User:Ynought/Playground

This is just my place of experimenting with notations.

\(I\) function
well first i have to explain my notation:

\(I^0(a...b_+)=I(a...b)\)

\(I^k(...b_+...)=I(...I^{k-1}(...b_+...)...)\) so here every number with a \(_+\) gets the same treatment

\(O\) is the sum of all the numbers in the array

now for \(Z\) lets say for \(I(10,10,10,10)\) then \(Z=I^O(10_+,10_+,10_+,9)\) but if the last entry is a zero (in the original array) then apply the rule for that case first

\(I(n)=n+1\)

\(I(a,b,...,c,0)=I^Z(a_+b_+..,c_+)\)

\(I(a,b,...,c)=I^Z(a_+,b_+,...,c-1)\)

\(G\) notation
$ is the rest of the notation

Start looking from right to left until you find a number then apply the following rules:

\(a G $b=(...((aG $b-1)G $b-1)...aG $b-1)G $b-1\) with a nests

except \(aG $0=a+1G $\)

a is always the number before the \(\#\)

\([\) \(]\) is any amount of brackets

with more than zero brackets \([b]=[b-1]...[b-1]\)

any with more than zero brackets \([0]=[a]\) with one less pair of brackets

Super \(G\) notation
new stuff :

\([ [ b ] \bullet] = [ [ b-1 ] \bullet]\dots[ [ b-1 ] \bullet]\) a nests

\( [b]^1= q\dots q\) with a nests.Where \(q=[...[b]...]\) with a nests

\( [b]^k= q\dots q\) with a nests.where \(q=[...[b]^{k-1}...]^{k-1}\)

here \($\) can also be a stack

\([b]^{$c}=o_{aG\dots$[b]^{$c-1}\dots}...o_{aG\dots[b]^{$c-1}\dots}\) with a nests where \(o_k=o_{k-1}...o_{k-1}\) with a nests and \(o_0=[b]^{$c-1}\dots[b]^{$c-1}\) with a nests

Hyper \(G\) notation
\(($,0)=($)...($)\) with a nests

and here \((_k\) means \((\dots(\) with k \((\)'s

and here \^$_k\) means \^$\dots)^$\) with k\^$\)'s

and \(k_n\) means replaceing k by the current system (without the \(a\#\) ) n times

\((b,c,d,e...f)=(y,y,y,y...,f-1)...(y,y,y,y...,f-1)\) a nests and \(y=((_kb_a)^{(b,c,d,e...f-1)}_k,(_kc_a)^{(b,c,d,e...f-1)}_k...,f-1)\) with all the element in the original array being effected.

Strong \(G\) notation W.I.P.
\($_k\) is just a labeled part of the notation and \(\{0\}\) is the same as a coma

here \($_1\) is a group of 0's and \($_2\) and has one less zero and \($_3\) is the rest of the notation \(($_1\{k\}b)=($_2a\{k-1\}b)\)

\(($_3b\{k\}0)=($_3b\{k-1\}a)^{($_3b\{k-1\}a)^{\dots^{($_3b\{k-1\}a) }}} \) with a tower of height a

\(($_3b\{k\}c)=G^{G^{\dots^G }} \) with a tower of height a and \(G=($_30...0,b\{k\}c-1)^{\dots^{($_30...0,b\{k\}c-1) }} \) with a tower of height a

Stronger operator notation (SON)
the normal \(+=+_0\)

and \(\#\) is any combination of hyper operators

\(a+b\) is just \(a+b\) but with more than one operator this happens:

\(a\#+b=a\#(...a\#(a)...)\) with a nests

\(a\#+_kb=a\#+_{k-1}(...(a\#+_{k-1})...)\) a nests

and \(\times_0=+_a\) \(\times_k=\times_{k-1}\dots\times_{k-1}\) with a copies

so you can also create stuff like \(+_{n,n}\) which follows the patern with the plus and the times but with the n-th symbol

and for 3 or more entries in the array at the bottom of the \(+\) this happens

\(a\#+_{b,c,...,d}e=a\#q...q e\) with a nests and where \(J_0=a\) and \(J_k\) gets solved by replacing the \(J_k\) with the current notation but with \(J_{k-1}\) and \(q=+_{J_b,j_c,...,d-1}...+_{J_b,j_c,...,d-1}\)

Analysis
where \(\alpha\) is the growth rate of \(a+_{a.,..,a}a\)

\(Q\) function
so like with every good function the start is a simple \(+\)

\(aQ(n)=n+a\)

\(aQ(n,0)=a+1Q(n)\)

\(aQ(0,n)=a+1Q(a,n-1)\)

\(C=a+1Q(b-1,c)\)

\(aQ^0(b_*,c)=a+1Q(\chi_{a+c}(a+c),c)\) \(\chi\) will be defined later

\(aQ^k(b_*,c)=a+1Q^{k-1}(a+1Q^{k-1}(...a+1Q^{k-1}(b_*,c),c)...)_*,c)\) a nests

\(aQ(b,c)=a+1Q^C(b_*,c-1)\)

\(\chi_0(c)=\chi(c)\)

\(\chi(0)=a\)

\(\chi(d)=\chi(d-1)+a\)

\(\chi_k(0)=\chi_{k-1}(a)\)

\(\chi_k(d)=\chi^a(\chi_k(d-1))\)

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

\((a)=a+1\)

\((a,0\#)=(a\#)\)

\((a,b)=(...((a,b-1),b-1)...,b-1)\) with a nests

\((0\#)=0\) Here \(\#\) isn't empty

\((\#0)\)=\((\#)\) Here \(\#\) isn't empty

\((a,b,1,1....,1,c,d,e...,f)=(a,b,b,b....,(a,b,b,b....,b,c-1,d,e...,f),c-1,d,e...,f)\)

\((a,b,c\#)=(a,(a,(...a,(a,b,c-1\#),c-1\#...),c-1\#),c-1\#),c-1\#)\) with b nests and \(\#\) contains no 0's or is empty

Seperated arrays
so \([0]\) is the same as a \(,\)

we need to define a, after other seperators so:

here \($\) is a group of \(1+\) seperator(s) and \($_k\) is the k-th step in the solving of \($\)

\((a\#_1$_i,0)=(a\#_1$_ia)\)

\((a\#_1$_k,b)=(((...(\#_1$_{k+1}b-1)...)\#_1$_{k+1}b-1)\#_1$_{k+1}b-1)\#_1,\#_1,\#_1...\#_1$_{k+1},b-1)\) with a \(\#_1\)'s and replace every entry in the array before the \($\) by the new array

and now we solve seperators by:

here \(\{\) is any amount of brackets and \(\}\) is also any amount of brackets

\(\{$[0]\}=\{$a\}=\{$\},\dots,\) with a ,'s and here \(\{\text{and} \} \) have more than zero brackets

and \(\{$_1[b]\}=\{$_1[b-1]...[b-1]\}\) with a b's

and \((a\#$_j0)=(a\#$_{j+1}a...$_{j+1}a)\) with a \($_{j+1}a\)'s where the last seperator in \($\) is no \(,\)

An attempt at an ordinal function
\(\Theta(0)[n]=n\) and \(\Theta_0=\Theta\) here \(\alpha[n]\) is the n-th term in the fundamental sequence of \(\alpha\)

\(\Theta(\alpha)[n]=\Theta(\alpha[n])\) Limit case

\(\Theta(\alpha+1)=\Theta(\alpha)\uparrow\uparrow n\) succesor case

to be continued

Fast growing array hierarchy
\(F_{\alpha}(a...,b)\) gets solved by the arry first (rules from my cosmic notation)

\(F_{\alpha+1}(b)=F_{\alpha}^{F_{\alpha}^{\dots^{F_{\alpha}^b(b...\text{b times},b)}\dots}(b...\text{b times},b)}(b...\text{b times},b)\) with b nests

\(F_{\alpha}(b)=F_{\alpha[F_{\alpha[\dots F_{\alpha[b]}\dots(b)]}(b)]}(b)\)