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

"4.\(((b)^x0)\rightarrow ((y_{b+a})^x)\) 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 arrays.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.\((\#,0)\rightarrow (\#)^{(\#)^{\dots^{(\#) }}} (\#)^{(\#)^{\dots^{(\#) }}} ...(\#)^{(\#)^{\dots^{(\#) }}} \) with the powertower of height \(a\) and \(a\) copies of that|undefined"

"2.\(\nabla\) is a string of 0's \((\nabla,0,b+1,\#)\rightarrow (\nabla,(\nabla,0,b,\#)_{(\nabla,0,b,\#)_{\dots_{(\nabla,0,b,\#) }}} ,b,\#)\) 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\(((\#a,0)^x\#)^{x}\rightarrow ((z_a)^{x-1}z_a\#z_a\#...z_a\#)^{x}\) 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\(((\#b,c)^x0)\rightarrow ((\#,((\#b,c-1)^x0)_{((\#b,c-1)^x0)_{\dots_{((\#b,c-1)^x0) }}} ,c-1)^x)^{x-1}0)\) with a nests|undefined"

"4.\((b+1\#_1\nabla\#_2)\rightarrow F_{F_{\dots_{(b\#_1\nabla\#_2) }}} \) where \(F=(0,\nabla,(0,\nabla,\#_2,0)^{(0,\nabla,(0,\nabla,\#_2,0)^{\dots^{(0,\nabla,(0,\nabla,\#_2,0) }}} ,0)\)|undefined"

"5.\((b+1,\#_1,c,\#_2,d+1)\rightarrow F_{F_{..._{F_a} }} \) with a nests|undefined"

"\(F_n\rightarrow (0,\#_1,(0,\#_1,(...(0,\#_1,,\#_2,d)^{F_{n-1 }}...)^{F_{n-1}} ,\#_2,d)^{F_{n-1 }} ,\#_2,d)^{F_{n-1 }} \) with a nests|undefined"

"\(F_0\rightarrow (0,\#_1,(0,\#_1,(...(0,\#_1,c,\#_2,d)...),\#_2,d),\#_2,d)\)"

Dimensional arrays(This will take a wile)
This is the 4rd part of ? parts and it gets kind of complex

Now i introduce a new type of array the \(.\) array.

It will be a way to define multiple new types of normal arrays to make it way stronger.But it will have really many rules.I will however try to make them the least amount of rules,while not loosing the growth

\(\Box\) is any seperator

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.\((\#\Box \{0\}0)\rightarrow (\#\Box a)\)

2.\((\#\{0\}0)\rightarrow (\#,a)\) here \(\#\) doesnt have a seperator at the end

3.\((\#\Box\{0\}b)\rightarrow (\#\Box(\#\Box b+a)^{(\#\Box b+a)^{...^{(\#\Box\{a\}b-1) }}} )\) with a \((\#\Box b+a)\)'s

4.\((\#\{0\}b)\rightarrow (\#,(\#,b+a)^{(\#,(\#,b+a)^{(\#, b+a)^{...^{(\#,b+a) }}} )^{...^{(\#,(\#,b+a)^{(\#, b+a)^{...^{(\#,b+a) }}} ) }}} )\) with a \((\#,(\#,b+a)^{(\#, b+a)^{...^{(\#,b+a) }}} )\)'s with a \(\#,a+b)\)'s and here \(\#\) doesn't have a seperator at the end

5.\((\#\Box\{b\}0)\rightarrow (\#\Box\{b-1\}...\{b-1\}a)^{(\#\Box\{b-1\}...\{b-1\}(\#\Box\{b-1\}...\{b-1\}a)^{...^{(\#\Box\{b-1\}...\{b-1\}a) }} )^{(\#\Box\{b-1\}...\{b-1\}a)^{..^{(\#\Box\{b-1\}...\{b-1\}a) }}} }\)