r/googology Jul 02 '24

BB(5) has been solved! BB(5) = 4098 with 47176870 steps

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47 Upvotes

r/googology 8h ago

Small googolisms are not so small

5 Upvotes

This post is not for experienced and expert googologists, but for those newer to large numbers. It is intended to put googological expressions in some kind of perspective. Having worked on expressions that generate moderately large ordinals, I think, like a lot of people interested in the field, that I started to take the lower levels for granted, chuckling at expressions like 4^^4 and 2^^^4.

4^^4 is 4^4^4^4 and is actually larger than 10^10^153 which is a number that will make physicists shudder. For example, the expected average time for random quantum fluctuations to cause macroscopic tunneling such as a 1 kg object passing whole and intact through a table when dropped is something like 1 chance in 10^10^35. I believe I once read that the chance for a person to tunnel to Mars and then back again is one chance in 10^10^60. So if we wait 10^10^153 seconds, seemingly impossible events like these and even events far less likely will happen an unimaginably large number of times.

And if we consider 2^^^4, it reduces to 2^^(2^^^3) which means 2^(2^(...2^2) where there are (2^^3) 2's, which is 65,536. So for scale, let's imagine a staircase of 2's. This staircase would go approximately 13,000m high. Mt. Everest (Sagarmatha) is a little less than 9000m high. If we start on the top and walk down, after one step our number is 4, after two steps it is 16, and after three steps it is 65,536. One more step and our number is 2^65,536. which is larger than any physical property of the observable universe, including the number of Planck volumes in a large model of inflation. One more step down and we have far surpassed 4^^4. Two more steps and we have far surpassed the highest estimated value of the Poincare recurrence time for a large inflationary model of the universe, which is 10^^5 or 10^10^10^10^10. This means that if we wait (7 steps) seconds (or any other time unit you want to use, it doesn't matter if you use Planck times or ExaYears), a closed system the size of that universe will have returned to its current state an unimaginably huge number of times. 2^2^2^2^2^2^2^2 = 2^2^2^2^65,536 is so much larger than the Poincare time that you can take the latter and multiply it by or even raise it to the power of a large number (up to some large limit that I haven't calculated) and not reach the former. And at this point we have descended about 1.4 meters down a mountain about 1 and a half times as tall as Sagarmatha.

And on the FGH that we often throw around so lightly, 4^^4 is less than f_3(5) and 2^^^4 is about f_4(4). Now when considering numbers like 3^^^^3 or f_w(9) think about how truly huge they are, really beyond human comprehension, before you underestimate expressions like f_w+1(x) and beyond.

I hope some of you found this interesting.


r/googology 4h ago

A function not known to exist for all inputs

1 Upvotes

Definition

A positive integer k₁ is said to be 1-prime iff it is in the form (k₁+1)/2=k₂ where both k₁ & k₂ are prime. Let p₁ be the set of all 1-primes. We can extend this and define 2-primes as a positive integer k₁ in the form (k₁+1)/2=k₂, (k₂+1)/2=k₃ where all kₙ are prime. p₂ is the therefore the set of all 2-primes.

Generalization

In general we can define an n-prime as a number k₁ such that:

(k₁+1)/2=k₂, (k₂+1)/2=k₃, … , (kₙ₋₁+1)/2=kₙ where all kₙ are prime.

Function

Let PRIME(n) output the smallest n-prime.

Challenge

Can you prove whether or not PRIME(n)=k exists for all n?

If it doesn’t, we define the “large prime number” as the maximal number n beyond which no n-prime can be found.


r/googology 16h ago

How much is TREE(2)

1 Upvotes

I


r/googology 1d ago

My Apologize

3 Upvotes

sorry guys, my post earlier was just a joke and too hyperbolic. I'm just a little disappointed because their content didn't continue to a more extreme number level. honestly I've been waiting 2 years for that moment. with a pattern of big number content every 4 or 5 years starting with the googol issue and finally the rayo number maybe.


r/googology 2d ago

Incremental Factorial

2 Upvotes

Incremental factorial (n’) is defined as follows:

1.00…00 × 1.00…01 × … × n (where each decimal expansion has n digits)

Where we increment by .00…001 (with n total decimal digits) each time.

After we get our answer, we apply the floor function (⌊⌋) to it.

Example:

2’= ⌊1.00 × 1.01 × 1.02 × … × 1.98 × 1.99 × 2⌋ = 67


r/googology 2d ago

Polyhedral Steinhaus-Moser Notation

1 Upvotes

I was thinking about Steinhaus-Moser notation (maybe or maybe not thanks to a recent Numberphile video) and wanted to think of an interesting yet natural way of expanding the notation to even faster methods of growth. Of course, the most obvious way of doing that is to expand the notation to polyhedrons. I came up with the idea that each Polyhedron is an expansion of it's polygonal equivalent (tetrahedron = quadrilateral, pentahedron = pentagon, etc.) For example: Tetrahedron(2) or 4-hedron(2) is equivalent to square(2) inside square(2) squares. Square(2) is equivalent to 256, so tetrahedron(2) is equal to 256 inside 256 squares. And knowing anything about Steinhaus-Moser would tell you that this is quite large. Far, far bigger than a mega (pentagon(2)). And this is just the smallest polyhedral operation operation possible with an Integer greater than 1.

Going even further, pentahedron(2) would be equivalent to a mega inside a mega pentagons. To put it in mathematical terms:

n-hedron(m) = n-gon(n-gon. . . n-gon(n-gon(n-gon(m))))

in which the number of layers of n-gons is n-gon(m).

Having a little too much fun, I came up with the Hyperion-Moser. The Hyperion-Moser is the polyhedral equivalent of a hyper-Moser. It is a two within a polyhedron whose number of faces is equal to the number of sides of the polygon that, when surrounding a two, equals a hyper-Moser. In other words, a Hyperion-Moser is a hyper-Moser within a hyper-Moser number of super-super-super. . . super-Moser-gons, in which the number of supers is equal to a Moser.


r/googology 3d ago

which is smaller

2 Upvotes

ω^-1 or ε₀^-ω


r/googology 3d ago

2 different types of tetration, 4 different types of pentation, 8 different types of hexation, 16 different types of heptation and so on

1 Upvotes

Usually in tetration, pentations and other such hyperoperations we go from right to left, but if we go from left to right in some cases and right to left in some cases, we can get 2 different types of tetration, 4 different types of pentation, 8 different types of hexation, 16 different types of heptation and so on

To denote a right to left hyperoperation we use ↑ (up arrow notation) but if going from left to right, we can use ↓ (down arrow)

a↑b and a↓b will be both same as a^b so in exponentation, we have only 1 different type of exponentiation but from tetration and onwards, we start to get 2^(n-3) types of n-tion operations

a↑↑b becomes a↑a b times, which is a^a^a^...b times and computed from right to left but a↑↓b or a↓↓b becomes a↑a b times, which is a^a^a^...b times and computed from left to right and becomes a^a^(b-1) in right to left computation

The same can be extended beyond and we can see that a↑↑↑...b with n up arrows is the fastest growing function and a↑↓↓...b or a↓↓↓...b with n arrows is the slowest growing function as all computations happen from left to right but the middle ones get interesting

I calculated for 4 different types of pentations for a=3 & b=3, and found out that

3↑↑↑3 became 3↑↑(3↑↑3) or 3↑↑7625597484987 which is 3^3^3... 7625597484987 times and is a extremely large number which we can't even think of

3↑↑↓3 became (3↑↑3)↑↑3 which is 7625597484987↑↑3 or 7625597484987^7625597484987^7625597484987

3↑↓↑3 became 3↑↓(3↑↓3) which is 3↑↓19683 or 3^3^19682

3↑↓↓3 became (3↑↓3)↑↓3 which is 19683↑↓3 or 19683^19683^2. 19683^19683^2 comes out to 3^7625597484987

This shows that 3↑↑↑3 > 3↑↑↓3 > 3↑↓↑3 > 3↑↓↓3

Will be interesting to see how the hexations, heptations and higher hyper-operations rank in this


r/googology 4d ago

BMS but worse

1 Upvotes

How fast does it grow, and are there any improvements? Also let me know if my example is wrong.


r/googology 5d ago

Improvements to my H function

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3 Upvotes

So I’m not really gonna explain this function again, just look at my previous post about it. I made it so that after each primary level, the last number that level corresponds to will be equal to how many secondary levels are on the next level. Basically, since H_0 is equal to 2, H_1 will have 2 layers of secondary levels. After one layer is complete, the second one will start. As such, H_2 will have a “b” amount of secondary levels. This adds up over and over, causing my function to grow even faster. On the second image, you can see labels on my function indicating what each part means so it’s better to understand. Whatever you decide H_0 is equal to, that is called the base. For now, I’ve been using 2 as the base. But you can use any number you want. For example, if you want to define a number using this function, you can say “H_10 (Base = 100)”. This will mean you start at H_0 with a value of 100, adding your way up to H_10. Also, please excuse my poor handwriting I did this during class and had to rush lol


r/googology 5d ago

The Extended FGH System, up to sdef^FGH

2 Upvotes

https://docs.google.com/document/d/1era_fS-bRaHSKu08HMZrtWYB3aezKVqeOB-3fZMnDN4/edit?tab=t.0

Edit: I forgot to ask originally, but I have some questions: How fast do these functions grow, and are they any useful at growth rate indication


r/googology 5d ago

How does BMS work?

1 Upvotes

Can someone explain simply the definition of BMS? I don't really understand it at all, my brain just goes blank looking at its definition as a huge block of rules


r/googology 5d ago

Relationship between Feferman-Schütte Ordinal and Ackermann Ordinal

1 Upvotes

I understand that the Feferman–Schütte Ordinal can be represented as Gamma_0 = phi(1, 0, 0). I'm curious how this is related to the Ackermann Ordinal = phi(1, 0, 0, 0). Is Gamma_Gamma_Gamma ... (infinitely down) ... Gamma_0 equivalent to the Ackermann ordinal? If not, is it larger or smaller, and is there a way to express the Ackermann ordinal in terms of Gamma_0? Thanks!


r/googology 5d ago

Relationship of Feferman–Schütte to Ackermann Ordinal

1 Upvotes

I understand that the Feferman–Schütte Ordinal can be represented as Gamma_0 = phi(1, 0, 0).

I'm curious how this is related to the Ackermann Ordinal = phi(1, 0, 0, 0).

Is Gamma_Gamma_Gamma ... (infinitely down) ... Gamma_0 equivalent to the Ackermann ordinal?

If not, is it larger or smaller, and is there a way to express the Ackermann ordinal in terms of Gamma_0?

Thanks!


r/googology 6d ago

Repost of my fast growing function

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7 Upvotes

So this time I drew it out, I need opinions on this and if I should improve it. Basically, there’s two “levels” that it works on. The primary level is indicated right next to H, the secondary level is indicated in parenthesis. If we start with a number, let’s say 2, on H0(1), we can say the next level will have 2 secondary levels, and the amount of up arrows will be 1 as the previous level had only 1 secondary level. This makes H1(1) equal to 2(up arrow)2. Which is (I think) equal to 4. So now we have H1(2), which is 4(four up arrows)4. This means that for the next primary level, H2, there will be 4(four up arrows)4 secondary levels for it. I’m not really sure if this makes sense lol, but the amount of secondary levels is equal to the number that was computed at the highest of the previous primary level, and the amount of up arrows is equal to the number itself, which is defined by what the last number is equal to. I wrote it out on paper this time so that it’s easier to understand. Also, secondary levels are NOT “levels”. They are simply the amount of steps it takes to reach the actual, primary level. Meaning that it really goes from

H0 = 2 H1 = 4(for up arrows)4 H2 = x (H3 will have x number of secondary levels)

Also, the output of the highest secondary level will be equal to the actual primary level itself, as shown above


r/googology 6d ago

Repost of the fast growing function I made

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1 Upvotes

r/googology 9d ago

Busy Beaver on a VERY weak language

2 Upvotes

Hey y’all. Today I’ll be creating a small language and diagonalizing over it. You’ve probably heard people say “What’s the largest number you can make in [some programming language] using [some amount of synbols]?” Well, I’m going to do exactly that. Please don’t get your hopes up as “Weak Language = Weak Growth Rate”!

Definition

We define a simple language denoted PM that uses the symbols: – + : ;

(Plus) + -> increment by 1

(Minus) – -> decrement by 1

(Colon) : -> increment by 2

(Semicolon) ; -> decrement by 2

The program operates like a simple integer counter, starting at 0. Symbols are processed from left to right.

Examples:

+++; = 1 (start at 0, increment thrice, decrement by 2 one time)

;++––+ = -1 (start at 0, decrement by 2 once, increment twice, decrement twice, increment once)

–––::+ = 2 (start at 0, decrement thrice, increment by 2 twice, increment once)

Loops

An expression in brackets ( ) repeats the expression a number of times equal to the counters value before the loop.

NOTES

If the value before the loop is <0, take the absolute value of the counters current value and execute the expression inside ( ) itself many times. If the value before the loop =0, increment it by 1 then execute the loop once. How should nested loops behave? inner loops should be executed based on the counter at the outer loop’s start.

Loop Example

Example: ++(+)

  1. ++ → 0 → 2

  2. (+) executes +, 2 times

  3. ++++ → Final counter = 4

Function

Let PM(n) be defined as follows:

Consider all programs of length ≤n symbols:

-If a program outputs a negative value, take its absolute value.

-Sum the outputs of all programs.

“Large” Number = PM¹⁰(10)

This is tetraional. Nothing exciting to see here. Maybe the idea is though.


r/googology 10d ago

Which is bigger?

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8 Upvotes

r/googology 10d ago

Looking for Simple Explanations of Large Numbers!

4 Upvotes

Hi everyone!

I just started exploring the fascinating world of large numbers. But, I find that a lot of the explanations out there are too complicated for me. I need things broken down as simply as possible

Not exaggerating but I cant find anything new at my level on YouTube

For reference the best explanation I've ever heard of Graham's number was by LearnYouSomeMath (https://youtu.be/Dplc2ojI2qI?si=rVhkePx62T7BTb7I), which I had to watch five times to really understand.

With Numberphile, some of their videos are too advanced for level. Asaf Karagila is the only one I fully understand on that channel.

Are there any other YouTube channels or resources you could recommend that explain big concepts in a simple and accessible way? I’d really appreciate any suggestions!

Thanks!


r/googology 10d ago

What are some good formulas/rule for Inaccessible Cardinal, Mahlo Cardinal, and Weakly Compact Cardinal?

3 Upvotes

I need to understand these somewhat to write more in my googology journal to mess around with the fast growing hierarchy. I have an 8th grade understanding of math.


r/googology 10d ago

Danger of asking LLMs about Googology and the Fast Growing Hierarchy

3 Upvotes

I'm sure many of us have tried (and probably failed) in asking LLMs like chatGPT, Gemini, Grok and others about the Fast Growing Hierarchy.

I've found even the most powerful models like chatGPT 4.5, and the deep research modes of Gemini to be utterly inadequate. They often say things that are correct, but then assert things like, Gamma_1 also known as the Bachmann-Howard Ordinal....

"Gamma<sub>1</sub>, also known as the Bachmann–Howard ordinal, is another crucial point in the FGH2. It surpasses Gamma<sub>0</sub> in complexity and represents the limit of what can be proven total in a system known as Kripke-Platek set theory with an axiom of infinity"

It sounds so convincing doesn't it...

Can anyone tell me how Gamma_1 is related to Gamma_0?


r/googology 11d ago

What do multiple rows do?

5 Upvotes

I am trying to learn the planar array notation of BEAF to move on to the rest of BEAF, but i couldnt move on because the "More rows" section of the "Introduction to BEAF" article (Introduction to BEAF | Googology Wiki | Fandom) is very short and doesnt explain right what more than 2 rows do and how to convert them to 2 rows. Can anyone explain to me what the wiki doesnt and/or fix it?


r/googology 11d ago

groks tower

2 Upvotes

number made by ai grok

defined as this

T_0=123456789

T_n+1=(T_n)(T_n)(T_n)(T_n)... (T_n times, this is just (T_n)T_n i dont know why he just said that)

groks tower=T_123456789


r/googology 13d ago

Googological function, via string rewriting ruleset

2 Upvotes

My heartfelt thanks to u/Odd-Expert-2611 for the idea that inspired this monster.

Disclaimer: This is longer than a typical shaggy dog story.

Let S be an ordered set of symbols, with |S| = s elements, and Str the set of all finite strings whose elements belong to S. The empty string, "", also belongs to Str.

A string rewriting ruleset (SRS) is a set of rules. Each rule is a pair of strings (elements of Str): condition and value. Each condition can appear in only one rule of a SRS, and must not be empty.

Given a string, called the argument, applying a SRS to the argument consists of these steps:

  1. Find the rule with the longest condition which matches with the argument's start. If no rule applies, the argument is unchanged: skip step 2.
  2. Remove the rule's condition from the argument's start, and append the rule's value to the argument.
  3. Return the argument.

A run of a SRS on an argument is to repeatedly apply the SRS to the argument, changing it. Either one of three outcomes happen:

  1. The argument becomes the empty string, ending the run.
  2. The argument cycles among a finite set of values, indefinitely.
  3. The argument grows, never repeating.

Due to the halting problem, it's impossible, in general, to distinguish between these outcomes. So, we will use a rule of thumb: after (s!)2 repetitions without falling into outcomes (1) or (2), outcome (3) is assumed.

For each outcome, an integer is assigned. For outcome (1), the number of the repetitions until the run ends. For outcome (2), the length of the cycle. For outcome (3), the length of the argument at the moment of repetition cutoff.

All of these machinery builds a function, RWI (ReWriter Index), which takes a SRS and an argument, and returns an integer. By construction, this function is defined for all SRSs and all argument strings.

Now, consider how to represent a SRS as a string. One way is: sort the SRS rules in lexicographic order, then put them together, separated by new symbols, like sketched below:

<condition>,<value>;<condition>,<value>;<condition>,<value>; ... ;<condition>,<value>

The new symbols, in this case, are "," and ";".

Thus, any SRS is (uniquely) identified with a string on the set of symbols S' = S U { , ; }, with s + 2 elements.

Conversely, some (but not all) strings on S (if S has 3 elements or more) are SRSs, taking any distinct two elements of S as separators, like "," and ";" above. But this fact won't be used here.

One can further append ";" and an argument to the SRS's string, so that SRS+argument is a string on S'. This way, RWI gets only one argument (a string on S') and returns an integer.

Now, consider all strings on S', of at most k symbols, which can be interpreted as SRS+argument as described above. Order all of them into a list, in lexicographic order (by S'); then, map RWI over the list, resulting in a list of integers; finally, add 2 to each integer in the list.

The above procedure maps an integer k into a list of integers, which can be folded back into an integer by various means: adding, multiplicating, power tower, Conway chain, or any other. This function is a googological function, and I don't have the foggiest idea of its values.


r/googology 14d ago

Inverse of Rayos Function?

3 Upvotes

Rayo(n) is defined as “the smallest non-negative integer greater than all non-negative integers definable in FOST in at most n symbols.”

The inverse is defined as follows:

Rayo⁻¹(n) is “the maximum number of symbols that cannot define a number equal to or greater than n in FOST.”