The modern mathematics detect even larger numbers, e.g. TREE(3) and SCG(13), and all these numbers have some reason in theorems and proofs, so Graham's number already is not the biggest. Ikosarakt1 (talk) 09:02, November 24, 2012 (UTC)

- The article says that Graham's number is
*celebrated*as the largest number used in a mathematical proof. You're right, and popular opinion is indeed wrong. Of course, we come from such an obscure corner of mathematics that there's not much we can do to change popular opinion :( FB100Z • talk • contribs 19:50, November 24, 2012 (UTC)

It's easy to find the last digits of Graham's number all over the Internet. A great addition to the article would be an explanation for how these digits are computed. FB100Z • talk • contribs 03:00, January 23, 2013 (UTC)

there will be g64. g1 is 3^^^^3 g2 is 3^...(the amount of ^ in the g2 will be 3^^^^3 which is g1.)...^3 (3^...g1 ^'s...^3) g3 is 3^...(g2 ^'s)...^3 Repeat from to the previous one till you get g64. g64 will be 3^...(g63^s)...^ Jiawheinalt (talk) 05:21, February 9, 2013 (UTC)

More formally, arrow notation:

\(a \uparrow^{1} b = a^b\) (c=1)

\(a \uparrow^{c} 0 = 1\) (b=0)

\(a \uparrow^{c} b = a \uparrow^{c-1} (a \uparrow^{c} (b-1))\) (otherwise)

And Graham's function:

\(G(1) = 3 \uparrow^{4} 3\) (n=1)

\(G(n) = 3 \uparrow^{G(n-1)} 3\) (otherwise)

Then Graham's number = G(64). This shows how easy definition can be used to create so big number. Ikosarakt1 (talk) 07:26, February 9, 2013 (UTC)

In the picture, it says that 3^^^3 = 3^^(3^^3) = 3^^7625597484987 (no problem so far) = 3^(7625597484987^7625597484987)!?!? shouldn't it be a power tower of 3's 7625597484987 high, which is far more than what it says? DrCeasium (talk) 16:53, May 19, 2013 (UTC)

You found a typo in serious mathematician article. I shall remove the photo. Ikosarakt1 (talk ^ contribs) 18:10, May 19, 2013 (UTC)

- Nevertheless, the article has historical significance. We can, of course, point out the error. FB100Z • talk • contribs 19:04, May 19, 2013 (UTC)

## Graham full name.

his full name is "Ronald Graham".^{[1]}

Sources:

Graham number is 3^[3^^^^3]^3 then 62 more times. Jiawheinalt (talk) 10:08, March 6, 2013 (UTC)

### 3 entry form

First, is it = to {3.5,65,1,2}?

then... {3,3,{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3{3,3}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}.

$ Jiawhein $^{\(a\)\(l\)\(t\)} 10:46, April 25, 2013 (UTC)

Graham's number is between {3,65,1,2} and {3,66,1,2}. Bowers' arrays aren't defined for nonintegers. Deedlit11 (talk) 18:12, April 25, 2013 (UTC)

- But the 3^^^^3 is not 3^^^3. $ Jiawhein $
^{\(a\)\(l\)\(t\)}00:15, April 26, 2013 (UTC) - We
*wish*they were defined. If a continuous analog were found, that would be a huge breakthrough. FB100Z • talk • contribs 22:12, April 26, 2013 (UTC)- A fully continuous analog may be tricky, but my "letter notation" happens to be equivalent to a Bowers Linear Array of the form {10,x,c,d,...} where x can be any positive noninteger. This actually happened by accident. My notation was simply meant to be a continuous ω
^{ω}-level function, but it turned out to be equivalent to Bowers Arrays for integer values. To make a long story short, in my notation you can approximate Graham's number as "K64.492" which translates to {10,64.492,1,2}. Or if you prefer base 3: {3,65.147,1,2} PsiCubed2 (talk) 03:58, December 15, 2016 (UTC)

- A fully continuous analog may be tricky, but my "letter notation" happens to be equivalent to a Bowers Linear Array of the form {10,x,c,d,...} where x can be any positive noninteger. This actually happened by accident. My notation was simply meant to be a continuous ω

Right, so Graham's number is not {3,65,1,2}, it is larger, like I said. Deedlit11 (talk) 03:07, April 26, 2013 (UTC)

There is no reasonable way to represent Graham's number in array notation. But we can bound it with {3,65,1,2} < G < {3,66,1,2}. FB100Z • talk • contribs 22:12, April 26, 2013 (UTC)

In the 3 entries form, Graham's number is written as: |
---|

{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,{3,3,4}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}}} |

$ Jiawhein $^{\(a\)\(l\)\(t\)} 08:18, May 8, 2013 (UTC)

## Стасплекс

Do you know that G(100) is called "Стасплекс"? [1] Ikosarakt1 (talk ^ contribs) 09:46, June 6, 2013 (UTC)

- Ok, lets add. Remember to put "See also". 220.255.2.158 10:46, June 6, 2013 (UTC)

## BEAF and array

It is written that "it is reasonably close to … {3,65,1,2} in BEAF" and "There is no reasonable way to represent Graham's number in array notation", but I understand that BEAF is exactly same as array notation in that level. It looks weird that BEAF can reasonably represent and array notation cannot represent well. Kyodaisuu (talk) 22:15, December 10, 2013 (UTC)

## Graham's Number = 13?

Graham's Number was originally an upper bound to a problem. The answer may be 13. No one knows but I personally guess that it's less than 20. BTD6 maker (talk) 19:39, January 28, 2015 (UTC)

- Currently, the term "Graham's number" is given to the most well-known upper bound to this problem. But it's indeed true that the answer to the original question posed might be as low as 13. LittlePeng9 (talk) 19:43, January 28, 2015 (UTC)

- Exoo, who proved that Graham's Solution (may as well call the answer to the original question something) was at least 11, conjectured that it was much higher, due to the fact that he could make a lot of random decisions and still find a counterexample at n=10. But I guess even 20 could be "much higher". Deedlit11 (talk) 19:59, January 28, 2015 (UTC)

## Last digits

222.97.145.127 20:35, November 1, 2017 (UTC) Although I have proven that TREE(3) is odd and the last 3 binary digits are 011, Graham's number can be calculated with more accuracy AND with decimal digits, which the last 1,001 could be found at https://is.gd/gn1001

## Asea-Graham

There are not that many Asea-Graham lifts in the world. --109.40.0.225 21:32, December 22, 2017 (UTC)

## Again, we numbered things visually written out.

WHY do we say like "2^2^2^2^2^2^2^2^2 with 9 2s" instead of just "2^2^2^2^2^2^2^2^2"? 80.98.179.160 13:33, January 9, 2018 (UTC)

## PEGG notation

What is the PEGG notation of this number? --84.61.221.211 10:35, March 25, 2018 (UTC)

## Extended Graham's problem

Let f(n) be the least dimension N of hypercube such that if the lines joining all pairs of corners are **n**-colored, a complete graph \(K_4\) of one color with coplanar vertices will be forced. So Graham's problem is to find f(2).

Let g(n) be the least dimension N of hypercube such that if the lines joining all pairs of corners are 2-colored, a complete graph \(K_{2^n}\) of one color with all vertices in one n-hyperplane will be forced. So Graham's problem is to find g(2).

Does f(n) and g(n) exist for all n? {hyp/^,cos} (talk) 13:17, April 6, 2018 (UTC)

- Correct. This is what Graham and Rothschild have proven. Indeed, they have proven more - fix numbers n, r and k<l. Then there is a dimension N = N(n,k,l,r) such that, if we consider the N-dimensional n x n x ... x n cube and color all of its k-dimensional affine subspaces with r colors, then there is an l-dimensional affine subspace with all k-dimensional subspaces of the same color. Original Graham's problem is to bound N(2,1,2,2). Existence of N(n,k,l,r) was proven in the same paper where N(2,1,2,2) was introduced. LittlePeng9 (talk) 13:28, April 6, 2018 (UTC)

## PEGG

Today, PEGG surpassed Graham's number. --84.61.221.211 06:14, April 22, 2018 (UTC)

## Leading digits

To my knowledge, there is no proof that first digits of Graham's number can't be calculated. Maybe we just don't know something yet. Triakula (talk) 10:44, December 26, 2019 (UTC)

- Right. I will correct the description. I also found several wrong unsourced statements on first digits and last digits in articles in this wiki these days. Therefore it is better to strictly check whether the descriptions have sources.
- p-adic 12:58, December 26, 2019 (UTC)

## Last digits

Surely there will be a better way than simply replacing the method of computing the last digits with "this is NOT how you calculate the last digits". How about actually adding the correct method there? -- ☁ I want more clouds! ⛅ 08:33, December 27, 2019 (UTC)

- I agree with that it better to write both "the method which was believed to be correct without any sources in this community is wrong" and the correct method. Then could you tell me the correct method with a source?
- p-adic 09:23, December 27, 2019 (UTC)
- I don't think the wrong method needs to be included on this article; a mention on the Tetration article is enough. I don't know the source about the correct method, but I remember that the Wikipedia article for Graham's number lists the last 500 digits of the number. Maybe check the sources of that article? (Unless that method is wrong too) -- ☁ I want more clouds! ⛅ 16:51, December 27, 2019 (UTC)

- I think that it is better to keep to mentioning the error, because it was believed so long time without sources. If we delete it, then people will just believe "the naive method works" even if there is no source.
- The "source" in wikipedia looks wrong, too. It explains the same wrong method. I note that it includes a correct explanation with a proof by another one, but it does not imply the way to compute last digits of Graham's number if I am correct. Of course, the method fortunately occasionarily outputs the correct answer. If there is a proof, then please tell me.
- p-adic 00:27, December 28, 2019 (UTC)
- I don't really get why the recursive modular exponentiation method is wrong, because even though plugging 10^n into the totient function gives 4*10^(n-1), the last n digits of powers of a fixed base always repeat sooner. Specifically, the last 3 digits of powers of 3 repeat after only 100, even though phi(1000) is 400, and the last 4 digits of powers of 3 repeat after not 4000 (that is, phi(10000)), but only 500. In general, the period of the last digits of 3^n is NOT phi(10^n) or 4*10^(n-1) which does not divide into 10^n evenly, but rather 5*10^(n-2) (which does divide into 10^n) for n >= 4. Allam948736 (talk) 02:05, December 28, 2019 (UTC)

- Please tell me what "the period of the last digits of x^n" precisely means in your context. Since the original description in the article of tetration was mentioning to an arbitrary base, your explanation is not generic to the whole issue. Also, could you tell me the exact method which you are using in order to compute last digits of tetration? I have asked you twice, but you are somewhy ignoring the requirement. Since you do not specify the precise method, it is difficult for others to state precise judgements.
- p-adic 06:38, December 28, 2019 (UTC)
- The period of the last digits of x^n is the n after which the last digits repeat. For example, the period of the last 3 digits of 3^n is 100 because 3^1 is just 3, and 3^101 = ...566003. For all bases I have checked, the last 3 digits repeat after only 100 powers (if not sooner) even though phi(1000) is 400. I had used the original "wrong" method to calculate last digits of tetrations. Allam948736 (talk) 16:09, December 28, 2019 (UTC)

- The statement for last 3 digits of 3^n in base 10 is correct, while it does not directly imply the method to compute the last digits of tetration in any base. Why do you keep your original method secret? Then it is impossible for us to argue on the correctness of your method. For example, you are stating that for
**any**(i.e. without exceptions) positive integers x, y, d, and b > 1, the last d digits N(y) of x↑↑y in base b can be computed in the following method in the original articke, right?

- The statement for last 3 digits of 3^n in base 10 is correct, while it does not directly imply the method to compute the last digits of tetration in any base. Why do you keep your original method secret? Then it is impossible for us to argue on the correctness of your method. For example, you are stating that for

- N(0) = 1
- N(y) = x^{N(y-1)} mod b^d

- Otherwise, please tell me how you intend to compute the last d digits of x↑↑y in base b. Hyding the precise method does not give a solution, because you can shift the goalpost as you wish.
- p-adic 23:19, December 28, 2019 (UTC)
- It seems to me that the recursive modular exponentiation method works in base 10 is because 10 is not prime. In a prime base, however, the recursive modular exponentiation method usually doesn't work (such as 3^n mod 19^d), but I'm sure your method works. Allam948736 (talk) 04:26, December 29, 2019 (UTC)

- What is "the recursive modular exponentiation method"...? At least, it is different from my recursive modular exponentiation method above according to you. Is that the best information which you can share with outhers? How could we argue on the correctness of your method which you are talking about? Why couldn't you specify your precise method?
- Anyway, even if you somewhy hesitate to clarify your method, you are wrong, because my method does not work in general.
- @Cloudy176
- As the user above is an example, just deleting the wrong method causes that people unreasonably believe that some secret methods in their own mind work. Please ask him to specify the method instead of me.
- p-adic 07:02, December 29, 2019 (UTC)
- The recursive modular exponentiation method is just repeatedly taking x^y mod 10^d (that is, taking x^n mod 10^d where n is any integer, then feeding that back into the "exponent", and repeating that process). This works in base 10 because 10 is not prime, but in prime bases (like 7, 13, or 17), it usually doesn't work. Allam948736 (talk) 16:09, December 29, 2019 (UTC)

- But you said "I'm sure your method works". What is my method then? What you explained is the same as I explained. Also you wrote "For all bases I have checked, the last 3 digits repeat after only 100 powers (if not sooner) even though phi(1000) is 400". t means that you have an alternative method applicable to
**any**bases which you have checked. Since you stated that the recursive modular exponentiation method does not work in general, it is different from yours. What is the precise method? Please tell us your method, which you believe to be correct for any bases. - Moreover, you wrote "This works in base 10 because 10 is not prime". Do you mean that the recursive modular exponentiation method works for any non-prime number? Please fix your statement in order to avoid shifting the goalpost. I am talking about your statements on
**any bases**, but not on base 10. (I note that the exact method to compute last digits of Graham's number in the article for base 10 is also wrong, even if you believe that it works. But before arguing the issue, please specify your method for**any bases**.) - p-adic 23:17, December 29, 2019 (UTC)
- By bases in that case, I meant the "base" value of the modular exponentiation (which in the case of Graham's number is 3), not the numeral base like I meant in my last message. I was assuming base 10 when I said that the period of the last 3 decimal digits is always 100 or less. Allam948736 (talk) 23:41, December 29, 2019 (UTC)
- Ok. You fixed one statement. Also, please answer about the sentense
- > Moreover, you wrote "This works in base 10 because 10 is not prime". Do you mean that the recursive modular exponentiation method works for any non-prime number? Please fix your statement in order to avoid shifting the goalpost."
- in order to argue on your explanation.
- p-adic 23:57, December 29, 2019 (UTC)
- It turns out that the recursive modular exponentiation method doesn't work for all composite numeral bases either, but it does work in base 10 because for d > 2 the period of the last d decimal digits of 3^n is always a divisor of 10^d (specifically, it is (10^d)/20 for d >= 4). (There is nothing special about the exponentiation base of 3, this happens with all other exponentiation bases).
- Also, the proof on the article that the recursive modular exponentiation method is wrong is itself wrong, because the recursive modular exponentiation method actually does stop working for the last digits of Graham's number at some point (however, the number of digits you would have to go to find the last non-convergent digit of Graham's number is itself extremely large). Allam948736 (talk) 01:19, December 30, 2019 (UTC)

- But you said "I'm sure your method works". What is my method then? What you explained is the same as I explained. Also you wrote "For all bases I have checked, the last 3 digits repeat after only 100 powers (if not sooner) even though phi(1000) is 400". t means that you have an alternative method applicable to

- You finally recognised your mistakes, right? It would cost less time if you had not kept your precise statements secret.
- > Also, the proof on the article that the recursive modular exponentiation method is wrong is itself wrong, because the recursive modular exponentiation method actually does stop working for the last digits of Graham's number at some point (however, the number of digits you would have to go to find the last non-convergent digit of Graham's number is itself extremely large).
- You are wrong. Read the original article. It describes a method which you are not talking about.
- p-adic 01:34, December 30, 2019 (UTC)

- @Cloudy176
- Finally, the user who persisted his or her own secret wrong method agreed that he or she was believing it without any proof. Then it is a good evidence that we need to keep the clarification of the incorrectness of the wrong method.
- Also, I added a precise condition, since I do not have to care about the shifting of goalpost by that user.
- p-adic 01:44, December 30, 2019 (UTC)

"How" wrong is the wrong algorithm shown in the main article? i.e. How many digits are there to the right of the rightmost digit that is different in Graham's Number and the leftward-infinite sequence of digits that the wrong algorithm suggest? --Nayuta Ito (talk) 03:08, February 2, 2020 (UTC)

- Although I do not know the answer for the algorithm in this article, at least we can see a descriotion of a correct algorithm in wikipedia. I note that the source linked from wikipedia includes both correct arguments and wrong arguments similar to the original description in this article, but the algorithm in wikipedia itself is based on a correct argument. Therefore if the written 500 digits are actually output by the corresponding algorithm, then it is correct. (I am not certain about whether it is actually output by the correct algorithm.) The estimation of how many correct digits it outputs is greater than g_63 according to wikipedia. (I have never verified the estimation.)
- p-adic 06:57, February 2, 2020 (UTC)

- The number of correct digits the method produces, as I have stated, is itself extremely large (meaning we will never actually reach that point) and, on this scale, is barely less than Graham's number itself. In fact, if N is the number such that 3^^N = G, then the number of correct digits is N-1. 3^^2 (27) has just 1 convergent digit, 3^^3 (7625597484987) has 2 convergent digits, 3^^4 has 3 convergent digits (...6100739387), and so on. Allam948736 (talk) 15:56, February 2, 2020 (UTC)

- > In fact, if N is the number such that 3^^N = G, then the number of correct digits is N-1.
- Please write down a proof of your statement that N(x) in the article is actually the last x-digits of Graham's number for any natural number x smaller than N. Also, I clarify that we are talking about the algorithm in the article, before you will change your statement by saying "no, I am talking about my secret great algorithm" as you prefer. A proof means a mathematical proof, but not an intuitive pattern matching of smaller values or an intuitive listing of unqunatified formulae.
- p-adic 22:35, February 2, 2020 (UTC)

- Yes, I was indeed talking about the recursive algorithm in the article. The last k digits of n determine the last k+1 digits of \(3^n\) since
**the period of the last d digits of 3^n in base 10 is equal to**\(5 \times 10^{d-2}\)**for**\(d \ge 4\),**which divides into**\(10^{d-1}\)**evenly**. For 2 digits, the period is 20: 3^20 = 34867844__01__. For 3 digits, the period is 100: 3^100 = 515,377,...,107,522,001 and multiplying by a number ending in ...001 has no effect on the last 3 digits. 3^500 indeed ends in ...610,001, 3^5000 ends in ...100,001, and so on. Each time we take the 10th power we add another zero before the final 1, confirming what I stated near the top of this message. This means that 3^...87 always equals ...387, confirming that 3^^4 ends in ...387 and thus has 3 convergent digits. Similarly, 3^...387 always equals ...5387, confirming that 3^^5 indeed ends in ...5387 and thus has 4 convergent digits. This confirms that Graham's number indeed ends in ...04,575,627,262,464,195,387. - P. S. Why doesn't LaTeX work on here? Allam948736 (talk) 23:56, February 2, 2020 (UTC)

- Yes, I was indeed talking about the recursive algorithm in the article. The last k digits of n determine the last k+1 digits of \(3^n\) since

- Your explanation is not a proof of your estimation. (Also, the argument itself is essentially written in the article of tetration by me using Chinese remainder theorem and Euler's theorem, and hence it is not what we are asking.) The point is how you derived your statement "the number of correct digits is N-1", which implies "N(x) in the article is actually the last x-digits of Graham's number for any natural number x smaller than N". (Your statement is a bit stronger, because it also implies that the last N-th digit differs.) I guess that you are forgetting some non-trivial estimation of the least x for the congruence relations 3^{Graham's number} - (Graham's number) = 0 mod 10^x or (Graham's number)^{Graham's number} - (Graham's number) = 0 mod 10^x.
- p-adic 00:38, February 3, 2020 (UTC)

- I derived the statement by observing that 27 has 1 convergent digit, 7625597484987 has 2 convergent digits, 3^^4 has 3, 3^^5 has 4, and so on, adding another constant terminating digit each time we add another 3 to the tower. The reason why one digit is added and not 2 is because the period of the last d digits of 3^n in base 10 evenly divides 10^(d-1) for d >= 3 but not 10^(d-2). Allam948736 (talk) 01:09, February 3, 2020 (UTC)

- I said:
**A proof means a mathematical proof, but not an intuitive pattern matching of smaller values or an intuitive listing of unqunatified formulae.**Please wirte down the proof of the closed formula "N(x) in the article is actually the last x-digits of Graham's number for any natural number x smaller than N" using mathematical equations. - p-adic 01:16, February 3, 2020 (UTC)

- I said:

- The reason that each iteration adds another convergent digit and not more is because the period of the last d decimal digits of 3^n for d >= 4 is 5*10^(d-2), which divides evenly into 10^(d-1) but is greater than 10^(d-2). This means that 3^...087 = ...387, 3^...187 = 387, 3^...287 = 387, 3^...387 = 387, 3^...487 = ...387 -- okay, you get the picture. Allam948736 (talk) 23:47, February 8, 2020 (UTC)

- You need to learn how to write mathematical equations... Have you ever written a mathematical proof? I am not saying that you are wrong, but saying that you are not writing down proofs. For example, if you want to show 3^3^n-3^n ≡ 0 mod 10^d for any d >= 4 and n greater than some fixed lower bound n_0(d), then you need to write something like "I prove it by induction on d. If d = 4, then I prove the statement by induction on n. If n = n_0(d)+1, then ～. If n > n_0(d) + 1, then by the induction hypothesys, ～. Now we consider the case d > 4. By induction hypothesis on d, we have ～." instead of showing smaller values or pictures. Could you understand that what you wrote is just a collection of sentences without appropriate logical connectives and quantifications? If you want to check your argument by yourself, then it is good to learn to write down proofs. Otherwise, you will enjoy to shift goalposts again and again as you did. Since your current statement (and "arguments" in your mind) looks correct, you can write it down. (And no, it does not mean that you have essentially written down the proof. What you actually need to study how to improve to write down proofs.)
- p-adic 00:12, February 9, 2020 (UTC)

@Allam

I clearly told you that your randomly written sentences can never be a formal proof. What you should do is to write down a formal proof, but not to add your original work without any written proof to the article just by ignoring the fact that you cannot actually write down a proof. Could you remember that you did the same by stating that your secret method works without any proof until I disproved the validity? Do you want to repeat this awful issue?

p-adic 23:44, February 10, 2020 (UTC)