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These are the definitions:

X:0 or more integers and 0 or more (n)'s 
Y:0 or more zeroes and 0 or more (n)'s
Z:1 or more zeroes and 0 or more (n)'s 
Commas can be replaced by (1). 
Let f(x)=x+1,
1.{Ya}=f(a)
2.{Xb+1,0}={Xb,1} 
3.{Xb+1,a+1}={Xb,{Xb+1,a}} 
4.{Xb+1,0Ya}={Xb,aYa} 
5.{Xb+1(n+1)a}={Xb(n+1)a(n)…a times a's…(n)a} 
6.{Xb+1(n+1)ZYa}={Xb(n+1)a(n)…a times a's…(n)aYa} (Z does not contain (n+1))

I will call these six rules basic rules for more extension.

This notation reaches up to f_{\omega^{\omega^{\omega}}}(x) and this is posted:

多次元アッカーマン配列

Before I extend it, I found that it has very interesting property.

This looks like an "base-omega" number like this:

\{ a_n, a_{n-1}, \cdots a_1, a_0, x \} \simeq f_{a_n \omega^n + a_{n-1} \omega^{n-1} +\cdots + a_1 \omega^1 + a_0}(x) 

This relation continues up to n dimension. For example, the rightmost number in the second row is \omega^{\omega}'s place and then \omega^{\omega+1}, \omega^{\omega+2}, and so on.

Next, multidimension in multidimension.

If you have dimension of array, ignore it but the last one and keep going on as possible as you can. You will get stuck at {X(X0)n}. Then, use basic rules as if it is \{ X(\omega)X0(\omega)n \}. You don't need to subtract 1 from omega, so it is well defined.

Therefore,

x(1,0)x is at \omega^{\omega^{\omega}}'s place.

x(1)(1,0)x is at \omega^{\omega^{\omega}+\omega}'s place.

x(1)(1)(1,0)x is at \omega^{\omega^{\omega}+\omega2}'s place.

x(2)(1,0)x is at \omega^{\omega^{\omega}+\omega^2}'s place.

x(x)(1,0)x or x(1,0)(1,0)x is at \omega^{\omega^{\omega}*2}'s place.

x(1,1)x is at \omega^{\omega^{\omega}*\omega} = \omega^{\omega^{\omega+1}}'s place.

x(1,x)x or x(2,0)x is at \omega^{\omega^{\omega*2}}'s place.

x(x,0)x or x(1,0,0)x is at \omega^{\omega^{\omega^2}}'s place.

x(1(1)0)x is at ^4 \omega's place.


Actually, I don't know what will be going on after that. This is just too big for me to understand.

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