User:Vel!/pu/supernotation

A new notation! Due to the corner my extended arrow notation has pushed me into, I will start developing this notation instead.

Extension 0 - basic super notation (BSN)
[1](n) = nn

[x](n) = [x-1]n(n) if x > 1

This notation is equivalent to steinhaus-moser notation.

Extension 1 - linear array super notation (LASN)
New rules are in blue.


 * 1) is any string, $ is any string of ones.

[1,#](n) = [#](n)

[1](n) = nn

[x](n) = [x-1]n(n) if x > 1

[#,x](n) = [#,x-1]n(n) if x > 1

[x,1](n) = [x-1,n](n) if x > 1

[x,1,@](n) = [x-1,n,@](n) if x > 1

[#,x,1](n) = [#,x-1,n](n) if x > 1

[#,x,1,@](n) = [#,x-1,n,@](n) if x > 1

This reaches $$\omega^{\omega}$$. The typical approach next is to have n-dimensional structures which diagonalize over the sizes of each previous one. But here I will take an approach more similar to alemagno12's hyper notation.

Extension 2 - subblock super notation (SBSN)
Arrays can now contain arrays. It would be too difficult to type up symbolic rules so I will rewrite the definition. First, here are some definitions of things:


 * The active entry of an array is the first non-one entry if that entry is a number, or that entry's active entry if it is an array. The active entry is that of the entire array.
 * The active array of an array is itself if the first non-one entry either does not exist or is a number, if it is an array the active array is the active array of that array. The active array is that of the entire array

n is the number inside the parentheses. The active entry should always be in the active array.

Now for the rules:


 * If the entire array has no non-one entries, the expression is equal to nn.
 * If the active array has no non-one entries and it is not the entire array, replace it, and the rest of the array IT is in, with 2,1,1,...,1,1 with n 1's
 * If the active entry is the last entry in the active array, and the active array is not the entire array, replace the active array and the rest of the array IT is in with n copies of it with the active entry decreased by one
 * Otherwise, decrease the active entry by one and replace the entry after it with n

This should get us to $$\varepsilon_0$$. Next stop, legion!