Talk:Graham's number

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The ten leftmost digits of Graham’s number

The ten rightmost digits of Graham’s number is 2464195387, but what are the ten leftmost digits of Graham’s number? I want to know them like (sequence A138866 in the OEIS) 2402:7500:901:F4A7:B569:939A:AC71:FAC9 (talk) 01:36, 17 November 2022 (UTC)[reply]

Somebody please put a banner at the top of this talk page with a warning related to this question that I'm sure people will keep asking over and over again. Georgia guy (talk) 02:02, 17 November 2022 (UTC)[reply]

Sorry, but I think that this article can show the leftmost 1000 digits and the rightmost 1000 digits for the Graham’s number. 2402:7500:901:F4A7:B569:939A:AC71:FAC9 (talk) 02:05, 17 November 2022 (UTC)[reply]

No, they can't. This is currently too difficult a problem to calculate. At least, it is too difficult in base 10. In base 3 the answer is 1000000000... SpinningSpark 14:21, 17 November 2022 (UTC)[reply]

Reference 6

What does Iteror.org have to do with Graham's number? Chasrob (talk) 10:52, 10 October 2023 (UTC)[reply]

It is a part of the chain which lies on the path to Wayback machine's link to the Scientific American article. A part of the service the reader needs to read the reference. When one part of the chain is broken, the ref is broken, so we need the chain, to read the Martin Gardner article. --Ancheta Wis (talk | contribs) 13:09, 10 October 2023 (UTC)[reply]

I've reformatted the reference so now the link in the title points to the archived version rather than to the usurped / dead link. --JBL (talk) 21:42, 10 October 2023 (UTC)[reply]

64 vs 63

Christ this is stupid. The relevant edits are [1] [2] [3] from September, before [4] [5] [6] from the last two days. Pinging @DASL51984, Joajohball MSAE, Rickardou, and DragonflySixtyseven:. --JBL (talk) 23:45, 3 November 2023 (UTC)[reply]

Graham's Number is g(64), not g(63), most of the mathematical youtubers can confirm that Graham's Number is g(64) Joajohball MSAE (talk) 23:56, 3 November 2023 (UTC)[reply]

When a brand new editor makes a change like that without citing a source, it might be legit, but I can't assess it myself. "Most of the mathematical youtubers" - point to one, please.

Also, is this just a "start at 0" vs "start at 1" thing? DS (talk) 00:32, 4 November 2023 (UTC)[reply]

Numberphile, Dr. Trefor Bazett , Travis Richardson, a brazilian youtuber called "Ciência Todo Dia" and more, most of the people don't actually mention a number g(0). Joajohball MSAE (talk) 04:19, 4 November 2023 (UTC)[reply]

The number is almost always defined as g(64) instead of g(63). Source —Panamitsu (talk) 00:42, 4 November 2023 (UTC)[reply]

I don't really care whether this article uses zero indexing or not, whether is starts with g0 or g1, or whether Graham's number is g63 or g64.

But it needs to be consistent, and hopefully consistent with the body of literature it is based on.

Right now, we have the following text that disagrees with the mathematical notation that precedes it:

In other words, G is calculated in 64 steps: the first step is to calculate g(0) with four up-arrows between 3s; the second step is to calculate g(1) with g(0) up-arrows between 3s; the third step is to calculate g(2) with g(1) up-arrows between 3s; and so on, until finally calculating G = g(63) with g(62) up-arrows between 3s.

Earlier, g(0) is defined as equal to four, so not need to "calculate" it. And the final step is computing g(64) not g(63). Could someone please fix the text? Or the formulae? I'm not sure I want to wade into the dispute at this juncture. Mr. Swordfish (talk) 01:43, 13 November 2023 (UTC)[reply]

Ok. Nobody else stepped up so I made this edit myself.

There are still some other places in the article where there is a similar discrepancy with the stated definition. I'll take a look at addressing that. Mr. Swordfish (talk) 15:22, 15 November 2023 (UTC)[reply]

I don't want to bother right now, but g₆₄ (rather than g(64)) seems to be universal everywhere else. Looks like that got changed in the same revision that shifted the numbering convention by 1. Patallurgist (talk) 01:56, 7 November 2023 (UTC)[reply]

(Incidentally, neither Graham nor Gardner used this notation. Does anyone know where it came from?) Patallurgist (talk) 02:15, 7 November 2023 (UTC)[reply]

Really? I wonder why people use that notation then. Waylon111 (talk) 21:32, 16 November 2023 (UTC)[reply]

Because often early attempts at notation when the mathematician is still figuring it all out is cumbersome, and more elegant notation is developed later. For example, Newton's Method_of_Fluxions notation for calculus - Leibniz's notation is what we use today. Or the Ackerman function, to cite just two examples.

Anyway, we have a reliable source - [7] that uses that notation, or rather it uses subscript notation g_n instead of function notation g(n). I'll leave making that change to the article as an exercise for the reader. (c: Mr. Swordfish (talk) 14:37, 17 November 2023 (UTC)[reply]

I reverted the notation to the subscript (g₆₄) notation used in the revision of 17 August 2023, which was stable and had no issues. If anyone wishes to argue for changing to function notation they are welcome to post here. Arcorann (talk) 06:04, 22 November 2023 (UTC)[reply]

Rightmost decimal digits

Edited the "Rightmost decimal digits" section, fixing some mistakes (i.e., the wrong assumption that any integer tetration of height d should have at least d-1 stable digits, which is false... consequently, I added the trivial example of the base 2 that has d-2 stable digits only, given the fact that 2^^2=4 is not congruent modulo 10 to 2^^3=16), explaining which is the exact number of stable digits of G (providing also proper peer-reviewed references) and improving the overall quality of the page. Please, let me know if something is not clear or needs further edits. P.S. Still trying to learn how to sign my comments properly. --Marcokrt (talk) 16:35, 5 January 2024 (UTC)[reply]

Feedback from New Page Review process

I left the following feedback for the creator/future reviewers while reviewing this article: Please remember to tag redirects that you create per WP:REDCAT.

voorts (talk/contributions) 20:07, 4 July 2024 (UTC)[reply]

Candidate to disprove the Collatz conjecture?

WP:NOTFORUM

The following discussion has been closed. Please do not modify it.

Powers of 2 become exponentially rare at large numbers and Graham's number is a huge number, that is not a power of 2. Is Graham's number a candidate to disprove the Collatz conjecture? 94.31.89.138 (talk) 12:16, 18 August 2024 (UTC)[reply]

It can be difficult. The last 3 digits of Graham's number are 387. So multiplying it by 3 and adding 1 will give you a number ending in 162, which is divisible by 2 but not by 4. Then divide it by 2. I can't continue the sequence right now. Georgia guy (talk) 14:21, 18 August 2024 (UTC)[reply]