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Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 2:21 pm UTC
by Vytron
Prove they're different (I challenge your challenge.)

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 3:11 pm UTC
by mike-l
Nope. I just wanted to see if anyone was actually checking. I got my answer.

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 3:19 pm UTC
by Vytron
Okay, added to the rules:

4c All claims are true unless challenged

So I can claim:

2+2=5

And it'd be true by the rules of the thread.

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 3:25 pm UTC
by mike-l
Spectacular.

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 3:37 pm UTC
by Vytron
And what are you doing here if you don't wish to play?

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 4:03 pm UTC
by Daggoth
Don't shoo him off. we need more people in here not less

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 4:06 pm UTC
by Daggoth
In fact i challenge you to prove him wrong

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 4:10 pm UTC
by Vytron
We need more people playing (which includes sending numbers and/or showing why other people's notations are wrong, not just empty claiming other people's notations are wrong. mike-l's "Spectacular" is just trolling.)

In fact i challenge you to prove him wrong


I proved him wrong in this post - both notations produce the same values at the same points, therefore, both are equivalent and if one reaches epsilon_0 at one point so does the other one.

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 5:05 pm UTC
by mike-l
I don't think you know what trolling is. But then again, you just redefined "true", so I guess you can call anything anything you want.

You haven't even remotely shown that the two notations are the same. The linked post has different rules than the previous post that was confirmed to reach epsilon_0. Saying they are the same doesn't make them the same.

But you clearly don't want me here, so I'll let you continue to live in your delusional world where "all claims are true unless challenged" and you've driven off everyone who actually challenges anything you say. Here's a hint, putting those two together simply becomes "all claims are true". Have fun with that.

Re: Your number is, in fact, not bigger!

Posted: Wed Feb 18, 2015 5:18 pm UTC
by Vytron
I want you here playing mike-l, actually showing what you claim, not just claiming it.

Suppose I go and tell WarDaft:

What Vytron would say wrote:WarDaft, I'm pretty sure 2|(0[;](0[;]0)) doesn't get anywhere near f{ψ(Ω^Ω^(ψ(Ω^Ω^ω)))}(2) as you claim. I put it around w^w^w.

<End post>


See how's that trolling? What makes it different from what you're doing?

Challenges need the "How", not the "What", you fail at this game when you leave your claim as an exercise to the reader.

But you will always be welcome to play.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 3:30 am UTC
by Daggoth
if you really want him here why do you
a) call him names
b) resort to inventing faulty logic mechanisms to disregard his claims.

All claims are true unless challenged?
Then i am the eternal undisputed champion of this and any future number comparing games you come up with

feel free to challenge my claim. I will simply challenge all your challenges so it remains true.
according to your use of the rules this is completely valid gameplay

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 11:55 am UTC
by Vytron
Daggoth wrote:a) call him names


I haven't called him a troll, just said he had trolled.

Daggoth wrote:b) resort to inventing faulty logic mechanisms to disregard his claims.


Prove it's faulty logic. The thing here is he hasn't backed up his claims. A simple example of how his rules and my rules behave differently would suffice, but apparently he can't find such an example.

Daggoth wrote:All claims are true unless challenged?
Then i am the eternal undisputed champion of this and any future number comparing games you come up with


I challenge that: Your number is, in fact, not bigger than mine.

Daggoth wrote:feel free to challenge my claim. I will simply challenge all your challenges so it remains true.
according to your use of the rules this is completely valid gameplay


You're meant to backup your claims, doing otherwise would be trolling.

Show how you are the eternal undisputed champion of this and any future number comparing games I come up with or to challenge my challenge.

The "how" on mike-l's is missing too.

The "how" in my challenge isn't missing, you claim to reach some ordinal, I claim to surpass it, for your claim to hold you have to show how my notation is wrong or claim you reach a higher ordinal. You seem to assume mike-l must be right, but don't show any comparison to support his argument either.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 3:47 pm UTC
by Daggoth
I challenge that: Your number is, in fact, not bigger than mine.

Prove it
Prove it's faulty logic.

Prove that it isn't

You're meant to backup your claims, doing otherwise would be trolling.


No i am not. According to your faulty logic rules:
All claims are true unless challenged

Therefore the only requisite for a claim to be true is that it be unchallenged.
Since your challenge has been challenged, it is invalid, thus my claim remains valid and i am Still the champion.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 4:01 pm UTC
by Vytron
Daggoth wrote:
I challenge that: Your number is, in fact, not bigger than mine.

Prove it


Your number goes up to φ(102,100)100, mine goes up to f{ψ(Ω^Ω^(ψ(Ω^{{Ω^ω}2})))}(5). Mine reaches a higher ordinal and it follows that is should be higher.

Daggoth wrote:
Prove it's faulty logic.

Prove that it isn't


It's not faulty logic because if two series of sets reach the same value every time they should be equivalent.

You're meant to backup your claims, doing otherwise would be trolling.


No i am not. According to your faulty logic rules:
All claims are true unless challenged


But that doesn't apply to challenges.

Therefore the only requisite for a claim to be true is that it be unchallenged.


A claim is not a challenge.

Since your challenge has been challenged, it is invalid, thus my claim remains valid and i am Still the champion.


Only if you actually provide an argument that what I'm saying in this post is wrong. Saying "prove it" is not an argument.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 7:55 pm UTC
by Daggoth
I challenge everything on your previous post , i argument that all of it is either a lie or a wrong statement.

as this invalidates everything you just said until you prove it, i remain the champion

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 8:38 pm UTC
by Vytron
Daggoth wrote: i argument that all of it is either a lie or a wrong statement.


That's not an argument, it's a claim.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 8:49 pm UTC
by mike-l
Alright, this was the notation that I said got to e_0:

New reduction rule: always reduce the rightmost element unless it is immediately preceded by a smaller element. If the rightmost is 0, remove it and iterate the function. Otherwise if it is nonzero and being reduced, change m to some number of m-2s followed by m-1.some number of 0's followed by 1, 2,... m-1. If it is preceded by a lower element, reduce the preceding string with the same rules, except instead of iterating (if needed),replace the final element with a string of [0,0,...,0,n] where there are as many 0s as there are at the end of the reduced preceding string.

----
So I need to modify the rules slightly, as implied by my last post. When reducing something that is preceded by something smaller, look at the preceding string up until the next entry that is as large as the ultimate element. Reduce that string, call it X. Now replace the above preceding string and the ultimate element (call this Y) with a string of X,Y,X,Y...X,Y. (This is actually only necessary if the preceding string ended in 0, but it makes little difference)
----
The general idea of what's happening is that if an index represents an ordinal b, and if a is an ordinal less than b (thus having a representation with indices less than b's) then, by abuse of notation, [a,b] reduces a and multiplies by w, thus grows like b + w^a. Once a is large enough (this was my error), this is w^a.

Conversely, the old setup I gave b*a

Your problem is that you aren't caring about the preceding string other than the first element, while mine uses the full reduction rules on it.

Let's look at n,3 and vs my n[3]. They look the same when you start reducing
n,3 = n,0,0,...,0,1,2
n[3] = n[0,0,...,0,1,2]

but they diverge immediately thereafter. In your reduction, you say 'This ends in 2, now go all the way left until hitting another number at least as big as 2 (in this case all the way) and reduce that so you get
n,0,...,0,1,2,0,....,1,2, ~~~ 0,...,0,1,2.

In mine, however, you take the preceding string and start the rules over on that, instead of just looking at the starting element. So I'm looking at the 0,1 right before the 2, which reduces to a bunch of 1s, then I take a bunch of everything. So I get
n[0,0,...,1,1,....1,2, 0,0,...,1,1,....,1,2, ~~~, 0,0,...,1,1,...,1,2]
See how mine duplicates the 1s as well. At no point in yours will you ever get more than a single 1 preceding a 2, which is vital to get the proper growth.

Your growth looks like:
n,0 ~ 2
n,0,0 ~ 3
n,0,0,...0 with m 0's ~ m+1
n,1 ~ w
n,1,0 ~ w+1
n,1,0,...0 with m 0s ~ w+m
n,1,1 ~ w2
n,0,1 ~ w^2
n,0,0,1 ~w^3
n,2 ~ w^w <--- so far pretty much the same
n,0,2 ~ w^(w+1)
n,0,0,2 ~ w^(w+2)
n,1,2 ~ w^(w2 + 1)
n,0,1,2 ~ w^(w2+2)
n,0,0,1,2 ~ w^(w2+3)
n,3 ~ w^(w3) <--- way lower

Also, n,1,1,2 is actually bigger than n,3, as this is w^(w3+1) (it reduces to n,0,0,...0,1,2,0,0,...,0,1,2 ~~~ 0,0,...,0,1,2).


On a meta level, the reason I'm hesitant to post this at all is because you are being incredibly hypocritical. You say "don't just say I'm wrong, tell me why", but you haven't made any argument to why your values are what you claim them to be. You assert that this notation is the same as another, but you don't give any reason why, and clearly you haven't done any examples to check it, because if you had compared your example 1 in your notation to the same thing in the other, you would have seen the difference. And then you make possibly the most ridiculous rule ever, "All claims are true unless challenged", especially ridiculous when one of the few people who was actually looking at your notation had just said they weren't checking it anymore.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 8:51 pm UTC
by mike-l
Vytron wrote:
Daggoth wrote: i argument that all of it is either a lie or a wrong statement.


That's not an argument, it's a claim.

So it's true by your rule :P.

Re: Your number is, in fact, not bigger!

Posted: Thu Feb 19, 2015 9:34 pm UTC
by Vytron
Oh, I see, thanks mike, that does take down my number. I'm taking down rule 4c now.

I apologize for the hypocrisy, but on a side-by-side comparison I'd just have shown a wrong expansion for [0,0,...,0,1,2] so it wouldn't have helped much.

Re: Your number is, in fact, not bigger!

Posted: Fri Feb 20, 2015 11:50 pm UTC
by Daggoth
Resizing accounting for my mistake also i'm eliminating the comma as the base separator. (LOSING alot of power to gain consistency and confidence)

Rule 2 has been modified, so have references to rule 2
GFDQM The Final cut
Spoiler:
Red parts are the ones that get deleted or reduced

Rule 1: Base case: ¿0?n = n↑nn | ¿X?n = ¿X-1?nn (where ¿X-n?3=¿X-n?¿X-n?¿X-n?)

Rule 2: Base separator ¡0!
2.1: ¿...x¡0!y¡0!z?n where all arguments = 0 Reduces to ¿...x¡0!y?nn where all arguments = n
2.2 ¿...x¡0!0¡0!0..¡0!0¡0!0?n where X isnt 0. Reduces to ¿...x-1¡0!n¡0!n...¡0!n¡0!n?nn
2.3 ¿...x¡0!y¡0!z?n where Z isnt 0 reduces to ¿...x¡0!y¡0!z-1?nn

Rule 3: A single Nth separator,
3.1 ¿0¡X!0?n = ¿Y¡X-1!Y....Y¡X-1!Y?nn, (with n ¡X-1! separators) Where Y = Z¡X-2!Z...¡X-2!Z? (with n ¡X-2! separators)continue untill X=0
3.2 ¿Q¡X!0?n = ¿Q-1¡X!Y....Y¡X-1!Y?nn
3.3 ¿Q¡X!P?n = ¿Q¡X!P-1?nn

Rule 4: Chains.
4.1: The number most to the right isnt 0
¿#¡*!...Q¡X!P?n = ¿#¡*!...Q¡X!P-1?nn
4.2: All instances of the same separator, all 0's
¿0¡Q!0....0¡Q!0?n = ¿X...¡Q!X?nn where X = Y¡Q-1!Y....Y¡X-1!Y, (with n ¡Q-1! separators), Where Y = Z¡Q-2!Z...¡Q-2!Z? (with n ¡Q-2! separators)continue untill Q=0 the final value in this sequence is n
4.3: All instances of the same separator, the rightmost number is 0 but some or all of the others aren't.
From right to left, P is the first number that isnt 0. ¿0¡Q!...P...¡Q!0¡Q!0...?n = ¿0¡Q!...P-1...¡Q!X¡Q!X?nn Where X = Y¡Q-1!Y...Y¡Q-1!Y with n ¡Q-1! separators...Where Y = Z¡Q-2!Z...¡Q-2!Z? (with n ¡Q-2! separators)
4.4: Mixed Separators.
Locate the lowest ranking of separator. Call this ¡L! Its surrounding values are P and Q Adjacent separators of the same rank are called a group.
4.4.1: If there is only one ¡L! ¿#¡*!#¡*!...P¡L!Q...#¡*!#¡*!?n then
4.4.1.1: ¿#¡*!#¡*!...0¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...X...#¡*!#¡*!?nn Where X = Y¡L-1!Y...Y¡L-1!Y with n ¡L-1! separators where Y... ( same as with rule 3.1 )
IF ¡L! is ¡0! reduce according to rule 2
4.4.1.2: ¿#¡*!#¡*!...Q¡L!0...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q-1¡L!X...#¡*!#¡*!?nn X is the same as in the previous rule.
4.4.1.3: ¿#¡*!#¡*!...Q¡L!P...#¡*!#¡*!?n = ¿#¡*!#¡*!...Q¡L!P-1...#¡*!#¡*!?nn
Adjacent separators of the same rank are called a group.
4.4.2: A single group of ¡L! (the lowest ranking separator).
¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!?n
Reduce this group in the same way as 4.1, 4.2 and 4.3. the rest of the chain is untouched untill this group is reduced to 0
4.4.3: Multiple groups of ¡L!.
¿#¡*!#¡*!...P¡L!Q¡L!R...#¡*!#¡*!S¡L!T¡L!U...?n
Reduce the rightmost group first. even if the rightmost group only contains a single ¡L!

Rule 5: Nesting

All Nested separators rank above all ¡X!
Comparing two nested separators:
¡0...¡0!...0! with n nestings outranks ¡0...¡X!...0! with n-1 nestings
¡0¡R!0! outranks ¡x¡R-1!X!
¡R¡Q!0! outranks ¡R-1¡Q!X!
¡P¡Q!R! outranks ¡P¡Q!R-1!

As with rules 1-4, the lowest ranking group of separators gets reduced first.


5.1 ¿P¡#!X?n = ¿P¡#!X reduced?nn
5.2 ¿Q¡#!X?n = ¿Q reduced¡#!X?nn
5.3 ¿0¡#!0?n = ¿X¡# reduced!X?nn
5.4 if any of these three rules above would reduce 0¡0!0, it reduces to n

Rule 6:
Double Separator. Ranks above all nesteds. Rank according to reductions. If a Separator could eventually reduce to another, it ranks above it..
6.1 ¿0¡0!¡0!0?n = ¿X¡X...X...X!X?nn Where X = ¡Y...Y¡n!Y...Y! where Y = ¡Z...Z¡n-1!Z...Z! so on until ¡n-n=0! reduces to , Use n separators each step of the way
6.2 ¿P¡*!¡*!S?n = ¿P¡Q!¡R!S-1?nn
6.3 ¿P¡*!¡*!0?n = ¿P-1¡Q!¡R!X?nn Where X = Y¡#!Y... with n ¡#! separators where # is a separator immediately inferior in rank, Where Y is Z¡#-1!Z... with n ?¡#-1! Separators where #-1 is immediately inferior to #. Continue until # is the base separator
6.4 ¿0¡%!¡*!0?n = ¿X¡X...¡%!¡@n!...!X!X?nn Where @n =Y¡Y...¡%!¡@n-1!Y...!Y ..so on untill @n-n = Z¡Z...¡%!¡#!...Z!Z Each set of ellipses hides n nestings. X Y Z work as in 6.3
6.5 ¿0¡*!¡0!0?n = ¿X¡X...¡#!¡@n!...!X!X?nn @ Works the same as in 6.4
Triple and quadruple separators
¿0¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!...X!X?nn Here @ Works the same as in 6.4 taking * to be n
¿0¡0!¡0!¡0!¡0!0?n = ¿X¡X...¡@!¡@!¡@!...X!X?nn Here @ Works the same as in 6.4 taking * to be n


Sizing up to ω^^5
Red simply highlights the active part (by active i mean the next one to be reduced

¿0¡0!0?10 > fω2(10)
¿0¡0!9?10 > fω2+9(10)
¿1¡0!0?10 > fω3(10)
¿8¡0!9?10 > fω9+9(10)
¿0¡0!0¡0!0?10 > fω^2(10)
¿5¡0!9¡0!15?10 > f{ω^2}5+ω9+15(10)
¿0¡0!0¡0!0¡0!0?10 > fω^3(10)
¿a¡0!b¡0!c¡0!d¡0!e?f > fω^4a+ω^3b+ω^2c+ωd+e(f)
¿0¡1!0?10 > fω^ω(10)
¿0¡1!0¡1!0?10 > f{ω^ω}*{ω^ω}(10) > fω^ω2(10)
¿0¡2!0?10 > fω^ω^2(10)
¿0¡3!0?10 > fω^ω^3(10)
¿0¡0¡0!0!0?10 > fω^ω^ω(10)
¿0¡0¡0!0!0¡0¡0!0!0?10 > f{ω^ω^ω}^2(10)
¿0¡0¡0!1!0?10 > f{ω^ω^ω}^ω(10) or fω^{ω^ω}*ω or fω^ω^{ω+1}(10)
¿0¡0¡0!2!0?10 > fω^ω^{ω+2}(10)
¿0¡1¡0!0!0?10 > fω^ω^{ω2}(10)
¿0¡2¡0!0!0?10 > fω^ω^{ω3}(10)
¿0¡0¡0!0¡0!0!0?10 > fω^ω^ω^2(10)
¿0¡0¡0!0¡0!1!0?10 > fω^ω^{ω^2+1}(10)
¿0¡0¡0!1¡0!0!0?10 > fω^ω^{ω^2+ω}(10)
¿0¡1¡0!0¡0!0!0?10 > fω^ω^{ω^2}2(10
¿0¡0¡0!0¡0!0¡0!0!0?10 > fω^ω^ω^3(10)
¿0¡1¡0!1¡0!0¡0!0!0?10 > fω^ω^{{ω^3}2+{ω^2}2(10)
¿0¡0¡0!0¡0!0¡0!0¡0!0!0?10 > fω^ω^ω^4(10)
¿0¡0¡1!0!0?10 > fω^ω^ω^ω(10)
¿0¡0¡1!1!0?10 > fω^{ω^ω^ω}*ω(10) or fω^ω^{ω^ω+1}(10)
¿0¡0¡1!0¡0!0!0?10 > fω^ω^{ω^ω+ω}(10)
¿0¡0¡1!0¡0!1!0?10 > fω^ω^{ω^ω+ω+1}(10)
¿0¡0¡1!1¡0!0!0?10 > fω^ω^{ω^ω+ω2}(10)
¿0¡0¡1!0¡0!0¡0!0!0?10 > fω^ω^{ω^ω+ω^2}(10)
¿0¡0¡1!0¡0!0¡0!0¡0!0!0?10 > fω^ω^{ω^ω+ω^3}(10)
¿0¡1¡1!0!0?10 > fω^ω^{ω^ω}2 (10)
¿0¡0¡0!0¡1!0!0?10 > fω^ω^{ω^ω}ω (10) or fω^ω^ω^{ω+1} (10)
¿0¡0¡0!1¡1!0!0?10 > fω^ω^{ω^{ω+1}+{ω^ω}} or fω^ω^{ω^ω}{ω+1} (10)
¿0¡1¡0!0¡1!0!0?10 > fω^ω^{ω^{ω+1}}2 or fω^ω^{ω^ω}{ω2} (10)
¿0¡0¡0!0¡0!0¡1!0!0?10 > fω^ω^ω^{ω+2} or fω^ω^{ω^ω}{ω^2} (10)
¿0¡0¡1!0¡1!0!0?10 > fω^ω^ω^ω2 or fω^ω^{ω^ω}^2 (10)
¿0¡0¡1!0¡1!1!0?10 > fω^ω^{ω^ω2+1}(10)
¿0¡0¡1!0¡1!0¡0!0!0?10 > fω^ω^{ω^ω2+ω}(10)
¿0¡0¡1!0¡1!0¡0!0¡0!0!0?10 > fω^ω^{ω^ω2+ω^2}(10)
¿0¡0¡1!1¡1!0!0?10 > fω^ω^{ω^ω2+ω^ω}(10)
¿0¡0¡1!0¡0!0¡1!0!0?10 > fω^ω^{ω^ω2+ω^{ω+1}}(10)
¿0¡0¡1!0¡0!0¡0!0¡1!0!0?10 > fω^ω^{ω^ω2+ω^{ω+2}}(10)
¿0¡1¡1!0¡1!0!0?10 > fω^ω^{ω^ω2}2(10)
¿0¡0¡0!0¡1!0¡1!0!0?10 > fω^ω^ω^{ω2+1}(10)
¿0¡0¡0!0¡0!0¡1!0¡1!0!0?10 > fω^ω^ω^{ω2+2}(10)
¿0¡0¡1!0¡1!0¡1!0!0?10 > fω^ω^ω^ω3(10)
¿0¡0¡2!0!0?10 > fω^ω^ω^ω^2(10)
¿0¡1¡2!0!0?10 > fω^ω^{ω^ω^2}2(10)
¿0¡0¡0!0¡2!0!0?10 > fω^ω^ω^{ω^2+1}(10)
¿0¡0¡1!0¡2!0!0?10 > fω^ω^ω^{ω^2+ω}(10)
¿0¡0¡1!0¡1!0¡2!0!0?10 > fω^ω^ω^{ω^2+ω2}(10)
¿0¡0¡2!0¡2!0!0?10 > fω^ω^ω^{ω^2}2(10)
¿0¡0¡3!0!0?10 > fω^ω^ω^ω^3(10)
¿0¡0¡0¡0!0!0!0?10 > fω^ω^ω^ω^ω(10)

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 7:01 am UTC
by Vytron
Okay, I want to go back to the +s idea. It was abandoned previously because I had things like 0+0=huge number, and other nonsense, but I have thought of a way to fix it:

n+ = n+n+...n+n for n +s

Values:
0+ = 0
1+ = 1+1 = 2
2+ = 2+2+2 = 6
3+ = 3+3+3+3 = 12

n+ n+ = ((n)+)+

Values:

0+ 0+ = 0
1+ 1+ = (1+1)+ = 2+ = 6
2+ 2+ = (2+2+2)+ = 6+ = 42
3+ 3+ = (3+3+3+3)+ = 12+ = 156

n+ n+ n+ = ((n+ n+)+ (n+ n+)+)+

Values:

0+ 0+ 0+ = 0
1+ 1+ 1+ = ((1+ 1+)+ (1+ 1+)+)+ = ((6)+ (6)+)+ = (1806)+ = 3263442
2+ 2+ 2+ = ((2+ 2+)+ (2+ 2+)+)+ = ((42)+ (42)+)+ = (3263442)+ = 10650056950806
3+ 3+ 3+ = ((3+ 3+)+ (3+ 3+)+)+ = ((156)+ (156)+)+ = (592265232)+ = 350778105628279056

n+ n+ n+ n+ = (((n+ n+ n+)+ (n+ n+ n+)+ (n+ n+ n+)+)+ ((n+ n+ n+)+ (n+ n+ n+)+ (n+ n+ n+)+)+ ((n+ n+ n+)+ (n+ n+ n+)+ (n+ n+ n+)+)+)+

n+ n+ n+ n+ n+ =
Spoiler:
((((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+)+ (((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+)+ (((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+)+ (((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+ ((n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+ (n+ n+ n+ n+)+)+)+)+


And so on.

n+0+ = n+ n+ ... n+ n+ for n spaces.

1+0+ = 1+ 1+ = 6
2+0+ = 2+ 2+ 2+ = 10650056950806
3+0+ = 3+ 3+ 3+ 3+ = (((n+ n+ n+)+ (n+ n+ n+)+ (n+ n+ n+)+)+ ((n+ n+ n+)+ (n+ n+ n+)+ (n+ n+ n+)+)+ ((n+ n+ n+)+ (n+ n+ n+)+ (n+ n+ n+)+)+)+ Where each n is 350778105628279056.

n+0+ = f_ω(n)

n+0+ n+0+ = ((n)+0+)+0+
n+0+ n+0+ n+0+ = ((n+0+ n+0+)+0+ (n+0+ n+0+)+0+)+0+
n+0+ n+0+ n+0+ n+0+ = (((n+0+ n+0+ n+0+)+0+ (n+0+ n+0+ n+0+)+0+ (n+0+ n+0+ n+0+)+0+)+0+ ((n+0+ n+0+ n+0+)+0+ (n+0+ n+0+ n+0+)+0+ (n+0+ n+0+ n+0+)+0+)+0+ ((n+0+ n+0+ n+0+)+0+ (n+0+ n+0+ n+0+)+0+ (n+0+ n+0+ n+0+)+0+)+0+)+0+

And so on. I.e. n+x+ n+x+... follows that progression of nesting.

n+0+0+ = n+0+ n+0+ ... n+0+ n+0+ for n spaces. f_ω+1(n)
n+0+0+0+ = n+0+0+ n+0+0+ ... n+0+0+ n+0+0+ for n spaces. f_ω+2(n)
n+0+0+0+0+ = n+0+0+0+ n+0+0+0+ ... n+0+0+0+ n+0+0+0+ for n spaces. f_ω+3(n)

n+{0}+ = n+0+0+...+0+0+ for n 0s. f_ω2(n)

n+{0}+0+ = n+{0}+ n+{0}+ ... n+{0}+ n+{0}+ for n spaces. f_ω2+1(n)
n+{0}+0+0+ = n+{0}+0+ n+{0}+0+ ... n+{0}+0+ n+{0}+0+ for n spaces. f_ω2+2(n)
n+{0}+0+0+0+ = n+{0}+0+0+ n+{0}+0+0+ ... n+{0}+0+0+ n+{0}+0+0+ for n spaces. f_ω2+3(n)

n+{0}+{0}+ = n+{0}+0+...+0+0+ for n 0s. f_ω3(n)
n+{0}+{0}+{0}+ = n+{0}+{0}+0+...+0+0+ for n 0s. f_ω4(n)
n+{0}+{0}+{0}+{0}+ = n+{0}+{0}+{0}+0+...+0+0+ for n 0s. f_ω5(n)

n+0+{0}+ = n+{0}+...+{0}+ for n 0s. f_ω^2(n)

n+0+{0}+0+ = n+0+{0}+0+ n+0+{0}+0+ ... n+0+{0}+0+ n+{0}+0+ for n spaces. f_ω^2+1(n)
n+0+{0}+0+0+ = n+0+{0}+0+0+ n+{0}+0+0+ ... n+0+{0}+0+0+ n+{0}+0+0+ for n spaces. f_ω^2+2(n)
n+0+{0}+0+0+0+ = n+0+{0}+0+0+ n+0+{0}+0+ ... n+0+{0}+0+0+ n+0+{0}+0+ for n spaces. f_ω^2+3(n)

n+0+{0}+{0}+ = n+0+{0}+0+0+...+0+ = f_ω^2+ω(n)
n+0+{0}+{0}+{0}+ = n+0+{0}+{0}+0+0+...+0+ = f_ω^2+ω2(n)
n+0+{0}+{0}+{0}+{0}+ = n+0+{0}+{0}+{0}+0+0+...+0+ = f_ω^2+ω3(n)

n+0+{0}+0+{0}+ = n+0+{0}+{0}+...+{0}+{0}+ for n+0+00+ 00s = f_{ω^2}2(n)

n+0+{0}+0+{0}+0+ = f_{ω^2}2+1(n)
n+0+{0}+0+{0}+0+0+ = f_{ω^2}2+2(n)
n+0+{0}+0+{0}+0+0+0+ = f_{ω^2}2+3(n)

n+0+{0}+0+{0}+{0}+ = f_{ω^2}2+ω(n)
n+0+{0}+0+{0}+{0}+{0}+ = f_{ω^2}2+ω2(n)
n+0+{0}+0+{0}+{0}+{0}+{0}+ = f_{ω^2}2+ω3(n)

n+0+{0}+0+{0}+0+{0}+ = f_{ω^2}3
n+0+{0}+0+{0}+0+{0}+0+{0}+ = f_{ω^2}4
n+0+{0}+0+{0}+0+{0}+0+{0}+0+{0}+ = f_{ω^2}5

n+0+0+{0}+ = n+0+{0}+...+0+{0}+ for n+0+{0}+ 0+{0}s = f_ω^3

n+0+0+{0}+0+ = f_ω^3+1(n)
n+0+0+{0}+0+0+ = f_ω^3+2(n)
n+0+0+{0}+0+0+0+ = f_ω^3+3(n)

n+0+0+{0}+{0}+ = f_ω^3+ω(n)
n+0+0+{0}+{0}+{0}+ = f_ω^3+ω2(n)
n+0+0+{0}+{0}+{0}+{0}+ = f_ω^3+ω3(n)

n+0+0+{0}+0+{0} = f_ω^3+ω^2
n+0+0+{0}+0+{0}+0+{0} = f_ω^3+{ω^2}2
n+0+0+{0}+0+{0}+0+{0}+0+{0} = f_ω^3+{ω^2}3

n+0+0+{0}+0+0+{0} = f_{ω^3}2
n+0+0+{0}+0+0+{0}+0+0+{0} = f_{ω^3}3
n+0+0+{0}+0+0+{0}+0+0+{0}+0+0+{0} = f_{ω^3}4

n+0+0+0+{0} = f_ω^4
n+0+0+0+0+{0} = f_ω^5
n+0+0+0+0+0+{0} = f_ω^6

n+{0+0}+ = n+0+0+...+0+0+{0} for n+{0}+ 0s = f_ω^ω

n+{0+0}+0+ = f_{ω^ω}+1
n+{0+0}+0+0+ = f_{ω^ω}+2
n+{0+0}+0+0+0+ = f_{ω^ω}+3

n+{0+0}+{0}+ = f_{ω^ω}+ω
n+{0+0}+{0}+{0}+ = f_{ω^ω}+ω2
n+{0+0}+{0}+{0}+{0}+ = f_{ω^ω}+ω3

n+{0+0}+0+{0}+ = f_{ω^ω}+ω^2
n+{0+0}+0+{0}+0+{0}+ = f_{ω^ω}+{ω^2}2
n+{0+0}+0+{0}+0+{0}+0+{0}+ = f_{ω^ω}+{ω^2}3

n+{0+0}+0+0+{0}+ = f_{ω^ω}+ω^3
n+{0+0}+0+0+0+{0}+ = f_{ω^ω}+ω^4
n+{0+0}+0+0+0+0+{0}+ = f_{ω^ω}+ω^5

n+{0+0}+{0+0}+ = f_{ω^ω}2
n+{0+0}+{0+0}+{0+0}+ = f_{ω^ω}3
n+{0+0}+{0+0}+{0+0}+{0+0}+ = f_{ω^ω}4

n+0+{0+0}+ = f_{ω^(ω+1)}
n+0+0+{0+0}+ = f_{ω^(ω+2)}
n+0+0+0+{0+0}+ = f_{ω^(ω+3)}
n+{0}+{0+0}+ = f_{ω^ω2}
n+0+{0}+{0+0}+ = f_{ω^ω^2}
n+0+0+{0}+{0+0}+ = f_{ω^ω^3}

n+{0+0+0}+ = n+0+0+...+0+0+{0}+{0+0} for n+{0+0}+ 0s = f_ω^ω^ω
n+{0+0+0+0}+ = n+0+0+...+0+0+{0}+{0+0}+{0+0} for n+{0+0}+ 0s = f_ω^ω^ω^ω
n+{0+0+0+0+0}+ = n+0+0+...+0+0+{0}+{0+0}+{0+0+0} for n+{0+0}+ 0s = f_ω^ω^ω^ω^ω

n+{{0}}+ = n+{0+0+...+0+0}+ for n+{0}+ 0s = f_ε_0
(n+1 0s would suffice)

n+{{0}}+0+ = f_ε_0+1
n+{{0}}+0+0+ = f_ε_0+2
n+{{0}}+0+0+0+ = f_ε_0+3

n+{{0}}+{0}+ = f_ε_0+ω
n+{{0}}+{0}+{0}+ = f_ε_0+ω2
n+{{0}}+{0}+{0}+{0} = f_ε_0+ω3

n+{{0}}+{0+0} = f_ε_0+ω^ω
n+{{0}}+{0+0}+{0+0}+ = f_ε_0+{ω^ω}2
n+{{0}}+{0+0}+{0+0}+{0+0}+ = f_ε_0+{ω^ω}3

n+{{0}}+{0+0+0}+ = f_ε_0+{ω^ω^ω}
n+{{0}}+{0+0+0+0}+ = f_ε_0+{ω^ω^ω^ω}
n+{{0}}+{0+0+0+0+0}+ = f_ε_0+{ω^ω^ω^ω^ω}

n+{{0}}+{{0}}+ = n+1+00...00+ for n+1+ 0s = f_{ε_0}2
n+{{0}}+{{0}}+{{0}}+ = f_{ε_0}3
n+{{0}}+{{0}}+{{0}}+{{0}}+ = f_{ε_0}4

n+0+{{0}}+ = f_{ε_0}ω
n+0+{{0}}+0+{{0}}+ = f_{ε_0}ω2
n+0+{{0}}+0+{{0}}+0+{{0}}+ = f_{ε_0}ω3
n+0+0+{{0}}+ = f_{ε_0}ω^2
n+0+0+{{0}}+0+0+{{0}}+ = f_{ε_0}*{ω^2}2
n+0+0+0+{{0}}+ = f_{ε_0}*ω^3
n+{0}+{{0}}+ = f_{ε_0}*ω^ω
n+{0+0}+{{0}}+ = f_{ε_0}*ω^ω^ω

n+{{0}+0}+ = f_{ε_0}^2
n+{{0}+0+0}+ = f_{ε_0}^3
n+{{0}+0+0+0}+ = f_{ε_0}^4

n+{{0}+{0}}+ = f_{ε_0}^ω
n+{{0}+{0}+{0}}+ = f_{ε_0}^ω2
n+{{0}+{0}+{0}+{0}}+ = f_{ε_0}^ω3
n+{0+{0}}+ = f_{ε_0}^ω^2
n+{0+{0}+0+{0}}+ = f_{ε_0}^{ω^2}2
n+{0+0+{0}}+ = f_{ε_0}^{ω^3}
n+{0+0+0+{0}}+ = f_{ε_0}^{ω^4}
n+{{0+0}}+ = f_{ε_0}^{ω^ω}
n+{{0+0+0}}+ = f_{ε_0}^{ω^ω^ω}
n+{{0+0+0+0}}+ = f_{ε_0}^{ω^ω^ω^ω}

n+{{{0}}}+ = f_ω^{ε_0+1}
n+{{{{0}}}}+ = f_ω^ω^{ε_0+1}
n+{{{{{0}}}}}+ = f_ω^ω^ω^{ε_0+1}

n+00+ = {...{0}...} = ε_1
n+00+00+ = ε_1*2
n+0+00+ = ε_1*ω
n+0+00+0+00+ = ε_1*ω2
n+{0}+00+ = ε_1*ω^ω
n+{0+0}+00+ = ε_1*ω^ω^ω
n+{{0}}+00+ = ε_1*ε_0

n+{00}+ = n+{...{0}...}+00+ = ε_1^2
n+{00+0}+ = ε_1^3
n+{00+0+0}+ = ε_1^4
n+{00+{0}}+ = ε_1^ω
n+{00+{0+0}}+ = ε_1^ω^ω
n+{00+{{0}}}+ = ε_1^ε_0
n+{00+{{{0}}}}+ = ε_1^ε_0^ε_0
n+{00+00}+ = n+{00+{...{0}...}}+ f_ω^{ε_1+1}
n+{00+00+00}+ = f_ω^ω^{ε_1+1}
n+{00+00+00+00}+ = f_ω^ω^ω^{ε_1+1}

n+{0+00}+ = ε_2
n+{0+00}+{0+00}+ = ε_2*2
n+0+{0+00}+ = ε_2*ω
n+0+{0+00}+0+{0+00}+ = ε_2*ω2
n+{0}+{0+00}+ = ε_2*ω^ω
n+{0+0}+{0+00}+ = ε_2*ε_0
n+00+{0+00}+ = ε_2*ε_1
n+{0+00+0}+ = n+{00+...+00}+{0+00}+ = ε_2^2
n+{0+00+0+0}+ = ε_2^3
n+{0+00+{0}}+ = ε_2^ω
n+{0+00+{{0}}}+ = ε_2^ε_0
n+{0+00+00}+ = ε_2^ε_1
n+{0+00+00+00}+ = ε_2^ε_1^ε_1
n+{0+00+0+00}+ = f_ω^{ε_2+1}
n+{0+00+0+00+0+00}+ = f_ω^ω^{ε_2+1}
n+{0+0+00}+ = f_{ε_3}
n+{0+0+0+00}+ = f_{ε_4}
n+{{0}+00}+ = f_{ε_ω}
n+{{{0}}+00}+ = f_{ε_ω^ω}
n+{{00}}+ = n+{{~{0}~}+00}+ = f_{ε_ε_0}
n+{{{00}}}+ = f_{ε_ε_ε_0}
n+{{{{00}}}}+ = f_{ε_ε_ε_ε_0}

n+000+ = n+{...{00}...}+ = φ(2,0)
n+000+000+ = φ(2,0)*2
n+0+000+ = φ(2,0)*ω
n+0+000+0+000+ = φ(2,0)*ω2
n+{0}+000+ = φ(2,0)*ω^ω
n+{{0}}+000+ = φ(2,0)*ε_0
n+00+000+ = φ(2,0)*ε_1
n+{{00}}+000+ = φ(2,0)*ε_ε_0
n+{000}+ = {...{00}...}+000+ = φ(2,0)^2
n+{000+0}+ = φ(2,0)^3
n+{000+{0}}+ = φ(2,0)^ω
n+{000+{{0}}}+ = φ(2,0)^ε_0
n+{000+{{00}}}+ = φ(2,0)^ε_ε_0
n+{000+{{{00}}}}+ = φ(2,0)^ε_ε_ε_0
n+{000+000}+ = f_ω^φ(2,0)
n+{000+000+000}+ = f_ω^ω^φ(2,0)
n+{0+000}+ = f_ε_φ(2,0)
n+{0+000+000}+ = f_ω^ε_φ(2,0)
n+{0+000+000+000}+ = f_ω^ω^ε_φ(2,0)
n+{0+000+0+000}+ = f_ε_ε_φ(2,0)
n+{0+000+0+000+0+000}+ = f_ε_ε_ε_φ(2,0)
n+{0+0+000}+ = f_φ(2,1)
n+{0+0+000+000}+ = f_ω^φ(2,1)
n+{0+0+000+000+000}+ = f_ω^ω^φ(2,1)
n+{0+0+000+0+000}+ = f_ε_φ(2,1)
n+{0+0+000+0+000+0+000}+ = f_ε_ε_φ(2,1)
n+{0+0+0+000}+ = f_φ(2,2)
n+{0+0+0+0+000}+ = f_φ(2,3)
n+{{0}+000}+ = f_φ(2,ω)
n+{{{0}}+000}+ = f_φ(2,ε_0)
n+{{{00}}+000}+ = f_φ(2,ε_ε_0)
n+{{000}} = {{...{00}...}+000} = f_φ(2,φ(2,0))
n+{{{000}}} = {{{...{00}...}+000}} = f_φ(2,φ(2,φ(2,0)))
n+{{0000}}+ = n+{{{{...{000}...}}}+0000}+ = f_φ(3,0)
n+{{00000}}+ = n+{{{{{...{0000}...}}}}+00000}+ = f_φ(4,0)
n+1+ = n+{00...00}+ = f_φ(ω,0)
n+1+1+ = f_φ(ω,0)*2
n+0+1+ = n+1+1+...+1+1 = f_φ(ω,0)*ω
n+{0}+1+ = f_φ(ω,0)*ε_0
n+{00}+1+ = f_φ(ω,0)*φ(2,0)
n+{0000}+1+ = f_φ(ω,0)*φ(3,0)
n+{1}+ = n+{00...00}+1+ = f_φ(ω,0)^2
n+{1+0}+ = f_φ(ω,0)^3
n+{1+0+0}+ = f_φ(ω,0)^4
n+{1+{{0}}}+ = f_φ(ω,0)^ε_0
n+{1+{{00}}}+ = f_φ(ω,0)^φ(2,0)
n+{1+{{000}}}+ = f_φ(ω,0)^φ(3,0)
n+{1+1}+ = f_ω^φ(ω,0)
n+{1+1+1}+ = f_ω^ω^φ(ω,0)
n+{0+1}+ = f_ε_φ(ω,0)
n+{0+1+0+1}+ = f_ε_ε_φ(ω,0)
n+{0+0+1}+ = f_φ(2,φ(ω,0))
n+{0+0+0+1}+ = f_φ(3,φ(ω,0))
n+{{0}+1}+ = f_φ(ω,1)
n+{{{0}}+1}+ = f_φ(ω,ε_0)
n+{{{00}}+1}+ = f_φ(ω,φ(2,0))
n+{{{000}}+1}+ = f_φ(ω,φ(3,0))
n+{{{0000}}+1}+ = f_φ(ω,φ(4,0))

n+{{1}}+ = n+{{{00...00}}+1}+ = f_φ(ω,φ(ω,0))
n+{{{1}}}+ = n+{{{{00...00}}+1}}+ = f_φ(ω,φ(ω,φ(ω,0)))
n+10+ = n+{...{1}...}+ = f_φ(ω+1,0)
n+{10}+ = f_φ(ω+1,0)^2
n+{10+{{0}}}+ = f_φ(ω+1,0)^ε_0
n+{10+1}+ = f_φ(ω+1,0)^φ(ω,0)
n+{10+0+1}+ = f_φ(ω+1,0)^ε_φ(ω,0)
n+{10+{0}+1}+ = f_φ(ω+1,0)^ε_φ(ω,1)
n+{10+{1}}+ = f_φ(ω+1,0)^φ(ω,φ(ω,0))
n+{10+10}+ = n+{10+{...{1}...}}+ = f_ω^φ(ω+1,0)
n+{10+10+10}+ = f_ω^ω^φ(ω+1,0)
n+{0+10}+ = f_ε_φ(ω+1,0)
n+{0+10+0+10}+ = f_ε_ε_φ(ω+1,0)
n+{0+0+10}+ = f_φ(2,(φ(ω+1,0))
n+{0+0+0+10}+ = f_φ(3,(φ(ω+1,0))
n+{{0}+10}+ = f_φ(ω,(φ(ω+1,0))
n+{{0}+0+10}+ = f_φ(ω+1,1)
n+{{0}+{0}+10}+ = f_φ(ω+1,2)
n+{{{0}}+10}+ = f_φ(ω+1,ε_0)
n+{{{00}}+10}+ = f_φ(ω+1,φ(2,0))
n+{1+10}+ = f_φ(ω+1,φ(ω,0))
n+{{1}+10}+ = f_φ(ω+1,φ(ω,φ(ω,0)))
n+{{10}}+ = f_φ(ω+2,0)
n+{{{10}}}+ = f_φ(ω+3,0)
n+100+ = n+{~{10}~}+ f_φ(ω2,0)
n+1000+ = f_φ(ω3,0)
n+10000+ = f_φ(ω4,0)
n+11+ = n+100...00+ f_φ(ω^2,0)
n+110+ = f_φ(ω^2+1,0)
n+1100+ = f_φ(ω^2+2,0)
n+111+ = f_φ({ω^2}2,0)
n+1111+ = f_φ({ω^2}3,0)
n+01+ = n+11...11+ f_φ(ω^3,0)
n+001+ = n+11...11+ f_φ(ω^4,0)
n+0001+ = n+11...11+ f_φ(ω^5,0)
n+2+ = n+00...001+ f_φ(ω^ω,0)
n+3+ = n+00...0012+ f_φ(ω^ω^ω,0)

n+0{0}+ = n+n+ f_φ(ε_0,0)
n+0{0}+0{0}+ = n+n+ f_φ(ε_0,0)*2
n+0+0{0}+0{0}+ = n+n+ f_φ(ε_0,0)*ω
n+{0{0}}+ = n+n+ f_φ(ε_0,0)^2
n+{0{0}+{{0}}}+ = n+n+ f_φ(ε_0,0)^ε_0
n+{0{0}+{{00}}}+ = n+n+ f_φ(ε_0,0)^φ(2,0)
n+{0{0}+1}+ = n+n+ f_φ(ε_0,0)^φ(ω,0)
n+{0{0}+0{0}}+ = n+{0{0}+n}+ f_ω^φ(ε_0,0)
n+{0+0{0}}+ = f_ε_φ(ε_0,0)
n+{{{0}}+0{0}}+ = f_φ(2,φ(ε_0,0))
n+{1+0{0}}+ = f_φ(ω,φ(ε_0,0))
n+{{0{0}}}+ = n+{n+0{0}}+ φ(ε_0,1)
n+{{0{0}+0}}+ = φ(ε_0,2)
n+{{0{0}+{{0}}}}+ = φ(ε_0,ε_0)
n+{{0{0}+1}}+ = φ(ε_0,φ(ω,0))
n+{{0{0}+0{0}}}+ = n+{{0{0}+n}}+ φ(ε_0+1,0)
n+{{0+0{0}}}+ = φ(ε_0+ω,0)
n+{{{{0}}+0{0}}}+ = φ(ε_0*2,0)
n+{{{00}+0{0}}}+ = φ(ε_ε_0,0)
n+{{{{000}}+0{0}}}+ = φ(φ(2,0),0)
n+{{1+0{0}}}+ = φ(φ(ω,0),0)
n+{{{1+0{0}}}}+ = φ(φ(φ(ω,0),0),0)
n+{{{{1+0{0}}}}}+ = φ(φ(φ(φ(ω,0),0),0),0)
n+0{0+0}+ = n+{...{0{0}...}+ Γ_0
n+0{0+0}+0{0+0}+ = Γ_0*2
n+0+0{0+0}+ = Γ_0*ω
n+{0{0+0}}+ = Γ_0^2
n+{0{0+0}+0{0+0}}+ = Γ_0^Γ_0
n+{{{0}}+0{0+0}}+ = ε_Γ_0
n+{1+0{0+0}+0}+ = f_φ(ω,Γ_0+1)
n+{0{0}+0{0+0}+0}+ = f_φ(φ(ω,0),Γ_0+1)
n+{{0{0}}+0{0+0}+0}+ = f_φ(φ(φ(ω,0),0),Γ_0+1)
n+{{0{0+0}}}+ = f_φ(Γ_0,1)
n+{{0{0+0}+0}}+ = f_φ(Γ_0,2)

n+{{0{0}}}+ = n+{n+0{0}}+ φ(ε_0,1)
n+{{0{0}+0}}+ = φ(ε_0,2)
n+{{0{0}+{{0}}}}+ = φ(ε_0,ε_0)
n+{{0{0}+1}}+ = φ(ε_0,φ(ω,0))
n+{{0{0}+0{0}}}+ = n+{{0{0}+n}}+ φ(ε_0+1,0)
n+{{0+0{0}}}+ = φ(ε_0+ω,0)
n+{{{{0}}+0{0}}}+ = φ(ε_0*2,0)
n+{{{00}+0{0}}}+ = φ(ε_ε_0,0)
n+{{{{000}}+0{0}}}+ = φ(φ(2,0),0)
n+{{1+0{0}}}+ = φ(φ(ω,0),0)
n+{{{1+0{0}}}}+ = φ(φ(φ(ω,0),0),0)
n+{{{{1+0{0}}}}}+ = φ(φ(φ(φ(ω,0),0),0),0)
n+0{0+0}+ = n+{...{0{0}...}+ Γ_0
n+0{0+0}+0{0+0}+ = Γ_0*2
n+0+0{0+0}+ = Γ_0*ω
n+{0{0+0}}+ = Γ_0^2
n+{0{0+0}+0{0+0}}+ = Γ_0^Γ_0
n+{{{0}}+0{0+0}}+ = ε_Γ_0
n+{1+0{0+0}+0}+ = f_φ(ω,Γ_0+1)
n+{0{0}+0{0+0}+0}+ = f_φ(φ(ω,0),Γ_0+1)
n+{{0{0}}+0{0+0}+0}+ = f_φ(φ(φ(ω,0),0),Γ_0+1)
n+{{0{0+0}}}+ = f_φ(Γ_0,1)
n+{{0{0+0}+0}}+ = f_φ(Γ_0,2)
n+{{0{0+0}+{{0}}}}+ = f_φ(Γ_0,ε_0)
n+{{0{0+0}+1}}+ = f_φ(Γ_0,φ(ω,0))
n+{{0{0+0}+0{0+0}}}+ = f_φ(Γ_0+1,0)
n+{{0+0{0+0}}}+ = f_φ(Γ_0+ω,0)
n+{{{0{0+0}}}}+ = n+{{{...{0{0}...}+0{0+0}}}+ = f_φ(Γ_0*2,0)
n+{{{0{0+0}+0}}}+ = f_φ(Γ_0*3,0)
n+{{{0{0+0}+{0}}}}+ = f_φ(Γ_0*ω,0)
n+{{{0{0+0}+{{0}}}}}+ = f_φ(Γ_0*ε_0,0)
n+{{{0{0+0}+0{0+0}}}}+ = f_φ(Γ_0^Γ_0,0)
n+{{{0+0{0+0}}}}+ = f_φ(ε_Γ_0,0)
n+{{{1+0{0+0}}}}+ = f_φ(φ(ω,Γ_0+1),0)
n+{{{0{0}+0{0+0}}}}+ = f_φ(φ(φ(ω,0),Γ_0+1),0)
n+{{{0{0+0}}}}+ = n+{{{{...{0{0}}...}+0{0+0}}}}+ = f_φ(Γ_0,1)
n+{{{{0{0+0}}}}}+ = f_φ(φ(Γ_0,1),0)
n+{{{{{0{0+0}}}}}}+ = f_φ(φ(φ(Γ_0,1),0),0)
n+0{0+0+0}+ = n+{...{0{0+0}...}+ = Γ_1
n+0{0+0+0+0}+ = n+{...{0{0+0+0}...}+ = Γ_2
n+0{{0}}+ = n+0{0+0+...+0+0}+ = Γ_ω
n+0{{{0}}}+ = Γ_ε_0
n+0{{{00}}}+ = Γ_φ(2,0)
n+0{1}+ = Γ_φ(ω,0)
n+0{0{0+0}}+ = Γ_Γ_0
n+0{0{0{0+0}}}+ = Γ_Γ_Γ_0

n+0{0}{0}+ = n+0{0...{0+0}...}+ f_φ(1,0,0)
n+0{0}{0}+0{0}{0}+ = f_φ(1,0,0)*2
n+{0{0}{0}}+ = f_φ(1,0,0)^2
n+{0{0}{0}+0{0}{0}}+ = f_φ(1,0,0)^φ(1,0,0)
n+{0+0{0}{0}}+ = f_ε_φ(1,0,0)
n+0{0{0}{0}}+ = Γ_φ(1,0,0)
n+0{0{0{0}{0}}}+ = Γ_Γ_φ(1,0,0)
n+0{0}{0+0}+ = n+0{...0{0}{0}...}+ = φ(1,0,1)
n+0{0}{0+0+0}+ = n+0{...0{0}{0}...}+ = φ(1,0,2)
n+0{0}{0{0}{0+0}}+ = φ(1,0,φ(1,0,1))
n+0{0}{0{0}{0{0}{0+0}}}+ = φ(1,0,φ(1,0,φ(1,0,1)))
n+0{0+0}{0}+ = n+0{0}{0{0}...{0{0}{0+0}...}}+ φ(1,1,0)

n+0{0+0}{0}+0{0+0}{0}+ = f_φ(1,1,0)*2
n+{0{0+0}{0}}+ = f_φ(1,1,0)^2
n+{0{0+0}{0}+0{0+0}{0}}+ = f_φ(1,1,0)^φ(1,1,0)
n+{0+0{0+0}{0}}+ = f_ε_φ(1,1,0)
n+0{0{0+0}{0}}+ = Γ_φ(1,1,0)
n+0{0{0{0+0}{0}}}+ = Γ_Γ_φ(1,1,0)

n+0{0}{0{0+0}{0}}+ = n+0{0{0{0+0}{0}}}+ φ(1,0,φ(1,1,0))
n+0{0}{0{0}{0{0+0}{0}}}+ = φ(1,0,φ(1,0,φ(1,1,0)))
n+0{0+0}{0+0}+ = f_φ(1,1,1)
n+0{0+0}{0+0}+ = f_φ(1,1,2)
n+0{0+0}{{0}}+ = f_φ(1,1,ω)
n+0{0+0}{{{0}}}+ = f_φ(1,1,ε_0)
n+0{0+0}{1}+ = f_φ(1,1,φ(ω,0))
n+0{0+0}{0{0+0}{0+0}}+ = f_φ(1,1,φ(1,1,1))
n+0{0+0}{0{0+0}{0{0+0}{0+0}}}+ = f_φ(1,1,φ(1,1,φ(1,1,1)))
n+0{0+0+0}{0}+ = n+0{0+0}{0{0+0}{...0{0+0}{0+0}...}}+ f_φ(1,2,0)
n+0{0+0+0+0}{0}+ = f_φ(1,3,0)
n+0{{0}}{0}+ = f_φ(1,ω,0)
n+0{0{0+0}{0+0}}{0}+ = f_φ(1,φ(1,1,1),0)
n+0{0{0{0+0}{0+0}}{0+0}}{0}+ = f_φ(1,φ(1,φ(1,1,1),1),0)
n+0{0}{0}{0}+ = f_φ(2,0,0)
n+0{0{0}{0+0}{0+0}}{0}+ = f_φ(1,φ(2,1,1),1)
n+0{0+0}{0}{0}+ = f_φ(3,0,0)
n+0{{0}}{0}{0}+ = f_φ(ω,0,0)
n+0{1}{0}{0}+ = f_φ(φ(ω,0),0,0)
n+0{0{0}{0+0}}{0}{0}+ = f_φ(φ(1,0,1),0,0)
n+0{0{0{0}{0+0}}{0}{0}}{0}{0}+ = f_φ(φ(φ(1,0,1),0,0),0,0)
n+0{0}{0}{0}{0}+ = f_φ(1,0,0,0)
n+0{0}{0}{0}{0}{0}+ = f_φ(1,0,0,0,0)

n+{0}{0}+ = n+0{0}...{0}+ f_{ψ(Ω^Ω^ω)}
n+{0}{0}+{0}{0}+ = f_{ψ({Ω^Ω^ω}+1)}
n+0+{0}{0}+ = f_{ψ({Ω^Ω^ω}+ω)}
n+{{0}{0}}+ = f_{ψ({Ω^Ω^ω}2)}
n+{{0}{0}+{0}{0}}+ = f_{ψ({Ω^Ω^ω}^2)}
n+{0+{0}{0}}+ = f_{ψ(Ω^((Ω^ω)+1))}
n+{{{0}{0}}}+ = f_{ψ(Ω^((Ω^ω)2))}

n+0{{0}{0}}+ =n+{...{0}{0}...}+ f_{ψ(Ω^(Ω^{ω+1})))}

n+0{{0}{0}+0}+ =n+{...{0}{0}...}+ f_{ψ(Ω^(Ω^{ω+2})))}

n+{0}{0+0}+ = n+0{{0}{0}}{{0}{0}}...{{0}{0}}{{0}{0}}+ f_{ψ(Ω^Ω^{ψ(Ω^Ω^ω)})}

^ Matches WarDaft

n+{0}{{0}{0+0}}+ = n+0{{0}{0+0}}{{0}{0+0}}...{{0}{0+0}}{{0}{0+0}}+

Send:

6+{0}{{0}{0+0}}

For f_{ψ(Ω^Ω^{ψ(Ω^Ω^{ψ(Ω^Ω^ω)})})}+

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 7:04 am UTC
by Daggoth
You forgot to show how the notations expand for like 90% of your examples.
I got the rules tucked in the spoiler on MY notation. (which im still working on for proper sizing)

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 11:26 am UTC
by WarDaft
I'm going to take this time to introduce another notation I've been pondering for ages but never really explored.

This one doesn't even use numbers, just parenthesis.

(a,X,[Y],Z)-- = <(a--,X,(a--,X,[Y],Z)>
(a,X)-- = <(a--,X)>
[ ]-- = ()
[a,X]-- = <[a--,X]>
()-- = null, or nothing.

I have called it the hat-wearing-hydra. When you attack a head at the same level as a head that is wearing a hat, a new neck resembling the current hydra (post cutting) pops out, having being mischeviously hidden under the hat the whole time!

So, 3|(),(),() = 4|(),() = 5|() = 6
n|(()) = f{1}(n)
n|((),()) = f{2}(n)
n|((())) = f{ω}(n)
n|(((()))) = f{ω^ω}(n)
Etc, just like a PK Hydra.
But we also have [ ]
n|((),[]) = n|(([])) = f{ω}(n)
n|((),(),[]) = n(((),[])) = f{ω^ω}(n)
n|((()),[]) = f{ε0}(n)
n|((),(()),[]) = f{ω^(ε0+1)}(n)
n|((()),(()),[]) = f{ε1}(n)
n|(((),()),[]) = f{εω}(n)
n|( ((),(),()),[]) = f{εω^2}(n)
n|( ((())),[]) = f{εω^ω}(n)
n|( ((()),[]), []) = f{εε_0}(n)
n|( ( ((()),[]), []),[]) = f{εε_ε_0}(n)
n|([],[]) = n|(((),[]),[])
n|((),[],[]) = n|(([],[]),[])
n|((()),[],[]) = f{φ(2,0)}(n)
n|((),(()),[],[])=f{φ(1,φ(2,0)+1)}(n)
n|(((),()),[],[]) = f{φ(2,1)}(n)
n|(((),(),()),[],[]) = f{φ(2,2)}(n)
n|(((())),[],[]) = f{φ(2,φ(0,0))}(n)
n|( (()),[].[].[]) = f{φ(3,0)}(n)
n|( (()),[],[],[],[]) = f{φ(4,0)}(n)
n|([()]) = f{φ(φ(0,0),0)}(n)
So n|([[..[[()]]..]]) is f{φ(1,0,0)}(n)

Hmm, not bad for such a simple extension.

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 3:35 pm UTC
by Vytron
Daggoth wrote:You forgot to show how the notations expand for like 90% of your examples.


To the limit of the previous lines.

I.e. I have:

n+{1+{{00}}}+ = f_φ(ω,0)^φ(2,0)
n+{1+{{000}}}+ = f_φ(ω,0)^φ(3,0)
n+{1+1}+ = f_ω^φ(ω,0)

Which is actually f_ω^{φ(ω,0)+1}

So n+{1+1}+ expands to n+{1+{{00...00}}}+ for n+1 0s (n 0s doesn't suffice...).

Values:

0+{1+1}+ = 0
1+{1+1}+ = 1+{1+{{00}}}+ = f_φ(ω,0)^φ(2,0)(2) (Because at the base 1+ returns 2)
2+{1+1}+ = 2+{1+{{000}}}+ = f_φ(ω,0)^φ(3,0)(6)
3+{1+1}+ = 3+{1+{{0000}}}+ = f_φ(ω,0)^φ(4,0)(12)

And so on. Please tell me if any of this is unclear.

Note that I could expand to much stronger values, say:

n+{1+1}+ expands to n+{1+{{00...00}}}+ for n+ 0s

Values:

0+{1+1}+ = 0
1+{1+1}+ = 1+{1+{{00}}}+ = f_φ(ω,0)^φ(2,0)(2) (Produces the same)
2+{1+1}+ = 2+{1+{{000000}}}+ = f_φ(ω,0)^φ(6,0)(6)
3+{1+1}+ = 3+{1+{{000000000000}}}+ = f_φ(ω,0)^φ(12,0)(12)

Or any value I have access to at this point. But it wouldn't make any difference.

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 4:57 pm UTC
by mike-l
Vytron, you're using fixed numbers of iterations for your n+ n+ rules, so you never even get to the next ordinal. You start with n+ = n(n+1) which is between f_1 and f_2, so your n+0+ is only at f_2

Explictly, define g_m(n) to be n+ n+ ... n+ with m spaces. Then your rules are g_1(n) = n(n+1), g_{m+1}(n) = {g_1.g_m}^m(n). (Here . is the function composition operator).

Thus g_{m+1}(n) < g_m^{2m}(n) so by induction g_m(n) < g_1^{(2m)^m}}(n). Now g_1(n) < (n+1)^2 so g_1^m(n) < (n+1)^(2m) and hence n+0+ = g_n(n) < (n+1)^(2(2n)^n) < f_2^3(n) ~= 2^2^2^n.

Edit: actually you use n+0+ = g_{n+1}(n) but this doesn't affect the outcome in a material way.

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 5:29 pm UTC
by mike-l
How does

n+0+0+{0}+{0+0}+ reduce?

Incidentally, despite being at only f_2 when you said f_w, by n+{0+0}+ you actually are back to where you claim to be, as counting by w instead of counting by 1 makes no difference once you hit w^w

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 6:14 pm UTC
by WarDaft
That's one of the reasons I didn't nitpick the notation too much. Using the hyphen notation correctly, it would be the only thing that mattered and the end target would still be reached. I suppose I should have checked that the hyphen notation was working optimally.

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 6:25 pm UTC
by mike-l
I mean that's true. But my inclination is that major errors early on are indicative of problems later, and are much easier to spot.

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 7:49 pm UTC
by Vytron
You're right, I think I can have this at the base:

(Note: when I say "spaces" I mean to only count the ones between equal elements)

n+ = n+n+...+n+n for n plus signs

0+ = 0
1+ = 2
2+ = 6
3+ = 12

n+ 0+ = ((n+)+)+...)+)+ for n+ nestings

n+ 0+ 0+ = ((n+ 0+)+ 0+)+...)+ 0+)+ 0+ for n+ 0+ nestings

n+ 0+ 0+ 0+ = ((n+ 0+ 0+)+ 0+ 0+)+...)+ 0+ 0+)+ 0+ 0+ for n+ 0+ 0+ nestings

n+ a+ = ((n+ 0+...0+)+ 0+...0+)+...)+ 0+...0+)+ 0+...0)+ (a-1)+ for n+ nestings and n+ spaces in each nest

n+ a+ 0+ = ((n+ a+)+ a+)+...)+ a+)+ a+ for n+ a+ nestings

n+ a+ 0+ 0+ = ((n+ a+ 0+)+ a+ 0+)+...)+ a+ 0+)+ a+ 0+ for n+ a+ 0+ nestings

n+ a+ 0+ 0+ 0+ = ((n+ a+ 0+ 0+)+ a+ 0+ 0+)+...)+ a+ 0+ 0+)+ a+ 0+ 0+ for n+ a+ 0+ 0+ nestings

n+ a+ b+ = ((n+ a+ 0+...0+)+ a+ 0+...0+)+...)+ a+ 0+...0+)+ a+ 0+...0)+ a+ (b-1)+ for n+ a+ nestings and n+ a+ spaces in each nest

And:

n+0+ = n+ n+ ... n+ with n+ spaces.

Values:

1+0+ = 1+ 1+ 1+

1+ 1+ 1+ = ((1+ 1+ 0+...0+)+ 1+ 0+...0+)+...)+ 1+ 0+...0+)+ 1+ 0+...0)+ 1+ 0+ for 1+ 1+ nestings and 1+ 1+ spaces in each nest.

1+ 1+ = ((1+ 0+...0+)+ 0+...0+)+...)+ 0+...0+)+ 0+...0)+ 0+ for 1+ nestings and 1+ spaces in each nest.

1+ 1+ = ((1+ 0+ 0+ 0+)+ 0+ 0+ 0+)+ 0+

1+ 0+ 0+ 0+ = ((1+ 0+ 0+)+ 0+ 0+)+...)+ 0+ 0+)+ 0+ 0+ for 1+ 0+ 0+ nestings

1+ 0+ 0+ = ((1+ 0+)+ 0+)+...)+ 0+)+ 0+ for 1+ 0+ nestings

1+ 0+ = ((1+)+)+...)+)+ = ((1)+)+ = (2)+ = 6

1+ 0+ 0+ = (..((1+ 0+)+ 0+)+ 0+)+ 0+)+ 0+)+ 0+)+ 0+
= (..(6+ 0+)+ 0+)+ 0+)+ 0+)+ 0+)+ 0+

6+ 0+ = ((6+)+)+...)+)+ = ((6+)+)+)...42 nests...+)+)+
= (42+)+)...41 nests...+)+)+
= (1806+)+)...40 nests...+)+)+
= (3263442+)+)...39 nests...+)+)+
= (10650056950806+)+)...38 nests...+)+)+
= (113423713055400544247098830+)+)...37 nests...+)+)+

So I think this n+0+ goes over f_ω(n+1)

mike-l wrote:How does

n+0+0+{0}+{0+0}+ reduce?


You drop a 0+ and make n copies.

Values:

1+0+0+{0}+{0+0}+ = 1+0+{0}+{0+0}+0+{0}+{0+0}+
2+0+0+{0}+{0+0}+ = 2+0+{0}+{0+0}+0+{0}+{0+0}+0+{0}+{0+0}+
3+0+0+{0}+{0+0}+ = 3+0+{0}+{0+0}+0+{0}+{0+0}+0+{0}+{0+0}+0+{0}+{0+0}+

It's meant to reduce the same way as n[0,0,1,2] in your notation.

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 8:05 pm UTC
by mike-l
Didnt we just established that this isn't how my notation worked and that this only gets to w^w^w when you claim its at e_0?

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 8:06 pm UTC
by Vytron
Ohh, did I do it wrong again?

Should it be this way?:

String to reduce:

+0+0+{0}+{0+0}+

Find the string:

+(0+0+{0}+){0+0}+

Reduce the string

+(0+{0}+0+{0}+...n+1 copies...+0+{0}+0+{0}+){0+0}+

Now, make n+1 copies of that:

+0+{0}+0+{0}+...n+1 copies...+0+{0}+0+{0}+{0+0}+0+{0}+0+{0}+...n+1 copies...+0+{0}+0+{0}+{0+0}+0+{0}+0+{0}+...n+1 copies...+0+{0}+0+{0}+{0+0}+... (With n+1 instances)

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 8:16 pm UTC
by mike-l
That should work

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 9:48 pm UTC
by GoogologyMaster
$ contains only 1's and separators or nothing
# can be anything or empty
a [1] b = a^b
a [n #] b = a [n-1 #] a [n-1 #] ... [n-1 #] a only works if very first entry is greater than one
a [$ 1,n #] b = a [$ b, n-1 #] a works in nested separators
a [$ 1 [m] n #] b = a [$ 1 [m-1] 1 [m-1] ... [m-1] 1 [m-1] 2 [m] n-1 #] a works in nested separators
a [$ 1 \ n #] b = a [$ 1 [$ 1 [$ 1 [... [$ 1,2 \ n-1 #] ...] 2 \ n-1 #] 2 \ n-1 #] 2 \ n-1 #] a works in nested separators
a [# [1 ¬ 2] #] b = a [# \ #] b works anywhere

I will continue
{1 [1 \ 2 ¬ 2] n} = Γn-2
{1 [1 \ 2 ¬ 2] 1,2} = Γω
{1 [1 \ 2 ¬ 2] 1,3} = Γω2
{1 [1 \ 2 ¬ 2] 1,1,2} = Γω^2
{1 [1 \ 2 ¬ 2] 1 [2] 2} = Γω^ω
{1 [1 \ 2 ¬ 2] 1 [1 \ 2] 2} = Γε_0
{1 [1 \ 2 ¬ 2] 1 [1 [2 ¬ 2] 2] 2} = Γφ(ω,0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 2] 2} = ΓΓ_0
{1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 3] 2} = ΓΓ_0
{1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 2] 2] 2} = ΓΓ_Γ_0
{1 [1 \ 2 ¬ 2] 1 \ 2} = φ(1,1,0)

Now I’m going to try to beat them
{1 [1 \ 2 ¬ 2] 1 \ 1 \ 2} = φ(1,2,0)
{1 [1 \ 2 ¬ 2] 1 \ 1 \ 1 \ 2} = φ(1,3,0)
{1 [1 \ 2 ¬ 2] 1 [2 ¬ 2] 2} = φ(1,ω,0)
{1 [1 \ 2 ¬ 2] 1 [1,2 ¬ 2] 2} = φ(1,ω^ω,0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2] 2 ¬ 2] 2} = φ(1,ε0,0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 [2 ¬ 2] 2] 2 ¬ 2] 2} = φ(1,φ(ω,0),0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(1,Γ0,0)
I save the {1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} for later

{1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 3] 2 ¬ 2] 2} = φ(1,Γ1,0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 \ 2] 2 ¬ 2] 2} = φ(1,φ(1,1,0),0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [2 ¬ 2] 2] 2 ¬ 2] 2} = φ(1,φ(1,ω,0),0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2] 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(1,φ(1,ε0,0),0)
{1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(1,φ(1,Γ0,0),0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(2,0,0)

Now I’m starting to see a pattern. Each \ adds 1 to the second-last argument in the φ() function. Each [2 ¬ 2] adds ω to the second-last argument, each [1 \ 2 ¬ 2] adds 1 to the third-last argument.
By this method, I nest {1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [X] 2}s to make {1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 \ 2}

{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(2,0,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2] 2} = φ(2,0,φ(2,0,0))
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2] 2] 2} = φ(2,0,φ(2,0,φ(2,0,0)))
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 \ 2} = φ(2,1,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 \ 1 \ 2} = φ(2,2,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 \ 1 \ 1 \ 2} = φ(2,3,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [2 ¬ 2] 2} = φ(2,ω,0)

Now I nest into {1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [X ¬ 2] 2} to form {1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2}.
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [3 ¬ 2] 2} = φ(2,ω^2,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1,2 ¬ 2] 2} = φ(2,ω^ω,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [2] 2 ¬ 2] 2} = φ(2,ω^ω^ω,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2] 2 ¬ 2] 2} = φ(2,ε0,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(2,Γ0,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(2,φ(2,0,0),0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(2,φ(2,φ(2,0,0),0),0)

{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(3,0,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 3} = φ(3,0,1)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2] 2} = φ(3,0,φ(3,0,0))
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 \ 2} = φ(3,1,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 \ 1 \ 2} = φ(3,2,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [2 ¬ 2] 2} = φ(3,ω,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(3,Γ0,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(3,φ(3,0,0),0)

{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(4,0,0)
{1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(5,0,0)
and so on.
You will probably guess that I will go to [2 \ 2 ¬ 2]
{1 [2 \ 2 ¬ 2] 2} = φ(ω,0,0)

by rule: a [$ 1 [m] n #] b = a [$ 1 [m-1] 1 [m-1] ... [m-1] 1 [m-1] 2 [m] n-1 #] a

{1 [2 \ 2 ¬ 2] 3} = φ(ω,0,1)
{1 [2 \ 2 ¬ 2] 1 [1 [2 \ 2 ¬ 2] 2] 2} = φ(ω,0,φ(ω,0,0))
{1 [2 \ 2 ¬ 2] 1 \ 2} = φ(ω,1,0)
{1 [2 \ 2 ¬ 2] 1 \ 1 \ 2} = φ(ω,2,0)
{1 [2 \ 2 ¬ 2] 1 [2 ¬ 2] 2} = φ(ω,ω,0)
{1 [2 \ 2 ¬ 2] 1 [1 [1 \ 2] 2 ¬ 2] 2} = φ(ω,ε0,0)
{1 [2 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(ω,Γ0,0)
{1 [2 \ 2 ¬ 2] 1 [1 [1 [2 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(ω,φ(ω,0,0),0)

{1 [2 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω+1,0,0)
{1 [2 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω+2,0,0)
{1 [2 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω+3,0,0)
{1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 2} = φ(ω*2,0,0)
{1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 2} = φ(ω*3,0,0)
{1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 2} = φ(ω*4,0,0)

{1 [3 \ 2 ¬ 2] 2} = φ(ω2,0,0)
{1 [3 \ 2 ¬ 2] 3} = φ(ω2,0,1)
{1 [3 \ 2 ¬ 2] 1 [1 [3 \ 2 ¬ 2] 2] 2} = φ(ω2,0,φ(ω2,0,0))
{1 [3 \ 2 ¬ 2] 1 \ 2} = φ(ω2,1,0)
{1 [3 \ 2 ¬ 2] 1 [2 ¬ 2] 2} = φ(ω2,ω,0)
{1 [3 \ 2 ¬ 2] 1 [1 [1 [1 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(ω20,0)
{1 [3 \ 2 ¬ 2] 1 [1 [1 [3 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(ω2,φ(ω2,0,0),0)
{1 [3 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω2+1,0,0)
{1 [3 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω2+2,0,0)
{1 [3 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 2} = φ(ω2+ω,0,0)
{1 [3 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω2+ω+1,0,0)
{1 [3 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 2} = φ(ω2+ω2,0,0)
{1 [3 \ 2 ¬ 2] 1 [3 \ 2 ¬ 2] 2} = φ(ω2*2,0,0)
{1 [3 \ 2 ¬ 2] 1 [3 \ 2 ¬ 2] 1 [3 \ 2 ¬ 2] 2} = φ(ω2*3,0,0)

{1 [4 \ 2 ¬ 2] 2} = φ(ω3,0,0)
{1 [4 \ 2 ¬ 2] 3} = φ(ω3,0,1)
{1 [4 \ 2 ¬ 2] 1 [1 [4 \ 2 ¬ 2] 2] 2} = φ(ω3,0,φ(ω3,0,0))
{1 [4 \ 2 ¬ 2] 1 \ 2} = φ(ω3,1,0)
{1 [4 \ 2 ¬ 2] 1 [1 [1 [4 \ 2 ¬ 2] 2] 2 ¬ 2] 2} = φ(ω3,φ(ω3,0,0),0)
{1 [4 \ 2 ¬ 2] 1 [1 \ 2 ¬ 2] 2} = φ(ω3+1,0,0)
{1 [4 \ 2 ¬ 2] 1 [2 \ 2 ¬ 2] 2} = φ(ω3+ω,0,0)
{1 [4 \ 2 ¬ 2] 1 [3 \ 2 ¬ 2] 2} = φ(ω32,0,0)
{1 [4 \ 2 ¬ 2] 1 [4 \ 2 ¬ 2] 2} = φ(ω3*2,0,0)
{1 [4 \ 2 ¬ 2] 1 [4 \ 2 ¬ 2] 1 [4 \ 2 ¬ 2] 2} = φ(ω3*3,0,0)
{1 [5 \ 2 ¬ 2] 2} = φ(ω4,0,0)
{1 [6 \ 2 ¬ 2] 2} = φ(ω5,0,0)
{1 [1,2 \ 2 ¬ 2] 2} = φ(ωω,0,0)

Re: Your number is, in fact, not bigger!

Posted: Sat Feb 21, 2015 10:08 pm UTC
by Vytron
Okay, so let's see if I can write the rules of my notation in a comprehensive way once and for all.

Define:

nXY

Where XY is a string of something beginning with + and ending with +. Each element in the string is separated by a + itself

Like:

n+0+{0+0}+{0+0+0+0+0}+{0+0+0}+{0+0+0+0}+{0}+

Is a valid string.

X ends at the highest element and may be empty if the rightmost element is the highest one.

If the rightmost element in Y is a 0, remove it and return nXY nXY...nXY nXY for n spaces.

Otherwise, if the rightmost element in Y is preceded by an element equal or higher reduce as follows:

{0} = 0+0+0+0+0 (n plus signs)
{0+0} = 0+0+0+0+0+{0} (n plus signs between the 0s)
{0+0+0} = 0+0+0+0+0+{0}+{0+0} (n plus signs between the 0s)
{0+0+0+0} = 0+0+0+0+0+{0}+{0+0}+{0+0+0} (n plus signs between the 0s)

If the rightmost element in Y is preceded by a lower element, define a string from the beginning of Y to this element, and repeat the process until the rightmost element in this string is not preceded by a lower element.

If the rightmost element of this string is 0, drop it and make n+1 duplicates of the element to the right of this string, otherwise, reduce the string. Then return XYYY...YYY for n copies of Y.

Examples:

n+0+{0+0}+{0+0+0+0+0}+{0+0+0}+{0+0+0+0}+{0}+
=
n+0+{0+0}+{0+0+0+0+0}+{0+0+0}+{0+0+0+0}+0+0...0+0+

-------

n+0+{0+0}+{0+0+0+0+0}+{0+0+0}+{0+0+0+0}
X = 0+{0+0}+{0+0+0+0+0} Y = {0+0+0}+{0+0+0+0}

{0+0+0+0} is preceded by lower element, then make Y:

{0+0+0}

Reduce:

Y = 0+0+...+0+0+{0}+{0+0}+{0+0+0+0}

Return:

n+0+{0+0}+{0+0+0+0+0}+0+0+...+0+0+{0}+{0+0}+{0+0+0+0}+0+0+...+0+0+{0}+{0+0}+{0+0+0+0}+0+0+...+0+0+{0}+{0+0}+{0+0+0+0} (with n copies of Y)

{{0}} reduces to {0+0+...+0+0} with n{0}+ plus signs.

00 reduces to {...{0}...} with n nestings
000 reduces to {...{00}...} with n nestings
0000 reduces to {...{000}...} with n nestings

1 reduces to {00...00} with n+00 0s
10 reduces to {...{1}...}
100 reduces to {...{10}...}
1000 reduces to {...{100}...}

11 reduces to {100...00}
110 reduces to {...{11}...}
1100 reduces to {...{110}...}
11000 reduces to {...{1100}...}

01 reduces to {111...111} n+11 0s

010 reduces to {...{01}...}
0100 reduces to {...{010}...}
01000 reduces to {...{0100}...}

011 reduces to {010...00} n+01 0s

And the XY rules are applied to consecutive digits.

2 reduces to {00...001}
3 reduces to {00...0012}

0{0} reduces to 'n+{n}+' (where the 's are used to denominate an n of some value)

I.e. 2+0{0}+ = 2+'2+{n}+'+ = 2+{00...00123456789'10''11''12'...''2{n}+'-2'''2{n}+'-1'}+

0{0}{0} reduces to {...{0{0}}...}
0{0}{0}{0} reduces to {...{0{0}{0}}...}
0{0+0} reduces to 0{0}{0}...{0}{0}
0{0+0}{0} reduces to {...{0{0+0}}...}
0{0+0}{0}{0} reduces to {...{0{0+0}{0}}...}
0{0+0}{0+0} reduces to 0{0+0}{0}{0}...{0}{0}
0{0}{0+0} reduces to 0{0+0}{0+0}...{0+0}{0+0}

The XY rules go in there.

{0}{0} = 0{0{...{0{0}}...} with n nestings
{0}{0}{0} = 0{0{...{0{0}{0}}...} with n nestings
{0}{0}{0}{0} = 0{0{...{0{0}{0}{0}}...} with n nestings
{0}{0+0} = {0}{0}{0}...{0}{0} with n+{0}{0}+ {0}s

{0}XY rules happen here.

{0}{0{0}} = {0}{'n+{n}+'}{'n+{n}+'}{'n+{n}+'}{'n+{n}+'} ...
{0}{0{0{0}}} = {0}{{0}{'n+{n}+'}{'n+{n}+'}{'n+{n}+'}{'n+{n}+'}} ...

{0}{{0}{0}} = {0}{0{0...{0}...}}
{0}{{0}{0+0}} = {0}{{0}{0}{0}{0}}

And so:

n+{0}{{0}{0+0}}+ = n+{0}{{0}{0}...{0}{0}}+

Would reach f_{ψ(Ω^Ω^{ψ(Ω^Ω^{ψ(Ω^Ω^ω)})})}(n)

Re: Your number is, in fact, not bigger!

Posted: Sun Feb 22, 2015 1:48 pm UTC
by WarDaft
Okay, time to drag things into the 21st century!

< > are the repeater operator, such that <X> means the string X repeated n times
( ) denote an array, which is to be evaluated lazily, rather than immediately.
[;] is an advanced delimiter. : and ; will not work the same, so I'm not re-using :.
Capitals represent any number of terms (including zero), lower case represent exactly one term.
{ } represent special rules for decrementing a term.

n|0 = n
n|X = n+1|X--
(0)-- = 0
(b)-- = (b--)
(X,0,0)-- where X contains no non-zero terms = <(X,>0<)>
(X,0,b,Y)-- where X contains no non-zero terms = <(X,>0<,b--,Y)>
(a,X,0,b,Y)-- where X contains no non-zero terms, and b does not contain square brackets = <(a--,X,>(a--,X,0,b,Y)<,b--,Y)>
(a,b,X)-- = <(>(a--,b,X)<,b--,X)>
Terms containing [;] are considered non-zero.
(0[;]0,X) = (<0,>X)
(a,0[;]0,X)-- where X is all zero = (<0,>(a--,0[;]0,X),X)
(a,b[;]0,X)-- = (<0,>(a--,b[;]0,X),b--[;]0,X)
(0[;]b,X)-- = <(>0<[;]b--,X)>
(a,0[;]b,X)-- = <(>(a--,0[;]b,X)<[;]b--,X)>
(a[;]b,X)-- = <(>0<[;]b--,a--[;]b,X)>
(a,b[;]c,X)-- = <(>(a--,b[;]c,X)<[;]c--,b--[;]c,X)>
(0[a;]0,X)-- = <(0[a--;]>0<,X)>
(a,0[b;]0,X) = <(0[b--;]>(a--,0[b;]0,X)<,X)>

(0[0;]0) = ψ(Ω^Ω^Ω)
(0,0,0[0;]0) = ψ(Ω^(Ω^Ω+1))
(0,0,0,0[0;]0) = ψ(Ω^(Ω^Ω+Ω))
(0,0,0,0,0[0;]0) = ψ(Ω^(Ω^Ω+Ω^2))
(0[;]0,0[0;]0) = ψ(Ω^(Ω^Ω+Ω^ω))
(0,1[0;]0) = ψ(Ω^(Ω^Ω*2))
(0,0,1[0;]0) = ψ(Ω^(Ω^Ω*2+1))
(2[0;]0) = ψ(Ω^(Ω^Ω*3))
(0[0;]1) = ψ(Ω^(Ω^(Ω+1)))
(0[1;]0) = ψ(Ω^(Ω^(Ω*2)))
(0[0,0;]0) = ψ(Ω^(Ω^(Ω^2)))
(0[1,0;]0) = ψ(Ω^(Ω^(Ω^2+Ω)))
(0[0,0,0;]0) = ψ(Ω^Ω^Ω^3)
(0[0[0;]0;]0) = ψ(Ω^Ω^Ω^Ω)

Re: Your number is, in fact, not bigger!

Posted: Sun Feb 22, 2015 6:27 pm UTC
by mike-l
Hey WarDaft, I wonder if you might help me understand the psi function. I'm ok up to Psi(Omega^Omega) = Gamma_0.

The wiki page says Psi(Omega^Omega +Omega^alpha) = phi_(Gamma_0 + alpha)(0) and I don't follow this part. It's not a huge deal because by Omega^Omega *2 I agree again, but nonetheless I'm confused at this step

Let's just make alpha 1, so Psi(Omega^Omega + Omega). For sanity I'll abbreviate Psi = P,Omega = W, Gamma = G

P(W^W) = G_0
So P(W^W + 1) should be just what I can make by exponentiation with G_0, ie e_(G_0+1), and in general P(W^W+a) will be e_(G_0+a) until this reaches a fixed point. Since G_0 is a fixed point of all relevant functions here, this should get stuck at zeta_(G_0 + 1). It shouldn't come unstuck until we reach P(W^W + W), so I only get this equal to phi_2(Gamma_0 + 1), which is much less than phi_(Gamma_0 + 1)(0)

Re: Your number is, in fact, not bigger!

Posted: Sun Feb 22, 2015 6:34 pm UTC
by mike-l
A follow up question that may explain my confusion. Does C(alpha) contain any countable ordinals greater than Psi(alpha)? That is, are there gaps in the countable ordinals constructed?

Re: Your number is, in fact, not bigger!

Posted: Sun Feb 22, 2015 8:07 pm UTC
by WarDaft
No, there aren't any gaps in the countable ordinals constructed (unless you choose omega to be countable, but that's just nitpicking)

P(W^W+W) is indeed phi_2(G_0+1)
P(W^W+W*2) is phi_2(G_0+2)
P(W^W+W*a) is phi_2(G_0+a)
The fixed points of phi_2(G_0+a) is phi_3(G_0+1)
P(W^W+W^2) is phi_3(G_0+1) and P(W^W+W^2*a) = phi_3(G_0+a)
Following from this, P(W^W+W^3*a) = phi_4(G_0+a)
P(W^W+W^G_0) = sup{a->G_0 in phi_a(G_0+1)} = phi_G_0(1)

However, we can shove all these values into the function again.

P(W^W+W^(G_0)*a) = phi_G_0(a)
And so P(W^W+W^(G_0+1)*a) enumerates the fixed points of phi_G_0, so it is the function phi_(G_0+1).
By the time we reach P(W^W+W^(G_0*w)), it aligns with the generalization the Wikipedia article states (as G_0+G_0*w = G_0*w), but before that, it's off by a difference of G_0.

Re: Your number is, in fact, not bigger!

Posted: Sun Feb 22, 2015 8:16 pm UTC
by mike-l
Alright. That aligns with my understanding

Re: Your number is, in fact, not bigger!

Posted: Sun Feb 22, 2015 8:23 pm UTC
by WarDaft
Hmm, I think I also just noticed that I got part of my notation switched around from what I originally intended to go with. I'll fix it when it becomes relevant.