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Here's a new function I thought up a few days ago. I believe that it reaches so far up to epsilon_w. Here's something I've been playing around with for a while.

Up to f_w(n)

Let's introduce the question mark: "?". The "?" is essentially an extension to the hyperoperators. Let's start small: a[?]b=a^b. You'll notice that the ? is inside a pair of square brackets, and you'll soon see why. a[?][?]b=a[?]a[?]a[?]...[?]a with b a's. In general, a[?][?][?]...[?][?]b=a[?][?][?]...[?][?]a[?][?][?]...[?]a[?][?][?]...[?]a...a[?][?][?]...[?]a with b a's. That is, a[?][?]b=a^^b,a[?][?][?]=a^^^b, and in general: a[?][?][?]...[?]b with n [?]'s=a^nb. Nothing new so far. What happens however, if we put the question marks together, inside the brackets?

Up to f_w^w(n)

Clearly, what we need is a generalization of what is above. Therefore, let a[??]b=a[?][?][?]...[?]b with b [?]'s. Before we advance to [???] and such, let us consider what would happen with a[??][?]b. We have no rule for this, however we could use our rules above to make a[??][?]b=a[??]a[??]a[??]a...[??]a with b a's. It's easy to see what would happen with a[??][?][?]b, but what if we have a[??][??]b ? What we'd do in that case is first decompose the rightmost expression, and then "solve" the entry.
Therefore, a[??][??]b=a[??][?][?][?]...[?b with b [?]'s. From this we can quite easily reach a[???]b, a[????][??][???]b, or a[??...??]b with any amount of "?"'s. Can we extend this? Yes, with ease! Before I plunge straight into epsilon_0 territory however, I'll first give a formal definition up to here.

Formal definition so far

An entry is 1 or more consecutive question marks "?" encased in square brackets [].
A valid expression consists of 2 numbers with one or more entries between them.
@ means any amount of entries (including 0).
@@ means 1 or more entries.
# means any amount of question marks inside an entry.

  1. a[?]b=a^b
  2. a@@1=a
  3. a@@[?]b+1=a@@a@@[?]b
  4. a@[#?]b=a@[#][#][#]...[#]b with b [#]'s

Informal definition so far

An entry is 1 or more consecutive question marks "?" encased in square brackets [].
A valid expression consists of 2 numbers with one or more entries between them.
@@ means 1 or more entries.
In the expression a@@b, a is the base and b is the prime.

  1. If there is one entry in the expression and it is a single question mark encased in square brackets, then the expression is equal to the base^prime.
  2. If the last entry in the expression is a single question mark encased in square brackets, remove it, nest the expression repeatedly inside the prime (the number of nestings is equal to the prime) and change the prime to the base.
  3. Else, remove the last question mark from the last entry and copy that entry repeatedly to the right of the current last entry. The amount of copies is equal to the prime.

Now, onwards!

Up to f_e_0(n)

As I showed above, we can shorten repeated question marks into 2 consecutive question marks. How about shortening consecutive question marks? Well, firstly, we can shorten a[????]b=a[?(4)]b, a[?????]b=a[?(5)]b, and so forth. Essentially, [?(n)] means [???...??] with n ?'s. What would happen if we let the (n) become a question mark itself? a[?(?)]b=a[?(b)]b. The trivial extensions are obvious: a[?(?)][?(?)][?]b=a[?(?)][?(?)]a[?(?)][?(?)]a[?(?)][?(?)]a...a[?(?)][?(?)]a with b a's, a[?(?)][??]b=a[?(?)][?][?][?]...[?]b with b a's, and so forth. What would a[?(?)?]b be though? Well, a[?(?)?]b=a[?(?)][?(?)][?(?)]...[?(?)]b with b [?(?)]'s. Since entries are solved right to left, then a[?(?)?]b is far greater than a[??(?)]b, which merely means a[???...??]b with b+1 ?'s.

If a[?(?)?]b=a[?(?)?]b=a[?(?)][?(?)][?(?)]...[?(?)]b with b [?(?)]'s, then a[?(?)??]b=a[?(?)?][?(?)?][?(?)?]...[?(?)?]b with b [?(?)?]'s, and so forth for any number of question marks. a[?(?)?(?)]b=a[?(?)???...??]b with b ?'s after the (?). It's quite easy to see how to extend this to any amount of ?(?)'s. What would be above all those? a[?(??)]b=a[?(?)?(?)?(?)...?(?)]b with b ?(?)'s. From here it's simple to extend ?(??...??) with any amount of questions marks: a[?(??...??)]b with n+1 ?'s inside the brackets=a[?(??...?)?(??...?)?(??...?)...?(??...?)]b with b pairs of brackets and n ?'s in each pair of brackets. What would a[?(?(?))]b be? That would be a[?(??...?)]b with b ?'s inside the brackets. It's easy to see how to continue to more nested brackets.

But before I move to the rest of the rules I'll sum it up to here. It's very easy to say, "you just extrapolate from here and you reach epislon_0", but definitions are ultimately what matters.

Formal definition so far

An entry consists of 1 or more consecutive question marks in or out of parentheses encased in square brackets. If they are encased in parentheses then there must be at least one question mark before them.
& means any amount of question marks on the same level of nesting (that is, those question marks are encased in the same amount of parentheses as a whole).
A valid expression consists of 2 numbers with one or more entries between them.
@ means any amount of entries (including 0).
@@ means 1 or more entries.
# means the rest of the entry.
} means any amount of right parentheses ")". (This is to make the parentheses balanced).

  1. a[?]b=a^b
  2. a@@1=a
  3. a@@[?]b+1=a@@a@@[?]b
  4. a@[#?]b=a@[#][#][#]...[#]b with b [#]'s
  5. a@[#(&?)}]b=a@[#(&)?(&)?(&)?...?(&)}]b with b (&)'s
  6. a@[#?(?)}]b=a@[#???...?}]a with b ?'s

Informal definition so far

An entry consists of 1 or more consecutive question marks in or out of parentheses encased in square brackets. If they are encased in parentheses then there must be at least one question mark before them.
A valid expression consists of 2 numbers with one or more entries between them.
@@ means 1 or more entries.
In the expression a@@b, a is the base and b is the prime.

  1. If there is only one entry and it is a single question mark in square brackets, the expression is equal to base^prime.
  2. If the last entry in the expression is a single question mark encased in square brackets, remove it, nest the expression repeatedly inside the prime (the number of nestings is equal to the prime) and change the prime to the base.
  3. If the last question mark in the last entry is not encased in parentheses, remove it and copy the entry repeatedly after the last current entry. The amount of copies is equal to the prime.
  4. If the last question mark in the last entry is encased in parentheses, and there is more than one question mark inside the parentheses, remove the question mark and copy all of the question marks all of the question marks inside those two parentheses one level of nesting below (outsde those question marks) repeatedly. The amount of copies is equal to the prime.
  5. If the last question mark in the last entry is encased in parentheses, and there is only one question mark inside the parentheses, create repeated consecutive question marks. The amount of question marks is equal to the prime.

Up to f_e_w(n)

Reaching e_w from this is fairly straightforward. Let me introduce multiple parentheses. a[?((?))]b would solve to a[?(?(?(...?(?)...)))] b with b pairs of parentheses. In general, a[?(((...((?))...))]b with n parentheses would solve to a[?(((...((?(((...((? (((... ((?...(?)...)))...)))]b with b question marks and n parentheses between each question mark.

From now I will write a superscript next to each parenthesis to symbolize how many parentheses there are. That is, a[?(((...((?))...)))]b with n parentheses would be now a[?n(?)]b. Much cleaner.

Now, we have defined a[?n(?)]b for all a,n and b. What would a[?n(??)]b solve to? Well, in that case we'd use rule 5 (or rule 4 in the informal definition), so that would turn into a[?n(?)?n(?)?n(?)?...?n(?)?]b with b ?n(?)?'s.

So it turns out that we need to add only one rule to reach f_e_w(n). Yay!

Formal definition up to now

An entry consists of 1 or more consecutive question marks in or out of parentheses encased in square brackets. If they are encased in parentheses then there must be at least one question mark before them.
& means any amount of question marks on the same level of nesting (that is, those question marks are encased in the same amount of parentheses as a whole).
A valid expression consists of 2 numbers with one or more entries between them.
@ means any amount of entries (including 0).
@@ means 1 or more entries.
# means the rest of the entry.
} means any amount of right parentheses ")". (This is to make the parentheses balanced).
A superscript means that many consecutive parentheses.

  1. a[?]b=a^b
  2. a@@1=a
  3. a@@[?]b+1=a@@a@@[?]b
  4. a@[#?]b=a@[#][#][#]...[#]b with b [#]'s
  5. a@[#n(&?)}]b=a@[#n(&)?n(&)?n(&)?...?n(&)}]b with b n(&)'s
  6. a@[#?(?)}]b=a@[#???...?}]a with b ?'s
  7. a@[#?n+1(?)}]b=a@[#?n(?n(?n(?...n(?)))...))}] with b question marks and n right parentheses.

Informal definition so far

Coming soon!

Up to f_zeta_0(n)

The next step is to (quite obviously) index the superscript with a series of question marks. a[?[?](?)]b=a[?b(?)]b. What would a[?[?](?[?](?))]b be equal to? The weak option would be a[?b(?b(?))]b. A far stronger option would be a[?[?](?[?](?))]b=a[?[?](?b(?))]b. This way, we recurse over the question marks a lot of times (that would be equal to a[?[?](?b-1(?b-1(?...?b-1(?)))...))]b and so forth) which would eventually (after tons of recursion) reduce to a[?[?](?)]n=a[?n(?)]n for a ridiculously large n.

So we've defined it for a[?[?](?[?](?...?[?](?)))...))]b. Is that it? Not at all! So far we've used only the most basic symbol we have: a single isolated question mark. What if we used two question marks next to each other? a[?[?][?](?)]b=a[?[?](?[?](?...?[?](?)))...))]b with b [?]'s. Similarly, where & represents any amount of questions, a[?&[?](?)]b=a?&(?&(?&(?...?&(?)))...))]b with b &'s. Of course though, this isn't the limit. We can just use the rules from the notation, so a[?[??](?)]b=a[?[?][?][?]...[?](?)]b with b [?]'s, (skipping a considerable amount of steps) a[?[?(?(?))](?)]b=a[?[?(???...??)(?)]b with b ?'s inside the superscripted brackets, and so forth. This brings us up to f_e_e_w(n), not a bad improvement.

We can greatly boost the strength even more by letting the superscripts... have superscripts themselves! For example, a[?[?[??][?](?)](?)]b=a?[?[??](?[??](?[??](?...?([??]?)))...))](?)]b with b [??]'s. We can of course have superscripts for the superscripts for the superscripts, and so on and so forth. I will soon show how this reaches f_zeta_0(n), and more importantly, provide an accurate explanation.

Formal definition so far

Coming soon!

Informal definition so far

Coming soon!

Beyond

Coming?

FGH comparison

Here I will write only the entries, but the meaning is that the base and prime are n. Furthermore, I will only write the corresponding FGH ordinal.

[?]~2
[?][?]~3
[??]~w
[??][?]~w+1
[??][??]~w2
[???]~w^2
[?(m+1)]~w^m
[?(?)]~w^w
[?(?)][?]~w^w+1
[?(?)][?(?)]~(w^w)2
[?(?)?]~w^(w+1)
[?(?)?(m)]~w^(w+m)
[?(?)?(?)]~w^w2
[?(??)]~w^w^2
[?(?(?))]~w^w^w
[?(?(?)?)]~w^w^(w+1)
[?(?(?(...?(?)...)))] (with n question marks)~f_e_0(n)

As a matter of fact, I discovered a way to find the corresponding FGH ordinal for any few entries. Go over each entry. If it has no brackets, remove the last question mark in the entry. Next, change all question marks to omegas. If there is an entry without any question marks, write a 1 inside it. For example:

[?(?)][??][?] would turn into [w(w)][w][1]

Now, to calculate the ordinal: if there is a pair of brackets, exponentiate the outer omega by the inner expression, multiply all consecutive omegas, and add together all consecutive entries. For example:

[?(?)][??][?] would turn into [w(w)][w][1] would turn into [w^w][w][1] would turn into w^w+w+1, which is the actual best approximation for the FGH ordinal of this pair of entries.

Another example: [?(?(??)?)??][???][?][?] would turn into [w(w(ww)w)ww][ww][1][1] would turn into [(w^(w^w*w)*w)*w*w][1][1] would turn into (w^(w^w*w)*w)*w*w+2=w^(w^3+2)+2

This is because each extension repeatedly iterates the previous one. Placing entries next to each other merely iterates them, so it's like adding in the FGH. Placing question marks next to each other repeatedly iterates those iterations, so it is like multiplication. Placing question marks in brackets repeatedly iterates those repeated iterations of iterations, so it is like exponentiation. Similarly, my next extension is like tetrating omegas, as it is repeated iterations of that. A real advancement would be if I could find a way to diagonalize over (theoretically) applying the hyperoperators to omegas. From there the way to Gamma_0 is easy. Why is that?

Notice, that even though placing question marks next to each other is like multiplication, the FGH limit ordinal of that extension isn't w*w=w^2, it's w^w. Why is that? It's because I am repeating them, so it reaches w*w*w*w...=w^w. Similarly, my second extension, although it is equivalent to exponentiation, it reaches e_0=w^w^w... This is because each time I am iterating that operation. Similarly, even though it is similar to ordinal tetration, my next extension reaches w^^w^^w^^w...=zeta_0 (Don't worry, my claims of strength aren't built on that fact. I've done an analysis already). Therefore, if I reach phi(w,0), it should be very easy to iterate over that, as in phi(phi(phi(...),0),0)=phi(1,0,0)=Gamma_0.

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