Googology in Japan - exploring large numbers - Gyafun!

3

75 GDP

500

100

=10

10

79

14

10

68

1 6023 2

10

256

136

37

10 80000000000000000(8 10

) 63

4

20

Large Numbers 21

BEAF BEAF BEAF

2002

5

2013 2016

2007 2013 PDF 2

PDF

1

1

2 3 3 4

6

1 1 1

Kindle 2

Amazon 2

PDF -

4.0

3

15 PDF Limit of Empty Wiki

2017

1

6

MUPI

29

https://www.amazon.co.jp/dp/B01N4KCIJQ http://gyafun.jp/ln/ 3 CC BY-SA 4.0 https://creativecommons.org/licenses/by-sa/4.0/deed.ja 2

Kindle

noan

7

3 1 1.1

. . . . . . . . . . . . . . . . . . . . . . .

11 11

1.1.1

. . . . . . . . . . . . . . . . . . . .

11

1.1.2

. . . . . . . . . . . . . . . . . . . . . . . .

13

1.1.3

. . . . . . . . . . . . . . . . . . . . . . . .

17

1.1.4

. . . . . . . . . . . . . . . . . . . . . .

21

. . . . . . . . . . . . . . . . . . . . . . . .

25

. . . . . . . . . . . . . . . . . . . . . . .

27

1.2.1

. . . . . . . . . . .

28

1.2.2

. . . . . . . . . . . . . . . . . . . .

31

2

1.1.5 1.2

3

1.2.3

. . . . . . . . . . . . . . . . .

32

. . . . . . . . . . . . . . . . . . . .

33

. . . . . . . . . . . . . . . . . . . . .

34

. . . . . . . . . . . . . . . . . . . . . . . .

36

. . . . . . . . . . . . . . . . . . . . . . .

37

1.3.1

. . . . . . . . . . . . . . .

38

1.3.2

. . . . . . . . . . . . . . . . . . . . .

39

1.2.4

E

1.2.5 1.2.6 1.3

4

1.4

1.5

. . . . . . . . . . . . . . . . . . . .

40

1.4.1

5

. . . . . . . . . . . . . .

40

1.4.2

. . . . . . . . . . . . . . . . . . . . .

41

. . . . . . . . . . . . . . . . . . .

44

8 2

45

2.1

. . . . . . . . . . . . . . . . . . . . .

45

2.1.1

. . . . . . . . . . . . . . . . . . . . . . . .

45

2.1.2

. . . . . . . . . . . . . . . . . . . . . . . . . .

46

2.1.3

. . . . . . . . . . . . . . . . . . . .

49

2.1.4 2.1.5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49 54

2.2

. . . . . . . . . . . . . . . . . . . . .

57

2.3

. . . . . . . . . . . . . . . . . . . . . . . . .

61

3

2

3.1

65 . . . . . . . . . . . . . . . . . . . . . . .

65

. . . . . . . . . . . . . . . . . . . . . . . . .

68

3.3

. . . . . . . . . . . . . . . . . . . . . . . . . .

72

3.4

. . . . . . . . . . . . . . . . . . . . . . . . . .

75

3.5

. . . . . . . . . . . . . . . . . .

76

3.2

2

4

81

4.1

. . . . . . . . . . . . . . . . . . .

81

4.2

. . . . . . . . . . . . . .

84

4.3

. . . . . . . . . . . . . . . . . . . . . . .

86

4.4

. . . . . . . . . . . . . . . . . . . . . . . . .

87

4.4.1

1 . . . . . . . . . . . . . .

88

4.4.2

2 . . . . . . . . . . . . . .

100

4.4.3

3 . . . . . . . . . . . . . .

103

. . . . . . . . . . . . . . . . . . .

106

4.5 5

115

5.1

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

116

5.1.1

. . . . . . . . . . . . . . . . . . .

116

5.1.2

. . . .

120

5.1.3

. . . . . . . . . . . . . . . . .

122

5.1.4

. . . . . . . . . . . . . . . . .

124

9 5.2

. . . . . . . . . . . . . . . . . . . . . . . . . .

125

5.2.1

. . . . . . . . . . . . . . . . . . . . .

126

5.2.2

. . . . . . . . . . . . . . . . . . . . . .

130

5.2.3

. . . . . . . . . . . . . . . . . . . . . .

138

5.3

. . . . . . . . . . . . . . . . . . . . . . . . . .

139

5.3.1 5.3.2

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

140 140

5.3.3

. . . . . . . . . . . . . . . . . . . .

143

5.3.4

. . . . . . . . . . . . . . . . . . . . . .

145

6 6.1

147 . . . . . . . . . . . . . . . . . . . . . . . .

147

6.2

. . . . . . . . . . . . . . . . . . . .

149

6.3

5 . . . . . . . . . . . . . . . . . .

151

6.4

. . . . . . . . . . . . . . . . . . . . . . . . .

154

6.5

. . . . . . . . . . . . . . . . . . . . . . . . . .

160

6.6

. . . . . . . . . . . . . . . . .

167

6.7

. . . . . . . . . . . . . . . . . . .

169

. . . . . . . . . . . . . . . .

170

. . . . . . . . . . . . . . . . . . . .

175

6.8

F [ϵ0 ](n)

BEAF

6.9 7

177

7.1

. . . . . . . . . . . . . . . . . . . . . . . .

177

7.2

. . . . . . . . . . . . . . . . . . . . . . . .

181

. . . . . . . . . . . . . . . .

181

7.2.1 7.2.2

θ

7.2.3

ψ

7.2.4

ψ

7.2.5

ϑ

7.2.6 7.3

2 7.3.1 7.3.2

2

. . . . . . . . . . . . . . .

183

. . . . . . . . . . . . . . . .

187

. . . . . . . . . . . . . . . . . .

187

. . . . . . . . . . . . . . . .

192

. . . . . . . . . . . . . . . .

193

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

194

. . . . . . . . . . . . . . . . . . . . . . . . .

194

. . . . . . . . . . . . . . . . . . . . . . . .

194

10 7.3.3

. . . . . . . . . . . . . . . . . . . . .

196

. . . . . . . .

197

. . . . . . . . . . . . . . .

200

. . . . . . . . . . . . . . . . . . . . .

200

7.4.1

6 . . . . . . . . . . . . . .

201

7.4.2 7.4.3

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

205 215

7.4.4

. . . . . . . . . . . . . . . .

216

7.4.5

BAN . . . . . . . . . . . . . . . . . . . . . . . . . .

219

7.4.6

BEAF . . . . . . . . . . . . . . . . . . . . . . . . .

220

7.4.7

. . . . . . . . . . . . . . . . .

222

7.4.8

. . . . . . . . . . . . . . . . . . . . . .

224

7.4.9

7.3.4 7.3.5 7.4

2 C

. . . . . . . . . . . . . . . . . . . . . .

226

7.4.10

. . . . . . . . . . .

229

7.4.11

. . . . . . . . . . . . . . . . . . . .

232

7.4.12

. . . . . . . . . . . . . . . . . . .

234

8

239

8.1

. . . . . . . . . . . . . . . . . . . . .

239

8.2

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

243

8.3

. . . . . . . . . . . . . . . . . . . . . . . . . .

244

8.4

4 . . . . . . . . . . . . . . . . . .

245

8.5

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

248

8.6 8.7

7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

251 255

8.8

. . . . . . . . . . . . . . . . . . . . . . . . .

257 259 263 267

11

1

1.1

2

1.1.1 (Robert Munafo)

1996

1

0

1

3

2 2 2

2 0

(0–6): 2

1 Large

0

subitizing 1.1

Numbers http://www.mrob.com/pub/math/largenum.html E. L. et al. (1949). The discrimination of visual number. American Journal of Psychology 62 (4): 498–525. doi:10.2307/1418556 2 Kaufman,

12

1

1.1: 2 5 5 5

9

6

0

6

0 (6–106 ):

1

1 100

100

1

1

1 85

70 100

1 100 6

(106 –1010 ):

2 0

6

1 2

6

2

10

6

10 = 1000000

1010 = 101000000

2

1.1.

2

13

6

1010 = 10(10 10 6

(10 ) = 10

10×6

= 10

6

)

60

2

999999

111111 100 100

10

100

1.1.2

2

10 = 10

8

= 10

12

= 1024 = 1040

= 1056 = 1064

Wiki -

1 16

= 10

= 1032

= 1044

= 1052

3

= 10 = 1028

3

100

20

= 1036

= 1048 = 1060 = 1068

http://ja.googology.wikia.com/wiki/

14

1

1088

108 1068

1627 8

1088

11

✓

10

✒

1068

(1631) 1068

(1634)

68

✏ ✑

= 1068

4

10 4

=10

=10

5

6

=10

=10

1

14

1 1012

1012

1016 1044

1080

=108 = 102

2 =1012 =1028

=1031

=1056 2

7

=1015 =1060

128

2

8

4

=102 = 1016 5

=102 = 1032

=1024

=1040

=1048

6

=102 = 1064 256

2

10 = 10 =10 = 10 =10 11 12 =102 = 102048 =102 = 104096

4

=1020

9

= 10

=10127 512

=10

2

10

= 10

3

=

1024

=108191

http://www.sf.airnet.ne.jp/˜ts/language/largenumber.html

1.1.

2

15 100

(1299

2

)

1052 1052

2500 km 10000 km

km

4 km

2

10 km3

1m

0.05–2 mm

0.1 mm 50 % 100 %

2 100 %

10 km3 = 1010 m3

0.001 mm3 = 10−12 m3

16

1 1022

100

5

2

1 2

6

0.2nm

1

1

=500m

1 4 6

1

= 4 × 10 7

1

4000

=1

24

2

=500m

=1

0.5mm

= 6.4 × 1024 8

138

9

5 http://www.otani.ac.jp/yomu

1022 × 1024 = 1046

100

=100

page/b yougo/nab3mq0000000r5r.html http://d.hatena.ne.jp/inyoko/20111015/How long is kalpa2 7 http://d.hatena.ne.jp/inyoko/20111008/How long is kalpa 8 Universe as an Infant: Fatter Than Expected and Kind of Lumpy (New York Times) http://nyti.ms/YrtRRV 9 WikiArc: : http://bit.ly/1b4Myz9 (labo.wikidharma.org) 6

1.1.

2

17

1.1.3

(k, kilo-, 103 )

(M, Mega-, 106 )

(G, Giga-109 )

(T,

12

Tera-10 ) Mbps

1

1000

210 = 1024

2 1000 (IEC)

1024 2

2

10

2 (kibi)

Ki 20

2 12

10

(Mi) 2

(kilobinary) 1000

30

KB (Gi) 2

40

(Ti) 40

= 1, 000, 000, 000, 000

2

10 %

= 1, 099, 511, 627, 776 2

10 2

SI 18

(E, Exa-10 )

21

(Z, Zetta-10 )

(P, Peta-1015 ) (Y, Yotta-1024 ) =1024

18

1

(Sir Arthur Stanley Eddington, 1882–1944)

1938

(Tarner lecture) 136 × 2256

10

(Eddington number)

10

79

1.574 × 10

1080 –1085 10

million (100

)

billion (10

100

)

(Nicolas Chuquet, 1450 )

–1500

3 (Le triparty en la science des nombres) 1520 (Estienne de La Roche, 1470–1530) (l’Arismetique) 1870

(Aristide Marre, 1823–1918) 1880 6 million, byllion, tryllion, quadrillion, quyillion,

sixlion, septyllion, ottyllion, nonyllion nonyllion million

1054 6

3 million

100

10 http://ja.googology.wikia.com/wiki/

billion

10

1.1.

2

19 1

billion

billion

10

1

billion

1

trillion

million

billion, trillion, quadrillion,

100

million = 106 , billion = 109 , trillion = 1012 , quadrillion = 1015 , quintillion = 1018 , sextillion = 1021 , septillion = 1024 , octillion = 1027 , nonillion = 1030 , decillion = 1033 , undecillion = 1036 , duodecillion = 1039 , tredecillion = 1042 , quattuordecillion = 1045 , quindecillion = 1048 , sexdecillion (sedecillion) = 1051 , septendecillion = 1054 , octodecillion = 1057 , novemdecillion (novendecillion) = 1060 , (vigintillion) = 10

63

(

10 12

120

11

)

(centillion) = 10303 (

10600 ) 13

(googolplex) = 10 ✓ ✒

(googol) = 10100

10100

✏ ✑

= 10100

(Edward Kasner, 1878–1955) 11 http://ja.googology.wikia.com/wiki/ 12 http://ja.googology.wikia.com/wiki/ 13 http://ja.googology.wikia.com/wiki/

(Mathematics and the

20

1

Imagination)

14

9 1

(Milton Sirotta) 1911

0

100

9

1920 3 IT

Google

Google

googol Google

(Lawrence Edward “Larry” Page) (Sergey Mikhailovich Brin) 1997

BackRub

9 (Sean Anderson)

2004 (David Koller)

15

14 Kasner, E. and Newman, J. R. (1940) Mathematics and the Imagination. Simon & Schuster, New York. 15 http://graphics.stanford.edu/˜dk/google name origin.html

1.1.

2

21

googol

google.com google.com

googol

google.com

Google

1.1.4 16

(Shannon number)

10120

1950 17

(Claude Elwood Shannon, 1916–2001)

1949

EDSAC

30

16

Wiki http://ja.googology.wikia.com/wiki/ C. (1950) Programming a computer for playing chess. Philosophical Magazine 41 (314) 17 Shannon,

22

1

1

1

1

2

10

3

40

10120 1

1 1090

10120 ✓ ✒

10120 ✏ ✑

= 10120

64! 32!(8!)2 (2!)6

32 2

64 8

2

6 1043

40 (Godfrey Harold

1.1.

2

23 1080

Hardy, 1877–1947)

1010 18

50

3 10

3 1043 ×2

(10 )

43

= 10

3 6×1043

50

75

75 8848 26

8848

≈ 1012500 1043

1997 6

5

IBM 2

1

3

1997

1062 –1070 19

115 115

80

80 220

≈ 10

18 Hardy, G. H. (1940) Ramanujan: twelve lectures on subjects suggested by his life and work. Cambridge University Press. 19 (2008) , IPSJ Symposium Series 2008(11) 116–119.

24

1

20

10

10224

224 21

(2016

6

25

)

70

10

20 2010 2012 2013 5

5 1

3

1

2016 3

(AlphaGo) 4

1

10360 2

2016

319 ≈ 1.74 × 10172 1

2.081681994 × 10170

22

20

2006

2 20

http://ja.googology.wikia.com/wiki/ https://www.ipsj.or.jp/event/shogi.html 22 Number of legal Go positions https://tromp.github.io/go/legal.html 21

Wiki -

1.1.

2

25

6 )

million (100

7

billion (10

( )

)

trillion

8 (1063 )

1 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 23

103003

106000

(millinillion) = 103003

24

2

1.1.5 1.1

2

2

10 3.23 × 1021

10

10

23 Conway,

J. H. and Guy, R. K. (1996) The book of numbers. Copernicus. . 24 http://ja.googology.wikia.com/wiki/

26

1

1.1:

2 1046 1068 ≈ 1.57 × 1079 10100 10120 10303 103003 104096 101000000 = 1010

2

6

25

50

50 1.5747724136275002577605653961181555468044717914527 × 1079 80 25

(CASIO) http://keisan.casio.jp/

30

1.2.

3

27 26 27

1574772413 6275002577 6056539611 8155546804 4717914527 1167093662 3142507618 5631031296 2

(log)

10 e ≈ 2.71828

a

10a

exp(a) = ea 136 × 2256 log(136 × 2256 )

1079.197217798

≈

log(136) + 256 log(2)

≈

79.197217798

=

100.197217798+79

=

100.197217798 × 1079

≈

1.57477241 × 1079

100

2

100

1.2

3 3

6

(1010 –1010

106

):

2

10 2

26 Arbitrary

Precision Calculator http://apfloat.appspot.com/

27 http://www.javascripter.net/math/calculators/100digitbigintcalculator.htm

28

1 1010

2 ✓

6

1010

10

106

3 ✏

c(0) = 6, c(n + 1) = 10c(n)

c(n) n

(i) n = 0

c(0)

(ii) n > 0 ✒

c(n − 1)

✑

c(n)

1.2.1 (Archimedes; 287

? –

212

Reckoner;

) )

(The Sand

28

1 10

myriad (

)

8

108

108

1 2 10

2 10

16

10

8

108

2

8

1016

3 108 (108 )(10

8

)

= 108·10

108

8

1 2

1 2 28 http://ja.googology.wikia.com/wiki/

108·10

1 8

10

8

2 2

2

10

8

108

1.2.

3

29 3

108 !

1.

108

(108 )(10

40

8

)

"(108 )

108 = 108×10

1 403 = 64000

1

1 64000

6

1

1

4000

2

1 2 3.

16

3

1

2.

2

8

(108·10 ) = 1016·10

1

6

4000 (109 )

10 100

2

4. 1

107

1015

1 180 m

5.

180 m 1

40

10000

1/3

1 1

0.45 mm 0.02 mm

1

8

30

1

0.1 mm 3

5 = 125

6.

0.02 mm

2

1

100 107

2 3

100 1021

10

7. 300 100 1 10 2 8.

3

10

18

km

km

1010 1000

7 10

51

9.

(Aristarchus, 310

–

230

)

1

2 8

Harrison (2000)

29

1063

1000

1063

1080

29 Harrison, E. R. (2000) Cosmology - The Science of the Universe. Cambridge University Press. pp. 481–482.

1.2.

3

31

1.2.2

45 100

=10

1

1

1 1 1 1

1

=

7

10

2 =107

=107×2 = 1014 =107 2

× 3

2

= 1056

=107 2

✒

10

2

= 1028

2

1

107

n 30

✓

× 2

× n

2

122

7×2122

= 107×2

✏

122

31

✑

60

80 = 105×2 (Thomas Cleary)

120

32 33

30 http://ja.googology.wikia.com/wiki/ 31

(2016) (1957) 33 Cleary, T. (1993) The flower ornament scripture: a translation of the Avatamsaka Sutra. Shambhala, Colorado, USA. 32

32

1 unspeakable = 1010×2

120

, square untold = 1010×2

123

34

1.2.3 35

1010

✓

(googolplex)

(10googol )

10

100

✒

= 1010

✏

100

✑

9

1

1

1

0

0

1

0 1080 1

1085

0 10 10 3

4

3 7 × 2122

34 http://iteror.org/big/book/ch1/ch1 35 http://ja.googology.wikia.com/wiki/

7.html

1.2.

3

33

10x 7 × 2122

10100

7 × 2122 = 37218383881977644441306597687849648128 ≈ 3.72 × 1037 10100

1.2.4

E (Sbiis Saibian)

E

37

36

(hyper-E notation)

E

1

an

#

E(b)a1 #a2 #...#an

b

10

✓

✏

E

E(b)x

=

bx

E(b)a1 #a2 · · · #an #1

=

E(b)a1 #a2 · · · #an

=

E(b)a1 · · · #an−2 #(E(b)a1 · · · #an − 1)

✒

E(b)a1 #a2 · · · #an

bx

1. 1

2. 2

36 http://ja.googology.wikia.com/wiki/Sbiis 37 http://ja.googology.wikia.com/wiki/

1

Saibian E

✑

34

1 3. 3

1 z

2 (z)

Ea

=

E100 = E100#2 = E100#3 = Ea#b

Ea#1 = 10a E100#1 = 10100 = E(E100) = E10100 = 1010 E(E100#2) = 10 10

1010

a

E100#100#100#100 E

1

=

100

b

E

the Finite)

100

–

E

E

Web (One to Infinity: A Guide to

38

1.2.5 39

maxima)

N

N

N! 38 One

to Infinity https://sites.google.com/site/largenumbers/


(pro-

1.2.

3

35

10−35 m

1026 m (1026 /10−35 )3 = 10183 N

1010

185

N ! = 10183 ! = E185#2 10−43 N! (N !)t t

1 t

5 × 1017

(N !)t = 1010

245

= E245#2

(probability maximum)

1010 1010

245

1035

1081

1035 106.5446×10

245

343

36

1

1.2.6 40

(abundant number) 36

2 1, 2, 3, 4, 6, 9, 12, 18

55

36 12, 18, 20, 24, 30, 36, 40, 42, 48, ... 945 2

3 5391411025 = 52 × 7 × 11 × 13 × 17 × 19 × 23 × 29 11 2

13 × 172 × 19 × 23 × 29 × 31 × 37 × 41 × 43 × 47 × 53 × 59 × 61 × 67 × 71 × 73 × 79 × 83 × 89 × 97 × 101 × 103 × 107 × 109 × 113 × 127 × 131 × 137 × 139 × 149 × 151 × 157 × 163 × 167 × 173 × 179 × 181 × 191 × 193 × 197 × 199 × 211 × 223 × 227 ≈ 7.970466327 × 1087

2

3 n

σ(n)

σ(n) > 2n

σ(n) > 1000n

1000 N = ceil(e1000 )! ceil(x)

ceil

x

!

N N

N k

k

ceil(e1000 )

σ(N ) > N

#

k=1

(log

1 ≤ k ≤ ceil(e1000 )

1 > N log(ceil(e1000 )) > N log(e1000 ) = 1000N k

) N

1000

E446.94#2 40

Wiki -

ceil(e1000 )


N < 1010

446.94

=

1.3.

4

37

σ(n) > 1000n 1984

Guy Robin

n > 5040 σ(n) < eγ n log log n 41

γ

0.5772156649

σ(n) > 1000n 1000n < 1000 < eγ n >

σ(n) < eγ n log log n log log n ee

1000/eγ

> 1010

243.47

= E243.47#2

σ(n) > 1000n E243.47#2 < n < E446.94#2 3 m 1000n 1, 2, 4, 6, 12, 24,

36, 48, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 10080, 15120, 25200, 27720, 55440, 110880, 166320, 277200, 332640, 554400, 665280, 720720, 1441440, 2162160, 3603600, 4324320, 7207200, 8648640, 10810800, 21621600, ...

1.3

42

4 4

E6#4 = 1010 2 3

(E6#3–E6#4): 1010

4

6

41 Robin, G. (1984) Grandes valeurs de la fonction somme des diviseurs et hypothse de Riemann. Journal de Mathmatiques Pures et Appliques 63, 187–213. 42 https://oeis.org/A004394

38

1 10

10

1.3.1 10

100

10

10 10

N

N

(googolplexplex) (googolplusplex)

(gargoogolplex)

(googolplexian) (googolduplex) du 3

2 10

tri

E100

=

E100#2 = E100#3 = E100#4 = E100#5 = E100#6 = E100#7 =

(googoltriplex)

10100 = 1010 10

100

1010

(googol)

=

(googolplex)

100

=

(googolduplex) (googoltriplex) (googolquadriplex) (googolquinplex) (googolsextiplex)

E100#8 =

(googolseptiplex)

E100#9 =

(googoloctiplex)

1.3.

4

39

E100#10

=

(googolnoniplex)

E100#11

=

(googoldeciplex) 5

1.3.2 (prime counting function) π(x)

x x ̸= 1

li(x) li(x) =

$

x 0

dt log(t)

log

π(x) π(x) < li(x)

li(x) π(x) > li(x) π(x)

x

li(x) π(x) > li(x)

43 44

(Skewes number)

x

2 Sk1 = ee

1933 E(e)79#3 45

Sk2 = e

ee

e7.705

1 2

= E(e)7.705#4 46

43

=

π(x) > li(x) Sk1

1995

e79

π(x) > li(x)

x

Bays and Hudson (2000)

47

Wiki http://ja.googology.wikia.com/wiki/ - Skewes Number http://mathworld.wolfram.com/SkewesNumber.html 45 Skewes, S. (1933) On the difference pi(x)-li(x). Journal of the London Mathemetical Society s1-8(4): 277–283. doi:10.1112/jlms/s1-8.4.277 46 Skewes, S. (1955) On the difference pi(x)-li(x). II. Proceedings of the London Mathematical Society s3-5 (1): 48–70. doi:10.1112/plms/s3-5.1.48 47 Bays, C. and Hudson, R. H. (2000), A new bound for the smallest x with π(x) > li(x). Mathematics of Computation 69 (231): 1285–1296. doi:10.1090/S0025-5718-9901104-7 44 MathWorld

40

1 1.3983 × 10316

1010

10 ✓ ✒

103

Sk2

4

E963#3 = 1010

Sk1

10963

E34#3 = 1010

1034

E3#4 = ✏

2 = ee

2

7.705 ee

≈ 1010

1010

3

✑

= E3#4

48

Hypercalc

(Hypercalc)

2

: eêêê^7.705 : 10 ^ [ 10 ^ ( 3.299943221955 x 10 ^ 963 ) ] 2

1.4

5

1.4.1 ✓

10^10^10^10^10^1.1 = E1.1#5 ✒ 49 50

Page)

✏ (Don Nelson

✑

51

48 Hypercalc

http://www.mrob.com/pub/perl/hypercalc.html

49 http://ja.googology.wikia.com/wiki/ 50 Longest

Possible Time http://www.numberphile.com/videos/longest time.html D. N. (1994) Information loss in black holes and/or conscious beings? http://arxiv.org/pdf/hep-th/9411193 51 Page,

1.4.

5

41 1 52

)

1000

(5.391 × 10−44

Page (1994)

5

4 1.1

1.4.2 (Jonathan Bow53

ers)

(Forever endeavor) 54

Wiki

10

100

1 10

52 https://twitter.com/astrophys

tan/status/391928927012134912 Wiki http://ja.googology.wikia.com/wiki/ 54 http://www.polytope.net/hedrondude/foreverendeavor.htm 53

42

1

10 10

0 9 1

1

2 1

V

1 0

1

10 0

2 2

0000000000 3 1 0000000001, 0000000002, 0000000003, ...

1

1

18

500

1.4.

5

43

550

100

3 10 3

000 . . 000( % .&' 100

4

10

3 4

100 1 4 5

1010

10000000000

10

10 55

✓ 1 + 10 +

✒

55

8 #

E10#i = 1 + 10 + 10

10

(Bentley’s Number)

+ 10

1010

+ ... + 10

1010

✏ 1010

i=1

9

Wiki -


1010

1010

✑

44

1

1.5

56 57 58 59

pixiv

2017

8

10 60

20

Googology Wiki

Googology Wiki

Wiki

62

Wiki Googology Wiki

56 pixiv

http://comic.pixiv.net/works/1505


http://www.geocities.co.jp/Technopolis/9946/

59 http://www.slideshare.net/DoomKobayashi/presentations 60

(2016) Wiki http://googology.wikia.com/ Wiki http://ja.googology.wikia.com/

61 Googology 62

61

45

2

2.1 2.1.1 TEX

(Donald Knuth) 1 2 3

TEX

1

(arrow notation) LATEX

Wiki http://ja.googology.wikia.com/wiki/ - Arrow Notation http://mathworld.wolfram.com/ArrowNotation.html 3 Knuth, D. E. (1976) Mathematics and Computer Science: Coping with Finiteness. Science 194, 1235–1242. doi:10.1126/science.194.4271.1235 2 MathWorld

46

2

✓ x

y

x x ✒

n

x

xy

=

2 =

x

2 =

x

y

✏

=

x, x

y=x

x, x

x(n

(y − 1))

(x

y=x )y = x

ab

n−1

(y − 1))

(x

(x

n

✑

(y − 1))

b a

b ˆ

a

b

a ˆ

a

b

b

a ˆˆ b

2.1.2 1 Hypercalc

3

3

3 =

33 = 27

3

3

3 =

33 = 327 = 7625597484987

3

3

3

3 =

33

3

3

3

3

3 =

3

3

3

3

3 =

Hypercalc

10

3

33

= 1.35 × 10

3638334640024

10

(6.46 × 10

3638334640023)

10

10

10

4 PT ( 6.46

10

10

(6.46 × 10 (6.46

10

3638334640023) 3638334640023)

10 ˆ 3638334640023 )

2.1.

47 10

6.46 × 10

4

4 PT

3638334640023 = 10

1

3

10

12.56

3

10

3

3 =

27

3

3

3 =

10

12.88

3

3

3

3 =

10

10

12.56

3

3

3

3

3 =

10

10

10

12.56

3

3

3

3

3

3 =

10

10

10

10

12.56

3

3

3

3

3

3 =

10

10

10

10

10

3

12.56

10

27,12.88,12.56 3

10

10

3

2

2 2=2

2

2=2

2=4 3

3 3 1

3

3

3 3

3

3

3

3

3

3

2=2

2 4 3

3

3

3

3

1 3

5

2

3

48

2 3

10 3x

x

10x

10x = (3log3 (10) )x = 3x log3 (10) ≈ 32.1x x

1010

2.1

x

log3 (1010 ) 1010 x

2.1 × 10x ≈ 2.1 × 32.1x

≈

x

32.1×3

≈

2.1

2.1x

= 33

2.1x+log3 (2.1)

2.1x+0.68

2.1

0.68

3

2

3

3

3

3 = 10

10

x = 12.56 3

x

3

1

3

≈ 33

33

0.68

10

3

x

10

10

12.56

2.1x + 0.68 = 27.056

3 = 27

2.1x + 0.68

Hypercalc

x = 10 10

10

10

10

10

10

2x

10

xx 2x ≈ x x

10 ≈

0 ≈

2x

xx

4

4

log(log(log(log(2x )))) ≈ log(log(log(log(xx )))) ≈ 10 2x ≈ x x

2.1.

49

2.1.3 4

2

x

y

(y − 1))

=

x

(x

=

x

(x

(x

=

x

(x

(x

=

x %

x y

5

(power tower)

&'

x

(y − 2))) (y − 3))))

(x x (

(tetration) 6

4

tetra9 # i=0

10 ↑↑ i

2.1.4 4 4

(hyper-4)

n

Hn (a, b)

7

4 MathWorld

- Power Tower http://mathworld.wolfram.com/PowerTower.html

5 http://ja.googology.wikia.com/wiki/ 6 Goodstein, R. L. (1947). Transfinite Ordinals in Recursive Number Theory. Journal of Symbolic Logic 12(4): 123–129. doi:10.2307/2266486 7 http://ja.googology.wikia.com/wiki/

50

2 n

Hn (a, b) =

Hn (a, b)

⎧ ⎪ b+1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ a

(n = 0) (n = 1, b = 0)

0 1

⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩

(n = 2, b = 0) (n ≥ 3, b = 0)

Hn−1 (a, Hn (a, b − 1)) (

)

H0 (a, b) = b + 1

H1 (a, b) = a + b H2 (a, b) = a × b H3 (a, b) = ab H4 (a, b) = a

b

H5 (a, b) = a

b

H6 (a, b) = a

4

Hn (a, b) = a

n−2

3

b b(n ≥ 3)

3

3 =

3

3

3 =

3

=

3 %

3

3 = 7625597484987 7625597484987

3

3

&'

7625597484987

3

7625597484986

Hypercalc

3ˆ

3 (

7 7625597484983 PT ( 6.46

10 ˆ 3638334640023 ) 3

3 3

3

3 3

3

3 3

3=3 % 3

3 3

3

&'

3 3

3

3 (

Hypercalc

2.1.

51 Hypercalc (1010 )

10

10

(3

3

3)

Hypercalc

✓

hc(n)

✏

hc(0) = 6, hc(n + 1) = c(hc(n))

c(n)

(p.27

1.2

)

n hc(0)

(i) n = 0 (ii) n > 0 ✒

hc(n − 1)

✑

hc(n)

hc(n+ c(n + 1) = 10c(n)

1) = c(hc(n)) 10

x

c(x) 1

6

1

2

N

3

3

N 3

2 3

2 3

3

3

3

1 3 3

3 1

3

Hypercalc

3

5 8

(pentation)


52

2 3

✓

9

3

=3

✒

3=3

3

✏

3

7625597484986 10

(tritri)

✑

2 10

100

100 4

6 11

3

(hexation) 3 =

3

3

3

= 3 = 3 %

3

&'

3

3 (

3

3 f (4)

Hypercalc 1

9 http://ja.googology.wikia.com/wiki/ 10 http://ja.googology.wikia.com/wiki/ 11 http://ja.googology.wikia.com/wiki/

2.1.

53

1.

2. 3.

100 100000

100000 100

3

3

100

3

100

100

1) % &' ( b

X →c 1 =

X

→c−1

3 http://peterhurford.com/ 4

Wiki -


4.4.

87 X →c 1 →c a

=

X

X →c (a + 1) →c (b + 1)

=

X →c (X →c a →c (b + 1)) →c b

3 2

2

1

3 →2 x 3 →2 x →2

=

3→3→3

3 →2 4

=

3→3→3→3

3 →2 5

=

3→3→3→3→3

x 3 2

→2

2 3 →x x

3 →2 3

4

3 →n x

3 3 →3 x 3

1

x n−1

+2

→2

4.4 5


(Fish number)

3 3

88

4

4.4.1

1 6

2ch

161

02/06/20 22:25 >> 156 3

3 3

1 3

2

3

1

3

63 63 1


4.4.

89

7

(2ch

317-319

m

)

f (x)

n

m, f (x), n

m

f (x)

n

g(x) S : [m, f (x)]

f (x) =

[n, g(x)]

x

f (4) f (f (4)) 64

64 g(x) = f (x)

f (4) 64

64

S : [m, f (x)]

f (x)

[f m (m), f m (x)] f 64 (64)

m = 64, f (x)

S >> 161 S

10 S

Ackermann

S 7 2ch

http://www.geocities.co.jp/Technopolis/9946/log/ln023.html

90

4

S Ackermann B(0, n)

=

f (n)

B(m + 1, 0) =

B(m, 1)

B(m + 1, n + 1)

=

B(m, B(m + 1, n))

g(x)

=

B(x, x)

S : [m, f (x)]

[g(m), g(x)]

[3, f (x) = x + 1] S

10

>> 161

S S

f (m)

S2

m, f (x), S

S2

SS : [m, f (x), S]

[n, g(x), S2]

g(x) = S2[m, f (x)], n = g(m) SS [3, x + 1, S]

SS

1

S 1

4 S

4.4.

91

S 8

1 1.

S S

S(m, f (x)) = (g(m), g(x)) g(x) B(0, n)

=

B(m + 1, 0) =

f (n) B(m, 1)

B(m + 1, n + 1)

=

B(m, B(m + 1, n))

g(x)

=

B(x, x)

2.

SS SS

SS(m, f, S) = (S f (m) (m, f ), S f (m) ) (( (

,

3. 3

,

,

)

(m0 , f0 , S0 )

),

)

3 m0 = 3, f0 (x) = x + 1, S0 SS 63 (m0 , f0 , S0 )

1

1

2

1


1

S

92

4

1 1

F1

1 N

N

F1 (x)

FN

FN (x)

63 3

3

0

63 63

S : [m, f (x)]

[g(m), g(x)]

SS : [m, f (x), S]

[n, g(x), S2]

S2 S2 : [m, f (x)]

=

S f (m) [n, g(x)]

9

329,331,377379 [3, f (x) = x + 1]

1 B(0, n)

=

B(m + 1, 0) =

n+1 B(m, 1)

B(m + 1, n + 1)

=

B(m, B(m + 1, n))

g(x)

=

B(x, x)

B(m, n)

g(x) =

A(x, x) S : [3, x + 1]

[A(3, 3), A(x, x)]


:Mikadukim/SS

4.4.

93

10

A(3, 3) = 61

1

2 B(0, n)

=

A(n, n)

B(m + 1, 0) =

B(m, 1)

B(m + 1, n + 1)

=

B(m, B(m + 1, n))

g(x)

=

B(x, x)

g(x) g(1) =

B(1, 1) = B(0, B(1, 0))

=

B(0, B(0, 1)) = B(0, A(1, 1))

=

B(0, 3) = A(3, 3) = 61

g(2) =

B(2, 2) = B(1, B(2, 1))

=

B(1, B(1, B(2, 0))) = B(1, B(1, B(1, 1)))

=

B(1, B(1, 61))

=

B(1, B(0, B(1, 60)))

B(1, 1) =

61

B(1, 2) =

A(61, 61)

B(1, 3) =

A(A(61, 61), A(61, 61)) B(1, 61)

g(2) = B(1, B(1, 61))

10

2

g(2)

http://comic.pixiv.net/viewer/stories/6995

94

4 g(2)

g(x)

x

g(61) 2

g(61)

B(0, n)

=

B(m + 1, 0) = B(m + 1, n + 1)

=

gg(x) =

g(n) B(m, 1) B(m, B(m + 1, n)) B(x, x)

gg(x) gg(1)

=

B(1, 1) = B(0, B(1, 0))

=

B(0, B(0, 1)) = B(0, g(1))

=

B(0, 61) = g(61)

=

B(2, 2) = B(1, B(2, 1))

=

B(1, B(1, B(2, 0))) = B(1, B(1, B(1, 1)))

=

B(1, g(61))

B(1, 1)

=

g(61)

B(1, 2)

=

g(g(61))

B(1, 3)

=

g(g(g(61)))

gg(2)

gg(2)

61

g(x)

g(61)

gg(3), gg(4)...

SS

gg(g(61))

SS S

S

m, f (x), S

f (m)

S f (m)

S f (m)

4.4.

95 SS

S

m

S f (x) S

S S

380

02/07/02 19:47

386

02/07/02 20:25

388

02/07/02 20:32

10

88

96

4

11

Code Ass (@aycabta)

12

1

Ruby

13

Code Ass 14

15

16

F1

S

f (x)

g(x)

F1 2

2

2

2

2

1 2

2 F1

F1

f (x)

Ax (z, z) g(z) = Ax (z, z) 3

11 2ch

S

2

g(x)

S

f (z) =

3 2

2 [3, x + 1]

2

F1 F1

S

i

Si (x)

http://www.geocities.co.jp/Technopolis/9946/log/ln024.html @aycabta https://twitter.com/aycabta 13 https://github.com/aycabta/fish-number 14 https://github.com/aycabta/ackermann 15 https://github.com/aycabta/arrow-notation 16 https://github.com/aycabta/chained-arrow-notation 12 Twitter

S

4.4.

97

Si+1 (0, n)

=

Si (n, n)

Si (m + 1, 0)

=

Si (m, 1)

Si (m + 1, n + 1)

=

Si (m, Si (m + 1, n))

3 F1

S

i

S

B(m, n) = A(i, m.n)

1

g(2)

A(1, 2, 2)

A(1, 1, 0)

=

A(1, 0, 1) = A(1, 1) = 3

A(1, 1, 1)

=

A(1, 0, 3) = A(3, 3) = 61

A(1, 1, 2)

=

A(1, 0, 61) = A(61, 61) > 3

A(1, 1, 3)

>

A(1, 0, f 2 (1) + 2) > 3

A(1, 1, x)

>

3

3

x

A(1, 1, 65)

>

3

3

65

3

3 3

2

2+2

2+2

2 2>

A(1, 1, 65) A(1, 2, 0)

=

A(1, 1, 1) = 61

A(1, 2, 1)

=

A(1, 1, 61) > 3

A(1, 2, 2)

> A(1, 1, 3 >

3

3

3 (3

3 61

3

61

2

2) 61

2)

2

A(1, 2, 2) F1

4

4

A(1, 0, 1, 0)

=

A(1, 0, 0, 1) = A(1, 0, 1) = A(1, 1) = 3

A(1, 0, 1, 1)

=

A(1, 0, 0, (A(1, 0, 1, 0))

98

4 = A(1, 0, 1, 1)

A(1, 0, 0, 3) = A(3, 0, 3) = A(2, 3, 3)

2

A(1, 0, 1, 2)

=

A(1, 0, 0, A(1, 0, 1, 1))

=

A(1, 0, 0, A(2, 3, 3))

=

(A(2, 3, 3), 0, (2, 3, 3))

A(1, 0, 1, 3) =

A(1, 0, 0, A(1, 0, 1, 2))

A(1, 0, 1, n + 1)

A(1, 0, 1, 2) SS

2

=

A(A(1, 0, 1, 2), 0, A(1, 0, 1, 2))

=

A(1, 0, 0, A(1, 0, 1, n))

=

A(A(1, 0, 1, n), 0, A(1, 0, 1, n))

A(2, 3, 3)

S

F1

2

(

2

2

2

2

3

SS

S A(1, 0, 1, n)

n

m m

1

SS

f (m)

f (m)

SS

2

1

S

4

A(1, 0, 1, 1) A(1, 0, 1, 2) F1

SS

2

2

S

SS SS

A(1, 0, 1, 63)

A(1, 0, 2, 0)

=

A(1, 0, 1, 1) = A(2, 3, 3)

A(1, 0, 2, 1)

=

A(1, 0, 1, (A(1, 0, 2, 0)))

2 63

4.4.

99 =

A(1, 0, 2, 1)

S

SS

F1 SSSS

A(1, 0, 1, (A(2, 3, 3))) A(2, 3, 3)

SSS

A(1, 0, 2, 1)

A(1, 0, 3, 1) A(1, 1, 1, 1) A(1, 1, 1, 1)

SSSS

=

A(1, 1, 0, A(1, 1, 1, 0))

=

A(1, 1, 0, A(1, 1, 0, 1))

=

A(1, 1, 0, A(1, 0, 1, 1))

=

A(1, 1, 0, A(2, 3, 3))

=

A(1, 0, A(2, 3, 3), A(2, 3, 3))

S

A(2, 3, 3)

F1

A(1, 0, 1, 63) < F1 < A(1, 0, 2, 1) < A(1, 1, 1, 1) F1 S

1 17

87

W7plq.175s 3 2

a

B(x, y)

17

log/ln032.html

a−1

bN 2

S

2

3 F2 (x)

3

SS 63

3

S

102

4

SS

a

S

b

ga,b (n) = Ba,b (0, n)

A(a, b, 0, n)

a = 0, b = 0

B0,0 (0, n) = n + 1 = A(0, 0, 0, n) Ba,b (m, n) = A(a, b, 0, n)

S

1

A(a, b + 1, 0, n) S Ba,b (0, n)

=

A(a, b, 0, n)

Ba,b (m + 1, 0) = Ba,b (m + 1, n + 1)

Ba,b (m, 1)

=

Ba,b (m, B(m + 1, n))

Ba,b+1 (0, n) =

Ba,b (n, n)

Ba,b (m, n) = A(a, b, m, n) A(a, b, 0, n)

=

A(a, b, m, 1)

A(a, b, m + 1, n + 1)

=

A(a, b, m, A(a, b, m + 1, n))

A(a, b, 0, n)

SS SS

A(a, b, 0, n)

A(a, b, m + 1, 0)

A(a, b + 1, 0, n) =

S

=

SS 1

A(a, b, n, n)

A(a, b + 1, 0, n)

a

Ba,0 (0, n)

Ba+1,0 (0, n)

S

n

Ba,n (0, n)

f (m) n

Ba+1,0 (0, n)

=

A(a + 1, 0, 0, n) =

Ba,n (0, n)

A(a, n, 0, n)

4.4.

103 ga,b (n) = Ba,b (0, n) ≈ A(a, b, 0, n)

SS

63

A(63, 0, 0, n)

A(1, 0, 0, 0, 63) = A(63, 0, 0, 63)

4.4.3

3 3(F3 )

21

3 1.

f (x)

s(1)f s(n)f 2.

3.

f (x)

g(x)

s(n) (n > 0)

g; g(x) = f x (x)[

:=

] x

g; g(x) = [s(n − 1) ]f (x)(n > 1)[n

:=

g(x)

]

ss(n) (n > 0)

ss(1)f

:=

g; g(x) = s(x)f (x)[

]

ss(n)f

:=

g; g(x) = [ss(n − 1)x ]f (x)(n > 1)[

]

F3 (x) F3 (x) := ss(2)63 f ; f (x) = x + 1

4.

F3 := F363 (3) 22

F3


http://ja.googology.wikia.com/wiki/ 3

3 :Mikadukim/

104

4

f∗

f f ∗ (x) = f x (x) T∗

T

(T ∗ f )(x) = (T x f )(x) s(n), ss(n), F3 s(1)f

=

f∗

s(n)

=

s(n − 1)∗

(ss(1)f )(x)

=

(s(x)f )(x)

ss(n)

=

ss(n − 1)∗

F3 F1

63

=

(n ≥ 2) (n ≥ 2)

63

(ss(2) f0 ) (3),

f0 (x) = x + 1

F2

F2

F3 (

F3 )

F3 F3

F3

F1 ,F2 ,

F3 ,

F1

F3

F3

4.1 F3 F3

F3

F3

F3

F3

2

1

SS S

S

S S

S

4.4.

105

4.1: F1

F1

F3

2

S

S F2

(SS

S

(2

)

)

SS

(3

)

(SS F3

)

F2

S

,SS

,...

s(n)

(

)

ss(1) F3

F3

s(n)

s(1)

2

SS

S

SS s(n) F3

2

S 2

2 2

3 F3

s(1)

s(2) 2

F3 s(1)f

:=

g; g(x) = f x (x)

s(2)f

:=

g; g(x) = [s(1)x ]f (x)

F1

F2

s(2)

S s(1)f := g; g(x) = f x+1 (1)

106 s(2)

4 F1

F3

F2

S

s(n)

f (x) = x + 1 [s(1)a1 ][s(2)a2 ]...[s(n)an ]f (x) ≈ A(an , ..., a2 , a1 , x) s(1)f := g; g(x) = f x+1 (1) F3 s(n)

s(1)

F3 ss(1)

s(n)

g(x) = s(x)f (x)

f (x)

x

4.5 (Chris Bird)

23

uglypc.ggh.org.uk 2002

n 2 3 4 23

n 2 x →x x

Wiki - Chris Bird http://ja.googology.wikia.com/wiki/Chris Bird

4.5.

107 4

4

2

4

A(1, 0, 1, 2, 2)

24

(Bird’s linear notation) BEAF

(Jonathan Bowers)

25 26

2002

(John Spencer) (Bowers Exploding Array Function)

24 Chris

BEAF 2007

Bird’s Super Huge Numbers http://mrob.com/users/chrisb/ Bowers 26 Hedrondude’s Home Page http://www.polytope.net/hedrondude/home.htm 27 Wiki - BEAF http://ja.googology.wikia.com/wiki/BEAF 25 http://ja.googology.wikia.com/wiki/Jonathan

27

108

4

2002

1 2016

2

352

2008 Web 2015

E 3

E

15610 (Lawrence Hollom) (Aarex Tiaokhiao)

Hyp cos

28

BEAF

(array notation) BEAF

28

Wiki -


BEAF

4.5.

109 BEAF A = (a1 , a2 , . . . , an ) 1

v(A) = {a1 , a2 , . . . , an }

1. {a} = a, {a, b} = ab 2. {a, b, c, . . . , n, 1} = {a, b, c, . . . , n} 3. {a, 1, b, c, . . . , n} = a 4. 3

1 {a, b, 1, . . . , 1, c, d, . . . , n} % &' ( x

=

1

{a, a, . . . , a, {a, b − 1, 1, . . . , 1, c, d, . . . , n}, c − 1, d, . . . , n} % &' ( % &' ( x+1

•

a

x

1

1

•

2

• 5.

1

1

1 4 {a, b, c, d, . . . , n} = {a, {a, b − 1, c, d, . . . , n}, c − 1, d, . . . , n} 3 {3, 3, 3}

=

{3, {3, 2, 3}, 2}

=

{3, {3, {3, 1, 3}, 2}, 2}

=

{3, {3, 3, 2}, 2} = {3, {3, {3, 2, 2}, 1}, 2}

=

{3, {3, {3, 2, 2}}, 2}

1

110

4 {3, 3{3,2,2} , 2}

=

{3, 3{3,{3,1,2},1} , 2}

=

{3, 3{3,3} , 2} = {3, 33 , 2}

=

{3, 3

3, 2} = {3, {3, 3

= 3

{3, 3

= 3

3

{3, 3

= 3

3

3

=

{3, 3, 3}

3 − 2, 2}

{3, 3 3

3 − 3, 2} 3

3

3

3

3 3

{a, b, c} = a

b

c

=

{a, {a, b − 1, c}, c − 1}

=

a

c−1

=

a

c−1

=

... = a %

{a, b − 1, c} = a c−1

b

c−1

&'

...

a

c−1

b

{a, {a, b − 2, c}, c − 1}

{a, b − 2, c} c−1

a

b

{a, b} = a

c

c−1

c

c=a

c−1

a

3 BEAF

c=1

{a, b, c}

3 − 1, 2}}

3 − 1, 2}

... = 3

= 3 {3, 3, 3}

3

=

a=a (

c

b

4 {3, 65, 1, 2}

=

{3, 3, {3, 64, 1, 2}, 1} = {3, 3, {3, 64, 1, 2}}

✷

4.5.

111 =

{3, 3, {3, 3, {3, 63, 1, 2}}}

=

{3, 3, {3, 3, {3, 3, {3, 62, 1, 2}}}}

{65, 2, 2, 2} = {65, {65, 1, 2, 2}, 1, 2} = {65, 65, 1, 2}

1.

1

0

2.

3.

{a, b} = ab

A(a) = a + 1

2 1

{a, b, 1, . . . , 1, c, d, . . . , n} =

4.

{a, a, a, . . . , {a, b − 1, 1, . . . , 1, c, d, . . . , n}, c − 1, d, . . . , n} a

, a) = A(X, b, a,

A(X, b+1, 0,

, a)

1 4

A(1, 0, 0, 0, 0, 5)

1

=

A(1, 5, 0, 0, 0, 5)

=

A(1, 4, 5, 0, 0, 5)

=

A(1, 4, 4, 5, 0, 5)

=

A(1, 4, 4, 4, 5, 5)

112

4

f (m) = {n + 1, m, a0 , X}(X {n + 1, 2, a0 + 1, X} {n + 1, 3, a0 + 1, X}

{n + 1, 4, a0 + 1, X} {n + 1, m + 1, a0 + 1, X} 2

)

=

{n + 1, {n + 1, 1, a0 + 1, X}, a0 , X}

=

{n + 1, n + 1, a0 , X} = f (n + 1)

=

{n + 1, {n + 1, 2, a0 + 1, X}, a0 , X}

=

{n + 1, f (n + 1), a0 , X}

=

f 2 (n + 1)

=

{n + 1, {n + 1, 3, a0 + 1, X}, a0 , X}

=

f 3 (n + 1)

=

f m (n + 1)

m+1

f

{n + 1, 2, a0 + 2, X}

= =

3

0

{n+1, 2, a0 +1, X} = f (n+1) 1

m

{n + 1, n + 1, a0 + 1, X} f n (n + 1)

{n+1, 2, a0 +2, X} = f n (n+1) f n (n)

3

{n + 1, m + 1, 1, a1 + 1, X} {n + 1, n + 1, {n + 1, m, 1, a1 + 1, X}, a1 , X}

= 4

{n + 1, m, 1, a1 +

1

1, X}

2

4 1

2 5

3 1

4

2 n+2

3 n

4.5.

113

n>1 {n + 1, n + 1, n + 1, n + 1} = {n + 1, 2, 1, 1, 2} ≈ {n + 1, ..., n + 1} = {n + 1, 2, 1, 1, 1, 2} ≈ % &' (

A(1, 0, 0, n) A(1, 0, 0, 0, n)

5

{n + 1, 2, a0 + 1, a1 + 1, ..., ak + 1} ≈

A(ak , ..., a1 , a0 , n)

f (n) = A(ak , ..., a2 , a1 , a0 , n) {n + 1, m + 1, a0 + 1, a1 + 1, a2 + 1, ..., ak + 1} ≈ f m (n)

{1, a0 , a1 , ..., ak }

= 1

{2, a0 , a1 , ..., ak }

= 4 (k > 1, a0 > 1, ak > 1

)

4

{2, a0 , a1 , ..., ak }

=

{2, {2, a0 − 1, a1 , ..., ak }, a1 − 1, ..., ak }

=

... = {2, X, 1, ..., ak }(

=

{2, 2, Y, ..., ak }(

=

{2, {2, 1, Y, ..., ak }, Y − 1, ..., ak }

=

{2, 2, Y − 1, a2 , ..., ak }

=

{2, 2, Y − 2, a2 , ..., ak }

=

{2, 2, 1, a2 , ..., ak }

=

{2, 2, 1, 1, a3 , ..., ak }

=

{2, 2, 1, 1, ..., 1}

=

{2, 2} = 4

3 2

2

X

)

Y = {2, X − 1, 1, ..., ak })

1 2

4

4 3

114

4 3

5

1

4

1

4 3

Googology Wiki

29

1

30

4 (a, b, c, d) a

b

d > {a, b, c, d}

c

a = 2, b > 2, c > 0, d > 1

n−1

n n−2

5

4

2 3 5

31 32

2

(Bird’s proof)

3 4

=4

(p. 219)

29

http://ja.googology.wikia.com/wiki/ :1793 http://googology.wikia.com/wiki/User blog:Wythagoras/Results of the First International Googological Olympiad 31 Wiki http://ja.googology.wikia.com/wiki/ 32 5 http://www.mrob.com/users/chrisb/Proof.pdf 30

1

115

5

2

2 •

•

116

5

5.1 (Georg Cantor, 1845–1918)

(ordinal number)

(transfinite number)

5.1.1 ✓ ✒

1 MathWorld 2

Wikipedia

1 2

✏ ✑

- Ordinal Number http://mathworld.wolfram.com/OrdinalNumber.html MathWorld ZF Quine

5.1.

117

3

(ordered set) 2

(totally ordered set) (partially ordered set)

2 a, b ⊂

X X

2

X ≤

≤

≤

(X, ≤)

1. X

≤

3

a≤a

a

2. x ≤ y

y≤z

x≤z

3. x ≤ y

y≤x

x=y

4. X

3 Wikipedia

x, y

-

x≤y

http://ja.wikipedia.org/wiki/

y≤x

X (X, ≤)

118

5 4 5

(well-order)

(well-ordered set)

X ≤

✓

≤

A

(well-founded)

X a0

A A

a0 ≤ a

a

X

✒

✑ 0

0 a

a/2 an = 1/n

(axiom of choice)6

7

theory)

(naive set theory)

4

Wiki http://ja.math.wikia.com/wiki/ http://ja.wikipedia.org/wiki/ 6 Wikipedia http://ja.wikipedia.org/wiki/ 7 http://alg-d.com/math/ac/ 5 Wikipedia

✏

(axiomatic set

5.1.

119

(Ernst Friedrich Ferdinand Zermelo, 1871–1953) (Abraham Halevi Adolf Fraenkel, 1891–1965) (Zermelo-Fraenkel set theory) ZFC ZF (choice)

ZF Zermelo

Fraenkel

C

8 9 10

ZFC

ZFC

(order type) (A, ≤)

2

A

✓ a1 , a 2

B

(A, ≤)

(order isomorphic)

(B, ≤)

A

✏

a1 ≤ a2 ⇐⇒ f (a1 ) ≤ f (a2 ) ✒ 8

A

B

f

Wiki http://ja.math.wikia.com/wiki/ http://ja.wikipedia.org/wiki/ 10 MathWorld - Order Type http://mathworld.wolfram.com/OrderType.html 9 Wikipedia

✑

120

5 (bijective)

1.

: f (A) = B

2.

:

11

A

f :A

a1 , a2

B

2

f (a1 ) = f (a2 )

a1 = a2

5.1.2 (finite ordinal) A k

B

k 0, 1, 2, 3, ...

(

k

) (inifinite ordinal) N = {0, 1, 2, 3, . . .}

ω

(transfinite number)

0, 1, 2, 3,

ω

0, 1, 2, 3,

(fundamental sequence) 1

ω 0, 2, 4, 6, ...

ω

ω ω

ω + 1, ω + 2, ... 1+ω

11 Wikipedia

-

http://ja.wikipedia.org/wiki/

5.1.

121

1 + 1, 1 + 2, 1 + 3, ...

1+ω =ω

(A, ≤)

α

α

(John von Neumann, 1903 - 1957)

0

=

{} (

1

=

{0} = {{}} (1

2

=

{0, 1} = {{}, {{}}} (2

3

=

{0, 1, 2}

4

=

{0, 1, 2, 3}

ω

=

{0, 1, 2, ...} (

=0

) ) )

)

ω+1 =

{0, 1, 2, ..., ω}

ω+2 =

{0, 1, 2, ..., ω, ω + 1}

ω+3 =

{0, 1, 2, ..., ω, ω + 1, ω + 2}

ω+ω

{0, 1, 2, ..., ω, ω + 1, ω + 2, ...}

=

(cardinality) A

B

A

ω {0, 1, 2, ..., ω} B

A

B

A = {0, 1, 2, ...} B

ω+1

B=

f

f (0) = ω, f (k) = k − 1(k > 0) a1 ≤ a2 ⇐⇒ f (a1 ) ≤ f (a2 )

A

122

5 a1 = 0, a2 = 1 f

f (k) = ω

k

f (k + 1)

ω

B

ω

A

(cardinal number)

ω+1 A ω

ω 0 ℵ0 )

ω+1

ω0

ω

5.1.3 α

β

α

α = β+1

(successor ordinal)

0

(limit ordinal)

ω

ω + a(a = 1, 2, 3, ...)

0

✓ α

ξ ✒

ζ

✏

ζ · · · >

ω βi c i

i=1 β1

ω c 1 + ω β2 c 2 + · · · + ω βk c k

✒

✑

ϵ 0 = ω ϵ0

ϵ0

ϵ0 ϵ0

ω

ϵ0 ϵ0

α

f (α)

f (α)

• f (α) = 1 : 0, 1, 2, ..., 9, ω • f (α) = 2 : 10, 11, 12, ...., 99, ω2, ω3, ..., ω9, ω 2 , ω 3 , ..., ω 9 , ω ω

5.2 α

✏

N→N

(ordinal hierarchy)

126

5

5.2.1 1904

(Godfrey Harold 13 14

Hardy, 1877–1947)

(Hardy function)

H[α] : N → N

α ✓ H[0](n)

=

n

H[α + 1](n)

=

H[α](n + 1)

H[α](n)

=

H[α[n]](n)(α

✒

✏

✑

)

(

+1

)

x+1

α H[α] α>β H[α] > H[β]

H[ 1

(canonical

fundamental sequence) α[0], α[1], α[2], ... 1 (Wainer hierarchy)

]

H[α] α ≤ ϵ0

13 http://ja.googology.wikia.com/wiki/ 14 Hardy, G.H. (1904) A theorem concerning the infinite cardinal numbers. Quarterly Journal of Mathematics 35: 87–94.

5.2.

127

✓

α ≤ ϵ0

α

α[0], α[1], α[2], ... ω[n]

=

n

[n]

=

ωα n

ω α [n]

=

ω α[n] (α

(ω α1 + ω α2 + · · · + ω αk )[n]

=

ω α1 + ω α2 + · · · + ω αk [n]

ω

α+1

✏

(α1 ≥ α2 ≥ · · · ≥ αk

✒

ϵ0 [0]

=

1, ϵ0 [n + 1] = ω ϵ0 [n]

ϵ0 [0] = 0

✑

ϵ0 ω[n] = n

ω

0, 1, 2, 3, ...

H[3](n) =

H[2](n + 1) = H[1](n + 2) = H[0](n + 3) = n + 3

H[m](n) =

n+m

H[ω](n) =

H[n](n) = n + n = 2n

H[ω + 2](n) =

H[ω + 1](n + 1) = H[ω](n + 2) = 2(n + 2)

H[ω + k](n) =

2(n + k)

H[ω × 2](n) =

H[ω + n](n) = 2(n + n) = 4n

H[ω × 2 + 2](n) = H[ω × 3](n) = H[ω × k](n) = H[ω 2 ](n) =

H[ω 2 + 3](n) = H[ω 2 + ω](n) = H[ω 2 + ω + k](n) =

H[ω × 2 + 1](n + 1) = H[ω × 2](n + 2) = 4(n + 2) H[ω × 2 + ω](n) = 4(n + n) = 8n = 23 n

2k n

H[ω × n](n) = 2n n

H[ω 2 ](n + 3) = 2n+3 (n + 3) H[ω 2 + n](n) = 2n+n (n + n) = 22n (2n) H[ω 2 + ω](n + k) = 22(n+k) 2(n + k)

128

5

H[ω 2 + ω × 2](n) =

H[ω 2 + ω + n](n) = 22(n+n) 2(n + n) = 24n (4n)

H[ω 2 + ω × 3](n) =

H[ω 2 + ω × 2 + n](n) = 24(n+n) 4(n + n)

H[ω 2 + ω × k](n) =

22 n (2k n)

3

= 28n (8n) = 22 n (23 n) k

H[ω 2 × 2](n) =

H[ω 2 + ω × n](n) = 22

n

n

(2n n)

✓

✏ H[α + β](n) = H[α](H[β](n))

✒

α + β = γ + β, α > γ H[α + β](n)

γ H[α](n)

H[β](n)

α = ω 2 + 2, β = ω α + β = ω2 + 2 + ω = ω2 + ω +2 α = ω2 , β = ω + 3

H[α + β](n) = H[α](H[β](n))

H[α + β](n) = H[ω 2 + ω + 3](n) = 22(n+3) 2(n + 3)

H[α](n)

=

H[ω 2 ](n) = 2n n

H[β](n)

=

H[ω + 3](n) = 2(n + 3)

H[α](H[β](n)) =

H[α × 2](n)

=

22(n+3) 2(n + 3)

H[α + α](n) = H[α](H[α](n)) = H[α]2 (n)

✑

5.2.

129 H[α × 3](n)

H[α × 2 + α](n) = H[α × 2](H[α](n))

= =

H[α]2 (H[α](n)) = H[α]3 (n)

H[α × 4](n)

=

H[α]4 (n)

H[α × k](n)

=

H[α]k (n)

H[α × ω](n)

=

H[α]n (n)

H[α × ω](n)

H[ω](n) = H[ω 2 ](n) =

H[ω 3 ](n)

H[α](n)

n

2n H[ω × ω](n) = H[ω]n (n) = 2n n

=

H[ω 2 × ω](n) = H[ω 2 ]n (n)

f (n) = 2n n H[ω 2 ](n)

=

f (n) = 2n n

H[ω 2 × 2](n)

=

f 2 (n) = 22

H[ω 2 × 3](n)

=

f 3 (n) = 22

n

n

(2n n)

2n n

(2n n) 2n n

2

(2n n)

f (n) = 2n H[ω 2 ](n)

> f (n) = 2n

H[ω 2 × 2](n)

> f 2 (n) = 2

2

n>2

H[ω 2 × 3](n)

> f 3 (n) = 2

2

2

n>2

H[ω × 4](n)

> f (n) = 2

2

2

2

H[ω 3 ](n)

2

2

4

H[ω 2 ](n) > 2n n

>

2 3

n>2

4

n H[ω 3 ](n) > 2

130

5

H[ω 4 ](n) > 2

n

m>1

H[ω m ](n)

>

m−1

2

m=2

n

m=k

m=k+1 H[ω k ](n)

>

k−1

2

n

H[ω k × 2](n) =

H[ω k ]2 (n) > 2

H[ω × 3](n) =

H[ω ] (n) > 2

k

H[ω k+1 ](n) =

k 3

k−1 k−1

H[ω k × ω](n) > 2

2

k−1

2

k−1

k

k

n>2 2

k−1

2

n>2

k

3

n ✷

5.2.2 15

hierarchy) ✓

16

(FGH; Fast-growing

F [α](n)

✏

(FGH)

✒

F [0](n)

=

n+1

F [α + 1](n)

=

F [α]n (n)

F [α](n)

=

F [αn ](n)(α

Wiki

✑

)

fα (n) F [α](n)

15 16

Wiki ,

,

,

http://ja.googology.wikia.com/wiki/ (1997)

5.2.

131

F [α](n) = fα (n) α < ϵ0 F [α](n) = H[ω α ](n) 1 F [0](n) = n + 1 = H[1](n) = H[ω 0 ](n) F [α](n) = H[ω α ](n)

2 F [α + 1](n)

=

F [α]n (n) = H[ω α ]n (n)

=

H[ω α × ω](n)

=

H[ω α+1 ](n)

F [α + 1](n)

α = ϵ0

α

F [α](n) = H[ω ](n) F [ϵ0 ](3)

ω

=

F [ω ω ](3)

=

H[ω ω

≈

H[ϵ0 ](4)

ωω

](3)

F [ϵ0 ](n) ≈ H[ϵ0 ](n + 1) = H[ϵ0 + 1](n) F [ϵ0 ](n) ≈ H[ϵ0 ](n)

+1

f n (n)

1953 gorczyk, 1922–2014) chy)

(Andrzej Grze(Grzegorczyk hierar-

17

17 Grzegorczyk, A. (1953) Some classes of recursive functions, Rozprawy Matematyczne 4: 1–45.

132

5 n

• E0

En

x + 1, x + 2, ...

• E1

x + y, 3x, ...

• E2

xy, x3 , ...

• E3

x y , 22

2x

, elementary recursive

function

• E4 En

n

E n−1

(extended Grzegorczyk hierarchy)

18

1904

1953 (John Edensor Littlewood, 1885–1977) 19

18

(2016) J. E. (1953) A mathematician’s miscellany. Methuen

19 Littlewood,

5.2.

133 20

H[ω m ](n) > 2 F [m](n) = H[ω m ](n) > 2

m−1 m−1

n

n 6

m=3

F [α](n) F [α + 1](n)

m

F [m](n)

n F [n](n)

2 F [ω](n) = F [n](n) F [ω](n)

2

n−1

F [ω](n) > 2

f (n) = 3

n

n = A(n + 1, n − 3) + 3 > A(n, n)

f 64 (4)

3

F [ω + 1](n) = F [ω + 1](64)

=

F [ω]n (n) ≈ f n (n) f 64 (64) ≈

f (n) = 3 ω+1 1 1 ω

1

2 3 20

( )

3

n

≈

( ) (1990)

f (n) ≈ F [ω](n)

F [ω](n)

n

3

ω

f n (n)

+1

134

5

3

3

3

3

n

≈

F [ω + ω](n) = F [ω × 2](n)

3

3

3

n

≈

F [ω × 2 + ω](n) = F [ω × 3](n)

✓

✏

3

3

2n

≈

3

23

2n

≈

3

23

23

2n

≈

23

23

23

2n

≈

3

3n

≈

3

4n

≈

3

nn

≈

✒

F [ω × ω](n) = F [ω 2 ](n) F [ω 2 + ω](n)

F [ω 2 + ω × 2](n) F [ω 2 + ω × 3](n)

F [ω 2 + ω × ω](n) = F [ω 2 × 2](n) F [ω 2 × 3](n)

F [ω 2 × ω](n) = F [ω 3 ](n) 1

2

ω

2

ω

3

n ω n−1

• n • n

F [ω n−1 an−1 + ω n−2 an−2 + ... + a0 ](n)

n

3 F3

s(n)

fb (n) = A(X, b, n) (X

0

fb+1 (n)

=

A(X, b + 1, n)

=

A(X, b, A(X, b + 1, n − 1))

=

fb (A(X, b + 1, n − 1))

=

0

)

fb2 (A(X, b + 1, n − 2))

✑

5.2.

135 = fbn (A(X, b + 1, 0))

= ≈

fbn (n) s(1)f (x) = f x (x)

s(1) 3 1.

fb (n) = A(X, b, n)

fb+1 (n) ≈ fbn (n)

s(1)f (n) = f n (n)

2. s(1)

F [α + 1](n) = F [α]n (n)

3.

2

1

s(1)

1

f (n) = n + 1 A(2, n)

≈

A(3, n)

≈

A(4, n)

≈

A(m + 1, n)

≈

A(n, n)

≈

s(1)f (n) = f n (n) = F [1](n) = 2n s(1)2 f (n) = F [2](n) = 2n n s(1)3 f (n) = F [3](n) ≈ 2 ↑2 n

s(1)m f (n) = F [m](n) ≈ 2 ↑m−1 n s(1)n f (n) = F [n](n)

s(1) A(1, 0, n) = A(n, n), s(2)f (n) = s(1)n f (n), F [ω](n) = F [n](n) A(1, 0, n)

≈

s(2)f (n) = F [ω](n)

3 fb+1 (n) ≈ fbn (n), s(1)

F [α]n (n) A(1, 1, n)

≈

A(1, 2, n)

≈

s(2) ω , F [α + 1](n) =

s(1)s(2)f (n) = F [ω + 1](n) s(1)2 s(2)f (n) = F [ω + 2](n)

136

5 A(1, 3, n)

≈

A(1, n, n)

≈

≈

s(2)2 f (n) = F [ω × 2](n)

A(1, 0, n)

≈

s(2)f (n) = F [ω](n)

A(2, 0, n)

≈

≈

A(4, 0, n)

≈

A(n, 0, n)

≈

A(1, 0, 0, n)

≈

A(2, 0, 0, n)

≈

A(n, 0, 0, n)

≈

A(1, 0, 0, 0, n)

≈

≈

s(2)2 f (n) = F [ω × 2](n)

s(2)3 f (n) = F [ω × 3](n)

s(2)4 f (n) = F [ω × 4](n)

s(2)n f (n) = F [ω × n](n) = F [ω 2 ](n) s(3)f (n) = F [ω 2 ](n)

s(3)2 f (n) = F [ω 2 × 2](n)

s(3)n f (n) = F [ω 2 × n](n) = F [ω 3 ](n) s(4)f (n) = F [ω 3 ](n)

s(2)s(3)2 s(4)3 f (n) = F [ω 3 × 3 + ω 2 × 2 + ω](n)

✓

✏

s(n)

A(ak , ..., a2 , a1 , a0 , n)

≈ =

✒

s(1)n s(2)f (n) = F [ω + ω](n)

A(2, 0, n)

A(3, 0, n)

A(3, 2, 1, 0, n)

s(1)3 s(2)f (n) = F [ω + 3](n)

A(1, 1, ..., 1) % &' (

≈

s(1)a0 s(2)a1 ...s(k + 1)ak f (n) F [ω k × ak + ... + ω 2 × a2 + ω × a1 + a0 ](n) s(n)f (n) ≈ F [ω ω ](n)

n

s(n)

ω

F [ω ](n)

✑

5.2.

137 F3

ss(n)

ss(1)f (n)

=

ss(1)2 f (x)

≈

ss(2)f (n)

=

ss(3)f (n)

=

ss(4)f (n)

=

ss(n)f (n)

≈

s(n)f (n) ≈ F [ω ω ](n)

F [ω ω + ω ω ](n) = F [ω ω × 2](n)

ss(1)n f (n) ≈ F [ω ω × ω](x) = F [ω ω+1 ](x) ss(2)n f (n) ≈ F [ω ω+2 ](n) ss(3)n f (n) ≈ F [ω ω+3 ](n) F [ω ω×2 ](n)

F3 (n)

=

F3

=

ss(2)63 f (n) ≈ F [ω ω+1 × 63](n) F363 (3) ≈ F [ω ω+1 × 63 + 1](63)

BEAF {n + 1, m + 1, a0 + 1, a1 + 1, a2 + 1, ..., ak + 1} ≈

f m (n),

f (n) = A(ak , ..., a2 , a1 , a0 , n) (n > 1) ✓

✏

n>1 {n + 1, m + 1, a0 + 1, a1 + 1, a2 + 1, ..., ak + 1} ≈ =

✒

{3, 3, ..., 3} ≈ % &' (

F [ω k × ak + ... + ω 2 × a2 + ω × a1 + a0 ]m (n) H[ω ω

k

F [ω ω ](n)

× m](n) ✑

n

1 F3

×ak +...+ω 2 ×a2 +ω×a1 +a0

2

138

5

✓

✒

3

F1

≈

F2

≈

F3

≈

✏

A(1, 0, 1, 63) ≈ {4, 64, 1, 1, 2} ≈ F [ω 2 + 1](63) A(63, 0, 0, 63) ≈ {4, 2, 1, 1, 64} ≈ F [ω 3 ](63) F [ω ω+1 × 63 + 1](63)

✑

5.2.3 21

1 ✓

✒

✏

(SGH)

G[0](n)

=

0

G[α + 1](n)

=

G[α](n) + 1

G[α](n)

=

G[αn ](n)(α

0

G[1](n) =

1

G[2](n) =

2

G[m](n) =

m

G[ω](n) =

n

G[ω2](n) =

2n

G[ω ](n) =

n2

G[ω ω ](n) =

nn

G[ω ω + ω](n) = Wiki -

)

G[0](n) =

2

21

(slow-growing hierarchy)

nn + n


✑

5.3.

139 ω

G[ω ω ](n) = ω

nn

n

n G[ϵ0 ](n) = n ↑↑ n

3