],0]]] = φ(1,Γ_0+1)
n[0-[1-0]-0] = φ(φ(ω,0),Γ_0+1)
n[0-[[1-0]-0]-0] = φ(φ(φ(ω,0),0),Γ_0+1)
n[0-[0-0-0]-0] = φ(Γ_0,1)


And from there:

n[[0-[0-0-0]-0]-0] = φ(φ(Γ_0,1),0)

n[[[0-[0-0-0]-0]-0]-0] = φ(φ(Γ_0,1),0)

n[[[[0-[0-0-0]-0]-0]-0]-0] = φ(φ(φ(Γ_0,1),0),0)

n[0,0-0-0] = n[[[[0-[0-0-0]-0]-0]...-0]-0] = Γ_1

n[0,0-0-0],0 = n[0,0-0-0] n[0,0-0-0]...n[0,0-0-0] = Γ_1+1

n[0,0-0-0],0,0 = n[0,0-0-0],0 n[0,0-0-0],0 ... n[0,0-0-0],0 = Γ_1+2

n[0,0-0-0],1 = n[0,0-0-0],0,0,...,0,0

Send:

4[0,0-0-0],1

Which is like f_{Γ_1+ω}(4).

Daggoth
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Tue Feb 03, 2015 1:34 am UTC

how do you even get from
Γ0
to
φ(Γ_0,φ(Γ_0,0))
What are the inbetween points?

(some how i feel thats gonna be even harder than going from φ(2,0) to φ(2,1)

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Tue Feb 03, 2015 1:45 am UTC

φ(Γ_0,0)
φ(Γ_0,1)
φ(Γ_0,2)
φ(Γ_0,...)
φ(Γ_0,ω)
φ(Γ_0,ω^ω)
φ(Γ_0,ω^ω^ω...)
φ(Γ_0,ε_0)
φ(Γ_0,ε_0^ε_0)
φ(Γ_0,ε_0^ε_0^ε_0..)
φ(Γ_0,ε_1)
φ(Γ_0,ε_ω)
φ(Γ_0,ε_ω^ω)
φ(Γ_0,ε_ω^ω^ω...)
φ(Γ_0,ε_ε_0)
φ(Γ_0,...ε_ε_ε_0)
φ(Γ_0,ζ_0) = φ(Γ_0,φ(2,0))
φ(Γ_0,φ(3,0))
φ(Γ_0,φ(4,0))
φ(Γ_0,φ(ω,0))
φ(Γ_0,φ(ε_0,0))
φ(Γ_0,φ(φ(2,0),0))
φ(Γ_0,φ(φ(ω,0),0))
φ(Γ_0,φ(φ(ε_0,0),0))
φ(Γ_0,φ(φ(φ(2,0),0),0))
φ(Γ_0,φ(φ(φ(ω,0),0),0))
...
φ(Γ_0,φ(Γ_0,0)) = φ(Γ_0,φ(φ(φ...φ(φ(φ(ω,0),0),0))

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Tue Feb 03, 2015 3:30 am UTC

Well, then it appears I have no choice but to go to [2|0[0[]0[]: :0[]] for f_{Γ2}(2)
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Tue Feb 03, 2015 4:04 am UTC

Still GFDQM Grade F

Clipping away rule 6 so that ¿0¡0!<0>0?n actually reduces to ¿n¡n...n...n!n?nn with n ¡! pairs.

Reasessing size.
It seems there was some errors in my previous comparisons

Increasing the righthand argument with a lesser chain: adding that chains ordinal
Ordinal stands for f_ordinal(n)
¿0¡0!0? > ω^ω
¿0¡0!0,0,0,0,0? > ω^ω+ω^5
Increasing the lefthand argument with a lesser chain: multiplying by that chains ordinal (+1)
¿0,0,0,0,0¡0! > ω^ω*ω^5 = > (ω^ω)*ω*ω*ω*ω*ω = ω^(ω+5)
n of the same separator: that separator's ordinal to the n
¿0¡0!0¡0!0? > ω^ω^2 ¿0¡0!0¡0!0¡0!0¡0!0? > ω^ω^4
Increasing +1 in the separators (right most, for nests) argument : that separator's ordinal to the omega
¿0¡1!0? > ω^ω^ω
¿3¡1!0? > (ω^ω^ω)4
¿0¡1!0¡1!0? (ω^ω^ω)^2
¿0¡2!0? > ω^ω^ω^ω
¿0¡0,0!0? > ε0
¿0¡0,0!0¡0,0!0? > ε0^2
¿0¡0,1!0? > ε0^ω
¿0¡1,0!0? > ε0^ε0
¿0¡2,0!0? > ε0^ε0^ε0
¿0¡0,0,0!0? > ε1
¿0¡0,0,1!0? > ε1^ω
¿0¡0,1,0!0? > ε1^ε0
¿0¡1,0,0!0? > ε1^ε1
¿0¡0,0,0,0!0? > ε2
¿0¡0¡0!0!0? > εω
¿0¡0¡0!1!0? > εω^ω
¿0¡0¡0!0,0!0? > εω^ε0
¿0¡0¡0!0,0,0!0? > εω^ε1
¿0¡1¡0!0!0? > εω^εω
¿0¡2¡0!0!0? > εω^εω^εω
¿0¡0,0¡0!0!0? > εω+1
¿0¡1,0¡0!0!0? > εω+1^εω+1
¿0¡0,0,0¡0!0!0? > εω+2
¿0¡0¡0!0¡0!0!0? > ε_ω2
¿0¡0¡1!0!0? > ε_ω^2
¿0¡0¡0,0!0!0? > ε_ω^ω
¿0¡0¡1,0!0!0? > ε_ε0
¿0¡0¡2,0!0!0? > ε_ε0^ε0
¿0¡0¡0,0,0!0!0? > ε_ε1
¿0¡0¡0¡0!0!0!0? > ε_εω
¿0¡0¡0¡1!0!0!0? > ε_ε_ω^2
¿0¡0¡0¡0¡0!0!0!0!0? > ε_ε_ε_ω
¿0¡0!<0>0?n > ζ0

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Tue Feb 03, 2015 4:22 am UTC

I can't make such jumps because I have to manually define each nesting (to match the FGH), but that should be within reach...

n[0,0-0-0],[0,0-0-0] = Γ_1*2
n,0,[0,0-0-0] = Γ_1*ω
n[0,0-0-0,0] = n[[[[0-[0-0-0]-0]-0]...-0]-0],[0,0-0-0] = Γ_1^2
n[0,0-0-[0,0-0-0]] = Γ_1^Γ_1
n[0,0-0,0-0] = n[0,0-0-[0,0-0-...[0,0-0-[0,0-0-0],0]]] = φ(1,Γ_1+1)
n[0,0-[1-0]-0] = n[0,0-0-[0,0-0-...[0,0-0-[0,0-0-0],0]]] = φ(φ(ω,0),Γ_1+1)
n[0,0-[[1-0]-0]-0] = n[0,0-0-[0,0-0-...[0,0-0-[0,0-0-0],0]]] = φ(φ(φ(ω,0),0),Γ_1+1)
n[0,0-[0,0-0-0]-0] = φ(Γ_1,1)
n[[0,0-[0,0-0-0]-0]-0] = φ(φ(Γ_1,1),0)
n[[[0,0-[0,0-0-0]-0]-0]-0] = φ(φ(φ(Γ_1,1),0),0)

n[0,0,0-0-0] = n[[[0,0-[0,0-0-0]...-0]-0]-0] = Γ_2

n[0,0,0-0-0],0 = n[0,0,0-0-0] n[0,0,0-0-0]...n[0,0,0-0-0]

4[0,0,0-0-0],0

Which is like f_{Γ_2+1}(4).

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Re: Your number is, in fact, not bigger!

Postby Daggoth » Wed Feb 04, 2015 5:18 am UTC

Busy week. I know you guys are way beyond but there is no reason not to play for 5th place.
GFDQM Grade F
Apply rules 1-5 as before.
Rule 6:
Double Separator. Ranks above all nesteds. Rank according to reductions. If a Separator could eventually reduce to another, it ranks above it..
6.1 ¿0¡0!¡0!0?n = ¿n¡n...n...n!n?nn > ζ0
6.2 ¿P¡*!¡*!S?n = ¿P¡Q!¡R!S-1?nn
6.3 ¿P¡*!¡*!0?n = ¿P-1¡Q!¡R!X?nn Where X = Y¡#!Y... with n ¡#! separators where # is a separator immediately inferior in rank, Where Y is Z¡#-1!Z... with n ?¡#-1! Separators where #-1 is immediately inferior to #. Continue until # is the base separator
6.4 ¿0¡%!¡*!0?n = ¿X¡X...¡%!¡@n!...!X!X?nn Where @n =Y¡Y...¡%!¡@n-1!Y...!Y ..so on untill @n-n = Z¡Z...¡%!¡#!...Z!Z Each set of ellipses hides n nestings. X Y Z work as in 6.3
6.5 ¿0¡*!¡0!0?n = ¿X¡X...¡#!¡@n!...!X!X?nn @ Works the same as in 6.4
Triple and quadruple separators
¿0¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!...X!X?nn Here @ Works the same as in 6.4 taking * to be n
¿0¡0!¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!¡@!...X!X?nn Here @ Works the same as in 6.4 taking * to be n

Sizing

¿0¡0!¡0!0¡0!¡0!0?n = ¿X¡...X...!X¡0!¡0!X?nn > ζ0^2
¿0¡0¡0!¡0!0!0?n = ¿0¡X¡....X...!X!0?nn > εζ0
¿0¡0¡0¡0!¡0!0!0!0?n = ¿0¡0¡X¡....X...!X!0!0?nn > ε_ε ζ0
¿0¡0!¡1!0?n = ¿X¡X...¡0!¡0!...X!X? > ζ1
¿0¡0!¡0,0!0?n = ¿X¡X...¡0!¡n!...X!X? > ζω
¿0¡0!¡0¡0!0!0?n = ¿X¡X.. .¡0!¡n,...n!...X!X? > ζε0
¿0¡0!¡0¡0¡0!0!0!0?n = > ζε_ε0
¿0¡0!¡0¡0!¡0!0!0?n > ζζ0
¿0¡0!¡0¡0!¡0¡0!¡0!0!0!0?n > ζζ0
¿0¡1!¡0!0?n > η0
¿0¡0¡1!¡0!0!0?n > εη0
¿0¡1!¡0¡1!¡0!0!0?n > ζη0
¿0¡2!¡0!0?n > η1
¿0¡0¡0!¡0!0!¡0!0?n > ηη0
¿0¡0!¡0!¡0!0?n > φ(4,0)
¿0¡0¡0!¡0!¡0!0!¡0!¡0!0?n > φ(4,φ(4,0))
¿0¡0!¡0!¡0!¡0!0?n > φ(5,0)

General sizing
¿0¡0¡#!0!0?n > ε_#
¿0¡0!¡#!0?n > ζ_#
¿0¡#!¡0!0?n > η_#
¿0¡#!¡0!...¡0!¡0!0?n > φ(...,#)

Hook for Grade E GFDQM:
¿0<0>0?n = ¿n¡n!...¡n!n?nn
I enter ¿0<0>0?100 which beats φ(ω,ω)100 beating vytron's 4,[0-0,[0-0],[0,0-0]],0
(Edit: WIP, edits might come in the next hour or so)

User avatar
Vytron
Posts: 429
Joined: Mon Oct 19, 2009 10:11 am UTC
Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

Re: Your number is, in fact, not bigger!

Postby Vytron » Wed Feb 04, 2015 5:31 am UTC

I made mistakes somewhere, there's no point at which my notation would contain something like "n,[", so that should be 4[0-0,[0-0],[0,0-0]],0

Anyway, against Daggoth:

4[[0]-0] = 4[4-[4-[4-[4-0]]]] = φ(ω^ω^ω^ω,φ(ω^ω^ω^ω,φ(ω^ω^ω^ω,φ(ω^ω^ω^ω,0))))(4)

GoogologyMaster
Posts: 5
Joined: Sat Jan 10, 2015 2:49 pm UTC

Re: Your number is, in fact, not bigger!

Postby GoogologyMaster » Wed Feb 04, 2015 10:24 pm UTC

Anyway, I'll continue my notation.

To continue I nest [x [1 \ 2 ¬ 2] 3]s to make [1 [1 \ 2 ¬ 2] 4]
a {1 [1 \ 2 ¬ 2] 4} b = a {1 [1 [1 [1 [... ... [1 [1 [1 [1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] ... ...] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} a

{1 [1 \ 2 ¬ 2] 3} = Γ1
{2 [1 \ 2 ¬ 2] 3} = Γ1+1
{1 [1 [1 \ 2 ¬ 2] 3] 2 [1 \ 2 ¬ 2] 3} = Γ1*2
{1 \ 2 [1 \ 2 ¬ 2] 3} = φ(1, Γ1+1)
{1 \ 1 \ 2 [1 \ 2 ¬ 2] 3} = φ(2, Γ1+1)
{1 [2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(ω, Γ1+1)
{1 [1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(ε0, Γ1+1)
{1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(φ(ω,0), Γ1+1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ0, Γ1+1)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(φ(Γ0,1), Γ1+1)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ1, 1)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 3} = φ(Γ1, 2)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 1,2 [1 \ 2 ¬ 2] 3} = φ(Γ1, ω)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 1 \ 2 [1 \ 2 ¬ 2] 3} = φ(Γ1+1, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ1*2, 0)
{1 [2 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ1*ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 3 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ1^2, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ1^ω, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ10, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ10, 0)
{1 [1 [1 [1 \ 2 ¬ 2] 3] 1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ11, 0)
{1 [1 [2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(Γ11^ω, 0)
{1 [1 [1 \ 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(ε(Γ1+1), 0)
{1 [1 [1 [2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(φ(ω, Γ1+1), 0)
{1 [1 [1 [1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(φ(ω^ω, Γ1+1), 0)
{1 [1 [1 [1 [1 \ 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(φ(ε0, Γ1+1), 0)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 3} = φ(φ(Γ1, 1), 0)

After that, I nest [x [1 \ 2 ¬ 2] 4]s to make [1 [1 \ 2 ¬ 2] 5]
a {1 [1 \ 2 ¬ 2] 5} b = a {1 [1 [1 [1 [... ... [1 [1 [1 [1,2 ¬ 2] 2 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4] ... ...] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} a
{1 [1 \ 2 ¬ 2] 4} = Γ2
{2 [1 \ 2 ¬ 2] 4} = Γ2+1
{1 \ 2 [1 \ 2 ¬ 2] 4} = φ(1, Γ2+1)
{1 [2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(ω, Γ2+1)
{1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(Γ0, Γ2+1)
{1 [1 [1 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(Γ2, 1)
{1 [1 [1 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 3 [1 \ 2 ¬ 2] 4} = φ(Γ2, 2)
{1 [2 [1 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(Γ2*ω, 0)
{1 [1 [2 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(Γ22^ω, 0)
{1 [1 [1 \ 2 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(ε(Γ2+1), 0)
{1 [1 [1 [1 [1 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4] 2 ¬ 2] 2 [1 \ 2 ¬ 2] 4} = φ(φ(Γ2, 1), 0)

{1 [1 \ 2 ¬ 2] 5} = Γ3
{1 [1 \ 2 ¬ 2] 6} = Γ4
{1 [1 \ 2 ¬ 2] n} = Γn-2

My number:
99 {1 [1 \ 2 ¬ 2] 99} 99

Daggoth
Posts: 52
Joined: Wed Aug 05, 2009 2:37 am UTC

Re: Your number is, in fact, not bigger!

Postby Daggoth » Thu Feb 05, 2015 5:15 am UTC

GFDQM (General form daggoth's question mark) Grade F


Concatenated, fixed and sized using constant values
Spoiler:
Red parts are the ones that get deleted or reduced

Rule 1: Base case: ¿0?n = n↑nn | ¿X?n = ¿X-1?nn (where ¿X-n?3=¿X-n?¿X-n?¿X-n?)

Rule 2: Base separator ,
2.1: ¿...x,y,z?n where all arguments = 0 Reduces to ¿...x,y?nn where all arguments = n
2.2 ¿...x,0,0..,0,0?n where X isnt 0. Reduces to ¿...x-1,n,n...,n,n?nn
2.3 ¿...x,y,z?n where Z isnt 0 reduces to ¿...x,y,z-1?nn

Rule 3: A single Nth separator,
3.1 ¿0¡X!0?n = ¿Y¡X-1!Y....Y¡X-1!Y?nn, (with n ¡X-1! separators) Where Y = Z¡X-2!Z...¡X-2!Z? (with n ¡X-2! separators)continue untill X=0, ¡0! reduces to the Base separator, the comma.
3.2 ¿Q¡X!0?n = ¿Q-1¡X!Y....Y¡X-1!Y?nn
3.3 ¿Q¡X!P?n = ¿Q¡X!P-1?nn

Rule 4: Chains.
4.1: The number most to the right isnt 0

¿#¡*!...Q¡X!P?n = ¿#¡*!...Q¡X!P-1?nn
4.2: All instances of the same separator, all 0's
¿0¡Q!0....0¡Q!0?n = ¿X...¡Q!X?nn where X = Y¡Q-1!Y....Y¡X-1!Y, (with n ¡Q-1! separators), Where Y = Z¡Q-2!Z...¡Q-2!Z? (with n ¡Q-2! separators)continue untill Q=0, ¡0! reduces to the Base separator, the comma, the final value in this sequence is n
4.3: All instances of the same separator, the rightmost number is 0 but some or all of the others aren't.
From right to left, P is the first number that isnt 0. ¿0¡Q!...P...¡Q!0¡Q!0...?n = ¿0¡Q!...P-1...¡Q!X¡Q!X?nn Where X = Y¡Q-1!Y...Y¡Q-1!Y with n ¡Q-1! separators...Where Y = Z¡Q-2!Z...¡Q-2!Z? (with n ¡Q-2! separators)
4.4: Mixed Separators.
Locate the lowest ranking of separator. Call this ¡L! Its surrounding values are P and Q Adjacent separators of the same rank are called a group.
4.4.1: If there is only one ¡L! ¿#¡*!#¡*!...P¡L!Q...#¡*!#¡*!?n then
4.4.1.1: ¿#¡*!#¡*!...0¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...X...#¡*!#¡*!?nn Where X = Y¡L-1!Y...Y¡L-1!Y with n ¡L-1! separators where Y... ( same as with rule 3.1 )
IF ¡L! is , reduce according to rule 2.1. i.e. 0,0 reduces to n and 0,0,0 to n,n
4.4.1.2: ¿#¡*!#¡*!...Q¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q-1¡L!X...#¡*!#¡*!?nn X is the same as in the previous rule.
4.4.1.3: ¿#¡*!#¡*!...Q¡L!P...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q¡L!P-1...#¡*!#¡*!?nn
Adjacent separators of the same rank are called a group.
4.4.2: A single group of ¡L! (the lowest ranking separator).
¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!?n
Reduce this group in the same way as 4.1, 4.2 and 4.3. the rest of the chain is untouched untill this group is reduced to 0
4.4.3: Multiple groups of ¡L!.
¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!S¡L!T¡L!U...?n
Reduce the rightmost group first. even if the rightmost group only contains a single ¡L!

Rule 5: Nesting

All Nested separators rank above all ¡X!
Comparing two nested separators:
¡0...¡0!...0! with n nestings outranks ¡0...¡X!...0! with n-1 nestings
¡0¡R!0! outranks ¡x¡R-1!X!
¡R¡Q!0! outranks ¡R-1¡Q!X!
¡P¡Q!R! outranks ¡P¡Q!R-1!

As with rules 1-4, the lowest ranking group of separators gets reduced first.

5.1 ¿P¡#!X?n = ¿P¡#!X-1?nn
5.2 ¿Q¡#!X?n = ¿Q-1¡#!X?nn
5.3 ¿0¡#!0?n = ¿X¡# reduced as in rules 1-4!X?nn IF # is not a nested separator
5.4 ¿0¡#!0?n = ¿X¡# reduced as in rules 5.1, 5.2 and 5.3!X?nn IF # is a nested separator

Rule 6:
Double Separator. Ranks above all nesteds. Rank according to reductions. If a Separator could eventually reduce to another, it ranks above it..
6.1 ¿0¡0!¡0!0?n = ¿X¡X...X...X!X?nn Where X = ¡Y...Y¡n!Y...Y! where Y = ¡Z...Z¡n-1!Z...Z! so on until ¡n-n=0! reduces to , Use n separators each step of the way
6.2 ¿P¡*!¡*!S?n = ¿P¡Q!¡R!S-1?nn
6.3 ¿P¡*!¡*!0?n = ¿P-1¡Q!¡R!X?nn Where X = Y¡#!Y... with n ¡#! separators where # is a separator immediately inferior in rank, Where Y is Z¡#-1!Z... with n ?¡#-1! Separators where #-1 is immediately inferior to #. Continue until # is the base separator
6.4 ¿0¡%!¡*!0?n = ¿X¡X...¡%!¡@n!...!X!X?nn Where @n =Y¡Y...¡%!¡@n-1!Y...!Y ..so on untill @n-n = Z¡Z...¡%!¡#!...Z!Z Each set of ellipses hides n nestings. X Y Z work as in 6.3
6.5 ¿0¡*!¡0!0?n = ¿X¡X...¡#!¡@n!...!X!X?nn @ Works the same as in 6.4
Triple and quadruple separators
¿0¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!...X!X?nn Here @ Works the same as in 6.4 taking * to be n
¿0¡0!¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!¡@!...X!X?nn Here @ Works the same as in 6.4 taking * to be n

Resizing from scratch using constant values
¿0,0?10 > fω2(10)
¿0,0,0?10 > fω^2(10)
¿0,0,0,0,0?10 > fω^5(10)
¿0¡0!0?10 > fω^ω(10)
¿0¡0!1?10 > fω^ω+1(10)
¿0¡0!0,0?10 > fω^ω+ω(10)
¿0¡0!0,0,0,0,0?10 > fω^ω+ω^5(10)
¿1¡0!0?10 > fω^ω*2(10)
¿0,0¡0!0?10 > f(ω^ω)*ω(10)
¿0,0,0,0,0,0,0,0,0,0¡0!0?10 > f(ω^ω)*(ω^ω)(10)
¿0¡0!0¡0!0?10 > fω^ω^2(10)
¿0¡0!0¡0!0¡0!0?10 > fω^ω^3(10)
¿0¡1!0?10 > fω^ω^ω(10)
¿0¡1!0¡1!0¡1!0¡1!0¡1!0¡1!0¡1!0¡1!0¡1!0¡1!0?10 > fω^ω^ω^10(10)
¿0¡2!0?10 > fω^ω^ω^ω(10)
¿0¡99!0?10 > fω^^99(10)
¿0¡0,0!0?10 > fε_0(10)
¿0¡0,99!0?10 > fε_0^(ω^^99)(10)
¿0¡1,0!0?10 > fε_0^ε_0(10)
¿0¡99,0!0?10 > fε_0^^99(10)
¿0¡99,0,0!0?10 > fε_1^^99(10)
¿0¡0¡0!0!0?10 > fε_ω(10)
¿0¡0¡0!99!0?10 > fε_ω^(ω^^99)(10)
¿0¡0¡0!0,0!0?10 > fε_ω^ε_0(10)
¿0¡0¡0!99,0!0?10 > fε_ω^(ε_0^^99)(10)
¿0¡0¡0!0,0,0!0?10 > fε_ω^ε_1(10)
¿0¡0¡0!0,0,0,0!0?10 > fε_ω^ε_2(10)
¿0¡1¡0!0!0?10 > fε_ω^ε_ω(10)
¿0¡99¡0!0!0?10 > fε_ω^^99(10)
¿0¡0,0¡0!0!0?10 > fε_{ω+1}(10)
¿0¡1,0¡0!0!0?10 > fε_{ω+1}^ε_{ω+1}(10)
¿0¡0,0,0¡0!0!0?10 > fε_{ω+2}(10)
¿0¡0¡0!0¡0!0!0?10 > fε_{ω2}(10)
¿0¡0¡1!0,0,0!0?10 > fε_{ω^ω}^ε_1(10)
¿0¡0¡1!0¡0!0!0?10 > fε_{ω^ω}^ε_ω(10)
¿0¡1¡1!0?10 > fε_{ω^ω}^ε_{ω+1}(10)
¿0¡0,0¡1!0?10 > fε_{ω^ω}^ε_{ω2}(10)
¿0¡0,0,0¡1!0?10 > fε_{ω^ω}^ε_{ω^2}(10)
¿0¡0,0,0,0¡1!0?10 > fε_{ω^ω}^ε_{ω^3}(10)
¿0¡0¡0!0¡1!0?10 > fε_{ω^ω}^ε_{ω^ω}(10)
¿0¡1¡0!0¡1!0?10 > fε_{ω^ω}^ε_{ω^ω}^ε_{ω^ω}(10)
¿0¡0,0¡0!0¡1!0?10 > fε_{(ω^ω)+1}(10)
¿0¡0,0,0¡0!0¡1!0?10 > fε_{(ω^ω)+2}(10)
¿0¡0¡0!0¡0!0¡1!0?10 > fε_{(ω^ω)2}(10)
¿0¡1¡0!0¡0!0¡1!0?10 > fε_{(ω^ω)2+1}(10)
¿0¡2¡0!0¡0!0¡1!0?10 > fε_{(ω^ω)2+1}^(ε_{(ω^ω)2)^^10}(10)
¿0¡0¡0!0¡0!0¡0!0¡1!0?10 > fε_{(ω^ω)3}(10)
¿0¡0¡1!0¡1!0?10 > fε_{ω^ω^2}(10)
¿0¡0¡1!0¡1!0¡1!0?10 > fε_{ω^ω^3}(10)
¿0¡0¡2!0!0?10 > fε_{ω^ω^ω}(10)
¿0¡0¡0,0!0!0?10 > fε_ε_0(10)
¿0¡0¡10,0!0!0?10 > fε_ε_1(10)
¿0¡0¡10,0,0!0!0?10 > fε_ε_2(10)
¿0¡0¡0¡0!0!0!0?10 > fε_ε_ω(10)
¿0¡0!¡0!0?10 > fζ0(10)
To show you its beyond zeta zero power:, The Initial Breakdown of ¿0¡0!¡0!0?10
¿X¡X¡X¡X¡X¡X¡X¡X¡X¡X¡10!X¡10!X¡10!X¡10!X¡10!X¡10!X¡10!X¡10!X¡10!X¡10!X!X!X!X!X!X!X!X!X!X!X?1010
Where X = Y¡Y¡Y¡Y¡Y¡Y¡Y¡Y¡Y¡Y¡9!Y¡9!Y¡9!Y¡9!Y¡9!Y¡9!Y¡9!Y¡9!Y¡9!Y¡9!Y!Y!Y!Y!Y!Y!Y!Y!Y!Y!
Where Y = ...
¿0¡0!¡0!0¡0!¡0!0?10 > fζ0^2(10)
¿0¡0¡0!¡0!0!0?10 > fε_{ε_ε_ε_ε_ε_ε_ε_ε_ε_ε_ω}(10) > fε_{(ε_ε_ε_ε_ε_ε_ε_ε_ε_ε_0)+1} > fε_{ζ_0+1}
¿0¡0¡0¡0!¡0!0!0!0?10 > fε_ε_{ζ0+1}
¿0¡0!¡1!0?10 > fζ_1(10)
¿0¡0!¡0,0!0?10 > fζ_ω(10)
¿0¡0!¡0¡0!0!0?10 > fζ_{ω^ω}(10)
¿0¡0!¡0¡0¡0!0!0!0?10 > fζ_εω(10)
¿0¡0!¡0¡0!¡0!0!0?10 > fζ_ζ0(10)
¿0¡1!¡0!0?10 > fη_0(10)
¿0<0>0?100 > fφ(100,100)(100)

User avatar
WarDaft
Posts: 1583
Joined: Thu Jul 30, 2009 3:16 pm UTC

Re: Your number is, in fact, not bigger!

Postby WarDaft » Thu Feb 05, 2015 6:09 am UTC

Aha, a leap ahead!

[2|0[2: :0[]]] for f_{Γω2}(2)
Last edited by WarDaft on Thu Feb 05, 2015 5:02 pm UTC, edited 1 time in total.
All Shadow priest spells that deal Fire damage now appear green.
Big freaky cereal boxes of death.

User avatar
Vytron
Posts: 429
Joined: Mon Oct 19, 2009 10:11 am UTC
Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

Re: Your number is, in fact, not bigger!

Postby Vytron » Thu Feb 05, 2015 9:24 am UTC

n[0,0,0-0-0],[0,0,0-0-0] = Γ_2*2
n,0,[0,0,0-0-0] = Γ_2*ω
n[0,0,0-0-0,0] = n[[[[0-[0-0-0]-0]-0]...-0]-0],[0,0-0-0] = Γ_2^2
n[0,0,0-0-[0,0,0-0-0]] = Γ_2^Γ_2
n[0,0,0-0,0-0] = n[0,0,0-0-[0,0,0-0-...[0,0,0-0-[0,0,0-0-0],0]]] = φ(1,Γ_2+1)
n[0,0,0-[1-0]-0] = n[0,0,0-0-[0,0,0-0-...[0,0,0-0-[0,0,0-0-0],0]]] = φ(φ(ω,0),Γ_2+1)
n[0,0,0-[[1-0]-0]-0] = n[0,0,0-0-[0,0,0-0-...[0,0,0-0-[0,0,0-0-0],0]]] = φ(φ(φ(ω,0),0),Γ_2+1)
n[0,0,0-[0,0,0-0-0]-0] = φ(Γ_2,1)
n[[0,0,0-[0,0,0-0-0]-0]-0] = φ(φ(Γ_2,1),0)
n[[[0,0,0-[0,0,0-0-0]-0]-0]-0] = φ(φ(φ(Γ_2,1),0),0)

n[0,0,0,0-0-0] = n[[[0,0,0-[0,0,0-0-0]...-0]-0]-0] = Γ_3
n[0,0,0,0,0-0-0] = n[[[0,0,0,0-[0,0,0,0-0-0]...-0]-0]-0] = Γ_4
n[0,0,0,0,0,0-0-0] = n[[[0,0,0,0,0-[0,0,0,0,0-0-0]...-0]-0]-0] = Γ_5

n[1-0-0] = n[0,0,...n[0-0-0] ,s...,0,0-0-0] = Γ_ω

n[1-0-0],[1-0-0] = Γ_ω*2
n,0,[1-0-0] = Γ_ω*ω
n[1-0-0,0] = n[0,0,...n[0-0-0] ,s...,0,0-0-0],[1-0-0] = Γ_ω^2
n[1-0-[1-0-0]] = Γ_ω^Γ_ω
n[1-0,0-0] = n[1-0-[1-0-...[1-0-[1-0-0],0]]] = φ(1,Γ_ω+1)
n[1-[1-0]-0] = n[1-0-[1-0-...[1-0-[1-0-0],0]]] = φ(φ(ω,0),Γ_ω+1)
n[1-[[1-0]-0]-0] = n[1-0-[0,0,0-0-...[1-0-[1-0-0],0]]] = φ(φ(φ(ω,0),0),Γ_ω+1)
n[1-[1-0-0]-0] = φ(Γ_ω,1)
n[[1-[1-0-0]-0]-0] = φ(φ(Γ_ω,1),0)
n[[[1-[1-0-0]-0]-0]-0] = φ(φ(φ(Γ_ω,1),0),0)

n[1,0-0-0] = n[[[1-[1-0-0]-0]...-0]-0] = Γ_{ω+1}
n[1,0,0-0-0] = n[[[1,0-[1,0-0-0]-0]...-0]-0] = Γ_{ω+2}
n[1,0,0,0-0-0] = n[[[1,0,0-[1,0,0-0-0]-0]...-0]-0] = Γ_{ω+3}

n[1,1-0-0] = n[1,0,0,...n[1-0-0] ,s...,0,0-0-0] = Γ_{ω2}

n[1,1-0-0],[1,1-0-0] = Γ_{ω2}*2
n,0,[1,1-0-0] = Γ_{ω2}*ω
n[1,1-0-0,0] = n[1,0,0,...n[0-0-0] ,s...,0,0-0-0],[1,1-0-0] = Γ_{ω2}^2
n[1,1-0-[1,1-0-0]] = Γ_{ω2}^Γ_{ω2}
n[1,1-0,0-0] = n[1,1-0-[1,1-0-...[1,1-0-[1,1-0-0],0]]] = φ(1,Γ_{ω2}+1)
n[1,1-[1,1-0]-0] = n[1,1-0-[1,1-0-...[1,1-0-[1,1-0-0],0]]] = φ(φ(ω,0),Γ_{ω2}+1)
n[1,1-[[1,1-0]-0]-0] = n[1,1-0-[0,0,0-0-...[1,1-0-[1,1-0-0],0]]] = φ(φ(φ(ω,0),0),Γ_{ω2}+1)
n[1,1-[1,1-0-0]-0] = φ(Γ_{ω2},1)
n[[1,1-[1,1-0-0]-0]-0] = φ(φ(Γ_{ω2},1),0)
n[[[1,1-[1,1-0-0]-0]-0]-0] = φ(φ(φ(Γ_{ω2},1),0),0)

n[1,1,0-0-0] = n[[[1,1-...[1,1-0-0]-0]-0]-0] = φ(φ(φ(Γ_{ω2},1)...,0),0)

Send:

4[1,1,0-0-0]

For Γ_{ω2+1}(4)

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Thu Feb 05, 2015 4:17 pm UTC

It seems like gamma subindexes aren't too hard for your notations, so I'm skipping up to [2|0[0[0[]]: :0[]] for f{Γω^ω}(2)

EDIT: I keep accidentally typing 1 instead of 0[], which really matters in some places.
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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Fri Feb 06, 2015 1:14 am UTC

Well, I just noticed some redundancy in my notation, I have two possible ways to represent some value, but never use the second one, which means I have an incoming surprise that will be unfolded at some point :D (for short, I can save an extension because I have enough symbols to represent bigger values without needing to introduce extensions, so I can represent something that would require 4 -s in [0-0-0-0] with less -s).

n[1,1,0-0-0] = n[[[1,1-[1,1-0-0]-0]...-0]-0] = Γ_{ω2+1}
n[1,1,0,0-0-0] = n[[[1,1,0-[1,1,0-0-0]-0]...-0]-0] = Γ_{ω2+2}
n[1,1,0,0,0-0-0] = n[[[1,1,0,0-[1,1,0,0-0-0]-0]...-0]-0] = Γ_{ω2+3}

n[1,1,1-0-0] = n[1,1,0,0,...,0,0-0-0] with n[1,1-0-0]s = Γ_{ω3}
n[1,1,1,1-0-0] = n[1,1,1,0,0,...,0,0-0-0] with n[1,1,1-0-0]s = Γ_{ω4}
n[1,1,1,1,1-0-0] = n[1,1,1,1,0,0,...,0,0-0-0] with n[1,1,1,1-0-0]s = Γ_{ω5}

n[0,1-0-0] = n[1,1,...,1,1-0-0] (with n[1-0-0] ,s) = Γ_{ω^2}

n[0,1,0-0-0] = n[[[1,1-[1,1-0-0]-0]...-0]-0] = Γ_{ω^2+1}
n[0,1,0,0-0-0] = n[[[1,1,0-[1,1,0-0-0]-0]...-0]-0] = Γ_{ω^2+2}
n[0,1,0,0,0-0-0] = n[[[1,1,0,0-[1,1,0,0-0-0]-0]...-0]-0] = Γ_{ω^2+3}

n[0,1,1-0-0] = n[0,1,0,0,...,0,0-0-0] with n[1,1-0-0]s = Γ_{ω^2+ω}
n[0,1,1,1-0-0] = n[0,1,1,0,0,...,0,0-0-0] with n[1,1,1-0-0]s = Γ_{ω^2+ω2}
n[0,1,1,1,1-0-0] = n[0,1,1,1,0,0,...,0,0-0-0] with n[1,1,1,1-0-0]s = Γ_{ω^2+ω3}

n[0,1,0,1-0-0] = n[0,1,1,...,1,1-0-0] (with n[0,1-0-0] ,s) = Γ_{ω^2}2
n[0,1,0,1,0,1-0-0] = n[0,1,0,1,1,...,1,1-0-0] (with n[0,1,0,1-0-0] ,s) = Γ_{ω^2}3
n[0,1,0,1,0,1,0,1-0-0] = n[0,1,0,1,0,1,1,...,1,1-0-0] (with n[0,1,0,1,01-0-0] ,s) = Γ_{ω^2}4

n[0,0,1-0-0] = n[0,1,0,1,...,0,1-0-0] (with n[0,1-0-0] ,s) = Γ_{ω^3}
n[0,0,1,0,1-0-0] = n[0,0,1,1,...,1,1-0-0] (with n[0,1-0-0] ,s) = Γ_{ω^3+ω^2}
n[0,0,1,0,0,1-0-0] = n[0,0,1,0,1,...,0,1-0-0] (with n[0,0,1-0-0] ,s) = Γ_{ω^3}2
n[0,0,1,0,0,1,0,0,1-0-0] = n[0,0,1,0,0,1,0,1,...,0,1-0-0] (with n[0,0,1,0,0,1-0-0] ,s) = Γ_{ω^3}3

n[0,0,0,1-0-0] = n[0,0,1,0,0,1,...,0,0,1-0-0] (with n[0,0,1-0-0] ,s) = Γ_{ω^4}
n[0,0,0,0,1-0-0] = n[0,0,0,1,0,0,0,1,...,0,0,0,1-0-0] (with n[0,0,0,1-0-0] ,s) = Γ_{ω^5}
n[0,0,0,0,0,1-0-0] = n[0,0,0,0,1,0,0,0,0,1,...,0,0,0,0,1-0-0] (with n[0,0,0,0,1-0-0] ,s) = Γ_{ω^6}

n[2-0-0] = n[0,0,...,0,0,1-0-0] with n[1-0-0] ,s = Γ_{ω^ω}

n[2,0-0-0] = Γ_{ω^ω+1}
n[2,0,0-0-0] = Γ_{ω^ω+2}
n[2,1-0-0] = Γ_{ω^ω+ω}
n[2,1,1-0-0] = Γ_{ω^ω+ω2}
n[2,0,1-0-0] = Γ_{ω^ω+ω^2}
n[2,0,1,0,1-0-0] = Γ_{ω^ω+{ω^2}2}
n[2,0,0,1-0-0] = Γ_{ω^ω+ω^3}

n[2,2-0-0] = n[2,0,0,...,0,0,1-0-0] with n[1-0-0] ,s = Γ_{ω^ω}2
n[2,2,2-0-0] = n[2,2,0,0,...,0,0,1-0-0] with n[1-0-0] ,s = Γ_{ω^ω}3
n[2,2,2,2-0-0] = n[2,2,2,0,0,...,0,0,1-0-0] with n[1-0-0] ,s = Γ_{ω^ω}4

n[0,2-0-0] = n[2,2,2,...,2,2,2-0-0] with n[2-0-0] ,s = Γ_{ω^{ω+1}}
n[0,0,2-0-0] = n[0,2,0,2,...,0,2,0,2-0-0] with n[0,2-0-0] ,s = Γ_{ω^{ω2}}
n[1,2-0-0] = n[0,0,...,0,2-0-0] with n[0,2-0-0] ,s = Γ_{ω^ω^2}
n[0,1,2-0-0] = n[1,2,...,1,2-0-0] with n[1,2-0-0] ,s = Γ_{ω^ω^3}
n[0,0,1,2-0-0] = n[0,1,2,...,0,1,2-0-0] with n[0,1,2-0-0] ,s = Γ_{ω^ω^4}

n[3-0-0] = n[0,...,0,1,2-0-0] with n[2-0-0] ,s

Send:

4[3-0-0]

For f_{Γ_{ω^ω^ω}}(4)

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Fri Feb 06, 2015 2:02 am UTC

Well then, I simply must see how you respond to:

[2|0[0[][0[]]: :0[]], for f{Γ_(ε_1)}(2)
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Vytron
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Location: The Outside. I use She/He/Her/His/Him as gender neutral pronouns :P

Re: Your number is, in fact, not bigger!

Postby Vytron » Fri Feb 06, 2015 3:28 am UTC

Whew, it goes like this...

n[[0]-0-0] = n[n-0-0] = Γ_(ε_0)
n[[0],0-0-0] = n[[[[0]-[[0]-0-0]-0]...-0]-0] = Γ_(ε_0+1)
n[[0],1-0-0] = n[[0],0,...,0,0-0-0] = Γ_(ε_0+ω)
n[[0],2-0-0] = n[[0],0,...,0,0,1-0-0] = Γ_(ε_0+ω^ω)
n[[0],3-0-0] = n[[0],0,...,0,0,1,2-0-0] = Γ_(ε_0+ω^ω^ω)
n[[0],[0]-0-0] = n[[0],n-0-0] = Γ_(ε_0*2)
n[[0],[0],[0]-0-0] = n[[0],[0],n-0-0] = Γ_(ε_0*3)
n[[0],[0],[0],[0]-0-0] = n[[0],[0],[0],n-0-0] = Γ_(ε_0*4)
n[[0],[0],[0],[0],[0]-0-0] = n[[0],[0],[0],[0],n-0-0] = Γ_(ε_0*4)
n[0,[0]-0-0] = n[[0],[0],...,[0],[0]-0-0] = Γ_(ε_0*ω)
n[0,0,[0]-0-0] = n[0,[0],0,[0],...,0,[0],0,[0]-0-0] = Γ_(ε_0*{ω+1})
n[0,0,0,[0]-0-0] = n[0,0,[0],0,0,[0],...,0,0,[0],0,0,[0]-0-0] = Γ_(ε_0*{ω+2})
n[1,[0]-0-0] = n[0,...,0,[0]-0-0] = Γ_(ε_0*ω2)
n[[0,0]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0]-0-0] = Γ_(ε_0^2)
n[[0,0,0]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0],[0,0]-0-0] = Γ_(ε_0^3)
n[[0-0]-0-0] = n[[n]-0-0] = Γ_(ω^{ε_0+1})
n[[0-0,0]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0]-0-0] = Γ_(ω^ω^{ε_0+1})
n[[0-0,0,0]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0].[0-0,0]-0-0] = Γ_(ω^ω^ω^{ε_0+1})
n[[0-1]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0],...,[0-0,0,...,0,0]-0-0] = Γ_(ε_1)
n[[0-1,0]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0],...,[0-0,0,...,0,0],[1]-0-0] = Γ_(ε_1^ε_0)
n[[0-1,0,0]-0-0] = n[0,...,0,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0],...,[0-0,0,...,0,0],[1],[1,0]-0-0] = Γ_(ε_1^ε_0^ε_0)
n[[0-0,1]-0-0] = n[[0-1,1,...,1,1]-0-0] = Γ_(ω^{ε_1+1})
n[[0-0,0,1]-0-0] = n[[0-0,1,0,1,...,0,1,0,1]-0-0] = Γ_(ω^ω^{ε_1+1})
n[[0-0,0,0,1]-0-0] = n[[0-0,0,1,0,0,1,...,0,0,1,0,0,1]-0-0] = Γ_(ω^ω^ω^{ε_1+1})

n[[0-2]-0-0] = n[[0-0,0,...,0,0,1]-0-0]

Send:

4[[0-2]-0-0]

For f_{Γ_(ε_2)}(4)

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Fri Feb 06, 2015 3:52 am UTC

That was quick, clearly I am not being sufficiently evil challenging.

[2|0[0[][][0[]]: :0[]] for f{Γζ_1}(2)

If you folks want the progressions for these, by the way, the progression for [2|0[X: :0[]]] = f{Γδ}(2) looks like the progression for [2|X] = f{δ}(2)
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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Fri Feb 06, 2015 6:03 am UTC

Okay, so a problem with my notation is I can't do that because "with n ,s" or "with n elements" doesn't allow me to reach the next ordinal, I need n groups of elements, but I haven't defined such groups.

So I'll introduce a new symbol in my notation:

~

~ Rules: If you find a ~ from left to right, look for a string of something at the left of the ~, until you find a -, or a [. If you don't find a - or a [, use the leftmost ,

Now, you'll have one of these:

-X~
[X~
,X~ (where , is the leftmost ,)

Return for each case:

-X~ = -XX...XX with n Xs
[X~ = [XX...XX with n Xs
,X~ = ,XX...XX with n Xs

Where X is the string.

Exception: if you find a "]" then ignore the next "[" that you find, with the effects of ]s being accumulable

Example:

[0,1,2,3,~ = [0,1,2,3,0,1,2,3,0,1,2,3,... with n 0,1,2,3,s

[0,1,2,[0],~ = [0,1,2,[0],0,1,2,[0],0,1,2,[0],... with n 0,1,2,[0],s

There we go.

Now, I actually made a mistake on my previous post, because with these rules I get access to f_{Γ_(ε_ε_0)}(4) instead of f_{Γ_(ε_ω)}(4), bear with me:

n[[0]-0-0] = n[n-0-0] = Γ_(ε_0)
n[[0],0-0-0] = n[[[[0]-[[0]-0-0]-0]...-0]-0] = Γ_(ε_0+1)
n[[0],1-0-0] = n[[0],0,~-0-0] = Γ_(ε_0+ω)
n[[0],2-0-0] = n[[0],0,~1-0-0] = Γ_(ε_0+ω^ω)
n[[0],3-0-0] = n[[0],0,~1,2-0-0] = Γ_(ε_0+ω^ω^ω)
n[[0],[0]-0-0] = n[[0],n-0-0] = Γ_(ε_0*2)
n[[0],[0],[0]-0-0] = n[[0],[0],n-0-0] = Γ_(ε_0*3)
n[[0],[0],[0],[0]-0-0] = n[[0],[0],[0],n-0-0] = Γ_(ε_0*4)
n[[0],[0],[0],[0],[0]-0-0] = n[[0],[0],[0],[0],n-0-0] = Γ_(ε_0*4)
n[0,[0]-0-0] = n[[0],~-0-0] = Γ_(ε_0*ω)
n[0,0,[0]-0-0] = n[0,[0],~-0-0] = Γ_(ε_0*{ω+1})
n[0,0,0,[0]-0-0] = n[0,0,[0],~-0-0] = Γ_(ε_0*{ω+2})
n[1,[0]-0-0] = n[0,~[0]-0-0] = Γ_(ε_0*ω2)
n[[0,0]-0-0] = n[0,~,1,2,3,...,n-1,n,[0]-0-0] = Γ_(ε_0^2)
n[[0,0,0]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0]-0-0] = Γ_(ε_0^3)
n[[0-0]-0-0] = n[[n]-0-0] = Γ_(ω^{ε_0+1})
n[[0-0,0]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0]-0-0] = Γ_(ω^ω^{ε_0+1})
n[[0-0,0,0]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0].[0-0,0]-0-0] = Γ_(ω^ω^ω^{ε_0+1})
n[[0-1]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0],...,[0-0,~]-0-0] = Γ_(ε_1)
n[[0-1,0]-0-0] = n[0,~1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0],...,[0-0,~],[1]-0-0] = Γ_(ε_1^ε_0)
n[[0-1,0,0]-0-0] = n[0,~1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0],...,[0-0,~],[1],[1,0]-0-0] = Γ_(ε_1^ε_0^ε_0)
n[[0-1,1]-0-0] = n[[0-1,0,...,0,0]-0-0] = Γ_(ω^{ε_1+1})
n[[0-1,1,1]-0-0] = n[[0-1,1,0,...,0,0]-0-0] = Γ_(ω^ω^{ε_1+1})
n[[0-1,1,1,1]-0-0] = n[[0-1,1,1,0,...,0,0,]-0-0] = Γ_(ω^ω^ω^{ε_1+1})

New:

n[[0-0,1]-0-0] = n[[0-1,~]-0-0] = Γ_(ε_2)
n[[0-0,0,1]-0-0] = n[[0-0,1,~]-0-0] = Γ_(ε_3)
n[[0-0,0,0,1]-0-0] = n[[0-0,0,1,~]-0-0] = Γ_(ε_4)
n[[0-2]-0-0] = n[[0-0,...,0,1]-0-0] = Γ_(ε_ω)
n[[0-2,0]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0]-0-0],...,[0-1],...,[0-1,0,...,0],...,[0-1,~],...,[0-0,~1]-0-0] = Γ_(ε_{ω+1})
n[[0-2,1]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0]-0-0],...,[0-1],...,[0-1,0,...,0],...,[0-1,~],...,[0-0,~1],...,[0-2,0,...,0]-0-0] = Γ_(ε_{ω2})
n[[0-2,0,1]-0-0] = n[0,~,1,2,3,...,n-1,n,[0],[0,0],...,[1],[2],[3],...[n-1],[n],[0-0],[0-0,0],[0-0,0,0]-0-0],...,[0-1],...,[0-1,0,...,0],...,[0-1,~],...,[0-0,~1],...,[0-2,0,...,0],...,[0-2,1,...,1]-0-0] = Γ_(ε_{ω^2})
n[[0-2,2]-0-0] = n[0,~,...,[0-2,0,0,...,1]-0-0] = Γ_(ε_{{ω^2}2})
n[[0-2,2,2]-0-0] = n[0,~,...,[0-2,2,0,0,...,1]-0-0] = Γ_(ε_{{ω^2}3})
n[[0-0,2]-0-0] = n[0,~,...,[0-2,~]-0-0] = Γ_(ε_{ω^3})
n[[0-0,0,2]-0-0] = n[0,~,...,[0-0,2,~]-0-0] = Γ_(ε_{{ω^3}2})
n[[0-0,0,0,2]-0-0] = n[0,~,...,[0-0,0,2,~]-0-0] = Γ_(ε_{{ω^3}3})
n[[0-0,0,0,2]-0-0] = n[0,~,...,[0-0,0,0,2,~]-0-0] = Γ_(ε_{{ω^3}4})
n[[0-0,1,2]-0-0] = n[0,~,...,[0-1,2,~]-0-0] = Γ_(ε_ω^4)
n[[0-0,0,1,2]-0-0] = n[0,~,...,[0-0,~2]-0-0] = Γ_(ε_ω^5)
n[[0-3]-0-0] = n[0,~,...,[0-0,~1,2]-0-0] = Γ_(ε_ω^ω)
n[[0-4]-0-0] = n[0,~,...,[0-0,~1,2,3]-0-0] = Γ_(ε_ω^ω^ω)
n[[0-[0]]-0-0] = n[0,~,...,[0-n]-0-0] = Γ_(ε_ε_0)
n[[0-[0-[0]]]-0-0] = n[[0-[0,~,...,[0-n]]-0-0] = Γ_(ε_ε_ε_0)
n[[0-[0-[0-[0]]]]-0-0] = n[[0-[0-[0,~,...,[0-n]]]-0-0] = Γ_(ε_ε_ε_ε_0)
n[[0,0-0]-0-0] = n[[0-[0-...[0-[0]]]]-0-0] = Γ_(ζ_0)
n[[0,0-0],0-0-0] = Γ_(ζ_0+1)
n[[0-[0,0-0],0]-0-0] = Γ_(ε_ζ_0+1)
n[[0-[0-[0,0-0],0]]-0-0] = Γ_(ε_ε_ζ_0+1)
n[[0-[0-[0-[0,0-0],0]]]-0-0] = Γ_(ε_ε_ε_ζ_0+1)
n[[0,0-0,0]-0-0] = n[[0-[0-[0-...[0,0-0],0]]]-0-0]

Send:

4[[0,0-0,0]-0-0]

For f_{Γ_(ζ_1)}(4)

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Sun Feb 08, 2015 9:18 am UTC

Okay, sorry for the lame +1ing, I have only my notation to blame, because it's so hard to maintain.

The plan was:

Reach f_{Γ_(ζ_X)}(4) in 4[X-0-0], this way I can plop my value for Γ_0 directly on X and go up to f_{Γ_(ζ_ζ_X)}(4). I currently can't do that because I don't have full equivalence in symbols, I have 1 being ω in X, but I originally reached ω in [1], so I could just plop the same values inside a box, but then that'd waste symbols that I couldn't use for later.

So, I'm dropping my notation once again :P

But I'll switch for a notation that is just like the last one but solves all those problems.

The concept:

I have:

nX

Where X is a string beginning with [

Or

n,X

Where X is a string beginning with some integer.

Now, I can make X match any ordinal that I want, which will allow me to replace X for those ordinals once I get to the boxes.

As follows:

n,0 = n+ n+...n+ n+n with n plus signs
Where you solve the rightmost one to know the value of the previous one.

0,0 = 0
1,0 = n+n = 2
2,0 = n+ 2+2 = 4+4 = 8
3,0 = n+ n+ 3+3 = n+ 6+6 = 12+12 = 24
4,0 = n+ n+ n+ 4+4 = n+ n+ 8+8 = n+ 16+16 = 32+32 = 64
5,0 = n+ n+ n+ n+ 5+5 = n+ n+ n+ 10+10 = n+ n+ 20+20 = n+ 40+40 = 80+80 = 160
6,0 = n+ n+ n+ n+ n+ 6+6 = n+ n+ n+ n+ 12+12 = n+ n+ n+ 24+24 = n+ n+ 48+48 = n+ 96+96 = 192+192 = 384

n,0,0 = n,0 n,0...n,0 n,0 with n spaces

0,0,0 = 0
1,0,0 = n,0 1,0 = 2,0 = 8
2,0,0 = n,0 n,0 2,0 = n,0 8,0 = 2048,0 = 6x10^619
3,0,0 = n,0 n,0 n,0 3,0 = n,0 n,0 24,0 = n,0 402,653,184,0 = 10^10^8,0 = 10^(3*10^99999999)

n,0,0,0 = n,0,0 n,0,0...n,0,0 n,0,0 with n spaces
n,0,0,0,0 = n,0,0,0 n,0,0,0...n,0,0,0 n,0,0,0 with n spaces

n,1 = n,0,~ Where 0,~ = n copies of "0,"

n,1 = f_ω(n)

And from here onwards:

n,X where X is equivalent:

1,0 = 1 1 ... 1 1 (n spaces) = f_ω+1(n)
1,0,0 = 1,0 1,0 ... 1,0 1,0 (n spaces) = f_ω+2(n)
1,0,0,0 = 1,0,0 1,0,0 ... 1,0,0 1,0,0 (n spaces) = f_ω+3(n)

1,1 = 1,0,~ = f_ω2(n)
1,1,0 = 1,1 1,1 ... 1,1 1,1 (n spaces) = f_ω2+1(n)
1,1,0,0 = 1,1,0 1,1,0 ... 1,1,0 1,1,0 (n spaces) = f_ω2+2(n)
1,1,0,0,0 = 1,1,0,0 1,1,0,0 ... 1,1,0,0 1,1,0,0 (n spaces) = f_ω2+3(n)

1,1,1 = 1,1,0,~ = f_ω3(n)
1,1,1,1 = 1,1,1,0,~ = f_ω4(n)
1,1,1,1,1 = 1,1,1,1,0,~ = f_ω5(n)

0,1 = 1,~ Where 1,~ = n copies of "1," = f_ω^2(n)
0,1,0 = 0,1 0,1 ... 0,1 0,1 (n spaces) = f_{ω^2}+1(n)
0,1,1 = 0,1,0,~ = f_{ω^2}+ω(n)
0,1,1,1 = 0,1,1,0,~ = f_{ω^2}+ω2(n)
0,1,1,1,1 = 0,1,1,0,~ = f_{ω^2}+ω3(n)

0,1,0,1 = 0,1,~ Where 1,~ = n copies of "1," = f_{ω^2}2(n)
0,1,0,1,0,1 = 0,1,0,1,~ Where 1,~ = n copies of "1," = f_{ω^2}3(n)
0,1,0,1,0,1,0,1 = 0,1,0,1,~ Where 1,~ = n copies of "1," = f_{ω^2}4(n)

0,0,1 = 0,1,~~ Where ,0,1,~~ = n copies of "0,1," = f_ω^3(n)
0,0,1,0 = 0,0,1 0,0,1 ... 0,0,1 0,0,1 (n spaces) = f_{ω^3}+1(n)
0,0,1,1 = 0,0,1,0,~ = f_{ω^3}+ω(n)
0,0,1,0,1 = 0,0,1,~ = f_{ω^3}+{ω^2}(n)
0,0,1,0,1,0,1 = 0,0,1,0,1,~ = f_{ω^3}+{ω^2}2(n)
0,0,1,0,0,1 = 0,0,1,~~ = f_{ω^3}2(n)
0,0,1,0,0,1,0,0,1 = 0,0,1,0,0,1,~~ = f_{ω^3}3(n)
0,0,1,0,0,1,0,0,1,0,0,1 = 0,0,1,0,0,1,0,0,1,~~ = f_{ω^3}4(n)

0,0,0,1 = 0,0,1,~~~ Where 0,0,1,~~~ = n copies of "0,0,1" = f_{ω^4}(n)
0,0,0,0,1 = 0,0,0,1,~~~~ Where 0,0,0,1,~~~~ = n copies of "0,0,0,1" = f_{ω^5}(n)
0,0,0,0,0,1 = 0,0,0,0,1,~~~~~ Where 0,0,0,0,1,~~~~~ = n copies of "0,0,0,0,1" = f_{ω^6}(n)

2 = 0,~1 = f_{ω^ω}(n)
2,0 = 0,~1 = f_{ω^ω}+1(n)
2,1 = f_{ω^ω}+ω(n)
2,1,1 = f_{ω^ω}+ω2(n)
2,0,1 = f_{ω^ω}+{ω^2}(n)
2,0,1,0,1 = f_{ω^ω}+{ω^2}2(n)
2,0,0,1 = f_{ω^ω}+{ω^3}(n)
2,2 = 2,0,~1 = f_{ω^ω}2(n)
2,2,2 = 2,0,~1 = f_{ω^ω}3(n)
2,2,2,2 = 2,0,~1 = f_{ω^ω}4(n)
0,2 = 2,~ Where 2,~ = n copies of "2," = f_{ω^{ω+1}}(n)
1,2 = 0,~2 = f_{ω^ω2}(n)
0,1,2 = 1,2,~~ Where 1,2,~ = n copies of "1,2" = f_{ω^ω^2}(n)
0,1,2,0,1,2 = 0,1,2,1,2,~~ = f_{ω^ω^2}2(n)
0,0,1,2 = 0,1,2,~~~ Where 0,1,2,~~~ = n copies of "0,1,2" = f_{ω^ω^3}(n)
0,0,0,1,2 = 0,0,1,2,~~~~ Where 0,0,1,2,~~~~ = n copies of "0,0,1,2" = f_{ω^ω^4}(n)

3 = 0,~1,2 = f_{ω^ω^ω}(n)
4 = 0,~1,2,3 = f_{ω^ω^ω^ω}(n)
5 = 0,~1,2,3,4 = f_{ω^ω^ω^ω^ω}(n)

[0] = n = ε_0

That is:

0[0] = 0
1[0] = 1,1 = 1,0 = 2

2[0] =
2,2 =
2,0,0,0,1 =
2,0,0,1,0,0,1,0,0,1 =
2,0,0,1,0,0,1,0,1,0,1,0,1 =
2,0,0,1,0,0,1,0,1,0,1,1,1,1 =
2,0,0,1,0,0,1,0,1,0,1,1,1,0,0,0 =
n,0,0,1,0,0,1,0,1,0,1,1,1,0,0 n,0,0,1,0,0,1,0,1,0,1,1,1,0,0 2,0,0,1,0,0,1,0,1,0,1,1,1,0,0
Etc.

n[0],0 = n[0] n[0]...n[0] with n spaces = ε_0+1
n[0],0,0 = n[0],0 n[0],0...n[0],0 with n spaces = ε_0+2
n[0],1 = n[0],0,~ = ε_0+ω
n[0],2 = n[0],0,~ = ε_0+ω^ω
n[0],3 = n[0],0,~ = ε_0+ω^ω^ω

n[0],[0] = n[0],n = ε_0*2
n[0],[0],[0] = n[0],[0],n = ε_0*3
n,[0],[0],[0],[0] = n,[0],n = ε_0*4

n,0,[0] = n[0],~ = ε_0*ω
n,0,[0],0,[0] = ε_0*{ω+1}
n,0,[0],0,[0],0,[0] = ε_0*{ω+2}
n,0,0,[0] = n,0,[0],~~ = ε_0*{ω2}
n,1,[0] = n,0,~[0] = ε_0*{ω^ω}
n,2,[0] = n,0,~1,[0] = ε_0*{ω^ω^ω}
n[0]~[0] = n,n,[0] = ε_0^2
n[0],0~[0] = n,[0]~n,[0] = {ε_0^2}2
n[0],0,0~[0] = n,[0],0~n,[0] = {ε_0^2}3
n[0],1~[0] = n,[0],0~n,[0] = ε_0^3
n[0]~[0]~[0] = n[0],n~[0]~[0] = ω^{ε_0+1}
n[0]~[0]~[0]~[0] = n[0],n~[0]~[0]~[0] = ω^ω^{ε_0+1}
n[0,0] = n[0]~[0]~ = ε_1
n[0,0,0] = n[0,0]~[0,0]~ = ε_2
n[1] = n[0,~] = ε_ω
n[2] = n[0,~1] = ε_{ω^ω}
n[3] = n[0,~1,2] = ε_{ω^ω^ω}
n[[0]] = n[n] = ε_ε_0
n[[[0]]] = ε_ε_ε_0

n[0-0] = n[~[0]~] =ζ_0
n[0-0],0 = ζ_0+1
n[[0-0],0] = ε_{ζ_0+1}
n[[[0-0],0]] = ε_ε_{ζ_0+1}
n[0-0,0] = n[~[0-0],0~] ζ_1
n[0-[0-0]] = ζ_ζ_0
n[0,0-0] = n[0-[~0-0]~] = φ(3,0)
n[0,0,0-0] = n[0,0-[~0,0-0]~] = φ(4,0)
n[1-0] = n[0,~-[~0,~-0]~] = φ(ω,0)
n[[0]-0] = n[n-[~n-0]~] = φ(ε_0,0)
n[[0-0]-0] = n[[~[0]~]-[~[~[0]~]-0]~] = φ(ζ_0,0)
n[[0,0-0]-0] = n[[0-[~0-0]~]-[~[0-[~0-0]~]-0]~] = φ(φ(3,0),0)
n[[1-0]-0] = n[[0,~-[~0,~-0]~]-[~[0,~-[~0,~-0]~]-0]~] = φ(φ(ω,0),0)
n[[[1-0]-0]-0] = n[[[0,~-[~0,~-0]~]-[~[0,~-[~0,~-0]~]-0]~]-[~[[0,~-[~0,~-0]~]-[~[0,~-[~0,~-0]~]-0]~]-0]~] = φ(φ(φ(ω,0),0),0)

n[0-0-0] = n[~[1-0]-0~] = φ(Γ_0,0)

n[0-0-0],0 = Γ_0+1
n[0-0-0],[0-0-0] = Γ_0*2
n,0,[0-0-0] = Γ_0*ω
n[0-0-0]~[0-0-0] = Γ_0^2
n[[0-0-0],0] = φ(1,Γ_0+1)
n[0-[0-0-0],0] = φ(2,Γ_0+1)
n[0,0-[0-0-0],0] = φ(3,Γ_0+1)
n[1-[0-0-0],0] = φ(ω,Γ_0+1)
n[[1-0]-[0-0-0],0] = φ(φ(ω,0),Γ_0+1)
n[[[1-0]-0]-[0-0-0],0] = φ(φ(ω,0),Γ_0+1)
n[[[[1-0]-0]-0]-[0-0-0],0] = φ(φ(φ(ω,0),0),Γ_0+1)

n[0-0-0,0] = n[[~[1-0]-0~]-[0-0-0],0] = φ(Γ_0,1)

n[[0-0-0,0]-0] = φ(φ(Γ_0,1),0)
n[[[0-0-0,0]-0]-0] = φ(φ(φ(Γ_0,1),0),0)

n[0-0-0,0,0] = n[~[0-0-0,0]-0~] = Γ_1

n[0-0-0,0,0],0 = Γ_1+1
n[0-0-0,0,0],[0-0-0,0,0] = Γ_0*2
n,0,[0-0-0,0,0] = Γ_1*ω
n[0-0-0,0,0]~[0-0-0,0,0] = Γ_1^2
n[[0-0-0,0,0],0] = φ(1,Γ_1+1)
n[0-[0-0-0,0,0],0] = φ(2,Γ_1+1)
n[0,0-[0-0-0,0,0],0] = φ(3,Γ_1+1)
n[1-[0-0-0,0,0],0] = φ(ω,Γ_1+1)
n[[1-0]-[0-0-0,0,0],0] = φ(φ(ω,0),Γ_1+1)
n[[[1-0]-0]-[0-0-0,0,0],0] = φ(φ(ω,0),Γ_1+1)
n[[[[1-0]-0]-0]-[0-0-0,0,0],0] = φ(φ(φ(ω,0),0),Γ_1+1)

n[0-0-0,0,0,0] = n[~[1-0]-0~]-[0-0-0,0,0],0] = Γ_2
n[0-0-0,0,0,0,0] = n[~[1-0]-0~]-[0-0-0,0,0,0],0] = Γ_3
n[0-0-0,0,0,0,0,0] = n[~[1-0]-0~]-[0-0-0,0,0,0],0] = Γ_4

n[0-0-1] = n[0-0-0,~] = Γ_ω
n[0-0-[0]] = n[0-0-n] = Γ_ε_0
n[0-0-[[0]]] = n[0-0-[n]] = Γ_ε_ε_0
n[0-0-[0-0]] = n[0-0-[~[0]~]] = Γ_ζ_0

n[0-0-[0-0],0] = n[~[1-0]-0~]-[0-0-[0-0]] = ζ_0+1
n[0-0-[[0-0],0]] = Γ_ε_{ζ_0+1}
n[0-0-[[[0-0],0]]] = Γ_ε_ε_{ζ_0+1}
n[0-0-[0-0,0]] = n[0-0-[~[0-0],0~] = Γ_ζ_1
n[0-0-[0-0,0,0]] = n[0-0-[~[0-0,0],0~] = Γ_ζ_2
n[0-0-[0-0,0,0,0]] = n[0-0-[~[0-0,0,0],0~] = Γ_ζ_3

n[0-0-[0-1]] = n[0-0-[0-0,~]]

Send:

4[0-0-[0-1]]

For f_{Γ_ζ_ω}(4)

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Sun Feb 08, 2015 4:49 pm UTC

Ah yes, making sub sections act like substitutable ordinals is an excellent way forward.

On that note, I feel I must leap to [2|0[ :0[]:0[]]]. This operates at the first fixed point of a -> Γa.

I'm developing a made general extension from colons, part of that is making a repetition operator, < >.

In this manner, <(>0<,0)> for n=3 would become (((0,0),0),0) for example, or (<0<|>0>) would become (0|||00|||00|||0). I believe it will be useful for complicated nestings, and everyone is free to borrow it, since it's not a number notation, it's a notation describing tool.

Daggoth
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 08, 2015 5:53 pm UTC

Is that fixed point the limit of Γ_ Γ_.... Γ_0?

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Sun Feb 08, 2015 6:30 pm UTC

Yes.
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Daggoth
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Sun Feb 08, 2015 7:13 pm UTC

ok here's something i don't quite fully grasp and i'd like some help on
I get that ζ1 is limit of ε...ε_(ζ_0+1) but
if you use a sequence such as ζ_0+X, ε_(ζ_0+X), ε_ε_(ζ_0+X), ε_ε_ε_(ζ_0+X)
where X is a number higher than 1 or some ordinal lesser than ζ_0. Does this suffice to pass ζ1

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Sun Feb 08, 2015 7:29 pm UTC

If X is less than ζ_1, no. The supremum (least ordinal greater than all of the elements) of the sequence ζ_0 + 7, ε_(ζ_0+7), ε_ε_(ζ_0+7), ε_ε_ε_(ζ_0+7)... is also ζ_1.

The reason for this is because for every ordinal δ less than ζ_1, there is an ordinal in that sequence that is greater than δ, yet every ordinal in the sequence is still less than ζ_1, therefore the supremum is still only ζ_1. Another way of thinking about it, is that if you were 'counting up' to ζ_1, you would pass each of those ordinals on your way.

It doesn't even have to be a consistent X. You could have the sequence ζ_0 + 1, ε_(ζ_0+3), ε_ε_(ζ_0 + ε_3), ε_ε_ε_(ζ_0ε_ε_2+3)... and while we can no longer predict what exactly each term will look like from the example, it's still perfectly reasonable for such a sequence to have a limit of ζ_1.
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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 09, 2015 2:16 am UTC

Okay, so I have here:

n[0-0-[0-0-0]] = n[0-0-[~[1-0]-0~]] = Γ_Γ_0

n[0-0-[0-0-[0-0-0]]] = n[0-0-[0-0-[~[1-0]-0~]]] = Γ_Γ_Γ_0

n[0-0-[0-0-[0-0-[0-0-0]]]] = n[0-0-[0-0-[0-0-[~[1-0]-0~]]]] = Γ_Γ_Γ_Γ_0

n[0-0,0-0] = n[0-0-[~0-0-[0-0-0]~]]

(2[0-0,0-0] should match WarDaft)

Send

4[0-0-[0-0,0-0]]

For f_{Γ_(first fixed point of a -> Γa)}

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 09, 2015 2:29 am UTC

That's not how ordinals work. It's a fixed point, the point where δ = Γδ. So Γδ is still just δ. If your notation works so that you can truly substitute ordinals like that, then that should be equivalent to 5[0-0,0-0].
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Mon Feb 09, 2015 2:48 am UTC

Its like raising omega to epsilon0

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 09, 2015 3:28 am UTC

Well, I think I can do away with FGH comparisons and beat you old style:

n[0-0,0-0] = f_{Γ_Γ_...Γ_Γ_0}(n)

n[0-0,0-0],0 = n[0-0,0-0] n[0-0,0-0] ... n[0-0,0-0]
n[0-0,0-0],0,0 = n[0-0,0-0],0 n[0-0,0-0],0 ... n[0-0,0-0],0
n[0-0,0-0],0,0,0 = n[0-0,0-0],0,0 n[0-0,0-0],0,0 ... n[0-0,0-0],0,0

n[0-0,0-0],1 = n[0-0,0-0],0,~
n[0-0,0-0],1,1 = n[0-0,0-0],1,0,~
n[0-0,0-0],0,1 = n[0-0,0-0],1,~
n[0-0,0-0],0,0,1 = n[0-0,0-0],0,1,~~
n[0-0,0-0],0,0,0,1 = n[0-0,0-0],0,0,1,~~~

n[0-0,0-0],2 = n[0-0,0-0],0,~1
n[0-0,0-0],3 = n[0-0,0-0],0,~1,2

n[0-0,0-0],[0] = n[0-0,0-0],n

n[0-0,0-0],[0],2 = n[0-0,0-0],[0],0,~1
n[0-0,0-0],[0],3 = n[0-0,0-0],[0],0,~1,2
n[0-0,0-0],[0],[0] = n[0-0,0-0],[0],n
n[0-0,0-0],0,[0] = n[0-0,0-0],[0],~
n[0-0,0-0],0,0,[0] = n[0-0,0-0],0,[0],~~
n[0-0,0-0],1,[0] = n[0-0,0-0],0,~[0]
n[0-0,0-0],2,[0] = n[0-0,0-0],0,~1,[0]
n[0-0,0-0],[0]~[0] = n[0-0,0-0],n,[0]
n[0-0,0-0],[0,0] = n[0-0,0-0],[0]~[0]~
n[0-0,0-0],[0,0,0] = n[0-0,0-0],[0,0]~[0,0]~
n[0-0,0-0],[1] = n[0-0,0-0],[0,~]
n[0-0,0-0],[2] = n[0-0,0-0],[0,~1]
n[0-0,0-0],[[0]] = n[0-0,0-0],[n]
n[0-0,0-0],[[[0]]] = n[0-0,0-0],[[n]]
n[0-0,0-0],[0-0] = n[0-0,0-0],[~[0]~]
n[0-0,0-0],[0-0,0] = n[0-0,0-0],[~[0-0],0~]
n[0-0,0-0],[0,0-0] = n[0-0,0-0],[0-[~0-0]~]
n[0-0,0-0],[0-0-0] = n[0-0,0-0],[~[1-0]-0~]
n[0-0,0-0],[0-0-0,0] = n[0-0,0-0],[[~[1-0]-0~]-[0-0-0],0]
n[0-0,0-0],[0-0-[0-0-0]] = n[0-0,0-0],[0-0-[~[1-0]-0~]]
n[0-0,0-0],[0-0,0-0] = n[0-0,0-0],[0-0-[~0-0-[0-0-0]~]]
n[0-0,0-0],[0-0,0-0],[0-0,0-0] = n[0-0,0-0],[0-0,0-0],[0-0-[~0-0-[0-0-0]~]]

n,0,[0-0,0-0] = n[0-0,0-0],~
n,0,0,[0-0,0-0] = n,0,[0-0,0-0],~~
n,1,[0-0,0-0] = n,0,~[0-0,0-0]
n,2,[0-0,0-0] = n,0,~1,[0-0,0-0]
n[0],[0-0,0-0] = n,n,[0-0,0-0]
n,0,[0],[0-0,0-0] = n[0],~[0-0,0-0]
n[0]~[0],[0-0,0-0] = n,n,[0],~[0-0,0-0]
n[0,0],[0-0,0-0] = n[0]~[0]~,[0-0,0-0]
n[0,0,0],[0-0,0-0] = n[0,0]~[0,0]~,[0-0,0-0]
n[1],[0-0,0-0] = n[0,~],[0-0,0-0]
n[2],[0-0,0-0] = n[0,~1],[0-0,0-0]
n[[0]],[0-0,0-0] = n[n],[0-0,0-0]
n[[[0]]],[0-0,0-0] = n[[n]],[0-0,0-0]
n[0-0],[0-0,0-0] = n[~[0]~],[0-0,0-0]
n[0-0,0],[0-0,0-0] = n[~[0-0],0~],[0-0,0-0]
n[0,0-0],[0-0,0-0] = n[0-[~0-0]~],[0-0,0-0]
n[0-0-0],[0-0,0-0] = n[~[1-0]-0~],[0-0,0-0]
n[0-0-0,0],[0-0,0-0] = n[[~[1-0]-0~]-[0-0-0],0],[0-0,0-0]
n[0-0-[0-0-0]],[0-0,0-0] = n[0-0-[~[1-0]-0~]],[0-0,0-0]
n[0-0,0-0]~[0-0,0-0] = n[0-0-[~0-0-[0-0-0]~],[0-0,0-0]
n[0-0,0-0]~[0-0,0-0]~[0-0,0-0] = n[0-0-[~0-0-[0-0-0]~]~[0-0,0-0]

n[[0-0,0-0]] = n[0-0,0-0]~[0-0,0-0]~

(i.e. n[0-0,0-0]~[0-0,0-0]~ is n[0-0,0-0]~[0-0,0-0]...~[0-0,0-0] with n ~s)

I bolded the important parts.

Now, to get to n[[0-0,0-0]] (which is still far away from n[0-0-[0-0,0-0]]) one had to undergo limits of the [0-0-[~0-0-[0-0-0]~] kind several times, so how could 4[0-0-[0-0,0-0]] equal 5[0-0,0-0]?
Last edited by Vytron on Mon Feb 09, 2015 3:31 am UTC, edited 1 time in total.

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Vytron
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Re: Your number is, in fact, not bigger!

Postby Vytron » Mon Feb 09, 2015 3:30 am UTC

Anyway, rewriting:

n[0-0,0-0] = n[0-0-[~0-0-[0-0-0],0~]]

Send

4[0-0-[0-0,0-0]]

Now this is like raising omega to e_0+1.

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Mon Feb 09, 2015 6:26 pm UTC

Alright, I'll accept that.

In return, I challenge thee with [2|0[1:0[]:0[]]], being the ωth fixed point of the gamma function.
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Daggoth
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Re: Your number is, in fact, not bigger!

Postby Daggoth » Tue Feb 10, 2015 5:19 am UTC

is there any mainstream way to represent these fixed points / ordinals?

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WarDaft
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Re: Your number is, in fact, not bigger!

Postby WarDaft » Tue Feb 10, 2015 7:25 am UTC

Yes, we extend Veblen's φ function to multiple arguments.

Thus, φ(1,0,a) = Γa,
φ(a+1,0,b) is the b'th fixed point of c -> φ(a,c,0)
φ(a,b+1,c) is the c'th fixed point of d -> φ(a,b,d)
And some other bookkeeping needed to deal with limit ordinals, but this is the general idea.

So my latest entry is φ(1,1,ω).
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