User blog:Ynought/: notation up to linear arrays

\(:\) 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) \(a\) \((b-1)\)'sThe 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
<p data-parsoid="{"dsr":[1310,1375,0,0]}">This is the 2nd part of ? parts and it still is relativly simple.

<p data-parsoid="{"dsr":[1377,1451,0,0]}">Now i introduce a new type of brackets with exponentiation.The process is:

<p data-parsoid="{"dsr":[1453,1745,0,0]}">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 \(a\) brackets2. \((b)^c\rightarrow (...(b)^{c-1}...)^{c-1}(...(b)^{c-1}...)^{c-1}...(...(b)^{c-1}...)^{c-1}\) with a nests3.\(((0)^x\#)^{x}\rightarrow ((z_a)^{x-1}z_a\#z_a\#...z_a\#)^{x}\) 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=== <p data-parsoid="{"dsr":[2344,2408,0,0]}">This is the 3rd part of ? parts and it gets slighty more complex

<p data-parsoid="{"dsr":[2410,2448,0,0]}">Now i introduce arrays.The process is:

<p data-parsoid="{"dsr":[2450,2742,0,0]}">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 that2.\(\nabla\) 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's3.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 \(a\) brackets3.2.\((\#)^c\rightarrow (...(\#)^{c-1}...)^{c-1}(...(\#)^{c-1}...)^{c-1}...(...(\#)^{c-1}...)^{c-1}\) with a nests3.3\(((\#a,0)^x\#)^{x}\rightarrow ((z_a)^{x-1}z_a\#z_a\#...z_a\#)^{x}\) 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 nests4.\((b+1\#_1\nabla\#_2)\rightarrow F_{F_{\dots_{(b\#_1\nabla\#_2)}}}\) where \(F=(0,\nabla,(0,\nabla,\#_2,0)^{(0,\nabla,(0,\nabla,\#_2,0)^{\dots^{(0,\nabla,(0,\nabla,\#_2,0)}}},0)\)5.\((b+1,\#_1,c,\#_2,d+1)\rightarrow F_{F_{..._{F_a}}}\) with a nests\(F_n\rightarrow (0,\#_1,(0,\#_1,(...(0,\#_1,,\#_2,d)^{F_{n-1}}...)^{F_{n-1}},\#_2,d)^{F_{n-1}},\#_2,d)^{F_{n-1}}\) with a nests\(F_0\rightarrow (0,\#_1,(0,\#_1,(...(0,\#_1,c,\#_2,d)...),\#_2,d),\#_2,d)\)<p data-parsoid="{"stx":"html","dsr":[4519,4620,12,13]}">This is it for now i will post part 2 relativly soon but definetly not in april