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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.)