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Primary dropping array notation (pDAN) is the fifth part of my array notation. In this part, it will surpass all the BEAF (well, strictly speaking BEAF is not well-defined), BAN, ExE and HAN. The most important thing we need in this part is the 2-separator.

# Up to a 2-separator

The multiple grave accents are 1-separators, and the double comma (,,) is the first 2-separator. Well, we don’t know what’s a “1-separator” and what’s a “2-separator” now, but they will be defined later. Here we just can say that the double comma is a **high leveled** separator.

In this step, all the grave accents disappear, and the only basis separators are the comma and the double comma. The grave accent is just a shorthand for {1 ,, 2} so the dot is {1 {1 ,, 2} 2}. The double grave accent is a shorthand for {1 ,, 3}, the triple grave accent is a shorthand for {1 ,, 4}, and so on. The {A^{` `…`}} with m-1 grave accents is replaced by {A ,, m} now.

When we meet {1 ,, 1… ,, 1 ,, 1 ,, n #}, search for the innermost separator such that this {1 ,, 1… ,, 1 ,, 1 ,, n #} is inside it and it has lower level than {1 ,, 1… ,, 1 ,, 1 ,, n #}. If it has lower level than {1 ,, 1… ,, 1 ,, 1 ,, n-1 #}, we need to add something like what we do in mEAN. If not, it’ll expands like in mEAN. e.g. s(a,b {1 ,, 1 {1 ,, 1 ,, 3} 3 ,, 2} 2) = s(a,b {1 ,, 1 …{1 ,, 1 {1 ,, 1,2 {1 ,, 1 ,, 3} 2 ,, 2} 2 {1 ,, 1 ,, 3} 2 ,, 2}… 2 {1 ,, 1 ,, 3} 2 ,, 2} 2) with b-1 {1 ,, 1 ,, 3}′s.

And here is the first time the concept of “levels” really matters, not only to help the rule 4.

## Process

Note that case B1, B2 and B4 are terminal but case A and B3 are not. And note that b, n, B_{t} and K are parts of the original array but d, t, i, S_{i}, A_{t}, X, Y, P and Q are not. Before we start, let A_{0} be the whole array. First start from the 3rd entry.

- Case A: If the entry is 1, then you jump to the next entry.
- Case B: If the entry n is not 1, look to your left:
- Case B1: If the comma is immediately before you, then
- Change the “1,n” into “b,n-1” where n is this non-1 entry and the b is the iterator.
- Change all the entries at base layer before them into the base.
- The process ends.

- Case B2: If the double comma is immediately before you, then
- Let d and t be such that the double comma is at layer d = t. And let separator B
_{0}be the whole array now. - Repeat this:
- Subtract t by 1.
- Let separator B
_{t}be such that it’s at layer t, and the double comma is inside it. - If t = 1, then break the repeating, or else continue repeating.

- Find the maximum of t such that lv(A
_{t}) < lv(A_{d-1}). - Let string X and Y be such that A
_{d-1}= “X ,, n Y”. - If lv(A
_{t}) ≥ lv(“X A_{d-1}2 ,, n-1 Y”), then- Let string P and Q be such that B
_{t}= “P A_{d-1}Q”. - Change B
_{t}into S_{b}, where b is the iterator, S_{1}is comma, and S_{i+1}= “P S_{i}Q”. - The process ends.

- Let string P and Q be such that B
- If lv(A
_{t}) < lv(“X A_{d-1}2 ,, n-1 Y”), then- Let string P and Q be such that B
_{t}= “P B_{t+1}Q”. - Change B
_{t}into “P X A_{t+1}2 ,, n-1 Y Q”. - The process ends.

- Let string P and Q be such that B

- Let d and t be such that the double comma is at layer d = t. And let separator B
- Case B3: If a separator K neither comma nor double comma is immediately before you, then
- Change the “K n” into “K 2 K n-1”.
- Set separator A
_{t}= K, here K is at layer t. - Jump to the first entry of the former K.

- Case B4: If an lbrace is immediately before you, then
- Change separator {n #} into string S
_{b}, where b is the iterator, S_{1}= “{n-1 #}” and S_{i+1}= “S_{i}1 {n-1 #}”. - The process ends.

- Change separator {n #} into string S

- Case B1: If the comma is immediately before you, then

## Level comparison

First, all arrays have the lowest and the same level. Then, the double comma has the highest level. And note that the same separators have the same level. To compare levels of other separators A and B, we follow these steps.

- Apply rule 2 to A and B until rule 2 cannot apply any more.
- Let A = {a
_{1}A_{1}a_{2}A_{2}…a_{k-1}A_{k-1}a_{k}} and B = {b_{1}B_{1}b_{2}B_{2}…b_{l-1}B_{l-1}b_{l}} - If k = 1 and l > 1, then lv(A) < lv(B); if k > 1 and l = 1, then lv(A) > lv(B); if k = l = 1, follow step 4; if k > 1 and l > 1, follow step 5 ~ 10
- If a
_{1}< b_{1}, then lv(A) < lv(B); if a_{1}> b_{1}, then lv(A) > lv(B); if a_{1}= b_{1}, then lv(A) = lv(B) - Let , and .
- If lv(A
_{maxM(A)}) < lv(B_{maxM(B)}), then lv(A) < lv(B); if lv(A_{maxM(A)}) > lv(B_{maxM(B)}), then lv(A) > lv(B); or else – - If |M(A)| < |M(B)|, then lv(A) < lv(B); if |M(A)| > |M(B)|, then lv(A) > lv(B); or else –
- Let A = {#
_{1}A_{maxM(A)}#_{2}} and B = {#_{3}B_{maxM(B)}#_{4}} - If lv({#
_{2}}) < lv({#_{4}}), then lv(A) < lv(B); if lv({#_{2}}) > lv({#_{4}}), then lv(A) > lv(B); or else – - If lv({#
_{1}}) < lv({#_{3}}), then lv(A) < lv(B); if lv({#_{1}}) > lv({#_{3}}), then lv(A) > lv(B); if lv({#_{1}}) = lv({#_{3}}), then lv(A) = lv(B)

## Explanation

Case B2 changes here. The double comma is at layer d, so A_{d-1} is the separator that contains the double comma – it’s the {1 ,, 1… ,, 1 ,, 1 ,, n #}. So step 5 is the expanding rule, and step 6 is the adding rule.

## Comparison

Here’re comparisons between my array notation and FGH. If separator A has recursion level α, then s(n,n A 2) has growth rate .

{1 ,, 2} has recursion level

{1 ,, 3} has recursion level

{1 ,, 4} has recursion level

{1 ,, 1,2} has recursion level

{2 ,, 1,2} has recursion level

So {1,2 ,, 1,2} has recursion level .

{1 {1 ,, 1,2} 2} has recursion level

{2 {1 ,, 1,2} 2} has recursion level

{1,2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 1,2} 2} 1,2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 1,2} 2} 1 {1 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {2 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 2} 1 {1 {1 ,, 1,2} 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 2} 1 {1 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {2 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2} 2 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 2} 2 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {2 {1 ,, 2} 2 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 2} 3 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 2} 1 {1 ,, 2} 2 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 2} 2 ,, 2} 2 ,, 2} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 3} 3 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 3} 1,2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 3} 1 {1 ,, 3} 1,2 {1 ,, 1,2} 2} has recursion level

{1 {2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 3} 2 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 3} 3 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 3} 1 {1 ,, 3} 2 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {2 ,, 3} 2 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 {1 ,, 2} 2 ,, 3} 2 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 {1 {1 ,, 3} 2 ,, 2} 2 ,, 3} 2 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 {1 ,, 3} 2 ,, 3} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 4} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 5} 2 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 1,2} 3} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 2} 2 ,, 2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 2} 3 ,, 2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 2} 1 {1 ,, 2} 2 ,, 2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 {1 ,, 2} 2 ,, 2} 2 ,, 2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 3} 3} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 3} 1,2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 3} 1 {1 ,, 3} 1,2} has recursion level

{1 {1 ,, 1,2} 1 {2 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 2} 2 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 {1 ,, 3} 2 ,, 2} 2 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 {1 ,, 3} 3 ,, 2} 2 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 {1 ,, 3} 1 {1 ,, 3} 2 ,, 2} 2 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 3} 2 ,, 3} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 4} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 5} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 3} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 1 {1 ,, 2} 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 1 {1 ,, 1,2} 2} has recursion level

{1 {2 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2} 2 ,, 1,2} 2} has recursion level

{1 {1 {1 {2 ,, 1,2} 2} 2 ,, 1,2} 2} has recursion level

{1 {1 {1 {1 {1 {1 ,, 1,2} 2} 2 ,, 1,2} 2} 2 ,, 1,2} 2} has recursion level

So {1 {1 ,, 2} 2 ,, 1,2} has recursion level .

{1 {1 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 ,, 1,2} 2 ,, 2} 3} has recursion level

{1 {1 {1 ,, 1,2} 2 ,, 2} 1 {1 ,, 2} 2} has recursion level

{1 {1 {1 ,, 1,2} 2 ,, 2} 1 {1 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 ,, 1,2} 2 ,, 2} 1 {1 {1 ,, 1,2} 2 ,, 2} 1 {1 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {2 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 ,, 2} 2 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2 ,, 2} 2 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2 ,, 2} 3 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2 ,, 2} 1 {1 {1 ,, 1,2} 2 ,, 2} 2 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 {1 {1 {1 ,, 1,2} 2 ,, 2} 2 {1 ,, 1,2} 2 ,, 2} 2 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 {1 ,, 3} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2 ,, 2} 2 ,, 3} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 ,, 3} 2 ,, 3} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 ,, 3} 3 ,, 3} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 ,, 3} 1 {1 ,, 3} 2 ,, 3} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 {1 ,, 3} 2 ,, 3} 2 ,, 3} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 ,, 4} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 ,, 4} 3 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 ,, 4} 1,2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {2 ,, 4} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 ,, 3} 2 ,, 4} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 {1 ,, 4} 2 ,, 3} 2 ,, 4} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 ,, 4} 2 ,, 4} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 ,, 5} 2 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 3 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 1,2} 2 ,, 2} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {2 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 2} 2 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 {1 ,, 1,2} 2 ,, 2} 2 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 3} 2 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 4} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {2 ,, 4} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 {1 ,, 4} 2 ,, 3} 2 ,, 4} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 {1 ,, 4} 2 ,, 4} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 5} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 3 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 1 {1 ,, 2} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 1 {1 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 1 {1 {1 ,, 3} 2 ,, 3} 2 ,, 2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 1 {1 ,, 1,2} 2 ,, 2} has recursion level

{1 {2 ,, 1,2} 2 ,, 2} has recursion level

{1 {1 {1 ,, 2} 2 ,, 1,2} 2 ,, 2} has recursion level

So{1 {1 ,, 3} 2 ,, 1,2} has recursion level .

{1 {1 ,, 4} 2 ,, 1,2} has recursion level

{1 {1 ,, 5} 2 ,, 1,2} has recursion level

{1 {1 ,, 1,2} 2 ,, 1,2} has recursion level

{1 {1 ,, 1,2} 3 ,, 1,2} has recursion level

{1 {1 ,, 1,2} 1 {1 ,, 1,2} 2 ,, 1,2} has recursion level

{1 {1 {1 ,, 1,2} 2 ,, 1,2} 2 ,, 1,2} has recursion level

{1 {1 {1 ,, 1,2} 1 {1 ,, 1,2} 2 ,, 1,2} 2 ,, 1,2} has recursion level

{1 {1 {1 {1 ,, 1,2} 2 ,, 1,2} 2 ,, 1,2} 2 ,, 1,2} has recursion level

{1 ,, 2,2} has recursion level

So s(n,n {1 ,, 2,2} 2) eventually outgrows any function provably recursive in . It’s comparable to Buchholz hydra (BH(n)), and it’s the upper bound for growth rate of subcubic graph number (SCG(n)) and simple subcubic graph number (SSCG(n)). Then,

{1 ,, 3,2} has recursion level

{1 ,, 4,2} has recursion level

{1 ,, 1,3} has recursion level

{1 ,, 1,1,2} has recursion level

{1 ,, 1 {1 {1 ,, 2} 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1,2} 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1,3} 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1 {1 {1 ,, 1,2} 2} 2} 2} 2} has recursion level

{1 ,, 1 {1 ,, 2} 2} has recursion level , so s(n,n {1 ,, 1 {1 ,, 2} 2} 2) is comparable to the new S(n) in BAN, and it’s the limit of BAN.

{1 ,, 2 {1 ,, 2} 2} has recursion level

{1 ,, 1 {1 ,, 2} 3} has recursion level

{1 ,, 1 {1 ,, 2} 1 {1 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 2} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 2} 1 {1 ,, 2} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 {1 ,, 2} 2 ,, 2} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 3} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1,2} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1 {1 ,, 2} 2} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1 {1 {1 ,, 1,2} 2 ,, 2} 2} 2 ,, 2} 2} has recursion level

{1 ,, 1 {1 ,, 3} 2} has recursion level

{1 ,, 1 {1 ,, 3} 3} has recursion level

{1 ,, 1 {1 ,, 3} 1 {1 ,, 3} 2} has recursion level

{1 ,, 1 {1 {1 ,, 3} 2 ,, 3} 2} has recursion level

{1 ,, 1 {1 {1 ,, 4} 2 ,, 3} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1,2} 2 ,, 3} 2} has recursion level

{1 ,, 1 {1 ,, 4} 2} has recursion level

{1 ,, 1 {1 ,, 5} 2} has recursion level

{1 ,, 1 {1 ,, 1,2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 2,2} 2 ,, 1,2} 2} has recursion level

{1 ,, 1 {1 {1 ,, 1 {1 ,, 1,2} 2} 2 ,, 1,2} 2} has recursion level

{1 ,, 1 {1 ,, 2,2} 2} has recursion level

{1 ,, 1 {1 ,, 1,3} 2} has recursion level

{1 ,, 1 {1 ,, 1 {1 {1 ,, 2} 2} 2} 2} has recursion level

{1 ,, 1 {1 ,, 1 {1 {1 ,, 1 {1 ,, 1,2} 2} 2} 2} 2} has recursion level

{1 ,, 1 {1 ,, 1 {1 ,, 2} 2} 2} has recursion level

{1 ,, 1 {1 ,, 1 {1 ,, 3} 2} 2} has recursion level

{1 ,, 1 {1 ,, 1 {1 ,, 1,2} 2} 2} has recursion level

{1 ,, 1 {1 ,, 1 {1 ,, 1 {1 ,, 1,2} 2} 2} 2} has recursion level

{1 ,, 1 {1 ,, 1 ,, 2} 2} has recursion level C(C(Ω_{2}2,0),0), so s(n,n {1 ,, 1 ,, 2} 2) outgrows any function provably recursive in

More comparisons (vs. Taranovsky’s ordinal notation) are shown here. Result:

- s(n,n {1 ,, 2 ,, 2} 2) eventually outgrows any function provably recursive in KPi.

# Arrays on double comma

Now the primary dropping array notation comes. First the double comma is just a shorthand for {1^{,,}} with two commas at the left-superscript position of the rbrace (not double quotation mark). Then {2^{,,}} works in the exAN way. All these separators with two commas at the left-superscript position of the rbrace have higher level than separators without commas at the left-superscript position of the rbrace.

What’s notable is {1,,2^{,,}} – this time the A_{d-1} in case B2 is {1,,2^{,,}}, if it expands, then it’ll become {1{1…{1{1,2^{,,}}2^{,,}}…2^{,,}}2^{,,}} with b-1 2′s, and it reduces to {1{1…{1,,2^{,,}}…2^{,,}}2^{,,}} then expands again – that causes problems. So we need to change the rules. In EAN, to solve arrays on the grave accent, we search for the innermost separator without grave accents at the left-superscript position of the rbrace that the grave accent is inside it. In the same way, to solve arrays on the double comma, we search for the innermost separator without commas at the left-superscript position of the rbrace that the double comma is inside it.

But a separator without commas at the left-superscript position of the rbrace means that this separator has lower level than the double comma, so “searching for the innermost separator that has lower level than the double comma and the double comma is inside it” is OK.

## Process

Note that case B1, B2 and B4 are terminal but case A and B3 are not. And note that b, n, B_{t} and K are parts of the original array but t, u, v, i, S_{i}, A_{t}, X, Y, P and Q are not. Before we start, let A_{0} be the whole array. First start from the 3rd entry.

- Case A: If the entry is 1, then you jump to the next entry.
- Case B: If the entry n is not 1, look to your left:
- Case B1: If the comma is immediately before you, then
- Change the “1,n” into “b,n-1” where n is this non-1 entry and the b is the iterator.
- Change all the entries at base layer before them into the base.
- The process ends.

- Case B2: If the double comma (or equivalently, the {1
^{,,}}) is immediately before you, then- Let t be such that the double comma is at layer t. And let separator B
_{0}be the whole array now. - Repeat this:
- Subtract t by 1.
- Let separator B
_{t}be such that it’s at layer t, and the double comma is inside it. - If t = 1, then break the repeating, or else continue repeating.

- Find the maximum of u such that lv(A
_{u}) < lv(,,). - Find the maximum of t such that lv(A
_{t}) < lv(A_{u}). - Let string X and Y be such that B
_{u}= “X ,, n Y”. - If lv(A
_{t}) ≥ lv(“X A_{u}2 ,, n-1 Y”), then- Let string P and Q be such that B
_{t}= “P B_{u}Q”. - Change B
_{t}into S_{b}, where b is the iterator, S_{1}is comma, and S_{i+1}= “P S_{i}Q”. - The process ends.

- Let string P and Q be such that B
- If lv(A
_{t}) < lv(“X A_{u}2 ,, n-1 Y”), then- Find the minimum of v such that v > t and lv(A
_{v}) < lv(,,). - Let string P and Q be such that B
_{t}= “P B_{v}Q”. - Change B
_{t}into “P X A_{v}2 ,, n-1 Y Q”. - The process ends.

- Find the minimum of v such that v > t and lv(A

- Let t be such that the double comma is at layer t. And let separator B
- Case B3: If a separator K neither comma nor double comma is immediately before you, then
- Change the “K n” into “K 2 K n-1”.
- Set separator A
_{t}= K, here K is at layer t. - Jump to the first entry of the former K.

- Case B4: If an lbrace is immediately before you, then
- Change separator {n #} into string S
_{b}, where b is the iterator, S_{1}= “{n-1 #}” and S_{i+1}= “S_{i}1 {n-1 #}”. - The process ends.

- Change separator {n #} into string S

- Case B1: If the comma is immediately before you, then

## Explanation

The step 3 of case B2 is the searching-out step. But the step 7 of case B2 (the adding rule) looks very strange, so let me explain more about it.

Before it, we use A_{t+1} and B_{t+1} instead of A_{v} and B_{v} for the adding. If we use “t+1”, consider this separator: {1{1{1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}2^{,,}}2} – here u = 5, t = 1, so A_{u} = {1{1,,3^{,,}}2} and A_{t} = {1{1{1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}2^{,,}}2}. Then add “{1{1” and “2,,2^{,,}}2}” outside A_{t+1}, so it becomes {1{1{1{1{1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}2^{,,}}2,,2^{,,}}2}2}. What we get is “something {1{1 something 2^{,,}}2,,2^{,,}} something” – with a very high level separator. It can be reduced to separators even higher leveled than {1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}, and the same thing repeats, so it’ll never be solved.

To solve this problem, we use such a “v” that A_{v} is the outermost separator that has level lower than the double comma. For example,

{1{1{1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}2^{,,}}2}: A_{u} = {1{1,,3^{,,}}2}, A_{t} = {1{1{1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}2^{,,}}2}, A_{v} = {1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}, so it becomes {1{1{1{1{1{1{1{1,,3^{,,}}2}2,,3^{,,}}2}2,,2^{,,}}2}2^{,,}}2}.

{1{1{1{1,,3^{,,}}2}2^{,,}}2}: A_{u} = {1{1,,3^{,,}}2}, A_{t} = {1{1{1{1,,3^{,,}}2}2^{,,}}2}, A_{v} = {1{1,,3^{,,}}2} = A_{u}, so it becomes {1{1{1{1{1{1,,3^{,,}}2}2,,2^{,,}}2}2^{,,}}2}.

{1{1{1,,3}2,,3}2}: A_{u} = {1,,3}, A_{t} = {1{1{1,,3}2,,3}2}, A_{v} = {1{1,,3}2,,3}, so it becomes {1{1{1{1,,3}2,,3}2,,2}2}. Here v = t+1 and it happens to be the same as mEAN.

The range of v is t+1 ≤ v ≤ u. In example 2, v = u, and in example 3, v = t+1.

## Comparison

Comparisons (vs. Taranovsky’s ordinal notation) are shown here. Result:

- s(n,n {1 ,, 2 {1 ,, 2
^{,,}} 2} 2) eventually outgrows any function provably recursive in KPM. - s(n,n {1 ,, 2 {1,,1,,…1,,2
^{,,}} 2} 2) (with m double commas in the blue part, m > 1) eventually outgrows any function provably recursive in KP + Π_{m+1}reflection. - s(n,n {1 ,, 2 {1 {1 ,, 2
^{,,}} 2^{,,}} 2} 2) eventually outgrows any function provably recursive in KP + stable ordinal.

I know {1{1{1{1,,2^,,}2}2^,,}1,,2} = I(1,0,0) and here is why:

Let take a look for {1{1{1{1,,2^,,}2}2^,,}3}:

,, is in separator layer 5

u = 3 because {1{1,,2^,,}2} level is less than ,,

t =/ 2 because ,, separators including {1{1{1,,2^,,}2}2^,,} is greater level than {1{1,,2^,,}2}

t = 1 because {1{1{1{1,,2^,,}2}2^,,}3} level is less than {1{1,,2^,,}2}

Therefore, {1{1{1{1,,2^,,}2}2^,,}3} is 1-separator-like of {1{1_2^,,}2{1{1{1,,2^,,}2}2^,,}2}

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Wait, not ‘is 1-separator-like’, ‘expands to nests of’.

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Question: Which is the 1-separator of {1{1{_}2^,,}2}? {1{1{1{1,,2^,,}2}2^,,}2} or {1{1{1{1{1{1,,2^,,}2}2^,,}2}2^,,}2}?

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{1{1,,2^,,}2} or {1{1{1{1,,2^,,}2}2^,,}2}*

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The 1-separator for {1{1 { ____ } 2^,,}2} equals the 1-separator for { ____ } – that’s the grave accent.

But the 1-separator for {1{1 ____ 2^,,}2} is much higher – it’s {1{1,,2^,,}2}.

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Then PNAN (R Function) >> pDAN.

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pDAN is still as strong as PNAN in R function. The double comma in pDAN is a 2-separator; the comma in PNAN needs to search out for an array, then the array is a “diagonalizer” – corresponding to 1-separator in pDAN, so the comma in PNAN corresponds to the double comma in pDAN.

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No. In PNAN, {0{0,1*}1} works like M in chi(). In pDAN, {1{1,,2^,,}2} works like I(1,0,0).

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Therefore, pDAN is slightly stronger than PNAN.

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But I might not understand correctly the concept of 1-separators

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Wait, now that I think about it, {0{0,1*}1} might work like I(1,0,0), and you didn’t understand correctly how it works.

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Hyp cos’ analysis results pDAN is stronger than PNAN in R function.

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Hello.

I studied your notation and want to write a popular article about this in Russian. But because the Taranovsky’s notation is too difficult for popular exposition, I would like to compare pDAN with Inaccessible collapsing function. Could you help me compare, at least some of the steps from this table:

http://lihachevss.ru/pDAN.html

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“Inaccessible collapsing function” series can’t reach the limit of pDAN, but Taranovsky’s ordinal notation can.

Certainly you can use more extended “inaccessible collapsing function” series, but they suddenly become more difficult than Taranovsky’s ordinal notation, such as the Pi-4 reflecting one, the Pi-n reflecting one, or even higher.

So, if you want to explain the strength of pDAN more simply, Taranovsky’s ordinal notation is still a better choice than “inaccessible collapsing function” series.

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Ok, but still I must first show how strong this notation is on simpler examples. If it’s not difficult for you, please make at least a few comparisons from my table.

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Scorcher, I can help you with some of the comparisons.

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I don’t understand TON beyond C(C(C(W_2+C(…)),C(W_2*2,0))) = C(C(C(W_2*2,0),C(W_2*2,0))).

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I don’t understand some things.

What is the difference between a {1,,1,,2} seperator and a {1,,1{1,,1,,2}2} seperator?

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It’s the difference between the grave accent and {1`2}.

The double comma is the 2-separator on { ____ }, then {1,,1,,2} is the 1-separator on {1,, ____ }, so {1,,1{1,,1,,2}2} expands to {1,,1{1,,1…{1,,1,2}…2}2}.

Although a single {1,,1,,2} reduces to {1,,1{1,,1,,2}2} by adding then expands, it can also appear inside other separators such as {1,,1{1,,1,,2}3}, which expands to {1,,1{1,,1…{1,,1,2{1,,1,,2}2}…2{1,,1,,2}2}2{1,,1,,2}2}.

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So then all seperators within the {} braces are lower-ranked than the (,,) double comma?

(as long as they don’t contain higher-ranked seperators than (,,))

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Yes. In pDAN they are.

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Is the {1,,,2} -seperator equal to (,,) double comma?

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Yes. In sDAN and DAN it is. But in NDAN {1(3)△2} is not fully identical to (2)△ or (2){1(3)△2}.

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Hypcos where is X&&&&&&…&&&&&&X array space in your Strong Array Notation?

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There is ill-defined after linear array spaces, so we can’t do that.

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So are you saying X&&&…&&&X arrays don’t exist, because of I’ll definition? Like lugion marks (\)?

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Hypcos could you show me the steps to go from {1,2,,1,2} to {1{1,,1,2}2}?

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No. We don’t need steps from {1,2,,1,2} to {1{1,,1,2}2}. They’re both steps for a further target – {1{1,,1,2}2,,1,2}.

{1{1,,1,2}2,,1,2} reduces to {1{1,,b}2,,1,2}. If b=1, then it’s {1,2,,1,2}; if not, then we add a layer of { ____ ,,b-1}, and it becomes {1{1{1,,b}2,,1,2}2,,b-1}, and the beginning step of it is {1{1,,1,2}2,,b-1}.

So, to get {1{1,,1,2}2,,1,2}, we need

{1,,1,2}, as the beginning of {1,2,,1,2}

{1{1,,1,2}2} (has the same recursion level as {1,,1,2}), as the beginning of {1{1,,2}2,,1,2}

{1{1,,1,2}2,,2} (has the same recursion level as {1,,1,2}), as the beginning of {1{1,,3}2,,1,2}

{1{1,,1,2}2,,3} (has the same recursion level as {1,,1,2}), as the beginning of {1{1,,4}2,,1,2}

etc.

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So {1{1,,1,2}2,,#} is {_,,#} over {_,,1,2}? Like 1-separators over {1,,1,2}.

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What separator {1,,1{1,,1,,2}3} expands to?

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{1,,1{1,,1…{1,,1{1,,1,2{1,,1,,2}2}2{1,,1,,2}2}…2{1,,1,,2}2}2{1,,1,,2}2}

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I think {1,,1,,3} will have the recursion level C(W_2*2+W).

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Any chance that you could continue your comparisons with FGH and Taranovsky’s notation? I would say that you don’t need so much quantity of comparisons, but perhaps more intuition and more general rules of how things compare, even if those rules are not exact.

Is the limit of this notation the same as linear array notation for the R function? (which you claimed had a limit of psi_0(psi_{I(omega,0)}(0)) )

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What is the limit recursion level of pDAN?

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