# Talk:Large numbers

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OK, this is a first stab at getting all the large number topics together, please feel free to kick this into shape. The Anome

Can I suggest that we include only pure numbers in this article, not distances and other measurements? Would anyone object if I deleted the astronomical distances, since they are only large numbers when expressed in small units? I suppose I should go further and say that Avogradro's number is also just an arbitrary unit, but I shan't, because I feel I'm on a slippery slope towards excluding everything! -- Heron

Then why do people call large numbers "astronomical", as the article informs us? Perhaps it's because astronomical distances are large when expressed in any human-sized scale. I think the concept of "largeness" needs to be explained. The whole article is subjective anyway -- I wouldn't call 1010 large, I deal with those sorts of numbers every day. $10^{10^{10}}$ is more like it. -- Tim Starling 09:26 18 Jun 2003 (UTC)
I agree with you about 1010. I wouldn't call the number of bits on a hard disk particularly large, either. It is certainly subjective. My point was that measurements of distance etc. are different from pure numbers. Measurements are, by definition, relative, whereas at least pure numbers are absolute. Largeness is another thing. Perhaps one definition would be "a number considered as large at a particular time by a particular culture". For example, I seem to remember that the Old Testament uses the number 40 as a generic large number in several places (e.g. "40 days and 40 nights"). -- Heron
Let me put this another way. I think the present article should be, as it mostly is, about the mathematics of large numbers. Other large quantities, such as astronomical distances, already have a place on the orders of magnitude pages (1e10 m etc.) Perhaps we should just link to them. -- Heron
Yes, you're quite right. Well, about most things. I could argue that physically distance is dimensionless but that would just be arrogant pedantry. The page title is "large number" not just "large", and the order of magnitude pages are pretty good for comparing distances. BTW did you see my reply for you on Wikipedia:Reference desk? -- Tim Starling 13:53 18 Jun 2003 (UTC)

## World Almanac

What's special about the World Almanac year 2000?? Do any of you Wikipedians have an edition of the World Almanac year 2004?? Try it. 66.32.95.180 01:52, 27 May 2004 (UTC)

There's nothing at all special about it. It's a pretty lousy source, actually. But it's a source. But it was the only source I happened to have at hand for quattuordecillion, etc. If they're in the 2000 edition they're probably in the 2004 edition, too, but I didn't think it was appropriate to say "World Almanac" without identifying which edition, and I certainly didn't think it was appropriate to reference an edition I hadn't consulted.
I think these are in the Merriam-Webster Third and probably lots of other places. I may get around to making a trip to the library this weekend and finding out. Hopefully someone else will do it first. Dpbsmith 14:29, 27 May 2004 (UTC)

## Examples

• $10^{\,\!100^{10}} < 10^{10^{20}}$
• $100^{\,\!10^{10}} < 10^{10^{10.3}}$

Also compare:

• $1.1^{\,\!1.1^{1.1^{1000}}} > 10^{10^{1.02*10^{40}}}$

I'm reverting the changes that were made to these equations, the discussion surrounding the equations clearly delineates the purpose of each and each is construed to show a certain aspect of "power towers", the revision is completely misleading when reading the text (besides making the equations wrong).

$10^{\,\!100^{10}}=10^{10^{11}} << 10^{10^{20}}$

$100^{10^{\,\!10}} = 100^{10000000000}$

$10^{10^{\,\!10.3}} \approx\ 10^{19952623149} = 100^{19952623149}$ not even close

In the 1.1 problem, I simplified the last exponent, thus:

$1.1^{\!1.1^{\,\!2.4699*10^{41}}} ? 10^{10^{1.02*10^{40}}}$

$1.1^{24} \approx\ 10$ $1.1^{1.1^{22}} \approx\ 10$

thus:

$10^{\!10^{\,\!2.4699*10^{41}}} > 10^{10^{1.02*10^{40}}}$

note that $2.4699*10^{41}$ is not larger than $2.4699*10^{41} - 46$ by enough to change the outcome of the main problem. Dusty78 03:38, 12 May 2005 (UTC)

Dusty, I think you need to bone up on your high school algebra. (a^b)^c = a^(b*c) In particular,
$10^{\,\!100^{10}}=10^{10^{11}}$
is wrong. 100^10 is not equal to 10^11, it's equal to 10^20.
Revolver 04:05, 12 May 2005 (UTC)
oops, right I've gone nuts spent too much time working out the rest to bother with checking all my work, still, the example doesn't really fit with the explination... oh well, I'm quitting while I'm behind Actually, I think I'll be reworking the probs untill I'm sure I'm right on the others..Dusty78 04:14, 12 May 2005 (UTC)
Ahh... Pride goeth before a fall... and screwy math before a bad post.... don't know what I was smoking ;) I'm just going to revert back to when it was actually right and correct the text 2nd example is wrong for same reason as first, 3rd is wrong but it took some actual number crunching to evaluate.Dusty78 04:25, 12 May 2005 (UTC)

## Sorry

Sorry, I wasn't thinking for a second about the relative errors of really large numbers. Hopefully the current explanation explains it well. Revolver 05:23, 13 May 2005 (UTC)

## New Comment

I don't see any reference to the enormous numbers you get when you calculate permutations and combinations? How many permutations are there in a googolplex? (I hope that's grammatically correct.) {roger} 29 June 2005

You're on the right track! Check out the article on Combinatorics referenced at the beginning of the "even larger numbers" section, or better yet the Permutations and combinations article. Then you'll know to ask "How many permutations are there of 1 googolplex objects without replacement?", and that the the answer is googolplex factorial (written "googolplex!"). Lunkwill 29 June 2005 06:33 (UTC)

## Accumulated Error In Binary vs Decimal Exponents

The examples in the 'rule of thumb for converting between scientific notation and powers of two' section are misleading I think. The small error inherent in $2^{10} \approx 10^3$ is magnified immensely at these scales, not to mention the horrible single step of $100 \approx 128$. The former is covered in Binary Prefix but it bears noting here as well if this section is to remain. The examples indicates that $10^{47} \approx 2^{157}$ and $2^{101} \approx 2 \times 10^{30}$, which have errors of 82.7% and 26.76% respectively. $\approx$ might be appropriate with a 27% error, but 83% is very far outside the acceptable bounds of 'approximately equal to' imho. rules of thumb like that are what can put a crater instead of a lander on mars

Good point. I've added a disclaimer/correction for larger exponents. Feel free to reword it. Owen× 17:00, 5 December 2005 (UTC)
Changed your example. I hope you'll be OK with it and find it more instructive. 75.4.107.1 06:23, 17 November 2006 (UTC)
what's wrong with logarithms?
—Preceding unsigned comment added by 70.82.193.47 (talkcontribs) 15:26, 2 April 2006
Indeed. From a logarithmic point of view, not unreasonable considering the size of the numbers we're dealing with, an error of 82.7% may still be small. Yes, the difference between 10^999 and 10^1000 is huge from a strict percentage point of view, but it's obviously a tiny difference, from a logarithmic point of view, compared to the difference between 10^500 and 10^1500. The article actually goes at length into meaning of "accuracy" in large numbers under the section heading that is titled, appropriately, "Accuracy." —Lowellian (reply) 19:19, 8 April 2009 (UTC)

## About The Universe Computer Analogy...

How many possible characters are there to be chosen from? one human language? all human languages? ascii? ansi? the entire argument is invalid if we don't know how many individual characters there are to choose from. not trying to be mean, just a friendly request for clarification :)

Robby Shadowrunner340 02:02, 15 September 2006 (UTC)

It's explained in the article: we're assuming each character is one byte, so 2^8 = 256 possible characters. —Lowellian (reply) 19:14, 8 April 2009 (UTC)

## From article page

The heading "Uncomputably large numbers" is misleading, as no integer is uncomputably large (every value of a busy beaver function is an integer). It's a busy beaver function that's noncomputable, not the values in its range. Also, although Rado's sequences might have been the first ones, it's very easy to produce noncomputable functions that grow faster than every computable function. --r.e.s. 04:14, 14 October 2006 (UTC)

Tom Harrison Talk 13:32, 14 October 2006 (UTC)

Thanks for moving my comment (above) from the article page to here, where it belongs. Sorry for the mix-up.--r.e.s. 19:16, 14 October 2006 (UTC)

## Citations?

Many sections of this article, e.g. Systematically creating ever faster increasing sequences; Standardized system of writing very large numbers, have no citations. Although they are very interesting, it appears they are original research which is not allowed by Wikipedia policy. True? Mytg8 20:44, 30 October 2007 (UTC)

## Systematically creating ever faster increasing sequences

Moved from page (original research):

Let a strictly increasing integer sequence $f_0(n)$ (n≥1) be given (written as a function for conveniently writing the functional powers), with $f_0(1)>1$. Then a sequence of sequences can be found by $f_k(n)=f_{k-1}^n(1)$, from which we can select the "cross-sequence" $f_0(10), f_1(10),..$ [1]

Together this is a process of creating a new sequence from a given one. This can be repeated (i.e., we can apply recursion), and again we can select from the matrix of numbers a single sequence, by taking the 10th element of each. We can apply the same whole process again and again.

Starting from $f_0(n)=10^n=(10 \to n)$ this process corresponds to adding an element 10 in the Conway chain before the variable n, which is at the end of the chain: we get $f_k(n)=f_{k-1}^n(1)=10 \uparrow^{k+1} n=(10 \to n \to (k+1))$, and the new sequence selected from the matrix is that of which the kth element is $f_{k-1}(10)=f_k^{10}(1)=10 \uparrow^k 10=(10 \to 10 \to k)=(10 \to 10 \to k \to 1)$. (See also Knuth's up-arrow notation and Conway chained arrow notation). [2]

Repeating this process we get $(10 \to 10 \to k \to n)$ for successive values of n, and selecting k=10 we get a single sequence $(10 \to 10 \to 10 \to n)$.

Repeating this whole process we get ever longer chains. Selecting n=10 we get the sequence of (10→10), (10→10→10), (10→10→10→10),... This can be used as starting sequence to apply the process again, etc. Even the value $f_1(2)$ for this sequence is already a Conway chain of length 10 billion plus one.

Each sequence in this whole process can be identified by its order type in the process:

• (10→nk) is the sequence with index n with order type k - 1
• (10→10→nk) is the sequence with index n with order type ω + k - 1
• (10→10→10→nk) is the sequence with index n with order type 2ω + k - 1
• (10→10), (10→10→10), (10→10→10→10),... is the sequence with order type ω²

In a similar way this can be continued, and we get a set of sequences, well-ordered by the procedure of construction.

• zero case: $f_0(n)=10^n$
• successor ordinal: $f_{x+1}(n)=f_x^n(1)$
• for limit ordinals: $f_{a}(n)=f_{a_{n-1}}(10)$ for a suitable sequence of ordinals $a_{n-1}$ tending to a.

Thus we have transfinite recursion as far as we define for each limit ordinal a suitable sequence of ordinals tending to it.

Consider the Cantor normal form $\omega^{\beta_1} c_1 + \omega^{\beta_2}c_2 + \cdots + \omega^{\beta_k}c_k$, where k is a natural number, $c_1, c_2, \ldots, c_k$ are positive integers, and $\beta_1 > \beta_2 > \ldots > \beta_k$ are ordinal numbers. The limit ordinals cover the case $\beta_k>0$. For order types less than ε0 (epsilon nought) $\beta_k$ is less than the ordinal itself. The following rules provide for each limit ordinal a suitable sequence of ordinals:

If $a=(\omega\uparrow)^p (u+v)$ where p ≥ 0, uv, and v is a limit ordinal, we take $a_n=(\omega\uparrow)^p (u+v_n)$ for a suitable sequence of ordinals $v_n$ tending to v.

For $(\omega\uparrow)^p v$ where p ≥ 0, and v is a limit ordinal, we take the sequence $(\omega\uparrow)^pv_n$ for a suitable sequence of ordinals $v_n$ tending to v.

For $(\omega\uparrow)^{p+1} (a+1)$ where p ≥ 0 and a ≥ 0, we take the sequence $(\omega\uparrow)^p (\omega^a n)$.

For example, this procedure defines for the order type $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot5+578$ a particular, very rapidly increasing, sequence of integers, and to specify a particular large integer, we can refer e.g. to the 37th element in this sequence. The sequence is defined in 578 steps from the sequence with order type $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot5$, which in turn is defined from the sequences with order types $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot4+\omega^{\omega^n}$. The twelfth element in this sequence is defined from the sequences with order types $\omega^{\omega^{\omega6+42}\cdot1729+\omega^9+88}\cdot3+\omega^{\omega^\omega}\cdot4+\omega^{\omega^{11}n}$, etc.

For ε0 we can take the sequence $\omega\uparrow\uparrow n$, and for ε0(k+1) the sequence ε0k + $\omega\uparrow\uparrow n$, etc. We cannot reach omega-one1), the set of all countable ordinal numbers, and the smallest uncountable ordinal number: no sequence of ordinal numbers below ω1 has that ordinal as limit. It is also clear from the fact that we define sequences of which each element is defined in a finite number of steps, so making use of only a finite number of auxiliary sequences. Therefore for any of our sequences the set of auxiliary sequences is countable.

We have $f_{a}(n), except that we have "=" for n = 1. Note however that for a < b we do not always have $f_{a}(n). For example:

• $f_{3}(n) for n = 1, 3, 4, 5, 6, .., but we have "=" for n = 2.
• $f_{4}(n) for n = 1, 4, 5, 6, .., but we have ">" for n = 2, 3.

I would like to see this page become an A-class mathematics article. Might I suggest that one of the changes that would improve this article is to split it up into Large numbers and Notation of large numbers? AJRobbins (talk) 02:50, 20 November 2007 (UTC)

You need a notation to talk about things. There is only a need for a separate page if there is enough to say about alternative notations which is not needed for talking about the numbers themselves.--Patrick (talk) 12:53, 20 November 2007 (UTC)
Only the second part of "Systematically creating ever faster increasing sequences" was taken out, and is has not been put back.
Examples clarify notation, so perhaps they can better be put together.
Patrick (talk) 23:40, 22 November 2007 (UTC)

## Graham's number?

I noticed that the page said that Graham's number is between $10 \rightarrow 10 \rightarrow 64 \rightarrow 2$ and $10 \rightarrow 10 \rightarrow 65 \rightarrow 2$. Shouldn't the 10's be 3's? And if that's true, where does it go? ZtObOr 22:50, 22 January 2008 (UTC)

That's covered in footnote 4, where it is explained "regarding the comparison with the previous value: $10\uparrow ^n 10 < 3 \uparrow ^{n+1} 3$, so starting the 64 steps with 1 instead of 4 more than compensates for replacing the numbers 3 by 10".

Also related to Graham's Number, I'd like to keep it in "See Also" as was recently added -- even though it's linked by the article text -- either that or, perhaps we can diminish that long list of example numbers. If we keep 10 → 5 → 2 and 10 → 9 → 2, can't we remove the ones in between? (etc.) —Preceding unsigned comment added by Mrob27 (talkcontribs) 22:45, 13 February 2008 (UTC)

Re:[[1]], yes, 3 → 3 → 65 → 2 is larger than 10 → 10 → 64 → 2, but AIUI, Graham's Number is somewhere between 3 → 3 → 64 → 2 and 3 → 3 → 65 → 2. It is still no proof whether GN is larger than 10 → 10 → 64 → 2, is it? Chasrob (talk) 23:56, 11 December 2011 (UTC)
Footnote 4 in the article gives the basic intuition as to why 10 → 10 → 64 → 2 < Graham's number, but of course it's not a proof. To sketch how a proof could go, define the functions h(n) = 10↑n10 and f(n) = 3↑n3, and note that h64(1) = 10 → 10 → 64 → 2 and f64(4) = Graham's number. Then h64(1) < f64(4) can be proved by repeatedly applying the fact that h(n) < f(n+1) - 1 (which fact can be proved by induction):
h(1) < f(2) - 1
h(h(1)) < h(f(2) - 1) < f(f(2)) - 1
h(h(h(1))) < h(f(f(2))) - 1) < f(f(f(2))) - 1
...
h64(1) < h(f63(2) - 1)< f64(2) - 1 < f64(4) ■
r.e.s. (talk) 05:05, 12 December 2011 (UTC)
So, h(h(1)) < h(f(2) - 1) < f(f(2)) - 1. Thanks. Chasrob (talk) 16:22, 12 December 2011 (UTC)
You're quite welcome; however, I don't see why you're mentioning that particular step in the proof-sketch. In any case, here's a link to an induction proof of a change-of-base inequality that easily establishes h(n) < f(n+1) - 1, which is driving the proof-sketch. In terms of Knuth arrows, the change-of-base inequality is bkn < 2↑k((b-1)n), giving h(k) = 10k10 < 2↑k90 < 3↑k90 < 3↑k + 13 - 1 = f(k+1) - 1 for k ≥ 2 (for k = 1, the result may be verified directly).
r.e.s. (talk) 18:32, 12 December 2011 (UTC)

I cannot deal with this article. —Preceding unsigned comment added by 68.46.238.3 (talk) 21:59, 31 May 2008 (UTC)

## Merge to accompanying article of Leviathan number

Leviathan number is a long-standing stub on a number apparently invented by one author for the sole purpose of having a sexy example of a large number, and never used for any non-pedagogical purpose, but only as a curiosity or a guaranteed stumper for anyone who has not made a point to pay attention to useless information. Barring evidence that the British Pickover is an employee or owner of the US-based Mathematica, it does seem to have found slight recognition so i'm not inclined to AfD. But i think it would be more valuable as a section in the accompanying article (with a section lk from its current title) than standing alone.
--Jerzyt 22:30, 26 February 2009 (UTC)

## Computational Complexity

THe section on computational complexity (the part that talks about "what if the universe could be one big computer" seems to include some original research or conclusions drawn by inference from the cited paper. Particularly where cryptology is concerned. I would recommend using additonal sourcing for this section. HappyJake (talk) 18:33, 1 September 2009 (UTC)

This section is much too long and to 95% out of subject, (doubtful) "explanations" on quantum computers etc belong elsewhere. It should be condensed to a dozen of lines giving the numbers and what they refer to, conveniently linked to the article featuring further reading. (as a side note, i don't know anybody using 40 character passwords exhausting the full range of 8 bits. 8 characters ranging from ascii 32 to 97 is much more realistic, and considering the number of printable characters in the latin-n char sets should put us definitely on the safe side) — MFH:Talk 15:04, 29 May 2011 (UTC)

## Order type of sequences

From the article:

Each sequence in this whole process can be identified by its order type in the process:

• (10→nk) is the sequence with index n with order type k - 1
• (10→10→nk) is the sequence with index n with order type ω + k - 1
• (10→10→10→nk) is the sequence with index n with order type ω2 + k - 1
• (10→10), (10→10→10), (10→10→10→10),... is the sequence with order type ω²

Don't all increasing sequences have order type ω? For example, all sets with order type k < ω have exactly k elements, but the sequence an = (10→nk) is an infinite sequence.

What is the statement supposed to mean? Something about the complexity of the algorithm used to compute the numbers??? — sligocki (talk) 02:41, 19 August 2010 (UTC)

## Locating chained arrows in the fast-growing hierarchy

The section "Systematically creating ever faster increasing sequences" was recently revised (improved, imo) to focus exclusively on the fast-growing hierarchy. In the last group of examples now given there, concerning Conway chained arrow notation, the final one is the following:

• fω2(n) = fωn(n) cannot be expressed finitely with Conway arrows.

This is misleading, I think, because for any fixed n > 1, fω2(n) < nn → ... → nn for a sufficiently large finite number of n's in the chain. On the other hand, the following result would be consistent with the other examples:

• fω2(n) = fωn(n) > nn → ... → nn (Chain of n+2 n's).

Although these examples otherwise seem correct, it would be excellent if a "citable" source could be given. Lack of such a source is what prevented me from inserting similar examples in the article on Conway chained arrow notation, as mentioned in talk-page comments at Locating n-arrow chains in the fast-growing hierarchy.)
r.e.s. (talk) 16:43, 20 August 2010 (UTC)

Sorry about the hand-waving and uncited statements. What I mean by fω2(n) cannot be expressed finitely with Conway arrows is actually that it cannot be expressed by a fixed notation in conway arrows as a function of n. Alternatively that there is not bounded Conway arrow notation as n → ∞. I believe this is pretty straightforward, but I could be wrong. Sorry to include original research here, but I felt that it was better than the OR that was already present and at least more based in more commonly studied notations. Cheers, — sligocki (talk) 05:40, 21 August 2010 (UTC)
(BTW, it looks like you've omitted an apostrophe in writing fω2, inadvertently italicising the remainder of your paragraph.)
Oy, wiki syntax makes me crazy, thanks.
(1) It seems likely that "cannot be expressed" is weaker than what you intend to say, because "most" positive integers cannot be expressed by a Conway chain except by using the number itself in the chain (as the first element); e.g., the number 2 can be expressed only by using 2 itself as an element (i.e., 2, or 2→1, or 2→1→X). In this sense, only perfect powers (4, 8, 9, 16, 25, ...) are nontrivially expressible. Thus, any function whose range includes a number that's not a perfect power cannot be expressed nontrivially in Conway notation.
Hm, I don't mean to say anything about a number not being able to be represented with arrows without using that number.
(2) You seem to be saying that, as a function of n, no chain of the form nn → ... → nn (with a fixed number of 'n's) is an upper bound on the function fω2. Although that's true, it's a much weaker statement than the second bulleted statement above (which follows from the inequalities you posted — but see (3) below), in which the number of 'n's varies with n. (If the number of 'n's in the chain is given by a function g(n), then for g(n) = n + 2 we have the bulleted statement, but obviously there exist functions g such that the resulting "chain function" is an upper bound on fω2.)
Fair enough, I thought it was notable to point out the limitations of chain arrows, because they seemed to be used in this article as the example of large number notation, where-as I think that they are unnecessarily complicated to compare and annoying to deal with.
(3) Looking more-closely at the inequalities you posted, some are evidently too strong. (Also, you seem to have mistakenly treated pqr as pq r. For example, 2 ↑n - 1 n = 2 → n → n-1 ≠ 2 → n-1 → n.) Note that a counterexample to
• fωk(n) > nn → ... → nn (Chain of k+2 n's)
is obtained with k = 1, n = 3. On the other hand, if I'm not mistaken, a valid inequality has one less n in the chain:
• fωk(n) > nn → ... → nn (Chain of k+1 n's).
This leads to the following slightly weaker version of the second bulleted inequality in my opening comment above:
• fω2(n) > nn → ... → nn (Chain of n+1 n's).
I know this isn't the place to discuss OR, but I think this is worth the brief mention.
r.e.s. (talk) 18:35, 22 August 2010 (UTC)
Oops, that's what I get for being to hand-wavey. I'll look into it. Thanks, — sligocki (talk) 02:43, 23 August 2010 (UTC)

## A Billion, Anyone?

The article seems to use two different conventions for the meaning of a "billion" in a single sentence: 13.7 billion years (4.3 × 10^17 seconds) old, ... and contains about 5 × 1022 stars, organized into around 125 billion (1.25 × 10^11) galaxies". Tkuvho (talk) 19:51, 9 January 2011 (UTC)

The same, 10^9.--Patrick (talk) 23:57, 9 January 2011 (UTC)
OK, thanks. I missed the "seconds". Tkuvho (talk) 01:15, 10 January 2011 (UTC)

## Accuracy section

The statement "This seems like extremely poor accuracy, but for such a large number it may be considered fair (a large error in a large number may be 'relatively small' and therefore acceptable)" in the context of its previous statement is incorrect. The error in the exponent implies the error is already "relative". Thus, even though the number is large, the "relative error" is also very large. Subh83 (talk) 08:08, 29 January 2011 (UTC)

## Delete BOX_M~

Pending some discussion here, I recommend deleting $BOX_{-}\widetilde{M}$ from the examples of large numbers — it was edited into the article by its inventor, and is apparently only self-published online. The author defines this number in terms of a hodgepodge of functions, claiming it to be an "upper limit for ... the set of all integers that have been written/used/imagined by man to date" ("limite superiore per ... l’insieme di tutti gli interi che sono stati scritti/utilizzati/immaginati dall’essere umano fino ad oggi"). Whatever that might mean, it may be worth noting that $f_{\omega^2 + 1}(2) \ll BOX_{-}\widetilde{M} \ll f_{\omega^2 + 2}(3)$ with reference to the fast-growing hierarchy. — r.e.s. (talk) 01:46, 2 February 2012 (UTC)

Hello, I have no interest into adding bad informations to Wiki, so I'll try to explain/translate better the meaning of $BOX_{-}\widetilde{M}$ and, if that number is not good for the page, it will be ok to delete it for me too. BOX_M is based on a numbers-philosophy related issue, concerning the relation between "human problems" and natural (plus real) numbers. This is a quite thin point and you (R.E.S.) have forgotten to take a look at the other condition stated into the article: the set of "human problems" which BOX_M is concerning about (at the date of the article) doesn't include the "find a bigger number problems" (i.e. busy beaver, Turing Oracle, etc). In short, it represents an upper limit for real world/mind related problems with an immaginable link to external world reconnected into human eyes. It could be useful for upper limits of that kind, due to it constructions too (you can find "small" upper limits for large solutions using the right piece of the number definition at the right step).

No objection about the inequality on the pure math based "fast-growing hierarchy" stated above. The point is that $f_{\omega^2 + 2}(3)$ hasn't been appliyed to a specific problem outside the "find a bigger number than... problem" before the date of the BOX_M article and no one gave a "first name" to the specific inputs of the funtion ---> no one gave a specific name to the result/number of it.

If this kind of mental constructions is "outside the BOX", IMHO it's not a problem to delete BOX_M, but I haven't ever stated that it's the biggest number ever thought by a human being, I've stated that it was the biggest one at the moment of the article publication concerning human problems, related to thinkable problems concerning human dimension, excluding the "find a big number ones". I only claimed to be the first one to find this empty space into that kind of speculations... I have not the intention to create a big number only.
The first part of the article is inspired from a published book of mine with an ISBN number etc...
The best, Marcokrt (talk) 01:30, 7 February 2012 (UTC)

I'm removing BOX_M according to r.e.s.'s comment and WP:SPS. Your book, like your website, appears to be self-published and does not meet the standards of reliability for a WP source. 12:23, 4 July 2012 (UTC)
• ^ Another number instead of 10 could also be used; it has to be more than 1, and if $f_0(1)=2$ it has to be more than 2 to get an increasing sequence.
• ^ Alternatively, selecting the "cross-sequence" $f_0(2), f_1(2),..$, we get $f_{k-1}(2)=f_k^{2}(1)=10 \uparrow^k 2=(10 \to 2 \to k=10 \uparrow^{k-1} 10=(10 \to 10 \to (k-1))$, so the sequence is only shifted one position.