You might be wondering what an article on Googology is doing on a website for writing about games. This article is here because of my interest in math and googology, as well as defining fractorials and such better, with more powerful extensions than before.
First things first. What is addition? You probably already know. As a function, it can be represented as:
F(a,b) = S^b(a), Where S denotes the successor function.
You might be wondering what I mean by “successor function”, and what the small “b” represents. Simply,
S(n) = n+1
f^b (a)=(f(f(f(⋯(a)⋯), with b f’s
Now, multiplication is still pretty simple. As a function:
M(a,b)=A^(b-1)(a,a)=A^(a-1)(b,b), where M represents multiplication, and A represents addition. You may remember addition needs two terms, so:
f^c (a,b)=f(a,f(a,f(a,⋯(a,b)⋯) (c f’s) or f^c (a,b)=f(f(f…(a,b),b),b…,b)with c f’s.
These would be completely different and contradict each other, but for this, it doesn’t matter.
Exponentiation is a bit more interesting, as unlike multiplication and addition, a^b ≠ b^a.
E(a,b)=M^(b-1)(a,a)≠M^(a-1)(b,b)
All of this can also be represented as:
A(a,b)=a+b
M(a,b)=a×b
E(a,b)=a^b=a↑b
You might be wondering why I chose “a↑b” as a representation of E(a,b). This is called knuth’s up arrow notation, and it will be useful later on.
The first Hyperoperator is usually referred to as tetration. It has several formats, including:
a↑↑b
ba
a[4]b
And several others.
As a function:
T(a,b)=E^((b-1)) (a,a)=a^(a^(a^(a…^a)…), with b copies of a.
The second Hyperoperator is usually referred to as pentation. It also has several formats.
Operator notation is a notation that is used to denote hyperoperators and regular operators
a{c}b=a↑↑⋯↑↑b (c arrows)
Examples:
a{1}b= a^b
a{2}b= ba
a{3}b=a↑↑↑b
a{0}b=a×b
a{-1}b=a+b
a{10}b=a↑↑↑↑↑↑↑↑↑↑b
a{c+1}b=a{c}a⋯a{c}a (b copies of a) = a{c}(a{c+1}b-1)
Factorials are a simple function, denoted as x!
As a function, they can be expressed as:
f(x)=x×f(x-1), f(0)=1
This can also be expressed as:
f(x)=M(x,f(x-1))
, using our previous multiplication function.
This can be expanded, to:
f_2 (x)=E(x,f_2 (x-1))= x^(f_2 (x-1))
This is also known as expofactorial.
Also, f_2 (0)=1.
This can be extended to hyperoperators, which we discussed earlier.
f(a,b)=a{b}f(a-1,b)
In one of my notations, this is denoted a?b.
There are many ways to extend factorials. The way I chose to extend it in my specific notation, Fractorials (no, not factorials), is:
a??b=a{b{b⋯{b}⋯b}b}(a-1)??b, with b {‘s on each side.
This can be extended, to:
a???b=a{b{b⋯{b}⋯b}b}(b{b{b⋯{b}⋯b}b}b {‘s on each side (b {‘s on each side))(a-1)???b
You might be seeing where I’m going with this. Now, for upside-down ?’s.
a¿b=a??⋯??b (b copies of ?)
a¿¿b = a¿a¿…¿a, with b ¿’s
Now, Koppa. This symbol, ϟ, determines the number of ¿’s.
aϟb=a¿¿⋯¿¿b (b copies of ¿)
OR
aϟb=a¿bϟ(b-1)
The “A” Function
The A function is a simple function
A(n)=A(n-1){A(n-1)}A(n-1). A(0)=base
“base” is usually 10, but it will be called “base” or “b” to make it easier later on.
This might seem off-topic, but the fast-growing hierarchy (FGH) is a comparison that can be made. It is simple.
f_0 (n)=n+1
f_1 (n)=(f_0)^n (n)=n+1+1⋯+1 (n +1’s)=2n
f_(α+1) (n)=f_α^n (n)
f_α (n)=f_α[n], if α is a limit ordinal. Which means take the nth term along the diagonal.
Next, we need to learn about ordinals, specifically limit ordinals. You might be wondering why I’m using α instead of x, and this is simply because α represents an ordinal whereas x represents a variable.
Limit ordinals are a type of ordinal that represents the limit of something. Example: ω, the first limit ordinal, is the limit of finite ordinals, or the sequence (1,2,3,…). The next limit ordinal, ω2, is the limit of the sequence (ω+1,ω+2,…). Eventually you reach ωω, or ω², which is the limit of the sequence (ω,ω2,ω3,…). ω²2 is next. This is the limit of the sequence (ω²,ω²+ω2,ω²+ω3,…). I can continue with this, but for now, going past ω² shouldn’t be required. Note: I probably got this wrong.
FGH on certain limit ordinals
This section will mostly be Math, and not English. Note: FGH is ill-defined so this is just my interpretation.
f_ω (n)=f_n (n)
f_(ω+1) (n)=(f_ω)^n (n)≠(f_n)^n (n)
f_ω2 (n)=f_(ω+n) (n)
f_ω(α+1) (n)=f_(ωα+n) (n)
f_(ω^2 ) (n)=f_ωn (n)
The “AA” Function and the “B” Function
The AA function is quite simple, and is as follows:
AA(n)=A^A(n)(n)
The AAA function is simple as well, and is as follows:
AAA(n)=AA^AA(n)(n)
Continuing:
AA⋯AA(n) (m copies of A)=AA⋯AAm^AA⋯AA(n)(n), with m-1 copies of A
If we keep going, we can get a two argument function,
f(a,b)=AA⋯AA(a) (b copies of A)
This is not what I am looking for, however. Instead,
B(n)=AA⋯AA(n), with n copies of A
BB(n)=B^B(n) (n)
(not to be confused with the Busy Beaver Function).
Generally:
α_a (b)=α_(a-1) α_(a-1)⋯α_(a-1) α_(a-1) (b), with b copies of α_(a-1)
Where α_n denotes the nth letter.
New Nuclear’s Array Notation (NNAN) is a system I came up with as a vague shorthand for letter functions. It can display huge numbers, like Graham’s Number, easily.
With 1 Entry:
{a} = a
With 2 Entries:
{a,b}=α_b (a)
With 3 entries:
{a,b,c}=(α_b α_b⋯α_b α_b )c (a), c copies of α_b
Including base:
[x]{a,b,c}=(α_b α_b⋯α_b α_b ) (a), A(0)=x, c copies of α_b
Example: Graham’s number, G64
As a function, graham’s number can be expressed like so:
G(n)=3{G(n-1)}3, G(0)=4.
In NNAN, G64 can be approximated to [3]{65,1,1}, or A(65) when A(0)=3.
Numbers
Of course, I still need some numbers. Firstly, The Googol factorial Series:
10¹⁰⁰! (Googolbang, I believe)
10¹⁰⁰.> (10¹⁰⁰?1) (Probably has a name but I can’t remember it)
10¹⁰⁰?10¹⁰⁰ (Googolmark. Once again this probably already has a name so yeah)
10¹⁰⁰??10¹⁰⁰ (Gooqulmark. I am unsure if this has a name already or not)
10¹⁰⁰ ¿10¹⁰⁰ (Googoldown. I think this one has an original name)
10¹⁰⁰ ¿¿10¹⁰⁰ (Googoldudown. Same Here)
10¹⁰⁰ϟ10¹⁰⁰ (Koogol. A large number that probably has an original name)
Next up, some array numbers:
[3]{3,3} (I don’t really have a name for this one)
[10]{4,3} (This one is Explosive (C(4)))
[10]{64,7} (Not Graham’s Number (G(64)))
And a few other numbers featuring fractorials:
3?(3?(3?…5)+1)…+1), with 64 3’s = Graham’s number
3¿(3¿(3¿…5)+1)…+1), with 64 3’s = Anti Graham’s number
Notes:
I do not have much experience in googology, so most of this stuff is probably ill-defined. I probably got a lot of stuff wrong.
I’m too dumb to understand this, good job though