User blog:Pteriforever/Pteriforever's Relatively Boring Notation

I find the higher hyperoperators to be a bit confusing, so for a while now I've been trying to develop a system which will allow me to define them more easily.

My first attempt was "arrayspace functions" -- functions with of a base, an index in the form of an array, and an "arrayspace" consisting of a hierarchy of indices, with each index being used to define another function with different indices in a lower "arrayspace". This is an interesting concept I might return to, but my implementation ended up extremely weak.

Eventually I came up with this. It's nothing too new or interesting, and it doesn't grow particularly fast, but it's defined to make things convenient.

\[\circledast [\&Z](X)\]

One-entry Level
The most basic rule is:

\[\circledast [Z](X) = ZX\]

This is the only rule necessary for a one-entry Z-array.

A useful feature of PRBN as opposed to W-notation is that things will always work out so that X and functions thereof can be placed in the index and its meaning will still be easy to write down in terms of X.

\[\circledast [1](X) = X\]

\[\circledast [2](X) = 2X\]

\[\circledast [4](X) = 4X\]

\[\circledast [a](X) = aX\]

\[\circledast [X](X) = X^2\]

\[\circledast [X+5](X) = X^2+5X\]

\[\circledast [2X](X) = 2X^2\]

\[\circledast [X^2](X) = X^3\]

\[\circledast [X^3](X) = X^4\]

Obviously, we're hitting a limit pretty fast, so we need to introduce the second entry.

Two-entry Level
It is worth noting that: The rules for two-entry arrays are:

\[\circledast [\&Z,1](X) = X\]

\[\circledast [Z_2,Z_1](X) = \circledast [Z_2-1,\circledast [Z_2,Z_1-1](X)](X)\]

\[\circledast [0,\&Z](X) = \circledast [\&Z](X)\]

Therefore:

\[\circledast [1,1](X) = X\]

\[\circledast [1,2](X) = X^2\]

\[\circledast [1,3](X) = X^3\]

\[\circledast [1,a](X) = X^a\]

\[\circledast [1,X](X) = {}^2X\]

\[\circledast [2,1](X) = X\]

\[\circledast [2,2](X) = {}^2X\]

\[\circledast [2,3](X) = {}^3X\]

\[\circledast [2,a](X) = {}^aX\]

\[\circledast [2,X](X) = X\uparrow\uparrow\uparrow2\]

\[\circledast [3,2](X) = X\uparrow\uparrow\uparrow2\]

\[\circledast [3,3](X) = X\uparrow\uparrow\uparrow3\]

\[\circledast [3,X](X) = X\uparrow\uparrow\uparrow\uparrow2\]

\[\circledast [a,2](X) = X\{a\}2\]

\[\circledast [a,b](X) = X\{a\}b\]

\[\circledast [a,X](X) = X\{a\}X\]

\[\circledast [X,a](X) = X\{X\}a\]

\[\circledast [X,X](X) = X\{X\}X\]

\[\circledast [X\{X\}X,X](X) = X\{X\{X\}X\}X\]

\[\circledast [X\{X\{X\}X\}X,X](X) = X\{X\{X\{X\}X\}X\}X\]

At this point, we are approaching another limit.

Multiple-entry level
\[\circledast [\&Z_2,1,\&Z_1](X) = X

\[\circledast [Z_n,Z_{n-1},Z_{n-2},...,Z_2,Z_1](X) = \circledast [Z_n-1,[\circledast [Z_n,Z_{n-1}-1,Z_{n-2},...,Z_2,Z_1](X),Z_{n-2},...,Z_2,Z_1](X)\]

Summary of rules:

1. If there's a single entry, multiply it by X.

2. If there's a 0 at the beginning of the array, delete it.

3. If there's a 1 in the array anywhere except the beginning, the result is just X.

4. If there are multiple entries, the first entry is greater than 0 and all others are greater than 1, then reduce the first entry by 1 and replace the second entry with the entire function with the second entry reduced by 1.

Therefore:

\[\circledast [1,1,1](X) = X]

\[\circledast [1,1,2](X) = X]

\[\circledast [1,2,2](X) = X\{X\}2\]

\[\circledast [1,2,a](X) = X\{X\}a\]

\[\circledast [1,2,X](X) = X\{X\}X\]

\[\circledast [1,3,X](X) = X\{X\{X\}X\}X\]

\[\circledast [1,a,X](X) = X\{\{1\}\}a\]

\[\circledast [1,X,X](X) = X\{\{1\}\}X\]

\[\circledast [2,2,X](X) = X\{\{1\}\}X\]

\[\circledast [2,3,X](X) = X\{\{1\}\}X\{\{1\}\}X\]

\[\circledast [2,X,X](X) = X\{\{2\}\}X\]

\[\circledast [3,X,X](X) = X\{\{3\}\}X\]

\[\circledast [a,X,X](X) = X\{\{a\}\}X\]

\[\circledast [X,X,X](X) = X\{\{X\}\}X\]

\[\circledast [1,2,X,X](X) = X\{\{X\}\}X\]

\[\circledast [1,3,X,X](X) = X\{\{X\{\{X\}\}X\}\}X\]

\[\circledast [1,4,X,X](X) = X\{\{X\{\{X\{\{X\}\}X\}\}X\}\}X\]

\[\circledast [1,a,X,X](X) = X\{\{\{1\}\}\}a\]

\[\circledast [1,X,X,X](X) = X\{\{\{1\}\}\}X\]

\[\circledast [2,3,X,X](X) = X\{\{\{1\}\}\}X\{\{\{1\}\}\}X\]

\[\circledast [2,X,X,X](X) = X\{\{\{2\}\}\}X\]

\[\circledast [3,X,X,X](X) = X\{\{\{3\}\}\}X\]

\[\circledast [a,X,X,X](X) = X\{\{\{a\}\}\}X\]

\[\circledast [X,X,X,X](X) = X\{\{\{X\}\}\}X\]

The limit growth of this I conjecture to be:

f_{\omega^2}(X)

Extension 1: Subblocks
A subblock is a shorthand entity indicating a string of Xs.

\[\circledast [1,[3]](X) = X\{\{\{1\}\}\}X\]

\[\circledast [a,[3]](X) = X\{\{\{a\}\}\}X\]

\[\circledast 4(X) = X\{\{\{X\}\}\}X\]

\[\circledast [1,[a]](X) = X\underbrace{\{\{...\{\{}_a1\}\}...\}\}X\]

\[\circledast [1,[X]](X) = X\underbrace{\{\{...\{\{}_X1\}\}...\}\}X\]

\[\circledast [X,[X]](X) = X\underbrace{\{\{...\{\{}_XX\}\}...\}\}X\]

\[\circledast [X,[\circledast [X,[X]](X)]](X) = X\underbrace{\{\{...\{\{}_{\circledast [X,[X]](X)}X\}\}...\}\}X\]

Extension 2: Array-Subblocks
Similar rules to the main array also apply to subblocks:

5. If there's a 1 in a subblock anywhere except the beginning, the answer is X.

6. If there's a 0 at the beginning of a subblock, it is deleted.

7. If there's a single entry, replace it with that many entries all equal to X in the main array.

8. Otherwise, reduce the subblock's first entry by 1 and replace the second with the entire expression with the subblock's second entry reduced by 1.

\[\circledast [X,[2,X]](X) = X\underbrace{\{\{...\{\{}_{\circledast [X,[X]](X)}X\}\}...\}\}X\]

\[\circledast [X,[3,X]](X) = X^\{_\{^\}_\}3\]

\[\circledast [X,[a,X]](X) = X^\{_\{^\}_\}a\]

\[\circledast [X,[X,X]](X) = X^\{_\{^\}_\}X\]

And we've finally reached megotion!

[More stuff coming soon]