--- Abandoned ---
Reason: I created this notation with no ideas whether the notation is intended to have more extensions.
Hex function is the function created by me. It is based on hexadecimal (base 16) numeral system.
An expression consists of different kinds of entries and separators at lower levels, and later levels also introduces some new separators, which diagonalizes over many groups consisting of one or more items, which may be either nested separators (limit of the level which diagonalizes some fundamental sequences over εα), star (*), ampersand (&), caret (^), or an hyper-explodion array (#), starting from nested separators first.
For empty expressions, H() = 16.
For H(n), H(n) = 16*n.
If the first entry is 1, H(1,x) = H(1) = 16 (x is the latter of the expression.)
If there are 1’s at the end of the input, get rid of 1.
For two-entry functions, H(a,b) = H(H(a-1,b),b-1).
For those the first entry is 2, and the second entry is greater than 1, H(2,x,…) = H(16,x-1,…)
For three or more entries:
Reuse the previous rules.
The final … is the latter of the expression.
If there are 1’s in the intermediate entries, H(a,1,c,…) = H(a,(a-1,1,c,…),c-1,…), H(a,1,1,…,1,1,1,d,…) = H(a,1,1,…,1,1,H(a-1,1,1,…,1,1,1,d,…),d-1,…)
If the second entry is greater than 1, H(a,b,c,…) = H(H(a-1,b,c,…),b-1,c,…)
Note that all the arguments must be natural numbers (not including 0 in the entries, but including 0 in the separators), otherwise the expression will be ill-defined.
The extended notation introduces separators, which can be denoted using curly brackets ({}).
Commas indicate {0} separator.
H(a{1}2) = H(a,1,1,1,…,1,1,1,2) with a entries of 1’s.
H(a{1}n,…) = H(a,1,1,1,…,1,1,1,2{1}n-1,…) with a entries of 1’s.
H(a{1}1,n,…) = H(a{1}H(a-1{1}1,n,…),n-1,…), H(a{1}1,1,…,1,1,1,n,…) = H(a{1}1,1,…1,1,H(a-1{1}1,1,…,1,1,1,n,…),n-1,…)
H(a{1}1{1}2) = H(a{1}1,1,1,…,1,1,1,2) with a entries of 1’s after {1}.
H(a{2}n,…) = H(a{1}1{1}1{1}…{1}1{1}1{1}n-1,…) with a-1 entries of 1’s separated by {1}.
H(a{2}1{1}n,…) = H(a{2}1,1,1,…,1,1,1,2{1}n-1) with a entries of 1’s between {2} and {1}.
H(a{k}n,…) = H(a{k-1}1{k-1}…{k-1}1{k-1}2{k}n-1,…) with a-1 entries of 1’s separated by {k-1}.
H(a{0,1}2) = H(a{a}2).
H(a{0,1}n,…) = H(a{a}2{0,1}n-1,…).
H(a{k,1}n,…) = H(a{k-1,1}1{k-1,1}…{k-1,1}1{k-1,1}2{k,1}n-1,…) with a-1 entries of 1’s separated by {k-1,1}.
H(a{0,k}2) = H(a{a,k-1}2).
H(a{0,0,0,…,0,0,0,k}2) = H(a{0,0,0,…,0,0,a,k-1}2).
H(a{0{1}1}2) = H(a{0,0,0,…,0,0,0,1}2) with a entries of 0’s inside {}.
H(a{0{1}n}2) = H(a{0,0,…,0,0,1{1}n-1}2) with a entries of 0’s inside {}.
H(a{0{1}0{1}2}2) = H(a{0{1}0,0,…,0,0,1}2) with a entries of 0’s after {1} inside {}.
H(a{0{n}x}2) = H(a{0{n-1}0{n-1}…{n-1}0{n-1}x-1}2) with a entries of 0’s separated by {n-1}.
And so on.
H(3,2)
= H(H(2,2))
= H(H(H(1,2)))
= H(H(16)))
= H(16*16)
= H(256)
= 256*16
= 4,096
4,096
H(3,3)
= H(H(2,3),2)
= H(H(16,2),2)
= H(H(H(15,2)),2)
= H(H(H(H(14,2))),2)
…
H(3,1,1,3)
= H(3,1,H(2,1,1,3),2)
= H(3,1,H(2,1,H(2,1,H(1,1,1,3),2),2),2)
= H(3,1,H(2,1,H(2,1,16,2),2),2)
= H(3,1,H(2,1,H(2,H(1,1,16,2),15,2),2),2)
= H(3,1,H(2,1,H(2,16,15,2),2),2)
= H(3,1,H(2,1,H(16,15,15,2),2),2)
= H(3,1,H(2,1,H(H(15,15,15,2),14,15,2),2),2)
= H(3,1,H(2,1,H(H(H(14,15,15,2),14,15,2),14,15,2),2),2)
…
H(4{2}3)
= H(4{1}1{1}1{1}1{1}2{2}2)
= H(4{1}1{1}1{1}1,1,1,1,2{2}2)
= H(4{1}1{1}1{1}1,1,1, H(3{1}1{1}1{1}1,1,1,1,2{2}2){2}2)
= H(4{1}1{1}1{1}1,1,1, H(3{1}1{1}1{1}1,1,1, H(2{1}1{1}1{1}1,1,1,1,2{2}2){2}2){2}2)
= H(4{1}1{1}1{1}1,1,1, H(3{1}1{1}1{1}1,1,1, H(2{1}1{1}1{1}1,1,1, H(1{1}1{1}1{1}1,1,1,1,2{2}2){2}2){2}2){2}2){2}2)
= H(4{1}1{1}1{1}1,1,1, H(3{1}1{1}1{1}1,1,1, H(2{1}1{1}1{1}1,1,1, 16{2}2){2}2){2}2){2}2)
…
H(2{0{2}1{1}1}2)
= H(2{0{1}0{1}1{2}0{1}1}2)
= H(2{0{1}0,0,1{2}0{1}1}2)
= H(2{0{1}0,2{2}0{1}1}2)
= H(2{0{1}2,1{2}0{1}1}2)
= H(2{0,0,1{1}1,1{2}0{1}1}2)
= H(2{0,2{1}1,1{2}0{1}1}2)
= H(2{2,1{1}1,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{1,1{1}1,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{2{1}1,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{1{1}1,1{2}0{1}1}1{1{1}1,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{1{1}1,1{2}0{1}1}1{0{1}1,1{2}0{1}1}1{0{1}1,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{1{1}1,1{2}0{1}1}1{0{1}1,1{2}0{1}1}1{0,0,1{1}0,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{1{1}1,1{2}0{1}1}1{0{1}1,1{2}0{1}1}1{0,2{1}0,1{2}0{1}1}2)
= H(2{1,1{1}1,1{2}0{1}1}1{0,1{1}1,1{2}0{1}1}1{1{1}1,1{2}0{1}1}1{0{1}1,1{2}0{1}1}1{2,1{1}0,1{2}0{1}1}2)
…
H(2{1}2)
= H(2,1,1,2)
= H(2,1,H(1,1,1,2))
= H(2,1,16)
= H(2,H(1,1,16),15)
= H(2,16,15)
= H(H(1,16,15),15,15)
= H(16,15,15)
= H(H(15,15,15),14,15)
…
We can see that the expressions with 2 in the first entry will never degenerate the notation.
The previous rules remain unchanged.
It is easy to show that H(a,b) is equal to 16 ↑[b-1] a, using Knuth's up-arrow notation.
Under the fundamental sequences associated to the Wainer hierarchy, note that the expressions like H(2,1,1,2) has the fast-growing ordinal of ω*n+1, H(2,1,1,1,2) has level ω^2*n+1, H(2{1}1,2) has level ω^ω*n+1, these do not match exactly over the terminal fundamental sequences of ω's without +1, under the totalities of each functions that are provably recursive within Peano arithmetic.
The pre-Alpha level of the notation has the fast-growing hierarchy ordinal of ε0 with respect to the Wainer hierarchy.