Nested Block Hyper Notation (NBHN) is the fifth part of HN.

Rules

An expression is of the form H<n,A>, where n is the base and A is an array.

1. If there is no array, H<n> = n^n
2. If the array's first entry is 1, delete it.
3. If the array is a sequence array, decrease it by one, then nest the whole expression into the base n times.
4. If the array is a limit array, change the array into the nth member of it's FS.

Arrays

First of all, all of the definitions of BaHN still apply, except that there can now be any array inside of blocks. So the arrays are now of the form A,[B],C, where A, B and C are arrays.

If the array is not of the form 1,[B],C, where C is a sequence array and B is a limit array, solve like normal.

Else, to get the FS of 1,[B],C, decrease C by 1, and you got the first member of the FS. The next member of the FS is 1,[A],[B],C-1 (For an array M, M-1 is decreasing M by 1), where A is the first member of B's FS, and the (n+1)th member of the FS is 1,[A],[B],C, where A is the nth member of B's FS.

Now, one more thing. In BaHN, we compared numbers to determine [x] in the array. In NBHN, we need to compare arrays to determine [B] in the array. But how do we compare arrays?

First, let A,B = A,[0],B, where A and B are arrays.

Then, A1 and A2 will be the arrays that we will be comparing.

First, search for the highest block (the block with the highest array inside of it) in A1 (call this block A1X)and A2 (call this block A2X). Then, compare the arrays that are inside of A1X and A2X, and that is your result, unless they're the same, then continue the comparison process.

Then, since A1 and A2 are of the form A,[B],C, compare the C of A1 and the C of A2, and that is your result, unless they're the same, then compare the A of A1 and the A of A2, then that is your result.