User blog:Ynought/Part 2 of 2 "The & System"

Here S(k&(#))=k+k-1&(#-1)

# - or + a is the array with every entry + or - a

# is the array in k&(#)

k is always the k from k&(#)

and the system takes the form of k&(#)    (# is solved from right to left)

here i define the a{b}c seperator

added rules:

if there is only one pair of brackets : 

if b=0 then you replace the seperator with an c layer [] seperator(meaning [...c times…[[[[c]]......]]] ) and place k new a entries with these [...c times…[[[[c]]......]]] inbetween and k+S(k&(#))

if b>0 then #+S(k&(#)  then you decrement b by one and place k new c entries with these seperators inbetween ( {b-1} ) 

the {a{b}c} gets solved like (a{b}c)

when you have more than one pairs of brackets :

if b=0 then you replace the seperator with an c layer  seperator(meaning {c{c{...c times…{c}c}...c} ) and place k new a entries with these {c{c{...c times...c{c}c}...c} inbetween and k+S(k&(#))

if b>0 then you add S(k&(#) to every entry and to k  then you decrement b by one and place k new c entries with these seperators inbetween ( {b-1} )