From early history, humans thought of an extensible naming system for grouping large numbers. That is, in English, we group with millions, billions, and trillions. Around the 20th century, mathematicians proposed new illions that haven't been seen from the real world before.
In early 2000s, Jonathan Bowers started his extensive lists of -illions. Later, Sbiis Saibian introduced systems of Bowers' -illions (up to multillion).
Out of curiosity, Douglas pointed out there was a small mistake that the petillion is ambiguous from his list. He fixed the problem by putting another e for tier 3 multipliers to -et root. He also published the LNGI page for Sbiis’ illions which is shown here.
With problems fixed, we could name numbers up to the limit of Bowers’ illions, which is the supremum of all tier 0 - 4 roots being used.
With my -illion system, I coined another name for thousand (1,000), called "killion", and also equal to R0 using Extensible-R function.
Everyone knows about numbers such as a million, billion, trillion, and quadrillion–but how far have these “illion” numbers been extended? You may have heard of the centillion (the 100th illion), and later the millillion (1000th illion) which are all relatively old terms, with the first being published by the Oxford Dictionary in 1852, and the second by Henkle in 1904. The nth -illion corresponds to 1000^(n+1) or 10^(3*n+3), so the centillion is equated with 10^303, and millillion with 10^3003. However, for a few years, no one had a standard naming system for illions indexed between 10 and 100, 100 and 1000, or greater than 1000.
Bower’s system in 2002 changed this. He divided each illion into “tiers”: with the first tier going from the first illion to the 999th illion, the second tier going from the 1000th illion to the [millillion/1000]th illion (killillion), and the third from the millillionth to the [killillionth/1000]th (kalillion). Intermediate numbers can also be constructed by adding dashes, meaning a “kila-millillion” is equal to the [millillion/1000+1000]th illion.
Bowers ran out of prefixes halfway through Tier 4, where he began using certain astronomical object names astillion (asteroid) through multillion (multiverse). However he stopped at multillion due to the lack of post-multiverse object terms. But googologists have coined much larger numbers than it, so I have attempted to extend the system to at least 6 Tiers.
Hang on… How I start at metillion? Guess what, the naming gaps were already filled. I was going meta for nonstandard illions whatsoever.
But, one ubiquitous issue. If I take met and put the first letter off, it would result the same as I do to bet. They are splitted into 2 identical "et."
(taking the first letter out: bet -> et, met -> et)
But did I stump on that? NO.
I want to do a similar resolution as what Douglas resolved. He decided to turn et and eet when the first letter (b) is removed.
This is consistent as you are adding Tier 3 multiplicatives to Tier 4.
(1st: Taking the first letter out: bet -> et,
2nd: Douglas’ resolution, adding another e: et -> eet,
3rd: Finally, add Tier 3 roots as multiplicative: hot + eet -> hoteet)
This doesn’t work for other tier multiplicatives and all additives:
(Tier 1 multiplicative: vigint + (i) + bet -> vigintibet)
(Tier 2 multiplicative: duec + (e) + bet -> duecebet)
(Tier 1 additive: bet + (o) + zept -> betozept)
(Tier 2 additive: bet + (a) + xenn -> betaxenn)
(Tier 3 additive: bet + mej -> betmej)
I eradicate met into ett for Tier 3 multiplicatives to met. In that case, take out the first letter and add another t after e.
(d + met - (m) + (t) -> dettillion)
(tr + met - (m) + (t) -> trettillion)
(t + met - (m) + (t) -> tettillion)
(sequence: metillion, dettillion, trettillion, tettillion, …)
Miserly, one collides with the existing zettillion…
(z + met - (m) + (t) -> zettillion: collision detected)
So the resolved variant for zettillion has to be degenerated to zeettillion. Just add another e, like Douglas!
(z + (e) + met - (m) + (t) -> zeettillion)
Now I can go hypercelestial and reach far out of natural cosmology! Let's blast at speeds of abstraction!
[10^27-th tier 3 -illion] R(1*10^27)#3 = 10^(3*10^(3*10^(3*10^27))+3) ==> betillion[2*10^27-th tier 3 -illion] R(2*10^27)#3 = 10^(3*10^(3*10^(6*10^27))+3) ==> deetillion[3*10^27-th tier 3 -illion] R(3*10^27)#3 = 10^(3*10^(3*10^(9*10^27))+3) ==> treetillion[4*10^27-th tier 3 -illion] R(4*10^27)#3 = 10^(3*10^(3*10^(12*10^27))+3) ==> teetillion[5*10^27-th tier 3 -illion] R(5*10^27)#3 = 10^(3*10^(3*10^(15*10^27))+3) ==> peetillion[6*10^27-th tier 3 -illion] R(6*10^27)#3 = 10^(3*10^(3*10^(18*10^27))+3) ==> exeetillion[7*10^27-th tier 3 -illion] R(7*10^27)#3 = 10^(3*10^(3*10^(21*10^27))+3) ==> zeetillion[8*10^27-th tier 3 -illion] R(8*10^27)#3 = 10^(3*10^(3*10^(24*10^27))+3) ==> yeetillion[9*10^27-th tier 3 -illion] R(9*10^27)#3 = 10^(3*10^(3*10^(27*10^27))+3) ==> neetillion[10*10^27-th tier 3 -illion] R(10*10^27)#3 = 10^(3*10^(3*10^(30*10^27))+3) ==> dakeetillion[11*10^27-th tier 3 -illion] R(11*10^27)#3 = 10^(3*10^(3*10^(33*10^27))+3) ==> hendeetillionWelcome to the next level… The list of illions go wild out!
Now eradicate the first letters and mutate it on higher roots! This goes legionically worse.
The four names of these are parts of Nirvana Supermind's extended -illions beyond multillion, we have:
R15#4 = 10^(3*10^(3*10^(3*10^45))+3) = metillion
(from Metaverse)
R16#4 = 10^(3*10^(3*10^(3*10^48))+3) = xevillion
(from Xenoverse)
R17#4 = 10^(3*10^(3*10^(3*10^51))+3) = hypillion
(from Hyperverse)
R18#4 = 10^(3*10^(3*10^(3*10^54))+3) = omnillion
(from Omniverse)
Also the nineteenth tier 4 -illion is called:
R19#4 = 10^(3*10^(3*10^(3*10^57))+3) = boxillion
(from The Box) (*Nirvana Supermind originally called outsillion (from The Outside))
After these, I will name these numbers from French numerals (which uses base-20 parts for up to 100):
R20#4 = 10^(3*10^(3*10^(3*10^60))+3) = vingtillion
(from French word "vingt" (meaning 20), not to be confused with vigintillion)
We can reuse some roots from Jonathan Bowers' -illions by:
R21#4 = 10^(3*10^(3*10^(3*10^63))+3) = kalingtillion
(from "kalillion" + "vingtillion")
R22#4 = 10^(3*10^(3*10^(3*10^66))+3) = mejingtillion
(from "mejillion" + "vingtillion")
R23#4 = 10^(3*10^(3*10^(3*10^69))+3) = gijingtillion
(from "gijillion" + "vingtillion")
R24#4 = 10^(3*10^(3*10^(3*10^72))+3) = astingtillion
(from "astillion" + "vingtillion")
R25#4 = 10^(3*10^(3*10^(3*10^75))+3) = luningtillion
(from "lunillion" + "vingtillion")
R26#4 = 10^(3*10^(3*10^(3*10^78))+3) = fermingtillion
(from "fermillion" + "vingtillion")
R27#4 = 10^(3*10^(3*10^(3*10^81))+3) = jovingtillion
(from "jovillion" + "vingtillion")
R28#4 = 10^(3*10^(3*10^(3*10^84))+3) = solingtillion
(from "solillion" + "vingtillion")
R29#4 = 10^(3*10^(3*10^(3*10^87))+3) = betingtillion
(from "betillion" + "vingtillion")
We can even continue further to:
R30#4 = 10^(3*10^(3*10^(3*10^90))+3) = trentillion
(from French word "trente" (meaning 30))
R31#4 = 10^(3*10^(3*10^(3*10^93))+3) = kalentillion
(from "kalillion" + "trentillion")
R32#4 = 10^(3*10^(3*10^(3*10^96))+3) = mejentillion
(from "mejillion" + "trentillion")
R33#4 = 10^(3*10^(3*10^(3*10^99))+3) = gijentillion
(from "gijillion" + "trentillion")
R34#4 = 10^(3*10^(3*10^(3*10^102))+3) = astentillion
(from "astillion" + "trentillion")
R35#4 = 10^(3*10^(3*10^(3*10^105))+3) = lunentillion
(from "lunillion" + "trentillion")
R36#4 = 10^(3*10^(3*10^(3*10^108))+3) = fermentillion
(from "fermillion" + "trentillion")
R37#4 = 10^(3*10^(3*10^(3*10^111))+3) = joventillion
(from "jovillion" + "trentillion")
R38#4 = 10^(3*10^(3*10^(3*10^114))+3) = solentillion
(from "solillion" + "trentillion")
R39#4 = 10^(3*10^(3*10^(3*10^117))+3) = betentillion
(from "betillion" + "trentillion")
R40#4 = 10^(3*10^(3*10^(3*10^120))+3) = quarantillion
(from French word "quarante" (meaning 40))
R50#4 = 10^(3*10^(3*10^(3*10^150))+3) = cinquantillion
(from French word "cinquante" (meaning 50))
R60#4 = 10^(3*10^(3*10^(3*10^180))+3) = soixantillion
(from French word "soixante" (meaning 60))
R70#4 = 10^(3*10^(3*10^(3*10^210))+3) = glocoixantillion
(from "glocillion" + "soixantillion")
R80#4 = 10^(3*10^(3*10^(3*10^240))+3) = quatre-vingtillion
(from French word "quatre-vingt" (meaning 80))
R90#4 = 10^(3*10^(3*10^(3*10^270))+3) = glocatre-vingtillion
(from "glocillion" + "quatre-vingtillion")
We continue further using roots for hundreds in tier 4 -illion system by following:
R100#4 = 10^(3*10^(3*10^(3*10^300))+3) = kentillion
(corruption of French word "cent" (meaning 100) to avoid confusion with "centillion" (10^303))
R101#4 = 10^(3*10^(3*10^(3*10^303))+3) = kalkentillion
("kalillion" + "kentillion") *resolved to avoid confusion with "kalentillion (R31#4)"
R102#4 = 10^(3*10^(3*10^(3*10^306))+3) = mejkentillion
("mejillion" + "kentillion") *resolved again and so on
R103#4 = 10^(3*10^(3*10^(3*10^309))+3) = gijkentillion
("gijillion" + "kentillion")
R104#4 = 10^(3*10^(3*10^(3*10^312))+3) = astkentillion
("astillion" + "kentillion")
...
R114#4 = 10^(3*10^(3*10^(3*10^342))+3) = multkentillion
("multillion" + "kentillion")
...
Unlike from R100 through R119, these do not need the letter "k" before "-ent" before joining within other tiers of -illions by following:
R120#4 = 10^(3*10^(3*10^(3*10^360))+3) = vingtentillion
("vingtillion" + "kentillion")
R130#4 = 10^(3*10^(3*10^(3*10^390))+3) = trententillion
("trentillion" + "kentillion")
R140#4 = 10^(3*10^(3*10^(3*10^420))+3) = quarantentillion
("quarantillion" + "kentillion")
R150#4 = 10^(3*10^(3*10^(3*10^450))+3) = cinquantentillion
("cinquantillion" + "kentillion")
R160#4 = 10^(3*10^(3*10^(3*10^480))+3) = soixantentillion
("soixantillion" + "kentillion")
R180#4 = 10^(3*10^(3*10^(3*10^540))+3) = quatre-vingtentillion
("quatre-vingtentillion" + "kentillion")
We can even join some system even more together:
R200#4 = 10^(3*10^(3*10^(3*10^600))+3) = deuxentillion
(corruption of "deux-cents" (meaning 200 in French))
R300#4 = 10^(3*10^(3*10^(3*10^900))+3) = troisentillion
(corruption of "trois-cents" (meaning 300 in French))
R400#4 = 10^(3*10^(3*10^(3*10^1200))+3) = quatrentillion
(corruption of "quatre-cents" (meaning 400 in French))
R500#4 = 10^(3*10^(3*10^(3*10^1500))+3) = cinquentillion
(corruption of "cinq-cents" (meaning 500 in French))
R600#4 = 10^(3*10^(3*10^(3*10^1800))+3) = sixentillion
(corruption of "six-cents" (meaning 600 in French))
R700#4 = 10^(3*10^(3*10^(3*10^2100))+3) = septentillion
(corruption of "sept-cents" (meaning 700 in French))
R800#4 = 10^(3*10^(3*10^(3*10^2400))+3) = huitentillion
(corruption of "huit-cents" (meaning 800 in French))
R900#4 = 10^(3*10^(3*10^(3*10^2700))+3) = neufentillion
(corruption of "neuf-cents" (meaning 900 in French))
Root Parts
1 kal (-al)
2 mej (-ej)
3 gij (-ij)
4 ast (-ast)
5 lun (-un)
6 ferm (-erm)
7 jov (-ov)
8 sol (-ol)
9 bet (-eet)
10 gloc (-oc)
11 gax (-ax)
12 sup (-up)
13 vers (-ers)
14 mult (-ult)
15 met (-ett)
16 xev (-ev)
17 hyp (-yp)
18 omn (-omn)
19 box (-ox)
20 vingt (-ingt)
30 trent (-rent)
40 quarant (-arant)
50 cinquant (-inquant)
60 soixant (-oixant)
80 quatre-vingt (-atre-vingt)
100 kent (-ent)
200 deuxent (-euxent)
300 troisent (-oisent)
400 quatrent (-atrent)
500 cinquent (-inquent)
600 sixent (-ixent)
700 septent (-eptent)
800 huitent (-uitent)
900 neufent (-eufent)
We can also form some complex tier 4 -illions by:
R574#4 = 10^(3*10^(3*10^(3*10^1722))+3) = multoixantinquentillion
(from "multillion" + "soixantillion" + "cinquentillion")
Credited to Aarex's -illions.
For tier 5 -illions, I will name some -illions after colors.
Unlike previous tiers, this tier is always trailed by "i-" before the next part of -illion in additive form.
We have:
R1000#4 = R1#5 = 10^(3*10^(3*10^(3*10^3000))+3) = blackillion
("black")
We can also name some extended early tier 5 -illions by following:
R1001#4 = 10^(3*10^(3*10^(3*10^3003))+3) = blackikalillion
("blackillion" + "kalillion")
R1002#4 = 10^(3*10^(3*10^(3*10^3006))+3) = blackimejillion
("blackillion" + "mejillion")
R1003#4 = 10^(3*10^(3*10^(3*10^3009))+3) = blackigijillion
("blackillion" + "gijillion")
...
R1020#4 = 10^(3*10^(3*10^(3*10^3060))+3) = blackivingtillion
("blackillion" + "vingtillion")
R1100#4 = 10^(3*10^(3*10^(3*10^3300))+3) = blackikentillion
("blackillion" + "kentillion")
...
R2000#4 = 10^(3*10^(3*10^(3*10^6000))+3) = mejiblackillion
("mejillion" + "blackillion")
R3000#4 = 10^(3*10^(3*10^(3*10^9000))+3) = gijiblackillion
("gijillion" + "blackillion")
R4000#4 = 10^(3*10^(3*10^(3*10^12000))+3) = astiblackillion
("astillion" + "blackillion")
R5000#4 = 10^(3*10^(3*10^(3*10^15000))+3) = luniblackillion
("lunillion" + "blackillion")
R6000#4 = 10^(3*10^(3*10^(3*10^18000))+3) = fermiblackillion
("fermillion" + "blackillion")
R7000#4 = 10^(3*10^(3*10^(3*10^21000))+3) = joviblackillion
("jovillion" + "blackillion")
R8000#4 = 10^(3*10^(3*10^(3*10^24000))+3) = soliblackillion
("solillion" + "blackillion")
R9000#4 = 10^(3*10^(3*10^(3*10^27000))+3) = betiblackillion
("betillion" + "blackillion")
R10000#4 = 10^(3*10^(3*10^(3*10^30000))+3) = glociblackillion
("glocillion" + "blackillion")
R11000#4 = 10^(3*10^(3*10^(3*10^33000))+3) = gaxiblackillion
("gaxillion" + "blackillion")
R12000#4 = 10^(3*10^(3*10^(3*10^36000))+3) = supiblackillion
("supillion" + "blackillion")
R13000#4 = 10^(3*10^(3*10^(3*10^39000))+3) = versiblackillion
("versillion" + "blackillion")
R14000#4 = 10^(3*10^(3*10^(3*10^42000))+3) = multiblackillion
("multillion" + "blackillion")
...
R20000#4 = 10^(3*10^(3*10^(3*10^60000))+3) = vingtiblackillion
("vingtillion" + "blackillion")
R30000#4 = 10^(3*10^(3*10^(3*10^90000))+3) = trentiblackillion
("trentillion" + "blackillion")
R40000#4 = 10^(3*10^(3*10^(3*10^120000))+3) = quarantiblackillion
("quarantillion" + "blackillion")
R50000#4 = 10^(3*10^(3*10^(3*10^150000))+3) = cinquantiblackillion
("cinquantillion" + "blackillion")
R60000#4 = 10^(3*10^(3*10^(3*10^180000))+3) = soixantiblackillion
("soixantillion" + "blackillion")
R80000#4 = 10^(3*10^(3*10^(3*10^240000))+3) = quatre-vingtiblackillion
("quatre-vingtillion" + "blackillion")
...
R100000#4 = 10^(3*10^(3*10^(3*10^300000))+3) = kentiblackillion
("kentillion" + "blackillion")
R200000#4 = 10^(3*10^(3*10^(3*10^600000))+3) = deuxentiblackillion
("deuxentillion" + "blackillion")
R300000#4 = 10^(3*10^(3*10^(3*10^900000))+3) = troisentiblackillion
("troisentillion" + "blackillion")
...
After blackillion, we continue to:
R2#5 = R1,000,000#4 = 10^(3*10^(3*10^(3*10^3,000,000))+3) = whitillion
("white")
R10,000,000#4 = 10^(3*10^(3*10^(3*10^30,000,000))+3) = glociwhitillion
("glocillion" + "whitillion")
R100,000,000#4 = 10^(3*10^(3*10^(3*10^300,000,000))+3) = kentiwhitillion
("kentillion" + "whitillion")
...
R3#5 = R1,000,000,000#4 = 10^(3*10^(3*10^(3*10^3,000,000,000))+3) = redillion
("red")
R4#5 = R1,000,000,000,000#4 = 10^(3*10^(3*10^(3*10^3,000,000,000,000))+3) = yellowillion
("yellow")
R5#5 = 10^(3*10^(3*10^(3*10^(3*10^15)))+3) = bluillion
("blue")
R6#5 = 10^(3*10^(3*10^(3*10^(3*10^18)))+3) = greenillion
("green")
R7#5 = 10^(3*10^(3*10^(3*10^(3*10^21)))+3) = orangillion
("orange")
R8#5 = 10^(3*10^(3*10^(3*10^(3*10^24)))+3) = purpillion
("purple")
R9#5 = 10^(3*10^(3*10^(3*10^(3*10^27)))+3) = brownillion
("brown")
We continue to tens by following:
R10#5 = 10^(3*10^(3*10^(3*10^(3*10^30)))+3) = grayillion
("gray")
R11#5 = 10^(3*10^(3*10^(3*10^(3*10^33)))+3) = grayiblackillion
("grayillion" + "blackillion")
R12#5 = 10^(3*10^(3*10^(3*10^(3*10^36)))+3) = grayiwhitillion
("grayillion" + "whitillion")
R13#5 = 10^(3*10^(3*10^(3*10^(3*10^39)))+3) = grayiredillion
("grayillion" + "redillion")
R14#5 = 10^(3*10^(3*10^(3*10^(3*10^42)))+3) = grayiyellowillion
("grayillion" + "yellillion")
R15#5 = 10^(3*10^(3*10^(3*10^(3*10^45)))+3) = grayibluillion
("grayillion" + "bluillion")
R16#5 = 10^(3*10^(3*10^(3*10^(3*10^48)))+3) = grayigreenillion
("grayillion" + "greenillion")
R17#5 = 10^(3*10^(3*10^(3*10^(3*10^51)))+3) = grayiorangillion
("grayillion" + "orangillion")
R18#5 = 10^(3*10^(3*10^(3*10^(3*10^54)))+3) = grayipurpillion
("grayillion" + "purpillion")
R19#5 = 10^(3*10^(3*10^(3*10^(3*10^57)))+3) = grayibrownillion
("grayillion" + "brownillion")
...
R20#5 = 10^(3*10^(3*10^(3*10^(3*10^60)))+3) = cyanillion
("cyan")
R30#5 = 10^(3*10^(3*10^(3*10^(3*10^90)))+3) = magentillion
("magenta")
R40#5 = 10^(3*10^(3*10^(3*10^(3*10^120)))+3) = indigillion
("indigo")
R50#5 = 10^(3*10^(3*10^(3*10^(3*10^150)))+3) = limillion
("lime")
R60#5 = 10^(3*10^(3*10^(3*10^(3*10^180)))+3) = pinkillion
("pink")
R70#5 = 10^(3*10^(3*10^(3*10^(3*10^210)))+3) = violetillion
("violet")
R80#5 = 10^(3*10^(3*10^(3*10^(3*10^240)))+3) = silvillion
("silver" (as a color))
R90#5 = 10^(3*10^(3*10^(3*10^(3*10^270)))+3) = goldillion
("gold" (as a color))
We define the hundreds of the tier of -illions from tertiary colors by following:
R100#5 = 10^(3*10^(3*10^(3*10^(3*10^300)))+3) = vermillillion
("vermilion")
R200#5 = 10^(3*10^(3*10^(3*10^(3*10^600)))+3) = ambillion
("amber")
R300#5 = 10^(3*10^(3*10^(3*10^(3*10^900)))+3) = chartrillion
("chartreuse")
R400#5 = 10^(3*10^(3*10^(3*10^(3*10^1200)))+3) = ochrillion
("ochre")
R500#5 = 10^(3*10^(3*10^(3*10^(3*10^1500)))+3) = tealillion
("teal")
R600#5 = 10^(3*10^(3*10^(3*10^(3*10^1800)))+3) = fuchsillion
("fuchsia")
R700#5 = 10^(3*10^(3*10^(3*10^(3*10^2100)))+3) = olivillion
("olive")
R800#5 = 10^(3*10^(3*10^(3*10^(3*10^2400)))+3) = ultramarillion
("ultramarine")
R900#5 = 10^(3*10^(3*10^(3*10^(3*10^2700)))+3) = crimsillion
("crimson")
And we get some roots by following:
Root Parts
1 black
2 whit
3 red
4 yellow
5 blu
6 green
7 orang
8 purp
9 brown
10 gray
20 cyan
30 magent
40 indig
50 lim
60 pink
70 violet
80 silv
90 gold
100 vermill
200 amb
300 chartr
400 ochr
500 teal
600 fuchs
700 oliv
800 ultramar
900 crims
Yes, there is no exception that combining the roots with that tier to have different form!
For tier 6 -illions, it has the same nomenclature as in Nirvana Supermind's -illion extensions from Greek quantitive prefixes, but uses some different etymologies from the previous tiers, and uses "i-" instead.
We have some example such as:
R1#6 = R1000#5 = 10^(3*10^(3*10^(3*10^(3*10^3000)))+3) = hapaxillion
(from "hapax", a Greek quantitive prefix meaning "one")
R1001#5 = 10^(3*10^(3*10^(3*10^(3*10^3003)))+3) = hapablackillion
(from "hapaxillion" + "blackillion" (different from "hepillion" as in "hapahepillion"))
R2000#5 = 10^(3*10^(3*10^(3*10^(3*10^6000)))+3) = whitihapaxillion
(from "whitillion" (different from "ottillion" as in "ottapaxillion") + "hapaxillion")
For those beyond tier 6 -illions, see part 2.