Alright, so this is my extended arrow notation!
a,b,c are all natural numbers, A[num] differentiates a part of an array, [] is a part of 1's and @ means anything.
@, A[num] and [] may or may not exist.
a@1=a
a@[A1,[]]b=a[A1]b
a[1]b=a^b
a@[1]b=a@a@[1](b-1)
a@[c,A1]b=a@[c-1,A1][c-1,A1]...b[c-1, A1]'s...[c-1,A1]a
a@[[],1,c,A1]b=a@[[],a@[[],1,c,A1](b-1),c-1,A1]a
Now, let's implement it!
3[1]3=3^3=27
3[1][1]3=3[1]3[1][1]2
=3[1]3[1]3[1][1]1
=3[1]3[1]3
=3[1]27
=7,625,597,484,987
3[2]3=3[1][1][1]3
3[1,2]3=3[3[1,2]2]3
=3[3[3[1,2]1]3]3
=3[3[3]3]3
3[2,2]3=3[1,2][1,2][1,2]3
=3[1,2][1,2][3[1,2][1,2][1,2]2]3
=3[1,2][1,2][3[1,2][1,2][3[1,2][1,2][1,2]1]3]3
=3[1,2][1,2][3[1,2][1,2][3]3]3
=3[1,2][1,2][3]3=3[1,2][1,2][2][2][2]3...
This notation has a pretty fast growth rate. But by...um...pretty fast...it goes to dimensional arrays in BEAF, to ω^^3!
That's...um...an extremely good benchmark for a linear notation based on Knuth up arrows. Here, is when things start to grow. I'm gonna expand 3[2,2]3 in Arial font, size 12.
3[2,2]3=3[1,2][1,2][1,2]3=3[1,2][1,2][3[1,2][1,2][1,2]2,1]3=3[1,2][1,2][3[1,2][1,2][1,2]2]3=3[1,2][1,2][3[1,2][1,2][3[1,2][1,2][1,2]1,1]3]3=3[1,2][1,2][3[1,2][1,2][3[1,2][1,2][1,2]1]3]3=3[1,2][1,2][3[1,2][1,2][3]3]3=3[1,2][1,2][3[1,2][1,2][2][2][2]3]3=3[1,2][1,2][3[1,2][1,2][2][2][1][1][1]3]3=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1][1][1]2]3=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1][1][1]1]3=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1][1]3]3=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2][1]2]3=...2 steps...=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2][1]3]3=...3 steps...
=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2]3[1,2][1,2][2][2]3]3=...10 steps...=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2]3[1,2][1,2][2][1][1]3[1,2][1,2][2][1]3[1,2][1,2][2]3[1,2][1,2][2]3]3=...10 steps...=3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2]3[1,2][1,2][2][1][1]3[1,2][1,2][2][1]3[1,2][1,2][2]3[1,2][1,2][1][1]3[1,2][1,2][1]3[1,2][1,2]3[1,2][1,2]3]3
Scary, right? And that's only the first [1,2] block.
3[1,2][1,2][3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2]3[1,2][1,2][2][1][1]3[1,2][1,2][2][1]3[1,2][1,2][2]3[1,2][1,2][1][1]3[1,2][1,2][1]3[1,2][1,2]3[1,2][1,2]3]3 is, infact 3[2,2]3.
And guess what?
Once we evaluate the 3[1,2][1,2][3]3, or
3[1,2][1,2][2][2][1][1]3[1,2][1,2][2][2][1]3[1,2][1,2][2][2]3[1,2][1,2][2][1][1]3[1,2][1,2][2][1]3[1,2][1,2][2]3[1,2][1,2][1][1]3[1,2][1,2][1]3[1,2][1,2]3[1,2][1,2]3
, we repeat this process, except, instead of 3 at the end, you have 3[1,2][1,2][3]3.
Holy shit! That will expand into a string 3[1,2][1,2][3]3 cm long assuming each character is 1cm!
And best of all... once we resolve this nightmare of a equation, we have this:
3[1,2][1,2]X where X is some sort of stupidly large number
And, we do Rule 6, embedding 3[1,2][3]3 extremely deeply.
A resolution to 3[1,2]N from 3[2,2]3, is already nigh impossible! And 3[1,2]N would also be impossible to resolve...
An insight into a higher order notation.
Now, we're done with linear. Of course, 3[3,3,3,3]3 is stupidly large (in fact as large as dimentri), but this notation is weak compared to bigger ones like Extensible-E or BEAF. This is my official debut notation though, so I'm going to have this one at least beat ε0 in the fast growing hierarchy (any less is a mistake)
Let's first have 3[1\]3. This equates to 3[3,3,3]3, like a[a,a,a,a...b a's...a,a,a,a]a from a[1/]b.
Now, we have 3[2\]3. This is equal to 3[1\][1\][1\]3. Expand by following rule 5.
3[3\]3? Rule 5. Now, here's a snag- how do we add more power to 3[n\]3?
Simple... 3[1,2\]3. We'd get 3[3[3\]3\]3, by applying rule 6. So far so good, only 1 rule added (ref. 3 lines up).
3[1,3\]3 creates 3[3[3,2\]3,2\]3, pretty good.
Now, we hit another one. We can keep adding arguments after argument, but we want diagonalization.
Again, simple! 3[1\,\]3. Refer to 6 lines up.
Now we can have 3[1\,\,\]3, 3[1\,\,\,\,\,\,\]3... then, 3[<1>,\]3. This expands into a[1\,\,\...,\,\,\]a w/ b \'s with a[<1>\]b. Now we can have 3[2<1>\]3. Oh wait, that's the same idea, do 3[\<1>\]3,umm...
3[<1><1>\]3 is the diagonalization, again! Now we can have 3[<1><1><1><1>\]3...
But wait, there is a one in the middle of <>! Let's diagonalize that with...
3[<2>\]3! (that's an exclamation mark)
We can do 3[<1><2>\]3, aka 3[1\,\,\<2>\]3, wait, we can do 3[<2><2>\]3, wait wait...
Oh my gosh! It seems like there's no limit! And once we reach...
3[<100>\]3
the numbers can be looped back, by...
3[<1,2>\]3.
Now, there's 3[<100,2>\]3, wait, that can be <1,3>, and then...
We find, 3[<1\>\]3. This expression is cursed af. And guess what, we can have loops within itself...
3[<2\>\]3, 3[<1,2\>\]3, 3[<1\,\>\]3, then do that thing again, 3[<<1>\>\]3, loop that, 3[<<1><1>\>\]3,
3[<<2>\>\]3, 3[<<1,2>\>\]3, 3[<<1\>\>\]3...
At this point, it is time to introduce Billy:
3[1\\]3.
This is a extremely large number, equating to 3[<<\>\>\]3 (it's based on \, not separator <>).
And we can do that all again, so 3[1\\\]3 would be there, as 3[<<1\\>\\>\\]3, with the top 1\\ resolved like 3[1\\]3.
The limit of this... is a[1\2]b. Yes, that thing grows faster than a[1\\\\\\\]b.
An analysis of EAN in the Fast-Growing Hierarchy
The highest well defined part of this notation is n[1\\]b, but let's first decipher the earlier parts.
n[1]b is exponential, n[1][1]b is a PR (primitive recursion) over that, aka tetration...
n[2]b diagonalizes over all that so it's w. And n[2][1]b does PR over that...
n[3]b is diagonalization over [2]. So, it is w^2. And so on...
n[1,2]b is diagonalization over n[k]b so it is w^w.
Notice when you connect two [] boxes, the ordinal numbers they stand for are added together.
So, n[2,2]b is actually w^w times w (due to replication), w^w+1.
n[1,3]b would be w^w2. n[1,1,2]b would be w^w^2.
n[1,1,1,2]b would be w^w^3
so n[1\]b is w^w^w, n[2\]b is w^(w^w)+1, n[1,2\]b is w^(w^w)+w, n[1,1,2\]b is w^(w^w)+w^2.
You can tell that the ordinals are multiplied together. So,
n[1\,\]b is w^(w^w)2. You can guess it, n[<1>\]b is w^w^w+1. And, we have the same structure times two for n[<1><1>\]b, thus it is w^(w^w+1)2, causing n[<1,2>\]b being w^w^w2.
n[<1,3>\]b is w^w^w3, n[<1,4>\]b is w^w^w4.
n[<1,1,2>\]b is w^w^w^2! And as these are additive, n[<1,2,2>\]b is w^w^(w^2+w). Ditto for n[<1,1,3>\]b.
And the legendary w^w^w^w is at n[<\>\]b (this is as valid as n[<1\>\]b). Of course, we have...
n[<2\>\]b is w^w^(w^w+1). n[<1,2\>\]b is w^w^(w^w+w), and of course... n[<\,\>\]b is w^w^(w^w)2.
See a pattern? Yep, n[<<1>\>\]b exists... it is w^w^w^w+1,
just like n[<1>\]b is w^w^w+1! (take it either way)
There's a pattern...
n[<<<......<<<\>\>......>\>\]b...
n[1\\]b! That's [drum roll] ε0! We've reached a benchmark for a decent notation! Of course, this is only gonna continue with...
psi(3,0), and for n[1\2]b... um, it's the Ackermann Ordinal. This is actually extremely good for a notation with no climbing method!
last edited 4/10/2023