After a previous attempt to make tier 5 and 6 -illions, the tier 5 -illions have an incorrect color order, so I will partially remake the tier 5 -illions right now.
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.
Welcome to the realm of tetrational! Hyperexponents now feel close between 2 layers, and illions aren't pretty important.
We still feel like simpler formulas need special treatments from illions, hence low Tier 1.
So, that doesn't matter. I will keep going infinitum.
The Tier 5 -illions are based on colors. The additive -illion suffix of the current tier is simply -oni and the multiplicative -illion suffix of the previous tier is -ee (also replace "c" with "k" before adding "-ee").
The inter-tier additives uses the infix "-ai-" for the tier 5 indexes 21 through 999.
Redillion = H1#5 = H1,000#4
Redoni-kalillion = H1,001#4
Redoni-mejillion = H1,002#4
Redoni-gijillion = H1,003#4
Redoni-glocillion = H1,010#4
Redoni-kentillion = H1,100#4
Mejeeredillion = H2,000#4
Gijeeredillion = H3,000#4
Asteeredillion = H4,000#4
Luneeredillion = H5,000#4
Fermeeredillion = H6,000#4
Joveeredillion = H7,000#4
Soleeredillion = H8,000#4
Beteeredillion = H9,000#4
Glokeeredillion = H10,000#4
Gaxeeredillion = H11,000#4
Supeeredillion = H12,000#4
Verseeredillion = H13,000#4
Multeeredillion = H14,000#4
Vingteeredillion = H20,000#4
Trenteeredillion = H30,000#4
Kenteeredillion = H100,000#4
Orangillion = H2#5 = H1,000,000#4
Yellowillion = H3#5 = H1,000,000,000#4
Greenillion = H4#5 = H1,000,000,000,000#4
Cyanillion = H5#5
Bluillion = H6#5
Violetillion = H7#5
Purpillion = H8#5
Magentillion = H9#5
Pinkillion = H10#5
Piredillion = H11#5
Pyorangillion = H12#5
Pyellowillion = H13#5
Pigreenillion = H14#5
Picyanillion = H15#5
Piblueillion = H16#5
Pivioletillion = H17#5
Pipurpillion = H18#5
Pimagentillion = H19#5
Orankillion = H20#5
Orankairedillion = H21#5
Orankaiorangillion = H22#5
Orankaiyellowillion = H23#5
Yellinkillion = H30#5
Greeninkillion = H40#5
Cyaninkillion = H50#5
Bluinkillion = H60#5
Violinkillion = H70#5
Purplinkillion = H80#5
Magentankillion = H90#5
Vermillillion = H100#5
Limillion = H200#5
Tealillion = H300#5
Indigillion = H400#5
Carminillion = H500#5
Brownillion = H600#5
Blackillion = H700#5
Grayillion = H800#5
Whitillion = H900#5
For tier 6 -illion they have the same etymologies and construction of intermediate -illions as like with Aarex Tiaokhiao's construction. These are based on Greek quantitive prefixes like with "hapaxillion" = H1#6 = H1,000#5, "disillion" = H2#6 = H1,000,000#5, "trisillion" = H3#6 = H1,000,000,000#5, and so on, except it uses the completely different tier 4 and 5 -illion etymologies.
We can even have some limit like:
(by DeepLineMadom) whitaimagentankaimagentoni-enneacosienenecontaennisyn-whitaimagentankaimagentoni-enneacosienenecontaoctisyn-...whitaimagentankaimagentoni-trisyn-whitaimagentankaimagentoni-dusyn-whitaimagentankaimagentoni-pa-whitaimagentankaimagentillion
(by Nirvana Supermind) sifa'zoza'brog'enneacosienenecontaennisyn-sifa'zoza'brog'enneacosienenecontaoctisyn-...sifa'zoza'brog'trisyn-sifa'zoza'brog'dusyn-sifa'zoza'brog'paki-sifa'zoza'brontillion
Now we reach tetration level!
Tier 6 multiplicative: -pa (from hapaxillion), -syn (>=2)
Tier 7 additive: -ioni-
Inter-tier additive root: -i-
+Teen additives for tiers 21 through 99
Hydrillion = H1#7 = H1,000#6
Hydrioni-hapaxillion = H1,001#7
Hydrioni-decakisillion = H1,010#7
Hydroni-hectatonakisillion = H1,100#7
Dykehydrillion = H2,000#7
Trykehydrillion = H3,000#7
Tetrkehydrillion = H4,000#7
Penkehydrillion = H5,000#7
Hexkehydrillion = H6,000#7
Epitkehydrillion = H7,000#7
Ocekehydrillion = H8,000#7
Enekehydrillion = H9,000#7
Decisehydrillion = H10,000#7
Hectisehydrillion = H100,000
Helillion = H2#7 = H1,000,000#6
Lithillion = H3#7 = H1,000,000,000#6
Berillion = H4#7
Borillion = H5#7
Carbillion = H6#7
Nitrillion = H7#7
Oxygillion = H8#7
Fluorillion = H9#7
Neonillion = H10#7
Sodillion = H11#7
Magnesillion = H12#7
Aluminillion = H13#7
Silicillion = H14#7
Phosphorillion = H15#7
Sulfillion = H16#7
Chlorillion = H17#7
Argonillion = H18#7
Potassillion = H19#7
Calcillion = H20#7
Calcisodillion = H21#7
Calcimagnesillion = H22#7
Calci'aluminillion = H23#7
Calcisilicillion = H24#7
Calciphosphorillion = H25#7
Calcisulfillion = H26#7
Calcichlorillion = H27#7
Calciargonillion = H28#7
Calcipotassillion = H29#7
Titanillion = H30#7
Chromillion = H40#7
Ferrillion = H50#7 (from Latin name "ferrum", equivalent to "iron" in English)
Cobaltillion = H60#7
Nickelillion = H70#7
Cuprillion = H80#7 (from Latin name "cuprum", equivalent to "copper" in English)
Zincillion = H90#7
Gallillion = H100#7
Arsenillion = H200#7
Selenillion = H300#7
Bromillion = H400#7
Kryptillion = H500#7
Rubidillion = H600#7
Strontillion = H700#7
Niobillion = H800#7
Molybdenillion = H900#7
We continue the next tier of -illion with the same concepts!
Tier 6 multiplicative: -aisi
Tier 7 additive: -iumi-
Inter-tier additive root: -i- (for tiers 21 through 99 as well as hundreds)
Ruthenillion = H1#8 = H1,000#7
Rutheniumihydrillion = H1,001#7
Rutheniumineonillion = H1,010#7
Rutheniumigalillion = H1,100#7
Heliaisiruthenillion = H2,000#7
Lithiaisiruthenillion = H3,000#7
Beryliaisiruthenillion = H4,000#7
Boraisiruthenillion = H5,000#7
Carbaisiruthenillion = H6,000#7
Nitraisiruthenillion = H7,000#7
Oxygiaisiruthenillion = H8,000#7
Fluoriaisiruthenillion = H9,000#7
Neonaisiruthenillion = H10,000#7
Galliaisiruthenillion = H100,000#7
Rhodillion = H2#8 = H1,000,000#7
Palladillion = H3#8 = H1,000,000,000#7
Silvillion = H4#8
Stannillion = H5#8 (from Latin "stannum", equivalent to "tin" in English)
Antimonillion = H6#8
Tellurillion = H7#8
Iodillion = H8#8
Caesillion = H9#8
Barillion = H10#8
From 11 to 19, we use the rare-earth metals as a name as follows:
Lanthanillion = H11#8
Cerillion = H12#8
Praseodymillion = H13#8
Neodillion = H14#8
Samarillion = H15#8
Terbillion = H16#8
Dysprosillion = H17#8
Erbillion = H18#8
Lutetillion = H19#8
We move to:
Hafnillion = H20#8
Hafnilanthanillion = H21#8
Hafnicerillion = H22#8
Hafnipraseodymillion = H23#8
...
Tantalillion = H30#8
Tungstillion = H40#8
Osmillion = H50#8
Iridillion = H60#8
Platinillion = H70#8
Goldillion = H80#8
Mercurillion = H90#8
Leadillion = H100#8
Bismuthillion = H200#8
Radillion = H300#8
Actinillion = H400#8
Thorillion = H500#8
Uranillion = H600#8
Plutonillion = H700#8
Americillion = H800#8
Curillion = H900#8
The limit of the elemental realm is:
Molybdenizincipotassiaisicurimercurilutetiumi-molybdenizincipotassiaisicurimercurierbiumi-
molybdenizincipotassiaisicurimercuridysprosiumi-molybdenizincipotassiaisicurimercuriterbiumi-
molybdenizincipotassiaisicurimercurisamariumi-molybdenizincipotassiaisicurimercurineodiumi-
molybdenizincipotassiaisicurimercuripraseodymiumi-molybdenizincipotassiaisicurimercuriceriumi-
molybdenizincipotassiaisicurimercurilanthaniumi-molybdenizincipotassiaisicurimercuriumi-
...
...
...
molybdenizincipotassiaisicaesiumi-molybdenizincipotassiaisi'iodiumi-
molybdenizincipotassiaisitelluriumi-molybdenizincipotassiaisi'antimoniumi-
molybdenizincipotassiaisistanniumi-molybdenizincipotassiaisisilviumi-
molybdenizincipotassiaisipalladiumi-molybdenizincipotassiaisirhodiumi-
molybdenizincipotassiaisirutheniumi-molybdenizincipotassillion