In this blog post, I will analyze 3-row Bashicu Matrix System. Bubby3 and Aarex are also making the analysis, but I will make a more careful and detailed one, and give explanations.

Edit: Should I close this blog post or not?


Up to ε(0)

Up to ω

First, adding a (0,0,0) adds 1 to the level.

Trio Sequence System Level
(0,0,0) 1
(0,0,0)(0,0,0) 2
(0,0,0)(0,0,0)(0,0,0) 3
(0,0,0)(0,0,0)(0,0,0)(0,0,0) 4
(0,0,0)(0,0,0)(0,0,0)(0,0,0)(0,0,0) 5

Up to ω2

We can add a (1,0,0) to copy (0,0,0)[rest of sequence]'s.

Trio Sequence System Level
(0,0,0)(1,0,0) ω
(0,0,0)(1,0,0)(0,0,0) ω+1
(0,0,0)(1,0,0)(0,0,0)(0,0,0) ω+2
(0,0,0)(1,0,0)(0,0,0)(0,0,0)(0,0,0) ω+3

Up to ω^2

However, (1,0,0) only copies what's between it and the nearest (1,0,0) to the left.

Trio Sequence System Level
(0,0,0)(1,0,0)(0,0,0)(1,0,0) ω2
(0,0,0)(1,0,0)(0,0,0)(1,0,0)(0,0,0) ω2+1
(0,0,0)(1,0,0)(0,0,0)(1,0,0)(0,0,0)(1,0,0) ω3
(0,0,0)(1,0,0)(0,0,0)(1,0,0)(0,0,0)(1,0,0)(0,0,0)(1,0,0) ω4

Up to ω^ω

Groups of (1,0,0)s are treated like a single one.

Trio Sequence System Level
(0,0,0)(1,0,0)(1,0,0) ω2
(0,0,0)(1,0,0)(1,0,0)(0,0,0) ω2+1
(0,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0) ω2
(0,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0)(0,0,0)(1,0,0) ω2+ω2
(0,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0)(1,0,0) ω22
(0,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0)(1,0,0) ω23
(0,0,0)(1,0,0)(1,0,0)(1,0,0) ω3
(0,0,0)(1,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0)(1,0,0) ω32
(0,0,0)(1,0,0)(1,0,0)(1,0,0)(0,0,0)(1,0,0)(1,0,0)(1,0,0) ω32
(0,0,0)(1,0,0)(1,0,0)(1,0,0)(1,0,0) ω4
(0,0,0)(1,0,0)(1,0,0)(1,0,0)(1,0,0)(1,0,0) ω5
(0,0,0)(1,0,0)(1,0,0)(1,0,0)(1,0,0)(1,0,0)(1,0,0) ω6

Up to ω^ω^ω

(2,0,0) copies (1,0,0)[rest of sequence]s in a similar way that (1,0,0) copies (0,0,0)[rest of sequence]s.

Trio Sequence System Level
(0,0,0)(1,0,0)(2,0,0) ωω
(0,0,0)(1,0,0)(2,0,0)(0,0,0) ωω+1
(0,0,0)(1,0,0)(2,0,0)(0,0,0)(1,0,0) ωω
(0,0,0)(1,0,0)(2,0,0)(0,0,0)(1,0,0)(2,0,0) ωω2
(0,0,0)(1,0,0)(2,0,0)(1,0,0) ωω+1
(0,0,0)(1,0,0)(2,0,0)(1,0,0)(1,0,0) ωω+2
(0,0,0)(1,0,0)(2,0,0)(1,0,0)(2,0,0) ωω2
(0,0,0)(1,0,0)(2,0,0)(1,0,0)(2,0,0)(1,0,0)(2,0,0) ωω3
(0,0,0)(1,0,0)(2,0,0)(2,0,0) ωω2
(0,0,0)(1,0,0)(2,0,0)(2,0,0)(1,0,0)(2,0,0)(2,0,0) ωω22
(0,0,0)(1,0,0)(2,0,0)(2,0,0)(2,0,0) ωω3
(0,0,0)(1,0,0)(2,0,0)(2,0,0)(2,0,0)(2,0,0) ωω4
(0,0,0)(1,0,0)(2,0,0)(2,0,0)(2,0,0)(2,0,0)(2,0,0) ωω5

Up to ε(0)

And finally, (x+1,0,0) copies (x,0,0)[rest of sequence]s in a similar way as (1,0,0) and (2,0,0).

Trio Sequence System Level
(0,0,0)(1,0,0)(2,0,0)(3,0,0) ωωω
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(0,0,0)(1,0,0)(2,0,0)(3,0,0) ωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(1,0,0) ωωω+1
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(1,0,0)(2,0,0)(3,0,0) ωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(2,0,0) ωωω+1
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(2,0,0)(3,0,0) ωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(3,0,0) ωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(3,0,0)(3,0,0) ωωω3
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0) ωωωω
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0)(3,0,0)(4,0,0) ωωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0)(4,0,0) ωωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0)(5,0,0) ωωωωω
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0)(5,0,0)(5,0,0) ωωωωω2
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0)(5,0,0)(6,0,0) ωωωωωω
(0,0,0)(1,0,0)(2,0,0)(3,0,0)(4,0,0)(5,0,0)(6,0,0)(7,0,0) ωωωωωωω

Up to ζ(0)

Getting ε(0)

The limit of (0,0,0)(1,0,0)(2,0,0)(3,0,0)... is (0,0,0)(1,1,0). How can we explain this?

We need to introduce the new concept, which I call omega-sequences.

An omega-sequence is a sequence of sequences that has a sequence as it's limit and also a rule to get to the next term of the omega-sequence.

In this case, the omega-sequence

  1. (0,0,0)
  2. (0,0,0)(1,0,0)
  3. (0,0,0)(1,0,0)(2,0,0)
  4. (0,0,0)(1,0,0)(2,0,0)(3,0,0)
  5. ...

has a limit of (0,0,0)(1,1,0). You'll see why later in the analysis.

The rule for this omega-sequence is ...(x,0,0) -> ...(x,0,0)(x+1,0,0), where ... represents the rest of the term of the omega-sequence.

This type of omega-sequence is called a corresponding omega-sequence, because we can get it by making the last part of one of the terms correspond to the first part. In the omega-sequence of (0,0,0)(1,1,0), we can make (x,0,0) correspond to (0,0,0), and by adding (x+1,0,0), (x,0,0)(x+1,0,0) corresponds to (0,0,0)(0+1,0,0) = (0,0,0)(1,0,0).

Also, the first term can be important in determining the limit. You'll see why later in the analysis.

Up to ε(0)^2

We can continue the analysis by doing changes that we already know to (0,0,0)(1,1,0).

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