So in the notation, we currently have 4 symbols:
"s" for representing that this is a strong array
"(" and ")" for showing where the array starts and ends
and "," for separating entries
s(a,b) = a^b
s(a,b,c #) = s(a,s(a,b,c),c-1 #)
s(# 1) = s(#)
If none of these apply, start this process, starting from the third entry:
If the entry is 1, jump to the next entry.
If the entry is not 1 then
Change the 1,n to b,n-1 where b is the second entry
Change all entries before these two into the first entry
The process ends.
also, 1,n = b,n-1 where 1 is placed anywhere after first entry
and s(a,b,1,1,...,1,1,2) = s(a,a,a,...,a,a,a)
where the number of entries in first array is 1 more than second.
e.g
s(5,6,1,1,4,3)
= s(5,5,5,6,3,3)
= s(5,5,s(5,4,5,6,3,3),4,6,3,3)
= s(5,5,s(5,5,s(5,3,5,6,3,3),4,6,3,3),4,6,3,3)
= s(5,5,s(5,5,s(5,5,s(5,2,5,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3)
= s(5,5,s(5,5,s(5,5,s(5,5,s(5,1,5,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3)
= s(5,5,s(5,5,s(5,5,s(5,5,s(5,5,4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3)
= s(5,5,s(5,5,s(5,5,s(5,5,s(5,5,s(5,4,4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3),3,6,3,3)
= s(5,5,s(5,5,s(5,5,s(5,5,s(5,5,s(5,5,s(5,3,4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3),4,6,3,3),3,6,3,3),3,6,3,3)
...
and now for next part,
s(a,b{2}2) = s(a,a,1,1,...,1,1,2) w/ b-1 1's
s(a,b,1,1,1,...,1,1,1{2}2) = s(a,a,1,1,...,1,1,2) w/ b+(the number of ones in the unsolved array)-1 1's
so s(5,5,1,1,1{2}2) = s(5,5,1,1,1,1,1,1,1,2) w/ 7 1's
s(a,b{2}c) = s(a,a,1,1,1,...,1,1,2{2}c) w/ b-1 1's
tbc