Large Numbers
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Inventing New Operators and Functions
The concept of the "classes" described so far does quite well at handling everything that can be done with exponents, which are the most powerful operator known to most people. To proceed further we begin to invent new operators. This practice of inventing new operators continues over and over again as you go to higher and higher large numbers. The new operators overcome the limits of the old operators, limits that are reached as the old notation becomes unwieldy.
For example, class1 numbers are written in traditional placevalue notation, which is essentially abbreviated addition and multiplication. For example:
3158 = ((3 × 10 + 1) × 10 + 5) × 10 + 8
Although we don't normally think of it that way, the placevalue notation avoids the unwieldy use of lots of symbols.
When expressing larger numbers, like Avogadro's number and googol, one usually uses exponents and power towers, as discussed above:
6.02 × 10^{23}, 10^{100}, 10^{10100}, 27^{256312546656}, etc.
but after a while that becomes unwieldy too. Eventually there are so many exponents that it cannot be written on a page. Then it becomes a good idea to invent a new new shorthand, which amounts to defining a new operator.
Beyond Exponents: the hyper4 Operator
(most commonly called "tetration")
(my early name for this was "powerlog")
The first new operators used by those seeking large numbers are usually higher dyadic operators. A dyadic operator is one that has two arguments — two numbers that it acts on. Usually in notation the operator is placed between the two numbers.
The most common higher dyadic operators follow the pattern set by the wellknown three (addition, multiplication and exponentiation). These operators come up a lot in the definitions of large numbers that are to follow.
Following an obvious pattern in the three common operators, the new operator can be defined as shown here:

Note that for the last operator, there are two ways to interpret the absolute and inductive definitions, producing different hyper4 operators. In common practice, the first one is used because the other one can be reduced to a combination of two exponent operators: a_{④}b=a^{a(b1)}, and thus it does not really count as a new operator.
The names tetration, superpower, and superdegree have also been used to refer to the hyper4 operator. (As a child I used the somewhat misleading name powerlog for hyper_{4}, as in 2 powerlog 5 is 65536.)
Extension to reals : Now, suppose you want to calculate 2^{④}2.5 or pi^{④}e. The above definition isn't too useful because the number after the ^{④} has a fractional part. What we would need is a way to "extend" the hyper4 operator to real numbers. Unfortunately, this is tough to do in a way that meets the types of standards mathematicians generally want such things to have. I also know of no proof that such extension is impossible. A lot of people have worked on this over the years, and if you're interested, I suggest you check my notes here, and the Tetration FAQ.
A "logarithm" for hyper4 : Another common question about hyper4 is how to perform the inverse operations — the equivalent of the "hyper4 logarithm" and "hyper4 root". There is no good answer for either one, until the problem of extending hyper4 to the reals is solved. The "hyper4 root" can be evaluated for fixed integer "root" values using Newton's method. For example, to take the "2nd hyper4 root", use this algorithm:
(given: number X, we want to find R such that R^{④}2 = X. Note that R^{④}2 = R^{R}.)

The hyperlogarithm is intuitively similar to the "class number" (see my description of classes above) along with a fraction indicating how far through the class we are. It is very similar to the levelindex representation and to the internal format used by my hypercalc program. Here are some hyperlogarithm values (to base 10) using a definition from Trappman's Tetration FAQ^{15}:
hyperlog(2) ≈ 0.39
hyperlog(100) ≈ 1.39
hyperlog(10^{100}) ≈ 2.39
hyperlog(10^{10100}) ≈ 3.39
...
The function "below" addition : Some people have also developed a hyper0 function. If you think about it, addition is a shortcut for counting, in much the same way multiplication is shortcut for addition. The following definition for a hyper0 function was developed by Constantin Rubtsov:
a^{⓪}b = a (if b = ∞)
a^{⓪}b = b (if a = ∞)
a^{⓪}b = a+2 = b+2 (if a = b)
a^{⓪}b = a+1 (if a > b)
a^{⓪}b = b+1 (if b > a)
This function, appropriately enough, is also the "successor" function used as the primitive computational element in algorithms defined in the Church theory of computation, which includes the original Ackermann function. For more on how this is done see my page on functional computation.
The Hyperfactorial and Superfactorial Operators
These are singleargument functions like the factorial but producing higher values.
N.J.A. Sloane and Simon Plouffe use hyperfactorial to refer to the integer values of the Kfunction, a function related to the Riemann Zeta function, the Gamma function, and others. It is
H(n) = n^{n} (n1)^{n1} ... 3^{3} 2^{2} 1^{1}
For example, H(3) = 27×4×1 = 108 and H(5) = 86400000. This function does not really grow much faster than the normal factorial function.
In 1995, Pickover defined the superfactorial n$ (think of the dollar sign as a factorial sign with an S for "super" drawn on top of it) as follows:
n$ = n!^{n!n!....n!}
where there are n! repetitions of n! on the right hand side. Using the hyper4 operator, n$ is equivalent to:
n$ = n! ^{④} n!
There are other ways to define a higher version of the factorial, such as this and this.
To get an idea how big the hyperfactorial of a pretty normal number can be, read Wayne Baisley's wonderful article "Quantity Has A Quality All Its Own" (and bring your towel).
More Bowers Names
Jonathan Bowers, mentioned above, has many names covering this area. For example, in analogy to googol and googolplex he refers to 10^{④}100 as giggol and 10^{④}(10^{④}100) as giggolplex.
Higher hyper operators
Of course, the pattern of dyadic operators is easily continued:

and so on.
Bowers has several named numbers in this area, including trisept, 7^{⑦}7; tridecal, 10^{⑩}10; and the aptly named boogol, the frighteningly large 10^{(100)}10.
The first triadic operator
Since the dyadic operators all fall into a pattern, it is logical to define a triadic operator that combines them all. A triadic operator is an operator that acts on three numbers, just as a dyadic operator acts on two numbers.
This new triadic operator is represented as a function with three arguments, and defined as follows:
hy(a,n,b) = { 1 + b for n = 0 { { a + b for n = 1 { { a * b for n = 2 { { a ^ b for n = 3 { { a ^ hy(a,4,b1) for n = 4 { { hy(a,n1,hy(a,n,b1)) for n > 4 { { a for n > 1, b = 1the following definition is equivalent:
hy(a,n,b) = { 1 + b for n = 0 { { a for n = 1, b = 0 { { a for n > 1, b = 1 { { hy(a,n1,hy(a,n,b1)) for n > 0and also note that:
hy(a,3,b) = a↑b = a^{b}
hy(a,4,b) = a↑↑b
hy(a,5,b) = a↑↑↑b
hy(a,6,b) = a↑↑↑↑b
etc.
Bowers' Array Notation (3element Subset)
At this point we return to the work of Jonathan Bowers to introduce his array notation. This notation is elegant, powerful, relatively easy to use and covers a greater range than any other discussed on these pages, within the limits of functional formal systems.
We will start by showing a very reduced version of the notation, which uses arrays of only 1, 2, or 3 elements. The rules for converting the notation into a number are:
1. For one and twoelement arrays, just add the elements. [a] = a
and [a,b] = a+b
2. If rule 1 does not apply, and if there are any trailing 1's,
remove them: [a,b,1] = [a,b] = a+b; [a,1,1] = [a].
3. If neither previous rule applies,
and the 2nd entry is a 1, remove all but the first element:
[a,1,n] = [a] = a.
4. There is no rule 4 (there will be when we get to bigger arrays).
5. Otherwise replace the array [a,b,n] with [a,[a,b1,n],n1], then go
back and repeat the rules to expand it further.
With just a little effort you can see that these rules make [a,b,n] equivalent to hy(a,n,b) except for the special case of n=0. Compare the formula of rule 5:
[a,b,n] = [a,[a,b1,n],n1]
with the general case of the definition of the hyper function:
hy(a,n,b) = hy(a,n1,hy(a,n,b1))
They are the same except the order of the arguments is different. Bowers arranges the arguments in order of increasing "growth potential" — the operator has higher growth potential than b, so it goes last.
So, all 3element Bowers arrays are equivalent to the normal hyper operators. [3,2,2] = 3^{②}2 = 3×2 = 6; [3,2,3] = 3^{③}2 = 3^{2}, [4,5,6] = 4^{⑥}5, etc.
hyper operator variant: Knuth's Uparrow Notation
The use of two or more carets (as in "a^^b" or "a^^^b") resembles a notation defined by Donald Knuth^{17} in 1976 ("a↑↑b" and "a↑↑↑b" respectively), and is equivalent to the hyper operator. Carets are commonly seen in old ASCII sources such as mailing lists from the early days of USENET, but Knuth used real arrows: a↑↑b and a↑↑↑b instead of a^^b or a^^^b.
a ↑↑ b = hy(a,4,b)
a ↑↑↑ b = hy(a,5,b)
a ↑↑↑↑ b = hy(a,6,b)
(etc.)
using the hy() function allows for a more compact representation of really large numbers that would otherwise take a lot of arrows. For example, hy(10,20,256) is equivalent to
10 ↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑↑ 256
and hy(256,625,4096) would be very unwieldy. Bigger numbers like hy(256,4294967296,256) can't be written at all.
This uparrow notation is used in defining the Ackermann numbers
A(n) = n ↑↑↑...↑↑↑ n (with n uparrows) = hy(n, n+2, n)
which are related to the Ackermann function described below.
In 2010 Knuth informed me [50] that he has found "the Ackermannlike 'arrow notation' in a 19^{th} century encyclopedia."
Partial Ordering for Knuth UpArrows
One may speculate on the general problem of determining which is the larger of two values a↑↑↑...↑↑↑b and x↑↑↑...↑↑↑y. We can begin to make answer that question for small numbers of uparrows. In particular (for later discussion) we care about the answer when a, b, x and y are positive integers.
First, note that if a is 1, a↑↑↑...↑↑↑b is just a power of 1, which is always 1. Also, if a and b are 2 then the value of a↑↑↑...↑↑↑b is 4, regardless of the number of arrows.
With a single arrow a↑b is the familar exponentiation operation.
2^{3} is smaller than 3^{2}
2^{4} is the same as 4^{2}
for any other a and b, if b is greater than a then a^{b} is
greater than b^{a}.
in general, to compare a^{b} to c^{d} we can probably calculate both
directly, so long as all four numbers are class 1.
With two arrows, a↑↑b is a "powertower" of height b. Using Hypercalc it is relatively easy to compile a list of a↑↑b for all the smaller values of a and b, and larger values of a. Here I'll also show the smaller values of c↑d that are not expressible as in the form a↑↑b, to see where they fit in:
1↑↑a = 1, for all a
2↑↑1 = 1
3↑↑1 = 3
2↑↑2 = 2^{2} = 4 = 4↑↑1
5↑↑1 = 5
6↑↑1, 7↑↑1, etc. (a↑↑1 = a, for all a)
2^{3}
3^{2}
2↑↑3 = 2^(2^{2}) = 2^{4} = 4^{2} = 16
5^{2}
3↑↑2 = 3^{3} = 27
2^{5}
6^{2}, 7^{2}
2^{6} = 8^{2}
3^{4} = 9^{2}
10^{2}, 11^{2}
2^{7}
12^{2} through 15^{2} and 3^{5}
4↑↑2 = 4^{4} = 2^{8} = 256
17^{2}, etc.; 2^{8}, etc.; 3^{6}, etc.
5↑↑2 = 5^{5} = 3125
56^{2}, etc.; 15^{3}, etc.; 2^{12}, etc.; 3^{8}, etc.
6↑↑2 = 6^{6} = 36^{3} = 216^{2} = 46656
217^{2}, etc.
2↑↑4 = 2^{16} = 4^{8} = 16^{4} = 256^{2} = 65536
7↑↑2 = 7^{7} = 823543
.. 8↑↑2, 9↑↑2, through 11↑↑2
3↑↑3 = 3↑(3^{3}) = 3^{27} = 7625597484987
12↑↑2 = 12^{12} = 8916100448256
.. 13↑↑2 through 80↑↑2
4↑↑3 = 4↑(4^{4}) = 4^{256} ~ 1.3408×10^{154}
81↑↑2 = 81^{81} ~ 3.8662×10^{154}
.. 82↑↑2 through 758↑↑2
5↑↑3 = 5↑(5^{5}) = 5^{3125} ~ 1.911×10^{2184}
759↑↑2 = 759^{759} ~ 1.269×10^{2186}
.. 760↑↑2, etc.
2↑↑5 = 2↑(2↑↑4) = 2^{65536} ~ 2.004×10^{19728}
5298↑↑2 = 5298^{5298} ~ 2.214×10^{19730}
.. 5299↑↑2, etc.
6↑↑3 = 6↑(6^{6}) = 6^{46656} ~ 2.659×10^{36305}
7↑↑3 = 7↑(7^{7}) ~ 3.76×10^{695974}
.. 8↑↑3 through 11↑↑3
3↑↑4 = 3↑(3↑↑3) ~ 1.35×10^{3638334640024}
12↑↑3 = 12↑(12↑↑2) ~ 5.85×10^{9622088391634}
.. 13↑↑3 through 80↑↑3
4↑↑4 = 4↑(4↑↑3) ~ 10^{8.0723×10153}
.. 81↑↑3 through 758↑↑3
5↑↑4 = 5↑(5↑↑3) ~ 10^{1.336×102184}
.. 759↑↑3, etc.
2↑↑6 = 2↑(2↑↑5) ~ 10^{6.03×1019727}
6↑↑4 = 6↑(6↑↑3) ~ 10^{2.07×1036305}
.. 7↑↑4 through 11↑↑4
3↑↑5 = 3↑(3↑↑4) ~ 10^{6.46×103638334640023}
...
A pattern emerges: except when a is 2 or when b is 2, the values of a↑↑b generally follow the rule:
If y is larger than b, x↑↑y will be larger than a↑↑b.
However there are exceptions for smaller b or moderately larger a: as 12↑↑2 is larger than 3↑↑3; 81↑↑2 is larger than 4↑↑3, and similar things happen further along in the list.
But even including these smaller b or larger a cases, a more general pattern is seen, namely that increasing b by one always gives a value that is about 10 to the power of whatever we had before: 4↑↑3 ~ 1.3408×10^{154}, and 4↑↑4 ~ 10^{8.0723×10153}. This is related to the "power tower paradox".
It is also generally true that if b is 3 or more, all of the numbers of the form a↑↑b are larger than anything of the form c↑d (with one arrow, and with "reasonablysized" c and d). The smallest c↑d bigger than 3↑↑3 is 12↑12; in order for c↑d to be bigger than 4↑↑3 you need to go up to 81↑81, and so on.
Now let's make a similar list of a↑↑↑b examples, and showing how the a↑↑b values fit in:
2↑↑↑2 = 2↑↑2 = 2↑2 = 4
2↑↑3, 3↑↑2 through 6↑↑2
2↑↑↑3 = 2↑↑2↑↑2 = 2↑↑4 = 2↑2↑2↑2 = 2↑2↑4 = 2↑16 = 65536 = 2↑↑4
7↑↑2 through 11↑↑2
3↑↑↑2 = 3↑↑3 = 3↑27 = 7625597484987
12↑↑2, etc.; 4↑↑3 through 80↑↑3; and 3↑↑4
4↑↑↑2 = 4↑↑4 = 4↑4↑4↑4 ~ 10^{8.0723×10153}
all the rest of the a↑↑b in the list above
5↑↑↑2 = 5↑↑5 = 5↑5↑5↑5↑5 ~ 10^{101.33574×102184}
2↑↑↑4 = 2↑↑(2↑↑(2↑↑2)) = 2↑↑(2↑↑4) = 2↑↑16, a tower of height 16 (or
10↑10↑...6.03×10^{19727} with eleven 10's at the beginning,
which in Hypercalc is written "11pt6.03×10^{19727}")
3↑↑↑3 = 3↑↑(3↑↑3), a tower of height 7625597484987
4↑↑↑3 = 4↑↑(4↑↑4), a tower of height 10^{8.0723×10153}
5↑↑↑3 = 5↑↑(5↑↑5), a tower of height 10^{101.33574×102184}
6↑↑↑3 = 6↑↑(6↑↑6), a tower of height 3pt2.0692×10^{36305}
7↑↑↑3 = 7↑↑(7↑↑7), a tower of height 4pt3.177×10^{695974}
8↑↑↑3 = 8↑↑(8↑↑8), a tower of height 5pt5.43×10^{15151335}
9↑↑↑3 = 9↑↑(9↑↑9), a tower of height 6pt4.09×10^{369693099}
10↑↑↑3 = 10↑↑(10↑↑10), a tower of height 7pt10^{10000000000}
.. 8↑↑↑3 through 13↑↑↑3
2↑↑↑5 = 2↑↑(2↑↑↑4), a tower of height 2↑↑16 ~ 11pt6.03×10^{19727}
14↑↑↑3 = 14↑↑(14↑↑14), a tower of height 12pt1.2735782×10^{16}
.. 15↑↑↑3 through 7625597484980↑↑↑3 and (perhaps 7625597484981↑↑↑3)
3↑↑↑4 = 3↑↑(3↑↑↑3), a tower of height 3↑↑↑3 ~ 7625597484984pt3638334640023.8
4↑↑↑4 = 4↑↑(4↑↑↑3), a tower of height 4↑↑↑3
.. 5↑↑↑4 through 13↑↑↑4
2↑↑↑6 = 2↑↑(2↑↑↑5), a tower of height 2↑↑↑5
.. 14↑↑↑4 through 7625597484980↑↑↑4 ...
Once again a pattern emerges: except when a is 2, the ordering is determined first by b and then a. It shouldn't be hard to believe that the same thing happens again for a↑↑↑↑b, a↑↑↑↑↑b, and so on for larger numbers of arrows. The exception when a is 2 really continues all the way, for example:
2↑↑↑↑3 = 2↑↑↑(2↑↑↑2) = 2↑↑↑4, a tower of height 16,
but 3↑↑↑↑2 = 3↑↑↑3 = 3↑↑(3↑↑3) = 3↑↑(3↑3↑3) = 3↑↑(3↑27), a
tower of height 3^{27}, which is much larger
And so we have the:
General Rule for Partial Ordering of the hyper Operator:
If a, b, c, x, y and z are all "of reasonable size", then
with few exceptions, when comparing hy(a,b,c) to hy(x,y,z):
the one with more arrows (b versus y) is larger;
when b = y, the one with the larger number on the right (c versus
z) is larger;
when b=y and c=z, the one with the larger number on
the left (a versus x) is larger.
Detailed Rules for Partial Ordering of the hyper Operator:
When comparing hy(a,b,c) to hy(x,y,z):
if a = x = 1, they are equal,
if a is 1 and x is larger, then hy(x,y,z) is larger
if x is 1 and a is larger, then hy(a,b,c) is larger
if a = c = x = z = 2, they are equal,
if y is larger than b, then hy(x,y,z) is larger
if b is larger than y, then hy(a,b,c) is larger
if b and y (the number of uparrows) is the same, and if
a and x are both larger than 2, then hy(a,b,c) is larger if
c is larger than z, or hy(x,y,z) is larger if z
is larger than c
Proof Becomes Difficult
At this point we begin to encounter functions and definitions that are difficult to compare to one another, either because they are not very thoroughly worked out, or because it takes so much work to actually prove whether one grows more quickly than another.
Gödel Numbers
The Gödel number of G, Gödel's undecidable sentence, is probably around here somewhere (its value depends highly on what operators, functions, etc. are available to construct primitiverecursive statements in the formalised number theory system that the Gödel technique is applied to)
Goodstein sequences
Almost certainly higher than this (but who can say?) are numbers related to the Goodstein sequence.
(For more detailed descriptions, see the Wiki entry and this page by Justin Miller)
G(n) = highest value of the sequence constructed as in Goodstein's Theorem from starting value of n.
The hereditary baseb representations are analogous to ordinal infinities and subtracting one works like the successor function on the ordinal infinities — except that "w" is always finite no matter how big it gets, so every individual "w" term is guaranteed to eventually become bound to a finite value and therefore will eventually go down to zero.
The convergence of a series with no secondlevel exponents is easy to see:
v^{2} + 2
v^{2} + 1
v^{2}
v^{2}  1 = (a1) × v + (a1)
where a = value of v at this step
...
(a1) × v
(a2) × v + (b1)
where b = value of v at this step
...
(a2) × v
(a3) × v + (c1)
where c = value of v at this step
...
2 v
v + (d1)
where d = value of v at this step
...
v
e  1 where e = value of v at this step
When higherlevel exponents are involved, the series will get longer each time a higherlevel exponent has to be decremented. Each time the series will become enormously longer, but will still be of finite length. Therefore, the same principle applies.
For example, let's start with 27 (in base 2, 11011_{2}). We get:
11011_{2}
11011_{3} = 112_{10}. 11010_{3} = 111_{10}
11010_{4} = 324_{10}. 11003_{4} = 323_{10}
11003_{5} = 753_{10}. 11002_{5} = 752_{10}
11002_{6} = 1514_{10}. 11001_{6} = 1513_{10}
11001_{7} = 2745_{10}.
11000_{8} = 4608_{10}. 10777_{8} = 4607_{10}
10777_{9} = 7198_{10}.
10776_{10}
10775_{11}
10774_{12}
10773_{13}
10772_{14}
10771_{15}
10770_{16} = 67440_{10}. 1076F_{16} = 67439_{10}
1076F_{17}
...
lg(2^{n}) ≈ hy(2, 2+n, 2^{n})
Consider the lower Goodstein sequence, and look at just one of the exponents in that sequence, and call it "c". As we have already shown, "c" will eventually get decreased to a lower number. Call that number "d" (which is c minus one). At this point the iteration continues for an even longer time with no change to "d" or any of the other "exponents", but eventually as before, the lowest exponent will have to get dinimished again. So in this way we see that each exponent will eventually get replaced with a lower one. Each step takes massively longer than the previous step, but all steps are still of a finite length (not an infinite length) so eventually even the highest exponent will get decreased.
Other Triadic Operators
A common trick that clearly generates fastergrowing functions involves defining functions that take more than two arguments. We have seen how the hyper operator, our first triadic operator, easily covers everything all the dyadic operators can handle. This trend continues. Of course, all operators can be referred to as functions, and the dyadic operators are actually functions with two arguments.
The SteinhausMoserAckermann operators
The Ackermann function and the SteinhausMoser notation are both equivalent to a triadic operator that is somewhat more powerful than the hy(a,b,c) function above. The Ackermann function and SteinhausMoser are roughly equivalent to each other so we'll discuss them together.
Ackermann's Function
A recursive function first described by W. Ackermann in 1928 to demonstrate a property of computability in the field of mathematics, and also used more recently as an example of pathological recursive functions in computer science. There are many different versions of the function; for a complete description of each go here. I will use the version that is the simplest to convert to the hyper operators:
ackrm(a,b) = { 2b for a = 1 { { 2 for b = 1, a > 1 { { ackrm(a1, ackrm(a,b1)) for a,b > 1which yields to analysis as follows:
ackrm(1,b) = 2b ackrm(a,1) = 2 ackrm(2,b) = ackrm(1, ackrm(2,b1)) = 2*ackrm(2,b1) and by induction, ackrm(2,b) = 2^b ackrm(3,b) = ackrm(2, ackrm(3,b1)) = 2^ackrm(3,b1) and by induction, ackrm(3,b) = 2^{(#4#)}b ackrm(4,b) = ackrm(3, ackrm(4,b1)) = 2^^ackrm(4,b1) and by induction, ackrm(4,b) = 2^{(#5#)}b and by induction, ackrm(a,b) = hy(2,a+1,b)The example value most commonly cited is ackrm(3,5), 2^{④}5 which is 2^{65536}, a large class2 number. Of course, as with SteinhausMoser notation it is easy to transcend the classes entirely.
At this point it is tempting to try to avoid the "triadic function requirement" noted above by defining a singlevariable function, such as:
a1(n) = ackh(n,n,n)
While it is true that a1(x) grows just as fast as the ackh() function, and therefore serves as a good way of defining large numbers as a function of one variable, actually computing those numbers involves the recursive definition of the function. If x>1, we have:
a1(x) = ackh(x,x,x) = ackh(x1, x, ackh(x,x,x1))
note that the arguments of the two ackh functions on the right are not equal to each other, and therefore we can't substitute from the definition of a1(n) to make the right side be in terms of the a1() function.
However, as seen above it is possible to reduce the Ackermann function to two arguments. Furthermore, it is the fastestgrowing function you can get using two arguments, if the function is defined only in terms of calls to itself and the "successor function" f(x)=x+1.
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