Talk:Graham's number

Smallest factor of G-128

Is there any project anywhere online that reveals people are searching for the smallest prime factor of G-128?? For example, there's http://www.oddperfect.org where people are searching for odd perfect number requirements. (Note: G means Graham's number.) Georgia guy (talk) 14:45, 28 December 2011 (UTC)

By "G-128" do you mean "Graham's number minus 128"? If so, why 128? If not, what's the definition? — r.e.s. (talk) 02:46, 29 December 2011 (UTC)

Yes, that's what I mean. In an archived part of this talk page, it reveals that G-H has known prime factors for all 0 <= H < 128. Georgia guy (talk) 13:52, 29 December 2011 (UTC)

Ah, I'd forgotten about that archived section. I'm unaware of any projects of that kind, though. — r.e.s. (talk) 16:28, 29 December 2011 (UTC)

Rightmost (frozen) digits calculation

In my book (unfortunately in Italian) on tetration and others hyperoperators, it is described the "frozen digits" distribution for every natural base. A preview (introduction) of the book can be found here: http://www.uni-service.it/images/stories/product_uploaded_file/Coda%20Preview.pdf So, it's possible to calculate the p-adic convergence range of Graham's number too [p. 13].

In this case (n:=3), the convergence speed is equal to the height of the iterated tetration tower minus one (say [hyper-4 exponent]-1). Let this height be "k+1", for example, G(mod 10^k)==G^^G(mod 10^k). So "k", the number of "frozen" (calculable) digits have to satisfy the following inequality:

$g(63) << k < g(64)$

Could I have the permission to add this info/reference to the Graham's number page?

Best regards, Marco Ripà — Preceding unsigned comment added by Marcokrt (talk • contribs) 15:47, 19 January 2012 (UTC)

I suggest we limit to 500, because:

• We don't want Wikipedia articles to be huge.
• Somewhere deep along the line, the digits will start to mismatch the infinite 3^3^3^3^3...3^3^3^3^3 sequence; we don't have a way to tell where this point is.

Georgia guy (talk) 15:50, 19 January 2012 (UTC)

Marco:

What you're calling "frozen digits" are evidently the ones I've called "stable digits", e.g. in the recently-archived section Rightmost digits of G (cont'd) that was at the top of this very talk-page when you posted. Note that the upper bound in your inequality can be made tighter, as explained there. (I think "Georgia guy" may have misunderstood your intention, if I'm correct that you merely want to include in the article such a bounding inequality on the number of stable/frozen digits.) I haven't yet looked at your publication, so I have no comment about whether it is adequate as a source for citing such a result.
— r.e.s. (talk) 16:13, 19 January 2012 (UTC)

Hello again, My book is about the randomness of the tetrates digits too (i.e. caos theory and so on)... not only about perfect convergence (a very small amount of info can be found here http://math.eretrandre.org/tetrationforum/showthread.php?tid=720). BTW, IMHO, we could only add the point that the stable digits correspond (exactly) to k, where k+1 is the height of the equivalent hyper-4 hyperexponent value. This is a specific natural number, not only an upper bound. Let me know if this info could be found interesting or not for the page.

Marcokrt (talk) 18:26, 19 January 2012 (UTC) — Preceding unsigned comment added by Marcokrt (talk • contribs) 16:27, 19 January 2012 (UTC)

Visualizing Graham's number

The number of times you have to perform the "How many Knuth arrows are needed??" operation on Graham's number is finally small enough to write out; it is only 64. But have numbers for which the number of times you have to perform the "How many Knuth arrows are needed??" operation on a number too large to write out ever been used?? There have been a few references to numbers larger than Graham's number in some sections of the talk page. Do any of these numbers meet this criterion?? Georgia guy (talk) 16:52, 19 January 2012 (UTC)

For example, my BOX_M~ (a specific natural number involving "human problems" solutions) satisfies this criterion: http://www.scribd.com/doc/77714896/The-largest-number-ever-il-numero-dei-record The busy beaver function "Σ" is even faster (non computable functions). BTW I don't find this outcome very interesting for this section :)

Marcokrt (talk) 18:24, 19 January 2012 (UTC) — Preceding unsigned comment added by Marcokrt (talk • contribs) 18:19, 19 January 2012 (UTC)

Georgia guy:

In the fast-growing hierarchy we have f_ω⁶⁴(6) > G, where f_ω is a version of Ackermann's function and corresponds to the operation you're asking about. If we look at a much larger ordinal index, say ε₀ instead of ω, then we'll encounter numbers like f_ε0(2), which certainly meets your criterion -- even with the small argument of 2, it requires f₈(8) > 2↑⁷8 iterations of f_ω.

Marko:

I think it's not hard to show that your number satisfies $Box_{-}\widetilde{M} \ll f_{\epsilon_0}(3)\,\!$, where f_ε0 is as just described (linked) above.

(Please sign your posts by typing four tildes (~~~~) at the very end.)
— r.e.s. (talk) 18:30, 20 January 2012 (UTC)

Ok r.e.s., even if probably $Box_{-}\widetilde{M} > f_{\epsilon_0}(3)\,\!$, it's easily possible to surpass its size, for example considering the one at the bottom of this page: http://www.polytope.net/hedrondude/array.htm My number is based on a set of philosophical/empiric assumptions and its definition could (abstracly) be useful to construct upper limits for particular kinds of "human condition related problems"... it's quite similar to a universal outlet, even if far less interesting ;)

Best regards Marcokrt (talk) 00:08, 21 January 2012 (UTC)

Parsing?

There was something in the article on Knuth's up-arrow notation which made me pause -- a single arrow is ordinary exponentiation? So... if I've parsed that -- and consequently the contents of this article -- correctly, then I think the following expansion defines G1.

(I'm using 'U' for 'Up Arrow')

G1 = 3 U U U U 3

3 U 3 = 3^3 = 27

3 U U 3 = 3 U (3U3) = 3^27 ~= 7.6 e12

3 U U U 3 = 3 U (3 U (3U3)) = 3 U (3^27) = 3^(3^27) ~= 3^(7.6e12) = some monster that is ~3.6*10^12 digits long ~= 10^(10^(10^1.1))

3 U U U U = 3 U (3U (3U (3U3))) = 3 ^ (some monster) = G1, "a jillion"

and then 3 (a jillion instances of U) 3 = G2, "a kajillion", etc etc up to G64.

Is this correct? DS (talk) 14:05, 11 September 2012 (UTC)

Yup! I think you got it. Owen× ☎ 14:37, 11 September 2012 (UTC)

Yes and No. It is correct that g1 = 3↑↑↑↑3, g2 = 3↑↑...↑3 (with g1 ↑s), and so on up to g64; however, your evaluations of 3↑↑↑3 and 3↑↑↑↑3 are way too small.

For example, 3↑↑↑3 = 3↑↑3↑↑3 = 3↑3↑3↑...3↑3 where the number of 3s is 3↑↑3 = 3↑27. This is much larger than your 3↑3↑3↑3, which is actually just 3↑↑4.
— r.e.s. (talk) 01:15, 12 September 2012 (UTC)

R.e.s. is right; my mistake. Owen× ☎ 11:54, 12 September 2012 (UTC)

Improved bound

Look at this article. There are presented much stronger upper bound for that problem where Graham's number was used. 31.42.233.14 (talk) 22:23, 18 March 2013 (UTC)