User:Ynought/: notation

\(:\) notation
This will (hopefully) be my most refined (recursive) notation

\(\#_k\) is the labeled rest of the notation

\(\{ \text{and} \}\) is any amount of brackets

\(a\) will always denote the number before the \(:\)

\(\#_i\rightarrow\#_j\) means \(\#_i\) turns into \(\#_j\)

\(f_0(c)= (c+1)^2\) and \(f_k(c)= f_{k-1}^c(c)\) where \(f^0_k(c)=c\) and \(f^{d}_k(c)=f_k(f_k^{d-1}(c))\)

Simple notation
This is the 1st part of ? parts and it is the most simple

\(a:=a\)

\(0:\#=0\)

\(1:\#=1\)

\(a:\{\#_1b\}\rightarrow f_a^a(b):\{\#_1\}\) here \(\#_1\) isnt empty if \(\{ \text{ and } \}\) have \(1+\) brackets and b isnt inbetween

\(a:\#_1b\rightarrow f_a^a(b):\#_1\)

Start looking from right to left until you find a number inside a pair of bracket.Call that number \(b\).Then:

"1.If \(b=0\) then \((b)\rightarrow a\)"

"2.If \(b>0\) then \((b)\rightarrow(b-1)(b-1)...(b-1)\) with (in total) \(a\) \((b-1)\)'s"

The limit of \(n:(...(n)...)\approx f_{\varepsilon_0}(n)\) with \(n\) brackets

Examples
\(2:2=f_2^2(2)=f_2(f_2(2))=f_2(f_1(f_1(2)))=f_2(f_1(f_0(f_0(2))))=f_2(f_1(f_0(9)))=f_2(f_1(100))\)

\(2:(1)=2:(0)(0)=2:(0)2=f_2^2(2):(0)=f_2^2(2):f_2^2(2)=f^{f_2^2(2)}_{f_2^2(2)}(f_2^2(2))\)

Expanding brackets
This is the 2nd part of ? parts and it still is relativly simple.

Now i introduce a new type of brackets with exponentiation.The process is:

start looking from right too left and apply the rules above,until you find a bracket that has an,or more,exponents.If you start such a bracket look at the top most bracket(if the bracket is a bracket(with out exponents)then apply the normal rules too that bracket).And then start this process

"1. \((b)^0\rightarrow (...(b)...)...(...(b)...)\) with a \((...(b)...)\)'s where there are \(a\) brackets"

"2. \((b)^c\rightarrow (...(b)^{c-1}...)^{c-1}(...(b)^{c-1}...)^{c-1}...(...(b)^{c-1}...)^{c-1}\) with a nests"

"3.\(((0)^x\#)^{x}=((z_a)^{x-1}z_a\#z_a\#...z_a\#)^{x}\) with \(a\) \(z_a\)'s where \(z_0=a\) and \(z_n=(((z_{n-1})^{x-1}z_{n-1})d)^{x}\)"

"4.\(((b)^x0)=((y_{b+a})^x)\) where \(y_0=a\) and \(y_n=((...((y_{n-1})^x)...)^x)\) with a nests"

Simple arrays (w.i.p.)
This is the 3rd part of ? parts and it gets slighty more complex

Now i introduce a new type of brackets with exponentiation.The process is:

start looking from right too left and apply the rules above,until you find a bracket that has an,or more,exponents.If you start such a bracket look at the top most bracket(if the bracket is a bracket(with out exponents)then apply the normal rules too that bracket).And then start this process

"1.\((\#,0)=(\#)^{(\#)^{\dots^{(\#) }}} (\#)^{(\#)^{\dots^{(\#) }}} ...(\#)^{(\#)^{\dots^{(\#) }}} \) with the powertower of height \(a\) and \(a\) copies of that|undefined"

"2.\(\nabla\) is a string of 0's \((\nabla,0,b+1,\#)=(\nabla,(\nabla,0,b,\#)_{(\nabla,0,b,\#)_{\dots_{(\nabla,0,b,\#) }}} ,b,\#)\) with \(a\) b's|undefined""3.If there is a power on any of the arrayed brackets then proceed but with the arrays implemented in the following way:""3.1 \((\#)^0\rightarrow (...(\#)...)...(...(\#)...)\) with a \((...(\#)...)\)'s where there are \(a\) brackets""3.2.\((\#)^c\rightarrow (...(\#)^{c-1}...)^{c-1}(...(\#)^{c-1}...)^{c-1}...(...(\#)^{c-1}...)^{c-1}\) with a nests""3.3(((\#a,0)^x\#)^{x}=((z_a)^{x-1}z_a\#z_a\#...z_a\#)^{x}\) with \(a\) \(z_a\)'s where \(z_0=a\) and \(z_n=(((\#,a+z_{n-1})^{x-1}z_{n-1})d)^{x}\)""3.4\(((\#)^x0)=((y_{[+a})^x)\) where \(y_0=a\) and \(y_n=((...((y_{n-1})^x)...)^x)\) with a nests"