View source for Talk:Knuth's up-arrow notation

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==Untitled==
This could do with conversion to use [[Wikipedia:TeX notation]]

I'll get right on it! [[User:Dysprosia|Dysprosia]]

Can someone move this to "Knuth's up-arrow notation", please?

If it's gonna be moved, I'd vote for [[Knuth arrow]]. BTW I'd removed Ackermann function since Ackermann page no longer refers to it. [[user:kwantus|Kwantus]]

''use Knuth arrow for relatively small numbers'' LMAO! =) [[user:kwantus|Kwantus]]

It was either that or ''relatively small large numbers''! :) (Maybe smaller magnitude...) [[User:Dysprosia|Dysprosia]] 05:49, 30 Aug 2003 (UTC)


I tried to set it so that 3(arrow arrow)5 was shown as also being equal to 3^(3^7625597484987), but couldn't get the notation right. Can someone fix that? [[User:DragonflySixtyseven|DS]] 14:15, 7 May 2005 (UTC)

:Done. - [[User:Gauge|Gauge]] 23:03, 2 October 2005 (UTC)

== 3<sup>27</sup>=27<sup>3</sup> ==

Why is <math>3^{3^3}</math> = 3<sup>27</sup>, not 27<sup>3</sup>? Thanks, --[[User:Abdull|Abdull]] 23:12, 30 September 2005 (UTC)

:27<sup>3</sup> = (3<sup>3</sup>)<sup>3</sup> = 3<sup>9</sup> &ne; 3<sup>27</sup>. Writing the exponentials as <math>3^{3^3}</math> = pow(3, pow(3, pow(3, 1))), we see that we evaluate these functions from the inside out, so pow(3, pow(3, pow(3, 1))) = pow(3, pow(3, 3)) = pow(3, 27) = 3<sup>27</sup>, rather than 27<sup>3</sup>.  - [[User:Gauge|Gauge]] 22:59, 2 October 2005 (UTC)

::Very big difference... 3<sup>27</sup> = 7,625,597,480,000 &ne; 27<sup>3</sup> = 19,683 [[User:Frozen Mists|Frozen Mists]] 21:59, 28 June 2007 (UTC)

Left-associative exponents and higher operators ARE used in some applications (although I can't site one offhand), but if you define higher operators based on left-association, you don't get essentially new operators, since in that case a^^b would = <math>a^{(a-1)^b}</math>, rather than a^a^...^a^a (b a's).  <small>—Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/24.165.184.37|24.165.184.37]] ([[User talk:24.165.184.37|talk]]) 23:12, 11 March 2008 (UTC)</small><!-- Template:UnsignedIP --> <!--Autosigned by SineBot-->

== Study (generalization to non-integer arguments) ==

As I said in the article on the closely related Hyper Operator, I think it would beinteresting to investigate the notation for a non-natural number of arrows, equivalent to: <math>a^{(n)}b|n\notin\mathbb{N}</math> I've already defined over all integers n, but I think it would be interesting to expand this to all reals, or possibly even all complex numbers.
[[User:He Who Is|He Who Is]] 01:02, 23 April 2006 (UTC)

:Sounds like hard work!  I hope you can do it. [[User:Kaimiddleton|Kaimiddleton]] 07:54, 22 April 2006 (UTC)

It can't be done unfortunately, at least not with any degree of rigour or consistency, because unlike addition and multiplication exponentiation is *neither associative nor commutative; hence in general a^^(b^^c) =/= (a^^b)^^c and a^^b =/= b^^a. Hence you get anomalies such as 2^^(1/3) < 2^^(2/7) even though 1/3 > 2/7.

A few people have worked on the problem and found partial solutions, and these may be worth looking into, but a definitive answer is always likely to prove elusive.
 
*I am indebted to Dr. Roger Wheeler of Leicester University who pointed out to me a few years back that this is the reason any such attempts will fail.

[[User:Meltingpot|Meltingpot]] ([[User talk:Meltingpot|talk]]) 16:00, 6 September 2008 (UTC)

:This argument against noninteger arguments (sic/lol) seems fallacious. Usual exponentiation does not obey the rule (a^b)^c =?= a^(b^c) nor a^b =?= b^a, yet exponentiation is well defined for all positive real numbers (and more, specifically: complex numbers that are not negative reals).
:So, these apparent contradictions based on the premise that we have to have rules which we can't and therefore don't expect. I think various (e.g., "log-linear") interpolations could be possible - e.g.:
:* between 2^^2 = 4 = 2^2, 2^^3 = 16 = 2^4, 2^^4 = 65536 = 2^16 and 2^^5 = 2^65536, it seems quite natural to have
:* 2^^2.5 = 2^3, 2^^3.5 = 2^8 = 2^2^3 and 2^^4.5 = 2^256 = 2^2^8; with obvious generalisation
:* 2^^(2+x) = 2^(2+2x), 2^^(3+x) = 2^2^(2+2x), 2^^(4+x) = 2^2^2^(2+2x), etc., when 0 <= x <= 1.
:(I don't particularly like this idea, which came first to my mind. But the fractional derivatives are defined similarly, by separating the integer part and the fractional part. I guess that it should not be difficult to find a slightly "smoother" generalisation...) &mdash; [[User:MFH|MFH]]:[[User talk:MFH|Talk]] 22:31, 20 June 2024 (UTC)
::PS: a different, probably better generalisation comes to mind: append a final exponent 2^x where x is the fractional part: for x=0 it's 1 and does not change the value, for x=1 it's 2 and gives the next term in the series. This will obviously work for any other base (e.g., 3, 4, 5, ...) instead of 2. (It will *not* yield 2^^2.5 = 8 but rather 2^^2.5 = 2^2^sqrt(2) ≈ 6.3429... &mdash; [[User:MFH|MFH]]:[[User talk:MFH|Talk]] 22:46, 20 June 2024 (UTC)

== Formal definition ==

Okay, what's the formal definition of <math> a \uparrow^n b </math> when <math> n = 0 </math>? It was originally written as 1 in the formal notation, but as <math> ab </math> in the tables below. Now the line has been removed altogether, meaning the function is currently undefined for <math> n = 0 </math>. It should be either

:<math>
  a\uparrow^n b=
  \left\{
   \begin{matrix}
    a^b, & n \isin \{ 0, 1 \}; \\
    a\uparrow^{n-1}(a\uparrow^n(b-1)), & \mbox{otherwise}
   \end{matrix}
  \right.
 </math>

or 

:<math>
  a\uparrow^n b=
  \left\{
   \begin{matrix}
     ab, & n = 0; \\
     a^b, & n = 1; \\
    a\uparrow^{n-1}(a\uparrow^n(b-1)), & \mbox{otherwise}
   \end{matrix}
  \right.
 </math>.

[[User:Cbuckley|cBuckley]] ([[User talk:Cbuckley|Talk]] &#8226; [[Special:Contributions/Cbuckley|Contribs]]) 13:31, 21 May 2006 (UTC)

: The article had ''b''=0, '''not''' ''n'' = 0. As the article stands, the up-arrow is defined for ''n'' &ge; 1. [[User:Dysprosia|Dysprosia]] 13:40, 21 May 2006 (UTC)

:: Okay, my mistake. So what's with the first line in each of the tables? [[User:Cbuckley|cBuckley]] ([[User talk:Cbuckley|Talk]] &#8226; [[Special:Contributions/Cbuckley|Contribs]]) 14:26, 21 May 2006 (UTC)

I see no reason why it isn't defined for n = 0. The definition <math>a\uparrow^n b = a^{(n-2)} b</math> gives a perfect definition for all n &ge; 2; n <math>\in\mathbb{Z}</math>[[User:He Who Is|He Who I
s]] 20:24, 21 May 2006 (UTC)

: Well, what would be the point of that? The arrow notation should be used when it should be used, so there's no practical need to make such a definition when one can resort to conventional notation. [[User:Dysprosia|Dysprosia]] 22:35, 21 May 2006 (UTC)

I don't mean to say that there is a practical need for it, I merely mean to say that because of that definition, the Knuth up-arrow notation is defined for all n greater than of equal to -2.[[User:He Who Is|He Who Is]] 22:31, 22 May 2006 (UTC)

: I think you mean to say that the up-arrow is defined for all ''n'' &ge; 2: as it stands, the hyper operator isn't defined for negative ''n''. [[User:Dysprosia|Dysprosia]] 22:38, 22 May 2006 (UTC)

Actually I did mean negative two, because the standard notation is defined for all n greater than or equal to zero (or depending on your point of view, all integers), and <math>a\uparrow^n b = a^{n-2} b</math>
. But, as you said, anyone can resort to the standard notation if they have to, so there's no reason to argue this point.[[User:He Who Is|He Who Is]] 19:45, 23 May 2006 (UTC)

I haven't come across a source with a definition for zero or less ''n'' arrows.  There may be an inductive argument that <math>a\uparrow^0 b = a</math> if ''b'' is non-zero.  Unfortunately, this would need to be backed up by a citation.  We need to avoid original research.  --[[User:Ashawley|Ashawley]] ([[User talk:Ashawley|talk]]) 18:29, 28 March 2020 (UTC)

We currently have...
-------------------------
The general definition of the notation (by [[Recursion (mathematics)|recursion]]) is as follows (for integer ''a'', positive integer ''n'' and non-negative integer ''b''):
:<math>
  a\uparrow^n b=
   \begin{cases}
    a^b, & \text{if }n=1; \\
    1, & \text{if }b=0; \\
    a\uparrow^{n-1}(a\uparrow^{n}(b-1)), & \text{otherwise }
   \end{cases}
 </math>
-------------------------

I feel that this would look better as:
:<math>
  a\uparrow^n b=
   \begin{cases}
    1, & \text{if }b=0; \\
    a^b, & \text{if }n=1; \\
    a\uparrow^{n-1}(a\uparrow^{n}(b-1)), & \text{otherwise }
   \end{cases}
 </math>
So that the three expressions <math> 1, a^b, a\uparrow^{n-1}(a\uparrow^{n}(b-1)) </math> are in order of 'increasing complexity'.
Also, any tiny ambiguity about the value of <math> 0\uparrow 0 </math> would be immediately resolved. [[User:Clive tooth|Clive tooth]] ([[User talk:Clive tooth|talk]]) 17:14, 29 March 2020 (UTC)

== Bogus table row? ==

The first row in "Tables of Values" is for m = 0, but the definition above says that that number must be greater than or equal to one.  This seems inconsistent.

-- [[User:Offby1|Offby1]] ([[User talk:Offby1|talk]]) 00:05, 29 March 2010 (UTC)

: I agree.  Seems like there's no way to have zero arrows... --[[User:Ashawley|Ashawley]] ([[User talk:Ashawley|talk]]) 22:06, 19 March 2019 (UTC)

How much is <math>a\uparrow\uparrow 0</math>?  0 or 1 or another number? -- [[User:Athanatophobos|Athanatophobos]] August 4th 2016

:Well as usual it is up to us to decide what we want it to mean, but the sensible choice would be 1 to preserve the equation <math>a\uparrow\uparrow n=a\uparrow(a\uparrow\uparrow(n-1))</math>. [[User:Kestrelsummer|Kestrelsummer]] ([[User talk:Kestrelsummer|talk]]) 23:10, 30 October 2016 (UTC)

== OR for citation ==

I understand that this is [[WP:OR]], but it's the "proof" for the citation (or an estimate).

let's define T=SQRT(10). (T is approximately equal to 3.16, which is close enough to 3 for something like "moderately sized hard drives"). Using T as a base, 10 (in decimal) would be expressed as 100<sub>T</sub>. 100<sub>10</sub> is expressed as 10000<sub>T</sub>. The amount of space on a hard drive (d) it would take to store a number (using ASCII) on a hard drive is one byte per digit. For the powers of 10, let N be the power. the desired number to store is 10<sup>N</sup>. Expressed in N, the number of bytes it takes to store a number is N+1. If Expressed as base T, it would be 2N+1. Let's Examine T<sup>7625597484987</sup>. (I know it isn't equal to 3<sup>7625597484987</sup>, but comparatively, they're close). Such a beast would have 7625597484988 digits when expressed in Base T. As a result, it would have approximately 3812798742494 digits when expressed in base 10. assuming each digit takes one byte, the number can be stored in 3 812 798 742 494 bytes, or approximately 4 Terabytes (hard drive manufacturers terabyte). If using a binary storage mechanism (as might be used to most efficiently store any number, say as in a twos compliment storage), 3 decimal digits, can approximately be stored in 10 binary digits. at 8 bits to a byte, this would take approximately 1 588 666 142 705 bytes, or 1.5 terabytes. Moderately large hard drives these days are about 300gb, (large ones at 500gb), which is 12ish (ascii) or  5ish (binary) "moderately large hard drives"? [[User:Mckaysalisbury|McKay]] 06:38, 19 January 2007 (UTC)

== Error in table? ==

It seems to me that there is an error in the first table under table of values.  First, we have <math>2\uparrow\uparrow 6 = 2^{2^{65536}}\approx 10^{6.0 \times 10^{19,728}}</math> 

Then, in the very next table we have <math>2\uparrow\uparrow 7 = 2^{2^{2^{65536}}}\approx 10^{10^{6.0 \times 10^{19,728}}}</math> .

Assuming the first one is correct, shouldn't this be <math>2\uparrow\uparrow 7 = 2^{2^{2^{65536}}}\approx 2^{10^{6.0 \times 10^{19,728}}}</math>  since we are simply looking at 2 raised to the power in the previous column?  I'm not sure which of these is incorrect, but I thought I would just bring this to the attention of someone who is more certain. [[User:GreekHouse|GreekHouse]] 02:11, 12 October 2007 (UTC)

:Compared to the effect of the rounding error in the 6.0, replacing 2 by 10 at the bottom has a negligible effect: a factor a little more than 3 at the second level, i.e. a term 0.5 at the third level.--[[User:Patrick|Patrick]] 06:15, 12 October 2007 (UTC)

== Small issue on convention ==

This article starts with the example of multiplication. Although I think that the example of 3x2 is the correct one, it is not consistent with the general example ab (a copies of b instead of b copies of a). However the general trend in this article is of course b copies of a. So should we make the first example ba or change the explicit example to two copies of three, which is counter-intuitive? [[User:Kiliman|Paul]] ([[User talk:Kiliman|talk]]) 08:15, 25 January 2008 (UTC)

→==Numeration systems based on the hierarchy (+, *, ^, ^^, ^^^, ...)==

Knuth arrows denote the 3rd and higher operations in the hierarchy <math>( +, \ \times, \ \uparrow, \ \uparrow\uparrow, \ \uparrow\uparrow\uparrow, \ \dots)</math> which was introduced by Goodstein[1947] (he used other notation, but apparently also introduced the terms "tetration", "pentation", and so on).  I wrote a description of Goodstein's base-''b'' level-''k'' system of uniquely representing any nonnegative integer (it uses digits 0,...,''b''-1 and the first ''k'' operators in the hierarchy), but I wasn't quite sure of the best WP article in which to enter it.  Maybe it would be appropriate here (?), but because the family of [[hyper operator|hyper operations]] is precisely this hierarchy, I posted it over there, at [[Hyper_operator#Numeration_systems_based_on_hyper_operations]]. Nevertheless, I used mostly Knuth arrow notation in the examples. <br>--[[User:R.e.s.|r.e.s.]] ([[User talk:R.e.s.|talk]]) 21:23, 14 August 2008 (UTC)

== Googolplex ==

The article claims "Note: (10\uparrow)^k denotes a functional power of the function f(n) = 10^n (the function also expressed by the suffix -plex as in googolplex)." Now, sorry to be picky, but this needs a citation. I know a googol is 10^100 and a googolplex is 10^10^100, but this alone does not imply that "-plex" generally means 10^n. And there are no other number names ending -plex to use as data points. The words "googol" and "googolplex" were introduced by Kasner as described in the [[googol]] article, and should be regarded as indivisible [[morpheme]]s unless either (1) Kasner originally and explicitly intended "-plex" to be a generalisable suffix with this meaning, or (2) it has since come into widespread use with this meaning. Is either of those the case? I think not, but I'll put the question here in case anyone can answer more definitely. [[Special:Contributions/91.105.25.16|91.105.25.16]] ([[User talk:91.105.25.16|talk]]) 18:15, 14 January 2009 (UTC)

== Incorrect Information ==

The base-2 representations at the end of the article seem to be wrong.  Earlier in the same section it says that a base-b representation should only contain digits 0,1,...,b-1 and so a base-2 representation should only contain the digits 0 and 1 and yet all the representations contain a 2 digit.  I do not understand the representations enough to fix them, but I was able to understand enough to see this was wrong.  Could someone more knowledgeable fix this apparent inconsistency? --[[User:SurrealWarrior|SurrealWarrior]] ([[User talk:SurrealWarrior|talk]]) 17:01, 22 August 2009 (UTC)

:I think it's a matter of terminology, in the sense that although the "complete hereditary representation" of an integer in base ''b'' does involve ''b'', only {0,1,...,''b''-1} are called "digits". To some extent this can be compared to calling the expression 2*10<sup>2</sup> + 6*10<sup>1</sup> + 6*10<sup>0</sup> (which contains "10") the base-10 representation of the number usually written as 266. Probably this ought to be clarified in the article.<br>— [[User:r.e.s.|r.e.s.]] ([[User talk:r.e.s.|talk]])  13:27, 3 September 2011 (UTC)

== Up-arrow notation defined on ordinals? ==

The [[ε₀]] page uses Knuth's up-arrow notation extended to ordinals, but the definition for ordinals appears nowhere on this page, and perhaps it should. [[Special:Contributions/75.22.201.232|75.22.201.232]] ([[User talk:75.22.201.232|talk]]) 22:10, 24 April 2010 (UTC)

== Graham's Number too large? ==

The article currently reads:

''"Some numbers are so large that multiple arrows of Knuth's up-arrow notation become too cumbersome; then an n-arrow operator  is useful (and also for descriptions with a variable number of arrows), or equivalently, [[hyper operator]]s.

''Some numbers are so large that even that notation is not sufficient. [[Graham's number]] is an example."''

To my non-mathematician eye, this means hyper operators describe bigger numbers than Knuth's notation does, and Graham's number is even beyond the conventional scope of hyper operators. This seems to contradict the Graham's number page, which states, "it can be easily described by recursive formulas using Knuth's up-arrow notation or the equivalent, as was done by Graham."

Maybe this makes sense at some level, but to a non-mathematician like me this appears contradictory, and at the very least isn't clear enough.--[[User:Atkinson 291|Atkinson]] ([[User talk:Atkinson 291|talk]]) 04:40, 27 July 2012 (UTC)

== How do you say it? ==

What's the verbal way to express up-arrow notation? [[User: Kbog|'''<span style="color:black;">K</span>''']].[[User talk:Kbog|'''<span style="color:blue;">Bog</span>''']] 20:08, 22 April 2014 (UTC)
:There is a video of Ron Graham explaining Graham's (ie his) number, in which he pronounces it several different ways.  But they are all quite close to things like "three four arrows three" for <math>3\uparrow\uparrow\uparrow\uparrow 3</math>. [[User:Kestrelsummer|Kestrelsummer]] ([[User talk:Kestrelsummer|talk]]) 18:20, 30 October 2016 (UTC)

== The definition of up-arrow notation ==

Why <math>a\uparrow^n b</math> is <math>H_{n+2}(a,b)</math> instead of <math>H_n(a,b)</math> ?  <span style="font-size: smaller;" class="autosigned">— Preceding [[Wikipedia:Signatures|unsigned]] comment added by [[Special:Contributions/140.113.136.218|140.113.136.218]] ([[User talk:140.113.136.218|talk]]) 08:12, 13 May 2014 (UTC)</span><!-- Template:Unsigned IP --> <!--Autosigned by SineBot-->

        Here's a number for you:
10^10^10^100 ^(10^10^10^100) 10^10^10^100
          10^10^10^100 layers

== Severe error right at the beginning ==

before even the table of contents, we see https://wikimedia.org/api/rest_v1/media/math/render/svg/7bde1c371dfcf9daeaabd7284d046e92bebae399
which is soon followed by https://wikimedia.org/api/rest_v1/media/math/render/svg/7e2e1cbace2adb782550d7a5c3fffbc4e5add62d
THESE ARE INCONSISTENT.
If the first is correct, then we need to replace an 'n' with 'n-1' in the second.
If the second is correct, then an extra arrow is needed in the first.
I have added an  ERROR remark.  DO NOT REMOVE THE ERROR MARKER until the error is fixed.[[User:Jamesdowallen|Jamesdowallen]] ([[User talk:Jamesdowallen|talk]]) 12:41, 2 November 2017 (UTC)

Second definition is correct (note that the recursive definition works the same as the Ackermann function's recursive case). I have fixed the first definition to match the second. [[Special:Contributions/70.72.216.53|70.72.216.53]] ([[User talk:70.72.216.53|talk]]) 22:37, 23 November 2017 (UTC)

== 2^^^3 vs 2^^^4 ==

I think it would be better to have the opening example show direct correlations to different examples of Arrows.  Instead of using 2^4 2^^4 and 2^^^3 is it better to use 2 xUparrow 3 for all three to properly illustrate the difference between the 3 examples and to show the scale at which additional ^'s increase the results?  [[User:Apriestofgix|Apriestofgix]] ([[User talk:Apriestofgix|talk]]) 18:43, 28 November 2017 (UTC)

== associativity ==

The article claims that Knuth's arrow notation be right-associative, but I can't find any associativity rule in the cited links; they always use parentheses. Isn't it probably like exponentiation, where the writer has to specify the association (e.g. with parentheses or by using superscript notation)? Edit: In the first source right-associativity was used, but I'll check the source's sources.[[User:Ninjamin|Ninjamin]] ([[User talk:Ninjamin|talk]]) 16:29, 3 April 2019 (UTC)
:I found Conway using right-associativity for Knuth's arrow notation, but most sources use parentheses, so there's no reason to see a general convention here. Also the article claims that the notation be defined to be associative, but Knuth's paper makes no mention of that, thus I'm going to remove that passage.[[Special:Contributions/141.99.250.138|141.99.250.138]] ([[User talk:141.99.250.138|talk]]) 09:30, 4 April 2019 (UTC)

== Down-arrow notation ==

There are a few sources on the internet, including Googology Wiki, for down-arrows which are a kind of inverse of Knuth's up-arrow. I'm not qualified to add it myself, but I think down-arrows merit a section in this article. [[User:Glumblebee|Glumblebee]] ([[User talk:Glumblebee|talk]]) 06:25, 25 June 2019 (UTC)

== Definition ==
The formal definition in the first line uses parenthesis to define a right-associative operator.
The last line says: "the operator is SOMETIMES used right-associatively". This is confusing.
For a left associative operator the definition would have to be

:<math>
a\uparrow^n b=(a\uparrow^n (b-1))\uparrow^{n-1} a
 </math>
[[Special:Contributions/2A02:A210:2142:6C00:D5BA:D812:1452:F501|2A02:A210:2142:6C00:D5BA:D812:1452:F501]] ([[User talk:2A02:A210:2142:6C00:D5BA:D812:1452:F501|talk]]) 09:23, 12 November 2019 (UTC)

== Where does the definition for b = 0 come from? ==

See title. Knuth's original article never said anything about what happens when either term is zero, and I find the definition of 0 \uparrow^n 0 particularly suspect -- after all, isn't 0^0 indeterminate? [[User:Arcorann|Arcorann]] ([[User talk:Arcorann|talk]]) 13:33, 19 May 2020 (UTC)

: I've also raised the same concern.  It doesn't come from anywhere.  See discussion under [[#Formal definition]] above. --[[User:Ashawley|Ashawley]] ([[User talk:Ashawley|talk]]) 19:37, 30 June 2020 (UTC)

== English syntax question ==

Shouldn't this sentence in the Definition section :
:Remark however that Knuth did not define the "nil-arrow" (<math>\uparrow^0</math>).
more properly be
:Note, however, that Knuth did not define the "nil-arrow" (<math>\uparrow^0</math>).

I would make that change, but I'm unsure whether the first version, using "Remark" is correct English as well and I'm just not aware of it.  Thanks! -- [[User:ArglebargleIV|ArglebargleIV]] ([[User talk:ArglebargleIV|talk]]) 19:02, 23 March 2021 (UTC)

== 2↑↑4 and 2↑↑↑4 are bad examples because it is a coincidence that 2↑↑↑4 happens to be equal to 2^2^...^2 with 2↑↑4 2's ==

The examples for tetration and pentation used at the top of the article (2↑↑4 and 2↑↑↑4) are bad because they make it easy for a layperson like me trying to learn what pentation is from the article to come to a false conclusion about what pentation is.

Specifically, [https://i.imgur.com/oK4Wgba.png I saw that 2↑↑4=65536 and 2↑↑↑4=2^2^...^2 with 65536 2's] and made the false assumption that the general version of this is true, i.e. that a↑↑↑b=a^a^...^a with a↑↑b a's (but this is false).

I highly doubt I'm the only person who came to this false conclusion about what pentation is when reading this example at the top of the article. I suggest a different example be used instead that doesn't have this coincidental feature, perhaps a=2 and b=3.

I leave it to a math person to make that judgment call and change it. Thank you! [[Special:Contributions/70.123.9.81|70.123.9.81]] ([[User talk:70.123.9.81|talk]]) 04:53, 5 March 2022 (UTC)

== what is said here, if anything? ==

:Since <math>6\uparrow\uparrow4</math> = <math>6^{6^{6^{6}}}</math> = <math>6^{6^{46,656}}</math>, Thus the result comes out with <math>\underbrace{6^{6^{.^{.^{.^{6}}}}}}_{4}</math>

If this is anything other than tautology, it is poorly phrased. —[[User:Tamfang|Tamfang]] ([[User talk:Tamfang|talk]]) 01:54, 23 December 2022 (UTC)

== Equations are off the page? ==

Some equals signs are misaligned. Some equations and tables are too horizontally long. [[User:Talib1101|1101]] ([[User talk:Talib1101|talk]]) 00:42, 9 April 2025 (UTC)

== Recent edits ==

Can someone take a look at the recent series of edits by 46.159.138.78 and my reverts?  I think the reverted edits are somewhere between disruptive and approaching nonsense, I'm not certain of myself on this, though, and would appreciate some feedback, either confirmation of my reverts or telling me I'm way off base.
I've also a note on the Wikiproject Mathematics talk page.  Thanks. [[User:ArglebargleIV|ArglebargleIV]] ([[User talk:ArglebargleIV|talk]]) 09:21, 12 April 2025 (UTC)

:Just looking at their edits to the table section I'd agree with your characterization. The zero part in particular is the sort of incoherence I'd expect from an LLM losing its place. [[User:Sesquilinear|Sesquilinear]] ([[User talk:Sesquilinear|talk]]) 21:39, 12 April 2025 (UTC)
:I haven't looked closely enough to figure out what's going on but fwiw I think that not all the edits by this person from April 6 have been reverted, so if they've been problematic then someone should check those, too.  --[[User:JayBeeEll|JBL]] ([[User_talk:JayBeeEll|talk]]) 23:22, 12 April 2025 (UTC)

: I just realized that their talk page was empty (which I should have realized earlier!), so I welcomed them and invited them to come here and discuss their changes. [[User:ArglebargleIV|ArglebargleIV]] ([[User talk:ArglebargleIV|talk]]) 16:02, 15 April 2025 (UTC)
::Looks like there was another set of edits after you sent the message; I reverted it. [[User:Sesquilinear|Sesquilinear]] ([[User talk:Sesquilinear|talk]]) 19:23, 15 April 2025 (UTC)
:::Note: I had semi-protection requested and it just expired, so if we see the same IP cluster editing the page we can go back [[User:Sesquilinear|Sesquilinear]] ([[User talk:Sesquilinear|talk]]) 00:36, 23 April 2025 (UTC)

== Values in Table Computing 4  ↑n b ==

The values in this table don't make any sense. They should have a similar pattern as the other tables "Computing 2 ↑n b", "Computing 3 ↑n" and "Computing 10 ↑n b". [[Special:Contributions/2A02:2455:32A:A600:BCB7:4DEE:B521:CF98|2A02:2455:32A:A600:BCB7:4DEE:B521:CF98]] ([[User talk:2A02:2455:32A:A600:BCB7:4DEE:B521:CF98|talk]]) 19:49, 22 April 2025 (UTC)

:Agreed, I was working on updating this but figuring out the shorthand/computation is above my paygrade. I was thinking of updating the easy parts (1st/2nd row and 1st/2nd column), but the rest of the table would still be wrong [[User:Zdjensen|Zdjensen]] ([[User talk:Zdjensen|talk]]) 00:33, 18 June 2025 (UTC)