Dhani irwanto Illion series

CHAPTER 1: Pillars, Walls, and cubes

As for the illion issue, I'm beyond any doubt that from childhood, children as of now know and are beginning to be inquisitive around expansive numbers, one of the characteristics is by naming the number illion but with the off-base definition, children as it were compose 999,999,999 and so many times indeed in spite of the fact that they can do it. streamlined to (10^10) - 1 exceptionally basic, but numerous individuals do not know what the definition of illion is since not all googolisms are illion, googol and googolplex are not illion

this is what we have to fix, in class 3 there is a number that is very close to the Archimedes number in the sand reckoner but the name is very disappointing, namely onionillion This is true illion because it complies with the definition of illion,

so what do you understand about the numbers 10 100 and 1000, the number 10 is usually depicted as a 1-dimensional vertical pillar consisting of 10 ones blocks, the number 100 is described as a two-dimensional wall consisting of 10 tens pillars, and the number 1000 is described as a 3-dimensional cube consisting of 10 walls

then we see once that this is repeated again, what is repeated? It turns out that the thousand block forms another block like the unit block, meaning we can construct a 10,000 pillar by placing 10 thousand blocks and we can construct a 100,000 wall by placing 10 10,000 pillars and finally we can make another bigger block, and that is the million block.

then the same goes on, finally we can make a bigger cube, namely the billion cube and from the billion cube we can make pillars, walls, and finally the trillion cube,

10^0 = cube

10^1 = pillar

10^2 = wall

10^3 = cube

10^4 = pillar

10^5 = wall

10^6 = cube

10^7 = pillar etc

while illion is just a number in the form of a cube, so for numbers in the form of pillars or walls it is not illion, and illion is when the number blocks finally form a cube, and see how many cubes it is (+2), 1000 is the zeroth cube (1000 does not count), 1,000,000 is the first cube, 10^9 is the second cube, 10^12 is the third cube, 10^15 is the fourth cube, so in essence any power number above 10 must be divided by 3 to produce a perfect cube,

The exact formula for illion is

$10^{3n+3}$

and why add 1 at the end because illion starts from 10^6 not 10^3, 10^3 because it can be divided by 3 and forms a cube I gave it the name zerollion because it is in the zeroth order, remember why you added 3 because illion starts from 10^6 not 10^3, we just need to adjust the n number to get illion, The number one is also included because 10^(3*0) will produce the number 1 as well (10^0) so everything is clear and the number 1 is a perfect cube

and at last, to title illion since numerous individuals do not keep in mind it, we fair got to supplant the n within the equation over with the Latin number add illion with prefix, for latin numbers can be seen here https://www.omniglot.com/language/numbers/latin.htm, that's the name for illion, this is the first illion to decillion this has been confirmed because the next illion is made by me and does not have a confirmed name all names are mine

1. $10^{3*1+3}=10^{6}$ = Million
2. $10^{3*2+3}=10^{9}$ = Billion
3. $10^{3*3+3}=10^{12}$ = Trillion
4. $10^{3*4+3}=10^{15}$ = Quadrillion
5. $10^{3*5+3}=10^{18}$ = Quintillion
6. $10^{3*6+3}=10^{21}$ = Sextillion
7. $10^{3*7+3}=10^{24}$ = Septillion
8. $10^{3*8+3}=10^{27}$ = Octillion
9. $10^{3*9+3}=10^{30}$ = Nonillion
10. $10^{3*10+3}=10^{33}$ = Decillion

it is now known why the decillion is 10^33 not 10^30 even though 3x10 = 30, this is because the illion starts from 10^6 not from 10^3, while the name illion comes from the number of n in the formula, not from the number of powers of 10, for example septillion means seven-illion, which means not 10^7 but 10^(3*7+3) = 10^24,

then what's next?

CHAPTER 2: First generation -illions

able to get out of decillion to deal with indeed greater numbers, since decillion has as of now entered tens, we should just jump straight to 20 rather than 11 since it'll be a squander of time afterward and is futile, this "jump" framework is broadly utilized in numerals Greek and Hebrew to characterize the units tens and hundreds,

remember that the numbering system is n -illions where n is n in the formula above or 10^(3n+3), n here is what we will use to name illions, so for example 10^10 is not a decillion, decillion is 10^(33) or 10^(3*10+3), where n = 10, this needs to be remembered carefully, sorry if there is anything that needs to be corrected regarding these Latin numbers, and below is a list of numbers from decillion to centillion

1. $10^{3*10+3}=10^{33}$ = Decillion
2. $10^{3*20+3}=10^{63}$ = Vigintillion
3. $10^{3*30+3}=10^{93}$ = Trigintillion
4. $10^{3*40+3}=10^{123}$ = Quadragintillion
5. $10^{3*50+3}=10^{153}$ = Quinquagintillion
6. $10^{3*60+3}=10^{183}$ = Sexagintillion
7. $10^{3*70+3}=10^{213}$ = Septuagintillion
8. $10^{3*80+3}=10^{243}$ = Octogintillion
9. $10^{3*90+3}=10^{273}$ = Nonagintilllion, and last
10. $10^{3*100+3}=10^{303}$ = Centillion

and we can see for ourselves that after Quadragintillion this number is already fast and surpassing Googol even though the growth is still linear and not very fast but this is the official name, and all the names of the numbers here have been registered on tungsten alpha and wiki googology, actually maybe other people have named these numbers differently, which is definitely the formula (10^{Latin numbers * 3 + 3}), for example sexagintillion which means sixty-illion, by multiplying 60 by 3 the result is 180 then adding 3 to 183, and putting it in the exponent section of scientific notationand result = 10^183

Centillion is sbiis saibian's favorite number when he is in 2nd grade which is often seen with its notation along with Googolgong.

after we have wrapped up the -illion tens which has come to Nonagintillion which implies we got to proceed once more to the -illion hundreds segment which may be a parcel I named it regularly mistakes, these numbers are:

1. $10^{3*100+3}=10^{303}$ = Centillion
2. $10^{3*200+3}=10^{603}$ = Duocentillion
3. $10^{3*300+3}=10^{903}$ = Trecentillion
4. $10^{3*400+3}=10^{1203}$ = Quadringentillion
5. $10^{3*500+3}=10^{1503}$ = Quingentillion
6. $10^{3*600+3}=10^{1803}$ = Sescentillion
7. $10^{3*700+3}=10^{2103}$ = Septingentillion
8. $10^{3*800+3}=10^{2403}$ = Octingentillion
9. $10^{3*900+3}=10^{2703}$ = Nongentillion, and finally
10. $10^{3*1000+3}=10^{3003}$ = Milinillion,

and this millinillion appears to be the final substantial -illion in this case since -illions bigger than this as a rule do not have names or are less substantial so I will provide the final title for numbers more noteworthy than here utilizing the definition 10^(3n+3) , for case, after you sort 10^(3*2000+3) into wolframalpha, the framework will not show the title of the number, as it were the scientific notation.

after this what, we just follow the Latin numbers that I found on Google Translate even though it looks like the limit has run out because Latin numbers only use mille as the highest unit, we can continue

• $10^{3*1001+3}=10^{3006}$ = Milli-unum-illion
• $10^{3*1002+3}=10^{3009}$ = Milli-duo-illion
• $10^{3*1003+3}=10^{3012}$ = Milli-tres-illion

see that the numbers you see now are very poorly named because unfortunately illion only stops at Millinillion, this is very annoying in naming, okay then we will increase this to thousands or hundreds so that it goes faster

• $10^{3*1100+3}=10^{3303}$ = Milli-centillion
• $10^{3*1200+3}=10^{3603}$ = Milli-duocentillion
• $10^{3*1400+3}=10^{4203}$ = Milli-quadrigentillion
• $10^{3*1800+3}=10^{5403}$ = Milli-octingentillion

To be quick, I just use the power of 2 above because the growth is fast and to save time, and maybe the names that I present here are all wrong and what so, basically they are a bit confusing and doubtful, I hope it's true but this number is smaller than 10^(3*2000+3)

finally we have reached the scale of thousands now and the names may have a lot of errors and this number is still very small and there really is nothing, this is still class 2

• $10^{3*2000+3}=10^{6003}$ = Dumillillion
• $10^{3*3000+3}=10^{9003}$ = Trimillillion
• $10^{3*4000+3}=10^{12003}$ = Quadrimillillion
• $10^{3*5000+3}=10^{15003}$ = Quintimillillion
• $10^{3*6000+3}=10^{18003}$ = Seximillillion???
• $10^{3*7000+3}=10^{21003}$ = Septimillillion
• $10^{3*8000+3}=10^{24003}$ = Octimillillion
• $10^{3*9000+3}=10^{27003}$ = Non-millillion??? and finally
• $10^{3*10000+3}=10^{30003}$ = decemmillillion??? or I will also call it myrillion because 10,000 is a thousand in Greek even though it doesn't exist in Latin

in the general dictionary the largest -illion number is the centillion as seen by sbiis.pdf although this can be beaten with Googolplex, other numbers are formed using the formula 10^(3n+3) with Latin numeral construction although it is limited to 9000-illion

This is what is called the first generation illion because later there will be a second generation illion

So finally we can move on to 20,000, 30,000, then move on to 40,000 and so on. I can't name these numbers one by one because there are too many until the end.

we will be facing a number that is quite large. I don't know the name but just try

$10^{3*100,000+3}=10^{300003}$ = Centimillillion, If you want a number based on the Googology Wiki, that's are a lakhullion or whatever you want (I didn't know that this number was created by sbiis.pdf . Previously I only used Latin numbers heard via Google Translate), then

$10^{3*100,000+3}=10^{1500003}$ = Quingentimillion, then (This is starting to approach the second generation)

$10^{3*900,000+3}=10^{2700003}$ = Novacentimillion, a little more (maybe wrong name)

$10^{3*999,999+3}=10^{300000}$ = Novacentinovantanovemillanovacentonovantanovillion, the name is a bit long because the digits are 999,999, this is not very good and the name may be wrong, okay, I'll just give it a name: Prathama Pendhino Anta, which means "end of the first generation" because this is the end of this first illion series

but that is not the last number which is almost close to the second generation series, yes it is true that it is a power of 10 but what if the number n is a fractional decimal number which is very close to the second generation, It is

$10^{3*(999,999+(1-10^{-10^{3*999,999+3}}))+3}=???$ = Te ketalum nanum che teni kyareya kalpana kari nathi

this number is almost

CHAPTER 3: Second generation illions

Second generation First "year"

Finally, now we are starting to enter the second generation of illions, maybe this is starting to be a little difficult to understand because it looks like illions within illions or nested illions, this means that one of the illion numbers that I mentioned before becomes n and is entered into the formula 10^(3n +3) generates recursive and nested formulas, The smallest and earliest number in the second illion generation is:

$10^{3*(10^{3*1+3})+3}=10^{3*1,000,000+3}=10^{3000003}$

This number is the smallest second generation illin number. Why do I say second generation, you see above that the number million has been replaced with the variable n in the formula 10^(3n+3) which makes this illion nested. This can also be written 10^(3 *million+3), don't know what to call this number, or you can also write the number millionth illion number, how is that?

For this naming system we no longer use the Latin numeral system but we will use the SI prefix system, remember that Greek numbers only go up to a myriad or 10,000, the word "mega" comes from the word 'Great', it doesn't mean 1 million, the word "Giga" means "Giant " and this is NOT the Greek or Latin name for illions

while we will start from 10^3000003 or the number I mentioned above is called Megallion or Megillion, whatever, maybe this number has another name, Mega is the prefix of 1 million which means Great, this is because the number 1 million is added to n in the -illion formula and the number produces a megallion, although a better name is decicenetamillillion (Latin)

To understand this number, just assume that this number is written in a book and the font size is around 11 px, while the largest number that can be written in 1 Microsoft Word page with normal margins in Portrait orientation without images and textboxes or any smart art only uses one column and uses Callibri font without subscript or bold with size 11 is (10^3570)-1, if the size of the letters is then the number is (10^7016) -1, let's just say if we write in a book with paper size $21\times (21\times {\sqrt {2}})cm$then how many pages will it take to write Megallion

when we divide the number 3000003 by the number 3652 if the font size is 11 px then the result is this long number

821.46851040525739320920043811610076670317634173055859802847754654

So it took around 821 pages if rounded up to write Megallion

though a typical book is as of now 500 pages, it is as of now exceptionally thick, ordinary reading material are ordinarily as it were 300 pages, let's accept we'll utilize a 500 page book, the fifth version of the Great Indonesian Dictionary is distributed as thick as 2,040 pages, around 2 times the number of pages required to print megallion utilizing 11px text font,

from now on I will not use books to represent the digits of this number, I will only use the volume of the number of digits or powers of 10 and assume that we can write the digits of this number in the air in the form of a cube with a size of 1 cm, for example if we write googol (10^ 100) then we will get a cube with a volume of 100 cm if we want to write centillions we will get a cube with a volume of 303 cm, this is just for representation and remember, this is the number of digits, which means it's small if the googol is only 100 cm^3, it means small if a googol is only 100 cm^3, not so, the centillion measure is 303 cm^3, the millinillion measure is 3003 cm^3 this is a small growth because this is to represent the number of digits, because a megallion has 3000003 digits, a cube with a size of 3000003 cm^3 is formed, or with a cubic root of 144,225 cm. With the conversion we will get a result of 1.4 m long, wide and high

After the this number, of course you don't need to add a zero, but because this is the second generation, we have to be different, so in the end we just have to make the billionth illion number. (but even if you add million+1 above it will still be a million million because it is already included in the formula, this is only to accelerate growth)

$10^{3*(10^{3*2+3})+3}=10^{3*1,000,000,000+3}=10^{3000000003}$

So what if the digits are 3000000003, this means the number is 3 times a thousand times plus three bigger than a megallion, if you calculate the volume then the volume becomes 3000^m3 (plus 3 cm^3) remember what is meant by 3000 meters^3 is not 3 kilometers^3 because this is cubed, if you look for the square root of 3 from here this 14 meters is 10 times a megallion, and this is still small, don't be fooled yet

because this number appears after the megallion, this number will be named Gigallion, giga means "giant

Comparison of the number of digits of this number at the Gelora Bung Karno Stadium

then what if it is 10 times the length (cube root) or 1000 times the volume of a Gigallion, this produces a very large number which has 3000000000003 digits, this is the same as adding the third illion, namely a trillion, and entering it into the illion formula, this will produces a trillionth illion or 1 followed by 3000000000003 zeros, this number in cubic roots becomes 144 meters long, wide and high, meaning that the digits of this number have filled Bung Karno's stadium.

$10^{3*(10^{3*3+3})+3}=10^{3*1,000,000,000,000+3}=10^{3000000000003}$

as a continuation of gigalllion this number is named Terallion following the SI prefix

then what if we make a cube containing the number of other digits whose square root is 10 times the square root of terillion or 1000 times plus three big Terillion this will make another number which is a continuation, namely Petallion

$10^{3*(10^{3*4+3})+3}=10^{3*1,000,000,000,000,000+3}=10^{3000000000000003}$

Comparing the number of petallion digits with the local parliament complex, the stadium near is the Gelora Bung Karno Stadium

and if this cube is placed and starts from the center of the Gelora Bung Karno Stadium, then this cube will cover the entire parliamentary complex which is next to the Gelora Bung Karno Stadium even though the entire parliament complex is already 80,000 meters in size. This is already very large, the picture below shows a comparison of the number of petallion digits compared to the parliamentary complex

then what if we make a cube whose width and length are 10 times the length and width of a pertallion, the answer is very surprising, it is MUCH bigger, I'm not saying how big the area is, no, this cube already fills the entire city of Jakarta which is densely populated from the south all the way to the north, and its height is far higher than Mount Everest, this is unimaginable, maybe it will be seen covering the sky by people who live nearby, see the picture below, remember THIS IS THE NUMBER OF DIGITS, it's like if you write 1 digit on 1 cm paper then this is the number of piles of paper

Exallion digits have filled the entire city

because this number is a continuation of Petallion according to the prefix: Exallion

$10^{3*(10^{3*5+3})+3}=10^{3*1,000,000,000,000,000,000+3}=10^{3000000000000000003}$

This exallion is a quintillion which is put into the formula 10^(3n+3) then what if instead of a quintillion but we will upgrade it again to a sextillion, then the number of digits can fill almost the entire island of Java and is even bigger, and the height is unimaginable Maybe people from all over Southeast Asia will see the cube on the horizon even though it is tens of kilometers away,

this number will be named Zettalion

Zettalion digit

$10^{3*(10^{3*6+3})+3}=10^{3*1,000,000,000,000,000,000,000+3}=10^{3000000000000000000003}$

and finally is the largest extension of this series that can be accommodated on earth, if the cube is bigger than this then it must be in outer space, it seems, the cube here will be visible from the entire eastern hemisphere and will cover the entire Indonesian and Philippine islands as wide as from Dublin to Warsaw

I will not upload it here because I have uploaded many pictures which are certain that the cube has reached fourteen hundred kilometers in length, width and height.

this number is Yottalillion

$10^{3*(10^{3*7+3})+3}=10^{3*1000000000000000000000000+3}=10^{3000000000000000000000003}$

and the last 2 numbers

• Ronnallion =$10^{3*(10^{3*8+3})+3}=10^{3*1000000000000000000000000000+3}=10^{3000000000000000000000000003}$and last
• Quettallion = $10^{3*(10^{3*9+3})+3}=10^{3*1000000000000000000000000000+3}=10^{3000000000000000000000000000003}$

keep in mind that quetta is the last SI prefix that exists to indicate nonillon, I don't know what this quetta actually means but what is clear is that the Greeks never knew this number at all, the largest number is myriad,

so now the list of numbers first year second generation illions is

1. Megallion = $10^{3*(10^{3*1+3})+3}=10^{3*1,000,000+3}=10^{3000003}$
2. Gigallion = $10^{3*(10^{3*2+3})+3}=10^{3*1,000,000,000+3}=10^{3000000003}$
3. Terallion = $10^{3*(10^{3*3+3})+3}=10^{3*1,000,000,000,000+3}=10^{3000000000003}$
4. Petallion = $10^{3*(10^{3*4+3})+3}=10^{3*1,000,000,000,000,000+3}=10^{3000000000000003}$
5. Exallion = $10^{3*(10^{3*5+3})+3}=10^{3*1,000,000,000,000,000,000+3}=10^{3000000000000000003}$
6. Zettallion = $10^{3*(10^{3*6+3})+3}=10^{3*1,000,000,000,000,000,000,000+3}=10^{3000000000000000000003}$
7. Yottalion = $10^{3*(10^{3*7+3})+3}=10^{3*1000000000000000000000000+3}=10^{3000000000000000000000003}$
8. Ronnallion =$10^{3*(10^{3*8+3})+3}=10^{3*1000000000000000000000000000+3}=10^{3000000000000000000000000003}$and finally
9. Quettallion = $10^{3*(10^{3*9+3})+3}=10^{3*1000000000000000000000000000+3}=10^{3000000000000000000000000000003}$

there is a basic division of "Generations" here, each generation is divided into several "Years", for the first million generations there are only 6 "Years" namely unit years, tens, hundreds, thousands, ten thousands, and one hundred thousands, while this is still is in the first "year", only the first year is the official prefix name from SI while the others are not, and why is the list as it were up to Quettallion since quetta is the final prefix on the list, Ronna and quetta will as it were be included in 2022,

numerous googologist are making extra prefixes, sbiis.pdf made the Venda prefix for decillion, for the second generation of the second year illions we will use the sbiis.pdf prefix

Second generation Second "year"

there are several prefix systems created above nonillion by several googologists, the system that sbiis.pdf created is called SEPS or Sbiis Saibian's Extended Prefix System which is uploaded on this site

for numbers up to 10^24 it still uses the old si prefix system before 2022, sbiis.pdf wrote this before SI released a larger pefix, so the octillion and nonillion prefixes use Xenna and Wecca, how did this name come about? this is a modification of Latin numbers, Xenna is a modification of x - enn(e)a and wecca is a modification of w - (d)ec(i) - ca but this is not a nonillion not a decillion, for now we only use prefixes which is a representation of the second year illion such as decillion vigintillion and so on and prefixes that do not comply, for example uada (10^36) will be kicked because it will only be a waste of time, the selected prefix is:

1. Venda: 10^33
2. Unca: 10^63 (vigintillion)
3. untra: 10^93 (Trigintillion)
4. unsara: 10^123 (Quadragintillion)
5. unpana: 10^153 (quinquagintillion)
6. unexia: 10^183 (sexagintillion)
7. unhepa: 10^213
8. unocia: 10^243
9. unenea: 10^273

see that it seems here we don't follow the rules, remember that this prefix is a multiple of 3, but because we are using the standard illion, the final result above the exponent must be added by 3 because illion starts from 10^6 or million. This causes a big error and cannot be harmonized between the sbiis prefixes. pdf with illion, but for simplicity and for dependability I announced that I would use the previous prefix, By default we will use two numbers for every 1 number, and these are the second generation illions of the second year with the prefix sbiis.pdf

1. Vendallion = $10^{3*(10^{3*10+3})+3}=10^{3*10^{33}+3}=10^{3000000000000000000000000000000003}$
2. Uncallion = $10^{3*(10^{3*20+3})+3}=10^{3*10^{63}+3}=10^{3000000000000000000000000000000000000000000000000000000000000003}$
3. Untrallion = $10^{3*(10^{3*30+3})+3}=10^{3*10^{63}+3}=10^{3000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000003}$
4. Unsarallion = $10^{3*(10^{3*40+3})+3}=10^{3*10^{123}+3}$
5. Unpanallion = $10^{3*(10^{3*50+3})+3}=10^{3*10^{153}+3}$
6. Unexiallion = $10^{3*(10^{3*60+3})+3}=10^{3*10^{183}+3}$
7. Unhepalion = $10^{3*(10^{3*70+3})+3}=10^{3*10^{213}+3}$
8. Unociallion = $10^{3*(10^{3*80+3})+3}=10^{3*10^{243}+3}$
9. Uneneallion = $10^{3*(10^{3*90+3})+3}=10^{3*10^{273}+3}$

for the "second year" it seems to only stop at uneneallion which can represent nonagintillion in this sbiis.pdf prefix system because the sbiis.pdf prefix only stops at ecetta which is a representation of 10^300 or 10^(3*99+3) almost 1 another step towards centillion, but unfortunately sbiis.pdf doesn't have it

Ecemollion = $10^{3*(10^{3*99+3})+3}=10^{3*10^{300}+3}$

but sbiis.pdf in the closing sentence of the article he said something like this

So now we have a truncated system that goes as far as EMPS. Can we go further with this ? Not without some problems. I have considered extending it further. It's prefectly logical to continue with unecetta- (10^303), unecemo- (10^-303), duecetta- (10^306), duecemo- (10^-306), etc.

  —https://sites.google.com/site/largenumbers/home/2-2/2-2-4-sbiis-saibians-extended-prefixes-2009/seps

this is what also ends and is the largest number in the second generation "second year" and this is called the centillionth illion, this number is unecettallion based on the prefix sbiis.pdf or perhaps it could also be called centillionillion because there is a centillion for n in the illion formula (10^(3*(centillion)+3)

unecettallion

$10^{3*(10^{3*100+3})+3}=10^{3*10^{303}+3}$

This number is 10 raised to the power of 3 followed by 303 zeros plus 3 at the end. This looks like a palindrome because illion if it is a power of 10 will definitely look most dominant with the number 3 followed by many 0s and followed by 3 at the end, for example decillion is 10^ 33 centillion 10^303, Millinillion 10^3003 and like that but we can still duecettallion and the exponent is still the palindrome 3 followed by n 0 plus 3

Duecettallion = $10^{3*(10^{3*101+3})+3}=10^{3*10^{306}+3}$

Prefix symbols would become groups of 3 letters instead of 2, with the last letter representing the groups of a hundred place. For now I think the system is a sufficient demostration of how far the SI prefixes could potentially be extended without too much trouble.

now it's time for us to switch "years" and migrate from prefix sbiis.pdf to another system but we are still in the second generation now it is hundreds

Second generation Third and fourth year

we now know that the prefix sbiis.pdf only goes up to 10^300 even though it can be added at any time but this will still result in a perfix that is not good and too long, we agree that we will use the prefix sbiis.pdf until here only and we have to look for alternative prefixes for the third year besides the sbiis.pdf prefix, there are many other alternatives created by googology wiki users like me, and we have to find the prefix for ducentillion first (third year) or 10^603,but mostly it will start at 10^600 or 10^(3*199+3) because the exponent is even, this only changes because the first illion is 10^6 so you will definitely see the number 3 at the end of the illion (illion which is a multiple of 10) , this is certain,

some users suggested a prefix for 10^600

1. Aarex Tiaokiao proposed prefix Megiheatia for 10^600
2. Cookiefonster proposed prefix Beta for 10^600
3. MultiSoul proposes prefix Twodreda (source)

there is one other user who proposed an accurate prefix for duocentillion which is 10^603, I don't know where this usercame from and he only reached 992 edits but according to the page he made a degentili prefix but unfortunately the user has already deleted his wordpress

from now on for naming names I will no longer use the usual official names because this prefix is not available, so to mark the example 10^(3*10^603 + 3) then I will write ducentillionth illion because the prefix has not been found,

the prefix that is greater than the first centillion that you see in the googological affixes template on the googology wiki is 10^600, clicking on it will take you to the page with 3 people nominating prefixes, for now I will use the prefix aarex for naming, aarex prefix for 10^600 is Megiheatia, meaning Megiheatiallion

Megiheatiallion = $10^{3*(10^{3*199+3})+3}=10^{3*10^{600}+3}$

Megiheatiallion which comes from one of the prefixes aarex can also be represented as follows:

$10^{\overbrace {30000000000...0000000003} ^{600}}$

This is not appropriate, because remember ducentillion is 10^603 not 10^600 because 3 is added, this causes a prefix and this naming does not correspond to 10^(3*199+3) which is entered into the illion formula, What if the ducentillion is above, this produces an unnamed number

ducentillionth - illion = $10^{3*(10^{3*200+3})+3}=10^{3*10^{603}+3}$

This number has not been named because I have not found a suitable prefix,

The next prefix found on the Googology Wiki after the prefix 10^900 is a multiple of pervious prefix, namely the prefix 10^900, this prefix is called Gigiheatia by Aarex or Gamma by Cookie fonster which comes from the Greek letter

Gigiheatiallion = $10^{3*(10^{3*299+3})+3}=10^{3*10^{900}+3}$

then there was an unnamed version adapted to add 3 to the illion system

Trecentillionth - illion = $10^{3*(10^{3*300+3})+3}=10^{3*10^{903}+3}$

and so on so we will continue and be consistent with the aarex prefix and remember this is only up to 10^3000, so as not to waste time let's just start and I will start from 10^600 based on aarex prefix

1. Megiheatiallion= $10^{3*(10^{3*199+3})+3}=10^{3*10^{600}+3}$
2. Gigiheatiallion= $10^{3*(10^{3*299+3})+3}=10^{3*10^{900}+3}$
3. Teriheatiallion= $10^{3*(10^{3*399+3})+3}=10^{3*10^{1200}+3}$
4. Petiheatiallion= $10^{3*(10^{3*499+3})+3}=10^{3*10^{1500}+3}$
5. Exiheatiallion= $10^{3*(10^{3*599+3})+3}=10^{3*10^{1800}+3}$
6. Zettiheatiallion= $10^{3*(10^{3*699+3})+3}=10^{3*10^{2100}+3}$
7. Yottiheatiallion= $10^{3*(10^{3*799+3})+3}=10^{3*10^{2400}+3}$
8. Xotiheatiallion (Ronnihetiallion) = $10^{3*(10^{3*899+3})+3}=10^{3*10^{2700}+3}$
9. Hilatiunitallion (Quettihetiallion) = $10^{3*(10^{3*999+3})+3}=10^{3*10^{3000}+3}$

This is the limit of the templates on the Googology Wiki, if you want to go further you will be shown the Prefix 10^3000 and higher page which contains many prefixes created by many users, and the aarex prefix stops at 10^3000, so we have to change the system for the fourth year,

now the fourth year member actually only contains 1 googolism member. If we make a googolism based on the article https://googology.fandom.com/wiki/Prefix_10%5E3000_and_higher, this happens because the next prefix has entered the third generation or new level, 1 member that's just it

1. Tegomegallion = $10^{3*(10^{3*3999+3})+3}=10^{3*10^{1200}+3}$

Tegomega prefix does not come from aarex but comes from another user Cookiefonster according to the googology wiki, this number is the only number that is between Hilatiunitallion and level 3 illion numbers,

then we can create numbers again even though I haven't created a name yet, Note that the number at the end of the power is a multiple of the previous number causing its rapid growth

1. $10^{3*(10^{3*7999+3})+3}=10^{3*10^{2400}+3}$
2. $10^{3*(10^{3*1599+3})+3}=10^{3*10^{4800}+3}$
3. $10^{3*(10^{3*3199+3})+3}=10^{3*10^{9600}+3}$
4. $10^{3*(10^{3*6399+3})+3}=10^{3*10^{19200}+3}$
5. Fifth year:$10^{3*(10^{3*12799+3})+3}=10^{3*10^{38400}+3}$
6. $10^{3*(10^{3*25599+3})+3}=10^{3*10^{76800}+3}$
7. $10^{3*(10^{3*51199+3})+3}=10^{3*10^{153600}+3}$
8. Sixth year:$10^{3*(10^{3*102399+3})+3}=10^{3*10^{307200}+3}$
9. $10^{3*(10^{3*204799+3})+3}=10^{3*10^{614400}+3}$
10. $10^{3*(10^{3*409599+3})+3}=10^{3*10^{12288}+3}$
11. $10^{3*(10^{3*819199+3})+3}=10^{3*10^{2457600}+3}$