Large Numbers
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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, and similar definitions with hyper4 in place of exponentiation.
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: The Generalised "Hyper" Function
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 a function 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.
Generalising the Hyper Operator for NonInteger n
Previously we looked at efforts to extend tetration to noninteger values, which in terms of this new "hy(,,)" function means computing hy(a,4,b) for noninteger values of b. Since there is now a third parameter to the function (the middle parameter n) it is natural to consider noninteger values of that too. For example, hy(a,2.5,b) would be a function "between" multiplication and exponentiation.
Gottfried Helms describes an approach to this interesting problem on Maths StackExchange here but not explaining how to compute it. (He does refer vaguely to a "Schrödermechanism" and refers the reader to the tetration forum, without a specific link, and was possibly thinking of this discussion which also relates the problem to the Hyperoperation#Commutative hyperoperations of Albert Bennett.)
Folkman's Number
In the same article introducing the much more famous Graham's Number, Martin Gardner described a Ramsey theory problem investigated by Jon Folkman. Its value is 2↑↑↑(2^{901}), and it is much smaller than any of the variants of Graham's Number.
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 [51] 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
Composed UpArrow Notation
When it is necessary to repeat the same number and uparrow(s) multiple times, it is common to appreviate the repetition with a superscript. Take for example:
10↑↑10↑↑10↑↑4↑7↑↑7↑↑3↑27
and the usual convention that everything is rightassociative. This can be abbreviated:
(10↑↑)^{3}4↑(7↑↑)^{2}3↑27
For reasons similar to the power tower paradox, any time something with fewer arrows is followed by something with more arrows, the thing with fewer arrows can be ignored, which reduces the example to:
(10↑↑)^{3}(7↑↑)^{2}3↑27
and (again because of the power tower paradox) we can change the 7's to 10's (or vide versa) without making a significant change to the value, so it reduces further to
(10↑↑)^{5}3↑27
Usually the parentheses are left out to make it
10↑↑^{5}3↑27
One then typically reduces the final part to using base 10, and this time the base cannot be ignored so it becomes
10↑↑^{5}10↑12.88227387743 (approximately)
This approach is the best for actual practical use, such as in the OmegaNum library which handles such numbers with ease.
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.
However, in many cases we know how to compare things because someone has done the work to reconcile a given invented system with a rigidly specified function hierarchy.
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).
For many years I had this topic here, but it has been moved to here, which is a lot closer to where it belongs.
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} ≈ 2×10^{19728}, 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) = ackrm(n,n)
While it seems that a1(n) grows "just as fast" as the ackrm() function, this is not actually true. Each value of the first argument a in ackrm{a,b) corresponds to a different finite ordinal in the fast growing hierarchy, while ackrm(n,n) eventually exceeds all of those. This is similar to how x^{2} eventually exceeds the linear functions kx for any constant k. So a1(n) grows faster than all of the f_{k}(n) functions for finite k, and instead matches the ωindexed function f_{ω}(n).
a1(n) is a convenient way of defining large numbers as a function of one variable, but actually computing those numbers involves the recursive definition of the function. When x>1, we have:
a1(x) = ackrm(x,x) = ackrm(x1, ackrm(x,x1))
The problem here is that the arguments of the two ackrm functions on the right are not equal to each other, and therefore we can't substitute from the definition of a1(n) to put the right side in terms of the a1() function. So this means you always need the twoargument version in order to actually get anywhere: the growth rate of the oneargument a1(x) depends on the existence of a twoargument 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 a "successor function" f(x)=x+1.
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