Sbiis Saibian's

ULTIMATE LARGE NUMBER LIST

Total Entries: 1439

            Welcome to my Ultimate Large Numbers List (ULNL). If your here for the first time check out the introduction link which describes the purpose and formatting of this page.

>INTRODUCTION

ULTIMATE LARGE NUMBERS LIST

PART I - Hyper-E Notation

Numbers expressible with Hyper-E Notation, Knuth-arrows, Trientrical Arrays, or Steinhaus-Moser Polyogons, ...

0

zero

            "Zero" can be thought of as the smallest quantity possible. After all you can't have less than nothing, or can you? Sometimes "negatives" are thought of being "smaller" than zero, but this seems to defy logic since you can't have something smaller than something which is infinitely small to begin with! It is better to think of negatives as "less than zero" rather than "smaller than zero". 

            Negative numbers do not relate so much to the concept of quantity as to "position". An axis can have a "central point" at zero, with negatives to one side and positives to another. Quantity however can only travel in one direction from zero, namely, towards positive infinity.

            The exclusion of negatives from this list much easier to justify than the exclusion of zero (although I've decided to include them none the less). This is because the negatives really serve almost no purpose in the large numbers field. In order for algorithms to terminate it is necessary to have a minimum value for every argument. This means we have to choose a least number allowable as an argument. Common choices for the minimum value are 1 and 0. If however we allow any integer value, including negatives, we kind of drop the bottom out so to speak and the result is either a function which does not terminate for all values, some values, or requires at least 3 rules (a base case, a less-than-base case, and a more-than-base case). None of these options provides any advantages over simply deciding on a minimum integer value for the function. This is why zero has some importance in the large number field, because it serves as a beginning value. Some of the uses of zero in googology are as the minimum value of an argument in the Ackermann function, and the minimum order of a separator in array notation. It is also the minimum arity of an array. In cantor's system of ordinals, it is the smallest possible ordinal.

            Zero crops up when attempting to extend the hyper-operators to all integers. By definition a^^1=a. Since logaa^^n = a^^(n-1), for n>2, we can define this as a law for all integer values. Thus we obtain that a^^0 = logaa^^1 = logaa=1. Thus a^^0=1. Next let a^^(-1) = logaa^^0 = loga1 = 0. Thus we find that any positive integer>1, a, that a^^(-1)=0. 

            Zero is equal to the double logarithm of 10: loglog10=0. It is also equal to log1.

            Zero sometimes leads to undefines as in 1/0 and log0. For this reason I have occasionally been wary of zero. However we rarely have such problems in googology since the functions we work with are usually integral and not continuous.

            In some respects zero serves as a natural starting point for large numbers. We can think of it as the starting line, and any positive real becomes part of the race coarse. The end goal can be thought to be at infinity, although it is probably better to think of this as a race with no end goal! We can begin this race, but we can never finish it! Better get going then ...

1

one

            This is a number SO AMAZINGLY LARGE ... THAT it's not small! In fact, it's the smallest ... (Read More)

Next up, the Large Numbers ...

10^10^-Rayo(10^100)

rayominexiplex

We begin the Ultimate Large Number List with a number so close to 1 it requires a function that diagonalizes over first order set theory to define. The function is often called simply Rayo(n), and Rayo's Number, or simply "rayo" in this case, is defined as Rayo(10^100). Rayo(n) is the least positive integer not definable with n or less characters in the language of first order set theory. So Rayo is the least positive integer not definable with 10^100 (googol) symbols or less in first order set theory. Rayo's Number at one point held the distinction as the largest number number recognized in googology. This title was lost several times to other numbers, but these typically became dethroned once it was realized they were ill defined for one or another reason.  Because of this numbers lauded status it often finds itself at the extremes of these sorts of number lists, such as here. However in addition to Rayo's number, this number has an extra trick up it's sleeve that technically makes it the closest real number to 1 to ever show up on a list. Typically the approach here would be to start with 1 + 1/Rayo(10^100). However the number rayominexiplex is technically even closer to 1. To understand why we need to recall that 10^x = 1 + x*ln10 if x is very close to 0. Since 10^-Rayo(10^100) is very very close to 0, it follows that this number is appoximately 1 + (ln10)*10^-Rayo(10^100). The ln10 is just a small ordinary constant equal to about 2.302585, so we can mostly ignore that. Instead we note that we can rewrite this as 1 + ln(10)/10^Rayo(10^100). So we have 2.3 the reciprocal of 10 to the Rayo's Number (rayo-plexed). Even though I normally make a point to say that plexing a number so large does virtually nothing, here that difference matters and means that this number is necessarily much closer to 1. Firstly we can note that 10^x > x for all x. Now the ln10 might be of concern, if x was sufficiently small, however we can round it up to 10 and show this number is still closer to 1 than 1 + 1/Rayo. 10/10^Rayo(10^100) = 1/10^(Rayo(10^100)-1). So now what we want to show is when 10^(x-1) > x. Note that the slope of x is always 1. The slope of 10^(x-1) is 10^(x-1)*ln10, which is always positive and strictly increasing. We note that when x = 2, 10^(2-1) = 10, which means even for such a modest value the function is already greater. Next we note that the derivative at this point is 10^(2-1)*ln10 = 10*ln10 ~ 23.02585. The slope is already greater than 1 and always increasing. Therefore we have proven for x >= 2 , 10^(x-1) > x. This means 10^(Rayo-1) > Rayo. Which means 1/10^(Rayo-1) < Rayo. Since 1/10^(Rayo-1) = 10/10^Rayo > ln(10)/10^Rayo, we have proven our contention that rayominexiplex < 1 + 1/Rayo. This number would have 1.00000... with a Rayo's Number - 1 0s followed by 2302585.

1 + 1/Rayo(10^100)

One and a Rayo-minutia

This number is notable for being the smallest real number greater than 1 to appear on Cookie Fonster's well known Pointless Gigantic List of Numbers. This is because Rayo's Number has very commonly been held as the largest number in googology. Taking a large number and adding it's reciprocal to 1 is the first idea people come up with when trying to define numbers in this range. However there is actually a variety of tricks that can be employed as we will see.

1 + 1/{{L100,10}10,10&L,10}10,10

One and the reciprocal of a meameamealokkapoowa oompa

This is the second entry after 1 on Cookie Fonster's Pointless Gigantic List of Numbers. The meameamealokkapoowa oompa is easily one of Jonathan Bowers' most infamous googolisms, so its fitting to have it here. Although it would be computable, it's quite large for a computable number. The precise formulation of this number is not known, but estimates have been placed on its likely size given a reasonable interpretation of BEAF.

10^10^-E100{#,#,1,2}100

blasphemorgulminexiplex

This is an honest to goodness googolism formed by combining a "blasphemorgulus", a googolism created by me, and applying conway's minex suffix followed by the plex suffix. 

It's a googolism that is large...er than 1 ... barely :/

Perhaps it looks like this number should be huge! Well it IS larger than 1 at least, but of course you mean googologically large. To understand why its so small look at the function 10^10^x, and let x approach negative infinity. As x approaches negative infinity 10^x approaches 0, and therefore 10^10^x approaches 10^0 which equals 1. Stop it at any point however and we find the result has to be greater than 1, since 10^x is always greater than zero, it follows that 10^10^x is always greater than 10^0 or 1. This number begins 1.000000..... with a blasphemorgulus minus one zeroes and is then followed by 230258509299... matching up with ln10 perfectly, up to some point , and then following up with some mysterious digits we may never know.

This is only one example of the weird things that can happen from mixing up the -minex and -plex prefixes. For a full proof of the convergence of the digits as well as a further discussion of some of the interesting properties of these mixed typed googolisms (mixing macro and micro operators) click here.

1 + 1/(E100#^^#100)

one plus the reciprocal of a tethrathoth

                Let's begin our exploration of large numbers with a really really REALLY small example. We're in no rush and we have plenty of time to get to the real whoppers. Consider the number "one plus the reciprocal of tethrathoth". This number is inconceivably close to one, yet just ever so slightly larger. It starts out 1.000000... but it's an inconceivably long time until you get anything other than zero, though eventually it must reach ...000000001, on account of the fact that a tethrathoth is just a mind bogglingly large power of 10. You would have to raise this number to a tethrathoth to get a number just shy of e. Despite how small of a large number this is, keep in mind that there is an infinite number of smaller large numbers in the infinitesimal space between 1 and 1+1/(E100#^^#100).

(1+10^-10^100)^G(64)^-1

Graham's root of the sum of one and a googolminex

            Let's continue the discussion with a real whopper of a "large number"! This is a number so mind bogglingly large that by raising it to the power of a mere Graham's Number you get the ginormous result of a googolminex and one! What's a googolminex and one? It's 1 plus the reciprocal of a googolplex! More impressed by how close this would be to one than by how "large" it is? That's understandable. It's a stretch to call this large, except to say that it's definitely larger than 1. None the less, 1 to any finite power is still 1, so in comparison this number is quite amendable to exponential growth! And just think of the infinite number of yet smaller large numbers whose Graham's power doesn't even come close to the massive googolminex and one. Even if we raised the previous entry to a godgahlah it would still be way closer to 1 than "Graham's root of googolminex and one"!!!

            This was the smallest example I gave on my "Very Small Very Large Numbers" blog entry as an extremely small Large Number. The number must begin with 1.000000000000000000000............ but we can not compute what the first non-zero digit is after the decimal point, just as we can't compute the leading digit of Graham's Number. We also can't compute the exact number of zeroes before the first non-zero digit, though it must be about Graham's Number. All that being said it might be a bit tricky to remember that this number is virtually equal to 1!

            Now I know your brain is still reeling from the sheer massiveness of this number, but wait until you see what comes next!

10^10^-E100#100

grangolminexiplex

This entry combines my googolism grangol, the 100th member of the googol sequence, with the -minexiplex. The result is a number of the form 1.00000... with a (grangol-1) 0s before 2302585 etc. This is also the grangolplexth root of 10. That is grangolminexiplex^grangolplex = 10. This value is exact. See the next entry for more details.

10^10^-10^10^100

googolpleximinexiplex

As crazy as it sounds. This number takes a googolplex, creates an extremely small number called googolpleximinex which is the reciprocal of googolduplex, then, as if that wasn't enough, takes 10 and raises it to the miniscule power of a googolpleximinex. This results in a number almost but not quite equal to 1, being ever so slightly larger. How much slightly? Well it is approximately 1 + 2.302585*10^-10^10^100. What this means is that its 1.0000000000.... with a googolplex-1 0s and then followed by the digits 2302585 etc. This is also the googolduplexth root of 10. 

Proof. (10^10^-10^10^100)^10^10^10^100 = 10^(10^-10^10^100 * 10^10^10^100) = 10^10^(-10^10^100 + 10^10^100) = 10^10^0 = 10^1 = 10. QED.

10^10^-10^10^10^10^10^-10^100

googolminexiquadripleximinexiplex

This is one of many combination googolisms we can create by combining googol along with combinations of minex and plex. We can prove this number is exceptionally close to a trialogiaminexiplex. We note that googolminexiplex must be close to but greater than 1. It follows that googolminexiduplex is slightly larger than 10, googolminexitriplex is slightly larger than 10^10, and googolminexiquadriplex is slightly larger than 10^10^10. It follows that googolminexiquadripleximinexiplex is approximately 10^10^-10^10^10. Furthermore we know it is slightly closer to 1 than trialogiaminexiplex since googolminexiquadriplex is slightly larger than trialogue, meaning the negative exponent is slightly larger, causing this number to be slightly closer to 1. It can be shown that it is in fact close to trialogiaminexiplex than it is to 1. Specifically it will agree with trialogiaminexiplex for up to it's first approximately googol non-zero digits after 1. 

There is more ... (n)-minexiquadripleximinexiplex forms a function tailor made to continue to approach trialogiaminexiplex from below. The larger n is the closer it is. This leads to ...

10^10^-10^10^10^10^10^-E100{#,#,1,2}100

blasphemorgulminexiquadripleximinexiplex

As crazy as it sounds. This is likely one of the most insane googolisms I have ever bothered to explicitly coin. It came up in a discussion of mixing minexes and plexes I had with CookieFonster on Discord on September 2nd 2021. One notable thing about this particular number is that it can be said to arise naturally from messing with existing naming conventions in googology. As I pointed out to CF, when mixing minexes and plexes, if we can get arbitrarily close to 1 then we can get arbitrarily close to any member of the -logue series simply by continuing to add plexes. Since blasphemorgulminexiplex is googologically close to 1 it follows that:

blasphemorgulminexiduplex ~ 10

blasphemorgulminexitriplex ~ 10^10

blasphemorgulminexiquadriplex ~ 10^10^10

etc.

From there we can get arbitrarily close to certain numbers that are themselves extremely close to 1. For example the trialogiaminexiplex would be 10^10^-10^10^10. If we were to place a number extremely close to trialogue into -minexiplex we would necessarily get a number super close to trialogiaminexiplex. In other words:

blasphemorgulminexiquadripleximinexiplex 

= 10^10^-blasphemorgulminexiquadriplex

~ 10^10^-10^10^10

Furthermore, we know this number must be ever so sightly closer to 1 than trialogiaminexiplex (that is, it is slightly less), because we know that since blasphemorgulminexiplex > 1, it follows that blasphemorgulminexiduplex > 10, blasphemorgulminexitriplex > 10^10, and blasphemorgulminexiquadriplex > 10^10^10. Since it is slightly larger this means we are raising to a slightly less negative number, which means we have a number which is strictly closer to 1! 

Trying to imagine this number is kind of crazy. Imagine a number line. There is a number so close to 1 that it is barely indistinguishable from it. You need to zoom in a trialogue orders of magnitude ... before you notice 1 and trialogiaminexiplex differentiate themselves. At this point you see that trialogiaminexiplex is slightly to the right of 1 on the number line. But we aren't done ... next we need to zoom in approximately another blasphemorgulus orders of magnitude ... far far far far more zooming before ... and then finally we see the blasphemorgulminexiquadripleximinexiplex differentiate itself from trialogiaminexiplex. Blasphemorgulminexiquadripleximinexiplex turns out to be slightly to the left of trialogiaminexiplex. From the blasphemorgulminexiquadripleximinexiplexes point of view, the difference between 1 and trialogiaminexiplex is googologically vast ... even while from our point of view they are googologically indistinguisably close!!!

Another thing that might make this number seem as loopy as all-hell is how long it is. The blasphemorgulminexiquadripleximinexiplex contains exactly 40 letters! 

10^10^-10^10^10

trialogiaminexiplex

This curious number came up in a discussion I had with CookieFonster back in September of 2021. It's 1 point followed by a (trialogue-1) 0s and then 2302585. The significance of trialogue in this case is that power towers will converge to members of the -logue sequence, if all the bases are 10. To elaborate 10^x converges to 0 as x approaches negative infinity. 10^10^x approaches 1, 10^10^10^x approaches 10, 10^10^10^10^x approaches 10^10, and 10^x#5 approaches 10^10^10. Because of this, we can combine minex and plex in ways to actually arrive at numbers googologically close to trialogiaminexiplex even while trialogiaminexiplex is itself googologically close to 1. See previous two entries for examples of this phenomena!

1 + 10^-10^100

one and a googolminex

            This number is MASSIVE! It's is equal to 1 plus the reciprocal of a googolplex. It's a number so staggeringly gargantuan that if you raise it to the miniscule power of a googolplex you get a value just shy of the unfathomably large number e! Consider that up until now, raising the previous entries to a googolplex would not even come close to reaching 1.0000000000000000000000000000000000001 let alone a number larger than 2! Still not impressed?! Hmm, time to bring out the big guns ... 

10^10^-10^100

1.0000000000 ... ... ... ... ... ... ... ... ... ... 000000000230258509299...

w/(10^100-1) 0s

1 + 2.30258509299 * 10^-10^100

googolminexiplex

It was Milton Sirotta, nephew of Edward Kasner, who coined the googol for "1 followed by 100 zeroes", which may be denoted as 10^100. Kasner then defined the googolplex as "1 followed by a googol zeroes", which may be denoted as 10^10^100. These two numbers largely form the basis of early "googology". The sequence was continued with numbers such as googolduplex for 10^10^10^100, googoltriplex for 10^10^10^10^100, and so on. This makes sense if one's goal is to make larger and larger numbers. The idea clearly suggests that when appending a "googolism", (n), to the stem -plex, the result is always (n)-plex = 10^n. Conway, in his Book of Numbers decided to take this in another direction, devising (n)-minex to mean 10^-n. This allows us to describe extremely small numbers, in much the same way that plex allows us to describe extremely large numbers. To this end Conway coined googolminex as 10^-googol = 10^-10^100. Due to the laws of exponents this is also equal to 1/googolplex. What happens when we attempt to mix these suffixes together? It depends on the order. If we evaluated googolpleximinex applying suffixes from left-to-right we get 10^-10^10^100, which is also 1/googolduplex. This number is what we get if we apply them in the reverse order. googolminex is 10^-10^100, so it follows that googolminexiplex is 10^10^-10^100. Note this is not the same as evaluating 10^10^(-10)^100. Instead we evaluate exponents as well as signs from right-to-left. In otherword, this is 10^10^-googol. Since 10^-googol is an extremely small number we get 10^0.00000...00001 w/(googol-1) 0s after the decimal point. Since 10^0 = 1, this results in a number extremely close to 1. In fact it turns out that 10^10^-x is approximately equal to 1 + 2/x for sufficiently large x. What this means is that a googolminexiplex is approximately the square of one and a googolminex. That is (1+10^-10^100)^2.302 ~ 10^10^-10^100. In a strange way that makes both of these "extremely close" to each other, though not simply in the ordinary sense. Any two numbers in the Googological Superuniary Epoch (GSE) are necessarily arithmetically close. They must be a distance less than 0.01 from each other. But one number in this range may be raised to a googologically large number to get another number in this range. It is in this sense that the previous entry and this one are virtually in the same "ballpark" in terms of closeness to 1. They have the same number of trailing 0s, just with a different string of digits. This can not be said for most of the numbers in this range. Googolminexiplex is notable for combining 3 elements that are perhaps the most quintessential to early googology. Using (n)-minexiplex we can create many numbers in this range, but these become increasingly obscure. See googolpleximinexiplex for an even closer example.

2^2^-1,000,000

1.0000000000 ... ... ... ... ... ... ... ... ... ... 000000000700102597078...

w/301,030 0s

1 + 7.00102597078 * 10^-301,031

Millionth Square Root of two

Begin with 2 and apply the square root. You get approximately 1.4142. Now apply the square root to that and we obtain 1.189207115. Call this the second square root of 2. Take the square root a third time and we obtain 1.09050773267, a fourth time and we obtain 1.04427378243. Continue in this way until you have applied the square root a million times. We can show this is equal to 2^2^-1,000,000. Firstly we note that taking the square root is the same as raising to the half power. So the square root of 2 may also be denoted 2^(1/2). This may also be written as 2^2^-1. Now applying the square root again we obtain, (2^(1/2))^(1/2) = 2^(1/4) = 2^4^-1 = 2^(2^2)^-1 = 2^2^-2. In general if we have the kth square root of 2, then applying the square root again gives us (2^2^-k)^(1/2) = 2^((2^-k)/2) = 2^2^-(k+1). Thus the millionth square root can be written more simply as 2^2^-1,000,000. To compute this number we first convert the 2's into 10s. This gives 10^10^(loglog2-1,000,000*log2). This gives 10^10^-301,030.517054. Next we use the fact that 10^x ~ 1 + x * ln10 for x very close to 0. Thus we have 10^10^-301,030.517054 ~ 1 + ln10 * 10^-301,030.517054. Next to obtain the digits we separate out the decimal part of the fraction and obtain ln10 * 10^(-0.517054) = 0.700102597078. The placement then can be figured out with some simple manipulation of exponents.

One way to look at this is, this number is so small, that you have to square it a million times just to get 2. This is equivalent to raising this number to the 2^1,000,000 power just to get 2, which makes it unimaginably close to 1. We can demonstrate this as follows: (2^2^-1,000,000)^2^1,000,000 = 2^(2^-1,000,000 * 2^1,000,000) = 2^2^(-1,000,000 + 1,000,000) = 2^2^0 = 2^1 = 2. 

Interestingly if we then square 2 a million times, the number appears to "explode" in comparison. We get 2^2^1,000,000 which is about 10^10^301,029. This number is "slightly" larger than thrargoogolgong-plexed. See 2^2^1,000,000.

What if we raise 2^2^-1,000,000 to the power of 2^2^1,000,000. Do we get an extremely large number? Yes. We already know raising it to 2^2,000,000 will give us a large number. computing we have (2^2^-1,000,000)^2^2^1,000,000 = 2^(2^-1,000,000 * 2^2^1,000,000 ) = 2^2^(2^1,000,000 -1,000,000). 2^1,000,000 is so large that subtracting 1,000,000 on it has barely any effect, and so we would effectively get 2^2^2^1,000,000. Virtually indistinguishable from if the base had been 2. Much earlier numbers in this range are so close to 1 however that even raising to 2^2^1,000,000 would have no effect, even though this number is larger than a googolplex.

1.0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000001

1+10^-100

one and one googolminutia

            Alright how about this. Take 1 and add the reciprocal of a googol. This number is so B-I-G that you only have to square it a mere 332 times to get a value exceeding 2! Still not big enough?! Don't worry, we've just gotten started ...

1.0000000000000000000000000000000000000000000000000000000000000895510046885

1 + 8.95510046885 x 10^-62

Increase In Earth Mass By Single Solar Neutrino Collision

It is said that about 3x10^15 solar neutrinos pass through the Earth for every square meter of the Earth's surface every second. Although Neutrinos were originally believed to be massless, the detection of neutrino flavor fluctuations forced scientists to conclude that Neutrinos must have some small non-zero mass. Because of this, when a single neutrino passes through the Earth, it means the earth's total mass is in fact increased by some miniscule amount. The mass of the Earth is approximately 5.972x10^24 kg. The neutrino on the other hand, is extremely light even amongst sub-atomic particles. For point of comparison, the proton has a mass of 1.627x10^-27 kg. The neutrino on the other hand has a mass of approximately 5.347x10^-37 kg, that is it is approximately 3,042,827,753 times lighter than a proton! To find the ratio of the mass of the earth after the neutrino collision versus before, we simply compute (5.972x10^24 + 5.347x10^-37)/(5.972x10^24), which works out to about 1 + 8.955x10^-62. This number is incredibly close to 1. To get 2 we would have to raise this number to the power of approximately 7.74x10^60. In terms of squares, we would only need to square this number 203 times to obtain a number exceeding 2. Note that, due to the nature of floating point storage, these numbers can not be computed directly on a calculator since it will round of 1+8.9x10^-62 to exactly 1. To learn how these values were obtained click here.

1.000000000000000010903970549325460813650942266345982807...

Time dilation factor of person who is walking versus standing still

            This is the time dilation factor in general relativity of a person who is walking verses a person who is standing still. I assume that a walking person travels at about 1.4 m/s which is about 1/214,137,470 the speed of light. One way to think about this is as follows:

            Imagine two nigh immortals who can live for billions and billions of years. One decides to go for a walk for the next billion years while the other decides to stand still for just as long. At the end of the billion years when the walker again stops to meet up with the one who stood the whole time the difference in the amount of time that elapsed for them due to time dilation would only differ by about 1/3 of a second! The ratio of the larger elapsed time to the smaller elapsed time will be the value of this entry.

            This number is so large you only need to square it 56 times to get a value exceeding 2.

1.0000000000000002220446049250313080847263336181640625

1+1/2^52

Smallest Possible Large Number Using Double-Precision Floating Points

This value is exact. It can be approximated as 1+2.222044604925x10^-16. This is the smallest number we can express greater than 1 using double-precision floating points. Anything closer to 1 (from above) than this, will simply be rounded to 1. For example if we add 1+1/2^53, it will be rounded to 1. We can see this difference by computing 1+1/2^52-1 versus 1+1/2^53-1. In the first case the sum of 1+1/2^52 can be stored and so we have 1+1/2^52-1 = 2^-52. On the other hand 1+1/2^53 will be rounded to 1, thus 1+1/2^53-1 would return 0 instead! There is also a largest small number that can be stored as a double equal to 1-1/2^52. So all real numbers in (1-2^-52,1+2^-52) will simply be rounded to 1. This makes these numbers, for the most part, inexpressible with most computers and calculators. In fact, while this number can be stored as a "double", it can not be stored on my TI-89. It will instead round this value to 1. 

1.0000000000001

1+10^-13

Smallest Possible Large Number Expressible on the TI-89 in Approximate Mode

Smallest Possible Large Number Expressible on the TI- 89 in approximate mode. This can be detected as follows. Compute 1+10^-13-1. The difference shows up as 10^-13. If we instead have 1+10^-14-1 we get 0. If we have 1+5*10^-14-1 we get 10^-13. This reason for this is that 1+5*10^-14 is getting rounded up to 1+10^-13. Anything smaller than 5*10^-14 leads to a difference of 0.

1.0000000001

1+10^-10

This number has somewhat arbitrarily been chosen as the boundary between googologically superuniary numbers, and ordinary superuniar

1.000000000230258509299...

1+ln(10)/10^10

At this point we can't compute enough digits of 10^10^-10 (monologiaminexiplex) to find out how many digits of convergence there are. None the less we still know this is a lower bound even if we don't know how the decimals differ.

1.0000000002303...

10^10^-10

monologiaminexiplex / dekaminexiplex

The monologiaminexiplex is part of special set, of recursively generated numbers constructed using the roots -logue, -minex, and -plex. It is given a special designation, [0,1]. This means it is part of a sequence [0,1], [0,2], [0,3],...etc. that converges towards [0] which is 1, from above. Each member of this sequence can itself be converged to by a sequence from above as well as below. Notice that this number begins to test the limits of what the TI-89 can store directly in approximate mode. It actually stores it as 1.0000000002303 internally. At this precision, the calculator is unable to distinguish between 10^10^-10 and 1+ln(10)/10^10, even though they are not equal. If one computes 1+ln(10)/10^10-(10^10^-10) the TI-89 returns 0. So at this point difficulties begin to emerge regarding calculating these numbers directly. It's here where simply approximating 10^10^-n as 1+ln(10)/10^n becomes useful.

 1.00000069315...

2^1,000,000^-1

millionth root of two

            Let's pick up the pace and talk about a real super giant! This is a number so large that if you raise it to the millionth power you get 2! You only need to square this number 20 times to get a value exceeding 2. That's still not large, you say? But just think of all the smaller numbers still greater than 1, whose millionth power doesn't even reach 2. In fact, the millionth power of every previous entry is not even as large as this entry!

1.0000230258509299...

1+ln(10)/10^5

An approximation for 10^10^-5.  Despite the usual pattern there is only 4 digits of convergence (2302).

1.00002302612...

10^10^-5

fiveminexiplex / pentaminexiplex

This is 5 orders of hyper magnitude below 10. pentaminexiplex to the power of 100,000 equals exactly 10. This one is also notable for being the first case where 1+ln(10)/10^n does not have n digits of convergence. The reason for this has to do with the 5th place after the decimal 2.30258, being 8. This make it easier for it to overflow into the next place value. Hence we find our first exception to the general rule. Regardless we can still say the digits of convergence increase linearly with n. This becomes important when dealing with googologically close numbers to 1 of the form 10^10^-x.

1.000230258509299...

1+ln(10)/10^4

An approximation for 10^10^-4. It has 4 digits of convergence (2302).

1.00023028502...

10^10^-4

fourminexiplex / tetraminexiplex

4 hyper-orders of magnitude below 10. What that means is this is the result of taking the 10th root 4 times starting with 10. Or put another way this is the 10,000th root of 10. It follows that you need to raise this number to the 10,000th power to get 10.  You need to raise number to approximately 3010.3 to get 2, and  4342.9 to get e. Even though this number is pretty darn close to 1, enough so that it would really be imperceptible to us humans already, it still can be stored on a calculator quite easily. Soon we will be reaching values so small they can not be stored as anything other than 1.

1.00024414063...

1+1/2^12

This is 1.000000000001x2^0 using the binary format used for double-precision floating point numbers. The mantissa is stored as a 52-bit binary-part. This excludes the 1 before the binary-point because, it never changes value. 11 bits are devoted to the exponent, and 1 bit to the sign. Thus there are a total of 64 bits or 8 bytes in a double-precision floating point number. This number is just slightly above -4 hyper orders of magnitude. This format however allows for the expression of much smaller superuniary numbers.

1.001

This is the ratio of a one permille (1‰) increase. That is a 0.1% increase in something. Most people are familiar with percents. The word percent breaks down to "cent" which you should recognize from "century", meaning 100. That is a percent, is a part per hundred. A permille, comes from the same root used in "millenium", meaning 1000. Thus a permille is a part per thousand. This number has a hypermagnitude of approximately -3.362. What this means in practical terms is we would need to raise this number to the power of 10^3.362 = 2303.7 to get 10. In other words, if you had a bank account with an APY of 0.1% it would take 2,303 years for it to multiply 10 fold. Good luck living long enough to reap the benefit of that long term investment. We can also compute what power we need to reach 2. This works out to approximately 693.5. A good rule of thumb for approximating how long it takes a very small rate like this is to first perform the following calculation: 0.7/(r-1). This yields 0.7/(1.001-1) = 0.7/0.001 = 700. This works because for numbers extremely close to 1, it turns out that (1+x)^y ~ 1+xy. For example if we take 1.001^7 we get 1.00702103504 which is pretty close to 1.007. This approximation gets better the smaller the product of xy is, and assuming that x itself is already relatively small. It is also known that 1.07177^10 = 2. From these two facts we can estimate the doubling time simply by figuring out a y such that we get approximately 1.07. So we compute (0.07/(r-1))*10 = 0.7/(r-1). The 700 we get is pretty close to the actual answer of 693.5.

1.001 also shows up on Munafo's site, not as an entry in his number list, but in an example regarding power tower paradoxes. Munafo asks which is larger: 1.001^2^2^10 or 1,000,000^2^10? The answer, perhaps surprisingly, is 1.001^2^2^10, which he proves with some exponential and logarithmic math. This result may seem less surprising if we bare in mind that 1.001^13,822.4 ~ 1,000,000. When we get into the tetrational range this conversion factor or 13,822.4 becomes negligible, and thus it matters less than the height of the power tower.

1.00112989063...

1.5/(2^(7/12))

The ratio of a perfect fifth to 7 equal temperament half-steps. This is an example of a small large number, or superuniary, that comes up in a normal application. This number is therefore in some sense "useful" and "ordinary". We have to go a bit closer to 1 before we go beyond any meaningful physical application. The hyper-exponent of this number would be -3.309 (this is found by taking the common logarithm twice). This is a measure we can use to describe how close we are to 1. If you are going up this list instead of down, you will eventually find that this hyper-exponent will reach googological size before long. If you are going down the list then -3.3 is how far we have left before we reach 10 and start with what we would more normally consider "large numbers". We are making progress.

1.0013784192...

Ratio of Neutron to Proton mass

            How's THAT for a large number! Although both Protons and Neutrons are have more or less the same mass, the Neutron is just slightly heavier than a Proton by a factor of 1.0013784192...etc. To put this in perspective you only need to square this number 9 times to exceed 2.

1.00230258509299...

1+ln(10)/10^3

A lower bound on 10^10^-3. The difference in value between these two is only about 0.000002652985.

1.00230523808...

10^10^-3

threeminexiplex / triaminexiplex

We are now a mere 3 hyper-orders of magnitude below 10. We can call this number threeminexiplex or triaminexiplex (not to be confused with the root -triminexiplex). As usual the number of convergent digits with ln(10) is equal to n.

1.0026654123...

Ratio of troposphere to diameter of the earth 

            This is the ratio of the radius of the "troposphere" (the lowest part of the earths atmosphere) to the radius of the earth. This means that the earth's radius is only increased by a mere 0.2% by the enveloping troposphere! Our "sky" is little more than a thin film on the earths surface! But that's still bigger than the ratio of the Neutron mass to the Proton mass, so that's got to count for something right?! Also you only need to raise this number to the 261st power to get a number exceeding 2.

1.00336413942...

6378.137/6356.752

Oblateness of the Earth

You may have heard is said that the earth is not a perfect sphere, but an oblate spheroid. What exactly does that mean? It means the earth is shaped kind of like a sphere that has been slightly flattened at the poles, emphasis on slightly. Due to the earth's rotation, the centrifugal force causes the earth to bulge slightly around the equator. The upshot of this is that the equatorial radius is slightly greater than the polar radius. The ratio of the equatorial radius to the polar radius can be used as a natural measure of the oblateness of the spheroid. The equatorial radius is 6,378.137 km (3963.191 mi), while the polar radius is 6,356.752 km (3949.903 mi). That's a difference of only 21.385 km (13.288 mi). This means that the equatorial radius is only about 0.33% greater than the polar radius. Taking the polar radius to be 1, you can graph the shape of the earth as x^2/1.0033^2+y^2=1. The bulge is so slight that it looks basically like a perfect circle. Thus the earth is pretty well approximated as a sphere, at least as far as human perception goes. An oblateness as little as 1.05 is in fact pretty easy to notice. Getting much smaller than this however becomes largely indistinguishable. This suggests that our ability to distinguish superuniary numbers from 1 probably stops somewhere around this range.

1.01

One point oh one

            This is the numeric value of a 1% increase. An increase of 1% is considered to be a very small improvement. However, let's consider how large it is. Let's say you had a bank account with a 1% annual interest rate. How many years would you have to wait for you investment to double? It would only take a mere 70 years! Not impressed? Well 70 years might make for a really long term investment, but in other contexts 70 years might be cause for concern. An inflation rate of 1% would mean the price of all commodities would double every 70 years, quadruple every 140 years, etc. If the price of gasoline were to increase exponentially at a 1% annual rate, in only 40 years you'd be spending 50% more on gas. This might seem like a slow rate, but it only takes a few extra percents annually to get a dramatic change in growth rate. See 1.07 as an example.

1.01364326477...

3^12/2^19

This number is notable for being the smallest real number greater than 1 on Robert Munafo's "Notable Properties of Specific Numbers". In other words, this is the smallest large number that is an entry on his number list. In other other words, the prior entry to this one on Munafo's list is 1. 1 probably feels like ages ago if you have been reading through every entry in order since then. It will also be a while until we reach Munafo's second smallest large number, which also appears on this list. The significance of this number, outside of being on Munafo's list, is that it's a ratio that comes from music. A just-tuned perfect 5th is exactly the ratio 3/2. In our western tonal system this is approximated by 7 equal temperament half-steps, where there are 12 half-steps to an octave. The idea is, if we cycle through 12 perfect fifths we should increase by exactly 7*12=84 half-steps which would be 84/12=7 octaves. An octave is the exact ratio 2/1. This works only if we assume 7 half-steps is 2^(7/12) = 1.49830707688...etc. In this case this works out perfectly. As you can see, equal temperament actually leads to 7half-steps being ever so slightly flat from a perfect-fifth. However for the average person, the difference is barely distinguishable. If however we use just intonation the exact result of a "circle of fifths" (12 fifths) is (3/2)^12 = 3^12/2^12. Because 3/2 is slightly sharper than 2^(7/12) however raising to the 12th power must give us a value slightly greater than 7 octaves. 2^7 = 128, while 3^12/2^12 = 129.746337891. This 1.013 is the ratio between the slight sharpness of 12 perfect fifths against 7 octaves. This would be exactly (3^12/2^12)/2^7 = 3^12/2^19, and can be worked out to the above value. See 2^(7/12).

1.02

1+1/50

A weak lower bound on 10^10^-2. In general 10^10^-n can be approximated as 1+2/10^n. In fact we know it will always be greater than 1+ln(10)/10^n. ln(10) = 2.302585...etc. thus 1+2/10^n will always be a little less than 10^10^-n.

1.0230258509299...

1+ln(10)/100

A handy approximation for 10^10^-2. This acts as a lower bound. In fact we have 1+ln(10)/10^x < 10^10^-x, holds for all real x. As x grows however, this becomes an increasingly good approximation.

1.02329299288...

10^10^-2

twominexiplex / diaminexiplex

twominexiplex (or diaminexiplex not to be confused with the combo suffix -duminexiplex) is a googolism that can be used to represent 2 hyper orders of magnitude below the "hyper-norm" of 10. It can be approximated fairly accurately with the slightly less value 1+ln(10)/100. To see why we brush up on a little elementary calculus. We use the linearization of 10^x at x=0 to create the approximation. The derivative of 10^x is 10^x*ln(10). Setting x=0, this gives a slope of ln(10) at 0. Next we have the point of tangency (0,1). Putting this together gives us the linearization L(x) = 1+ln(10)*x, for x close to 0. It should be clear, that since 10^x is a strictly increasing function, this linearization will always provide a lower bound on the actual value. Taking all this into account since x=10^-2 in this case we have L(10^-2) = 1+ln(10)*10^-2 = 1+ln(10)/100. Note that 1+ln(10)/100 = 1.02302585...etc. and here we have 2 digits of convergence (23), before the two numbers diverge. For 10^10^-n, we should see that the number of convergent digits is always equal to n. To prove this we might consider the slope at x=0.01. This would be 10^(0.01)*ln(10). We create a line with this slope that also passes through (0,1) in order to create an upper bound. This gives 1+10^(0.01)*ln(10)*0.01. Next we subtract 1 from both the upper and lower bound and take the ratio of the results:

(10^(0.01)*ln(10)*0.01)/(ln(10)*0.01) = 10^0.01

The ratio is always the hyper-order of magnitude. This means, the number of convergent digits will closely match the number of 0s after the decimal point before the first non-zero digit. 

1.0235621919...

1+(10^10^-2)*ln(10)/100

This number comes from a generalized upper bound for 10^10^-n. It turns out this will be less than:

1+(10^10^-n)*ln(10)/10^n

The argument for this can be developed from calculus. From here we can create successive upper bounds that allow us to provide an absolute upper bound on the amount of error that can be present in the lower bound 1+ln(10)/10^n. For example since we know 10^10^-2 < 1+(10^10^-2)*ln(10)/100 it follows that:

 1+(10^10^-2)*ln(10)/100 < 1+(1+(10^10^-2)*ln(10)/100ln)*ln(10)/100

This can continue to be iterated indefinitely leading to the series 1+ln(10)/100+ln(10)^2/100^2+ln(10)^3/100^3+...etc.

1.02356853654...

1+ln(10)/100+ln(10)^2/100^2+ln(10)^3/100^3+ln(10)^4/100^4+...

This series can be used as an upper bound on 10^10^-2. We note that the difference between this number and the lower bound 1+ln(10)/100 is ln(10)^2/100^2+ln(10)^3/100^3+ln(10)^4/100^4+...etc. Thus we know the difference between 1+ln(10)/100 and 10^10^-2 can be no greater than this series. This implies that the digits of convergence should roughly match the place value of the first non-zero digit after the decimal. Why? Because the first term is ln(10)^2/100^2 which is as small relative to ln(10)/100 as it is to 1. Thus the change in digits must happen at approximately twice the number of digits. Note however that the additional terms would add further effects. These become exponentially less significant. However we can use infinite series to convert the difference into a single term. Using the infinite geometric series formula a/(1-r). Here "a"=ln(10)^2/100^2 and "r"=ln(10)/100. This yields:

(ln(10)^2/100^2)/(1-ln(10)/100) = ln(10)^2/(100^2 - 100*ln(10))

Calculating this directly yields 0.000542685609...etc. What this means is the difference between 1+ln(10)/100 and 10^10^-2 can be no greater than about 0.00054. Since this change is in the 4th place and the non-zero digits begin in the 2nd place, this means we should get 1 or 2 digits of convergence. The actual difference is a bit smaller at 0.000267141351...etc. In any case this provides mathematical justification for the idea that the number of converging digits will usually be equal to n. We have seen this is actually the case for n=1,2.

1.05

1+1/20

one and one twentieth

1.05 is the value of 1 plus 1 twentieth. As an interest rate, this would represent 5% APY. It would take 14.2 years for an account at this rate to double. This is frankly a long time, and this interest rate is unusually high compare to those typically given by banks to ordinary savings accounts! So ... according to the banks ... this is indeed a "large number" ... :p

1.05256461054...

1.05^1.05

When we raise a superuniary number to itself, we necessarily get a larger superuniary number, because we are raising to a power greater than 1 ... but it can still be very very close. 1.05 = 1+1/20. Raising it to its own power is just shy of 1+1/19 coincidentally. 

1.052631578947...

1+1/19

one and one nineteenth

The next two numbers of the form 1+1/n begin 1.05. None the less we are slowly gaining steam at this point.

1.05269632384...

1.05^1.05^1.05

1.05 tetrated to the 3rd. This number is still only slightly greater than 1.05, but it's at least greater than 1+1/19. As we continue to stack 1.05s they will technically continue to grow ... but at an ever slower rate ... converging to 1.05270345509...etc. We quickly converge and get extremely close to this value ...

1.05270308882...

1.05^1.05^1.05^1.05

With a mere 1.05 tetrated to the 4th, we already have convergence up to 1.052703. 

1.05270345504...

1.05^1.05^1.05^1.05^1.05^1.05^1.05

This is 1.05^^7. It's the last number of the form 1.05^^n that my TI-89 can distinguish from an infinite power tower of 1.05s. 

1.05270345509...

1.05^1.05^1.05^1.05^1.05^... ... ... ... ...

This is the convergence value of an infinite power tower of 1.05s. Unlike exponentiation, where any number greater than 1, raised to a sufficiently large power, can grow to any number, we find very different behavior for tetration. Here 1.05^^n, no matter how large n is, will always be less than 1.05270345509...etc. This seems quite remarkable. It means for example you can have a power tower like: 1.05^1.05^1.05^1.05^1.05^1.05^1.05^1.05^1.05^1.05, and it still doesn't even get out of the superuniary range. Normally if we had a power tower of 10 terms it would be huge. What's even more confusing is we can have 1.05 in the base and still get a very large number, such as a mere 1.05^10^10. Example power towers like these can be seen in the later epochs.

1.0555555555555...

1+1/18

one and one eighteenth

The repeated 5s imply this is also equal to 1+(5/9)(1/10). Simplifying we have:

 1+(5/9)(1/10) = 1+5/90 = 1+1/18

1.05882352941...

1+1/17

one and one seventeenth

We are moving quite slowly it seems, but things are picking up pace as well. We only need to raise this number to the power of approximately 12.126 to reach 2. This place us approximately within 1 hyper-magnitude of 2. To find the hyper-magnitude of a number take the logarithm twice. 2 = 10^10^-0.521390227654. 1+1/17 = 10^10^-1.60513552019. As you can see the hyper-exponent has a difference of approximately 1.

1.05946409436...

2^(1/12)

twelfth root of two

            The average pitch ratio between successive half-steps in western 12-tone music. This number can be computed as the 12th root of 2: 2^(1/12). This is a number large enough that pitches in this ratio can be easily distinguished. One way of looking at this numbers size is that you only have to raise it to the 12th power to get 2. You only need to square it 4 times to get a value exceeding 2.

1.0625

1+1/16

one and one sixteenth

1+1/16. This number can be represented exactly in common floating point number formats, as is the case for numbers of the form 1+1/2^n up to a certain limit. Eventually you get a number "too close to one" to actually express. 

1.0666666666666...

1+1/15

one and one fifteenth

You may notice the repeating 6s look like 2/3rds. This is not a coincidence. One point zero six recurring would be equal to 1+(2/3)(1/10). Simplifying we have 1+(2/3)(1/10) = 1+2/30 = 1+1/15.

1.06937505609...

71,492/66,854

Oblateness of Jupiter

This is the oblateness factor of Jupiter. As mentioned on the entry for 1.0033, an oblateness of around 1.05 is already visible. Thus although the bulge is still small, for jupiter it is noticable with an equatorial radius about 7% greater than it's polar radius. Jupiter's equatorial radius is 71,492 km, while it's polar radius is 66,854 km. As impressive as it is, Saturn has an even higher oblateness factor, that is unmistakable when seen.

1.07142857142...

1+1/14

one and a fourteenth

A 1/14th is pretty small, at least by ordinary standards, so 1 and 1/14th is a pretty tame "large number". Even if we raise this number to the 10th power it still returns a "superuniary number", that is to say, it's still less than 2. That is about to change with our next entry ...

1.07177346254...

2^(1/10)

tenth root of two

            This is the numeric value of a 7% increase. This sounds small as in "the population is increasing by 7% annually". However this value is notoriously deceptive. At a 7% rate it would only take 10 years for the initial value to double! After 20 years the value will be 4 times as large as originally, and after 30 years 8 times as large! All of this from a mere 7% annually.

1.07692307692...

14/13 = 1+1/13

1 and 1/13th. The first number of the form (n+1)/n with n a positive integer, greater than 2^(1/10). This means this simple ratio is large enough that if we raise it to the 10th power we will get a number greater than 2.

1.08333333333333...

13/12 = 1+1/12

one and a twelfth

A simple ratio of 13 to 12, equal to 1 and 1/12th. 

1.090909090909...

12/11 = 1+1/11

one and one eleventh

A simple improper fraction, 12/11. This number is pretty close to 1 by ordinary standards, but from all our previous entries this now feels like we've come a long long way.

1.09407190229...

ratio of radius of exosphere to radius of earth 

            This is the ratio of the radius of the exosphere to the radius of the earth. The exosphere is the very last layer of our planet's atmosphere. Beyond this one enters into true outer space. The exosphere only increases the radius of the planet by a mere 9.4%. That's sizable enough that it would be visible if it were highlighted. However the end of the exosphere can't really be seen clearly since the exosphere is very very thin in comparison to the lower and denser layers of the atmosphere.

1.1

11/10 = 1+1/10

one point one

            This is the numeric value of a 10% increase. This is a change significant enough that it is noticeable almost regardless of what is considered. Getting 10% extra for the same price is touted as a bargain. Still too small? We're getting there. Before long we'll be blasting off to the stars. Enjoy the smallish numbers while they last!

1.10860128026...

60,268/54,364

Oblateness of Saturn

The oblateness of Saturn exceeds 10%. 

1.111111111111...

10/9 = 1+1/9

one and a ninth

A simple ratio of 10 to 9, also 1 and 1/9th.

1.11178201104...

convergence value of iterated exponentiation of one point one

            This value is the limit of the infinite power tower of base 1.1. To approximate it begin with 1.1, and let this be Stage 1. Next compute 1.1^1.1 and let this be Stage 2. Next compute 1.1^1.1^1.1 and let this be Stage 3. For each successive stage just take 1.1 and raise it to the power of the previous result. The limit of this infinite sequence is 1.11178201104...

1.125

9/8 = 1+1/8

one and an eighth

A simple ratio of 9 to 8. Also 1 and 1/8th. Because 8 = 2*2*2, it is compatible with base 10, and thus we get a terminating decimal in this case. Unlike the previous and next members of the sequence (n+1)/n.

1.142857142857...

8/7 = 1+1/7

one and a seventh

A ratio of 8 to 7. Also 1 and 1/7th. This number will repeat the sequence of digits 142857 indefinitely after the decimal point. A simple example of an ordinary superuniary number. 

1.1666666666666...

7/6 = 1+1/6

one and a sixth

A simple ratio of 7 to 6. A simple example of an ordinary superuniary number.

1.2

6/5 = 1+1/5

one and a fifth

A simple ratio of 6 to 5. It is also one of the simple terminating decimal fractions. This may also be used as a very simple approximation of 10^10^-1, which is actually a little larger.

1.21

one point two one

            This is the result of 1.1^2. This is very easy to compute by hand as 1.1+.11 = 1.21. The digits match up with the second row of pascals triangle. They are also the coeffients of x^2+2x+1 which is the product of (x+1)(x+1).

1.2302585093...

1+ln(10)/10

This is an approximation for 10^10^-1. It turns out this value is actually smaller than the actual value of this expression. Both agree in the first non-zero digit after the decimal point. 

1.25

5/4 = 1+1/4

one and a quarter

A very ordinary number. One and a quarter is a simple ratio of 5 to 4. As such it is likely to occur in many situations. For example, dividing 5 cookies between 4 people. Each would get one and a quarter cookies in this case. In addition to this a "quarter" or 25 "cents" is a common coin in US currency. Thus 1.25 is likely to come up when one needs a "dollar and quarter" for something. Quarters are common in measurements, such as liquid and volume measures, because it is compatible with our base 10 system. Any number which contains only 2s and 5s in it's prime factorization, will form a unit fraction with a terminating decimal. As such 1/4 = 0.25 and 5/4 = 1.25 are commonly known fractions.

Incidentally this number is also very close to the tenth root of ten. See next entry.

1.25892541179...

10^10^-1

oneminexiplex / monominexiplex

This number has some significance as a kind of hyper-order-of-magnitude. Orders of magnitude can be expressed as 10^x, where as hyper-orders of magnitude can be expressed as 10^10^x. If we begin with the real numbers, x, then we can say that x < 0 corresponds to the negative numbers, x > 0 corresponds to the positive numbers, and x = 0 corresponds to the midpoint of the number line. With 10^x, and orders of magnitude, x < 0 corresponds to "small numbers", x > 0 corresponds to "large numbers", and x = 0, corresponds to the number 1, the only number which is neither larger nor small. The only "average" sized number. The "norm" by which all other numbers may be measured. We now extrapolate this to 10^10^x. Here, allowing x to range over the interval (-∞,∞), gives us numbers in the interval (1,∞), the so called "interval of the large numbers".  x = 0, therefore becomes a new kind of divide, not between negative and positive, nor small and large, but a sub-division of the large numbers into two sub-classes. In my article "Very Small Very Large Numbers", I posit the view that the large numbers should considered any number that is strictly greater than 1. This naturally leads to the idea that not only is there no "largest large number" but also no "smallest large number". It also suggests the idea of "very small large numbers" when we get arbitrarily close to 1. At the end of the article I posit a further delineation of the reader for when a "very small large number" is no longer "very small" that is to say, no longer very close to 1. With hyper-orders of magnitude we can create a mathematically justified dividing point. At x = 0, or 10^10^0 = 10^1 = 10, we can say this is neither a "small large number" nor a "large large number". Thus hyper-orders of magnitude that are negative are "small large numbers" and hyper-orders of magnitude that are positive are "large large numbers". The hyper-magnitudes have a nice consistent property, just like with normal orders of magnitude. With normal orders of magnitude, multiplying by 10, increases the order of magnitude by 1, and dividing by 10 decreases the order of magnitude by 1. Here, raising to the 10th power increases the hyper-order by 1, and taking the 10th root decreases the hyper-order by 1. Thus this number represents exactly one order of hyper-magnitude below the midpoint of large numbers, 10. It's the point opposite 10^10^1 on the hyper-magnitude scale. This could be used as the definition for when a "very small large number", is no longer "very small". 

One other reason to pay mind to this number is it illustrates how the general negative hyper-orders work. 10^10^-n is always close to 1+2*10^-n. We can approximate 10^10^-x as 1+ln(10)x10^-x. Thus 10^10^-1 ~ 1 + ln(10)*(10^-1) = 1.2302585093...etc. Note that the actual value is actually larger than this. In any case, the rule of thumb here is if we want to know the size of (n)-minexiplex, it is roughly 1 point followed by (n-1) zeroes and then 2 followed by some other digits.  

1.331

one point three three one

            This is the result of 1.1^3. This is also very easy to compute. Since 1.1^2 = 1.21, 1.1^3 = 1.21+0.121=1.331. Again the digits match up with the third row of pascals triangle and the coefficients of x^3+3x^2+3x+1 which is the result of (x+1)^3.

1.333333333333...

4/3 = 1+1/3

one and a third

A simple ratio of 4 to 3. Among the simplest possible fractions, and a simple an ordinary example of a superuniary number (a number greater than 1 but less than 2). It is also an example of an infinitely repeating decimal. We can form a sequence of simple fractions of the form (n+1)/n to continue to get superuniary numbers. This is equivalent to adding the reciprocal of a positive integer, n, to 1, to get 1 + 1/n. See 3/2 for the simplest superuniary fraction.  

1.4142135623...

sqrt(2)

square root of two

            1.41421... better known as the square root of two. It's a number "so large" that its square is equal to 2, the first integer after 1. Still feeling underwhelmed?! Tough crowd! Well, we still got a loooooooong way to go. The square root of two is notable for being the first number proven to be irrational. That is, it can be shown that it is not a ratio of integers. It lies between 1 and 2, since 1^2=1 and 2^2=4. It represents the logarithmic half-point between 1 and 2. It is also a definite tipping point on this list as entries begin to accelerate rapidly after this...

1.44466786101...

e^e^-1

e to the e to the negative one

            This is the number e^e^-1. It has a number of interesting properties. It is a solution of x^e = e, which seems impossible. But notice: (e^e^-1)^e = (e^e^0) = e^1 = e. This is the largest real, greater than 1 for which an infinite power tower converges. Interestingly, an infinite power tower of this number has the value of e. Another important property, for the study of large numbers, is that b^x > x for all real x, provided b > e^e^-1. Thus 2^3 > 3, but 1.1^3 < 3 because 1.1^3 = 1.331. Another way of thinking about it is that b^^n will grow without bound provided b > e^e^-1. You can think of it as a number, "so large" that a power tower with a base of anything higher will grow without bound as the number of terms increase.

1.4641

one point four six four one

            This is equal to 1.1^4. Since 1.1^3 = 1.331, it follows that 1.1^4 = 1.331+0.1331 = 1.4641. The numbers 1,4,6,4, and 1 are the numbers on the 4th row of pascals triangle, and these are also the coefficients of x^4+4x^3+6x^2+4x+1, the result of (x+1)^4.

1.49830707688...

2^(7/12)

This is the ratio of an equal temperament fifth, to the tonic. This number comes up in western music. Simple ratios such as 3:2 or 4:3 produce pleasant harmony between pairs of notes. It therefore makes sense to develop a musical system where by some interval is exactly 3/2. The problem is if one tries to tune a piano in this way, transposing will distort the ratios for other keys. The solution arrived at in western music was to tune all the half-step intervals equally. Thus each one has an exact ratio of 2^(1/12). It turns out that 7 of these half-steps is sufficiently close to 1.5 that the difference can largely be ignored. None the less, there is indeed a slight difference in value. The ratio of the larger to smaller produces the relatively small superuniary value of 1.00112989063...etc.

1.5

3/2 = 1+1/2 = 1/1 + 1/2

one and a half / second harmonic number 

            Better known as 3/2. This number is the sum of the first two terms of the harmonic series. This makes it the 2nd Harmonic number. That is, it is equal to 1/1+1/2. It's is also the number exactly half way between 1 and 2. This halfway point is larger than the square root of two, which is the logarithmic half point.

            1.5 might seem small but consider this: if you saw someone 50% taller than you, you'd think they were tall regardless of their actual size.

1.6180339887...

[1+sqrt(5)]/2

golden ratio 

            This number is known as the golden ratio. One way to explain it is as follows: cut a line segment such that the ratio of the larger part to the smaller is the same as the larger to the whole. It turns out that there is a solution to this problem, and that solution is the golden ratio. If the ratio between the larger and smaller part is the golden ratio, then so will be the ratio between the larger and the whole. The golden ratio can be expressed exactly as [1+sqrt(5)]/2. That is, half the sum of the square root of 5 and 1. The golden ratio has a rather unusual property. It's square is exactly one more than itself. The golden ratio is a number "so large" that the distance between itself and its square is 1!

1.83333333333...

1/1+1/2+1/3

third harmonic number

            This is 11/6 or 1+5/6. It is the sum of the first 3 terms of the harmonic series. ie. 1/1+1/2+1/3.

2

two

            I like to say that "two" is the very first large number, since it's at least larger than one. In truth 2 is a number that rests comfortably in the mind. It is actually one of only a handful of truly small whole numbers. By "truly small" I mean that these are the few whole numbers that psychologically seem small to us. These are numbers that we can grasp with our innate number sense...(READ MORE)

2 + ln(2)/2^10^10^100

This is a lower bound on F(-10^10^100) = (2^2^-10^10^100)+1. To understand this, we first conjecture that in binary 2^2^-n has convergent bits, in the same way that 10^10^-n has convergent digits. This isn't too hard to see, since we know that halving the exponent will result in a value half the distance to 1. Thus in binary the digits should appear to converge to some value. Much like we derive that we converge to ln(10) in decimal, we can show we converge to ln(2) in binary. Simply observe that:

ln(1+x) ~ x for small x --> e^x ~ 1+x for small x

2^x = e^(x*ln(2)) ~ 1 + ln(2)*x for small x

2^2^-x ~ 1 +ln(2)*2^(-x) = 1 + ln(2)/2^x for large x

The actual value is slightly larger than this, as can be seen by a few examples:

1+ln(2)/2 = 1.34657359028... < 1.41421356237... = sqrt(2) = 2^2^-1

1+ln(2)/4 = 1.17328679514... < 1.189207115... = 2^2^-2

1+ln(2)/8 = 1.08664339757... < 1.09050773267... = 2^2^-3

1+ln(2)/16 = 1.04332169879... < 1.04427378243... = 2^2^-4

etc.

This approximation gets better the larger n is. Thus we can say with high confidence that (2^2^-10^10^100)+1 is approximately 2 + ln(2)/2^10^10^100. 

F(-10^10^100) = (2^2^-10^10^100)+1

This is F(-10^10^100) where, F is the function for the Fermat Numbers. The second exponent here is so large that we can not know what the first non-zero digits are of this after the first 2. We know the number begins 2.00000... and then there is some first non-zero digit at a position approximately a googol digits in. To see this we do the following:

2^2^-10^10^100 = 10^10^( -10^10^100*log(2)+log(log(2)) )

~ 1 + ln(10)*10^(-10^10^100*log(2)+log(log(2)) )

To find the first non-zero digits we would need to find the decimal part of 10^10^100*log(2)-log(log(2)). We won't be able to do this however, because it would require about a googol digits of log(2). This is far more than we can compute. So the best we can do is estimate and say this value is approximately 2 + 10^-10^10^100. This means the first non-zero digit is about at the googolplex place value after the decimal point. Although we can get arbitrarily close to 2 by substituting a large googolism for googolplex, we wouldn't be able to know what the non-zero digits were, only about where they would occur. 

F(-10^100) = (2^2^-10^100)+1

~ 2 + 2.716x10^(-3.01x10^99)

It can be easily seen that this number must be extremely close to 2, since 2^-infinity = 0 and 2^0=1, it follows (2^2^-10^100)+1 must be very close to 1+1 = 2. How close? To calculate this we first convert to base 10:

2^2^-10^100 = 10^10^( -10^100*log(2)+log(log(2)) )

Next we use the identity 10^10^(-n) ~ 1 +ln(10)*10^(-n).

10^10^( -10^100*log(2)+log(log(2)) ) = 1 + ln(10)*10^( -10^100*log(2)+log(log(2)) )

We run immediately into a difficultly. To actually find the digits, it is necessary to know the values of log(2) to more than 100 decimal places. This is because we need to compute the decimal part of the exponent. This can be one with a large number calculator. Using one such calculator I obtain:

3010299956639811952137388947244930267681898814621085413104274611271081892744245094869272521181861720.928234999568634757057203092757

The bold part is the decimal part. Computing ln(10)/10^0.92823499956...etc. we can obtain the decimal part of: 0.271631848527. Next we take the exponent part to obtain the following approximation:

2 + 2.71631848527x10^(-3.0102999566x10^99)

Amazingly we can still know some of the non-zero digits after the 2 are. However this would easily become impossible with a sufficiently large googological number. 

2.000000000000000000000000000000546796712248...

~ 2+5.46796712248x10^-31

F(-100) = (2^2^-100)+1

This is a number extremely close to but slightly greater than 2, created by extrapolating the Fermat numbers to -100. Finding it's decimal digits presents something of a little mathematical challenge, but nothing a little work with logarithms can't solve. We know that 10^10^-n ~ 1 + ln(10)/10^n. Thus we can compute 2^2^-n by simply converting the 2s into 10s. We have:

2^2^-n = 10^(log(2) * 2^-n) = 10^( 10^log(log(2)) * 10^(-n*log(2)) ) = 10^10^( -n*log(2)+log(log(2)) )

From here we compute: -100*log(2)+log(log(2)) = 10^10^-30.6243897941 ~ 1 + ln(10)/10^30.6243897941

We can then find the decimal digits by simply dividing ln(10) by 10 to the power of the decimal part of the exponent. This gives us the result: 1 + 0.546796712248*10^-30 = 1+5.46796712248*10^-31. Adding 1 to this we obain 2+5.46796712248*10^-31. This is small enough to write out in full.

I believe this exact value came up while I was tutoring and thinking about Fermat Numbers. This number is too close to 2 for it to be able to be computed directly by repeatedly taking the sqrt 100 times. Thus I tried to find a way to compute the digits. I'm not sure if the above is the method I employed, but none the less this should be correct.

This idea can be extended much further. See entries above but still greater than 2. A decent approximation can be obtained by beginning with (2^2^-10)-1 = 0.000677130693. One then simply divides this by 2 90 more times to obtain a decent approximation of the difference from 2. Computing (2^2^(-10)-1)/2^90 ~ 5.46981817868x10^-31. The first 3 digits (546) appear to be correct.

2.00067713069...

F(-10) = (2^2^-10)+1

A value that can be generated that is close to 2. This is closer to 2 than Munafo's closest entry to 2 from above.

2.00231930436153+-53x10^-14

This is the smallest real number larger than 2 on Robert Munafo's Notable Properties of Specific Numbers. He describes it as the "gryomagnetic ratio of the electron". It's a physical constant believed to have special properties by some, just like the Fine-Structure Constant. Munafo also says it's notable for being known to such a high accuracy. It's amusing to note, this number is much closer to 2, than Munafo's closest number above 1 is to 1. See 1.013.

2.04427378243...

F(-4) = (2^2^-4)+1

An extrapolation of the Fermat Numbers to F(-4). See F(-1) = 2.414 for more details.

2.08333333333...

fourth harmonic number

            This the value of the 4th harmonic number, the sum of 1/1+1/2+1/3+1/4. It's notable for being the smallest harmonic number to exceed 2. The harmonic series grows perpetually slower. Getting to 3 in this manner proves to be a bit more difficult.

2.09050773267...

F(-3) = (2^2^-3)+1

An extrapolation of the Fermat numbers to F(-3). See F(-1) = 2.414 for more details.

2.189207115...

F(-2) = (2^2^-2)+1

A number obtained by expolating the Fermat Numbers to negative integer exponents. In otherwords, this can be thought of as F(-2). This can trivially be computed as the 4th root of 2 plus 1. See 2.41421356237 for more details. To see more information of the digits of (2^2^-n)+1 for large n, continue to numbers googologically close to 2 from above.

2.25

(1+1/2)^2

This is a lower bound on e created by plugging 2 into formula (1+1/n)^n. As n approaches infinity, (1+1/n)^n approaches e. One way to think about this number is, if one had 100% APY, but it was compounded twice a year at 50% each compounding, this is the return.

2.283333333333...

fifth harmonic number

            The sum of 1/1+1/2+1/3+1/4+1/5.

2.30258509299...

ln(10)

ln(10) comes up when working with numbers of the form 10^10^-n. This comes up in googology when appending a googolism to -minexiplex. Such numbers, such as googolminexiplex are extremely close to 1, and have the general form 1.00000...00002302... , where the first non-zero digits are the 1 converge to the digits of ln(10) up to a certain point. The number of zeroes and the number of "convergent" digits is directly proportional to the googolism appended to -minexiplex. For example, in the case of googolminexiplex, there are exactly (googol-1) zeroes after the decimal point, following likely by a googol convergent digits of ln(10), with a possible offset of 1. The reason we can't know exactly how many convergent digits there are is because this pattern is not consistent and depends on the googolth digit. If it's too large it can overflow and cause the number of convergent digits to be one less than the googolism being appended. See 10^10^-5 for the smallest known example of this phenomena. We can not calculate a googol digits of ln(10) and as far as I know there is no known way to compute the googolth place value independently. Thus we can not say for sure. We can't know what the last convergent digit is, nor the first non-convergent digit. We can however say confidently that any number of digits we could compute for ln(10) will be included in googolminexiplex, as well as higher versions such as trialogiaminexiplex, googolpleximinexiplex, grangolminexiplex, etc.

2.4142135624...

(2^2^-1)+1 = sqrt(2)+1

This number, most simply expressed as sqrt(2)+1, came from the idea of extending the "Fermat Numbers". The Fermat Numbers are positive integers of the form (2^2^n)+1 where n is a non-negative integer. The smallest Fermat Number is (2^2^0)+1 = (2^1)+1 = 2+1 = 3. The next smallest would be (2^2^1)+1 = (2^2)+1 = 4+1 = 5. Next we have (2^2^2)+1 = (2^4)+1 = 17. It is very obvious that the first 3 are prime numbers. These are known as the Fermat Primes. The Fermat Numbers may be denoted by F(n) = (2^2^n)+1. Thus we say F(0)=3, F(1)=5, and F(2)=17 are Fermat Primes. Is F(3) a Fermat Prime? Well, F(3) = (2^2^3)+1 = (2^8)+1 = 256+1 = 257. This too turns out to be prime. What about F(4)? F(4) = (2^2^4)+1 = 2^16+1 = 65,536+1 = 65,537. This too turns out to be prime. What about F(5)? F(5) = (2^2^5)+1 = (2^32)+1 = 4,294,967,296+1 = 4,294,967,297. F(5) however turns out to not be prime. In fact, you can confirm it is factorable as: 641x6,700,417. Simply multiply them together to confirm. In these are it's only two prime factors. The sequence roughly squares the previous entry to reach the next entry, thus it exhibits hyper-exponential growth. This means it quickly goes beyond what we can confirm directly. F(6) is already equal to 18,446,744,073,709,551,617. This some effort my TI-89 obtains the factorization: 274,177x67,280,421,310,721. This will quickly exceed the abilities of brute force calculation. F(0), F(1), F(2), F(3), and F(4) are in fact the only known Fermat Primes. It is not known if there are any more further along the sequence. Fermat himself conjectured that all Fermat Numbers were Fermat Primes. Fermat's conjecture was refuted by Leonhard Euler in 1732 when he successfully factored F(5). As of 2014, F(n) is known to be composite for 5<=n<=32. Complete factorizations for F(n) are only known for 0<=n<=11, and there are no known prime factors for n=20 or n=24 (despite the fact they are still known to be composite). The largest Fermat Prime known to be composite is F(18,233,954), and it has a large known prime factor equal to 7x(2^18,233,956)+1. One can easily see why the sequence of Fermat Numbers might be of interest to googologists, as it involves hyper-exponentially large numbers that were actually of some theoretical interest to mathematicians. 

One day while I was tutoring however, I considered what would happen if we allowed the exponent to continue downwards towards the negative integers. In this case we approach (2^2^-infinity)+1 = (2^0)+1 = 1+1 = 2. So this sequence can be used to approach 2 from above, much in the same way that we used 10^10^(-n) to approach 1 from above. My particular theoretical interest however was to find what the string of digits would be after 2.000... for given large negative integers. Plugging in -1 into the formula is trivial to compute as 2^(-1) = 1/2 and 2^(1/2) is simply the sqrt(2). Thus we obtain sqrt(2)+1 in this case. The digits are as easy to compute as the sqrt(2) in this case. This number has another curious property. No matter how many times we square it, we will always have an irrational number. If we square sqrt(2) we get 2. After that we get 4,16, 256, 65536, etc. Once we obtain an integer it remains an integer indefinitely. But if we begin with sqrt(2)+1 we know from multiplying binomials that (sqrt(2)+1)^2 = (sqrt(2)+1)(sqrt(2)+1) = 2+2sqrt(2)+1 = 2sqrt(2)+3 = 5.82842712475. Squaring again we now have (2sqrt(2)+3)(2sqrt(2)+3) = 12sqrt(2)+17. Each time we apply binomial multiplication we are left with a number of the form Asqrt(2)+B. And each time we square it we have: 2ABsqrt(2)+2A^2+B^2. If at any point it were rational then we would have Asqrt(2)+B = a/b, and we could solve for the sqrt(2) and obtain: (a/b-B)/A and we would get a rational expression for sqrt(2). Since sqrt(2) is irrational this is impossible, and thus we can square it forever and always obtain an irrational number. This can theoretically be used then to generate a large irrational number. 

We can continue to get closer to 2 from above by simply choosing a larger negative number, but the decimal expansion becomes difficult to compute since it doesn't converge like 10^10^(-n). However this is mainly due to expressing it in decimal instead of binary. In binary the digits would converge just as 10^10^(-n) converges in decimal. It is possible to find the digits but this requires some trickery. I do believe I was able to figure out the digits that day, but in any case I promptly forgot the result. I have since rediscovered how to do the same thing to find the digits. See earlier 2^2^(-n)+1. 

2.45

6th Harmonic Number

            This number is notable for reducing to a relatively simple terminating decimal. It is equal to 1/1+1/2+1/3+1/4+1/5+1/6. This sum equates to 147/60 or 2 + 9/20.

2.59285714286...

7th Harmonic Number

The 7th Harmonic Number. This is 1/1+1/2+1/3+1/4+1/5+1/6+1/7. The harmonic series, is a series which, while diverging to infinity, grows slower as it goes. One way this is related to googology is one can get a very large number by choosing a bound, say 100, and then asking what the minimum number of terms to exceed it are. It will take quite a while for the Harmonic series to find it's way beyond the palpable epoch. This value is strangely close to 2.598 which is 3*sqrt(3)/2, a bad lower bound for pi using Archimedes method with an equilateral triangle. See 2.7178.

2.59807621135...

3*sqrt(3)/2

This is a very bad lower bound for pi based on creating an inscribed equilateral triangle, thus taking Archimedes Method to one of it's logical extremes. See 2.828.

2.666666666666...

4/1-4/3 = 8/3

This is the first two terms of the Leibniz Series, the first infinite series used to express pi. It comes from the arctan formula that says:

arctan(x) = x^1/1 - x^3/3 + x^5/5 - x^7/7 + ...

Substituting x=1, we obtain the series for arctan(1), which is the angle where the tangent returns 1. In other words the right triangle in which the legs are equal. This happens at 45 degrees or pi/4 radians. This trigonometric series, like all trigonometric series, is in radians, so arctan(1) = pi/4. so we get:

pi/4 = arctan(1) = 1/1-1/3+1/5-1/7+...

pi = 4*arctan(1) = 4/1-4/3+4/5-4/7+...

4*arctan(1) is a common value used by programmers to obtain pi. It can be used to obtain the nearest double floating point number less than pi. See 3.14159265358979311.

2.71785714286...

8th Harmonic Number

This is 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8. It is surprisingly close to the value e.

2.718281828459...

e

        This is the transcendental constant, e. The value can be obtained by taking the sum of the reciprocals of the factorials:

e = 1/0! + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! + ...

Another important definition is:

e = lim n-->oo (1+1/n)^n

        It is the base of the natural logarithm. The derivative of the function e^x is e^x. This means that the slope at any point along the curve is equal to the y coordinate of that point. An interesting consequence of this is that the value of e^n, where n is a counting number, must be larger than the nth triangular number plus 1. ie. e^1 > 2, e^2 > 4, e^3 > 7 etc. "e" is really quite a small magnitude, lying between 2 and 3. "e" comes into play in the large number field in the definition of Skewes' Number, Ballium's Number, and in studying infinite power towers.

2.82842712475...

2*sqrt(2)

This is a bad lowerbound created by using an inscribed square. Note that, if we set the sides of the square to 1, then it's diagonal is the diameter of the circle and equal to sqrt(2). We then take the perimeter of the square and divide it by the diagonal to obtain: 4/sqrt(2) = 4*sqrt(2)/2 = 2*sqrt(2). This is one of a handful of lower bounds on pi that are actually less than 3. See 2.598... and 2.666... for some other examples.

2.82896825397...

9th Harmonic Number

The 9th harmonic number exactly equal to 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9. It is slightly greater than 2*sqrt(2).

2.89523809524...

4/1-4/3+4/5-4/7 = 304/105

The fourth partial sum of the Leibniz Series, and the second underestimate in the series. See 3.1805238095.

2.92896825397...

10th Harmonic Number

The 10th harmonic number exactly equal to 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10. It is slightly less than the lower bound 2.938 for pi using an inscribed regular pentagon.

2.93892626146...

5*sin(π/5)

This value is a lower bound for pi created from an inscribed regular pentagon. Take a n-sided regular polygon and chop it into n congruent triangles, then cut each of these into two right triangles. If we set the hypotenuse to 1, then the radius of the circle is 1, and the diameter is 2. Next we use the angle around the center of circle, which would be 2pi/(2n) = pi/n. The side opposite this angle if half the length of a side of the polygon. It would be equal to sin(pi/n). Thus we have 2n*sin(pi/n)/2 = n*sin(pi/n). Combined with the formula for the upperbound (see 3.632) we have:

n*sin(pi/n) < pi < n*tan(pi/n)

In this way we can easily get any of the lower bounds or upper bounds created by Archimedes method. This allows us to get polygons that would otherwise be tricky like pentagons, heptagons, etc.

3

M2

three

        "3" is the 2nd triangular number. It is one of the few numbers that can be understood with our number sense alone. It is one of the larger examples of a "truly small number".

            3 is also the 2nd prime prime, and the smallest mersenne prime (primes of the form 2^p-1 where p is prime. The next mersenne prime is 7). 3 is the sum of 1+2, 2+1, the product 1*3, 3*1, and also is equal to the expressions 3^1, 3^^1, 3^^^1, 3^^^^1, etc.

3 can be thought of as a very primitive approximation of pi. In fact it is one of the 3 most commonly used approximations of antiquity, the other two being 25/8 and 22/7, entries also in the ULNL. The figure of 3 for pi seems to be implied by the bible, suggesting a rather crude mathematics, but it can also be obtained by an extremely simple means using Archimedes method. It is one of the simplest cases as well. Begin with a regular hexagon inscribed in a circle. Note that if we slice it into 6 equilateral triangles, which we can do because of the interior angles being 120 degrees each, that two sides of the equilateral triangle are also the diameter of the circle. The perimeter of the hexagon would be 6. Thus we can say since the perimeter of the hexagon must be smaller than the perimeter of the circle, since straight lines are necessarily shorter than the arcs connecting the end points. Thus we can say pi > 6/2 = 3. In a similar way a circumscribed hexagon can be used to find the upperbound sqrt(12). See 3.464. Thus this is one side of the first level of precision in archimedes polygon method for finding bounds for pi. That is 3 < pi < 3.464. The next set of bounds would use 12-sided polygons.

3.01987734488...

11th Harmonic Number

            This is the smallest harmonic number greater than 3.

3.10321067821...

12th Harmonic Number

This is 1/1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+1/9+1/10+1/11+1/12. It is the last Harmonic Number less than pi.

3.125

25/8 = 3 + 1/8

03.07:30

A babylonian approximation of pi. Unlike 22/7, 25/8 is a lower bound. Thus pi lies between 3 + 1/8th and 3 + 1/7th. This approximation is just slightly lower than our current modern rule of thumb approximation, 3.14.

Recall that the babylonians used a sexagesimal system. 1/8th is easily expressed in the sexagesimal system as 3.07:30. That is 3+7/60+30/3600. Upon simplifying we have 3+7/60+30/3600 = 3+7/60+1/120 = 3+14/120+1/120 = 3+15/120 = 3+1/8. Despite the historical significance of the approximation of 25/8, almost on par with 22/7, Munafo's number list does not contain it. It has perhaps fell out of favor since 22/7 has the distinct advantage of beginning with the same 3 digits as our most common approximation of pi, 3.14.

Another much closer sexagesimal value used by the greeks was 3.08:30, which yields the much better value 3.141666...etc. This approximation is even better than 22/7 in fact.

One last note, this babylonian approximation of pi is one of the few that can be expressed exactly in binary as 11.001. Thus  it can be represented precisely as a double-floating point precision number.

3.13274336283...

354/113

355/113 is a ratio that is much closer to pi than it's denominator would imply. Subtract or add 1 to the numerator and suddenly you have a ratio that is approximately 1/113 away from pi. This gives a better idea of the expected precision for a denominator of 113. One would expect about 2 decimal digits of accuracy. Actually we only get 1 decimal digit here. See 3.15044.

3.14

three point one four

These 3 digits are the ones most strongly associated with pi. The digits are commemorated on pi day with the date 3/14.

A very common approximation for pi used in school. Most people have pi memorized up to the first two decimal places. It's so commonly cited, school aged children are likely to confuse this approximation with pi itself. In actuality, it is slightly less. Since this is basically just pi chopped off at the second place value (and pi is irrational), it must necessarily be a lower bound. Just for fun, I'll note that the relative power between 3.14 and pi is 1.00044317062. This means that 3.14^1.00044317062... = pi. This is simply to say, the two values are relatively close together. Relative powers will become more relevant later in the list where numbers no longer tend to be this close, arithmetically, logarithmically, or even by relative powers

It's worth noting that Archimedes bounds of 3+10/71 = 3.1408... < pi < 3.1428... = 22/7, amounts to (in modern notation) confirming the first two decimal digits of pi (3.14...).

3.14084507042...

3+10/71 = 223/71

This is Archimedes rational approximation of the inscribed 96-sided regular polygon. Combined with Archimedes upperbound of 3+1/7 for pi, archimedes would have confirmed the first two decimal digits of pi.

3.14103195089...

96*sin(π/96)

This is the exact value obtained when using an inscribed 96-sided polygon to create a lower bound for pi. It's worth noting however that Archimedes didn't use trigonometry and would not have been able to calculate this exact value. This value would involve nested radicals. Since direct computation of the radicals was out of the question, Archimedes would have instead used rational approximations of them. Furthermore, to ensure the rational expression was less than pi, Archimedes would ensure that the rational approximation of this expression was less. And that is exactly what we find. Archimedes lowerbound for the exact perimeter of the inscribed 96-sided regular polygon with diameter of 1, is the rational value 3+10/71 = 223/71.

3.14138888888888888888888888888888888888888...

03.08:29

The value of pi correct to two sexagesimal digits. Sexagesimal base is notable for being the number system used in babylon, and is responsible for our division of an hour in 60 minutes and a minute into 60 seconds. We also have 360 degrees, with each degree equal to 60 minute arcs, and each minute arc equal to 60 second arcs, thanks to them. Thusly there are two approximations of pi based on in base 60. These are 3.07:30 = 3.125 = 3+1/8, and 3.08:30 = 3.14666... = 3+1/8+1/60. But the correct value up to the first two sexagesimals would simply be 03.08:29. If one has the digits of pi in decimal, and a calculator, these sexagesimal places can found quite easily by taking pi, subtracting 3, and multiplying by 60, then repeatedly removing the integer part and multiplying by 60. See 3.1415959259.

3.14141414141414141414141414141414141414141414141414141414...

3+14/99 = 311/99

A fun little rational approximation of pi one can create by simply repeating the first two decimal digits over and over again forever. This works out pretty well since the next two digits of 15, only slightly less. This would make this a general pretty good rational approximation of pi, since it actually snags an extra correct digit, with a denominator of only 2 digits.

3.14150943396...

333/106

Like 3/1, 22/7, and 355/113, 333/106 is a "best rational approximation of pi" for the size of it's denominator. What this means technically, is that it will be better than any rational approximation with a smaller denominator.  Wait, isn't that always the case? No. Let's start with d=1. In this case 3/1 is the best we can do, in terms of arithmetic distance. What about d=2? Well 6/2 is better than 7/2, but 6/2=3/1. So halves are no better than wholes in this case. What about thirds? 10/3 - pi = 0.19 while pi - 9/3 = 0.14. Thus thirds still don't give us a better approximation. With d=4, we have 13/4 - pi = 0.108, so this gives us the first case of a better approximation for pi than any smaller denominator. So the new champion is 3.25 and the new difference to beat is about 0.108. With fifths we get 3.2 which is obviously closer, then 19/6 = 3.1666..., then 22/7 = 3.142857...etc. The difference here however is extremely good. 22/7 - pi = 0.00126. Eighths can't compete with this, nor ninths, nor tenths, etc. So 333/106 is one of these special cases of "good rational approximation" based on the size of it's denominator. In fact it's so good that it beats out Archimedes inscribed 96-sided regular polygon! With 1000 sides however ...

3.14158748588...

1000*sin(π/1000)

This is a lower bound for pi one can obtain using an inscribed chiliagon (1000-sided polygon). Essentially this is taking Archimedes method even further. Even so we only get 4 decimal digits of accuracy. If we use the inscribed and circumscribed chiliagon we have:

3.14158748588... < pi < 3.14160298906...

This would only confirm the first 3 decimal digits (3.141), despite having a 1000 sides. 

3.14159259259...

03.08:29:44

The first 3 sexagesimals after the sexagesimal point. This gives us a surprisingly good decimal approximation of pi, correct up to 6 decimal places.

3.14159265358

First 11 decimals of Pi

The correct first 11 digits of pi. Because the next digit after 8 is 9, on a 12 digit display this would be rounded up to 3.14159265359. Because of rounding it's important not to assume that the last digit of a display is actually a digit of pi. This mostly applies to when a display is smaller than the actual storage of the number. Most floating points are in binary, so rounding happens with bits instead of digits. This leads to many many additional digits that do not add to the decimal precision of the number. In the case of the double format, for example, the most common value of pi actually gets the first 15 decimal digits correct. See 3.14159265358979311 below.

3.1415926535897

First 13 decimals of Pi

The first 13 decimal digits of pi. This would be the closest value less than pi, that the TI-89 could store in approximate mode. Unlike the actual value of pi stored on the TI-89, this one has the correct first 13 decimal places, even though, rounding up the last 7 to 8 is arithmetically closer to pi than this number is.

3.141592653589793

First 15 decimals of Pi

The first 15 decimal digits of pi. I have read that Newton computed the first 15 decimal digits of pi. Another significance to this particular approximation is that in Double Precision Floating point representation one can only get a maximum of 15 correct decimal places of pi stored. The digits beyond this are sort of "garbage data". An artifact of converting the binary storage into a decimal display. Digits are the first 15 can not actually be manipulated individually, because despite containing 51 decimal digits, only 15 can be manipulated individually. In reality the number is stored as 52 bits, and the first bit is dedicated to before the "binary point". 

This is also the digits of pi that I'm most familiar with. My reasoning was, it would be good to know pi up to the maximum precision allowable by double floating point, which theoretically, is enough digits of pi to perform any sensitive physical calculation one would want to perform. Even NASA engineers admit that they don't need a lot of digits of pi in their calculations. 15 decimals would probably be enough to perform any desired maneuver within say our solar system. Having more digits of pi is mostly an obsession of mathematicians and math enthusiasts, then something that preoccupies engineering types.

3.141592653589793115997963468544185161590576171875000

Closest Lower Bound of Pi using Double Precision

This value is exact! You may recognize the beginning digits as those of pi ... and they are ... up to a point. This is the largest number less than pi using double precision floating point representation. Double precision floating point representation uses 8 bytes or 64 bits, with 1 bit for the sign, 11 bits for the exponent, and 52 bits of precision. In essence this is the closest a standard computer representation can get to the actual value of pi! In binary this is exactly equal to:

11.001001000011111101101010100010001000010110100011000

Despite containing 51 decimal digits (including the last 3 zeroes), only the first 15 decimal digits match that of pi.  Everything after 3.141592653589793 is essentially garbage digits that result from the number being expressed in binary.

3.1415926535897932384626433832795028841971693993751058209749445923078164... 

10^50*sin(π/10^50)

Pi via Perimeter of Inscribed Gogolgon

What if we were to take Archimedes polygon method, and take it to googological extremes? Well the first thing to note is that Archimedes is not actually the best method to estimate pi. It's convergence is relatively slow. You need about 10^n sides to get n digits. So let's say we have an inscribed regular gogolgon (10^50 sides). Surely then we get 50 decimal digits correct, right? Wrong! We would actually get the first 99 decimal digits of pi. Here is the inscribed gogolgon perimeter:

3. 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170674 6537680850 8285303722 488593491

and here is pi:

3. 1415926535 8979323846 2643383279 5028841971 6939937510 5820974944 5923078164 0628620899 8628034825 3421170679 8214808651 3282306647 0938446095

The bolded digits are where they finally diverge. Being the inscribed polygon it's a slight and I do mean slight, underestimate. But the change only happens beginning at the 100th decimal place.

This may seem strange. Why are we getting twice as many as you might expect? This however requires a deeper investigation to find out why. To understand how close Archimedes inscribed and circumscribed polygons of the same number of sides will bound pi, we can take their difference. This would be n*tan(pi/n) - n*sin(pi/n). Factoring out n and sin we obtain nsin(pi/n)(1/cos(pi/n)-1). We note that cos(x) ~ 1 - x^2/2. This is simply the first two terms of cosine's infinite series. It follows from this that cos(pi/n) = 1 - (pi/n)^2/2. Next, since pi/n is extremely small we can approximate the reciprocal of 1-(pi/n)^2/2 as 1+(pi/n)^2/2. Furthermore sin(pi/n) ~ pi/n, since again pi/n is very small. So we get the following:

n*sin(pi/n)(1+pi^2/(2n^2) - 1 )

n*(pi/n)(pi^2/(2n^2))

(pi^3/2)/n^2

In this case n is a power of 10, and so this tells us the difference between the two will be the square of the power of 10. We also have a digital convergence that will be approximately equal to pi^3/2 = 15.5031383402. So plugging in n=10^50 we expect that the range between the inscribed and circumscribed gogolgon will be 15.503/10^100 = 1.5503/10^99, which suggests a disagreement in the 99th place. We don't see this because this difference is not divided evenly amongst the lower bound and upper bound. It appears that about 1/3rd of that goes to the lower bound and 2/3rds to the upperbound. If this is correct then the error from pi should be -0.516/10^99 = -5.16/10^100, which shows a change of about 5 in the 100th decimal place. And lo and behold, we see that the 9 was changed to a 4! So we can accurately predict how much these inscribed and circumscribed googological polygons will differ from pi. So great! That means we can use this method to easily calculate googologically many place values of pi, right?! Well of course not. We can compute the error with high precision, but this doesn't actually tell us what the value we will get is. The main problem is computing sin(pi/10^50) accurately. It will be extremely close to 0 and extremely sensitive in value. This value was obtained using a high precision online calculator that offers the ability to go up to 130 digits of precision. However going much further will lead to something we could not actually calculate. None the less here we have definitely a value that would have been a great approximation of pi throughout most of history, and we can even say where it should start differing from pi without knowing the value of pi. Even so ... this is still nothing compare to the number of digits that have been calculated by computer. So let's go even further towards pi!

ENIAC Pi Computation

In 1949, the ENIAC computer was used to compute the first 2037 decimal digits of pi. This was the first time that a computer, had been used to calculate digits of pi. It took 70 hours of computation to perform. It easily surpassed the furthest human achievements, 707 digits of Shanks, only 527 of which were correct, and the most digits ever calculated by hand, which was 620 by a Mr. Ferguson. This more than tripled the former record!

IBM 704 Pi Computation

In 1958 10,000 digits of pi were computed for the first time by an IBM 704. Unlike the ENIAC this was achieved in a mere 1.7 hours of computation!

10^100,000*sin(π/10^100,000)

~ π - 5/10^200,000

Pi Estimated by perimeter of Inscribed Googolgongagon

Surprisingly this is NOT the first context under which I have considered the googolgongagon (googolgong-sided polygon where a googolgong = 10^100,000). When I first conceived of what would become Extensible-E Notation (to be seen later in Part I of this number list) I imagined it as a series of polygon notations, where each regular polygon was cut up into n isosceles triangles, and numbers were written inside each one. The way to make larger numbers was to have more sides. So naturally I conceived of polygons of googologically many sides, more as a notational side effect. My favorite number was a googolgong, so naturally I considered the case of a googolgong sided polygon (what I now call whimsically a googolgongagon) and imagined filling it with 10s and a single 303 for the third argument. The details of this are somewhat tangential to our current discussion, but it will be visited again later. In any case, what would happen if we consider this googolgongagon in the context of Archimedes estimation of pi? Well if we are using an inscribed googolgongagon this leads exactly to the value 10^100,000*sin(pi/10^100,000). A perfectly reasonable question is ... just how close would this be to pi. Keep in mind that a googolgongagon has an absolutely unimaginable number of sides. This puts a chiliagon, myriagon, gogolgon, or even a googolgon to shame. Despite this we can know with a pretty how certainty that it has at most 199,999 correct digits of pi. The reason I don't know quite for sure is that the known approximate difference of 5/10^200,000 could potentially roll back multiple digits if they were all 0s. It might not be likely, but without knowing the exact digits of pi at that point in the sequence there is no way to know for sure. We can only say it's likely to be good up at least 199,998 decimal digits, with about a 1% chance of doing worse than that, assuming pi is normal. This level of precision is quite remarkable still. And yet as impressive as it is to think of 200,000ish correct digits of pi, and the ridiculous polygon we used to get there ... we still wouldn't have even come close to the best computations of pi. But ... we are getting there ...

CDC 6600 Pi

In 1967, the CDC6600 had computed 500,000 decimal digits of pi for the first time. This figure is the last one cited in Petr Beckmann's A History of (PI), which was published in 1971. As he wryly observes, even though it was the furthest he was aware of, he acknowledged it could have easily gone further in his own time, and surely would be surpassed soon regardless. This value is so accurate that even using an inscribed googolgongagon would not be enough to calculate pi this accurately. We would need an inscribed polygon of approximately 10^250,000 sides to get an accuracy of this level. Such a method was not employed for this computation, instead being performed by much faster and more reliable modern formula with very rapid convergences. The computation of these digits only took 28 hours.

CDC 7600 Pi

Only 2 years after the publication of Petr Beckmann's book, in 1973 the CDC 7600 topped the previous record, and got 1,001,250 digits of pi. So it took as late as 1973 for the first million decimals of pi to be known! This was only about 10 years before I was born. To compute 1,000,000 digits of pi would take a polygon of approximately 10^500,000 sides.

IBM 3090 Pi

In June of 1989 The IBM 3090 became the first computer to compute the 1 billionth (with a B!) decimal place of pi. Specifically it computed 1,011,196,691 decimal places. This was only 16 years after the millionth digit had been found for the first time, quite a remarkable leap in capability.

HITACHI SR8000/MPP Pi

In November of 2002, the HITACHI SR8000/MPP computed pi to cover 1 trillion (10^12) decimal places for the first time. In all 1,241,100,000,000 decimal places were computed. This was only 13 years since 1 billion decimal places were computed.

Emma Haruka Iwao Pi

On March 14th, 2019, it's a pi day, one of the most enigmatic computations of pi was performed. 31,415,926,535,897 digits of pi were computed, where the digits in the number of digits are themselves the first 14 digits of pi! How cool is THAT!

They were computed by Emma Haruka Iwao, a Japanese Computer Scientist and Cloud Developer for Google.

This suggests a recursive continuation. If we compute the number of decimal digits computed, we can then look at how many decimal digits it contains. In this case, the number contains 13 decimal digits. However the only reason Emma knew to compute 31,415,926,535,897 places is because the first 13 places were already known! Where might that lead ...

Team DAViS Pi

In August of 2021, post covid pandemic, the record of approximately 62.8 trillion digits of Pi was reached. As of March of 2022, this is the furthest record of computing pi. If we truncate pi at any finite point, there must be remaining non-zero digits, because the number is irrational and would otherwise be rational (though with an absurdly large denominator of around 62 trillion digits). The upshot of this is any truncation of pi must be a lower bound. That is assuming that only correct digits of pi are included and not a last rounded digit. I assume that these calculations are performed with a few extra digits to confirm which digits are definitely part of the sequence of pi. The official computation was 62,831,853,071,796 decimal digits of pi. These are actually the decimal sequence of 2pi = 6.28318530717...etc. This is one upmanship from Emma's Pi of 31.4 trilion digits of Pi. 

Where will pi calculation go from here? Apparently, despite taking up terabytes of data at this point and taking nearly up to a year of computation at this point, there are still those enthusiasts who keep pushing it ever further! It as already been 20 years (as of 2022) since 1 trillion digits was first computed, and yet we are still in the trillions. Here is a rough time table:

1949 one thousand decimals

1973 one million decimals

1989 one billion decimals

2002 one trillion decimals

These are leaps of 24, 16, and then 13 years. It is quite possible that one quadrillion decimal digits of pi, will not be reached until several years to come. The next big milestone would appear to be the 100 trillion decimal mark. It's a pretty safe bet this is likely to be carried out and achieved by someone within the next 2 years. Assuming a doubling time of about 2 years from there (using a simplified version of Moore's law) this suggests that we might reach the one quadrillion decimal digit some time by around 2032. That's still a potentially long way off.

It's crazy to think about, but the current record is googologically smaller than the actual value of pi. Just how close will be ever become? There are practical limitations. Obviously the next entry will be pi ... after all where else could we go ... but wait...

floor(π*10^10^15)/10^10^15

One Quadrillion Decimal Digits of Pi

This benchmark has not been reached yet (as of 2022). But we don't need to wait until it's actually reached to have it as a number on our number list because ... it already exists, mathematically. Take pi and multiply it by a power of 10. say pi*100 = 314.159...etc. Then floor it. Then divide by that power of 10. You get pi up to some number of decimal places. Pi is well defined, so this is well defined. In other words, even if we haven't actually gotten there yet, and we don't know what the quadrillionth decimal of pi actually is, we can still imagine it. With googology we can simply imagine these values as long as they are well defined. We can imagine pi to any integral number of decimal places ... as long ... as we can already imagine that integer. A quadrillion is an easily knowable integer, and so we can imagine a decimal string starting 3.14159 etc. and continuing for a quadrillion digits after the decimal. This imagined quantity would be far closer than any decimal approximation of pi we have actually computed. Will we ever reach a quadrillion decimal digits? I would say it is quite likely at this point. However ... there are some practical considerations that start kicking in at this level and beyond. The computation would require the storage of about a petabyte of data. We are already getting close to the limits of miniaturization. So at some point we bump up against a hard limit of how much we can shrink the data and the total physical size of that data. A petabyte might be getting a bit difficult to store. However ... this would just be getting started with the infinite digits of pi ...

floor(pi*10^10^100)/10^10^100

One Googol Decimal Places of Pi

This would pi truncated to a googol decimal places. Here we have gone way beyond anything we could really expect to ever actually compute.  There is simply not enough data or computational power or time to compute this many digits of pi. We can only imagine such a thing. It's well defined, and it's googologically close yet less than pi. One thing we can say is that an inscribed googolplexagon would almost certainly (99.9999999...etc%) give us a better approximation of pi than to a googol decimal places, and should reasonably give 2 googol decimal places. 

10^10^100*sin(pi/10^10^100)

Inscribed Googolplexagon

An inscribed googolplexagon, if it could somehow be computed exactly, would give us 2x10^100 digits of pi approximately (always about twice the common logarithm of the number of sides).

(E100#100)*sin(pi/(E100#100))

Inscribed Grangolgon

If we can imagine a googolgong sided polygon, there is absolutely nothing stopping us from imagining a grangol sided polygon, which I have dubbed a grangolgon. A grangol = 10^10^...^10^10^100 w/100 10s, evaluated from right-to-left. It's a named number we will encounter much later in the number list. This number is so huge, that the poor convergence of polygons barely matters. This would theoretically give us pi accurate to 2*E100#99 decimals. Basically it's unimaginable. We are talking about more than a googol digits (E100), more than a googolplex digits (E100#2), more than a googolduplex digits (E100#3), more than a googoltriplex digits (E100#4) etc. etc. The number of correct decimal places would not fit in this universe, nor even if you scaled up from atoms to the observable universe a googolplex times, it still would be basically nothing. Does anything fantastic happen this far into the decimal expansion? Here is a almost unimaginable question to ponder. What if there were a run of 0s in pi that was proportional to the number of digits up to that point. I don't just mean like, statistically at the googol digit mark there is likely to be a run of 100 zeroes somewhere. What I mean is, is there an N so large such that after the first N digits, the next N digits are all 0s? That sounds like it would be impossible ... but for infinity? How could we prove it would never happen in all of infinity of the decimal expansion of pi. Think about that ... and maybe you will get an inkling of what we are dealing with here at the grangolgon scale. We could imagine scaling up exponentially, very rapidly, and each time we don't see such a run of 0s we just double the number of digits. We could do this for an inconceivably long time on the order of E100#98. In all of those doublings would we never find such a case? It could well be, for reasons we can't explain right now, such a thing is gauranteed to never happen in all of the infinity of the digits of pi, but ... what if such farlands of pi exist? That would certainly be something. Perhaps every so often there are deserts of 0s like this far out. Even at that particular scale the gap of 0s would be noticable and would like something like 3.1415926535...0000000000...7...etc. that would be really something. Even if we could mathematically prove such deserts exist ... we would likely never reach the first one ... something to screw your mind with ... and finally ...

3.14159265358979323846264338327950288419716939937510 ...

π

        A transcendental constant defined as the ratio of the circumference of a circle to its diameter. It is usually denoted "pi", and it lies somewhere between 3 and 4, making it a rather small quantity. It can be approximated as 22/7, but it can not be represented as the ratio of integers. It sometimes crops up mistakenly in large number discussions as an example of a large number. This is because the sequence of digits is never ending, and so it is assumed to have "more digits" than any other number. Usually someone will point out sarcastically that 3.2 is bigger, and an even more impressive number would be 4! A more clever use of pi for a large number might be to say "the largest number is pi with the decimal point removed". However such a number wouldn't be finite and therefore would not even count as a legitimate number (googologist's ban infinities from the discussion as it tends to be a game-breaker, and for other considerations of well-foundedness). Other than that pi doesn't come up too frequently in googology, though it is part of the definition of Ballium's Number.

3.14159265358979323846 ... 983367336244193 ... 0570985

Shanks Number

William Shanks was an amateur mathematician living in the 19th century (1812-1882). He is remembered best for his massive calculation of pi by hand. He lived right between the mathematical revival of pi following Newton's invention of calculus, but before the invention of the computer. Thus he had access to all the new formulas discovered for pi, but had to perform all the calculations through tedious tabulation. William Shanks was the last great pi hunter. No one would really try to push forward until mechanical calculation was made possible. As such this is where we reach an end of pencil&paper calculations of pi. In April 14th, of 1873, at the age of 61, Shanks published his 707 decimal digits of pi, the work of some twenty years of on and off calculation. The digits, as seen in print, are these:

3.

14159 26535 89793 23846 26433 83279 50288 41971 69399 37510 58209 74944

59230 78164 06286 20899 86280 34825 34211 70679 82148 08651 32823 06647

09384 46095 50582 23172 53594 08128 48111 74502 84102 70193 85211 05559

64462 29489 54930 38196 44288 10975 66593 34461 28475 64823 37867 83165

27120 19091 45648 56692 34603 48610 45432 66482 13393 60726 02491 41273

72458 70066 06315 58817 48815 20920 96282 92540 91715 36436 78925 90360

01133 05305 48820 46652 13841 46951 94151 16094 33057 27036 57595 91953

09218 61173 81932 61179 31051 18548 07446 23798 34749 56735 18857 52724

89122 79381 83011 94912 98336 73362 44193 66430 86021 39501 60924 48077

23094 36285 53096 62027 55693 97986 95022 24749 96206 07497 03041 23668

86199 51100 89202 38377 02131 41694 11902 98858 25446 81639 79990 46597

00081 70029 63123 77381 34208 41307 91451 18398 05709 85

To ensure that the digits up to the 707th place were correct, William actually computed 709 digits of the series he was using to compute pi. Thus, to William, these were the actual first 707 digits, no rounding or approximation involved. But there was a problem! The bold digits are all the digits that do not match up with pi. It is an often cited figure that William only got the first 527 decimals correct. However I actually found the original print of this number and compared it to pi, and this turns out to not be the case. William got pi correct up to the first 512 places, however there are 15 additional correct places after that, thus giving him a total of 527 correct places, just not consecutively. This subtle distinction has apparently been glossed over in discussions of this number! The digits Shanks got wrong I like to refer as gligits (glitch digits). These are digits that occur specifically from erroneous calculation, and they must be assumed to be correct at some point.

In any case this number, is technically a little larger than pi. Instead of the first bold digits being "193" they should be "065". If we take this decimal ending in 85 to be a terminating decimal we get a number mind-numblingly close to pi I shall call Shanks Number. It is technically distinct from pi, though barely so. But let's first offer some appreciation. Despite the error, Williams still managed to get the first 512 digits correct before the advent of computers. Compare this to the double format which only allows 15 decimal digits of precision!