Talk: TREE sequence

Archive 1

should we redirect TREE(3) to TREE sequence?

discuss here. i think yes. Cookiefonster (talk) 01:31, February 17, 2015 (UTC)

I saw that for example Fish's numbers has been added to the full list twice. So, could other values of TREE(k) be in the list? I mean there certainely is values of k at wich TREE(k) would overpower for example Meameamealokkapoowa Oompa of Loader's Number, that would be cool to add a such number 

Fluoroantimonic Acid (talk) 18:52, June 4, 2015 (UTC)Trifluoromethanesulfonic Acid

TREE(3) is the only value that is particularly notable, since it's the smallest nontrivial value of TREE(n) - TREE(1) and TREE(2) are 1 and 3 respectively. the next values aren't really in a different realm of numbers from TREE(3) if i'm not mistaken. the fish numbers are different, since each of them are really separately devised values and not just different outputs of a single function. Cookiefonster (talk) 19:24, June 4, 2015 (UTC)

Another thing is that other specific values of TREE(n) have never been (afaik) considered, so there is no reason to put different values of TREE into the list. LittlePeng9 (talk) 20:53, June 4, 2015 (UTC)

\(ACA_0+\Pi_2^1-BI\)

"How strong is \(ACA_0+\Pi_2^1-BI\)?" In another word, what's the least ordinal \(\alpha\) such that \(f_\alpha(n)\) eventually outgrows all functions provably recursive in \(ACA_0+\Pi_2^1-BI\)? {hyp/^,cos} (talk) 03:07, July 7, 2015 (UTC)

Let$\omega[n] = \Sigma(n)$. Now$f_\omega(n) [>] \Sigma(n)$ and since$f_{\omega }$ is above recursive, the answer is$\alpha = \omega$. To see that it's least such ordinal, observe that all ordinals less than$\omega$ are natural numbers and FGH for them doesn't depend on FS.

I think more interesting question would be: "What's the least ordinal existence of which cannot be proved by  $ACA_0+\Pi_2^1-BI$?"

Ikosarakt1 (talk ^ contribs) 12:14, August 2, 2015 (UTC)

Finding the PTO of \(ACA_0+\Pi_2^1-BI\) is a research-level problem for sure. -- vel! 13:20, August 2, 2015 (UTC)

I would guess that the PTO of \(ACA_0+\Pi_2^1-BI\) is \(\theta(\Omega^{\omega},0)\), but I do not have a reference for that. Deedlit11 (talk) 06:33, August 5, 2015 (UTC)

Where does TREE(3) fit into the heirarchy on Bowers' infinity scrapers page?

Jonathan Bowers (Hedrondude), best known for the BEAF notation has a webpage which expands upon the array notation to define large numbers: http://www.polytope.net/hedrondude/scrapers.htm

It's clear from reading the available online information that TREE grows more slowly than Rayo's and other functions which are arbitrarily defined as largest number that is theoretically writeable using an n-state Turing machine or a mathematical language with n symbols.  So, TREE(3) is smaller than Oblivion, Rayo's number, etc.  But what about large numbers that have been written clearly, and not based on a linguistic algorithm?  

I have read that TREE(3) is smaller than Loader's number .  Loader's number is computable; the README for Loader's code states that this is because because the Calculus of Constructions is not Turing-complete, so the program will eventually terminate.  But Loader's number is still defined using a concept of "The largest number that can be stated in the language L in less than N symbols", where, in this case L is the Calculus of Constructions.  Using "less than N symbols" in the definition of a function allows for larger numbers than any other known technique.

But what about numbers that use an already-defined mathematical language (like BEAF) and a fixed number of symbols?  Numbers such as meameamealokkapoowa oompa?  Is TREE(3) smaller than meameamealokkapoowa oompa, or any other number defined using an expansion of up-arrow notation without the "less than N symbols" trick?  I suspect that TREE(3) is smaller than oompa, but since I have trouble reading the notation of the fast-growing heriarchy, I would be curious to know approximately where it fits into Bowers' infinity scrapers page.  For example assuming meameamealokkapoowa oompa beats it, is TREE(3) also smaller than a Gongulus?  Bowers does not resort to using the phrase "defined in fewer than N symbols" until the definition of Oblivion at the very bottom of the page.

BEAF did not well-defined above the tetrational arrays. AarexWikia04 - 01:26, October 1, 2016 (UTC)

I think TREE(3) can be expressed in Sublegion BEAF \(\841 (Talk)\) 12:27, October 13, 2016 (UTC)

I can argue with AarexWikia04, saying that pentational arrays are also well-defined with work of Deedlit11 and Ikosarakt1. So, I dunno if pentational arrays should beat TREE (3), and I think that they don't. —Preceding unsigned comment added by Tetramur (talk • contribs) 11:23, May 2, 2019 (UTC)

Value of Weak Tree Function

I think that this section:

A larger lower bound has since been found. Define \(\text{tree}_2(n)=\text{tree}^n(n)\) and \(\text{tree}_3(n)=\text{tree}_2^n(n)\). Then \(\text{TREE}(3) > \text{tree}_3(\text{tree}_2(\text{tree}(8)))\). The latter expression is equal to tree^{treetree...tree(n)...(n)(n)}(n) with n layers, where \(n = \text{tree}^{\text{tree}(8)}(\text{tree}(8))\).

is more accurately (and helpfully) written as this:

A larger lower bound has since been found. Define \(\text{tree}_2(n)=\text{tree}^n(n)\) and \(\text{tree}_3(n)=\text{tree}_2^n(n)\). Then \(\text{TREE}(3) > \text{tree}_3(\text{tree}_2(\text{tree}(8)))\). The latter expression is equal to tree^{treetree...tree(n)...(n)(n)}(n) with \(n^2\) layers, where \(n = \text{tree}^{\text{tree}(8)}(\text{tree}(8))\).

I notice that \(\text{tree}_3(n)\) creates a tower of n \(\text{tree}_2(n)\)s, each level of which creates a tower of \(\text{tree}_2(n)\) \(\text{tree}\)s.  The pattern shows that \(\text{tree}_m(n)\) creates a tower of \(n^{m-1}\) \(\{tree}\)s following FGH's \(f_2\)(n) growth rate (in the number of \(\text{tree}\) iterations.

24.163.58.140 13:01, December 24, 2016 (UTC)

tree_m(n) equals tree_m-1ⁿ(n); if you then expand the latter expression out in terms of tree_m-2(n), you get a tower of functional exponents of height n. Expanding it further to tree_m-3(n) would get you an ENORMOUS tower of functional exponents, much higher than n^2 or n^3; the actual height of the tower would be the previous exponential tower of tree_m-2(n), but height one less (so n-1). The value of that exponential tower of n-1 tree_m-2's would get you how high the exponential tower of tree_m-3 has to be. So things are a little more complicated. Deedlit11 (talk) 14:30, December 25, 2016 (UTC)

How much larger is TREE(4) than TREE(3)?172.58.107.8 01:06, January 26, 2017 (UTC)

Approximately TREE(4) larger 217.194.207.20 09:43, January 27, 2017 (UTC)

Only TREE(3)?

Why not TREE(4) or even TREE_{TREE...(TREE10¹⁰⁰(1000)}(1000)? (TREE_n+1(k)=TREE_n^k(k)) 80.98.179.160 19:39, November 14, 2017 (UTC)

One can of course talk about the values of TREE(n) for larger inputs. TREE(3) was notable in that it was the first value which jumped up really high: TREE(1) = 1, TREE(2) = 3, and TREE(3) requires going beyond the Small Veblen Ordinal to bound it in the fast-growing hierarchy using a reasonable number of symbols. Deedlit11 (talk) 03:43, November 17, 2017 (UTC)

TREE(3) hits mainstream media!

Popular Mechanics recently published an article on TREE(3) here: http://popularmechanics.com/science/math/a28725/number-tree3 (note: partially paywalled) --Ixfd64 (talk) 06:00, January 15, 2018 (UTC)

TREE(3) had already hit mainstream media in my opinion after the Numberphile video, but it's nice to know it is sinking it's roots into other places, (pun intended). Edwin Shade (talk) 16:12, January 15, 2018 (UTC)

Growth rate of TREE function

It's solved here, page 39.

Definitions: Labeled tree A is homeomorphically embeddable into B if there exists such an injection f from nodes of A to nodes of B that

1. For any node s and t of A, f(the nearest common ancestor of s,t) is the nearest common ancestor of f(s), f(t).
2. For any node t of A, (the label of t, the label of f(t)) is "in certain ordering".

If the labels of nodes are in a well-partial ordering (i.e. the "in certain ordering"), then the trees are also in a well-partial ordering by homeomorphically embedding. Further, if the labels of nodes have order type \(\alpha\), then the corresponding trees have order type \(\le\theta(\Omega^\omega\alpha,0)\).

For TREE(n), the labels can be 1,2,...,n, and "in certain ordering" means "equal", so the labels form a well-partial ordering of type n. So the labeled trees have order type \(\le\theta(\Omega^\omega n,0)\). Here shows an ordering of \(\theta(\Omega^\omega n,0)\), so \(\text{TREE}(m,n)\approx f_{\theta(\Omega^\omega m,0)}(n)\) and \(\text{TREE}(n)\approx f_{\theta(\Omega^\omega\omega,0)}(n)\).

Now we can say about upper bounds of TREE(3), e.g. \(f_{\theta(\Omega^\omega\omega,0)}(3)\). {hyp/^,cos} (talk) 04:03, May 31, 2018 (UTC)

I have discussed it with Plain'N'Simple on the estimation using the upperbound of the ordinal types, but could not derive the upperbound of TREE(3) in your way. How could you verify the upperbound of the value from the upperbound of the ordinal types? Please read issues in the thread here starting from my comment "That function employs the 2-adic encoding...". The upperbound of the ordinal types do not give even an upperbound of the growth rate using a canonical system of fundamental sequences. If you have some unwritten reasoning, please tell me. If you just mistook the argument, please clarify it since there are many reasonless statements on the upperbound of TREE(3), which might have been spread from this community.

p-adic 04:13, October 25, 2019 (UTC)

Weak tree(3) bound by mgiroux and Deedlit

It origins here, but I have calculated again and result a different value.

4. (()()())

5. ((()())())

6. ((((()()))))

7. ((((())(()))))

8. ((((((())))()))) [Let tree 1 = () and n+1 = (n), then this can be written as (((4 1)))]

9. (((3 1)))

10. (((2 1)))

11. (((1 1)))

12. ((5 5))

13. ((7 4))

16. ((4 4))

17. ((12 3))

26. ((3 3))

27. ((23 2))

48. ((2 2))

49. ((46 1))

94. ((1 1))

95. (47 47)

96. (49 46)

99. (46 46) [From (47 47) to here it takes 4 steps]

100. (54 45)

109. (45 45) [From (46 46) to here it takes 10 steps]

110. (65 44)

131. (44 44) [From (45 45) to here it takes 22 steps]

Generally, from (48-n 48-n) to (47-n 47-n) it takes \(3\cdot2^n-2\) steps, so from (47 47) to (1 1) it takes \(\sum_{n=1}^{46}(3\cdot2^n-2)=422212465065886\) steps.

422212465065981. (1 1)

422212465065982. 422212465065982

844424930131963. 1

So my calculation results tree(3) ≥ 844424930131960. {hyp/^,cos} (talk) 15:02, October 29, 2018 (UTC)

What is the first source?

The article refers to the following two sources, but they are obviously non-first sources. In the archive page of this talk page, Ikosarakt1 states that Chapter 11 of "ENORMOUS INTEGERS IN REAL LIFE" might be the first source, but it seems to be a different sequence. Could someone tell us the first source? Otherwise, I will clarify in the article that there is no source.

p-adic 03:49, March 26, 2020 (UTC)

Do you mean this article? I'm not confident about how does this TR(n) compare with TREE(n), but if they're different, then possibly there is no explicitly given source by Friedman himself. At least, the Wikipedia article doesn't have the source for the definition. It's okay to clarify it, in my opinion. Ikosarakt1 (talk ^ contribs) 06:38, March 26, 2020 (UTC)

Thank you. Exactly, I meant it. In that case, is there any reason why "TREE" in the sense of Friedman is believed to be defined as the one in the article? In other words, is there a possibility that what Friedman calls "TREE" is different from the TREE function in the article? In that case, I am afraid that there is no justification of the results (e.g. the termination, the growth rate, and the specific estimations) of TREE.

p-adic 06:52, March 26, 2020 (UTC)

I think that this is the first source. Then the definition in the article is correct as long as the ambiguous "homeomorphically embeddability" means the other condition in the source, and I guess that the description in wikipedia (stating TREE is TR) is non-trivial because TR refers to wider cases. I will update the article later.

p-adic 07:12, March 26, 2020 (UTC)

Regarding the "proof"

It seems we have a full-on edit war here.

Please debate the validity of the "proof" here rather than repeatedly trying to add it to the page.

And to P-bot: You're not doing anything to dispell Goucher's accusations of being "surprisingly harsh". The tone you are adopting when editing this article comes off as extremely rude, at least to me. You are automatically discounting this proof even though the guy who made it presumably has some idea of what he's talking about. And even if it were valid, this type of petty complaining belongs somewhere other than a mainspace article.

continued attempts to change this article will result in a temporary protection due to repeated edit warring.

Username5243 (talk) 23:59, July 24, 2020 (UTC)

I wrote a question to you in this talk page related to this topic, but you deleted it. How can I discuss this topic, even if you delete my opinion here? Please revert the deletion. Also, please answer my question. It is your turn.

p-adic 00:03, July 25, 2020 (UTC)::

:: P-adic. Please. With all due respect, what you're doing has to stop. It's just not right for a wiki. This is a wiki article, an article used for information that people might see when they look up "TREE(3)". You really shouldn't put that type of things here. This isn't peer review against another person's paper, it's a site for information. The article could inlcude something like "this proof, however, isn't completely rigorous", not something that outright calls out another person for being wrong. There's more we want to say. We can talk about this somewhere else.

Antimony Star (talk) 00:10, July 25, 2020 (UTC)

What Username did is worse. He removed what I asked in order to discuss this. Do you think that it is allowed?

For the completeness, please read the source. It is not something "not completely rigorous", but is completely wrong, as I clearly pointed out. It is something like "Since f_ω(n) > BB(n) with respect to ω[n] = BB(n), f_{ω+1} in Wainer hhierarchy outgrows BB". We should not hide the reproducible reason. You can change how to point out the obvious errors, but hiding the reasons by saying "not completely rigorous" is dishonest. We need to feedback the author because he is believing that he is completely correct, as he stated "surprising".

p-adic 00:21, July 25, 2020 (UTC)

There are wasys to criticize the proof without calling out this guy in a MAIN SPACE ARTICLE. It's on someone's blog right? Can't you leave a comment on that blog with your objections?

Anyway, the mainspace of the wiki is not a place for personal attacks of this sort. The vast majority of people trying to learn about TREE(3) don't care if the proof of how strong it is is fully rigorous, and care even less about specific reasons why which they might not understand. Username5243 (talk) 00:55, July 25, 2020 (UTC)

"The vast majority of people trying to learn about TREE(3) don't care if the proof of how strong it is is fully rigorous." I don't think so. Actually if the proof is not valid it is not worth mentioning in the main article. I suggest writing the objection in p進大好きbot's blog post, and cite the blog post in this article as the objection to the proof. 🐟 Fish fish fish ... 🐠 02:01, July 25, 2020 (UTC)

Hello. To put it short, my opinion is that there was indeed some aggressive (or could be taken aggressively) wording. However, I also think that if there is an error in the proof, then that should not be ignored either, and it should not be presented as complete truth. I do not support reversion by it being "a perfectly good proof" or the edit being "petty". P.S. While I was writing this, Kyodaisuu posted a comment. I like their idea. --πaruyoko (Talk) 02:05, July 25, 2020 (UTC) Fixed few double negatives. --πaruyoko (Talk) 02:45, July 25, 2020 (UTC)

@Username

You have not answered the reason why you removed my opinion in this talk page. Please honestly answer it, and stop such an inappropriate attitude as I required you so many times.

Also, you made mistakes:

1. "There are wasys to criticize the proof without calling out this guy in a MAIN SPACE ARTICLE.": The one who firstly mentioned the guy is not I. Someone wrote his statement in the main space, and I added the lack of a proof.
2. "Can't you leave a comment on that blog with your objections?": It is meaningless in this case, because the problem is that the main space refers to his wrong argument as if it were correct.
3. "the mainspace of the wiki is not a place for personal attacks of this sort.": Pointing out actual errors in proofs is not a personal attack, while removing an opinion without any justification and ignoring a requirement to explain the reason is actually a personal attack.
4. "The vast majority of people trying to learn about TREE(3) don't care if the proof of how strong it is is fully rigorous,": I do not think so. Further this is not an issue of the full rigorousness, as I clarified. It is not a proof which is not completely regorous, but is a sentence which is completely wrong. As I wrote, please take a look at the source.
5. "and care even less about specific reasons why which they might not understand.": It cannot be a reason to allow incorrect arguments as if they were correct. If you believe that we do not need proofs, then just delete the guy's arguments, because they are know to be incorrect.

@fish

You are right. But I guess that Username will delete my blogpost in the same way without explaining reasons.

p-adic 02:52, July 25, 2020 (UTC)

You could post on Japanese Gwiki. We don't blame on posting in English. 🐟 Fish fish fish ... 🐠 03:01, July 25, 2020 (UTC)

Sure. But the serious issue is that many people believe this wiki, as they believed the well-definedness of and the correctness of analyses by BEAF, UNOCF, and so on. (Uh, you mean that I can refer to my Japanese blog post in the main space of this wiki, right? It might be true.)

p-adic 03:11, July 25, 2020 (UTC)

Yes. I think citing blog post in Japanese gwiki is allowed. If it is written in English there is no reason lo deny it. 🐟 Fish fish fish ... 🐠 03:19, July 25, 2020 (UTC)

It makes sense. (It does not change the fact that Username's attitude is inappropriate, though.)

p-adic 03:23, July 25, 2020 (UTC)

Now we have a direction given by Fish. How about only mentioning the incorrectness of the "proof", and put a reference to a blog post on the precise errors? Also, Username does not like to mention the original poster. Therefore how about replacing the name of the guy by "some googologist"? Also, it is meaningless to mention the exact statement by the original poster now. Therefore how about write something like "Some googologist write statements without proofs, and they were believed in this community. Due to the issue, the person updated his or her blog post, and submitted a new blog post including an explanation on 25/07/2020. However, it turned out to be incorrect. In that way, it is really difficult to estimate TREE and tree by fast-growing hierarchy."?

Please give me opinions. Also, please do not just remove my comment. Also, please do not revert my edit if you have no disagreement.

p-adic 11:44, July 25, 2020 (UTC)

I've decided to add a compromise to the article. Moooosey (talk) 13:02, July 25, 2020 (UTC)

I made a citation to his blog post and changed the wording because citation should be accurate. 🐟 Fish fish fish ... 🐠 13:19, July 25, 2020 (UTC)

@Moooosey

Please wait, and please do not ignore the arguments above. I am kindly asking you to give us opinions. The original poster did nothing bad against this community, but you are the one who added the description to the article by clarifying a "perfectly good proof" and insulted corrections, which costed time and effort, as "needlessly petty" thing which can be freely deleted. Please inform of what you think about my suggestion above. Please be more careful, because you are the very trigger of the arguments.

p-adic 13:24, July 25, 2020 (UTC)

Is there no opinion from Moooosey? Then I will remove the original poster's name and the "results" tomorrow as I suggested above. Please do not revert it if you will not give us opinions.

p-adic 22:58, July 27, 2020 (UTC)

How has it come to this...

Clarification: now I'm not good with wikia's inconvenientness so idk. I think anyone can delete a page, and anyone can reverse it unless Username put a extra setting or something.

Anyway Username said he tried to reverse something, and reversed the question you put on the talk page. Anyone can reverse that, but we already put too many things on this page so it would take deleting everything we said after that to reverse it.

So. In the beginning I already said this before: we need some way to tell Adam Goucher about this. I think he already knows. Anyway I always wanted to fix the stuff to my own ability, and explain it well enough so that it's factually correct. The wiki is about facts (for example, someone tired to prove TREE(3)>stuff but it's validity is doubted), it doesn't need to be too detailed in why we think the proof is wrong.

You could link stuff in the Japanese wiki, also I think even if username deletes a blog post you make you or anyone can simply reverse it?

Antimony Star (talk) 22:19, July 28, 2020 (UTC)

Maybe you have not read the arguments above. We have already discussed how to put the source. The remaining issue is in my comment in front of yours, i.e. whether Moooosey, who is the one who started this trouble, agrees with the decision.

> You could link stuff in the Japanese wiki, also I think even if username deletes a blog post you make you or anyone can simply reverse it?

No way. A page deleted by an admin cannot be reverted by a usual user... Admins actually deleted pages several times even after discussions/votings to keep them. Therefore it is actually the case. They ignore our decision when they do not like it. It is awfully bad. Therefore we should blame the attitude. Could you understand it?

p-adic 00:46, July 29, 2020 (UTC)

I have done, because there is no disagreement. If anybody have a new opinion on this issue, please write it down here before ignoring the argument by editing the article.

p-adic 02:12, July 30, 2020 (UTC)

Generalized tree function idea

Hi. I'm interested in trying to bridge the gap between functions like TREE(n) on one hand and functions like n! on the other, because I feel like it might help me and others understand those enormously fast growing functions a little bit better.

One idea for that is the generalized tree function GT(c, f), where the kth rooted tree may use up to c colours and up to f(k) nodes. GT(n, k) is the TREE(n) function, while GT(1, k+n) is the tree(n) function.

Suppose, for example, that f(1)=4, f(2)=f(3)=5, f(4)=f(5)=f(6)=f(7)=6. Then the first tree may contain up to 4 nodes, the second and third may contain up to 5, and the next four trees may contain up to 6 nodes. All else is the same as in the main tree(n) and TREE(n) functions.

If the function f grows too slow, it doesn't get very interesting: For any given integers c and s, there is a fixed number N of rooted trees with at most c colours and at most s nodes. If f(k) <= s for all k from 1 to N, then N is an upper bound for GT(c, f).

How fast does GT(1, ln(n*k)) grow?

How fast does GT(1, ln(n)*ln(k+1)) grow?

How fast does GT(1, ln(k+1)^ln(n)) grow?

How fast does GT(1, (n*k)^0.01) grow?

How fast does GT(1, (n*k)^0.5) grow?

Can we find a generalized tree function that grows about as fast as the Goodstein function?

And finally, for those who are still looking for bigger and bigger functions: How fast does GT(n, TREE(k)) grow? Iwer Sonsch (talk) 02:45, 9 December 2022 (UTC)

The growth of TREE function mainly comes from the amount of available colors. In GT(c,f(n+k)), if f has growth rate \(\approx f_\alpha\) in FGH, then GT(c,f(n+k)) has growth rate \(\approx f_{\alpha+\psi(\Omega^{\Omega^\omega\cdot c})}\). If f grows slower than TREE, GT(c,f(n+k)) (where c is a constant independent of n) will still grow slower than TREE. {hyp/^,cos} (talk) 00:03, 10 December 2022 (UTC)

And, GT(c,TREE(n+k)) is a nontrivial case of the function f, with growth rate \(\approx f_{\psi(\Omega^{\Omega^\omega\cdot\omega})+\psi(\Omega^{\Omega^\omega\cdot c})}\), and then GT(n,TREE(k)) has growth rate \(\approx f_{\psi(\Omega^{\Omega^\omega\cdot\omega})\cdot2}\), exactly "times 2" from TREE in the scale of FGH. {hyp/^,cos} (talk) 00:07, 10 December 2022 (UTC)

I mean sure, it grows slower than TREE. That's the point. I'm trying to bridge the gap from slow-growing functions towards tree and TREE, so if I was overshooting TREE then that would be quite the issue indeed.

But I've noticed that I don't know how to evaluate these generalized (and slower growing) tree functions, e.g. I don't know how to find lower and upper bounds for even something as basic as GT(1, 4+ln(k)). Is there a good way to prove whether the functions I named grow faster or slower than say \(\approx f_3\), \(\approx f_\omega\), \(\approx f_(epsilon_0)\) etc.?

Also, I'm not sure if I was clear on what GT does. GT(c, f(k)) is a constant that merely uses the values of f to determine how many nodes are allowed for each tree. So please be careful, GT(c, TREE(k)) doesn't grow at all because it is independent of n, and GT(c, TREE(n)) defeats the point of the generalized tree function because the number of available nodes doesn't depend on the index of the tree. A generalized tree function must depend on both k (the tree index) and n (the input along which GT grows) in order to be a sensible function. Iwer Sonsch (talk) 02:29, 10 December 2022 (UTC)

As long as f grows slower than tree, GT(1,f(n+k)) doesn't make significant change from tree. But once f is tree, GT(1,f(n+k)) will immediately jump into the growth of \(\approx f_{\psi(\Omega^{\Omega^\omega})\cdot2}\).

The same jumping phenomenon happens on 2-color "tree", 3-color "tree", ..., and TREE. {hyp/^,cos} (talk) 06:26, 10 December 2022 (UTC)

That's cool. I'm more interested in what happens when f(n+k) grows slower than identity(n+k) though (rather than faster than identity(n+k) but slower than TREE(n+k). As I said, I'm trying to bridge the enormous gap between the weak tree function and slower-growing functions like n!. Iwer Sonsch (talk) 16:56, 10 December 2022 (UTC)

TREE[2]

Using the alternative notation, what's wrong with this sequence:

1. ()
2. [[]]
3. [()()]
4. []

? ——SonataGreen (talk) 18:55, 6 March 2023 (UTC)

The first tree is embeddable in the third tree: [()()] --Naruyoko (Talk) 21:08, 6 March 2023 (UTC)

Section: Definition

Upon reading the definition of the TREE[n] function, I find some of the wording a bit problematic. Particularly, there are a lot of terms used that aren't defined on the wiki, nor are there any links connecting the terms to a source containing their definitions. I suggest rewriting it, or even just adding an alternative definition, so that the definition is more straight forward. When discussing TREE[n] in a graph theory course at my school, it was introduced in a much more convincing manner than is suggested in this article.  nnn6nnn likes this. (talk) 05:38, 8 April 2024 (UTC)

The labeled tree is linked to a Wikipedia article about Tree (graph theory), which includes an explanation of the basic concept of a tree. Therefore, readers can refer to this article on Wikipedia for information. I believe this is sufficient. However, it might be beneficial to also reference a notable textbook that explains this basic concept. While someone could write a page defining such a basic concept, if it is not superior to the Wikipedia article, there is little point in doing so. I do not think it is advisable to provide an alternative definition. Such an endeavor could be undertaken in a personal space, such as a blog post, and we could link to such a blog post if it proves to be helpful. 🐟 Fish fish fish ... 🐠 05:56, 8 April 2024 (UTC)

I see and understand. Perhaps I might consider doing this if I begin doing more with googology in the near future. :)

 nnn6nnn likes this. (talk) 06:40, 8 April 2024 (UTC)