User blog:Googology Noob/Question Mark Notation

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[?][?][?]...[?]a 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[??][?][?][?]...[?]a 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.
 * 1) 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@[#][#][#]...[#]a 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, copy that entry repeatedly to the right of the current last entry (where the amount of copies is equal to the prime) and change the prime to the base. 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)]a. The trivial extensions are obvious: a[?(?)][?(?)][?]b=a[?(?)][?(?)]a[?(?)][?(?)]a[?(?)][?(?)]a...a[?(?)][?(?)]a with b a's, a[?(?)][??]b=a[?(?)][?][?][?]...[?]a with b a's, and so forth. What would a[?(?)?]b be though? Well, a[?(?)?]b=a[?(?)][?(?)][?(?)]...[?(?)]a with b [?(?)]'s. Since entries are solved right to left, then a[?(?)?]b is far greater than a[??(?)]b, which merely means a[???...??]a with b+1 ?'s.

If a[?(?)?]b=a[?(?)?]b=a[?(?)][?(?)][?(?)]...[?(?)]a with b [?(?)]'s, then a[?(?)??]b=a[?(?)?][?(?)?][?(?)?]...[?(?)?]a with b [?(?)?]'s, and so forth for any number of question marks. a[?(?)?(?)]b=a[?(?)???...??]a 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[?(?)?(?)?(?)...?(?)]a 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[?(??...?)?(??...?)?(??...?)...?(??...?)]a with b pairs of brackets and n ?'s in each pair of brackets. What would a[?(?(?))]b be? That would be a[?(??...?)]a 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 any amount of right parentheses ")". (This is to make the parentheses balanced).
 * 1) means the rest of the entry.


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

Informal definition so far
Coming soon!

Up to f_e_w(n)
Coming soon!

Beyond
Coming?

FGH comparison
Coming soon!