Ascending With Up Arrows

3.2.3

^^^^ ASCENDING WITH UP ARROWS ^^^^

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Introduction

Humans have long had a penchant for large numbers, even if only within esoteric circles. It was a common claim in Archimedes time to say that the number of sands on the earth were "infinite"; that is to say "beyond counting". It was assumed that no number system conceivable by man could encompass such a number; in gist, that the world was much bigger than man could imagine. Archimedes showed such a notion is mathematically naive. He made a gross overestimate of the number of grains of sand on the earth, by filling up the entire universe (as understood at the time) with sand! This upper bound proved that there could not be more than 10⁶⁴ grains of sand on the entire earth. More to the point, the actual value would have to be significantly less! He then went on to create a numeration system that could be extended as far as 10^{8*10^16}_[1]. Thus Archimedes demonstrated once and for all that contrary to the folk wisdom that the numbers of man could never encompass the size of the world, that the opposite was in fact the case: the world could never hope to encompass the merest stretch of mans imagination!

Buddhist religious literature is also known for some very large numbers. The largest named number in buddhist literature is a number known as "bukeshuo bukeshuo zhuan" equal to exactly 10^{7*2^122}_[2]. This number is approximately equal to and greater than 10^{10^37}, making it even larger than Archimedes' Number!

This was more or less as far as humans had gone in antiquity. However it was not until the modern era when mathematicians were studying recursion that rapid progress was made in the creation of very large numbers. This advance was largely due to the improved notational and conceptual tools that mathematicians were working with, that made understanding large numbers a great deal easier than was previously possible. Archimedes used a very arcane place value system using the greek alphabet that was somewhat cumbersome by modern standards. Decimal notation vastly improved the ability of mathematicians to work with and describe very large numbers, but even this was not enough to really transcend numbers like those expressed by archimedes. Following the end of the dark ages, when much mathematical knowledge had been lost, an earnest attempt to revitalize mathematics began. An important part of this mathematical renaissance was the standardization of much mathematical notation we use today.

With the serious study of exponents underway in the 16th Century it was only natural that mathematicians would soon wonder what would happen if you continued the operations beyond exponents. It was this line of thinking, motivated by many other considerations, that eventually lead to some truly vast numbers that would make even the dreamers of the bukeshuo bukeshuo zhuan gasp in horror!

Let us now turn to the important precursors to the modern large number field...

Development of Exponential Notation

Modern mathematical notation is so ubiquitous that it comes as a bit of a surprise that it wasn't always so tidy. The "+" and "-" signs that come in so handy for expressing sums and differences first appeared in print in the year 1526_[3]. In this document they were used for accounting purposes to represent surpluses and deficits, but they soon took on the more general meaning of "sign" that we're familiar with. The "+" sign probably began as a shorthand for "et", which is latin for "and". For example, to write out something like 2 plus 3, mathematicians of the time might have written something like:

2 et. 3

suggesting "2 and 3". This eventually was shortened to the "2+3" that we are familiar with.

The use of the "cross" or "x" for multiplication comes a little later. It appears in William Oughtred's "Key to Mathematics" composed in 1628. It is tempting to speculate that the "x" for multiplication was the result of taking the "+" for addition and rotating it. Thus Oughtred would have written:

2 X 3

to mean 2 multiplied by 3. This is not that unfamiliar to most people since this notation is still used today in elementary school. However, even at the time the use of "x" as an unknown in algebra was widespread. Because of this Gottfried Leibniz complained that the use of "X" for multiplication could get confused with the "x" used in algebra. Leibniz suggested the use of the dot, "·", as early as 1698. Thus he would write:

2·3

to mean 2 multiplied by 3. He is also responsible for the omission of the dot in algebraic expressions, ie. "2x", "xy" etc. The Asterisk was also used as in "2*3" as early as 1659.

Mathematicians used exponents sparingly at first. Although it was sometimes necessary to speak of square numbers, cubes, and tesseracts, this could be specified simply by using repeated multiplication. For example "x squared" could be written simply as:

Rene Descartes advocated this notation for square numbers. Around 1634 Pierre Herigone was writing powers of x as x2, x3, x4, etc. The problem with this notation is it wouldn't work when the base was an integer rather than a variable. This could also be confused with multiplication as in "2x", "3x", "4x", etc.

The use of a superscript for exponents was first introduced in 1636 in Jame Hume's "The algebra of Vieta Method Novelle, clear and easy"_[4]. James used a raised roman numeral to represent an exponent. For example, to express x cubed, James would have used:

x^III

In 1637 Rene Descartes replaced the roman numerals with ordinary decimal numbers. Thus he would have written:

x³

like we would today. Oddly the idea of using raised exponents for any power took some time to take hold. Alternative notations were used when the power was a negative, or fraction. Eventually however the simple superscript notation prevailed for all real and complex powers.

In the 1960s when typewriters were already common, and computers were just getting started, the caret symbol, "^", came into vogue for writing exponents. The caret key was originally a proof reading symbol which served to insert something additional. In fact the name "caret" was derived from latin for "it lacks"_[5]. In order to change superscript notation into a linear format the caret was used in the following manner:

b^p = b^p

This usage has become quite popular on the internet to the present day.

What is Tetration & The Hyper-Operators?

In 1544 German mathematician Michael Stifel_[6] published an investigation into exponentiation. He was the first to discover logarithms, one of the two inverse functions of exponentiation. Logarithms were further studied by John Napier, who is often credited with their creation. The domain of the exponential operation was extended steadily in the following decades. Mathematicians eventually realized a connection between fractional powers and roots, and proceeded to extend powers to the real numbers. It was sometime in the 18th Century that mathematicians figured out how to raise numbers to complex powers thanks to Euler's formula.

With an understanding of exponentiation well underway it was only a matter of time until someone began to wonder about what the "next" operation would be. The next operation is what is commonly referred to as tetration.

The study of "tetration" can be said to have begun in 1758 with French mathematician Johann Heinrich Lambert. Specifically he was studying the infinite iterated exponential function. This function can be defined as the limit as n approaches infinity of the function:

exp(a,n,z) = a^a^...a^z w/n a's

A stack of exponents in this manner is often referred to as a "power tower". Surprisingly, the iterated exponential function does not always increase without bound when n goes to infinity. Sometimes it approaches a limit. This can be anticipated by the following observations:

If we begin with 2, then have 2² = 4, then 2²² = 2⁴ = 16, then 2²²² = 65,536, etc. it can be seen that as we feed back the exponent into itself the values increase without bound. However if we begin with 1, then have 1¹ = 1, then have 1¹¹ = 1, etc. we see that the value will never increase beyond 1. So the question is, what would happen if we use a value between 1 and 2? It turns out that if the chosen value is less than e^e^-1 (approximately 1.444) then the infinite power tower converges to some finite value.

As surprising as this is tetration largely remained at the outskirts of mathematical investigation until the 20th century. It was for the most part only of theoretical interest with few practical applications. Today most of the interest in tetration revolves around extending it to complex numbers, just as was done for exponentiation. For example, the background image at the top of this page was created by mapping tetration unto a complex number plane, producing fractal like images.

The status of tetration began to change due to developments within mathematics during the late part of the 19th century. Mathematicians were searching for a firmer foundation of mathematics. After much rapid development in mathematics during the last couple hundred years it was only natural to turn ones attention in the opposite direction and wonder where all of this development was coming from. An interest in foundational issues probably began with the invention of Calculus, because dealing with limits and the infinite involved very careful reasoning, thus raising the question of what the proper way to reason in mathematics is.

In the mist of this atmosphere, Georg Cantor devised "set theory", a new foundational theory of mathematics. The basic gist was that all mathematical objects could be understood as sets. In 1901 however Bertrand Russell discovered a devastating paradox, now known as Russell's paradox, within set theory itself!

Thus at the dawn of the 20th century mathematicians were very much absorbed with the problem of patching up such paradoxes and creating a more rigorous foundation for mathematics. This general imperative spread to all areas of mathematical research.

The beginnings of computer science were also underway and some mathematicians were looking for a rigorous definition of what it meant for something to be "computable". They had devised a set of rules for creating an infinite class of functions, now called the "primitive recursive functions". It was thought that this infinite class contained all "computable functions". However in 1928 Wilhelm Ackermann devised a 3-argument function that was not a member of the primitive recursive functions, thus proving computability was a broader concept than was thought. As it turns out, Ackermann's function describes a series of operations, and among them one that is very much like tetration. Better yet Ackermann's function can also be used to define an infinite number of operations past tetration.

The 3-argument Ackermann function can be written as:

A(m,n,p)

It can then be defined as follows:

A(m,n,0) = m+n

A(m,n,1) = m*n

A(m,n,2) = mⁿ

When p = 3, the Ackermann function reduces to tetration. When p > 3 , we get one of the higher so called hyper-operators. I'll use the term hyper-operator to refer to any operation along this sequence higher than exponentiation.

Although Ackermann didn't make his function expressly to define the hyper-operators, it none the less was soon adapted to that purpose.

Later mathematicians refined both the notations and terminology for the hyper-operators. The term "tetration" was coined for the 4th operator in 1947 by Reuben Louis Goodstein. He obtained it by combining "tetra" which is greek for 4, and "iteration". He also coined "pentation", "hexation", and so on using the greek prefixes.

The notation:

^pb

for b tetrated to the p, was popularized by Goodstein, but was first used in 1901 by Hans Maurer. While this notation is handy, there wasn't really such handy notations for any of the higher operators. This is where up-arrows come in...

Donald Knuth's Up-Arrow Notation

In 1976 Donald E. Knuth introduced his up-arrow notation for all the hyper operations. The basic idea is to use a special symbol, the up-arrow, "↑", to act as a tally mark for the operation level. He begins by letting the single up-arrow represent exponents:

b↑p = b^p

Tetration is represented by two up-arrows between arguments:

b↑↑p = ^pb

Each successive operation is represented by including one additional up-arrow between arguments. Thus:

b↑↑↑p is pentation

b↑↑↑↑p is hexation

b↑↑↑↑↑p is heptation

b↑↑↑↑↑↑p is octation

etc.

On the internet the "caret variant" is the more popular because of the universality of the caret symbol, where as the up-arrow isn't included in the standard ASCII character set. Using the "Caret notation" we have:

b^p is exponentiation

b^^p is tetration

b^^^p is pentation

b^^^^p is hexation

etc.

Knuth up-arrows, and especially the caret variety are currently the de facto standard for higher operations, at least within recreational mathematics. To further enhance the usefulness of the up-arrow notation a more compact form is also available. By using a superscript in front of a single up arrow we can indicate "any" number of arrows easily:

b↑^cp = b↑↑↑ ... ↑↑p w/c ↑s

provided of coarse that "c" is itself a number we can easily express.

Defining the Hyper-Operators

Knuth's Up-arrow notation is very convenient, and makes it very easy to explain how the hyper-operations actually work.

The basic idea of the hyper-operators can be understood as the continuation of a pattern established by the sequence of addition, multiplication, and exponentiation.

Firstly note that multiplication can be understood as "repeated addition", related by the following equation:

b*p = b+b+b+ ... +b+b w/p bs

Furthermore, exponentiation can be understood as "repeated multiplication":

b^p = b↑p = b*b*b* ... *b*b w/p bs

This implies that we could create another operation to represent "repeated exponentiation".

Using Up-arrow notation we can say that:

b↑↑p = b↑b↑b↑ ... ↑b↑b w/p bs

There is a problem with this definition however. Depending on the order that such an expression is evaluated, it may have different results. For example, take 3↑3↑3. If we evaluate it from left to right we obtain:

3↑3↑3 = 27↑3 = 19,683

If we evaluate it from right to left however we obtain:

3↑3↑3 = 3↑27 = 7,625,597,484,987

Note the radical difference in size of the results! In general, the largest result is always generated by working from right to left. Because of this, it has been generally accepted that tetration should refer specifically to "right to left resolution" of repeated exponentiation. To avoid ambiguity the following recursive definition can be used:

b↑↑1 = b

b↑↑p = b↑(b↑↑(p-1)) for p>1

This is equivalent to right to left resolution. Tetration is only the smallest of the hyper-operators however! A hyper-operator here will refer to any operation past exponentiation. The hyper-operator hierarchy can continue indefinitely by simply defining the next operator as repeated applications of the previous operator.

The next operation after tetration would be "repeated tetration" or "pentation". We can say that:

b↑↑↑p = b↑↑b↑↑b↑↑ ... ↑↑b↑↑b w/p bs

Whenever we resolve such expressions we will always resolve them from right to left to generate the largest possible result. Hexation can be defined as:

b↑↑↑↑p = b↑↑↑b↑↑↑b↑↑↑ ... ↑↑↑b↑↑↑b w/p bs

heptation as:

b↑↑↑↑↑p = b↑↑↑↑b↑↑↑↑b↑↑↑↑ ... ↑↑↑↑b↑↑↑↑b w/p bs

octation as:

b↑↑↑↑↑↑p = b↑↑↑↑↑b↑↑↑↑↑b↑↑↑↑↑ ... ↑↑↑↑↑b↑↑↑↑↑b w/p bs

etc.

In this way we can construct an infinite hierarchy of operations beginning with addition, multiplication, exponentiation, and continuing with tetration, pentation, hexation, heptation, octation, and beyond...

Alternative Hyper-Operator Notations

In addition to the Up-arrow notation, there are other notations that can be used to represent the hyper-operators. There has been attempts to extend the superscript notation used for exponents to arbitrary operation levels. Although the notation b^p for exponents and ^pb for tetration is convenient and generally accepted, continuing this pattern proves somewhat awkward.

The pattern established by these first two operations seems to suggest continuing by going counter-clockwise. Thus we might use _pb for pentation. Although the use of a smaller "p" distinguishes this from exponentiation, the difference is subtle and can easily be mistaken for p^b (p to the b).

In 2004, Mark Cutler_[7] introduced "bar notation" as a solution to this problem. Some general examples are presented below:

The basic idea is to use a series of bars over the base to distinguish the operation level. Pentation can be distinguished from exponents by the inclusion of a bar. Hexation can be distinguished from tetration in a likewise manner. Heptation (b^^^^^p) can be distinguished from pentation by using a double bar, and octation can be distinguished from Hexation using a double bar. In this manner, one can go around counter-clockwise as many times as one wishes, and simply add more bars to distinguish various operation levels.

I myself use a similar system when working with higher operators on paper. Instead of using bars, which can lead to some problems when mixing operations, I use arrows. An example of how I represent various operation levels is presented below:

I rarely use subscript/superscript notations past hexation, however one can continue by letting heptation be represented as b^-->p , and octation as ^p<--b. To continue, one can simply add additional arrow heads to distinguish them from earlier operations. Ennation would be _p<<--b, dekation would be b_-->>p, hendekation would be b^-->>p, and dodekation would be ^p<<--b.

Jonathan Bowers also has an infix notation which allows him to express the higher operators very easily. We can say that:

b<c>p = b^^^...^^p w/c ^s

Robert Munafo uses a very similar notation which uses a circle with the operator number inside. I will use parentheses to represent the circle. In Munafo's system we have:

b(1)p = b+p

b(2)p = b*p

b(3)p = b^p

b(4)p = b^^p

b(5)p = b^^^p

b(6)p = b^^^^p

etc.

Analogous to how we can repeat the caret symbol to continue the operations, we can also use the "+" symbol in the same way. Let:

b++p = b*p

b+++p = b^p

b++++p = b^^p

etc.

The guys over at the "big psi" website have a similar system, except they use an extension of the star notation:

b**p = b^p

b***p = b^^p

b****p = b^^^p

etc.

The use of stars suggests that addition should be considered as the zeroth operation, multiplication as the first, exponents as the 2nd, etc. Although this might at first seem unnatural consider that addition is unlike all the other operators. Unlike all the others it is not composed by repetitions of some lower operation. Multiplication is therefore the first operation built from repeating a lower one, exponentiation the 2nd, tetration the 3rd, and so on. In this way, the use of stars to count the operation level seems appropriate.

There is also other ways of representing the operator hierarchy. For example, it can be represented as a 3-argument function, involving a "base", a "polyponent" and an "operator rank". In Jonathan Bower's Array notation, the operators can be represented as:

<b,p,c> = b^^^...^^p w/c ^s

That covers all of the more popular ways of representing the various operators. There are of coarse more, but this should give you a good familiarity with most of the main notations used.

Tetrational Numbers

Now that we have some understanding of the definition, and common notations for the higher operators, let's explore the power of the smallest of the so called "hyper-operators", tetration. We will be using Donald Knuth's Up-arrow notation for the expressions, and my sub/superscript notations to help relate the sheer scale of these numbers.

Tetration is already powerful enough to exceed Numbers like those described by Archimedes with arguments less than ten! The number being repeated we will call the "base", and the number that tells us how many times to repeat it we will call the "polyponent" (a combination of the words "poly" and "exponent").

"b↑↑p" read as "b tetrated to p" is commonly described as "b raised to itself p times". This however is a terribly ambiguous phrase and is arguably incorrect. "b raised to itself" would naturally mean b↑b. "b raised to itself twice" should mean (b↑b)↑b. Therefore b raised to itself p times should mean:

((...((b↑b)↑b)...)↑b)↑b w/p+1 bs = b↑(b↑b)

This result can be obtained using the exponential law (b↑p)↑a = b↑(pa).

A better and less ambiguous description of b tetrated to p would be "a power tower of bs p terms high". This is generally understood without confusion. Using Up-arrows, let's evaluate some tetrational expressions to get a feel for how they work and how powerful they are.

Firstly, let's make some observations. Whenever the polyponent is 1, we find that the expression is equal to the base:

b↑↑1 = b

Another important observation is to consider what happens when the base is 1. Say we have 1↑↑p, where p>1. Observe:

1↑↑p = 1↑1↑1↑ ... ↑1↑1↑1 w/p 1s

But since 1↑1 = 1, it follows that :

1↑1↑1↑ ... ↑1↑1↑1

1↑1↑1↑ ... ↑1↑1

1↑1↑1↑ ... ↑1

... ... ... ... ... ...

1↑1↑1

1↑1

In other words, 1↑↑p will always reduce to 1, no matter how long the initial expression is! Thus if either the base or polyponent is equal to 1, we won't get very impressive values. Let use consider then the smallest possible non-trivial example, 2↑↑2.

2↑↑2 = 2² = 4

So far not very impressive. However, what if we have 2↑↑3:

2↑↑3 = 2²² = 2⁴ = 16

So far not too big a deal. How about 2↑↑4:

2↑↑4 = 2²²² = 2²⁴ = 2¹⁶ = 65,536

That's big as far as ordinary numbers go, but its nothing surprising. Now let's consider 2↑↑5:

2↑↑5 = 2²²²² = 2²²⁴ = 2²¹⁶ = 2^65,536 =

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7895905719156736

Woah!? Where the heck did that come from!!! It seems so out of left field. The resulting decimal expansion of 2↑↑5 has exactly 19,729 digits! It's approximately equal to 2E19,728. This might seem like an unexpected jump from the previous results ... unless you observe the pattern. Notice that each time we increase the polyponent by 1, we are plugging in the previous result into the exponent of the next. That is:

2↑↑p = 2²^↑↑^(p-1)

Since 2↑↑4 = 65,536, it follows that 2↑↑5 = 2^65,536. Now try to imagine the size of 2↑↑6:

2↑↑6 = 2↑2↑2↑2↑2↑2 = 2↑2↑2↑2↑4 =

2↑2↑2↑16 = 2↑2↑65,536 = 2^{20035299 ... ... ... ... 19156736 (w/19,729 digits)}

≈ 10^6*10^19,727

With an expression as small as 2↑↑6, we have already surpassed Archimedes number, and the largest number in Buddhist literature! Note that 2↑↑6 is so large that I can't even compute its complete decimal expansion, as it literally has more digits than can be stored in the entire observable universe! In fact, the number of digits in 2↑↑6 is roughly 10^19,727. Even if every computer on earth contained a 1000 terabytes and there were 10 billion computers (one for every person on earth and then some) this would still only amount to 10²⁵ bytes, and this would not be enough to even begin to write out the decimal expansion of 2↑↑6.

However, we can determine that the ones place digit must be a "6" because 2↑↑6 must be a power of 16, and every power of 16 ends in "6". This is simple to prove:

2↑↑6 = 2²^65,536 = (2⁴)⁽²^65,536^/4)= 16²^65,534

We can also obtain the tenths place digit fairly easily. Since 36¹⁶ ends in 36, it means that the 16th power of any number ending in 36 must end in 36. 65,536 ends in 36 and is equal to 2¹⁶. It follows that 2¹⁶^Nends in 36 for any positive integer, N. To prove 2↑↑6 ends in 36 observe:

2↑↑6 = 2²^65,536= 2⁽²⁴⁾^16,384 = 2¹⁶^16,384

We can also obtain the first few leading digits. To do so we first multiply the decimal expansion of 2↑↑5 by log2, accurate to at least 20,000 decimal places, to extract the first dozen or so digits after the decimal point. The can be done with a number of online big number calculators_[8]. I obtain 0.32634379468 for the decimal part of the result. Raising 10 to this power gives me 2.12. Therefore the first 3 leading digits of 2↑↑6 are "212". Thus I can say:

2↑↑6 = 2^{20035299 ... ... ... ... 19156736 (w/19,729 digits)} =

212 ... ... ... ... ... ... ... ... ... ... 36 (w/ approx. 6.03*10^19,727 digits)

Buried within this massive 10^19,727 digit number are digits no super computer will ever be able to compute! The best we can do is fray off the edges of the unknown. This is both an exhilarating and frustrating aspect of large numbers. Although we can never know its full decimal expansion, mainly because its just too large, we can get a feel for how large this number is. Here's another way of looking at 2↑↑6. Take the number 2, and square it. You get 4. Call this Stage 1. Square this result and you get 16. Call this Stage 2. Square this result and you get 256. Call this Stage 3. Continue in this manner and you create the following sequence:

256

65536

4294967296

18446744073709551616

340282366920938463463374607431768211456

...

The number of digits roughly doubles each time we square a member of the sequence. This sequence grows fairly rapidly. It's faster than any exponential function (in fact it exhibits the same growth rate as a function of the form f(x) = a^{b^x} ). To reach the number 2↑↑5, which is already insanely large, only takes a mere 16 stages. In order to reach 2↑↑6 however, one must square the number 2 a total of 65,536 times! The result is a number that makes the googolplex look tiny. There just is no wrapping your brain around a number like this. It's already well outside the scope of known reality. And yet we've only just begun, for now we can say that:

2↑↑7 ≈ 10¹⁰¹⁰^19,727

2↑↑8 ≈ 10¹⁰¹⁰¹⁰^19,727

2↑↑9 ≈ 10¹⁰¹⁰¹⁰¹⁰^19,727

2↑↑10 ≈ 10¹⁰¹⁰¹⁰¹⁰¹⁰^19,727

etc.

2↑↑7 is equivalent to taking 2 and squaring it 2^65,536times, 2↑↑8 is equivalent to taking 2 and squaring it 2^{2^65,536}times, or 2↑↑6 times! 2↑↑9 is equivalent to taking 2 and squaring it 2↑↑7 times, 2↑↑10 is equivalent to taking 2 and squaring it 2↑↑8 times, etc.

The growth of the sequence, 2↑↑1, 2↑↑2, 2↑↑3, 2↑↑4, etc. goes way beyond exponential growth. In fact its faster than the growth of iterated squaring. We can call it "tetrational growth". It's the same kind of growth exhibited by repeated applications of the "plex function" and its also roughly equivalent to the growth rate of the "megafuga".

Let's try a larger base now. Let's begin with 3↑↑2:

3↑↑2 = 3³ = 27

Again this is nothing spectacular, but its only the starting value. Next try 3↑↑3:

3↑↑3 = 3³³= 3²⁷ = 7,625,597,484,987

This is a number we will encounter again and again as we climb up the operator hierarchy. 7 trillion is already pretty big, being a small astronomical class number, but wait 'til we get to 3↑↑4:

3↑↑4 = 3³³³ = 3³²⁷ = 3^{7,625,597,484,987}

The result of 3^{7,625,597,484,987} is a number with about 3.6 trillion digits! This is much much larger than 2↑↑5. So large in fact that I can't even compute it with a big number calculator. The resulting number would take up about a Terabyte! I wouldn't even be able to upload such a number to this site because it would take up too much memory! That's huge already, and yet we've just begun. Next we have 3↑↑5:

3↑↑5 = 3³³³³ = 3³³²⁷ = 3³^{7,625,597,484,987}

This number is approximately 10^{10^(3.6*10^12)}. Now imagine how big 3↑↑6, 3↑↑7, 3↑↑8, 3↑↑9, 3↑↑10 etc. are!

You'll notice the values were greater than when we used a base of 2. What if we use 4 instead?

Next we can look at:

4↑↑2 = 4⁴ = 256

Not bad for a starting value. Next comes:

4↑↑3 = 4⁴⁴ = 4²⁵⁶ =

13,407,807,929,942,597,099,574,024,998,205,846,127,479,365,820,592,393,377,723,561,443,721,764,

030,073,546,976,801,874,298,166,903,427,690,031,858,186,486,050,853,753,882,811,946,569,946,

433,649,006,084,096

That's approximately 1.34E154 or 13 quinquagintillion. 4↑↑3 is already bigger than a googol. Compare it to 3↑↑3 (7,625,597,484,987) and 2↑↑3 (16).

Next up would be:

4↑↑4 = 4⁴⁴⁴ = 4⁴²⁵⁶ =

4^{13,407,807,929, ... ... ... ... ,006,084,096 (w/155 digits)}

This is a number already larger than a googolplex. 4↑↑4 is a number Jonathan Bower's used to refer to as tritet. However he has assigned that name to a much larger number now. We will therefore refer to 4↑↑4 as tritet Jr. to distinguish it from the larger version. A proof that Tritet Jr. is larger than a googolplex is fairly straight forward:

Tritet Jr. = 4^^4 = 4^4^4^4 = 4^4^256 = 4^16^(256/2) = 4^16^128 =

16^((16^128)/2) > 16^((16^128)/16) = 16^16^127 > 10^10^100 = Googolplex

Therefore: Googolplex < Tritet Jr.

4↑↑5 = 4↑4↑4↑256, is a number larger than Skewes' Number! And of coarse we can continue with 4↑↑6, 4↑↑7, 4↑↑8, 4↑↑9, 4↑↑10 etc.

Now let's try base 5:

5↑↑2 = 5⁵ = 3125

A sizable starting value. Next would be:

5↑↑3 = 5⁵⁵ = 5³¹²⁵ =

19110125979454775203564045597039645991980810489900943371395127892465205302426158030120

59386519739850265586440155794462235359212788673806972288410146915986602087961896757195

70183928166033804761122597553362610100148265112341314776825241149309444717696528275628

51967375143953575424790932192066418830117871691225524210700507090646743828708514499502

56586194461543183511379849133691779928127433840431549236855526783596374102105331546031

35372532574863690915977869032826645918298381523028693657287369142264813129174376213632

57303216452829794868625762453622180176732249405676428193600787207138370723553054463561

53946401185348493792719514594505508232749221605848912910945189959948686199543147666938

01303717616359259447974616422005088507946980448713320513316073913423054019887257003832

98012460501970134673971759090273894939238173157869968458997947810680428224360937839463

35265422815704302832442385515082316490967285712171708123232790481817268327510112746782

31741098588868370852200071173349225391332230075614718042900752767779335230620061828601

24552542430610068948054465847048206509826643193609603887362585107470743406362869765767

02699258649953557976318173902550891331223294743930343956161328334072831663498258145226

86200430779908468810380418736832480090387359621291963360258312078167367374253332287929

69072054905956214068888259912445818423795978634764843156737609236250903715117989414242

62270220066286486867868710182980872802560693101949280830825044198424796792058908817112

32719230145558291674679519743054802640464685400273399386079859446596150175258696581144

75685100415686877309037124825353438392853975987494584970500382250124892840018265900562

51286187629938044407340142347062055785305325034918189589707199305662188512963187501743

53596028220103821161604854512103931331225633226076643623668829685020883949614283048473

91139916696226499485636852347128732947966808845094058939511046509441379095022765456531

33018670633521323028460519434381399810561400652595300731790772711065783494174642684720

95613464732774858423827489966875505250439421823219135722305406671537337424854364566378

20457016545932181540535483936142506644985854033074664685418901481343477146503150379541

75778622811776585876941680908203125

This number contains exactly 2185 digits! Compare 5↑↑3 to 4↑↑3, which only had 155 digits. 3↑↑3 only had 13 digits, and 2↑↑3 was only had 2. As you can see the values have grown rapidly.

Next we can have:

5↑↑4 = 5⁵⁵⁵ = 5⁵³¹²⁵ =

5^{191101259 ... ... ... ... 908203125 (w/2185 digits)}

Next we would have 5↑↑5, 5↑↑6, 5↑↑7, 5↑↑8, 5↑↑9, 5↑↑10, etc.

Let's try base 6:

6↑↑2 = 6⁶ = 46,656

6↑↑3 = 6⁶⁶ = 6^46,656 =

265911977215322677968248940438791859490534220026992430066043278949707355987388290912134229290617558

303244068282650672342560163577559027938964261261109302039893034777446061389442537960087466214788422

902213385381919290542791575075927495293510931902036227198983057853932880763319683450709063994613113

899946027767197828941253221232925203296051182048791364008389549044365792095626712629192228922460944

103484957826646121969087967503992005139138817452525944319386504163034800032329572923169272580812038

862895645133020319941418621365459381247897039074948526861497196424842856278264872081854849393729902

569313271916554406043106913602901064055895953421212374607966076698330673506053248292555308212118869

609719907808556324604601321610265543147645240039660236473266424625276737459325658213465426520922004

237014324045664647970387863402847679951630130232058962063799462389599996237653062587414899860913010

693318793720947688749687131107213871394369489435728806106155770846190228969206308973600818603623114

117110987742542396241094684938897236922298685698434294496424701408029483172932386618629065996040198

226842823646538422586113704883857989007702229553937496834838684871215875596667442178077474619334121

028488932766083813638995310139077457475830177551325338144732131509087834262021410745896573377350678

674850098385414388602041091113739704476471248961781346576074509649789560993834854196961759483379704

895730837108476073243820752302407048567636674324794945261956856304309751928127990186556725142283682

860915312621646073453158008803215388643682616411219867012372209679528053501516815952615060246385651

031976562807101388014614012727625982704958770707410574826165853317280489979664766579951201612185779

455431473280794446393767971020284196670959037332691306485094879711657607592738752654869501755331152

983075423102199223873789151170394698211199449920074131308554214681213100553681303428933484193987068

952218348302678831827921107401559352065829092083220235853420468576169753257004117360848240587083162

510683198727366804395473509269321561136071328831038756345050322110155370605205773322174682858545420

089112607776943756235718685791083329489150388849193534368560214077259275017330081801022785506555813

374097953179402013552081883260678201008163982491139468985286586839525924516399022060889167034026307

259495537031036865710606177193275628159515876151135875206383384565803345010001967990900297897721047

625041503686189053938223646710134213940404719809893782173642263134508441550849287889447186844367723

211381684596533430978485492183483286280866059762937533235790152346913482052907574065476985428074987

463200630523320192960729488033220100471728132419843750933418994651827766195280398535457879210590938

725541731207476215251178380018544287270261438480751698862350923128502235087740792752699447916376634

508067462510897466136541982989588498885556218913512465902304156653376336871566688387251967036152900

780064068602888069931935207999310689140776424351948418730004965515832221710968997431459333784941295

229440309655511855867028445035198762119848288829295380202219936661078120427828843001276230841951275

304807279299943098104127984653113848966391547080166051152516695839546951857411342509029205658037023

718240037774832603354543093511500118479701534297400745136744846115854618011592988204941297108616076

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446000027476477191798414513254581650275367333019821278843172709465750692752171808594901217509501207

417421617586944497161282705033574579740947969818330006129341292987484417693746915156267819694419069

070551740834127011011653057480954762302376223721506011117278656836781687467642565660245054195922383

015063540983302976564585032232212348328492684438517686135813543777864727386998363224693825363151801

851106716453198841449354389812191334113521409843330236524538792076198600969222577026478681492227501

540566298351707639566416336862868467950597297476836923209472900922293702286186940890332878669151652

070026861441427046793265655884082960714097320267268049408177667445088496534818017853672361822688485

733913385400255456567934282631484871977114047875708058685436192476663298724761448156342842830402602

505021574204296564158930366078365481437442997463536675091000411699963360673064732160583745460680016

965456758566096290515270920164587535961300182886653830784412200942190677050444291047244003389456102

541814425739429044704330880008696148653836354976484820864386333590772821957611006776947872616515008

395331769033371947624815253698065061836502368055443145909315203440556148456886065074331669131896824

759958256508800018017148943114323386161400405917446760035859282703814428874771624962132036197765252

454933467861555487197397619695431510692383342974202667545414955466938030160673169706724462625855137

085164539132257730434810596984801783923003483105541564030100015743183630882716196703166554265650963

071038792205396803159204779044332369316517777859190655421620788215534975404890121605354486244302325

493234970592691383707828560134937722533674320536537516987734918377779951255422227597271611802930324

522204021758057752412497127455218252920942278428479471698539697813452696767907179228508491839775153

628228243052306992659549440199892772061036616473355839050070498310473487182064267545001577079206261

540203265717301064221019125068350266791249399179609291090367076356849258440718452955522072597481584

063311817403113771618731801958998128677729597378367094064062033675539166479397859978760723803175210

934594321121324518477664328229171706732939059061137733423725445425333376420568500011937009911220804

854369888817399953540372960084640042151028030142531122831961619571524522422266809290751616197961883

530242609434051902857830543763397061153243582494210503092681521241839983612388292362168558481080453

724926008984522786772923739257464771832909220710235406334318389309363173179239077529730921381408865

699869199075222777463146266117543342519940204770331943339078678783474761260772613750616583256430398

445294250039146099139399860652378000419745042078241181875123263807183683939638150929032237564582040

219925725544066194365465456072775046043019953993748322983683489516208150738424224196778813811196748

091317343045579356718550325307737095111433942009322636455898150323810716497918542945385958749141645

826113403821601639049146905546228789873935871271101782517043060004493333790352302613291297240328564

529779500158901992057657168980489272863072158461749965174823389901373567049766197477456695811834009

402652112653589115553592994292565814973047995369272297261017111119344511199618772851626644393751712

810887048244077398755032272241691310649437516764253759232572706022781934855221122174485770810461351

760722786628516143158455550642611470543692693710220729528959120340445210529788692903692277110929238

083192013051552699177390638979085275991136415947758803248065658708433324533317375936823619422963019

751271123958949266745154287327508514137369673627508864714146824134776235394625172665199539967765285

667592328910514509788672574445865951869081323812569736008479758269334749025290058433197876943284080

580745193611989429990468280533178199550083230163227944716082271056280731820677005355961868362024509

891022139820125623841787690585309928612267303595625824508436018544241421864042750273384716167606981

280931250329649599434752052971462973363672222035938392416581549006207453756828721700346944745735294

757240987973604996890528648783086812325888847405838796103442374669651862064044434887259551215107382

960629850522578340713349255321774853789028601004245977335322044392190426364996651123770623213960923

374849645936962093842457558760552195368235329700552845151069077200413182755865827314444969724970410

721026079430403668206368143382579398573856744555443907200016627368907470972273592217094439747656651

587759921449935310041011278838427588582684058312705649787648119492683806365037319800437060302992518

823221372565574622697024120942134211743868144784037217825939334570966080175296107167014761063573878

764298769872073630597860069568435695489708455471417296025446297551851567364487849328103367222701989

883802494786171436302954969918207348486513230712677788751869942315794571999554857157496720173838789

205258104548081079810451185601751030835725087918474499339432087503990193076470750547948775977855358

485409413012204167161008344913898947567902878642168826811118376878528543846712653149082404118060490

652755864988462101460370601811780023726130518568384806414483578827026111605055591597332741401022568

196421828157231909693624098089055609103548620026370881684170231338454875054885411126510070161725747

437966591249357866528886086706489298876392642744160118970606956928665435440332410837187184771741684

096117675991110268740224339044107121286235825797872641933884252895920171694272374852330986381424704

405554720432213808590250910052515845368806455661289841979888898496284938747083462857147394174768624

657638140589204233692069874340266435896620604197215859497775734519355596485093355986281552555553876

637144264112107050854226245879960823641485704807633869148606767226023183261796956540334262005758333

261045868501266693308281023681427721918822056924812008821204534830044654602066446687314813657144604

524271412597676737460939657352003890495077862128279292449414208247189079648329501330142509407727347

591066686369332364802156261024141032358374502783755634343861798500282951879660646660629481466668133

405040220522200280171108546969716343594555462039079692851957417002424888105893065048277613075277549

575313245632715147314669715412024040329823708192959224423808246692757409437349315576119847840425210

718798954425192388505562440936682434151237375513226295724498075849529979937732877694105147208871848

111189812916480877473371785496167602764408871411247754224503353946883319174406712144432837791420728

008577200624913032381852272196096837422198435262420299836299729271608009378269710389890410171747674

715111904373088567457435014594879091368705600688007101215820026528663115527646246147728140985412548

401456814934852708715538761594295805551267452651180429333684268516031849931062195799522678787157577

600563347342341331836322988591894412992525937987173318134193990210689518078163555907235238853815379

773691546246995995608375329977513075931839086826932047597565683018282299362258470622934630351415117

782501732398783298437154204578260937966327569667172864420534855709163219251858795348798337006617691

315353399699461860343190483637203756593186066419144028800227652609457125666376633516790109663620347

636975698740142905735080137176798745760015315507027808438442428818313304866101754151500082701256096

763943176106018568210388507516985167245380074711202913270160244940132970964370790189988317746584257

548400289775525880140679745537412020982395367711258261144075551696849205873413439885660637900372629

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625524292260940347541700734738562142047647305723053881410229273825705072603260789109530288429356116

848550927749075462006726604534197266327419654040613692101581093498295056301997299527017704165195710

476563419145822684533889701667936555956484762791914445656139676481352345551216386449858109988937265

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450900403092885758378292759533393050237320140953727613312702323636738951262974123205418288741397543

451587603552646624247785552013314031192733246010912730484816415238192195349735466749770728067928503

549656651656073233219678107283595565211925536596476285568332871576455097190154739727932800047918608

913057369146767557746792528779299399728183694415823499687330127087758737047474994989156500107530548

160976123883318437794772844111307075048213457629500443809360274691131620176241365042370710483640300

663273050001492189318116979891143885697328268812532306505266364178977237563209358836554989475411347

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077079583376130225517380758198285144994545668777118410072441336848088288912710026485935090558379495

040979631540762442690074319825071132295476718724755365322307056654838102078522716992464692185740996

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584094989943337553566464573425551343857891020619203587260757327436584898400993791510054147654356650

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942882325081614482406954734609046200523990247481209299536211933236590300256653077831592975673453496

541090621701855413666597584167024844986010971736908181470245722975460872681422559593960315974723533

197673148835514287190868362599834270400499224298567901025841345491864396262256475645201871059473180

234586620834044589586160820199442461923914785856105411868402806753113119614849401467281772004893587

038668657846417413716863007212681900774342602539212792220982908691560673641270058922493307705207693

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231544532257474116257191040537530672362313491705629432886056717863878656

That's a number with 36,306 digits! So far this is the largest decimal expansion I've displayed on this page, this number being almost as big as the square of 2↑↑5.

Next would be:

6↑↑4 = 6⁶⁶⁶ = 6⁶^46,656 =

6^{265911977 ... ... ... ... 7863878656 (w/36,306 digits)}

Of coarse we could continue with 6↑↑5, 6↑↑6, 6↑↑7, 6↑↑8, 6↑↑9, 6↑↑10, etc.

With base 7 we have:

7↑↑2 = 7⁷ = 823,543

7↑↑3 = 7⁷⁷ = 7^823,543

7↑↑4 = 7⁷⁷⁷ = 7⁷^823,543

etc.

With base 8 we have:

8↑↑2 = 8⁸ = 16,777,216

8↑↑3 = 8⁸⁸ = 8^16,777,216

8↑↑4 = 8⁸⁸⁸ = 8⁸^16,777,216

etc.

With base 9 we have:

9↑↑2 = 9⁹ = 387,420,489

9↑↑3 = 9⁹⁹ = 10^369,693,099

9↑↑4 = 9⁹⁹⁹ = 10¹⁰^369,693,099

etc.

Tetrational numbers, as you can see, are already staggeringly powerful. They are already enough to keep up with the largest numbers used even in the most esoteric realms of science. Yet these numbers pale in comparison to what the next hyper-operator can do ...

Pentational Numbers

Just like tetration we can eliminate trivial cases. First off b↑↑↑1 = b. Furthermore 1↑↑↑p = 1 based on the fact that 1↑↑p = 1, & 1↑↑↑p = 1↑↑1↑↑1↑↑ ... ↑↑1↑↑1. So we again begin with 2↑↑↑2 as our first non-trivial example:

2↑↑↑2 = 2↑↑2 = 2² = 4

Oddly, it turns out that 2↑↑↑2 is no bigger than 2↑↑2. Stranger still, although we are working with pentation, 2↑↑↑2 is trivially small. What about if we increase the polyponent by 1:

2↑↑↑3 = 2↑↑2↑↑2 = 2↑↑2²= 2↑↑4 = 2²²² = 65,536

Not bad, but not all that impressive either. Now prepare for a shock:

2↑↑↑4 = 2↑↑2↑↑2↑↑2 = 2↑↑2↑↑4 = 2↑↑2²²² = 2↑↑65,536 =

2²²²^...^...²²²² w/65,536 2s

That's a power tower of 2's 65,536 terms high!! We've just begun and we've already left every tetrational number in the previous section in the dust! Next would be ...

2↑↑↑5 = 2↑↑2↑↑2↑↑2↑↑2 = 2↑↑2↑↑2↑↑4 = 2↑↑2↑↑65,536 =

2↑↑(2²²²^...^...²²²² w/65,536 2s) =

2²²²^...^...²²²² w/2²²²^...^...²²²² w/65,536 2s 2s

That's a power tower of 2's with as many terms as a power tower of 2's 65,536 terms high. Hopefully you can now see the pattern. With pentation, each time we increase the polyponent by one we have a power tower with as many terms as the previous expression. So 2↑↑↑6 = 2↑↑2↑↑↑5, which means a power tower of 2's with 2↑↑↑5 terms. 2↑↑↑7 would be a power tower with 2↑↑↑6 terms, 2↑↑↑8 a power tower with 2↑↑↑7, etc.

Now let's try base 3:

3↑↑↑2 = 3↑↑3 = 3³³ = 7,625,597,484,987

There is that number again. Next up is a staggering number:

3↑↑↑3 = 3↑↑3↑↑3 = 3↑↑3³³= 3↑↑7,625,597,484,987

3³³^...^...³³³ w/7,625,597,484,987 3s

These are the kinds of numbers that usually break ones notion of infinity and greatly expand it when first encountered! And yet as large as a number like this seems, its really quite tiny when compared to the next one in the sequence:

3↑↑↑4 = 3↑↑3↑↑3↑↑3 = 3↑↑3↑↑3³³= 3↑↑(3³³^...^...³³³ w/7,625,597,484,987 3s) =

3³³^...^...³³³ w/3³³^...^...³³³ w/7,625,597,484,987 3s 3s

Just getting your head around the idea is tricky. The power tower doesn't have 7 trillion 3s, no ... it has as many 3's as a power tower of 7 trillion 3s! Dizzy much?!Good, 'cause we've only just begun:

3↑↑↑5 = 3³³^...^...³³³ w/3³³^...^...³³³ w/3³³^...^...³³³ w/3³³ 3s 3s 3s

3↑↑↑6 = 3³³^...^...³³³ w/3↑↑↑5 3s

3↑↑↑7 = 3³³^...^...³³³ w/3↑↑↑6 3s

3↑↑↑8 = 3³³^...^...³³³ w/3↑↑↑7 3s

etc.

Okay now for base 4:

4↑↑↑2 = 4↑↑4 = 4↑4↑4↑4 = 4↑4↑256

4↑↑↑3 = 4↑↑4↑↑4 = 4↑↑(4⁴²⁵⁶) =

4⁴^...⁴⁴ w/4⁴²⁵⁶ 4s

4↑↑↑4 = 4↑↑4↑↑4↑↑4 = 4↑↑(4⁴^...⁴⁴ w/4⁴²⁵⁶ 4s) =

4⁴^...⁴⁴ w/4⁴^...⁴⁴ w/4⁴²⁵⁶ 4s 4s

4↑↑↑5 = 4↑↑(4↑↑↑4) = 4⁴^...⁴⁴ w/4⁴^...⁴⁴ w/4⁴^...⁴⁴ w/4⁴²⁵⁶ 4s 4s 4s

4↑↑↑6 = 4↑↑(4↑↑↑5) = 4⁴^...⁴⁴ w/4↑↑↑5 4s

4↑↑↑7 = 4↑↑(4↑↑↑6) = 4⁴^...⁴⁴ w/4↑↑↑6 4s

4↑↑↑8 = 4↑↑(4↑↑↑7) = 4⁴^...⁴⁴ w/4↑↑↑7 4s

etc.

Now base 5:

5↑↑↑2 = 5↑↑5 = 5⁵⁵³¹²⁵

5↑↑↑3 = 5↑↑5↑↑5 = 5↑↑5⁵⁵³¹²⁵=

5⁵^...⁵⁵w/5⁵⁵³¹²⁵5s

5↑↑↑4 = 5↑↑5↑↑5↑↑5 = 5↑↑(5⁵^...⁵⁵w/5⁵⁵³¹²⁵5s) =

5⁵^...⁵⁵w/5⁵^...⁵⁵w/5⁵⁵³¹²⁵5s 5s

5↑↑↑5 = 5↑↑5↑↑5↑↑5↑↑5 =

5⁵^...⁵⁵w/5⁵^...⁵⁵w/5⁵^...⁵⁵w/5⁵⁵³¹²⁵5s 5s 5s

5↑↑↑6 = 5⁵^...⁵⁵w/5↑↑↑5 5s

5↑↑↑7 = 5⁵^...⁵⁵w/5↑↑↑6 5s

5↑↑↑8 = 5⁵^...⁵⁵w/5↑↑↑7 5s

etc.

I think you get the idea. Things get really insane with the next up-arrow ...

Hexational Numbers

As usual we can begin with the smallest non-trivial case:

2↑↑↑↑2 = 2↑↑↑2 = 2↑↑2 = 2² = 4

Notice that we keep getting 4, no matter how many up-arrows we have. This follows from the definition. It follows from:

b↑^cp = b↑^c-1b↑^c-1b↑^c-1... b↑^c-1b w/p bs

that...

2↑^c2 = 2↑^c-12

This means that the number of up-arrows will constantly tick down until we are left with 2↑2 = 4. Because of this 2↑^c2 can also be considered a trivial case of sorts. Let's try increasing the polyponent by one:

2↑↑↑↑3 = 2↑↑↑2↑↑↑2 = 2↑↑↑4 = 2↑↑2↑↑2↑↑2 = 2↑↑2↑↑4 =

2↑↑65,536 = 2²^...²²w/65,536 2s

That's a huge leap from 4! Of coarse we've already seen numbers like this with pentation. In order to transcend the previous operator class we need at least a polyponent of 4 for a base of 2, and at least a polyponent of 3 for any higher base. Get ready for something insane:

2↑↑↑↑4 = 2↑↑↑2↑↑↑2↑↑↑2 = 2↑↑↑(2²^...²²w/65,536 2s) =

2↑↑2↑↑2↑↑2↑↑2↑↑ ... ... ... ... ↑↑2↑↑2↑↑2↑↑2↑↑2 w/(2²^...²²w/65,536 2s) 2s =

2²^...²²w/2²^...²²w/2²^...²²w/ ... ... ... ... 2²^...²²w/2²²² w/2²w/2 2s ... ... 2s

where there are "2²^...²²w/65,536 2s" power towers, the 1st being "2", the 2nd being "2²" etc.

In case that's just a bit much, let me clarify. To get 2↑↑↑↑4 begin with "2", and call it Stage 1. Now have a power tower with that number of 2s: 2². That's Stage 2. 2^2 = 4. Now have a power tower of that many 2s: 2^2^2^2 = 65,536. That's Stage 3. Now have a power tower of 65,536 2s. That's Stage 4. Continue in this manner until you reach Stage "A power tower of 2's 65,536 terms high". At this point, as you can see, power towers are becoming cumbersome as an explanatory tool for these numbers! We can gain some clarity by switching to what I like to call tetra-towers.

First off recall that:

^pb = b↑↑p

Now try to imagine the following transformation:

^pb = b^b^...^b^bw/p bs

I like to imagine the p passing through the b, changing direction, and then leaving a trail of b's. This notation now provides us with a convenient shorthand for power towers. A "Tetra-tower" is analogous to a power tower, except that it is formed by iterated tetration instead of exponents, and it leans to the left instead of the right. Expanding a tetra-tower into its equivalent power tower is fairly intuitive. Here is an example:

²²²2 = ²²²2 = ⁴²2 = ²²²²2 =

^65,5362 = 2²^...2²w/65,536 2s

Now with that in mind, we can use tetra-towers to make comprehending 2↑↑↑↑4 a little easier:

2↑↑↑↑4 = 2↑↑↑2↑↑↑2↑↑↑2 = 2↑↑↑2↑↑↑4 = 2↑↑↑(2↑↑2↑↑2↑↑2) =

2↑↑↑(²²²2) =

²^...²2 w/ ²²²2 2s

The expression certainly looks a lot more tame. Try to remember however that the tetra-tower to the left has ²²²2 terms. As the tetra-tower is collapsed, the last few 2s on top will be replaced by more and more absurdly long power towers. By the time it reaches the end, we've gone beyond what we can really describe. Then of coarse the power tower must be collapsed, leaving in its wake decimal expansions of unfathomable size!!! Would Archimedes be thrilled or horrified?! Perhaps both. And of coarse we can increase the polyponent:

2↑↑↑↑5 = ²^...²2 w/²^...²2 w/²²²2 2s 2s

2↑↑↑↑6 = ²^...²2 w/²^...²2 w/²^...²2 w/²²²2 2s 2s 2s

etc.

Let's see what happens with base 3:

3↑↑↑↑2 = 3↑↑↑3 = 3↑↑3↑↑3 = ³³3 = ³³³3 =

^{7,625,597,484,987}3 = 3³^...³ w/7,625,597,484,987 3s

Now let's try the next one:

3↑↑↑↑3 = 3↑↑↑3↑↑↑3 = 3↑↑↑(³³3) =

³^...³3 w/³³3 3s

Here is a visual representation of 3↑↑↑↑3 using power towers:

The number 3↑↑↑↑3 has the special designation G₁. As you might guess, the polyponent for hexation determines the number of vertical layers in the above stack. So 3↑↑↑↑4 would require a series of power towers 3↑↑↑↑3 terms long, beginning with 3, 3^3^3, etc. 3↑↑↑↑5 would require a series of power towers 3↑↑↑↑4 terms long, etc.

Changing the base to 4 would simply mean replacing all instances of 3 with 4. Numbers this large are way beyond anything used outside of mathematics, and yet mathematicians have actually worked with numbers larger even than this! Now let's take this process of ascension to its natural conclusion...

Heptation, Octation, ...

We already know that 2↑↑↑↑↑2 = 4 and 2↑↑↑↑↑3 = 2↑↑↑↑4. These numbers are nothing new. So let's begin with:

2↑↑↑↑↑4 = 2↑↑↑↑2↑↑↑↑2↑↑↑↑2 = 2↑↑↑↑2↑↑↑↑4 = 2↑↑↑↑(2↑↑↑2↑↑↑2↑↑↑2) =

2↑↑↑↑(2↑↑↑2↑↑↑4) = 2↑↑↑↑(2↑↑↑(2↑↑2↑↑2↑↑2)) = 2↑↑↑↑(2↑↑↑(²²²2)) =

2↑↑↑↑(²^...²2 w/ ²²²2 2s) =

2↑↑↑2↑↑↑2↑↑↑2↑↑↑ ... ... ... ... 2↑↑↑2↑↑↑2↑↑↑2 w/(²^...²2 w/ ²²²2 2s) 2s =

²^...²2 w/²^...²2 w/ ... ... ²^...²2 w/ ²²²2 w/²2 w/2 2s ... ... 2s

where there are "²^...²2 w/ ²²²2 2s" tetra-towers beginning with "2", then "²2" etc.

Of coarse, things are again looking too complicated. So now I introduce my penta-towers. Oddly they aren't really towers, since they head downwards, but let's not get too hung up on terminology. Firstly I define:

_p<--b = ^b^...^bb w/p bs

We now have a convenient shorthand for tetra-towers!! Once again I imagine the p passing through the b, changing direction, and expanding to the upper left to create a tetra-tower. Here is a simple example of a penta-tower to help you get a feel for how this works:

_2<--_2<--_2<--2 = ²_2<--_2<--2 = _4<--_2<--2 =

²²²_2<-- 2 = ⁴²_2<--2 = ²²²²_2<--2 = ^65,536_2<--2 =

₍₂²^...²_{w/65,536 2s)<--} 2 =

²²^...² 2 w/2²^...² w/65,536 2s 2s

At this point things are beginning to get quite abstract, as we not only have to deal with expanding this example to tetra-towers, must must also use power towers to have a clear understanding of what we are doing. If you have read through the earlier examples however, some of the results should already be becoming familiar to you: For example, that 2²²² = 65,536 without any computation. Although my Counter-Clockwise notation is pretty effective at displaying the process of these numbers there are some difficulties with mixed hyper-operations. For example, when a power tower is the bottom most polyponent to a penta-tower. In this case the notation would technically overlap itself since power towers and penta-towers are traveling in mutually opposed directions. The power tower would get "written over" the penta-tower. In practice however, this can be avoided by using a set of parenthesis for the bottom most polyponent. In general however the notation does have the tendency to coil about itself. This becomes more apparent past hexation, but the beginnings of the problem can be seen as early as pentation. None the less, I have found this notation is the best way to extend the subscript/superscript notation established by b^p for exponents and ^pb for tetration.

Now let's try to apply our penta-towers to the problem of expanding 2↑↑↑↑↑4:

2↑↑↑↑↑4 = 2↑↑↑↑2↑↑↑↑2↑↑↑↑2 = 2↑↑↑↑2↑↑↑↑4 = 2↑↑↑↑(2↑↑↑2↑↑↑2↑↑↑2) =

2↑↑↑2↑↑↑2↑↑↑ ... ... ↑↑↑2↑↑↑2 w/(2↑↑↑2↑↑↑2↑↑↑2) 2s =

_2<--_2<--..._<--_2<--2 w/_2<--_2<--_2<--2 2s

These numbers are so big that they are barely comparable to what most people imagine as "big numbers". We can barely make out how the power tower is constructed, let alone what the decimal expansion would be like. Forget understanding these kinds of numbers on any terms but their own!!

Let me take this opportunity to point out a common rookie mistake. Some peoples reaction to this stuff when they first encounter it is to forestall shock or horror by preempting it with an "even larger number". What sometimes happens is that a person has a vague sense that something "very large" is being described, but they don't quite have their sea legs yet. That is to say, they don't fully understand how the notation works. They will then fall back on something more familiar that still seems "very large" to them. Usually it takes the following form: Take the largest number defined in what they've just read, for example 2↑↑↑↑↑4, and then come up with a retort of the form, '1 followed by THAT number of zeroes!'. That's got to HUGE right?! Yes and no. Obviously 10^N is always larger than N. However "how much larger" depends on the perspective we take. In a sense 1 followed by 2↑↑↑↑↑4 zeroes isn't really much larger than 2↑↑↑↑↑4. How is that possible?! If this seems inconceivable then you really haven't quite understood the vastness of 2↑↑↑↑↑4.

Firstly let me clarify what large number aficionados mean when they say a number is "slightly larger", and "much much larger", because it doesn't mean the same thing as in ordinary language. Contrary to what you may have thought, our perception of number is largely logarithmic, not linear. To help you grasp this, consider the difference between the numbers, 3 and 47. Would you say that 47 is much bigger than 3? Of coarse, it's "MUCH bigger". Now compare the numbers 54069489061 and 54069489105. Would you say that 54069489105 is MUCH bigger than 54069489061? Of coarse not, you'd say they are basically the same. That is to say, it is only "very slightly larger". However the difference between 3 and 47 and 54069489061 and 54069489105 is both 44. So in theory the difference in size should be the same, right? Our number sense however just doesn't work like that. We rate size based on relative scale and proportion. The best way to understand this is by using a ratio. The reason why we say 47 is much larger than 3 is because it's over 15 times its size. The ratio can be found by dividing 47 by 3 ie. 47/3. The reason why we say 54069489105 is only slightly larger than 54069489061 is because 54069489105 is only 1.0000000008 times larger than 54069489061. So basically when an ordinary person says a number is MUCH larger than another,they mean its a large multiple of that number. When an ordinary person says a number is slightly larger, they usually mean that the number is greater than it, but the difference is very small fraction of the original value. By this reckoning 10^N is always VASTLY MORE than N no matter how large N is. In fact the ratio (10^N)/N is ALWAYS increasing! Try it for the counting numbers {1,2,3,...}, and you'll see how rapidly it grows. The first few values you obtain are 10/1 = 10, 100/2 = 50, 1000/3 = 333+(1/3), 10000/4 = 2500,etc. Generally the value increases by 1 order of magnitude each time you increase N by 1. So if this is so, why can I say that 1 followed by 2↑↑↑↑↑4 zeroes isn't much larger than 2↑↑↑↑↑4, in any sense?!

The reason is because our logarithmic sense of number goes beyond mere ratios. Because very large numbers, like the ones were dealing with, can be no more than mathematical abstractions to us, we begin to judge size relative to the notations used. Now let's consider 1 followed by 2↑↑↑↑↑4 zeroes in a new light. Using the notations now at our disposal, we can say "1 followed by 2↑↑↑↑↑4 zeroes" is just 10↑2↑↑↑↑↑4 resolved from right to left. We've already established that 2↑↑↑↑↑4 is difficult to expand into a tetra-tower, let alone a power tower. Yet according to the definitions we know that eventually 2↑↑↑↑↑4 must expand to an unfathomably vast power tower of 2s. Although we can't say how many terms the power tower would have,(not even in terms of powers towers describing power towers!), we can say that it is some very vast number, M. Now consider this:

2↑↑↑↑↑4 := 2²^...² w/M 2s

implies...

10↑2↑↑↑↑↑4 = 10²^...² w/M+1 terms

In essence, having 1 followed by 2↑↑↑↑↑4 zeroes, is like adding a single 10 at the bottom of a VAST power tower of 2s. From this point of view, it barely makes a difference! In fact, it would be much much better to put the 10 way up at the top, but as far as we are concerned that still wouldn't make much of a difference. As you can see, it now becomes very very easy to imagine numbers way vaster than 1 followed by 2↑↑↑↑↑4 zeroes, or even 1 followed by THAT number of zeroes. How about a power tower of 2's 2M terms high, M^2 terms high, or M^M^...^M w/M Ms terms high! You've just jumped out of one class of thinking and into a completely new one. Now with the merest stroke of your imagination you can trump someone whose still only thinking in terms of decimal expansions. But wait, it get's worse, because even if you imagined a power tower of 2s 2↑↑↑↑↑4 terms high, it STILL wouldn't be much larger than 2↑↑↑↑↑4. What?! The reason is similar. Simply recall that 2↑↑↑↑↑4 must expand to some vast tetra-tower of 2s. Say such a tetra-tower has "m" terms. Now consider:

2↑↑↑↑↑4 := ²^...²2 w/m 2s

implies...

2↑↑2↑↑↑↑↑4 = ²^...²2 w/m+1 2s

Hopefully now, you are beginning to understand just how vast these numbers are. You'll also notice that their properties are very counter-intuitive. For example, even if we go with a number like (2↑↑↑↑↑4)↑↑↑(2↑↑↑↑↑4), it will still be less than 2↑↑↑↑↑5. This is because it would by definition be equal to 2↑↑↑↑(2↑↑↑↑↑4) = 2↑↑↑2↑↑↑2↑↑↑ ... 2↑↑↑2↑↑↑2 w/(2↑↑↑↑↑4) 2s.

Here is a table which shows how to construct base 3 heptational numbers:

To understand the above construction remember that every tetra-tower has as many 3's as the value of the tetra-tower to the immediate left. The left most tetra-tower always a lone 3. Each row of tetra-towers has as many towers as the value of the very last tetra-tower of the row immediately above it. The top most row contains only a lone tetra-tower. Using these rules one can extend the table arbitrarily. Keep in mind that each "tetra-tower" collapses to a power tower by repeatedly taking the two top most terms, resolving them to a power tower, evaluating, and then using this as the new polyponent for the next power tower.

And we can continue to octation. Simply replace the tetra-towers with penta-towers! Ennation can be explained with the above system using hexa-towers, dekation with hepta-towers, and so on! Eventually the up-arrow notation will become unwieldy once it becomes difficult to determine the number of up-arrows, ie. 3↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑3. When there is more up arrows then we can actually write out, we've gone beyond the practical limitations of the notation. It would seem that up-arrows provide us with a terrific system for describing large numbers. Yet we've really only just gotten started, as we'll see as we progress...

As you can see, Archimedes' point is made all the more vivid with Up-arrow notation. Archimedes dwarfed the known universe of his time with his notation. Since then the known universe has grown considerably. Archimedes' universe had a diameter of a hundred myriad myriad myriad stadia. However this only amounts to about 2 light years in modern units. We now know the observable universe to have a diameter of about 93 billion light years! Yet we still have no definite way to determine exactly how big the universe actually is, or whether it is indeed the only one. None the less indirect evidence has been used to make an educated guess. One of the most exotic theories is something called inflation. In this theory, the universe is supposed to have expanded very rapidly to smooth out inconsistencies in the background radiation. The most radical version claims that the universe may be as large as 10^{10^12} centimeters, meters, light years , or whatever (it makes little difference with numbers this big since there is only 18 orders of magnitude between centimeters and lights years, and this can easily be rounded off when we are dealing with a trillion orders of magnitude).

Much like how Archimedes' model universe was highly theoretical at the time, inflationary theory is also more guess work than fact. What is interesting however is that our fringe theories are now approaching the size of Archimedes' number of 10^{8*10^16}. It is conceivable that in coming years theories may inflate the "supposed universe" even beyond this olympian value. And yet the scale of the universe we can prove exists still trails pathetically behind the numbers we can imagine! For even the largest values in physics only amount to power towers with a handful of terms, and as we've seen, power towers are quickly dwarfed by tetra-towers, penta-towers, hexa-towers, and beyond!

Perhaps reality itself is infinite, but if this was so, how would we ever be able to observe this fact? If this is the case our "supposed universe" might grow indefinitely. Yet as our knowledge of the world has increased, so has our knowledge of number, and it has always out paced it by leaps and bounds. Our grasp of number surpassed our conception of the world a long time ago, and perhaps its a lead that will never again be closed. The universe may be larger than we suppose, but I doubt it will ever be larger than we could imagine!

Comparison to Hyper-E Notation

I first encountered Knuth Up-arrow notation in 2004 during my first year of college. One day while I was in the college library I was remembering the large numbers I had devised as a kid, and I was wondering if there was anything about large numbers on the internet. For all I knew, no one had bothered to explore the topic. Let me tell you, did that turn out to be so wrong! As I soon discovered, there was a lot of information on large numbers on the internet, and people had actually bothered quite a bit. The first notation that I encountered and understood was the Up-arrow notation. One of the first things I did was try to make a comparison between the numbers I had devised as a kid, and those expressible with Up-arrows. I soon realized that the two systems were essentially expressing the same class of numbers. It turns out, that adding an argument in Hyper-E is roughly equivalent to adding an Up-arrow. In fact, it can be proven inductively that:

E(b)p = b↑p

E(b)1#p = b↑↑p

E(b)1#1#p = b↑↑↑p

E(b)1#1#1#p = b↑↑↑↑p

etc.

More generally we can say that:

E(b)1#1# ... #1#p w/n arguments = b↑↑↑ ... ↑↑↑p w/n ↑s

The count of arguments here does not include the base. This above statement is relatively easy to prove using mathematical induction. Firstly by definition E(b)p = b^p. Also E(b)1#p becomes b^b^b^...^b^b^1 w/p bs, and this is equivalent to b^^p.

To prove the general progression first assume that E(b)1#1# ... #1#p has "n" arguments (not including the base), and that it is equal to b<n>p (using Bower's inline notation).

Now observe, E(b)1#1# ... #1#1#1 w/n+1 args = b. Therefore it is also equal to b<n+1>1=b.

Furthermore, E(b)1#1# ... #1#1#2 w/n+1 args = E(b)1#1# ... #1#(E(b)1#1# ... #1#1#1) = E(b)1#1# ... #1#b w/n args := b<n>b. Therefore it is also equal to b<n+1>2.

Now assume that E(b)1#1# ... #1#1#k w/n+1 args = b<n+1>k.

Observe that E(b)1#1# ... #1#1#(k+1) = E(b)1#1# ... #1#(E(b)1#1# ... #1#1#k) := E(b)1#1#...#1#(b<n+1>k) := b<n>b<n+1>k := b<n+1>(k+1).

What this implies is that E(b)1#1# ... #1#1#p w/n+1 args = b<n+1>p. Thus, if the general statement is true of the nth case, it must also be true of the n+1th case. This sets up a domino effect and ricochets all the way to infinity! For if E(b)1#p = b^^p, then it follows that E(b)1#1#p = b^^^p. This in turn implies E(b)1#1#1#p = b^^^^p, which implies E(b)1#1#1#1#p = b^^^^^p, and so on ad infinitum.

With this now proven, we can further show that:

b↑↑...↑↑p w/n ↑s ≤ E(b)a₁#a₂#...#a_n-1#p

Where all arguments a₁ ~ a₂ are counting numbers. To prove this first note that:

E(b)a₁#a₂# ... #a_k# ... #a_n-1#p < E(b)a₁#a₂# ... #(a_k+c)# ... #a_n-1#p

That is, if an arbitrary argument, the kth argument, where k can be anything from the 1st to the (n-1)th, is increased by some counting number , c, then the new expression is always greater. Combining this with our previous result we obtain the following important result:

b<n>p ≤ E(b)a₁#a₂# ... #a_n-1#pwhere n is a positive integer.

This statement shows that Hyper-E notation grows at least as fast as Knuth Up-arrows, and a little faster if the arguments a₁~a_n-1 are greater than 1. This implies for example that:

boogol < goolda

Given that:

boogol = <10,10,100> = 10<100>10

goolda = E100##100 = E100#100# ... #100#100 w/100 100s

For it follows that:

boogol = 10<100>10 < 10<100>100 =

E1#1#...#1#100 w/99 1s < E100#100# ... #100#100 w/100 100s =

E100##100 = goolda

It might seem that, if Hyper-E notation is larger than Knuth Up-arrows when the arguments (other than the base and the last argument) are greater than 1, then perhaps it eventually outclasses them. Does Hyper-E notation grow faster than Knuth Up-arrows?

Consider the following example:

E5#5#1 = 10^10^10^10^10^5 > 10^^^1 = 10

E5#5#2 = E5#(E5#5) = 10^10^...^10^5 w/10^...^10^5 w/5 10s 10s > 10^^^2 = 10^^10

By a similar token E5#5#3 > 10^^^3. So it would seem that in this case the Hyper-E notation is "growing faster" than the up-arrow notation. In fact we can easily prove that E5#5#p > 10^^^p for all p. Assume that the statement is true for k and every positive integer less than k. We can then show that...

Given: 10^^^k < E5#5#k

E5#5#(k+1) = E5#(E5#5#k) > E5#(10^^^k) > E1#(10^^^k) = 10^^10^^^k = 10^^^(k+1)

This is again mathematical induction, and the result is a domino effect that proves the statement for all positive integers.

Generally speaking, if we are give two functions, defined over the positive integers, call them f(n) and g(n), then saying g "grows faster" than f generally means that g eventually takes a lead on f that f never overcomes. Of coarse there is a more formal way we can state this:

Given two functions , f(n) and g(n), from the positive integers to the positive integers, g is said to "grow faster" than f if and only if there exists an M such that, f(n) < g(n) whenever n > M.

As a good example for how this definition works to define "faster than", consider the functions f(n) = 2n & g(n) = n² . Clearly we would say that g(n) is the faster growing function. We can see this in the sequence of values:

f(n) = 2,4,6,8,10,12, ... for n=1,2,3,4,5,6, ...

g(n) = 1,4,9,16,25,36, ... for n=1,2,3,4,5,6, ...

We can see that the numbers themselves grow more quickly. However this doesn't mean that g is always greater than f. For example g(1) < f(1) since 1 < 2. Eventually however g takes the lead over f, and f never "catches up" again. We can look at the graph to see how g overtakes f:

We can see that f(2) = g(2) since 4 = 4, but f(3) < g(3) since 6 < 9. After this g sky rockets further and further way from f. How do we know that g and f really never cross again? Recall that:

n² = 1+3+5+ ... +(2n-1)

That is, n² is the sum of the first n odd numbers. We know that f(2) = g(2). However:

g(3) = 5+g(2) = 5+4 = 9

g(4) = 7+g(3) = 7+9 = 16

g(5) = 9+g(4) = 9+16 = 25

etc.

This means the difference between consecutive terms is constantly increasing. In all these instances the difference is greater than 2, however with f, the difference is always 2 since:

f(3) = 2+f(2) = 2+4 = 6

f(4) = 2+f(3) = 2+6 = 8

f(5) = 2+f(4) = 2+8 = 10

etc.

Since the difference is always 2 for f, and the distance is constantly increasing for g, it follows that g overtakes f, and f can never again catch up. To satisfy the definition for a faster growing function we can say that M = 2. If n > 2 it follows that f(n) < g(n). Thus g grows faster than f.

The definition works pretty well at defining what it means to say one function grows faster than another. However there is certain limitations with such a definition. Take a pair of parallel lines, j and k, for example:

j is always greater than k. Because of this we could pick any value for M and show that j grows faster than k. For example, M=0. We can then say k(n) < j(n) when n > 0. So according to our definition, j "grows faster" than k. However common sense would say "they grow at the same rate". They should be considered to be growing at the same rate, it's just that "j" has a bit of a head start. This provides the motivation for devising a better definition. Notice, if we moved j so that it overlapped with k, the two functions would match. Moving a function is known as a "translation". So in our new definition we want all translations of any function to be of the same growth rate. When functions have a different "shape" however, it is possible that one grows faster than the other. Note that, in our previous example, no matter how we move f(n) = 2n, it will always be overtaken by g(n) = n^2 eventually. This is because, regardless of how great a head start f is given, since g is accelerating while f is increasing at a constant rate, g must eventually catch up and then overtake f. We therefore use the following revised definition to say a function "grows faster" than another:

Given f and g, g is said to grow faster than f if and only if for every translation of f there exists an M such that f(n) < g(n) when n>M.

Given this definition, it follows that to disprove that g grows faster than f it is sufficient to give a single translation in which M fails to exist. Returning to j and k, we can see that M fails to exist when k is translated to a position above j. However it also fails that k is greater than j for the same reason. If neither function can be established to grow faster than the other, they are declared of an "equivalent growth class". That is, they grow at the same rate.

Using this definition can we come up with a precise formulation of the conjecture that "Hyper-E grows as fast as Knuth Up-arrows"? The problem is that our definition applies specifically to unary (single argument) functions. Up-arrow notation however is a tertiary (3-argument) function, and Hyper-E can have an arbitrary number of arguments. To get around this, we must define unary functions to represent the growth rate of each notation.

I'll now define the following functions:

h(p) := E(b)a₁#a₂# ... #a_n-1#p w/n args

k(p) := b<n>p

The variable, p, is the argument of the functions, and all other variables are fixed constants. I will show that for every set of fix constants, h(p) is in the same growth class as k(p), as just defined. This will be symbolized as:

h(p) == k(p)

To begin, we first assume that all the arguments, a₁ through a_n-1 are equal to 1. If this is so then:

h(p) = E(b)1#1# ... #1#p = b<n>p = k(p)

It follows that...

h(p) = k(p) implies h(p) == k(p)

We therefore will assume that at least one argument, a₁ through a_n-1, is not equal to 1. As we proved earlier however, this means that:

k(p) < h(p)

because...

k(p) = b<n>p = E(b)1#1# ... #1#p < E(b)a₁#a₂# ... #a_n-1#p = h(p)

This would seem to imply that h(p) grows faster than k(p), since it is always greater. To show that they are of the same growth class we must show that there is a translation of k in which k is always greater than h. We will do this by giving k a head start.

Firstly assume that the arguments, a₁ through a_n-1, which we'll call the non-terminal arguments, are less than or equal to the base, b. If this is true it can be shown that:

E(b)a₁#a₂# ... #a_n-1#p < E(b)1#1# ... #1#(p+2) = b<n>(p+2) = k(p+2)

k(p+2) corresponds to a translation of k, two units to the left. To prove this we will use the following valid transformation:

E(b)a₁#a₂# ... #a_n-1#p = E(b)a₁#a₂# ... #X#(p-1) where X = E(b)a₁#a₂# ... #a_n-1

From this we can say that:

E(b)1#1# ... #1#(p+2) = E(b)1#1# ... #(E(b)1#1# ... #1)#(p+1) =

E(b)1#1# ... #b#(p+1) = E(b)1#1# ... #(E(b)1#1# ... #b)#p =

E(b)1#1# ... #(b<n-1>b)#p

We could continue in this manner until the last argument was reduced to 1. We could then drop it, and we would be left with:

E(b)1#1# ... #(b<n-1>b<n-1> ... <n-1>b w/p+1 bs)

To continue, we can subtract from the new terminal argument (last argument), and the 2nd to last argument will itself become an operator expression. Specifically:

E(b)1#1# ... #1#(b<n>(p+1)) = E(b)1#1# ... #b#(-1+b<n>(p+1)) =

E(b)1#1# ... #(b<n-2>b)#(-2+b<n>(p+1))

etc.

The next step is tricky to explain, so I'll employ some special notation. Let:

b^n-->p:= b<n>p

We now state that:

E(b)1#1# ... #b^n-2-->b#(b^n-1-->b^n-1-->^...^n-1-->b -2)

is greater than...

E(b)1#1# ... # b^n-2-->b#(b^n-1-->^...^n-1-->-2+b^n-1-->b)

is equal to...

E(b)1#1# ... # b^n-2-->b#(-2+b^n-1-->b)#p

In other words, if we decrease the (n-1)th argument by 2, and replace the (n-2)th argument with b^n-2-->b , this will still be less than the original expression. By a similar argument, since the (n-2)th argument would eventually reduce to an "arrow tower" of order n-2, if we decrease the (n-2)th argument by 2, and replace the (n-3)th argument with b^n-3-->b , the resulting expression will again be less than the original expression. The upshot of this is that we can generate the following expression:

E(b)b^b#(^bb)-2#(_b<--b)-2# ... #-2+b^n-2-->b#(-2+b^n-1-->b)#p

The -2 will have a very minor effect on the arguments, however we can overestimate their effect, by replacing every argument with b. We can be sure this works, because the lowest base possible would be 2. In this case 4 = 2²= ²2 = _2<--2 etc. Since 4-2 = 2, and the base is 2, it means the best case scenario is that the -2 will result in reducing the "arrow towers" to the base b. Note that if b>2, then the result will be larger than the base. For example:

3³-2 > 3³-3 > 3³-8*3= 3

implies...

3 < 3³-2

Because the final expression is a lower bound on the original expression it shows that:

E(b)b#b# ... #b#b#p < E(b)1#1# ... #1#1#(p+2)

Thus if the non-terminal arguments are less than or equal to the base, we can say it is still much less than E(b)1#1# ... #1#(p+2).

By using the same trick we can show that if a₁ =< b^b , a₂ =< ^bb , a₃ =< _b<--b , and in general a_k=< b^k-->b , then E(b)1#1# ... #1#(p+3) is greater than E(b)a₁#a₂# ... #a_n-1#p.

This works because:

(2^{2^2})-3 = 2⁴-3 > 2⁴-3*2² = 2²

This will only be more so the larger the base and operator. We can also create an upper bound for the non-terminal arguments that result in a value less than E(b)1#1# ... #1#(p+4), E(b)1#1#...#1#(p+5), etc.

The argument here rests on showing that (2^^k)-k > 2^^(k-1). This can be shown as follows:

(2^^k)-k > (2^^k)-(2^^(k-1))

Furthermore...

(2^^k)-(2^^(k-1)) > 2^{(2^^(k-1))-1}

This is because 2^^(k-1) is much less than half of 2^^k, and therefore it must have less of an effect than subtracting one from the exponent. Furthermore...

2^{(2^^(k-1))-1}> 2^{2^^(k-2)} = 2^^(k-1)

Thus, the negatives can never do more than reduce the arrow towers by 1 term.

Therefore we can now make the following assertion with confidence:

E(b)a₁#a₂#...#a_n-1#p < E(b)1#1#...#1#(p+c)

provided...

a_k =< b<k+1>(c-1)

This proves, that no matter how large the non-terminal arguments become, or how large the initial p, there must be an up-arrow expression which is larger. Essentially we have shown that:

h(p) < k(p+c)

This proves, that given the appropriate shift to the left, k will gain a lead over h. Furthermore, because p wasn't specified, the statement applies no matter how large the argument p gets in the above statement. In fact, although it might seem that since h is "faster" than k, it would eventually catch up, it in fact gets further and further away from k, when k has a head start of c units.

Now since

k(p) < h(p)

it follows k can not be a faster growing function than h, since M fails to exist because k is always less than h. Yet h can not be a faster growing function than k since h(p) < k(p+c) and M again fails to exist. Since neither is greater than the other it implies that they have the same growth rate. Thus we have h(p) == k(p) for every set of non-terminal arguments.

The consequence of this is that we can now say definitively that:

E(b)p has the same growth rate as b^p

E(b)a₁#p has the same growth rate a b^^p

E(b)a₁#a₂#p has the same growth rate a b^^^p

etc.

In general the growth rate of the nth argument is equivalent to the nth up-arrow operator with base b.

It now becomes a lot easier to show more generally that Hyper-E notation is equivalent to Knuth Up-arrow notation. Firstly we now define:

H(n) = E(b)a#a# ... #a#a#p w/n args

K(n) = b<n>p

Where a,b and p are arbitrary constants. We know that when a>1:

K(n) < H(n)

However we can show that with the appropriate shift, k leads h no matter how large a is.

We now know that there exists some c such that:

H(n) < k(p+c) = b<n>(p+c)

Further we know there must exist some d such that:

b<n>(p+c) < b<n+d>p = K(n+d)

Because c is a fixed value based on a, and d is a fixed value based on c, it means that:

H(n) < K(n+d)

Thus:

H(n) == K(n)

We have now proven that Hyper-E is equivalent to Knuth Up-arrow notation. That is not to say that they are the same function, but rather to say that they exhibit the same growth rate.

Conclusion

Knuth up-arrows represent one of the most natural and intuitive ways of expressing large numbers. However, there are many other notations we have to explore in up coming articles. Many of these are growth equivalent to Knuth Up-arrows.

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Works Cited

[1] http://en.wikipedia.org/wiki/The_Sand_Reckoner

[2] http://en.wikipedia.org/wiki/History_of_large_numbers

[3] http://jeff560.tripod.com/operation.html

[4] http://www.ehow.com/about_5339194_were-exponents-first-used.html

[5] http://en.wikipedia.org/wiki/Caret

[6] http://andydude.5gigs.net/tetration.co.cc/htdocs/hist.html

[7] http://en.wikipedia.org/wiki/Cutler%27s_bar_notation

[8] http://www.alpertron.com.ar/BIGCALC.HTM