
5
This is a continuation of the discussion here. Here I am attempting to write a reasonable set of rules for an ordinal notation up to ψ(ψᵢ(0)) without consulting any reference materials. If it works as intended, the functions here will be very similar to the various versions of ψ that pop up in various places on this wiki.
Ofcourse, feedback (and especially correction of mistakes) would be most welcome.
Anyway, here goes:
1. To avoid confusion with all the different versions of ψ on this wiki, I'll notate my functions with "A" instead of ψ.
2. A_{β} is an ordinal function whose output is always between Ω_{β} and Ω_{β+1}
3. For α,β
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12H is a super fastgrowing hierarchy.
\(A_0(n)=n+1 \\ A_{b+1}(n)=A^n_b(n) \\ A_{\alpha}(n)=A^n_{\alpha[n]}(n)\) if and only if \(\alpha\) is a limit ordinal.
\begin{eqnarray*} A_0(n) &=& n+1 \\ A_1(n) &=& 2n \\ A_2(n) &=& 2^nn \\ A_m(n) &>& 2\uparrow^{m1}n \\ A_{\omega}(n) &>& 2\uparrow^nn \end{eqnarray*}
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 Egg Notation, Red and Blue eggs
Motivation: The idea was to make a FGH clone without using math language like functions, iterations, ordinals, etc.. in order to make it the least mathy possible it i chose to make it completely without numbers
Why eggs?, because notations in googology sooner or later end up being mixed into salad numbers so i figured i would provide a tasty ingredient for any salad.

 Concepts
 Eggs
 A pair of grouping symbols
 Egg colors
 Red = ()
 Blue = []
 Syntax Rules
 There may be no red egg to the left of a nonred egg. Example: ()[] is invalid
 The above rule also applies to eggs inside other eggs
 There is at least one red egg
 Symbol Glossary
 R = Zero or more Red Eggs.
 B = Zero or more Blue Eggs.
 A/B = Zero or more eggs of colors A and/or B
 …

You wanted more.
I forgot for a while.
Wojowu helped.
This. Is Big Bigeddon.
Although Sasquatch fits better.
And I'm not going to be quite as rigorous as my last post. That said, we work in the language \((\in, \bar\in,
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Well I had to do it at some point.
\(b(a,a)\)
\(a^2\)
\(b(a,a,1)\) \(f_2(a)\)
\(b(a,a,n)\) \(f_{n+1}(a)\)
\(b(a,a,0,1)\) \(f_{\omega}(a)\)
\(b(a,a,n,1)\) \(f_{\omega +n}(a)\)
\(b(a,a,n,k)\) \(f_{\omega k + n}(a)\)
\(b(a,a,0,0,1)\) \(f_{\omega^2}(a)\)
\(b(a,a,n,k,i)\) \(f_{\omega^2 i + \omega k + n}(a)\)
\(b(a,a,0,0,0,1)\) \(f_{\omega^3}(a)\)
\(b(a,a,n,k,l,m)\) \(f_{\omega^3 m + \omega^2 l + \omega k + n}(a)\)
\(b(a,a,b,c,d,e,.....)\) \(f_{...... + \omega^3 e + \omega^2 d + \omega c +b}(a)\)
Limit is \(\omega^\omega\).
\(b(a,a\{1\}1)\) \(f_{\omega^\omega}(a)\)
\(b(a,a,1\{1\}1)\) \(f_{\omega^\omega +1}(a)\)
\(b(a,a,n\{1\}1)\) \(f_{\omega^\omega +n}(a)\)
\(b(a,a,0,1\{1\}1)\) \(f_{\omega^\omega + \omega}(a)\)
\(b(a,a,n,k\{1\}1)\) \(f_{\omega^\omega + \omega k +…
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In my last Blog on the Supremus Array Notation I introduced numbers constructed using 3 brackets within [ ]'s this time I'm going to introduce 4 brackets. We'll use the previous examples too give an idea of size [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9)(5,6,7,8,9)] means that you repeat the process described in my previous Blog [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9) times. With 4 brackets you must repeat the process not [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9) times but [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9)(5,6,7,8,9)] times. 5 brackets means that you have to repeat the process [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9)(5,6,7,8,9)(5,6,7,8,9] times also something to remember the values don't have to be 5,6,7,8,9 then be whatever you like. We continue inserting brackets into the…
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In this Blog post I am going to define a salad number That may or may not prove to be interesting.
 Start with a relatively small number say 1 billion.
 Apply the up arrow notation on 1 billion,1billion times too make (10^9)↑(10^9)10^9) or 1 billion followed by 1 billion up arrows 1 billion.
 Call this number N and input it into the Ackermann function so A(N,N).
 Produce out another number and call this N.
 Input this into the Busy beaver function BB(N)=N.
 Call this number N and repeat steps 15 again using this new number as a starting point.
 Input this new number N into the Psi function to produce another number N.
 Input this number into the fw+w (n) function to produce an even bigger number called N.
 Input this new number N into the Busy Beaver funct…

I have defined a new version of the 4 basic operators addition,subtraction,multiplication and division.
Addition is usually defined as take a and b add them together and produce c like so a+b=c. In my version you don't do this instead you combine the terms like so a+b=ab. Too give an example of this is 1+2=12 or 5+55=555. Subtraction using my new system is essentially just crossing out for example 2222=22 you and 3454=35. Division means that you split a number up into non equal or equal parts for example 566/2= 56,6 where the comma represents a space. Multiplication is just repeated addition using my method for example 5*5= 5+5=5555555555 another example of multiplication is 10*4=104104104104.
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Here are some improvements or extensions too my Array notation: [5,6,7,8,9(5,6,7,8,9)((5,6,7,8,9)] means that you take [5,6,7,8,9(5,6,7,8,9] and make that the number of arrows as well as the number of things in the chain. You then produce out of it an even bigger number and make that the number of things in the chain. You repeat this process [5,6,7,8,9(5,6,7,8,9] times. [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9)(5,6,7,8,9)] means that you repeat the process on step 2 [5,6,7,8,9(5,6,7,8,9)(5,6,7,8,9) times instead of [5,6,7,8,9(5,6,7,8,9] times.
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Supremus Array Notation is denoted with [] brackets instead of {} brackets because it is easier to understand [] brackets than it is {} brackets. [n]a single letter inside brackets just represents a number. [n,b] this is just exponentiation. [n,b,c] this uses up arrows for example [5,6,7] means 5↑(6)7 where there are 6 arrows between the 5 and 7. The 4th level uses Conway chained arrows as this is the Natural next step to go after Knuth up arrows. [5,6,7,8,9]=5→6→7→8→9. The 5th level uses repeated or iterated Conway chained arrows for example [5,6,7,8,9(5,6,7,8,9)] means 5→6→7→8→9...5 →6→7→8→9. This means that there are 5→6→7→8→9 number of arrows in the chain and also 5→6→7→8→9 things in the chain. Or in other words the number of arrows is…
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Start with a Gorgcal in Aarex's Graham generator and call say this number N. Place a Gorgcal number of arrows between two Gorgcal's like so N↑(n)N where there are a Gorgcal number of arrows between the N's. Now call this number N and say that this is the number of arrows. continue this process of producing a Larger number and saying that this is the number of arrows between the Gorgcal's a Gorgcal number of times.
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This high value positive integer value is a very larger positive integer that is defined as follows take a Googolquadriplex and call that X. Create a tower containing containing X , X's in the Knuth up arrow notation this is X↑↑X a very larger number. We now call this number Y and say that Y is the number of arrows between the two X's this can be written as X↑(y)X where there are Y number of arrows between the X's. We take this ridiculously big number and call it A1 We then say that this is the number of arrows between the X's X↑(A1)X Creating an even larger number called A2. We continue until We get To AA1 this is the high value positive integer also called the Y or AA1 number. Too Finnish off I have some questions do you think this numbe…
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1. Start with a Googolplex and make that X.
2. X↑(x)X The brackets indicate how many arrows there are. 3. Make the number you produced from step 2 the value for X and repeat step 2 with this new number. X↑(x)X. 4. Make the number you produced from step 3 X and repeat what you did for steps 2 and 3. X↑(x)X. 5. Repeat this process again making an even larger value for X. X↑(x)X. 6. Repeat again using the number you produced from step 5. X↑(x)X 7. Repeat again using the Number you produced from step 6. X↑(x)X. . . . . . 10^(10^100)Repeat this long process once again using the number you produced from the (googolplex1) step X↑(x)X. The number you produce from the Googolplexth step is the X number. Too end this blog post on the X number I wou…
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so i recently saw this blog post: http://karpathy.github.io/2015/05/21/rnneffectiveness/
there's this thing called an LSTM
you can give it training data, like, some wikipedia articles
then you let the LSTM train for a certain amount of time
then the LSTM will learn from that data, and it will try to output something like the training data
so if we used, say, googology wiki articles as the training data and we let the LSTM train for a long time, the LSTM will try to output something like a googology wiki article.
and we could use the LSTM to generate other stuff, like generate random ordinals between 0 and e^{0}, or do a comparison of two notations
even though it might not generate actual googology but just hallucinated googology, i would still like t…
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If we can define w in a programming language, then we can define larger ordinals by using that programming language and w.
In this blog post, I will extend Brainfuck to include w, and see what ordinals we can generate.
There is the tape with an infinite number of cells, and a pointer which start at cell 0. Each cell starts with a value of 0.
> moves the pointer 1 cell to the right
< moves the pointer 1 cell to the left
+ increases the value of the cell the pointer is on by 1
 decreases the value of the cell the pointer is on by 1
[ does nothing
] does nothing, unless the value of the cell the pointer is on is not 0, then find the rightmost [ at the left of the ] and continue running the code starting from the [ you found
. outputs the value of the…
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http://lihachevss.ru/ordinal.html
Credits to Aarex for finding it
And it also contains ordinal levels (the table)
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Well, might as well start writing...
I've lately seen a few interesting things on the wiki, the Little Biggedon being one of them. Ultimately, I admire that we finally have yet another wiki record for the highest coined number officially present on the site; yet a matter comes to mind when reading through it.
That matter, my friends and colleagues, is something many of you may probably remember; the norminals function. The norminals, by their very design, are variables built in a way such that they will always be able to ascend beyond their old selves, and diagonalise their entire structure in a single step, every time. This does, of course raise the question "well, are 'norminals' well defined?" and the answer is, to a degree, no.
Norminals …
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Iff we have a matrix such as
\(G\frown B_0\frown B_1\frown\cdots\frown B_n\)
which βreduced from \(G\frown B\frown N\) , then it is holds that
\(B_k\frown N_k
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When it encounters a firstorder comma, hit should search for a 0th order separator (a separator without a subscript). This should give the ordinal strength BachmannHoward ordinal instead of FefermanSchutte Ordinal. For example, in {0{0,_{1}1}_{1}1}, which has the strength of your system, S1 would be {0,_{1}1} and Sn+1 would be {0,{0 Sn 1}_{1}1}.
We now introduce nth order separators ({}_{n}) and nth order commas (,_{n}). An nth order comma would search for an n1th order separator, and if doesn't find one, it adds one. right inside the highest nested lower level separator For example, {0,_{2}2} becomes {0{0,_{2}2}_{1}1}. This has level psi(psi(0))
,_{n} should be shorthand for {0,,n} and {m}_{n} should be shorthand for {m,,n}. Each doublecomma containing separator search…
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We now come to the Q function or the 3rd layer in the fst(x) hierarchy what this Q function does is takes a function in the second layer and iterates a certain number of times. For example Q{5,5,5)(5) means that you take fst[5](5)#####(5) and make that the input of fst(x) You then produce an even larger number and put that into the input of x and produce an even larger number. You repeating this process of generating an even larger number and imputing it into the function 5 times all this is applied to the number 5. Another example is Q{10,10,10}(10) means that you take fst[10](10)##########(10) and make that the input of fst(x) you then produce an even larger number and input this new number into the function. You continue this process 10…
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The 4th level of the fst(x) hierarchy is denoted with ### it works in the same way as the second and third level the only difference to it and the second level is that it uses fst[x](n)##(p)as input of fst(x). For example fst[5](8)###(13)is is equal to the fst[4]##8th term of 13 in the fst(x) hierarchy for more details into this see my previous blogs. fst[x](n)####(p) or the 5th level in the hierarchy works in the same way as the 2nd,3rd or 4th level the only difference between it and the 4th level is because it uses fst[x](n)###(p) as the index of fst[x]. For example fst[6](10)####(100) is equal to the fst[5]###10th function or term applied to 100. We can continue the process on and on with the higher levels in hierarchy producing larger…
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fst[5](x) is defined to be repeated fst[4](x) example fst[5](3)=fst[4](3)^fst[4](3)^fst[4](3)^fst[4](2)^fst[4](2)^fst[4](1) if you would like to know more about fst[4](x) go check out my blog post entitled factorial summation tree function.fst[6](x) is defined to be repeated fst[5](x) for example fst[6](3) is equal to fst[5](3)^fst[5](3)^fst[5](3)^fst[5](2)^fst[5](2)^fst[5](1). As fst[6](x) is repeated fst[5](x) then it is safe to assume that fst[7](x) is repeated fst[6](x). The greater the input into the [ ] the greater the output will be for example fst[500](5)>fst[200](5). This first level of the fst(x) function is called the zeroth level as it is nothing compared to whats coming next. Beyond this the 2nd level these functions are den…
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In my previous blog post entitled slow growing trees I described a function called the factorial summation tree function or fst(x) for short. What this function does is it takes the sum of all the factorials from 1! to a given value x for example [fst(10)=1!+2!+3!+4!+5!+6!+7!+8!+9!+10! this comes to 4037913. This function is quite slow as you have to input some very large numbers to get a large number out. To fix that I have devised a function called fst[2](x) what this does is instead of taking the factorial of a number it takes the double factorial it also works through multiplication instead of addition. For example fst[2](5)=1!!*2!!*3!!*4!!*5!!=5.97670760389 × 10^225.See quite an improvement over the fst(x) function but still its prett…
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You start off with one node or point you then branch off twice meaning you now have 3 nodes this completes 2 steps. Next you branch off 3 nodes from the 2 nodes you branched off previously making 9 nodes in total and 6 on the 3rd level. After that you branch off 4 nodes from each of the six previous nodes meaning you will have 24 nodes on the 4th level and 33 nodes in total. You then branch of 5 nodes from each of the 24 making 120 nodes on the 5th level and 153 nodes in total. You branch off 6 nodes from each of the 120 nodes making 720 nodes for the 6th level and 873 nodes in total. After that you branch off 7 nodes from each of the 720 nodes making 5040 nodes for the 7th level and 5913 nodes in total. You branch off 8 nodes from each …
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The funky number is equal to : 10^(10^2)^(10^3)^(10^4)^(10^5)^(10^6)^(10^7)^(10^8)^(10^9)^(10^10)^(10^11)^(10^12)^(10^13)^(10^14)^(10^15)^(10^16)^(10^17)^(10^18)^(10^19)^(10^20)^(10^21)...^(10^(10↑↑↑↑10).
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The ultra billion is constructed in the following way take 1000,000,000 and apply the double up arrow to it 1000,000,000↑↑1000,000,000. Once you've got that number of which I have decided to name a pseudo ultra billion because of its small size of to the real number. You make that the number of arrows between the billions 1000,000,000↑(pseudo billion) 1000,000,000 you then make that the number of arrows and repeat the process again in total you must do this process described above 1000,000,000↑↑1000,000,000 times.
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Does anyone know if there are any good books on the Busy Beaver function as I would like to know more about this function?
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Using the fast growing Hierarchy what is the Highest number you can reach using Robert Munafo's Hyper calc.
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I am trying to create numbers for powers of ten with already existing number names, but no suffixes or prefixes, because they will create new number names. For 10^{n}:
 n=0, One
 n=1, Ten
 n=2, Hundred
 n=3, Thousand
 n=4, Myriad
 n=5, Lakh
 n=6, Million
 n=7, Crore
 n=8, Awk
 n=9, Billion
 n=10, Dialogue
 n=11, Kharab
 n=12, Trillion
 n=13, Neel
 n=14,
 n=15, Quadrillion
 n=16, Byllion
 n=17, Shankh
 n=18, Quintillion
 n=19, Mahasankh
 n=20, Gai
 n=21, Sextillion
 n=22, Vṛnda
 n=23,
 n=24, Septillion
 n=25, Minnow
 n=26,
 n=27, Octillion
 n=28, Rang
 n=29,
 n=30, Nonillion
 n=31,
 n=32, Tryllion
 n=33, Decillion
 n=34, Gobychunk
 n=35, Goby
 n=36, Undecillion
 n=37, Mahāpadma
 n=38,
 n=39, Duodecillion
 n=40, Zheng
 n=41,
 n=42, Tredecillion
 n=43,
 n=44, Zai
 n=45, Quattuordecillion
 n=46,
 n=47, Mahākharva
 n=48, Quindecillion
 n=49,…

f3(1)=f2(1)=2^1*1=2 an interesting note about 2 it is the only even prime number all the others are odd.
f3(2)=f2(f2(2)=f2(8)=2^8*(8)=2048 if you add the digits together you get a prime 2+0+4+8=2+12=2+1+2=3+2=5.
f3(3) also known as fw(3)= f2(f2(f2(3)=f2(f2(24)=f2(2^24*(24)=402653184 added together all the digits come to six which is the first perfect number. We can prove this very easily 4+0+2+6+5+3+1+8+4=4+8+8+9+4=4+16+13=4+29=33=3+3=6.
f3(4)=f2(f2(f2(f2(4)=f2(f2(f2(64)=f2(f2(1.180591620717 × 10^21)=f2( 2.676043612057 × 10^42)=1.710261754098 × 10^85.
f3(5)=f2(f2(f2(f2(f2(5)=f2(f2(f2(f2(160)=f2(f2(f2(2.338402619729 × 10^50)=f2(f2( 1.18262925860 × 10^101)=f2( 2.68445111670 × 10^202)=1.72566228676 × 10^405.
f3(6)=f2(f2(f2(f2(f2(f2(6)=f2(f2(f2(f2(…
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F2(500) is equal to 1.63669530395 × 10^153 I have decided to name this number Pencentaillion the Pen comes from Penta meaning five while the cent comes from a word meaning 100.
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The multi million number is constructed in the following way you first start off with one million 1000,000 You raise 1000,000 to the power of itself. 1000,000^(1000,000) this comes to 10^6000000. You raise 10^6000000 to the power of itself once this comes to 10^(10^6.7781513 × 10^6000000 you take that number of which I have decided to call the sand universe number of which , the reciprocal is related to probability of running into an identical clone of yourself in less than one minute. You take the sand universe number and raise it to the power of itself to make an even bigger number. you then take that number of which I have not decided a name and raise it to the power of itself you repeat this process of taking the number and raising it …
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Let's suppose you have an enourmous number like Graham's number
If G (Graham's number) is enourmous, so is G1 enourmous
If G1 is enourmous, so is G2 enourmous
...
So is 1 an enourmous number? There has to be a threshole between an enourmous and nonenourmous number
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I have defined the Elgoog number to be equal to 10↑(10↑(10)10 this means that you have two tens with 10↑↑↑↑↑↑↑↑↑↑10 arrows in between the tens.
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The exponential factor of 1000 is defined as 1000^999^998^997^996^995^994...^1.
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So I was watching the David Metzler videos on the fast growing hierarchy but felt like he didn't do enough examples so does anyone know if there are better introductions to the fast growing hierarchy.
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Circle Function
10 + 10 = 20
10 (+) 10 = 10^^^... (10+10 amount of ^'s) ...^^^10
10 ^ 10 = 1E10
10 (^) 10 = 10^^^... (10^10 amount of ^'s) ...^^^10
• Circles  ( )  can also have circles around them10 ((^)) 10 = 10^^^... (10^^^... (10^10 amount of ^’s) …^^^10 amount of ^’s) …^^^10
10 ((((((((((^)))))))))) 10 = [10,10] This can also be written as = [10,10] or [2^1] for short (10's are given)
Bracket Function
[2^1] = [10,10][3^1] = [10, [10, 10]
[2^2] = [[10,10],[10,10]
[5^1] = [10,[10,[10,[10,10]]]]
[2^1] = [10,10] [3^1] = [10, [10, 10] [2^2] = [[10,10],[10,10] [5^1] = [10,[10,[10,[10,10]]]]
You can also do [ [10^10] ^ [10^10] ]  which can be written as 2, 10, 1
2, 2, 2 10, 10, 10
You can also do  10, 10, 10, 10, 10, 10, 10, 10, 10  Which…
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[2^1] = [10,10] [3^1] = [10, [10, 10] [2^2] = [[10,10],[10,10] [5^1] = [10,[10,[10,[10,10]]]]
You can also do [ [10^10] ^ [10^10] ]  which can be written as 2, 10, 1
2, 2, 2 10, 10, 10
You can also do  10, 10, 10, 10, 10, 10, 10, 10, 10  Which is written as 3:3 (3 groups of 3, which leads to my next function)
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Here is my new circle Function:
Here are some new functions I have created in making large numbers:
10 + 10 = 20 10 (+) 10 = 10^^^... (10+10 amount of ^'s) ...^^^10
10 ^ 10 = 1E10 10 (^) 10 = 10^^^... (10^10 amount of ^'s) ...^^^10 10 ((^)) 10 = 10^^^... (10^^^... (10^10 amount) ...^^^10 amount) ...^^^10
I would move this to your Talk Page or Profile. One of the modifiers may delete it soon. Googology wiki has a strict rule about selfmade content so cpoy and paste it before it's gone! I don't want your work to be lost. Simon Weston 20:26, March 8, 2017 (UTC)
• 10 (+) 10 = deka21xis using Sbiis Saibian's naming scheme and is in between g(1) and gulgold.
This is the new Circle Function I have been working on:
10 + 10 = 20
10 (+) 10 = 10^^^... (10+…
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(How do you edit titles??)
Large Numbers I have created using my functions:
10 (+) 10 = 10^^^^^^^^^^^^^^^^^^^^10 = deka21xis using Sbiis Saibian's naming scheme and is in between g(1) and gulgold.
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I define this to be the incredibly small number 10^(10^60) or 1 followed by a novemdecillion zeros or 10^60 zeros this number is so small that it doesn't even reach the same level as a Googolplex.
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So I was watching some of the David Metzler videos people were telling me about and I came across a puzzle in the Video called Harvey Friedmans word puzzle. The rules are simple you have to make the longest string of letters/word possible without any substrings having any matches to another substring. For example abbaaabb is illegal as it has two matches but abbbaaaaaaa is legal it has been found that the longest string you can make with one letter/symbol is aaa or bbb etc because if you cross out or remove symbols you won't get pairs aaa=aa,a see no matches. My question is what is the longest string possible so that no matches occur using all 26 letters of the English alphabet?
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I have been wondering about why Grahams number uses 3's in the first place why can't it be 2 or 4 or 5 or 6 etc is there any reason that Graham used 3's?
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This number gets its name from the Greek for two as in the number 2 the reason I chose the name is because I wanted to pay respect the Greeks who in my opinion did more for mathematics than any other civilization.The dyo number is defined 2↑↑↑↑...↑↑↑↑2 where the number of up arrows between the 2's is 2↑↑↑↑2.
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I have been looking at the fast growing hierarchy and wish to know if I've got the answer right.
Does f2(4)=64?
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This will be the last of the non Googology style blogs for a while after this I will focus more on improving my Googolisms.
First of all I apologise if any one has been offended by my comments I didn't mean to troll or offend anyone this is the honest truth and I'm sorry. The second thing I wish to apologise about are my Jokes blogs of which seem to offend people all though I don't understand why. I also apologise for calling people names as well as slagging off people. To make things clear this blog is not intended as a way of trolling people no it is a blog intended as an apology to whoever I have wronged. After this blog I plan to focus just on my Googology but you need to remember something the reason I did it was because I was sick and…
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This morning I was thinking about a sequence that contains the perfect exponents now what does this mean well I tell you it means that you have a value x and the exponent x is equivalent to x.
For example 1^1 is a perfect exponent,2^2 is another perfect exponent,3^3 etc etc.
Here are some of the perfect exponents in answer form by this we mean 3^3=27 etc
1,4,27,256,3125,46656,823543,16777216,387420489,10,000,000,000,285311670611, 8916100448256,3.028751065923 × 10^14, 1.111200682556 × 10^16,4.378938903809 × 10^17,1.844674407371 × 10^19,8.272402618863 × 10^20,3.934640807530 × 10^22, 1.978419655660 × 10^24,1.048576000000 × 10^26 Here is the sequence up to 20^20. I don't think this has anything to do with Googology but I just thought that this w…
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I have been wondering about how I could improve the names of my numbers in future if you have any suggestions to what names I could use comment down below.
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Many of you believe that I am the second account of Cloudy176. This is not true, as the person behind the account is my semiidentical nontemporalduplicate twin brother.
Also at least I'm not stupid like he is, for example he doesn't know that everything is made out of paradoxes. In addition, my account is intended to be used primarily for joking, while his is intended to be used for serious purposes (and also joking).
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To make the tree for ordinal α, start with α at the top, then do this process to α, then do the process to the members of α, then do the process to the members of those members, and continue repeating.
1. If ordinal β is a sequence ordinal, nothing happens to β.
2. If ordinal β is a limit ordinal, the children of β will be the first n members of the FS of β. n is called the base.
That's it.
Also, I am redefining HN. You can check the HN definitions if you go to my site, there is a link in my userpage.
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