Talk: Pair sequence number

Analysis

I have just spent quite a bit of time analyzing this system, and so far, I'm getting similar results.

My basic results:

let |A| be the ordinal for sequence A.

Theorem 1. (0,0) basically separates the sequence into summands, so for example

|(0,0) A (0,0) B (0,0) C| = |(0,0) A| + |(0,0) B| + |(0,0) C|

Proof: The (0,0)'s serve as stopgaps; the reduction rule can never pass to the left of a (0,0). So each partition that the (0,0)'s get divided into get evaluated one after the other, so the final ordinal is just the sum.

Let [A - (b,c)] be sequence A with each term decreased by b in the first term and c in the second term. So for example [(1,1)(2,1)(3,1) - (1,1)] = (0,0)(1,0)(2,0)

Theorem 2. In a part of the sequence with no (0,0)'s, the (1,0)'s satisfy the following:

|(0,0) A (1,0) B| = |(0,0)A| * w^|[(1,0) B - (1,0)]|

Proof: We use induction on the exponent |[(1,0)B - (1,0)]|.

Successor case: From the previous theorem, |(0,0) B| + 1 = |(0,0) B (0,0)|. So we have

|(0,0) A| * w^|(0,0) B (0,0)| = |(0,0) A| * w^(|(0,0) B| + 1) = |(0,0) A| * w^|(0,0) B| * w = |(0,0) A (1,0) [B + (1,0)]| * w (by the induction hypothesis)

and

|(0,0) A (1,0) [B + (1,0)] (1,0)| = sup_n |{(0,0) A (1,0) [B + (1,0)]} repeated n times| = |(0,0) A (1,0) [B + (1,0)]| * w, as desired.

Limit case: Straightforward. QED

Note that we can chain the previous theorem, so for example

|(0,0) A (1,0) B (2,0) C (3,0) D| = |(0,0) A| * w^|(0,0) [B - (1,0)] (1,0) [C - (1,0)] (2,0) [D - (1,0)|

= |(0,0) A| * w^(|(0,0) [B - (1,0)]| * w^|(0,0) [C - (2,0)] (1,0) [D - (2,0)]|)

= |(0,0) A| * w^(|(0,0) [B - (1,0)]| * w^(|(0,0) [C - (2,0)]| * w^|(0,0)[D - (3,0)]|))

= |(0,0) A| * w^|(0,0) [B - (1,0)]| * w^w^|(0,0) [C - (2,0)]| * w^w^w^|(0,0)[D - (3,0)]|

Since we can partition out (0,0)'s and (1,0)'s, we will assume no interior such pairs from now on.

Theorem 3. |(0,0) A (1,1)| is the next epsilon number after |(0,0) A|

Proof:

|(0,0) A (1,1)| = sup_n |(0,0) A (1,0) [A + (1,0)] (2,0) [A + (2,0)] (3,0) [A + (3,0)] {n repeats}|

= sup_n (|(0,0) A| + w^|(0,0) A| + w^w^|(0,0) A| + w^w^w^|(0,0) A|...)

which will be the next epsilon number after |(0,0) A|.

Theorem 4. If B has no (1,1)'s, |(0,0) A| = epsilon_a, and |(0,0) A (1,1) B| = epsilon_(a+b), then |(0,0) A (1,1) B (2,0)| = epsilon_(a+b*w)

Proof:

|(0,0) A (1,1) B (2,0)| = sup_n |(0,0) A (1,1) B (1,1) B (1,1) B {n repeats}| = epsilon_(a+b*w)

at this point proofs get more and more difficult, but we can notice certain patterns. Once we partition out all the (0,0)'s and (1,0)'s (as well as all the new (1,0)'s that get formed by subtraction) the remaining partitions represent epsilon numbers. (1,1) increments the epsilon argument by 1, and (2,0) multiplies the part between the last (1,1) and the (2,0) by w. So for example

|(0,0)(1,1)(2,0)| = epsilon_w

|(0,0)(1,1)(2,0)(2,0)| = epsilon_(w^2)

|(0,0)(1,1)(2,0)(2,0)(2,0)| = epsilon_(w^3)

|(0,0)(1,1)(2,0)(2,0)(2,0)(1,1)| = epsilon_(w^3 + 1)

|(0,0)(1,1)(2,0)(2,0)(2,0)(1,1)(2,0)| epsilon_(w^3 + w)

Then (3,0) functions in the same way as (1,0) did before, representing the function w^a in the epsilon argument. So

|(0,0)(1,1)(2,0)(3,0)| = epsilon_(w^w)

|(0,0)(1,1)(2,0)(3,0)(3,0)| = epsilon_(w^(w^2))

|(0,0)(1,1)(2,0)(3,0)(3,0)(2,0)| = epsilon_(w^(w^2 + 1))

|(0,0)(1,1)(2,0)(3,0)(3,0)(2,0)(3,0)| = epsilon_(w^(w^2 + w))

so just like (0,0)(1,0)(2,0)(3,0)... goes up to epsilon_0, (0,0)(1,1)(2,0)(3,0)(4,0)(5,0)... goes up to epsilon_(epsilon_0). Thus we have

|(0,0)(1,1)(2,0)(3,1)| = epsilon_epsilon_0

|(0,0)(1,1)(2,0)(3,1)(4,0)(5,1)| = epsilon_epsilon_epsilon_0

|(0,0)(1,1)(2,1)| = phi(2,0)

Just as (0,0) and (1,0) took care of addition and exponentiating by w, (1,1) and (2,0) take care of epsilon arguments. So for example

|(0,0)(1,1)(2,1)(2,0)| = epsilon_(phi(2,0) * w)

|(0,0)(1,1)(2,1)(2,0)(3,0)| = epsilon_(w^(phi(2,0) * w))

|(0,0)(1,1)(2,1)(2,0)(3,0),(4,0)| = epsilon_(w^w^(phi(2,0) * w))

|(0,0)(1,1)(2,1)(2,0)(3,1)| = epsilon_epsilon_(phi(2,0) + 1)

At this point the patterns seem to get quite complicated.

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,0)(5,1)| = epsilon_epsilon_epsilon_(phi(2,0) + 1))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)| = phi(2,1)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(2,0)(3,1)(4,1)| = phi(2,2)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)| = phi(2,w)

It looks like appending (2,0)(3,1)(4,1) takes the ordinal to the next zeta value.

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(2,0)(3,1)(4,1)| = phi(2,w+1)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(2,0)(3,1)(4,1)(3,0)| = phi(2,w2)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(3,0)| = phi(2,w^2)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(3,0)(3,0)| = phi(2,w^3)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,0)| = phi(2,w^w)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)| = phi(2,epsilon_0)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,0)(6,1)| = phi(2,epsilon_epsilon_0)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)| = phi(2, phi(2,0))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(2,0)(3,1)(4,1)| = phi(2, phi(2,0)+1)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)| = phi(2, phi(2,0)*2)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(3,0)| = phi(2, phi(2,0)*w)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(3,0)(3,0)| = phi(2, phi(2,0)*w^w)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(3,0)(4,0)| = phi(2, phi(2,0)*w^w^w)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(3,0)(4,1)| = phi(2, phi(2,0)*epsilon_0)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(3,0)(4,1)(5,0)(6,1)| = phi(2, phi(2,0)*epsilon_epsilon_0)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(3,0)(4,1)(5,1)| = phi(2, phi(2,0)^2)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(4,0)| = phi(2, phi(2,0)^w) = phi(2, w^w^(phi(2,0)+1))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(4,0)(4,0)| = phi(2, w^w^(phi(2,0)+2))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(4,0)(5,0)| = phi(2, w^w^(phi(2,0)+w))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(4,0)(5,0)(6,0)| = phi(2, w^w^phi(2,0)+w^w))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(4,0)(5,1)| = phi(2, w^w^(phi(2,0) + epsilon_0))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)(4,0)(5,1)(6,1)| = phi(2, w^w^(phi(2,0)*2))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)| = phi(2, epsilon_(phi(2,0)+1))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(3,0)(4,1)(5,1)(4,1)| = phi(2, epsilon_(phi(2,0)+2))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(3,1)| = phi(2, epsilon_(phi(2,0)+w))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(4,0)(5,1)(6,1)| = phi(2, epsilon_(phi(2,0)*2))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(4,0)(5,1)(6,1)(4,0)(5,1)(6,1)| = phi(2, epsilon_(phi(2,0)^2))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(4,0)(5,1)(6,1)(5,0)(6,1)(7,1)| = phi(2, epsilon_(w^w^(phi(2,0)*2)))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(4,0)(5,1)(6,1)(5,1)| = phi(2, epsilon_epsilon_(phi(2,0)+1))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(4,1)| = phi(2, phi(2,1))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(3,1)(4,1)(3,1)(4,1)| = phi(2, phi(2,2))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(4,0)| = phi(2, phi(2,w))

This is getting quite unwieldy. I'll think about it more later. Deedlit11 (talk) 06:16, September 29, 2015 (UTC)

I tried to replicate your results, but got larger ordinals for everything beyond ζ₀:

|(0,0)(1,1)(2,1)| = ζ₀

|(0,0)(1,1)(2,1)(1,1)| = ε_(ζ₀+1)

|(0,0)(1,1)(2,1)(1,1)(2,0)| = ε_(ζ₀+ω)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)| = ε_(ζ₀+ε₀)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)| = ε_(ζ₀×2)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(2,0)| = ε_(ζ₀×ω)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(2,0)(3,1)| = ε_(ζ₀×ε₀)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(2,0)(3,1)(4,1)| = ε_(ζ₀²)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(3,0)| = ε_(ζ₀^ω)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(3,0)(4,1)| = ε_(ζ₀^ε₀)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(3,0)(4,1)(5,1)| = ε_(ζ₀^ζ₀)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(3,1)| = ε_ε_(ζ₀+1)

|(0,0)(1,1)(2,1)(1,1)(2,0)(3,1)(4,1)(3,1)(4,0)(5,1)(6,1)(5,1)| = ε_ε_ε_(ζ₀+1)

|(0,0)(1,1)(2,1)(1,1)(2,1)| = ζ₁

|(0,0)(1,1)(2,1)(1,1)(2,1)(1,1)(2,1)| = ζ₂

|(0,0)(1,1)(2,1)(2,0)| = ζ_ω = φ(2,ω)

|(0,0)(1,1)(2,1)(2,0)(2,0)| = φ(2,ω²)

|(0,0)(1,1)(2,1)(2,0)(3,1)| = φ(2,ε₀)

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)| = ζ_ζ₀ = φ(2,φ(2,0))

|(0,0)(1,1)(2,1)(2,0)(3,1)(4,1)(4,0)(5,1)(6,1)| = ζ_ζ_ζ₀ = φ(2,φ(2,φ(2,0)))

|(0,0)(1,1)(2,1)(2,1)| = η₀ = φ(3,0)

And I'm guessing the progression continues with:

|(0,0)(1,1)(2,1)(2,1)(2,1)| = φ(4,0)

|(0,0)(1,1)(2,1)(2,1)(2,1)(2,1)| = φ(5,0)

|(0,0)(1,1)(2,1)(3,0)| = φ(ω,0)

|(0,0)(1,1)(2,1)(3,0)(3,0)| = φ(ω²,0)

|(0,0)(1,1)(2,1)(3,0)(4,1)| = φ(ε₀,0)

|(0,0)(1,1)(2,1)(3,0)(4,1)(5,1)| = φ(ζ₀,0)

|(0,0)(1,1)(2,1)(3,0)(4,1)(5,1)(6,0)(7,1)(8,1)| = φ(φ(ζ₀,0),0)

|(0,0)(1,1)(2,1)(3,1)| = Γ₀ (which we already know from the article)

Of-course, the whole thing is very confusing and I could very well have made a mistake. So if anyone has the time and the will to verify/refute these results, I'll be most grateful. 84.229.92.191 22:38, November 24, 2016 (UTC)

Hmm, your analysis looks pretty good. I wrote mine over a year ago, so I don't remember exactly what I was thinking near the end.

It would great if you would continue your analysis, especially until or beyond$\psi(\Omega_\omega)$. Deedlit11 (talk) 00:19, December 6, 2016 (UTC)

Okay, here is my analysis up to BHO:

|(0,0)(1,1)(2,1)(3,1)| = Γ₀

|(0,0)(1,1)(2,1)(3,1)(1,1)| = ε_(Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(1,1)| = ε_(Γ₀+2)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)| = ε_(Γ₀+ω)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)| = ε_(Γ₀+ε₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(3,1)| = ε_(Γ₀+ε₁)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,0)| = ε_(Γ₀+ε_ω)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,0)(5,1)| = ε_(Γ₀+ε_ε₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)| = ε_(Γ₀+ζ₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)| = ε_(Γ₀×2)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(1,1)(2,0)(3,1)(4,1)(5,1)| = ε_(Γ₀×3)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(2,0)| = ε_(Γ₀×ω)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(2,0)(3,1)| = ε_(Γ₀×ε₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(2,0)(3,1)(4,1)| = ε_(Γ₀×ζ₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(2,0)(3,1)(4,1)(5,1)| = ε_(Γ₀²)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(3,0)| = ε_(Γ₀^ω)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(3,0)(4,1)| = ε_(Γ₀^ε₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(3,0)(4,1)(5,1)(6,1)| = ε_(Γ₀^Γ₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(3,0)(4,1)(5,1)(6,1)(4,0)(5,1)(6,1)(7,1)| = ε_(Γ₀^Γ₀^Γ₀)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(3,1)| = ε_ε_(Γ₀+1))

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,0)(3,1)(4,1)(5,1)(3,1)(4,0)(5,1)(6,1)(7,1)(5,1)| = ε_ε_ε_(Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)| = ζ_(Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(2,1)| = η_(Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(2,1)(2,1)| = φ(4,Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)| = φ(ω,Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)(4,1)| = φ(ε₀,Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)(4,1)(5,1)| = φ(ζ₀,Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)(4,1)(5,1)(6,1)| = φ(Γ₀,Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)(4,1)(5,1)(6,1)(4,1)| = φ(ε_(Γ₀+1),Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)(4,1)(5,1)(6,1)(4,1)(5,1)| = φ(ζ_(Γ₀+1),Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,0)(4,1)(5,1)(6,1)(4,1)(5,1)(6,0)(7,1)(8,1)(9,1)(7,1)(8,1)| = φ(φ(ζ_(Γ₀+1),Γ₀+1), Γ₀+1)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,1)| = Γ₁ (which was expected)

|(0,0)(1,1)(2,1)(3,1)(1,1)(2,1)(3,1)(1,1)(2,1)(3,1)| = Γ₂

|(0,0)(1,1)(2,1)(3,1)(2,0)| = Γ_ω = φ(1,0,ω)

|(0,0)(1,1)(2,1)(3,1)(2,0)(3,1)| = φ(1,0,ε₀)

|(0,0)(1,1)(2,1)(3,1)(2,0)(3,1)(4,1)(5,1)| = φ(1,0,Γ₀)

|(0,0)(1,1)(2,1)(3,1)(2,1)| = φ(1,1,0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(1,1)(2,1)(3,1)(2,1)| = φ(1,1,1)

|(0,0)(1,1)(2,1)(3,1)(2,1)(1,1)(2,1)(3,1)(2,1)(1,1)(2,1)(3,1)(2,1)| = φ(1,1,2)

|(0,0)(1,1)(2,1)(3,1)(2,1)(2,0)| = φ(1,1,ω)

|(0,0)(1,1)(2,1)(3,1)(2,1)(2,0)(3,1)| = φ(1,1,ε₀)

|(0,0)(1,1)(2,1)(3,1)(2,1)(2,0)(3,1)(4,1)(5,1)| = φ(1,1,Γ₀)

|(0,0)(1,1)(2,1)(3,1)(2,1)(2,0)(3,1)(4,1)(5,1)(4,1)| = φ(1,1,φ(1,1,0))

|(0,0)(1,1)(2,1)(3,1)(2,1)(2,0)(3,1)(4,1)(5,1)(4,1)(4,0)(5,1)(6,1)(7,1)(6,1)| = φ(1,1,φ(1,1,φ(1,1,0)))

|(0,0)(1,1)(2,1)(3,1)(2,1)(2,1)| = φ(1,2,0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(3,0)| = φ(1,ω,0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(3,0)(4,1)| = φ(1,ε₀,0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(3,0)(4,1)(5,1)(6,1)| = φ(1,Γ₀,0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(3,0)(4,1)(5,1)(6,1)(5,1)| = φ(1,φ(1,1,0),0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(3,1)| = φ(2,0,0)

|(0,0)(1,1)(2,1)(3,1)(2,1)(3,1)(2,1)(3,1)| = φ(3,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,0)| = φ(ω,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,0)(4,1)| = φ(ε₀,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,0)(4,1)(5,1)(6,1)| = φ(Γ₀,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,0)(4,1)(5,1)(6,1)(6,0)(7,1)(8,1)(9,1)| = φ(φ(Γ₀,0,0),0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)| = φ(1,0,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(1,1)(2,1)(3,1)(3,1)| = φ(1,0,0,1)

|(0,0)(1,1)(2,1)(3,1)(3,1)(1,1)(2,1)(3,1)(3,1)(1,1)(2,1)(3,1)(3,1)| = φ(1,0,0,2)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,0)| = φ(1,0,0,ω)

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)(3,1)| = φ(1,0,0,ε₀)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,0)(3,1)(4,1)(5,1)(5,1)| = φ(1,0,0,φ(1,0,0,0))

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)| = φ(1,0,1,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(2,1)| = φ(1,0,2,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,0)| = φ(1,0,ω,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,0)(4,1)| = φ(1,0,ε₀,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,0)(4,1)(5,1)(6,1)(6,1)(5,1)| = φ(1,0,φ(1,0,1,0),0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)| = φ(1,1,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(2,1)(3,1)| = φ(1,2,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,0)| = φ(1,ω,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,0)(4,1)| = φ(1,ε₀,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,0)(4,1)(5,1)(6,1)(6,1)(5,1)(6,1)| = φ(1,φ(1,1,0,0),0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,1)| = φ(2,0,0,0) (again - quite expected)

|(0,0)(1,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,1)(2,1)(3,1)(3,1)| = φ(3,0,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(3,0)| = φ(ω,0,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(3,0)(4,1)(5,1)(6,1)(6,1)| = φ(φ(1,0,0,0) ,0,0,0)

|(0,0)(1,1)(2,1)(3,1)(3,1)(3,1)| = φ(1,0,0,0,0) = ψ(Ω^Ω³)

And the pattern is obvious by now:

|(0,0)(1,1)(2,1)(3,1)(3,1)(3,1)(3,1)| = φ(1,0,0,0,0,0) = ψ(Ω^Ω⁴)

|(0,0)(1,1)(2,1)(3,1)(3,1)(3,1)(3,1)(3,1)| = φ(1,0,0,0,0,0) = ψ(Ω^Ω⁵)

|(0,0)(1,1)(2,1)(3,1)(4,0)|= ψ(Ω^Ω^ω) = SVO

|(0,0)(1,1)(2,1)(3,1)(4,0)(3,1)| = ψ(Ω^Ω^(ω+1))

|(0,0)(1,1)(2,1)(3,1)(4,0)(3,1)(3,1)| = ψ(Ω^Ω^(ω+2))

|(0,0)(1,1)(2,1)(3,1)(4,0)(3,1)(3,1)(3,1)| = ψ(Ω^Ω^(ω+3))

|(0,0)(1,1)(2,1)(3,1)(4,0)(3,1)(4,0)| = ψ(Ω^Ω^(ω×2))

|(0,0)(1,1)(2,1)(3,1)(4,0)(3,1)(4,0)(3,1)(4,0)| = ψ(Ω^Ω^(ω×3))

|(0,0)(1,1)(2,1)(3,1)(4,0)(4,0)| = ψ(Ω^Ω^ω²)

|(0,0)(1,1)(2,1)(3,1)(4,0)(4,0)(4,0)| = ψ(Ω^Ω^ω³)

|(0,0)(1,1)(2,1)(3,1)(4,0)(5,0)| = ψ(Ω^Ω^ω^ω)

|(0,0)(1,1)(2,1)(3,1)(4,0)(5,1)| = ψ(Ω^Ω^ε₀)

|(0,0)(1,1)(2,1)(3,1)(4,0)(5,1)(6,1)| = ψ(Ω^Ω^ζ₀)

|(0,0)(1,1)(2,1)(3,1)(4,0)(5,1)(6,1)(7,1)| = ψ(Ω^Ω^Γ₀) = ψ(Ω^Ω^ψ(Ω^Ω))

|(0,0)(1,1)(2,1)(3,1)(4,0)(5,1)(6,1)(7,1)(8,0)(9,1)(10,1)(11,1)| = ψ(Ω^Ω^ψ(Ω^Ω^ψ(Ω^Ω)))

|(0,0)(1,1)(2,1)(3,1)(4,1)| = ψ(Ω^Ω^Ω) = LVO

|(0,0)(1,1)(2,1)(3,1)(4,1)(1,1)(2,1)(3,1)(4,1)| = ψ(Ω^Ω^Ω × 2)

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)| = ψ(Ω^Ω^Ω × ω)

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)(3,1)| = ψ(Ω^Ω^Ω × ε₀) = ψ(Ω^Ω^Ω × ψ(0))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)(3,1)(4,1)| = ψ(Ω^Ω^Ω × ζ₀) = ψ(Ω^Ω^Ω × ψ(Ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)(3,1)(4,1)(5,1)| = ψ(Ω^Ω^Ω × Γ₀) = ψ(Ω^Ω^Ω × ψ(Ω^Ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)(3,1)(4,1)(5,1)(6,1)| = ψ(Ω^Ω^Ω × ψ(Ω^Ω^Ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,0)(3,1)(4,1)(5,1)(6,1)(4,0)(5,1)(6,1)(7,1)(8,1)| = ψ(Ω^Ω^Ω × ψ(Ω^Ω^Ω × ψ(Ω^Ω^Ω)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)| = ψ(Ω^Ω^Ω × Ω) = ψ(Ω^(Ω^Ω + 1))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(2,1)| = ψ(Ω^Ω^Ω × Ω) = ψ(Ω^(Ω^Ω + 2))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(2,1)(2,1)| = ψ(Ω^Ω^Ω × Ω) = ψ(Ω^(Ω^Ω + 3))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,0)| = ψ(Ω^Ω^Ω × Ω) = ψ(Ω^(Ω^Ω + ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,0)(4,1)| = ψ(Ω^Ω^Ω × Ω) = ψ(Ω^(Ω^Ω + ε₀))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,0)(4,1)(5,1)(6,1)(7,1)(5,1)| = ψ(Ω^Ω^Ω × Ω) = ψ(Ω^(Ω^Ω + ψ(Ω^(Ω^Ω + 1))))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)| = ψ(Ω^(Ω^Ω + Ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(2,1)(3,1)| = ψ(Ω^(Ω^Ω + Ω×2))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(3,0)| = ψ(Ω^(Ω^Ω + Ω×ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(3,0)(3,1)| = ψ(Ω^(Ω^Ω + Ω×ε₀))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(3,1)| = ψ(Ω^(Ω^Ω + Ω×Ω)) = ψ(Ω^(Ω^Ω + Ω²))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(3,1)(3,1)| = ψ(Ω^(Ω^Ω + Ω³))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(4,0)| = ψ(Ω^(Ω^Ω + Ω^ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(2,1)(3,1)(4,1)| = ψ(Ω^(Ω^Ω + Ω^Ω)) = ψ(Ω^(Ω^Ω × 2))

|(0,0)(1,1)(2,1)(3,1)(4,1)(3,0)| = ψ(Ω^(Ω^Ω × ω))

|(0,0)(1,1)(2,1)(3,1)(4,1)(3,1)| = ψ(Ω^(Ω^Ω × Ω)) = ψ(Ω^(Ω^(Ω+1)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(3,1)(3,1)| = ψ(Ω^(Ω^(Ω+2)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(3,1)(4,0)| = ψ(Ω^(Ω^(Ω+ω)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(3,1)(4,1)| = ψ(Ω^(Ω^(Ω+Ω))) = ψ(Ω^(Ω^(Ω×2)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(3,1)(4,1)(3,1)(4,1)| = ψ(Ω^(Ω^(Ω×3)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(4,0)| = ψ(Ω^(Ω^(Ω×ω)))

|(0,0)(1,1)(2,1)(3,1)(4,1)(4,1)| = ψ(Ω^(Ω^(Ω×Ω))) = ψ(Ω^Ω^Ω²)

|(0,0)(1,1)(2,1)(3,1)(4,1)(4,1)(4,1)| = ψ(Ω^Ω^Ω³)

|(0,0)(1,1)(2,1)(3,1)(4,1)(5,0)| = ψ(Ω^Ω^Ω^ω)

|(0,0)(1,1)(2,1)(3,1)(4,1)(5,1)| = ψ(Ω^Ω^Ω^Ω)

|(0,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,0)| =  ψ(Ω^Ω^Ω^Ω^ω)

|(0,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)| =  ψ(Ω^Ω^Ω^Ω^Ω)

|(0,0)(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)(7,1)| =  ψ(Ω^Ω^Ω^Ω^Ω^Ω)

|(0,0)(1,1)(2,2)| =  ψ(ψ₁(0)) 77.127.67.249 10:51, December 11, 2016 (UTC)

Heuoristic Analysis

This iso only heuristic, but I believe that the specified strength is reached through representing the pair sequence as a tree. If the first number in each pair is used to mark the depth of each node in the tree, with each layer down effectively representing addition, and the second number in the pair represents what function is passed represents a label on the node (effectively a different function for each input value), then this describes a tree with an ordering of order type

$\psi(\Omega_\omega)$ , and this also corresponds (as described above) rather well with the pair sequencing. I believe it is possible to make this representation more rigorous, but I can't do it right now. ~εmli 11:34, January 5, 2016 (UTC)

I can't quite follow what you are saying, but it sounds promising. One way to possibly simplify things: if you are trying to find the order type of set A, finding an order-preserving injection to a set B would tell you that A has order type at most that of B, and an order-preserving injection from B to A would tell you that A has order type at least that of B. That could potentially be easier than finding an exact correspondence between A and B. Deedlit11 (talk) 01:41, January 6, 2016 (UTC)

Just thought of something

What happens when you add a sequence inside another sequence?

Example:\(((3,3),4)\) Boboris02 (talk) 12:38, November 20, 2016 (UTC)

I think you will get a sequence of sequences. 109.236.81.168 20:47, November 20, 2016 (UTC)

Is a thing like (0,0)(1,2) defined?

Because this set looks like it ends at (0,0)(1,1)(2,2)... Same goes for (0,0)(1,3), (0,0)(1,4)..., (0,0)(2,0), (0,0)(2,1), (0,0)(2,2)... If so, is (0,1) defined? Alpha-ketoacid (talk) 15:00, April 4, 2018 (UTC)

Yes, it is extremely straight forward, firstthen just carry the 7 and that's it! I'm sure you understand.

If you have any more questions feel free to ask! Qaz809 (talk) 16:03, April 4, 2018 (UTC)

So \((0,0)(1,2)=\psi(\psi_\omega(0))\) or \((0,0)(1,2)=\psi(\psi_2(0))\)? Alpha-ketoacid (talk) 08:58, April 6, 2018 (UTC)

According to Bashicu matrix calculator, it's actually just ε₀. Nishada 11:30, April 6, 2018 (UTC)

And how about (0,0)(1,2)(2,4), (0,0)(1,2)(2,4)(3,6),...? How about (0,0)(2,2)? (0,1)? If that evaluates to a finite amount, (0,1)(1,1),...? Are sequences starting with (1,x) defined? Alpha-ketoacid (talk) 17:35, April 7, 2018 (UTC)

Fixing of the evaluation

The evaluation here is written by Bashicu's unspecified OCF and I think it should be fixed by well known notation.

Can I fix it by the Buchholz's psi notation? https://googology.fandom.com/ja/wiki/%E3%83%88%E3%83%BC%E3%82%AF:%E3%83%9A%E3%82%A2%E6%95%B0%E5%88%97%E6%95%B0#.E4.BF.AE.E6.AD.A3.E6.A1.88 Koteitan (talk) 18:56, 26 March 2022 (UTC)

It might be good to write both evaluations, because there is an external source of the automatic conversion by fish into expressions in Bashicu's unspecified OCF and hence Bashicu's unspecified OCF itself can be a target of googological study such as formalisation. Indeed, the existene of an OCF compatible with PSS is a very big topic (in order to discover a further correspondence for TSS).

p-adic 23:56, 26 March 2022 (UTC)

I'm replying here because I think it is related. The beginning section of the page (current version) says that \((0,0)(1,1)(2,2)(3,3)(2,2)\) corresponds to \(\psi_0(\Omega_3+\psi_2(\Omega_3+\Omega_2))\). However, with \(\textrm{Trans}\) in the proof of termination, we get \(\textrm{Trans}(((0,0),(1,1),(2,2),(3,3),(2,2)))=D_0\underline{(}D_30,D_20\underline{)}\) (calculated with my implementation). Since the map is an injection, the matrix must then correspond to the ordinal no more than \(\psi_0(\Omega_3+\Omega_2)\). Does anyone have any idea where the statement originated? Also, should this be fixed? --Naruyoko (Talk) 20:26, 1 August 2022 (UTC) | Fixed a typo --Naruyoko (Talk) 20:26, 1 August 2022 (UTC)

I think that I have ever edited sentences arround the analysis, and hence it is perhaps my bad edit ┐(ツ)┌ I guess that what I actually wanted to write is "(0,0)(1,1)(2,2)(3,3)(2,2)(2,1) corresponds to ψ_0(ψ_3(0)+ψ_2(0)+ψ_1(ψ_3(0)+ψ_2(0)+ψ_1(0)))", because I like this milestone explaining the difference of PSS and Buchholz's ψ.

p-adic 03:28, 2 August 2022 (UTC)

Wait a minute... I just realized that it says \((0,0)(1,1)(2,2)(3,3)({\color{red}3},2)\) instead of \((0,0)(1,1)(2,2)(3,3)({\color{red}2},2)\)... I don't know how I misread it... In this case, there was no mistake after all. Sorry for the stir. --Naruyoko (Talk) 21:00, 11 August 2022 (UTC)

Oops. Sorry for not checking the article itself!

p-adic 23:03, 11 August 2022 (UTC)