Hyperoperation

This article is about the arithmetic concept. For the group theory hyperoperation concept, seeHyperstructure.

Inmathematics,thehyperoperation sequenceis an infinitesequenceof arithmetic operations (calledhyperoperationsin this context)that starts with aunary operation(thesuccessor functionwithn= 0). The sequence continues with thebinary operationsofaddition(n= 1),multiplication(n= 2), andexponentiation(n= 3).

After that, the sequence proceeds with further binary operations extending beyond exponentiation, usingright-associativity.For the operations beyond exponentiation, thenth member of this sequence is named byReuben Goodsteinafter theGreek prefixofnsuffixed with-ation(such astetration(n= 4),pentation(n= 5), hexation (n= 6), etc.)and can be written as usingn− 2 arrows inKnuth's up-arrow notation. Each hyperoperation may be understoodrecursivelyin terms of the previous one by:

$a[n]b=\underbrace {a[n-1](a[n-1](a[n-1](\cdots [n-1](a[n-1](a[n-1]a))\cdots )))} _{\displaystyle b{\mbox{ copies of }}a},\quad n\geq 2$

It may also be defined according to the recursion rule part of the definition, as in Knuth's up-arrow version of theAckermann function:

$a[n]b=a[n-1]\left(a[n]\left(b-1\right)\right),\quad n\geq 1$

This can be used to easily show numbers much larger than those whichscientific notationcan, such asSkewes's numberandgoogolplexplex(e.g.$50[50]50$is much larger than Skewes's number and googolplexplex), but there are some numbers which even they cannot easily show, such asGraham's numberandTREE(3).

This recursion rule is common to many variants of hyperoperations.

Definition

Definition, most common

Thehyperoperation sequence$H_{n}(a,b)\colon (\mathbb {N} _{0})^{3}\rightarrow \mathbb {N} _{0}$is thesequenceofbinary operations$H_{n}\colon (\mathbb {N} _{0})^{2}\rightarrow \mathbb {N} _{0}$,definedrecursivelyas follows:

$H_{n}(a,b)=a[n]b={\begin{cases}b+1&{\text{if }}n=0\\a&{\text{if }}n=1{\text{ and }}b=0\\0&{\text{if }}n=2{\text{ and }}b=0\\1&{\text{if }}n\geq 3{\text{ and }}b=0\\H_{n-1}(a,H_{n}(a,b-1))&{\text{otherwise}}\end{cases}}$

(Note that forn= 0, thebinary operationessentially reduces to aunary operation(successor function) by ignoring the first argument.)

Forn= 0, 1, 2, 3, this definition reproduces the basic arithmetic operations ofsuccessor(which is a unary operation),addition,multiplication,andexponentiation,respectively, as

${\begin{aligned}H_{0}(a,b)&=1+b,\\H_{1}(a,b)&=a+b,\\H_{2}(a,b)&=a\times b,\\H_{3}(a,b)&=a\uparrow {b}=a^{b}.\end{aligned}}$

The$H_{n}$operations forn≥ 3 can be written inKnuth's up-arrow notation.

So what will be the next operation after exponentiation? We defined multiplication so that$H_{2}(a,3)=a[2]3=a\times 3=a+a+a,$and defined exponentiation so that$H_{3}(a,3)=a[3]3=a\uparrow 3=a^{3}=a\times a\times a,$so it seems logical to define the next operation, tetration, so that$H_{4}(a,3)=a[4]3=a\uparrow \uparrow 3=\operatorname {tetration} (a,3)=a^{a^{a}},$with a tower of three 'a'. Analogously, the pentation of (a, 3) will be tetration(a, tetration(a, a)), with three "a" in it.

${\begin{aligned}H_{4}(a,b)&=a\uparrow \uparrow {b},\\H_{5}(a,b)&=a\uparrow \uparrow \uparrow {b},\\\ldots &\\H_{n}(a,b)&=a\uparrow ^{n-2}b{\text{ for }}n\geq 3,\\\ldots &\\\end{aligned}}$

Knuth's notation could be extended to negative indices ≥ −2 in such a way as to agree with the entire hyperoperation sequence, except for the lag in the indexing:

$H_{n}(a,b)=a\uparrow ^{n-2}b{\text{ for }}n\geq 0.$

The hyperoperations can thus be seen as an answer to the question "what's next" in thesequence:successor,addition,multiplication,exponentiation,and so on. Noting that

${\begin{aligned}a+b&=1+(a+(b-1))\\a\cdot b&=a+(a\cdot (b-1))\\a^{b}&=a\cdot \left(a^{(b-1)}\right)\\a[4]b&=a^{a[4](b-1)}\end{aligned}}$

the relationship between basic arithmetic operations is illustrated, allowing the higher operations to be defined naturally as above. The parameters of the hyperoperation hierarchy are sometimes referred to by their analogous exponentiation term;soais thebase,bis theexponent(orhyperexponent),andnis therank(orgrade),and moreover,$H_{n}(a,b)$is read as "thebthn-ation ofa",e.g.$H_{4}(7,9)$is read as "the 9th tetration of 7", and$H_{123}(456,789)$is read as "the 789th 123-ation of 456".

In common terms, the hyperoperations are ways of compounding numbers that increase in growth based on the iteration of the previous hyperoperation. The concepts of successor, addition, multiplication and exponentiation are all hyperoperations; the successor operation (producingx+ 1 fromx) is the most primitive, the addition operator specifies the number of times 1 is to be added to itself to produce a final value, multiplication specifies the number of times a number is to be added to itself, and exponentiation refers to the number of times a number is to be multiplied by itself.

Definition, using iteration

Defineiterationof a functionfof two variables as

$f^{x}(a,b)={\begin{cases}f(a,b)&{\text{if }}x=1\\f(a,f^{x-1}(a,b))&{\text{if }}x>1\end{cases}}$

The hyperoperation sequence can be defined in terms of iteration, as follows. For all integers$x,n,a,b\geq 0,$define

${\begin{array}{lcl}H_{0}(a,b)&=&b+1\\H_{1}(a,0)&=&a\\H_{2}(a,0)&=&0\\H_{n+3}(a,0)&=&1\\H_{n+1}(a,b+1)&=&H_{n}^{b+1}(a,H_{n+1}(a,0))\\H_{n}^{x+2}(a,b)&=&H_{n}(a,H_{n}^{x+1}(a,b))\end{array}}$

As iteration isassociative,the last line can be replaced by

${\begin{array}{lcl}H_{n}^{x+2}(a,b)&=&H_{n}^{x+1}(a,H_{n}(a,b))\end{array}}$

Computation

The definitions of the hyperoperation sequence can naturally be transposed toterm rewriting systems (TRS).

TRS based on definition sub 1.1

The basic definition of the hyperoperation sequence corresponds with the reduction rules

${\begin{array}{lll}{\text{(r1)}}&H(0,a,b)&\rightarrow &S(b)\\{\text{(r2)}}&H(S(0),a,0)&\rightarrow &a\\{\text{(r3)}}&H(S(S(0)),a,0)&\rightarrow &0\\{\text{(r4)}}&H(S(S(S(n))),a,0)&\rightarrow &S(0)\\{\text{(r5)}}&H(S(n),a,S(b))&\rightarrow &H(n,a,H(S(n),a,b))\end{array}}$

To compute$H_{n}(a,b)$one can use astack,which initially contains the elements$\langle n,a,b\rangle$.

Then, repeatedly until no longer possible, three elements are popped and replaced according to the rules

${\begin{array}{lllllllll}{\text{(r1)}}&0&,&a&,&b&\rightarrow &(b+1)\\{\text{(r2)}}&1&,&a&,&0&\rightarrow &a\\{\text{(r3)}}&2&,&a&,&0&\rightarrow &0\\{\text{(r4)}}&(n+3)&,&a&,&0&\rightarrow &1\\{\text{(r5)}}&(n+1)&,&a&,&(b+1)&\rightarrow &n&,&a&,&(n+1)&,&a&,&b\end{array}}$

Schematically, starting from$\langle n,a,b\rangle$:

WHILEstackLength <> 1
{
POP3 elements;
PUSH1 or 5 elements according to the rules r1, r2, r3, r4, r5;
}

Example

Compute$H_{2}(2,2)\rightarrow _{*}4$.

The reduction sequence is

${\underline {H(S(S(0)),S(S(0)),S(S(0)))}}$

$\rightarrow _{r5}H(S(0),S(S(0)),{\underline {H(S(S(0)),S(S(0)),S(0))}})$

$\rightarrow _{r5}H(S(0),S(S(0)),H(S(0),S(S(0)),{\underline {H(S(S(0)),S(S(0)),0)}}))$

$\rightarrow _{r3}H(S(0),S(S(0)),{\underline {H(S(0),S(S(0)),0)}})$

$\rightarrow _{r2}{\underline {H(S(0),S(S(0)),S(S(0)))}}$

$\rightarrow _{r5}H(0,S(S(0)),{\underline {H(S(0),S(S(0)),S(0))}})$

$\rightarrow _{r5}H(0,S(S(0)),H(0,S(S(0)),{\underline {H(S(0),S(S(0)),0)}}))$

$\rightarrow _{r2}H(0,S(S(0)),{\underline {H(0,S(S(0)),S(S(0)))}})$

$\rightarrow _{r1}{\underline {H(0,S(S(0)),S(S(S(0))))}}$

$\rightarrow _{r1}S(S(S(S(0))))$

When implemented using a stack, on input$\langle 2,2,2\rangle$

the stack configurations	represent the equations
${\underline {2,2,2}}$	$H_{2}(2,2)$
$\rightarrow _{r5}1,2,{\underline {2,2,1}}$	$=H_{1}(2,H_{2}(2,1))$
$\rightarrow _{r5}1,2,1,2,{\underline {2,2,0}}$	$=H_{1}(2,H_{1}(2,H_{2}(2,0)))$
$\rightarrow _{r3}1,2,{\underline {1,2,0}}$	$=H_{1}(2,H_{1}(2,0))$
$\rightarrow _{r2}{\underline {1,2,2}}$	$=H_{1}(2,2)$
$\rightarrow _{r5}0,2,{\underline {1,2,1}}$	$=H_{0}(2,H_{1}(2,1))$
$\rightarrow _{r5}0,2,0,2,{\underline {1,2,0}}$	$=H_{0}(2,H_{0}(2,H_{1}(2,0)))$
$\rightarrow _{r2}0,2,{\underline {0,2,2}}$	$=H_{0}(2,H_{0}(2,2))$
$\rightarrow _{r1}{\underline {0,2,3}}$	$=H_{0}(2,3)$
$\rightarrow _{r1}4$	$=4$

TRS based on definition sub 1.2

The definition using iteration leads to a different set of reduction rules

${\begin{array}{lll}{\text{(r6)}}&H(S(0),0,a,b)&\rightarrow &S(b)\\{\text{(r7)}}&H(S(0),S(0),a,0)&\rightarrow &a\\{\text{(r8)}}&H(S(0),S(S(0)),a,0)&\rightarrow &0\\{\text{(r9)}}&H(S(0),S(S(S(n))),a,0)&\rightarrow &S(0)\\{\text{(r10)}}&H(S(0),S(n),a,S(b))&\rightarrow &H(S(b),n,a,H(S(0),S(n),a,0))\\{\text{(r11)}}&H(S(S(x)),n,a,b)&\rightarrow &H(S(0),n,a,H(S(x),n,a,b))\end{array}}$

As iteration isassociative,instead of rule r11 one can define

${\begin{array}{lll}{\text{(r12)}}&H(S(S(x)),n,a,b)&\rightarrow &H(S(x),n,a,H(S(0),n,a,b))\end{array}}$

Like in the previous section the computation of$H_{n}(a,b)=H_{n}^{1}(a,b)$can be implemented using a stack.

Initially the stack contains the four elements$\langle 1,n,a,b\rangle$.

Then, until termination, four elements are popped and replaced according to the rules

${\begin{array}{lllllllll}{\text{(r6)}}&1&,0&,a&,b&\rightarrow &(b+1)\\{\text{(r7)}}&1&,1&,a&,0&\rightarrow &a\\{\text{(r8)}}&1&,2&,a&,0&\rightarrow &0\\{\text{(r9)}}&1&,(n+3)&,a&,0&\rightarrow &1\\{\text{(r10)}}&1&,(n+1)&,a&,(b+1)&\rightarrow &(b+1)&,n&,a&,1&,(n+1)&,a&,0\\{\text{(r11)}}&(x+2)&,n&,a&,b&\rightarrow &1&,n&,a&,(x+1)&,n&,a&,b\end{array}}$

Schematically, starting from$\langle 1,n,a,b\rangle$:

WHILEstackLength <> 1
{
POP4 elements;
PUSH1 or 7 elements according to the rules r6, r7, r8, r9, r10, r11;
}

Example

Compute$H_{3}(0,3)\rightarrow _{*}0$.

On input$\langle 1,3,0,3\rangle$the successive stack configurations are

${\begin{aligned}&{\underline {1,3,0,3}}\rightarrow _{r10}3,2,0,{\underline {1,3,0,0}}\rightarrow _{r9}{\underline {3,2,0,1}}\rightarrow _{r11}1,2,0,{\underline {2,2,0,1}}\rightarrow _{r11}1,2,0,1,2,0,{\underline {1,2,0,1}}\\&\rightarrow _{r10}1,2,0,1,2,0,1,1,0,{\underline {1,2,0,0}}\rightarrow _{r8}1,2,0,1,2,0,{\underline {1,1,0,0}}\rightarrow _{r7}1,2,0,{\underline {1,2,0,0}}\rightarrow _{r8}{\underline {1,2,0,0}}\rightarrow _{r8}0.\end{aligned}}$

The corresponding equalities are

${\begin{aligned}&H_{3}(0,3)=H_{2}^{3}(0,H_{3}(0,0))=H_{2}^{3}(0,1)=H_{2}(0,H_{2}^{2}(0,1))=H_{2}(0,H_{2}(0,H_{2}(0,1))\\&=H_{2}(0,H_{2}(0,H_{1}(0,H_{2}(0,0))))=H_{2}(0,H_{2}(0,H_{1}(0,0)))=H_{2}(0,H_{2}(0,0))=H_{2}(0,0)=0.\end{aligned}}$

When reduction rule r11 is replaced by rule r12, the stack is transformed acoording to

${\begin{array}{lllllllll}{\text{(r12)}}&(x+2)&,n&,a&,b&\rightarrow &(x+1)&,n&,a&,1&,n&,a&,b\end{array}}$

The successive stack configurations will then be

${\begin{aligned}&{\underline {1,3,0,3}}\rightarrow _{r10}3,2,0,{\underline {1,3,0,0}}\rightarrow _{r9}{\underline {3,2,0,1}}\rightarrow _{r12}2,2,0,{\underline {1,2,0,1}}\rightarrow _{r10}2,2,0,1,1,0,{\underline {1,2,0,0}}\\&\rightarrow _{r8}2,2,0,{\underline {1,1,0,0}}\rightarrow _{r7}{\underline {2,2,0,0}}\rightarrow _{r12}1,2,0,{\underline {1,2,0,0}}\rightarrow _{r8}{\underline {1,2,0,0}}\rightarrow _{r8}0\end{aligned}}$

The corresponding equalities are

${\begin{aligned}&H_{3}(0,3)=H_{2}^{3}(0,H_{3}(0,0))=H_{2}^{3}(0,1)=H_{2}^{2}(0,H_{2}(0,1))=H_{2}^{2}(0,H_{1}(0,H_{2}(0,0)))\\&=H_{2}^{2}(0,H_{1}(0,0))=H_{2}^{2}(0,0)=H_{2}(0,H_{2}(0,0))=H_{2}(0,0)=0\end{aligned}}$

Remarks

• $H_{3}(0,3)=0$is a special case. See below.
• The computation of$H_{n}(a,b)$according to the rules {r6 - r10, r11} is heavily recursive. The culprit is the order in which iteration is executed:$H^{n}(a,b)=H(a,H^{n-1}(a,b))$.The first$H$disappears only after the whole sequence is unfolded. For instance,$H_{4}(2,4)$converges to 65536 in 2863311767 steps, the maximum depth of recursionis 65534.
• The computation according to the rules {r6 - r10, r12} is more efficient in that respect. The implementation of iteration$H^{n}(a,b)$as$H^{n-1}(a,H(a,b))$mimics the repeated execution of a procedure H.The depth of recursion, (n+1), matches the loop nesting.Meyer & Ritchie (1967)formalized this correspondence. The computation of$H_{4}(2,4)$according to the rules {r6-r10, r12} also needs 2863311767 steps to converge on 65536, but the maximum depth of recursion is only 5, as tetration is the 5th operator in the hyperoperation sequence.
• The considerations above concern the recursion depth only. Either way of iterating leads to the same number of reduction steps, involving the same rules (when the rules r11 and r12 are considered "the same" ). As the example shows the reduction of$H_{3}(0,3)$converges in 9 steps: 1 X r7, 3 X r8, 1 X r9, 2 X r10, 2 X r11/r12. The modus iterandi only affects the order in which the reduction rules are applied.

Examples

Below is a list of the first seven (0th to 6th) hyperoperations (0⁰is defined as 1).

n	Operation, Hn(a,b)	Definition	Names	Domain
0	$1+b$or$a[0]b$	$1+\underbrace {1+1+1+\cdots +1+1+1} _{\displaystyle b{\mbox{ copies of 1}}}$	Increment,successor,zeration, hyper0	Arbitrary
1	$a+b$or$a[1]b$	$a+\underbrace {1+1+1+\cdots +1+1+1} _{\displaystyle b{\mbox{ copies of 1}}}$	Addition,hyper1
2	$a\times {b}$or$a[2]b$	$\underbrace {a+a+a+\cdots +a+a+a} _{\displaystyle b{\mbox{ copies of }}a}$	Multiplication,hyper2
3	$a^{b}$or$a[3]b$	$\underbrace {a\times a\times a\times \;\cdots \;\times a\times a\times a} _{\displaystyle b{\mbox{ copies of }}a}$	Exponentiation,hyper3	breal, with some multivalued extensions tocomplex numbers
4	$^{b}a$or$a[4]b$	$\underbrace {a[3](a[3](a[3](\cdots [3](a[3](a[3]a))\cdots )))} _{\displaystyle b{\mbox{ copies of }}a}$	Tetration,hyper4	a≥ 0 or an integer,ban integer ≥ −1(with some proposed extensions)
5	$_{b}a$or$a[5]b$	$\underbrace {a[4](a[4](a[4](\cdots [4](a[4](a[4]a))\cdots )))} _{\displaystyle b{\mbox{ copies of }}a}$	Pentation,hyper5	a,bintegers ≥ −1
6	$a[6]b$	$\underbrace {a[5](a[5](a[5](\cdots [5](a[5](a[5]a))\cdots )))} _{\displaystyle b{\mbox{ copies of }}a}$	Hexation, hyper6

Special cases

H_n(0,b) =

b+ 1, whenn= 0

b,whenn= 1

0, whenn= 2

1, whenn= 3 andb= 0

0, whenn= 3 andb> 0

1, whenn> 3 andbis even (including 0)

0, whenn> 3 andbis odd

H_n(1,b) =

b,whenn= 2

1, whenn≥ 3

H_n(a,0) =

0, whenn= 2

1, whenn= 0, orn≥ 3

a,whenn= 1

H_n(a,1) =

a, whenn≥ 2

H_n(a,a) =

H_n+1(a,2), whenn≥ 1

H_n(a,−1) =

0, whenn= 0, orn≥ 4

a− 1, whenn= 1

−a,whenn= 2

⁠1/a⁠,whenn= 3

H_n(2, 2) =

3, whenn= 0

4, whenn≥ 1, easily demonstrable recursively.

History

One of the earliest discussions of hyperoperations was that of Albert Bennett in 1914, who developed some of the theory ofcommutative hyperoperations(seebelow).About 12 years later,Wilhelm Ackermanndefined the function$\phi (a,b,n)$,which somewhat resembles the hyperoperation sequence.

In his 1947 paper,Reuben Goodsteinintroduced the specific sequence of operations that are now calledhyperoperations,and also suggested the Greek namestetration,pentation, etc., for the extended operations beyond exponentiation (because they correspond to the indices 4, 5, etc.). As a three-argument function, e.g.,$G(n,a,b)=H_{n}(a,b)$,the hyperoperation sequence as a whole is seen to be a version of the originalAckermann function$\phi (a,b,n)$—recursivebut notprimitive recursive— as modified by Goodstein to incorporate the primitivesuccessor functiontogether with the other three basic operations of arithmetic (addition,multiplication,exponentiation), and to make a more seamless extension of these beyond exponentiation.

The original three-argumentAckermann function$\phi$uses the same recursion rule as does Goodstein's version of it (i.e., the hyperoperation sequence), but differs from it in two ways. First,$\phi (a,b,n)$defines a sequence of operations starting from addition (n= 0) rather than thesuccessor function,then multiplication (n= 1), exponentiation (n= 2), etc. Secondly, the initial conditions for$\phi$result in$\phi (a,b,3)=G(4,a,b+1)=a[4](b+1)$,thus differing from the hyperoperations beyond exponentiation.The significance of theb+ 1 in the previous expression is that$\phi (a,b,3)$=$a^{a^{\cdot ^{\cdot ^{\cdot ^{a}}}}}$,wherebcounts the number ofoperators(exponentiations), rather than counting the number ofoperands( "a" s) as does thebin$a[4]b$,and so on for the higher-level operations. (See theAckermann functionarticle for details.)

Notations

This is a list of notations that have been used for hyperoperations.

Name	Notation equivalent to$H_{n}(a,b)$	Comment
Knuth's up-arrow notation	$a\uparrow ^{n-2}b$	Used byKnuth(forn≥ 3), and found in several reference books.
Hilbert's notation	$\phi _{n}(a,b)$	Used byDavid Hilbert.
Goodstein's notation	$G(n,a,b)$	Used byReuben Goodstein.
OriginalAckermann function	${\begin{matrix}\phi (a,b,n-1)\ {\text{ for }}1\leq n\leq 3\\\phi (a,b-1,n-1)\ {\text{ for }}n\geq 4\end{matrix}}$	Used byWilhelm Ackermann(forn≥ 1)
Ackermann–Péter function	$A(n,b-3)+3\ {\text{for }}a=2$	This corresponds to hyperoperations for base 2 (a= 2)
Nambiar's notation	$a\otimes ^{n-1}b$	Used by Nambiar (forn≥ 1)
Superscript notation	$a{}^{(n)}b$	Used byRobert Munafo.
Subscript notation (for lower hyperoperations)	$a{}_{(n)}b$	Used for lower hyperoperations by Robert Munafo.
Operator notation (for "extended operations" )	$aO_{n-1}b$	Used for lower hyperoperations byJohn DonerandAlfred Tarski(forn≥ 1).
Square bracket notation	$a[n]b$	Used in many online forums; convenient forASCII.
Conway chained arrow notation	$a\to b\to (n-2)$	Used byJohn Horton Conway(forn≥ 3)

Variant starting froma

Main article:Ackermann function

In 1928,Wilhelm Ackermanndefined a 3-argument function$\phi (a,b,n)$which gradually evolved into a 2-argument function known as theAckermann function.TheoriginalAckermann function$\phi$was less similar to modern hyperoperations, because his initial conditions start with$\phi (a,0,n)=a$for alln> 2. Also he assigned addition ton= 0, multiplication ton= 1 and exponentiation ton= 2, so the initial conditions produce very different operations for tetration and beyond.

n	Operation	Comment
0	$F_{0}(a,b)=a+b$
1	$F_{1}(a,b)=a\cdot b$
2	$F_{2}(a,b)=a^{b}$
3	$F_{3}(a,b)=a[4](b+1)$	An offset form oftetration.The iteration of this operation is different than theiterationof tetration.
4	$F_{4}(a,b)=(x\mapsto a[4](x+1))^{b}(a)$	Not to be confused withpentation.

Another initial condition that has been used is$A(0,b)=2b+1$(where the base is constant$a=2$), due toRózsa Péter,which does not form a hyperoperation hierarchy.

Variant starting from 0

In 1984, C. W. Clenshaw and F. W. J. Olver began the discussion of using hyperoperations to prevent computerfloating-pointoverflows. Since then, many other authorshave renewed interest in the application of hyperoperations tofloating-pointrepresentation. (SinceH_n(a,b) are all defined forb= -1.) While discussingtetration,Clenshawet al.assumed the initial condition$F_{n}(a,0)=0$,which makes yet another hyperoperation hierarchy. Just like in the previous variant, the fourth operation is very similar totetration,but offset by one.

n	Operation	Comment
0	$F_{0}(a,b)=b+1$
1	$F_{1}(a,b)=a+b$
2	$F_{2}(a,b)=a\cdot b=e^{\ln(a)+\ln(b)}$
3	$F_{3}(a,b)=a^{b}$
4	$F_{4}(a,b)=a[4](b-1)$	An offset form oftetration.The iteration of this operation is much different than theiterationof tetration.
5	$F_{5}(a,b)=\left(x\mapsto a[4](x-1)\right)^{b}(0)=0{\text{ if }}a>0$	Not to be confused withpentation.

Lower hyperoperations

An alternative for these hyperoperations is obtained by evaluation from left to right.Since

${\begin{aligned}a+b&=(a+(b-1))+1\\a\cdot b&=(a\cdot (b-1))+a\\a^{b}&=\left(a^{(b-1)}\right)\cdot a\end{aligned}}$

define (with ° or subscript)

$a_{(n+1)}b=\left(a_{(n+1)}(b-1)\right)_{(n)}a$

with

${\begin{aligned}a_{(1)}b&=a+b\\a_{(2)}0&=0\\a_{(n)}1&=a&{\text{for }}n>2\\\end{aligned}}$

This was extended to ordinal numbers by Doner and Tarski,by:

${\begin{aligned}\alpha O_{0}\beta &=\alpha +\beta \\\alpha O_{\gamma }\beta &=\sup \limits _{\eta <\beta,\xi <\gamma }(\alpha O_{\gamma }\eta )O_{\xi }\alpha \end{aligned}}$

It follows from Definition 1(i), Corollary 2(ii), and Theorem 9, that, fora≥ 2 andb≥ 1, that

$aO_{n}b=a_{(n+1)}b$

But this suffers a kind of collapse, failing to form the "power tower" traditionally expected of hyperoperators:

$\alpha _{(4)}(1+\beta )=\alpha ^{\left(\alpha ^{\beta }\right)}.$

If α ≥ 2 and γ ≥ 2,^{[Corollary 33(i)]}

$\alpha _{(1+2\gamma +1)}\beta \leq \alpha _{(1+2\gamma )}(1+3\alpha \beta ).$

n	Operation	Comment
0	$F_{0}(a,b)=a+1$	Increment, successor, zeration
1	$F_{1}(a,b)=a+b$
2	$F_{2}(a,b)=a\cdot b$
3	$F_{3}(a,b)=a^{b}$
4	$F_{4}(a,b)=a^{\left(a^{(b-1)}\right)}$	Not to be confused withtetration.
5	$F_{5}(a,b)=\left(x\mapsto x^{x^{(a-1)}}\right)^{b-1}(a)$	Not to be confused withpentation. Similar totetration.

Commutative hyperoperations

Commutative hyperoperations were considered by Albert Bennett as early as 1914,which is possibly the earliest remark about any hyperoperation sequence. Commutative hyperoperations are defined by the recursion rule

$F_{n+1}(a,b)=\exp(F_{n}(\ln(a),\ln(b)))$

which is symmetric inaandb,meaning all hyperoperations are commutative. This sequence does not containexponentiation,and so does not form a hyperoperation hierarchy.

n	Operation	Comment
0	$F_{0}(a,b)=\ln \left(e^{a}+e^{b}\right)$	Smooth maximum
1	$F_{1}(a,b)=a+b$
2	$F_{2}(a,b)=a\cdot b=e^{\ln(a)+\ln(b)}$	This is due to theproperties of the logarithm.
3	$F_{3}(a,b)=a^{\ln(b)}=e^{\ln(a)\ln(b)}$	In afinite field,this is theDiffie–Hellman key exchangeoperation.
4	$F_{4}(a,b)=e^{e^{\ln(\ln(a))\ln(\ln(b))}}$	Not to be confused withtetration.

Numeration systems based on the hyperoperation sequence

R. L. Goodsteinused the sequence of hyperoperators to create systems of numeration for the nonnegative integers. The so-calledcomplete hereditary representationof integern,at levelkand baseb,can be expressed as follows using only the firstkhyperoperators and using as digits only 0, 1,...,b− 1, together with the basebitself:

• For 0 ≤n≤b− 1,nis represented simply by the corresponding digit.
• Forn>b− 1, the representation ofnis found recursively, first representingnin the form

b[k]x_k[k− 1]x_{k− 1}[k- 2]... [2]x₂[1]x₁

wherex_k,...,x₁are the largest integers satisfying (in turn)

b[k]x_k≤n

b[k]x_k[k− 1]x_{k− 1}≤n

...

b[k]x_k[k− 1]x_{k− 1}[k- 2]... [2]x₂[1]x₁≤n

Anyx_iexceedingb− 1 is then re-expressed in the same manner, and so on, repeating this procedure until the resulting form contains only the digits 0, 1,...,b− 1, together with the baseb.

Unnecessary parentheses can be avoided by giving higher-level operators higher precedence in the order of evaluation; thus,

level-1 representations have the form b [1] X, withXalso of this form;

level-2 representations have the form b [2] X [1] Y, withX,Yalso of this form;

level-3 representations have the form b [3] X [2] Y [1] Z, withX,Y,Zalso of this form;

level-4 representations have the form b [4] X [3] Y [2] Z [1] W, withX,Y,Z,Walso of this form;

and so on.

In this type of base-bhereditaryrepresentation, the base itself appears in the expressions, as well as "digits" from the set {0, 1,...,b− 1}. This compares toordinarybase-2 representation when the latter is written out in terms of the baseb;e.g., in ordinary base-2 notation, 6 = (110)₂= 2 [3] 2 [2] 1 [1] 2 [3] 1 [2] 1 [1] 2 [3] 0 [2] 0, whereas the level-3 base-2 hereditary representation is 6 = 2 [3] (2 [3] 1 [2] 1 [1] 0) [2] 1 [1] (2 [3] 1 [2] 1 [1] 0). The hereditary representations can be abbreviated by omitting any instances of [1] 0, [2] 1, [3] 1, [4] 1, etc.; for example, the above level-3 base-2 representation of 6 abbreviates to 2 [3] 2 [1] 2.

Examples: The unique base-2 representations of the number266,at levels 1, 2, 3, 4, and 5 are as follows:

Level 1: 266 = 2 [1] 2 [1] 2 [1]... [1] 2 (with 133 2s)

Level 2: 266 = 2 [2] (2 [2] (2 [2] (2 [2] 2 [2] 2 [2] 2 [2] 2 [1] 1)) [1] 1)

Level 3: 266 = 2 [3] 2 [3] (2 [1] 1) [1] 2 [3] (2 [1] 1) [1] 2

Level 4: 266 = 2 [4] (2 [1] 1) [3] 2 [1] 2 [4] 2 [2] 2 [1] 2

Level 5: 266 = 2 [5] 2 [4] 2 [1] 2 [5] 2 [2] 2 [1] 2

See also

• Large numbers

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