In this article, we will cover Steinhaus-Moser notation, a notation used for expressing certain very large numbers by putting numbers inside polygons. It gets its name because it was originally defined by Hugo Steinhaus, and later Leo Moser added hexagons, heptagons, octagons, and replaced the old circle with the pentagon.
n in a triangle is equal to nn, and n in a square is equal to n in n triangles, and n in a pentagon is equal to n in n squares, and n in a hexagon is equal to n in n pentagons, and so on. We will begin by covering the squaral numbers (not to be confused with square numbers).
Squaral Numbers
This section is about the sequence of numbers in a square, so if you came to this page to read about the mega, skip to the next section of the article, which covers the numbers that are a number inside a pentagon.
2[4] = 2[3][3] = (22)^(22) = 256
Not too surprising. 2 in a square is merely 2 in 2 triangles, and 2 in one triangle is only 4. Next up is:
3[4] = 3[3][3][3] = ((3^3)^(3^3))^((3^3)^(3^3)) = (27^27)^(27^27) = (3^81)^(3^81) = 3(3^85)
The result of 3^(3^85) is approximately 101.7137*10^40, meaning it is a number with over 17 duodecillion digits! The next one is not only greater than a googolplex but also greater than a googolduplex:
4[4] = 4[3][3][3][3] = ((256^256)^(256^256))^((256^256)^(256^256)) is approximately 10^(10^(1.99237*10^619))
It is now impossible for anyone to compute the leading digits of this number as it has over 10^(1.99237*10^619) digits! In fact, it is the second number that is n inside an n-sided polygon, and equal to 2 in 5 triangles as 2 in one triangle is 4.
Pentagon Numbers
The pentagon is the next polygon operator after the square. Below is the first pentagon number other than 1:
2[5] = 2[4][4] = 256[4] = 256[3][3][3][3][3][3][3]...[3][3][3][3][3][3] with 256 copies of [3]
This seems totally out of left field. Even just 2 in a pentagon is equal to 256 in 256 triangles! This number is the mega. And yet even just the first triangle of 256 is very hard to calculate if you don't have access to a big number calculator or similar. Its exact value is:
32,317,006,071,311,007,300,714,876,688,669,951,960,444,102,669,715,484,032,130,345,427,524,655,138,867,890,893,197,201,411,522,913,463,688,717,960,921,898,019,494,119,559,150,490,921,095,088,152,386,448,283,120,630,877,367,300,996,091,750,197,750,389,652,106,796,057,638,384,067,568,276,792,218,642,619,756,161,838,094,338,476,170,470,581,645,852,036,305,042,887,575,891,541,065,808,607,552,399,123,930,385,521,914,333,389,668,342,420,684,974,786,564,569,494,856,176,035,326,322,058,077,805,659,331,026,192,708,460,314,150,258,592,864,177,116,725,943,603,718,461,857,357,598,351,152,301,645,904,403,697,613,233,287,231,227,125,684,710,820,209,725,157,101,726,931,323,469,678,542,580,656,697,935,045,997,268,352,998,638,215,525,166,389,437,335,543,602,135,433,229,604,645,318,478,604,952,148,193,555,853,611,059,596,230,656
This number has exactly 617 digits, and can be expressed in base 2 as 22048, or 2^(2^11). The next number in the sequence is this number raised to the power of itself, which is:
398,443,420,064,016,039,.........,648,486,875,640,365,056 with about 1.9923739*10619 digits
This number is so large that it is impossible for anyone to compute its complete decimal expansion, and yet there are still 254 inscribed triangles to go! The result of the next inscribed triangle after that is equal to 4 in a square (a number that we encountered in the last section).
The final result is about 10^(10^(10^(...(10^(7.938*10^19923739028520154087706422945147014652916223529059455829739546236757445592829019852096549871643037231579555867729029727837739722687243833688041650758866703047684995147926044802500789969233229482277620428871361665114606086501621360310636409247822506979293012834235605892457887360583787492777424798206285182369042469497447438158240050711323245053205431372163355524614258748270064178183600550138767745559315784832858638844869498054620521042914198455705585134437206064557323165937735931605786380378378018264857422432758696743477636091751483267310595348292927018011128165226311150554708199087683524760666293693562405279021537))...))) with 255 10s including the "10" in 7.938*10^19923739...021537.
To read more about bounding the mega, click here.
The last few digits of the mega can be computed quite easily, and it is in fact possible to calculate thousands of terminating digits of the mega using methods such as modular exponentiation. The last 5000 digits of the mega are given below.
...95349023028618553749832119920226094055997377690296163291300803575045996441166806301502490450260656489250941844326130861876392462502884613118510014308714341419920580220488755977791245771686978621780403387279049316009315221282761396923062835402571774079871954531151266685756361190405786179863997344068152164745142524897730204092118143855405317944082521493422518317613074196462805648070013494883116882004797830412565696958279065952177095868847496615856875852827813700840223622561610205724003893286970820394311776709595607453373509304364840305069952253071508094565321578887063138658966575711049819432725968506965930299264055858177423532584243554764391610278179076245812181716331733159767873005927139753605775517120744004250432567177363584503824271363294656157510703027464952289280918078420433574244249786552393917572978712500322409413567808864770480362989064766191315200888400447492540383424696602021239736353891084877006003571681139812144518601466782766729076339952661466381190438945575188075171180135825803964011080368184144595766219179628343924571964117866881238351646791810054633514790090480044244685246024533808314928362155553765205438396723116142935075450090534837013734856677220282083148689064105138170868631895122129272060982997139875659986041345044350440927598486168740859881248566969923372060589951648114028836936704388949027171407618738222648780263600718103937678102460630405955327423976298361967379921490236772141189786313638175210409645244821740911631287628942078356239984976117168957007693658904438759589140649745855249646025304676914433806466179007002634722296651015289857094282501322900251659741108047304808503604808605953443391389201816835003736669323947661981562334664299378003823283506012908613626917253640755992580534561235403330513563433184311921752909314501366444647358186796261387016366273975550847865126346078704950794203263912243070102964503080187139456044715599312191926888336033853800136039906307961759929981034765131565665285357062755812508724646246576938803863584152146119129142215433310350605350043061109446660174055356688566632269360070882426362037823204020044131477637548964265797420406440203455139619314560448120115647321668061982480649019048385777227628441937345048054710943419343980811158314685075006484053366940174705907304079611684792646530235472006333076110574120487496348502866017575099422424941312865775622882350971868378969773121060297876975338019148889157716153492096851318552709076028059272548029438773875047381687939634495341685512643664231696478629048224472316041444252932601815776797792801595529986871529099021548381403098115162964634595550036775303387680861805861137545491529425401245054386393545831529415008237525623656592473044974441044445433588247192262911815940508229424374824427972571789678016455860632780682638159149431900407221261908828157935556386629092271527099527548495870693287431596032432813170791418070111288240997174197687045039388519282596515663904905899483642109930612002436419162932346974468120744933882539509389941962019047762216608957011722784055021500518051182386455695865138139265731364383557811531546972412222073847589808307671259289947493055201747562185914472471563537904639373432516604956957510953882072435930977225907249715370894200775843537944693315633809827093707371119887243789293950211589902345569778452807213111715048263369611842524703330048385057028466909305253336373281753304083288839157974079560719057087445555019809033629084808498482700958480836589924034313908129508178954403079678655369215555193487391832303113527535338604061475248064276862936758605367812980003425680369838450282475531576982183932638795992157158722571233321027011290298989159537809293240246808467427355220756060413176878244759327955244774312122618195057638789807285186306199600717614998210505569476805810678818980548760679753320874780578116884099700423716302102864814370267881993840332320606411659160486754107821615798064217137799684818996751279640380942073475112280910627589866684606623694380777240017114646807096674299884998468194716532630918779544419471882730607724108264936414916927895525116970379589815699860789428179752223540768603591111785116844998864190112486263849933610613465386927919192469831496598716697950848752582277438126636875521941759989636744054945416125552511322379108415541598503747293982471986260657665703465038601001235952116037944917901444627919809770467399267919231340786425321602602844904940428194927026211510168127986032737054183640162959410191530910128122872448951298789194912244209798914913155189119008412102279931826609234262961460122045725315785186425736461268034794021680658790786586247524764872803902837976103376819650464339061415489976640723044576837710810075626398219535580177406392400725774201348843029490585776329091934693098566888682178174163383443798637165141207014929599697879439430046777231331325642730394610854006632923911301838502897561606601450696228009791206968151401164151569201969226596586425189027828673870027190816633074193637388524012354109181117468059671715057145877044749894910112922449731993539660742656
The next one is:
3[5] = 3[4][4][4] = (3^(3^85))[4][4] is about 10^^(10^^(10^(10^40)))
This number is called the grand mega, or triton (I like the latter name better, mainly because it sounds more like something new than just a mega extension). It is approximately a power tower of 10s that is "a power tower of 10s with 3^3^85 terms" terms high, which is much larger than the mega.
The triton is approximately:
10^10^10^10^...^10^10^10^10^40
where the number of 10s is
10^10^10^10^...^10^10^10^10^40
where the number of 10s is 3^(3^85) + 2
Computing last digits of the triton is more difficult than calculating the last digits of the Mega, mainly because the number of operations required to find the Mega's last digits is fixed at 256 or so, but the number of operations required to find the last digits of this number grows exponentially with the number of desired digits. The same difficulty is encountered with all numbers other than Mega defined with an operator higher than the square operator.
We can find the last 2 digits of the Triton, however, which are ...27. The last digits of the iterated triangles of 3 alternate between ...03 and ...27, and the number of times we apply the triangle operator to get to 3 in a pentagon is 3 + 33^85 + (33^85)[4], which is odd.
To find the last 3 digits is slightly more difficult, but can still be done relatively easily. The last 3 digits of iterated triangles of 3 repeat with a period of 10, which means that the last 3 digits of Triton can be derived from the last digit of the number of times the triangle operator is applied:
3 + 33^85 + (33^85)[4] = 3 + XXX...XX7 + XXX...XX3 = XXX...XX3
Thus the last 3 digits of Triton are the same as 3[4], or ...627.
Finding the last 4 digits is significantly more difficult, but it can still be done by finding the number of times the triangle operator is applied mod 50. 3 is just 3, 33^85 is 27 mod 50, and (33^85)[4] is also 3 mod 50, so the number of times we have to apply the triangle operator is 3 + 27 + 3 = 33. It can be shown that 3[3]33 ends in ...1627, and so those are also the last 4 digits of Triton. With more effort, it can be shown that the last 5 digits are ...21627.
To go further, I wrote a program in Wolfram MathematicaTM that would iterate the modulo of the triangle operator, first to compute the last digits of 3[4]2, and then the Triton. Using this, I found that the last 6 digits of Triton are ...721627. I next computed the last 7 digits, which are ...3721627.
I eventually obtained the last 10 digits, which are ...6013721627. The time it takes to compute the last digits increases tenfold when the number of desired digits is increased by 1, though, which would impede much further progress beyond this. Nonetheless, by pattern matching I later found that the last 17 digits are ...49185706013721627.
5[5] = 5[4][4][4][4][4] = 5[3][3][3][3][3][4][4][4][4] = (55^15,631)[3][3][4][4][4][4]
This number is referred to as the gong mega. It is approximately 10^^(10^^(10^^(10^^(1010^10^10^10921)))), and its last 20 digits are ...31732654571533203125. Computing last digits of this number is easier, because the number of operations merely doubles with each additional digit, instead of multiplying 5 or 10-fold, and also, the last 4 digits remain fixed at 3125 with each triangle.
I named the bong mega based on the analogy googolgong:googolbong::gong mega:?. It is equal to 15 in a pentagon, and its last 23 digits are ...23516213893890380859375. Its last digits can be calculated more easily because 15 is a multiple of 5, and one less than a power of 2. It is approximately 10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^(10^^16)))))))))))))), or more accurately 10^^(14)(10^10^...^10^10^18 with 15 10s) (that means f^14(10^10^10^10^10^10^10^10^10^10^10^10^10^10^10^18) where f(n) = 10^^n.
I then also named the throng mega, equal to 25 in a pentagon, which is approximately 10^^(10^^(10^^(...(10^^(N))...))) with 24 10s (and N is about 10^10^10^...^10^34 with 25 10s), and the quadrigong (gandigang) mega, equal to 35 in a pentagon, which is approximately 10^^(10^^(10^^(...(10^^(10^^35))...))). The last digits of these numbers are ...447265625 and ...46435546875 respectively.
Last digits of pentagonal numbers (calculated using similar methods used for last digits of 2[5] and 3[5]):
1st: 1
2nd (mega): ...9660742656
3rd (triton): ...6013721627
4th (great mega): ...8726598656
5th (gong mega): ...1533203125
6th (hexomega): ...8726598656
7th (heptomega): ...2545463543
8th (octomega): ...5602159616
9th (nonomega): ...4451420489
10th (megiston): ...0000000000
Hexagon Numbers
The hexagon is the next polygon operator after the pentagon. And yet even the first one greater than 1 is the already gigantic mega in a pentagon:
2[6] = 2[5][5] = (mega)[5] = 2[5][4][4][4][4][4][4][4][4]...[4][4][4][4][4][4][4][4] w/ mega 4s
This is a number called A-ooga, and also megision. This number is approximated by 10^^(10^^(10^^(10^^...(10^^(10^^(10^^257)))...))) where the number of 10s is (2[5])+1.
Unlike with the mega, computing this number's last digits is harder, and in fact impossible after about a dozen digits. We can, however, find the rightmost few digits. We already know that the last 2 digits are ...56, because at the end of the day 2[6] is just 2 in a very large number of inscribed triangles. Finding the last 3 digits, though, is slightly harder. We already know that the last 3 digits repeat with a period of 5 each time the triangle operator is applied. Since 256, 256[4], 256[4][4], ... are all congruent to 1 mod 5, each time we apply the square operator, it is the same as applying the triangle operator once, which adds 400 to the last 3 digits. This means that 256[4][4] ends in ...056, 256[4][4][4] ends in ...456, and so on. Since the number of squares we apply (which is a mega) is 1 mod 5, 2[6] = ......056.
Obtaining the last 4 digits of A-ooga is still relatively easy. Simply observe that the last 4 digits repeat with a period of 25, and 256, 256[4], 256[4][4], ... are all congruent to 6 mod 25. This means that each time we apply the square operator, it is equivalent to applying the triangle operator 6 times. The number of squares we apply is itself 6 mod 25, meaning that we apply the triangle operator 6*6 = 36 == 11 (mod 25) times. Thus the last 4 digits of 2[6] are the same as those of 256[3]267, which are also the same digits that end 256[3]17, which are ...1056.
The last 10 digits of A-ooga (computed by me) are ...7079341056.
The difficulty in finding the A-ooga's last digits, just as with 3[5], stems not from the amount of space (memory) required, but that the time it takes to perform the computation grows exponentially with the number of desired digits. This happens with basically all numbers defined with Steinhaus-Moser notation with a polygon with at least 5 sides (an exception is the Mega, because the number of iterations remains fixed at 256 in that case).
Aarex Tiaokhiao didn't coin the name A-ooga, but he did coin the name megision.
3[6] = 3[5][5][5] = 3[4][4][4][5][5] = 3[3][3][3][4][4][5][5]
3 in a hexagon is equal to the grand mega inside two more pentagons! Its last 7 digits are ...4881627. This number can be called the grand megision in analogy with the grand mega, which we saw in the last section, and this number is also referred to as A-oogra.
4[6] = 4[5][5][5][5] = 4[4][4][4][4][5][5][5]
Aarex extended on the megision by defining the megisiduon, which is equal to 2 in a hexagon inside another pentagon.
Aarex extended on the A-ooga by defining the A-oogatiplex (which is 2[6] inside another hexagon, and (spoiler alert) a number we will encounter in the next section), the A-oogatiduplex (equal to 2[6][6][6]), the A-oogatitriplex (2[6][6][6][6]), and the A-oogatiquadriplex (2[6][6][6][6][6]).
Heptagons, Octagons, and Beyond...
We can continue with heptagons. 2[7] is a number that Aarex refers to as the betomega. It is equal to the A-ooga inside another hexagon, or approximately 10^^^^(10^^^(mega+1)), and its last digits are ...3395456. Aarex also refers to 3[7] as betogiga, extrapolating from mega being the SI prefix for a million, and giga being the SI prefix for a billion.
2[8] is the first number of the form 2 in a polygon that Aarex didn't coin a name for, and it is approximately 10 heptated to 2[7], and its last digits are ...1305856.
Next we will cover the Moser, sometimes referred to as Moser's number, a very large number equal to 2 in a megagon in Steinhaus-Moser notation, which we covered in the last section. The number is named after Leo Moser (who was believed to have extended the notation to include polygons with more than four sides).
Estimating the Moser
Estimating the Moser is much harder than estimating the Mega, but it can still be done to some extent.
First, we will try to create a lower-bound and an upper-bound for the Moser in terms of up-arrow notation. 2 in a square, or 256, is between 2^^3 and 2^^4, and the mega is between 2^^^3 and 2^^^4. This pattern continues: 2 in a hexagon is between 2^^^^3 and 2^^^^4, 2 in a heptagon is between 2^^^^^3 and 2^^^^^4, and so on. Therefore the Moser, or 2 in a megagon, is between 2^^^...^^^3 with mega-2 ^s and 2^^^...^^^4 with mega-2 ^s.
Using base 10 instead of 2, we get: 256 is between 10^^1 and 10^^2, the mega is between 10^^^2 and 10^^^3, 2 in a hexagon is between 10^^^^2 and 10^^^^3, and so on. Therefore the Moser is between 10^^^...^^^2 with mega-2 ^s and 10^^^...^^^3 with mega-2 ^s. However, the upper bounds are actually larger, so the best bounds for the Moser we have at the moment are:
10^(mega-2)2 < Moser < 2^(mega-2)4
We can actually do better by observing that mega is about 10^^256 (actually it is more than 10^^257), and 2 in a hexagon is a bit larger than 10^^^mega+1. So, the Moser can be lower-bounded by
10^^...(mega-3 ^s)...^^(10^^...(mega-4 ^s)...^^(10^^...(mega-5 ^s)...^^(10^^...(mega-6 ^s)...^^(...(10^^^^^(10^^^^(10^^^mega+1)))...)))) (w/ mega - 5 levels of parentheses)
To get an upper bound, just take that expression and replace the 10^^^mega+1 with 10^^^mega+2, as 2[6] is upper-bounded by 10^^^mega+2.
Last digits of the Moser
It is significantly harder to compute the last digits of the Moser than the Mega. However, it is still possible to find the last few terminating digits of the Moser. First of all we know it must end in ...56 since it reduces to 2 in a very large number of inscribed triangles.
If we look at the last 3 digits of a number that is 2 in a polygon, we see the following cycle with a period of 5: 256, 656, 056, 456, 856. We then observe:
2 in an n-gon == 256 mod 1000 if n leaves remainder 4 when divided by 5
2 in an n-gon == 656 mod 1000 if n leaves no remainder when divided by 5
2 in an n-gon == 56 mod 1000 if n leaves a remainder of 1 when divided by 5
2 in an n-gon == 456 mod 1000 if n leaves a remainder of 2 when divided by 5
2 in an n-gon == 856 mod 1000 if n leaves a remainder of 3 when divided by 5
Since the mega leaves a remainder of 1 when divided by 5, the Moser ends in ...056. The cycle period for the last 4 digits is, of course, 25. The mega leaves remainder 6 when divided by 25, and 2[6] ends in ...1056, and so the Moser also ends in ...1056. By going up another level of abstraction we can find that the last 5 digits of the Moser are ...01056. The last 8 digits of the Moser are ...80301056.
Extensions to the Moser
The grand moser is equal to 3 in a grand mega-gon. It is approximately 3^^^...^^^4 with grand mega - 2 ^s in up-arrow notation. Like the grand mega, great mega, gong mega, and so on, it was named by Aarex Tiaokhiao.
The last 5 digits of the grand Moser are ...81627.
The great moser is equal to 4 in a great mega-gon. It is approximately 4^^^...^^^5 with mega-2 ^s in up-arrow notation, and its last three digits are ...656.
In a discussion board somewhere on the Internet, I encountered much larger numbers based on the Moser. We will begin with the smallest of these, which is the mosiston. As you would guess from the name megiston, it is equal to 10 in a polygon with a megiston sides, and is approximately equal to 10^^^^^^...^^^^^^11 where the number of arrows is a megiston minus 2.
Next is the super-moser, defined as 2 in a polygon with Moser's number of sides, similar to how Moser's number is defined as 2 in a polygon with a mega sides. It is approximately equal to 10^^^^^^^^^...^^^^^^^^^^2 where the number of arrows is Moser's number minus 2. What about the last digits of that number, you might ask? Computing last digits of the super-moser is only about as complex as computing last digits of the Moser, and the last 8 digits of that number are ...02061056. The hyper-moser is equal to super-super-super-...-super-super-super Moser, where each "super" iterates the number of sides in the polygon, and the number of copies of "super" is Moser's number. These unspeakably huge numbers provide a transition into part 3...
NEXT >> Part 3