User blog:Ynought/Expansion system

BTW every single number in this system is an integer and larger or equall to 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

Now for the 1-entry system (1 entry in \(C\)


 * 1) if \(C\)=1 then you remove that entry and take That new system copy it and take \(B\) to the power of \(E\)  
 * 2) if \(C\)≠1 then you decrement the entry by one and take \(textrm{B}^E\) now place \(B\) entrys of value \(C\)-1
 * 3) Repeat the previous steps

For 2+ entry system 

Here \(F\) is the number of entrys in the system(per (a,b,c,d…,z) ,this referes to \(C\) )


 * 1) If every entry in \(C\) is bigger than 1 create a new entry where \(D\)(1,\(C\)) is reduced by one  now create a new entry where \(D\)(2,\(C\)) is reduced by one.Keep going like this until \(D\)(\(F\),\(C\)) and take \(B\) and take it to the power of \(E\)
 * 2) If there’s at least one non 1 entry and it at the front remove that first entry and take \(B\) and take it to the power of \(E\) now create \(B\) new entrys that are exactly the same as the new \(C\) 
 * 3) If there’s at least one non 1 entry and its not at the front take the entry before that means that if the 1st one is at place \(D\)(\(n\),\(C\)) now add one to \(D\)(\(n-1\),\(C\))and remove entry \(D\)(\(n\),\(C\)) now take \(B\) and take it to the power of \(E\) now create \(B\) new entrys that are exactly the same as the new \(C\)