User:Ynought

So i don't wanna annoy everybody with my expansion system but it still came up with an extension

this is the ☀Hyper Hyper extended expansion system

so let me define sepereator seperators and every 0 gets remove if [k]0[n]= [k+n]

and \(a[k]0\) = \(a-1(k+a)a-1\)

and \(0[k]a = a+k\)

terminology :

\(A\) is the rest of the seperator

\(B\) is the array without the current \(C\)

\(C\) is the leftmost cell

\(D\) is the n in [n] at the leftmost side (from (a,b,c,...)....(d,e,f,...)[n]

\(E\) is the whole array

 

if there are no commas(and more than one entry)  
 * 1) then \( [ \(A\)[1]k]\) gets solved with replacing one by k-1 and also decrementing k down by one and taking  \(D^B\) and creating \(D\) new \(C\) entries
 * 2) then  \( [ \(A\)[n]k]\) gets solved with taking  \(D^B\) with reducing k by one then n+ \(E\)

if there is a comma somewhere in the cell replace the comma with a seperetor whose k is one less than all the earliest non comma seperator to the left if there is none to the left then look towards the right

when there are arays in the  sepereator seperators they get solved the same