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'''Conway chained arrow notation''', created by mathematician [[John Horton Conway]], is a means of expressing certain extremely [[large numbers]]. It is simply a finite sequence of positive [[integers]] separated by rightward arrows, e.g. 2 → 3 → 4 → 5 → 6.


As with most [[combinatorial]] symbologies, the definition is [[recursion|recursive]]. In this case the notation eventually resolves to being the leftmost number raised to some (usually enormous) integer power.


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==Definition and overview==
A ''Conway chain'' (or ''chain'' for short) is defined as follows:
* Any positive integer is a chain of length 1.
* A chain of length ''n'', followed by a right-arrow → and a positive integer, together form a chain of length <math>n+1</math>.
 
Any chain represents an integer, according to the four rules below. Two chains are said to be equivalent if they represent the same integer.
 
If <math>p</math> and <math>q</math> are positive integers, and <math>X</math> is a subchain, then:
# The chain <math>p</math> represents the number <math>p</math>.
# <math>p \to q</math> represents the [[exponentiation|exponential]] expression <math>p^q</math>.
# <math>X \to p \to 1</math> is equivalent to <math>X \to p</math>.
# <math>X \to p \to (q + 1)</math> is equivalent to <math>X \to ( X \to ( \cdots (X \to ( X ) \to q)\cdots ) \to q ) \to q</math> <br />(with ''p'' copies of ''X'', ''p''&nbsp;&minus;&nbsp;1 copies of ''q'', and ''p''&nbsp;&minus;&nbsp;1 pairs of parentheses; applies for ''q''&nbsp;>&nbsp;0).
 
Note that the last rule can be restated recursively to avoid the [[ellipsis|ellipses]]:
:4a. <math>X \to 1 \to (q + 1) = X</math>
:4b. <math>X \to (p + 1) \to (q + 1) = X \to (X \to p \to (q+1)) \to q</math>
 
==Properties==
# A chain of length 3 corresponds to [[Knuth's up-arrow notation]] and [[hyper operator]]s:
#:<math>\begin{matrix}
p \to q \to r = \text{hyper}(p,r+2,q) = p \!\!\! & \underbrace{ \uparrow \dots \uparrow } & \!\!\! q = p\uparrow^r q.\\
& \!\!\! r \text{ arrows} \!\!\!
\end{matrix}</math>
# a chain ''X''&nbsp;→&nbsp;''Y'' is of the form ''X''&nbsp;→&nbsp;''p''; hence:
# a chain starting with ''a'' is a power of ''a''
# a chain 1&nbsp;→&nbsp;''Y'' is equal to 1
# a chain ''X''&nbsp;→&nbsp;1&nbsp;→&nbsp;''Y'' is equal to ''X''
# a chain 2&nbsp;→&nbsp;2&nbsp;→&nbsp;''Y'' is equal to 4
# a chain ''X''&nbsp;→&nbsp;2&nbsp;→&nbsp;2 is equal to ''X''&nbsp;→&nbsp;(''X'') (chain ''X'' with its value concatenated to it)
 
==Interpretation==
One must be careful to treat an arrow chain ''as a whole''. Arrow chains do not describe the iterated application of a binary operator. Whereas chains of other infixed symbols (e.g. 3&nbsp;+&nbsp;4&nbsp;+&nbsp;5&nbsp;+&nbsp;6&nbsp;+&nbsp;7) can often be considered in fragments (e.g. (3&nbsp;+&nbsp;4)&nbsp;+&nbsp;5&nbsp;+&nbsp;(6&nbsp;+&nbsp;7)) without a change of meaning (see [[associativity]]), or at least can be evaluated step by step in a prescribed order, e.g. 2<sup>3<sup>4</sup></sup> from right to left, that is not so with Conway's arrow.
 
For example:
* <math>2\rightarrow3\rightarrow2 = 2\uparrow\uparrow3 = 2^{2^2} = 16</math>
* <math>2\rightarrow\left(3\rightarrow2\right) = 2^{(3^2)} = 2^{3^2} = 512</math>
* <math>\left(2\rightarrow3\right)\rightarrow2 = \left(2^3\right)^2 = 64</math>
 
The fourth rule is the core: A chain of 3 or more elements ending with 2 or higher becomes a chain of the same length with a (usually vastly) increased penultimate element. But its ''ultimate'' element is decremented, eventually permitting the third rule to shorten the chain. After, to paraphrase [[Donald Knuth|Knuth]], "much detail", the chain is reduced to two elements and the second rule terminates the recursion.
 
==Examples==
Examples get quite complicated quickly, here are small examples:
 
''n''
:= ''n'' (by rule 1)
 
''p→q''
:= ''p<sup>q</sup>'' (by rule 2)
:Thus 3→4 = 3<sup>4</sup> = 81
 
1→(''any arrowed expression'')
:= 1 since the entire expression eventually reduces to 1<sup>number</sup> = 1. (Indeed, any chain containing a 1 can be truncated just before that 1; e.g. ''X''→1→''Y''=''X'' for any (embedded) chains ''X,Y''.)
 
4→3→2
:= 4→(4→(4)→1)→1 (by 4) and then, working from the inner parentheses outwards,
:= 4→(4→4→1)→1 (remove redundant parentheses [rrp])
:= 4→(4→4)→1 (3)
:= 4→(256)→1 (2)
:= 4→256→1 (rrp)
:= 4→256 (3)
:= 4<sup>256</sup> (2)
:= 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 exactly ≈ 1.34078079299 × 10<sup>154</sup>
 
With Knuth's arrows: <math>4 \uparrow \uparrow 3 = 4 \uparrow 4 \uparrow 4 = 4^{256}</math>
 
2→2→4
:= 2→(2)→3 (by 4)
:= 2→2→3 (rrp)
:= 2→2→2 (4, rrp)
:= 2→2→1 (4, rrp)
:= 2→2 (3)
:= 4 (2) (In fact any chain beginning with two 2s stands for 4.)
 
2→4→3
:= '''2'''→('''2'''→('''2'''→('''2''')→2)→2)→2 (by 4) ''The four copies of '''X''' (which is '''2''' here) are in bold to distinguish them from the three copies of '''q''' (which is also 2)''
:= 2→(2→(2→2→2)→2)→2 (rrp)
:= 2→(2→(4)→2)→2 (previous example)
:= 2→('''2→4→2''')→2 (rrp) ''(expression expanded in next equation shown in bold on both lines)''
:= 2→('''2→(2→(2→(2)→1)→1)→1''')→2 (4)
:= 2→(2→(2→(2→2→1)→1)→1)→2 (rrp)
:= 2→(2→(2→(2→2)))→2 (3 repeatedly)
:= 2→(2→(2→(4)))→2 (2)
:= 2→(2→(16))→2 (2)
:= 2→65536→2 (2,rrp)
:= 2→(2→(2→(...2→(2→(2)→1)→1...)→1)→1)→1 (4) with 65535 sets of parentheses
:= 2→(2→(2→(...2→(2→(2))...))) (3 repeatedly)
:= 2→(2→(2→(...2→(4))...))) (2)
:= 2→(2→(2→(...16...))) (2)
:= <math>2^{2^{\dots^2}}</math> (a tower with 2<sup>16</sup> = 65536 stories) = <sup>65536</sup>2  (See [[Tetration]])
 
With Knuth's arrows: <math>2 \uparrow \uparrow \uparrow 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=2 \uparrow \uparrow 2 \uparrow 2 \uparrow 2 \uparrow 2 = 2 \uparrow \uparrow 65536</math>.
 
2→3→2→2
:= 2→3→(2→3)→1 (by 4)
:= 2→3→8 (2 and 3)  With Knuth's arrows: 2 ↑<sup>8</sup> 3 (prop1)
:= 2→(2→2→7)→7 (1)
:= 2→4→7 (two initial 2's give 4 [prop6])  With Knuth's arrows: 2 ↑<sup>7</sup> 4 (prop1)
:= 2→(2→(2→2→6)→6)→6 (4)
:= 2→('''2→4→6''')→6 (prop6)
:= 2→('''2→(2→(2→2→5)→5)→5''')→6 (4)
:= 2→(2→('''2→4→5''')→5)→6 (prop6)
:= 2→(2→('''2→(2→(2→2→4)→4)→4''')→5)→6 (4)
:= 2→(2→(2→('''2→4→4''')→4)→5)→6 (prop6)
:= 2→(2→(2→('''2→(2→(2→2→3)→3)→3''')→4) →5)→6 (4)
:= 2→(2→(2→(2→(2→4→3)→3)→4)→5)→6 (prop6)
:= 2→(2→(2→(2→(2→65536→2)→3)→4)→5)→6 (previous example)
:= ''much larger than previous number''
 
With Knuth's arrows: <math>2 \uparrow^6 2 \uparrow^5 2 \uparrow^4 2 \uparrow^3 2 \uparrow^2 65536.</math>
 
3→2→2→2
:= 3→2→(3→2)→1 (4)
:= 3→2→9 (2 and 3)
:= 3→3→8 (4)
 
With Knuth's arrows: <math>3 \uparrow^8 3</math>.
 
===Systematic examples===
The simplest cases with four terms (containing no integers less than 2) are:
* <math>a \to b \to 2 \to 2 = a \to b \to 2 \to (1 + 1) = a \to b \to (a \to b) \to 1 = a \to b \to a^b = a \uparrow^{a^b} b</math>
: (also following from the last-mentioned property)
* <math>a \to b \to 3 \to 2 = a \to b \to 3 \to (1 + 1)</math><br /> <math> = a \to b \to (a \to b \to (a \to b) \to 1) \to 1 = a \to b \to (a \to b \to a^b) = a \uparrow^{a \to b \to 2 \to 2} b</math>
* <math>a \to b \to 4 \to 2 = a \to b \to (a \to b \to (a \to b \to a^b)) = a \uparrow^{a \to b \to 3 \to 2} b</math>
 
We can see a pattern here. If, for any chain ''X'', we let <math>f(p) = X \to p</math> then <math>X \to p \to 2 = f^p(1)</math> (see
[[Function composition|functional powers]]).
 
Applying this with <math>X = a \to b</math>, then <math>f(p) = a \uparrow^p b</math> and <math>a \to b \to p \to 2 = a \uparrow^{a \to b \to (p-1) \to 2} b = f^p(1)</math>
 
Thus, for example, <math>10 \to 10 \to 3\to 2 = 10 \uparrow ^{10 \uparrow ^{10^{10}} 10} 10 \!</math>.
 
Moving on:
* <math>a \to b \to 2 \to 3 = a \to b \to 2 \to (2 + 1) = a \to b \to (a \to b) \to 2 = a \to b \to a^b \to 2 = f^{a^b}(1)</math>
 
Again we can generalize. When we write <math>g_q(p) = X \to p \to q</math> we have <math>X \to p \to q+1 = g_q^p(1)</math>, that is, <math>g_{q+1}(p) = g_q^p(1)</math>. In the case above, <math>g_2(p) = a \to b \to p \to 2 = f^p(1)</math> and <math>g_3(p) = g_2^p(1)</math>, so <math>a \to b \to 2 \to 3 = g_3(2) = g_2^2(1) = g_2(g_2(1)) = f^{f(1)}(1) = f^{a^b}(1)</math>
 
==Ackermann function==
The [[Ackermann function]] may be expressed using Conway chained arrow notation:
 
:''A''(''m'', ''n'') = (2 → (''n'' + 3) → ''(m'' − 2)) − 3 for ''m'' > 2
 
hence
 
:2 → ''n'' → ''m'' = ''A''(''m'' + 2,''n'' &minus; 3) + 3 for ''n'' > 2
 
(''n'' = 1 and ''n'' = 2 would correspond with ''A''(''m'',&nbsp;&minus;2)&nbsp;=&nbsp;&minus;1 and ''A''(''m'',&nbsp;&minus;1)&nbsp;=&nbsp;1, which could logically be added).
 
==Graham's number==
[[Graham's number]] <math>G \!</math> itself can not be expressed accurately in Conway chained arrow notation, but by defining the intermediate function <math>f(n) = 3 \rightarrow 3 \rightarrow n \!</math>, we have:
<math>G = f^{64}(4)\, </math> (see [[Function composition|functional powers]]), and
<math>3 \rightarrow 3 \rightarrow 64 \rightarrow 2 < G < 3 \rightarrow 3 \rightarrow 65 \rightarrow 2\, </math>
 
'''Proof:''' Applying in order the definition, rule 3, and rule 4, we have:
 
<math>f^{64}(1)\, </math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 1))\cdots ))\, </math> (with 64 <math>3 \rightarrow 3</math>'s)
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3) \rightarrow 1) \cdots ) \rightarrow 1) \rightarrow 1\, </math>
:<math>= 3 \rightarrow 3 \rightarrow 64 \rightarrow 2;\, </math>
 
<math>f^{64}(4) = G;\, </math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 4))\cdots ))\, </math> (with 64 <math>3 \rightarrow 3</math>'s)
 
<math>f^{64}(27)\, </math>
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27))\cdots ))\, </math> (with 64 <math>3 \rightarrow 3</math>'s)
:<math>= 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (\cdots (3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow (3 \rightarrow 3)))\cdots ))\, </math> (with 65 <math>3 \rightarrow 3</math>'s)
:<math>= 3 \rightarrow 3 \rightarrow 65 \rightarrow 2\, </math> (computing as above).
 
Since ''f'' is strictly increasing,
:<math>f^{64}(1) < f^{64}(4) < f^{64}(27)\, </math>
which is the given inequality.
 
With chain arrows it is very easy to specify a much larger number. For example, note that
:<math> 3 \rightarrow 3 \rightarrow 3 \rightarrow 3 ~~ = ~~ 3 \rightarrow 3 \rightarrow (3 \rightarrow 3 \rightarrow 27 \rightarrow 2) \rightarrow 2\, </math>
which is much greater than Graham's number.
 
==See also==
* [[Steinhaus–Moser notation]]
* [[Ackermann function]]
* [[Large numbers#Systematically creating ever faster increasing sequences|Systematically creating ever faster increasing sequences]]
 
==External links==
* [http://www-users.cs.york.ac.uk/~susan/cyc/b/big.htm Factoids &gt; big numbers]
* [http://www.mrob.com/pub/math/largenum.html Robert Munafo's Large Numbers]
*[https://docs.google.com/viewer?a=v&q=cache:gv7MebfOp6oJ:futuretg.com/FTHumanEvolutionCourse/FTFreeLearningKits/01-MA-Mathematics,%2520Economics%2520and%2520Preparation%2520for%2520University/011-MA11-UN03-10-Number%2520Theory%2520and%2520Cryptography/Additional%2520Resources/J.H.%2520Conway,%2520R.K.%2520Guy%2520-%2520The%2520Book%2520of%2520Numbers.pdf+The+Book+of+Numbers+by+J.+H.+Conway+and+R.+K.+Guy&hl=en&pid=bl&srcid=ADGEEShgWcsuShpVnS-hYtNfbwOq4TEpkeQ7YOZGVEk-omzaiEs4VKdsXFz1Su-Uh1po2QEXnmSivKhRixbQK6puTsf92WYUWuAcxyeOpXvn4JcEs-wsAJ1aF1Bk5I4JU7WCKoOUQCTL&sig=AHIEtbT5_BLlXtiF0i6dMiG6hNP8C58zKw The Book of Numbers by J. H. Conway and R. K. Guy]
 
{{DEFAULTSORT:Conway Chained Arrow Notation}}
[[Category:Mathematical notation]]
[[Category:Large numbers]]

Revision as of 19:04, 19 November 2013

Conway chained arrow notation, created by mathematician John Horton Conway, is a means of expressing certain extremely large numbers. It is simply a finite sequence of positive integers separated by rightward arrows, e.g. 2 → 3 → 4 → 5 → 6.

As with most combinatorial symbologies, the definition is recursive. In this case the notation eventually resolves to being the leftmost number raised to some (usually enormous) integer power.

Definition and overview

A Conway chain (or chain for short) is defined as follows:

  • Any positive integer is a chain of length 1.
  • A chain of length n, followed by a right-arrow → and a positive integer, together form a chain of length n+1.

Any chain represents an integer, according to the four rules below. Two chains are said to be equivalent if they represent the same integer.

If p and q are positive integers, and X is a subchain, then:

  1. The chain p represents the number p.
  2. pq represents the exponential expression pq.
  3. Xp1 is equivalent to Xp.
  4. Xp(q+1) is equivalent to X(X((X(X)q))q)q
    (with p copies of X, p − 1 copies of q, and p − 1 pairs of parentheses; applies for q > 0).

Note that the last rule can be restated recursively to avoid the ellipses:

4a. X1(q+1)=X
4b. X(p+1)(q+1)=X(Xp(q+1))q

Properties

  1. A chain of length 3 corresponds to Knuth's up-arrow notation and hyper operators:
    pqr=hyper(p,r+2,q)=pq=prq.r arrows
  2. a chain X → Y is of the form X → p; hence:
  3. a chain starting with a is a power of a
  4. a chain 1 → Y is equal to 1
  5. a chain X → 1 → Y is equal to X
  6. a chain 2 → 2 → Y is equal to 4
  7. a chain X → 2 → 2 is equal to X → (X) (chain X with its value concatenated to it)

Interpretation

One must be careful to treat an arrow chain as a whole. Arrow chains do not describe the iterated application of a binary operator. Whereas chains of other infixed symbols (e.g. 3 + 4 + 5 + 6 + 7) can often be considered in fragments (e.g. (3 + 4) + 5 + (6 + 7)) without a change of meaning (see associativity), or at least can be evaluated step by step in a prescribed order, e.g. 234 from right to left, that is not so with Conway's arrow.

For example:

The fourth rule is the core: A chain of 3 or more elements ending with 2 or higher becomes a chain of the same length with a (usually vastly) increased penultimate element. But its ultimate element is decremented, eventually permitting the third rule to shorten the chain. After, to paraphrase Knuth, "much detail", the chain is reduced to two elements and the second rule terminates the recursion.

Examples

Examples get quite complicated quickly, here are small examples:

n

= n (by rule 1)

p→q

= pq (by rule 2)
Thus 3→4 = 34 = 81

1→(any arrowed expression)

= 1 since the entire expression eventually reduces to 1number = 1. (Indeed, any chain containing a 1 can be truncated just before that 1; e.g. X→1→Y=X for any (embedded) chains X,Y.)

4→3→2

= 4→(4→(4)→1)→1 (by 4) and then, working from the inner parentheses outwards,
= 4→(4→4→1)→1 (remove redundant parentheses [rrp])
= 4→(4→4)→1 (3)
= 4→(256)→1 (2)
= 4→256→1 (rrp)
= 4→256 (3)
= 4256 (2)
= 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 exactly ≈ 1.34078079299 × 10154

With Knuth's arrows: 43=444=4256

2→2→4

= 2→(2)→3 (by 4)
= 2→2→3 (rrp)
= 2→2→2 (4, rrp)
= 2→2→1 (4, rrp)
= 2→2 (3)
= 4 (2) (In fact any chain beginning with two 2s stands for 4.)

2→4→3

= 2→(2→(2→(2)→2)→2)→2 (by 4) The four copies of X (which is 2 here) are in bold to distinguish them from the three copies of q (which is also 2)
= 2→(2→(2→2→2)→2)→2 (rrp)
= 2→(2→(4)→2)→2 (previous example)
= 2→(2→4→2)→2 (rrp) (expression expanded in next equation shown in bold on both lines)
= 2→(2→(2→(2→(2)→1)→1)→1)→2 (4)
= 2→(2→(2→(2→2→1)→1)→1)→2 (rrp)
= 2→(2→(2→(2→2)))→2 (3 repeatedly)
= 2→(2→(2→(4)))→2 (2)
= 2→(2→(16))→2 (2)
= 2→65536→2 (2,rrp)
= 2→(2→(2→(...2→(2→(2)→1)→1...)→1)→1)→1 (4) with 65535 sets of parentheses
= 2→(2→(2→(...2→(2→(2))...))) (3 repeatedly)
= 2→(2→(2→(...2→(4))...))) (2)
= 2→(2→(2→(...16...))) (2)
= 222 (a tower with 216 = 65536 stories) = 655362 (See Tetration)

With Knuth's arrows: 24=2222=2222=224=22222=265536.

2→3→2→2

= 2→3→(2→3)→1 (by 4)
= 2→3→8 (2 and 3) With Knuth's arrows: 2 ↑8 3 (prop1)
= 2→(2→2→7)→7 (1)
= 2→4→7 (two initial 2's give 4 [prop6]) With Knuth's arrows: 2 ↑7 4 (prop1)
= 2→(2→(2→2→6)→6)→6 (4)
= 2→(2→4→6)→6 (prop6)
= 2→(2→(2→(2→2→5)→5)→5)→6 (4)
= 2→(2→(2→4→5)→5)→6 (prop6)
= 2→(2→(2→(2→(2→2→4)→4)→4)→5)→6 (4)
= 2→(2→(2→(2→4→4)→4)→5)→6 (prop6)
= 2→(2→(2→(2→(2→(2→2→3)→3)→3)→4) →5)→6 (4)
= 2→(2→(2→(2→(2→4→3)→3)→4)→5)→6 (prop6)
= 2→(2→(2→(2→(2→65536→2)→3)→4)→5)→6 (previous example)
= much larger than previous number

With Knuth's arrows: 262524232265536.

3→2→2→2

= 3→2→(3→2)→1 (4)
= 3→2→9 (2 and 3)
= 3→3→8 (4)

With Knuth's arrows: 383.

Systematic examples

The simplest cases with four terms (containing no integers less than 2) are:

(also following from the last-mentioned property)

We can see a pattern here. If, for any chain X, we let f(p)=Xp then Xp2=fp(1) (see functional powers).

Applying this with X=ab, then f(p)=apb and abp2=aab(p1)2b=fp(1)

Thus, for example, 101032=101010101010.

Moving on:

Again we can generalize. When we write gq(p)=Xpq we have Xpq+1=gqp(1), that is, gq+1(p)=gqp(1). In the case above, g2(p)=abp2=fp(1) and g3(p)=g2p(1), so ab23=g3(2)=g22(1)=g2(g2(1))=ff(1)(1)=fab(1)

Ackermann function

The Ackermann function may be expressed using Conway chained arrow notation:

A(m, n) = (2 → (n + 3) → (m − 2)) − 3 for m > 2

hence

2 → nm = A(m + 2,n − 3) + 3 for n > 2

(n = 1 and n = 2 would correspond with A(m, −2) = −1 and A(m, −1) = 1, which could logically be added).

Graham's number

Graham's number G itself can not be expressed accurately in Conway chained arrow notation, but by defining the intermediate function f(n)=33n, we have: G=f64(4) (see functional powers), and 33642<G<33652

Proof: Applying in order the definition, rule 3, and rule 4, we have:

f64(1)

=33(33((33(331)))) (with 64 33's)
=33(33((33(33)1))1)1
=33642;

f64(4)=G;

=33(33((33(334)))) (with 64 33's)

f64(27)

=33(33((33(3327)))) (with 64 33's)
=33(33((33(33(33))))) (with 65 33's)
=33652 (computing as above).

Since f is strictly increasing,

f64(1)<f64(4)<f64(27)

which is the given inequality.

With chain arrows it is very easy to specify a much larger number. For example, note that

3333=33(33272)2

which is much greater than Graham's number.

See also

External links