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 integers in the \(I\) function

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

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

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

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

\(G\) 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.

Stronger operator notation
first of all let the n-th hyperoperator be displayed as \(+_n\) so the normal \(+=+_0\)

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

\(a+b\) is just \(a+b\) but with moer than one hyper operator

\(a\#+b=a\#(...a\#(a)...)\) 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 \(q=+_{J_b,j_c,...,d-1}...+_{J_b,j_c,...,d-1}\)