User blog:Ynought/Expanded expansion system

BTW every single number in this system is an integer and larger than one                                                             terminology :

\(A\) is [n]

\(B\) is The n in \(A\)

\(C\) is the entry (a,b,c….,z)(up from one entry in the brackets)to the left of \(A\)

\(D\)(\(n\),\(C\))is the \(n\) - th entry in \(C\)

\(E\) is the new formed array

\(F\) is the number of entries in \(C\)

\(G\) is the seperator between the entries in \(C\)

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

a cell is (a,b,c,....,z)

Every single Cell has to have 2+ entries seperated by seperators in the form of [\(K\)] \(K\) is some integer 

when the cell has to entries when there are 3+ entries
 * 1) if \(K\) =  then (a[1]b) gets solved like so : first take \(B^H\) then create \(B\) (the new one)  new entries with b entries of value a
 * 2) If \(K\) ≠ then take \(B^H\) then create \(B\) (the new one)  new entries with length b and value a with [k-1] seperators in between.
 * 1) 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
 * 2) if there is no comma then you take \(B^H\) then you decrement \(D_F\) by 1 now you create \(B\) new entries that are exactly the same as \(C\) .repeat until \(D_F\)=0 then you decrement the leftmost seperator by 1 and then you replace  the 0 with the entry(not the 0 one) next to the seperator (the leftmost one)