Method of notation of very large integers
In mathematics , Knuth's up-arrow notation is a method of notation for very large integers , introduced by Donald Knuth in 1976.[ 1]
In his 1947 paper,[ 2] R. L. Goodstein introduced the specific sequence of operations that are now called hyperoperations tetration , pentation , etc., for the extended operations beyond exponentiation . The sequence starts with a unary operation (the successor function with n = 0), and continues with the binary operations of addition (n = 1), multiplication (n = 2), exponentiation (n = 3), tetration (n = 4), pentation (n = 5), etc.
Various notations have been used to represent hyperoperations. One such notation is
H
n
(
a
,
b
)
{\displaystyle H_{n}(a,b)}
↑
{\displaystyle \uparrow }
the single arrow
↑
{\displaystyle \uparrow }
exponentiation (iterated multiplication)
2
↑
4
=
H
3
(
2
,
4
)
=
2
×
(
2
×
(
2
×
2
)
)
=
2
4
=
16
{\displaystyle 2\uparrow 4=H_{3}(2,4)=2\times (2\times (2\times 2))=2^{4}=16}
the double arrow
↑↑
{\displaystyle \uparrow \uparrow }
tetration (iterated exponentiation)
2
↑↑
4
=
H
4
(
2
,
4
)
=
2
↑
(
2
↑
(
2
↑
2
)
)
=
2
2
2
2
=
2
16
=
65
,
536
{\displaystyle 2\uparrow \uparrow 4=H_{4}(2,4)=2\uparrow (2\uparrow (2\uparrow 2))=2^{2^{2^{2}}}=2^{16}=65,536}
the triple arrow
↑↑↑
{\displaystyle \uparrow \uparrow \uparrow }
pentation (iterated tetration)
2
↑↑↑
4
=
H
5
(
2
,
4
)
=
2
↑↑
(
2
↑↑
(
2
↑↑
2
)
)
=
2
↑↑
(
2
↑↑
(
2
↑
2
)
)
=
2
↑↑
(
2
↑↑
4
)
=
2
↑
(
2
↑
(
2
↑
…
)
)
⏟
=
2
2
⋯
2
⏟
2
↑↑
4
copies of
2
65,536 2s
{\displaystyle {\begin{aligned}2\uparrow \uparrow \uparrow 4=H_{5}(2,4)=2\uparrow \uparrow (2\uparrow \uparrow (2\uparrow \uparrow 2))\\&=2\uparrow \uparrow (2\uparrow \uparrow (2\uparrow 2))\\&=2\uparrow \uparrow (2\uparrow \uparrow 4)\\&=\underbrace {2\uparrow (2\uparrow (2\uparrow \dots ))} \;=\;\underbrace {\;2^{2^{\cdots ^{2}}}} \\&\;\;\;\;\;2\uparrow \uparrow 4{\mbox{ copies of }}2\;\;\;\;\;{\mbox{65,536 2s}}\\\end{aligned}}}
The general definition of the up-arrow notation is as follows (for
a
≥
0
,
n
≥
1
,
b
≥
0
{\displaystyle a\geq 0,n\geq 1,b\geq 0}
a
↑
n
b
=
H
n
+
2
(
a
,
b
)
=
a
[
n
+
2
]
b
.
{\displaystyle a\uparrow ^{n}b=H_{n+2}(a,b)=a[n+2]b.}
↑
n
{\displaystyle \uparrow ^{n}}
n arrows, so for example
2
↑↑↑↑
3
=
2
↑
4
3.
{\displaystyle 2\uparrow \uparrow \uparrow \uparrow 3=2\uparrow ^{4}3.}
The hyperoperations naturally extend the arithmetic operations of addition and multiplication as follows.
Addition by a natural number is defined as iterated incrementation:
H
1
(
a
,
b
)
=
a
+
b
=
a
+
1
+
1
+
⋯
+
1
⏟
b
copies of
1
{\displaystyle {\begin{matrix}H_{1}(a,b)=a+b=&a+\underbrace {1+1+\dots +1} \\&b{\mbox{ copies of }}1\end{matrix}}}
Multiplication by a natural number is defined as iterated addition :
H
2
(
a
,
b
)
=
a
×
b
=
a
+
a
+
⋯
+
a
⏟
b
copies of
a
{\displaystyle {\begin{matrix}H_{2}(a,b)=a\times b=&\underbrace {a+a+\dots +a} \\&b{\mbox{ copies of }}a\end{matrix}}}
For example,
4
×
3
=
4
+
4
+
4
⏟
=
12
3
copies of
4
{\displaystyle {\begin{matrix}4\times 3&=&\underbrace {4+4+4} &=&12\\&&3{\mbox{ copies of }}4\end{matrix}}}
Exponentiation for a natural power
b
{\displaystyle b}
a
↑
b
=
H
3
(
a
,
b
)
=
a
b
=
a
×
a
×
⋯
×
a
⏟
b
copies of
a
{\displaystyle {\begin{matrix}a\uparrow b=H_{3}(a,b)=a^{b}=&\underbrace {a\times a\times \dots \times a} \\&b{\mbox{ copies of }}a\end{matrix}}}
For example,
4
↑
3
=
4
3
=
4
×
4
×
4
⏟
=
64
3
copies of
4
{\displaystyle {\begin{matrix}4\uparrow 3=4^{3}=&\underbrace {4\times 4\times 4} &=&64\\&3{\mbox{ copies of }}4\end{matrix}}}
Tetration is defined as iterated exponentiation, which Knuth denoted by a “double arrow”:
a
↑↑
b
=
H
4
(
a
,
b
)
=
a
a
.
.
.
a
⏟
=
a
↑
(
a
↑
(
⋯
↑
a
)
)
⏟
b
copies of
a
b
copies of
a
{\displaystyle {\begin{matrix}a\uparrow \uparrow b=H_{4}(a,b)=&\underbrace {a^{a^{{}^{.\,^{.\,^{.\,^{a}}}}}}} &=&\underbrace {a\uparrow (a\uparrow (\dots \uparrow a))} \\&b{\mbox{ copies of }}a&&b{\mbox{ copies of }}a\end{matrix}}}
For example,
4
↑↑
3
=
4
4
4
⏟
=
4
↑
(
4
↑
4
)
⏟
=
4
256
≈
1.34078079
×
10
154
3
copies of
4
3
copies of
4
{\displaystyle {\begin{matrix}4\uparrow \uparrow 3=&\underbrace {4^{4^{4}}} &=&\underbrace {4\uparrow (4\uparrow 4)} &=&4^{256}&\approx &1.34078079\times 10^{154}&\\&3{\mbox{ copies of }}4&&3{\mbox{ copies of }}4\end{matrix}}}
Expressions are evaluated from right to left, as the operators are defined to be right-associative .
According to this definition,
3
↑↑
2
=
3
3
=
27
{\displaystyle 3\uparrow \uparrow 2=3^{3}=27}
3
↑↑
3
=
3
3
3
=
3
27
=
7
,
625
,
597
,
484
,
987
{\displaystyle 3\uparrow \uparrow 3=3^{3^{3}}=3^{27}=7,625,597,484,987}
3
↑↑
4
=
3
3
3
3
=
3
3
27
=
3
7625597484987
≈
1.2580143
×
10
3638334640024
{\displaystyle 3\uparrow \uparrow 4=3^{3^{3^{3}}}=3^{3^{27}}=3^{7625597484987}\approx 1.2580143\times 10^{3638334640024}}
3
↑↑
5
=
3
3
3
3
3
=
3
3
3
27
=
3
3
7625597484987
≈
3
1.2580143
×
10
3638334640024
{\displaystyle 3\uparrow \uparrow 5=3^{3^{3^{3^{3}}}}=3^{3^{3^{27}}}=3^{3^{7625597484987}}\approx 3^{1.2580143\times 10^{3638334640024}}}
etc. This already leads to some fairly large numbers, but the hyperoperator sequence does not stop here.
Pentation , defined as iterated tetration, is represented by the “triple arrow”:
a
↑↑↑
b
=
H
5
(
a
,
b
)
=
a
↑↑
(
a
↑↑
(
⋯
↑↑
a
)
)
⏟
b
copies of
a
{\displaystyle {\begin{matrix}a\uparrow \uparrow \uparrow b=H_{5}(a,b)=&\underbrace {a_{}\uparrow \uparrow (a\uparrow \uparrow (\dots \uparrow \uparrow a))} \\&b{\mbox{ copies of }}a\end{matrix}}}
Hexation , defined as iterated pentation, is represented by the “quadruple arrow”:
a
↑↑↑↑
b
=
H
6
(
a
,
b
)
=
a
↑↑↑
(
a
↑↑↑
(
⋯
↑↑↑
a
)
)
⏟
b
copies of
a
{\displaystyle {\begin{matrix}a\uparrow \uparrow \uparrow \uparrow b=H_{6}(a,b)=&\underbrace {a_{}\uparrow \uparrow \uparrow (a\uparrow \uparrow \uparrow (\dots \uparrow \uparrow \uparrow a))} \\&b{\mbox{ copies of }}a\end{matrix}}}
and so on. The general rule is that an
n
{\displaystyle n}
n
−
1
{\displaystyle n-1}
a
↑
↑
…
↑
⏟
n
b
=
a
↑
…
↑
⏟
n
−
1
(
a
↑
…
↑
⏟
n
−
1
(
…
↑
…
↑
⏟
n
−
1
a
)
)
⏟
b
copies of
a
{\displaystyle {\begin{matrix}a\ \underbrace {\uparrow _{}\uparrow \!\!\dots \!\!\uparrow } _{n}\ b=\underbrace {a\ \underbrace {\uparrow \!\!\dots \!\!\uparrow } _{n-1}\ (a\ \underbrace {\uparrow _{}\!\!\dots \!\!\uparrow } _{n-1}\ (\dots \ \underbrace {\uparrow _{}\!\!\dots \!\!\uparrow } _{n-1}\ a))} _{b{\text{ copies of }}a}\end{matrix}}}
Examples:
3
↑↑↑
2
=
3
↑↑
3
=
3
3
3
=
3
27
=
7
,
625
,
597
,
484
,
987
{\displaystyle 3\uparrow \uparrow \uparrow 2=3\uparrow \uparrow 3=3^{3^{3}}=3^{27}=7,625,597,484,987}
3
↑↑↑
3
=
3
↑↑
(
3
↑↑
3
)
=
3
↑↑
(
3
↑
3
↑
3
)
=
3
↑
3
↑
⋯
↑
3
⏟
3
↑
3
↑
3
copies of
3
=
3
↑
3
↑
⋯
↑
3
⏟
7,625,597,484,987 copies of 3
=
3
3
3
3
⋅
⋅
⋅
⋅
3
⏟
7,625,597,484,987 copies of 3
{\displaystyle {\begin{matrix}3\uparrow \uparrow \uparrow 3=3\uparrow \uparrow (3\uparrow \uparrow 3)=3\uparrow \uparrow (3\uparrow 3\uparrow 3)=&\underbrace {3_{}\uparrow 3\uparrow \dots \uparrow 3} \\&3\uparrow 3\uparrow 3{\mbox{ copies of }}3\end{matrix}}{\begin{matrix}=&\underbrace {3_{}\uparrow 3\uparrow \dots \uparrow 3} \\&{\mbox{7,625,597,484,987 copies of 3}}\end{matrix}}{\begin{matrix}=&\underbrace {3^{3^{3^{3^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{3}}}}}}}}} \\&{\mbox{7,625,597,484,987 copies of 3}}\end{matrix}}}
In expressions such as
a
b
{\displaystyle a^{b}}
b
{\displaystyle b}
a
{\displaystyle a}
programming languages and plain-text e-mail — do not support superscript typesetting. People have adopted the linear notation
a
↑
b
{\displaystyle a\uparrow b}
character set does not contain an up arrow, the caret (^) is used instead.
The superscript notation
a
b
{\displaystyle a^{b}}
a
↑
b
{\displaystyle a\uparrow b}
a
↑
n
b
{\displaystyle a\uparrow ^{n}b}
a
↑
4
b
=
a
↑↑↑↑
b
{\displaystyle a\uparrow ^{4}b=a\uparrow \uparrow \uparrow \uparrow b}
Writing out up-arrow notation in terms of powers [ edit ] Attempting to write
a
↑↑
b
{\displaystyle a\uparrow \uparrow b}
power tower .
For example:
a
↑↑
4
=
a
↑
(
a
↑
(
a
↑
a
)
)
=
a
a
a
a
{\displaystyle a\uparrow \uparrow 4=a\uparrow (a\uparrow (a\uparrow a))=a^{a^{a^{a}}}}
If
b
{\displaystyle b}
a
↑↑
b
=
a
a
.
.
.
a
⏟
b
{\displaystyle a\uparrow \uparrow b=\underbrace {a^{a^{.^{.^{.{a}}}}}} _{b}}
Continuing with this notation,
a
↑↑↑
b
{\displaystyle a\uparrow \uparrow \uparrow b}
a
↑↑↑
4
=
a
↑↑
(
a
↑↑
(
a
↑↑
a
)
)
=
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
a
{\displaystyle a\uparrow \uparrow \uparrow 4=a\uparrow \uparrow (a\uparrow \uparrow (a\uparrow \uparrow a))=\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{a}}}}
Again, if
b
{\displaystyle b}
a
↑↑↑
b
=
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
⋮
⏟
a
}
b
{\displaystyle a\uparrow \uparrow \uparrow b=\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}b}
Furthermore,
a
↑↑↑↑
b
{\displaystyle a\uparrow \uparrow \uparrow \uparrow b}
a
↑↑↑↑
4
=
a
↑↑↑
(
a
↑↑↑
(
a
↑↑↑
a
)
)
=
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
⋮
⏟
a
}
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
⋮
⏟
a
}
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
⋮
⏟
a
}
a
{\displaystyle a\uparrow \uparrow \uparrow \uparrow 4=a\uparrow \uparrow \uparrow (a\uparrow \uparrow \uparrow (a\uparrow \uparrow \uparrow a))=\left.\left.\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}a}
And more generally:
a
↑↑↑↑
b
=
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
⋮
⏟
a
}
a
a
.
.
.
a
⏟
a
a
.
.
.
a
⏟
⋮
⏟
a
}
⋯
}
a
⏟
b
{\displaystyle a\uparrow \uparrow \uparrow \uparrow b=\underbrace {\left.\left.\left.\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {a^{a^{.^{.^{.{a}}}}}} _{\underbrace {\vdots } _{a}}}\right\}\cdots \right\}a} _{b}}
This might be carried out indefinitely to represent
a
↑
n
b
{\displaystyle a\uparrow ^{n}b}
a
{\displaystyle a}
n
{\displaystyle n}
b
{\displaystyle b}
The Rudy Rucker notation
b
a
{\displaystyle ^{b}a}
tetration allows us to make these diagrams slightly simpler while still employing a geometric representation (we could call these tetration towers ).
a
↑↑
b
=
b
a
{\displaystyle a\uparrow \uparrow b={}^{b}a}
a
↑↑↑
b
=
a
.
.
.
a
a
⏟
b
{\displaystyle a\uparrow \uparrow \uparrow b=\underbrace {^{^{^{^{^{a}.}.}.}a}a} _{b}}
a
↑↑↑↑
b
=
a
.
.
.
a
a
⏟
a
.
.
.
a
a
⏟
⋮
⏟
a
}
b
{\displaystyle a\uparrow \uparrow \uparrow \uparrow b=\left.\underbrace {^{^{^{^{^{a}.}.}.}a}a} _{\underbrace {^{^{^{^{^{a}.}.}.}a}a} _{\underbrace {\vdots } _{a}}}\right\}b}
Finally, as an example, the fourth Ackermann number
4
↑
4
4
{\displaystyle 4\uparrow ^{4}4}
4
.
.
.
4
4
⏟
4
.
.
.
4
4
⏟
4
.
.
.
4
4
⏟
4
=
4
.
.
.
4
4
⏟
4
.
.
.
4
4
⏟
4
4
4
4
{\displaystyle \underbrace {^{^{^{^{^{4}.}.}.}4}4} _{\underbrace {^{^{^{^{^{4}.}.}.}4}4} _{\underbrace {^{^{^{^{^{4}.}.}.}4}4} _{4}}}=\underbrace {^{^{^{^{^{4}.}.}.}4}4} _{\underbrace {^{^{^{^{^{4}.}.}.}4}4} _{^{^{^{4}4}4}4}}}
Some numbers are so large that multiple arrows of Knuth's up-arrow notation become too cumbersome; then an n -arrow operator
↑
n
{\displaystyle \uparrow ^{n}}
hyper operators .
Some numbers are so large that even that notation is not sufficient. The Conway chained arrow notation can then be used: a chain of three elements is equivalent with the other notations, but a chain of four or more is even more powerful.
a
↑
n
b
=
a
[
n
+
2
]
b
=
a
→
b
→
n
(Knuth)
(hyperoperation)
(Conway)
{\displaystyle {\begin{matrix}a\uparrow ^{n}b&=&a[n+2]b&=&a\to b\to n\\{\mbox{(Knuth)}}&&{\mbox{(hyperoperation)}}&&{\mbox{(Conway)}}\end{matrix}}}
6
↑↑
4
{\displaystyle 6\uparrow \uparrow 4}
6
6
.
.
.
6
⏟
4
{\displaystyle \underbrace {6^{6^{.^{.^{.^{6}}}}}} _{4}}
6
↑↑
4
{\displaystyle 6\uparrow \uparrow 4}
6
6
6
6
{\displaystyle 6^{6^{6^{6}}}}
6
6
46
,
656
{\displaystyle 6^{6^{46,656}}}
6
6
.
.
.
6
⏟
4
{\displaystyle \underbrace {6^{6^{.^{.^{.^{6}}}}}} _{4}}
10
↑
(
3
×
10
↑
(
3
×
10
↑
15
)
+
3
)
{\displaystyle 10\uparrow (3\times 10\uparrow (3\times 10\uparrow 15)+3)}
100000...000
⏟
300000...003
⏟
300000...000
⏟
15
{\displaystyle \underbrace {100000...000} _{\underbrace {300000...003} _{\underbrace {300000...000} _{15}}}}
10
3
×
10
3
×
10
15
+
3
{\displaystyle 10^{3\times 10^{3\times 10^{15}}+3}}
Even faster-growing functions can be categorized using an ordinal analysis called the fast-growing hierarchy . The fast-growing hierarchy uses successive function iteration and diagonalization to systematically create faster-growing functions from some base function
f
(
x
)
{\displaystyle f(x)}
f
0
(
x
)
=
x
+
1
{\displaystyle f_{0}(x)=x+1}
f
2
(
x
)
{\displaystyle f_{2}(x)}
f
3
(
x
)
{\displaystyle f_{3}(x)}
f
ω
(
x
)
{\displaystyle f_{\omega }(x)}
Ackermann function ,
f
ω
+
1
(
x
)
{\displaystyle f_{\omega +1}(x)}
Graham's number , and
f
ω
2
(
x
)
{\displaystyle f_{\omega ^{2}}(x)}
These functions are all computable. Even faster computable functions, such as the Goodstein sequence and the TREE sequence require the usage of large ordinals, may occur in certain combinatorical and proof-theoretic contexts. There exist functions which grow uncomputably fast, such as the Busy Beaver , whose very nature will be completely out of reach from any up-arrow, or even any ordinal-based analysis.
Without reference to hyperoperation the up-arrow operators can be formally defined by
a
↑
n
b
=
{
a
b
,
if
n
=
1
;
1
,
if
n
>
1
and
b
=
0
;
a
↑
n
−
1
(
a
↑
n
(
b
−
1
)
)
,
otherwise
{\displaystyle a\uparrow ^{n}b={\begin{cases}a^{b},&{\text{if }}n=1;\\1,&{\text{if }}n>1{\text{ and }}b=0;\\a\uparrow ^{n-1}(a\uparrow ^{n}(b-1)),&{\text{otherwise }}\end{cases}}}
for all integers
a
,
b
,
n
{\displaystyle a,b,n}
a
≥
0
,
n
≥
1
,
b
≥
0
{\displaystyle a\geq 0,n\geq 1,b\geq 0}
[ nb 1]
This definition uses exponentiation
(
a
↑
1
b
=
a
↑
b
=
a
b
)
{\displaystyle (a\uparrow ^{1}b=a\uparrow b=a^{b})}
tetration
(
a
↑
2
b
=
a
↑↑
b
)
{\displaystyle (a\uparrow ^{2}b=a\uparrow \uparrow b)}
hyperoperation sequence except it omits the three more basic operations of succession , addition and multiplication .
One can alternatively choose multiplication
(
a
↑
0
b
=
a
×
b
)
{\displaystyle (a\uparrow ^{0}b=a\times b)}
exponentiation becomes repeated multiplication. The formal definition would be
a
↑
n
b
=
{
a
×
b
,
if
n
=
0
;
1
,
if
n
>
0
and
b
=
0
;
a
↑
n
−
1
(
a
↑
n
(
b
−
1
)
)
,
otherwise
{\displaystyle a\uparrow ^{n}b={\begin{cases}a\times b,&{\text{if }}n=0;\\1,&{\text{if }}n>0{\text{ and }}b=0;\\a\uparrow ^{n-1}(a\uparrow ^{n}(b-1)),&{\text{otherwise }}\end{cases}}}
for all integers
a
,
b
,
n
{\displaystyle a,b,n}
a
≥
0
,
n
≥
0
,
b
≥
0
{\displaystyle a\geq 0,n\geq 0,b\geq 0}
Note, however, that Knuth did not define the "nil-arrow" (
↑
0
{\displaystyle \uparrow ^{0}}
H
n
(
a
,
b
)
=
a
[
n
]
b
=
a
↑
n
−
2
b
for
n
≥
0.
{\displaystyle H_{n}(a,b)=a[n]b=a\uparrow ^{n-2}b{\text{ for }}n\geq 0.}
The up-arrow operation is a right-associative operation , that is,
a
↑
b
↑
c
{\displaystyle a\uparrow b\uparrow c}
a
↑
(
b
↑
c
)
{\displaystyle a\uparrow (b\uparrow c)}
(
a
↑
b
)
↑
c
{\displaystyle (a\uparrow b)\uparrow c}
Computing
0
↑
n
b
=
H
n
+
2
(
0
,
b
)
=
0
[
n
+
2
]
b
{\displaystyle 0\uparrow ^{n}b=H_{n+2}(0,b)=0[n+2]b}
0, when n = 0 [ nb 2]
1, when n = 1 and b = 0 [ nb 1] [ nb 3]
0, when n = 1 and b > 0 [ nb 1] [ nb 3]
1, when n > 1 and b is even (including 0)
0, when n > 1 and b is odd Computing
2
↑
n
b
{\displaystyle 2\uparrow ^{n}b}
2
b
{\displaystyle 2^{b}}
Values of
2
↑
n
b
{\displaystyle 2\uparrow ^{n}b}
H
n
+
2
(
2
,
b
)
{\displaystyle H_{n+2}(2,b)}
2
[
n
+
2
]
b
{\displaystyle 2[n+2]b}
2 → b → n
b
ⁿ
1
2
3
4
5
6
formula
1
2
4
8
16
32
64
2
b
{\displaystyle 2^{b}}
2
2
4
16
65536
2
65,536
≈
2.0
×
10
19,728
{\displaystyle 2^{65{,}536}\approx 2.0\times 10^{19{,}728}}
2
2
65,536
≈
10
6.0
×
10
19,727
{\displaystyle 2^{2^{65{,}536}}\approx 10^{6.0\times 10^{19{,}727}}}
2
↑↑
b
{\displaystyle 2\uparrow \uparrow b}
3
2
4
65536
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
2
.
.
.
2
⏟
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
2
.
.
.
2
⏟
2
2
.
.
.
2
⏟
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
↑↑↑
b
{\displaystyle 2\uparrow \uparrow \uparrow b}
4
2
4
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
.
.
.
2
2
⏟
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{2}.}.}.}2}2} \\\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
.
.
.
2
2
⏟
2
.
.
.
2
2
⏟
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{2}.}.}.}2}2} \\\underbrace {^{^{^{^{^{2}.}.}.}2}2} \\\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
.
.
.
2
2
⏟
2
.
.
.
2
2
⏟
2
.
.
.
2
2
⏟
2
2
.
.
.
2
⏟
65,536
copies of
2
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{2}.}.}.}2}2} \\\underbrace {^{^{^{^{^{2}.}.}.}2}2} \\\underbrace {^{^{^{^{^{2}.}.}.}2}2} \\\underbrace {2_{}^{2^{{}^{.\,^{.\,^{.\,^{2}}}}}}} \\65{,}536{\mbox{ copies of }}2\end{matrix}}}
2
↑↑↑↑
b
{\displaystyle 2\uparrow \uparrow \uparrow \uparrow b}
The table is the same as that of the Ackermann function , except for a shift in
n
{\displaystyle n}
b
{\displaystyle b}
We place the numbers
3
b
{\displaystyle 3^{b}}
Values of
3
↑
n
b
{\displaystyle 3\uparrow ^{n}b}
H
n
+
2
(
3
,
b
)
{\displaystyle H_{n+2}(3,b)}
3
[
n
+
2
]
b
{\displaystyle 3[n+2]b}
3 → b → n
b
ⁿ
1
2
3
4
5
formula
1
3
9
27
81
243
3
b
{\displaystyle 3^{b}}
2
3
27
7,625,597,484,987
3
7,625,597,484,987
≈
1.3
×
10
3,638,334,640,024
{\displaystyle 3^{7{,}625{,}597{,}484{,}987}\approx 1.3\times 10^{3{,}638{,}334{,}640{,}024}}
3
3
7,625,597,484,987
{\displaystyle 3^{3^{7{,}625{,}597{,}484{,}987}}}
3
↑↑
b
{\displaystyle 3\uparrow \uparrow b}
3
3
7,625,597,484,987
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
3
.
.
.
3
⏟
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
3
.
.
.
3
⏟
3
3
.
.
.
3
⏟
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
↑↑↑
b
{\displaystyle 3\uparrow \uparrow \uparrow b}
4
3
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
.
.
.
3
3
⏟
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{3}.}.}.}3}3} \\\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
.
.
.
3
3
⏟
3
.
.
.
3
3
⏟
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{3}.}.}.}3}3} \\\underbrace {^{^{^{^{^{3}.}.}.}3}3} \\\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
.
.
.
3
3
⏟
3
.
.
.
3
3
⏟
3
.
.
.
3
3
⏟
3
3
.
.
.
3
⏟
7,625,597,484,987
copies of
3
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{3}.}.}.}3}3} \\\underbrace {^{^{^{^{^{3}.}.}.}3}3} \\\underbrace {^{^{^{^{^{3}.}.}.}3}3} \\\underbrace {3_{}^{3^{{}^{.\,^{.\,^{.\,^{3}}}}}}} \\7{,}625{,}597{,}484{,}987{\mbox{ copies of }}3\end{matrix}}}
3
↑↑↑↑
b
{\displaystyle 3\uparrow \uparrow \uparrow \uparrow b}
We place the numbers
4
b
{\displaystyle 4^{b}}
Values of
4
↑
n
b
{\displaystyle 4\uparrow ^{n}b}
H
n
+
2
(
4
,
b
)
{\displaystyle H_{n+2}(4,b)}
4
[
n
+
2
]
b
{\displaystyle 4[n+2]b}
4 → b → n
b
ⁿ
1
2
3
4
5
formula
1
4
16
64
256
1024
4
b
{\displaystyle 4^{b}}
2
4
256
4
256
≈
1.34
×
10
154
{\displaystyle 4^{256}\approx 1.34\times 10^{154}}
4
4
256
≈
10
8.0
×
10
153
{\displaystyle 4^{4^{256}}\approx 10^{8.0\times 10^{153}}}
4
4
4
256
{\displaystyle 4^{4^{4^{256}}}}
4
↑↑
b
{\displaystyle 4\uparrow \uparrow b}
3
4
4
4
256
≈
10
8.0
×
10
153
{\displaystyle 4^{4^{256}}\approx 10^{8.0\times 10^{153}}}
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
↑↑↑
b
{\displaystyle 4\uparrow \uparrow \uparrow b}
4
4
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
.
.
.
4
4
⏟
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{4}.}.}.}4}4} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
.
.
.
4
4
⏟
4
.
.
.
4
4
⏟
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{4}.}.}.}4}4} \\\underbrace {^{^{^{^{^{4}.}.}.}4}4} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
.
.
.
4
4
⏟
4
.
.
.
4
4
⏟
4
.
.
.
4
4
⏟
4
4
.
.
.
4
⏟
4
4
.
.
.
4
⏟
4
4
256
copies of
4
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{4}.}.}.}4}4} \\\underbrace {^{^{^{^{^{4}.}.}.}4}4} \\\underbrace {^{^{^{^{^{4}.}.}.}4}4} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\\underbrace {4_{}^{4^{{}^{.\,^{.\,^{.\,^{4}}}}}}} \\4^{4^{256}}{\mbox{ copies of }}4\end{matrix}}}
4
↑↑↑↑
b
{\displaystyle 4\uparrow \uparrow \uparrow \uparrow b}
We place the numbers
10
b
{\displaystyle 10^{b}}
Values of
10
↑
n
b
{\displaystyle 10\uparrow ^{n}b}
H
n
+
2
(
10
,
b
)
{\displaystyle H_{n+2}(10,b)}
10
[
n
+
2
]
b
{\displaystyle 10[n+2]b}
10 → b → n
b
ⁿ
1
2
3
4
5
formula
1
10
100
1,000
10,000
100,000
10
b
{\displaystyle 10^{b}}
2
10
10,000,000,000
10
10
,
000
,
000
,
000
{\displaystyle 10^{10,000,000,000}}
10
10
10
,
000
,
000
,
000
{\displaystyle 10^{10^{10,000,000,000}}}
10
10
10
10
,
000
,
000
,
000
{\displaystyle 10^{10^{10^{10,000,000,000}}}}
10
↑↑
b
{\displaystyle 10\uparrow \uparrow b}
3
10
10
10
.
.
.
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\10{\mbox{ copies of }}10\end{matrix}}}
10
10
.
.
.
10
⏟
10
10
.
.
.
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\10{\mbox{ copies of }}10\end{matrix}}}
10
10
.
.
.
10
⏟
10
10
.
.
.
10
⏟
10
10
.
.
.
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\10{\mbox{ copies of }}10\end{matrix}}}
10
10
.
.
.
10
⏟
10
10
.
.
.
10
⏟
10
10
.
.
.
10
⏟
10
10
.
.
.
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\\underbrace {10_{}^{10^{{}^{.\,^{.\,^{.\,^{10}}}}}}} \\10{\mbox{ copies of }}10\end{matrix}}}
10
↑↑↑
b
{\displaystyle 10\uparrow \uparrow \uparrow b}
4
10
10
.
.
.
10
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\10{\mbox{ copies of }}10\end{matrix}}}
10
.
.
.
10
10
⏟
10
.
.
.
10
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\10{\mbox{ copies of }}10\end{matrix}}}
10
.
.
.
10
10
⏟
10
.
.
.
10
10
⏟
10
.
.
.
10
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\10{\mbox{ copies of }}10\end{matrix}}}
10
.
.
.
10
10
⏟
10
.
.
.
10
10
⏟
10
.
.
.
10
10
⏟
10
.
.
.
10
10
⏟
10
copies of
10
{\displaystyle {\begin{matrix}\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\\underbrace {^{^{^{^{^{10}.}.}.}10}10} \\10{\mbox{ copies of }}10\end{matrix}}}
10
↑↑↑↑
b
{\displaystyle 10\uparrow \uparrow \uparrow \uparrow b}
For 2 ≤ b ≤ 9 the numerical order of the numbers
10
↑
n
b
{\displaystyle 10\uparrow ^{n}b}
lexicographical order with n as the most significant number, so for the numbers of these 8 columns the numerical order is simply line-by-line. The same applies for the numbers in the 97 columns with 3 ≤ b ≤ 99, and if we start from n = 1 even for 3 ≤ b ≤ 9,999,999,999.
Primary Inverse for left argumentInverse for right argument Related articles
Examples Expression
Related
![{\displaystyle a\uparrow ^{n}b=H_{n+2}(a,b)=a[n+2]b.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/85c46ffdf75f03948d485eeffeea3486354eb211)
![{\displaystyle {\begin{matrix}a\uparrow ^{n}b&=&a[n+2]b&=&a\to b\to n\\{\mbox{(Knuth)}}&&{\mbox{(hyperoperation)}}&&{\mbox{(Conway)}}\end{matrix}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d66ddee4f3e087805a66045c64577585c46e685)
![{\displaystyle H_{n}(a,b)=a[n]b=a\uparrow ^{n-2}b{\text{ for }}n\geq 0.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/013f5aded05b23ab8552efd77dca57c4694d8b1e)
![{\displaystyle 0\uparrow ^{n}b=H_{n+2}(0,b)=0[n+2]b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c74b3c65305086a3ffe141df38df0ae9061806ef)
![{\displaystyle 2[n+2]b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eb4b0efd43c05147b6e2cf0f82e4e8578d9bb164)
![{\displaystyle 3[n+2]b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ba2687ff69ff6cb85e93440bbf3c34422f0c1399)
![{\displaystyle 4[n+2]b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7e886a59164bc509fe87631a45781e0fc21586a2)
![{\displaystyle 10[n+2]b}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5bb5b9086a31901fae5ebe46f61c0aba5b340ec4)
