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First, all superdimension notations starts with SD.

Then, brackets or non-negative integers come. First and last thing must be a number. If you want to write more than one number next to each other, use comma. Also, (0) is the same as a comma. So, these are some valid things:

SD3(2)3(8)4
SD5,0,6,2
SD3(3)3,0(2)6

So how is it calculated? Let's call the last number identifier. Small letters are numbers. Capital letters are any valid notations excluding the "SD."

  1. SD 0(a)B=SD B
  2. If identifier is zero,
    Subtract the rightmost non-zero number by one. Then, change the zero into the number (the rightmost number) was.
    If there are more than 1 zero between, fill the zeros with the new identifier, too.
    e.g. SD A,x,0=SD A,x-1,x
    SD a,0,0,0 =SD a-1,a,a,a
  3. If what it before the identifier is comma (by definition, a number must be before comma.)
    1. If the number before the comma is NOT zero
      SD Ax,y= SD Ax-1,SD Ax,y-1
      This is same as Ackermann.
    2. If the number before the comma IS zero
      Subtract the second rightmost non-zero number by one. Then, change the zero into the number (the rightmost number) was.
      If there are more than 1 zero between, fill the zeros with the new second-from-right number, too.
      e.g. SD A,b,0,x=SD A,b-1,b,x
      SD a,0,0,b =SD a-1,a,a,b
  4. If what it before the identifier is brackets
    It must be the form of "SD Ab(c_1)(c_2)...(c_n)d."
    SD Ab(c_1)(c_2)...(c_n)d = SD Ab-1(c_1)(c_2)...(c_n)d(c_1)(c_2)...((c_n)-1)d(c_1)(c_2)...((c_n)-1)...d(c_1)(c_2)...((c_n)-1)d(c_1)(c_2)...((c_n)-1)d

I think it will be {\omega^{\omega^{\omega}}} because it is made from recursive and multidimensional.


(Next: What about (1,0)? You can't subtract one from 1,0.)

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