User blog comment:DrCeasium/new hyperfactorial array notation/@comment-3427444-20130416122943/@comment-7484840-20130416165351

Certainly. Just to allow me to show some more of the rules more easily, I will change the example to  4![1,[1,2],1,2]. To start with, the active entry is an array. To solve this, you just use the array in the active entry's main entry as if it was the active entry itself. As in this example, if this (main entry of array in active entry) is one, evaluate this sub-array until either the main entry becomes non-1 or it just completely decomposes into a single number. Therefore, this example would go:

4![1,[1,2],1,2] = (evaluating sub array)  4![1,1,1],1],1,2] = (removing trailing 1's)  4![1,[[1,1,2] = (using fact that [] = n)  4![1,[4],1,2].

Now, the sub-array has a main entry > 1, so we can continue:

4![1,[4],1,2] =  4![ [1,[1],1,2] ,[3],1,2] =  4![ [1,4,1,2] ,[3],1,2] =  4![ [ [1,1,1,2] ,3,1,2] ,[3],1,2] =  4![ [ [ [1,1,1,1] ,1,1,1] ,3,1,2] ,[3],1,2] =  4![ [ [ 4 ] ,3,1,2] ,[3],1,2] = (now we can evaluate outermost array due to dual-nesting in the main entry) ((( 4![ [ [ 3 ] ,3,1,2] ,[3],1,2]) ![ [ [ 3 ] ,3,1,2] ,[3],1,2]) ![ [ [ 3 ] ,3,1,2] ,[3],1,2]) ![ [ [ 3 ] ,3,1,2] ,[3],1,2] and so on.