## FANDOM

10,266 Pages

Mashimo scale is defined as

$n = floor(M^{-1}(x))$

where $$M(x)$$ is Mashimo function[1][2]. Therefore, $$M(n) \le x < M(n+1)$$.

For example, Mashimo scale of $$x=10^{34}$$ is calculated as $$n = floor(M^{-1}(10^{34})) = floor(3.4) = 3$$. Therefore, $$10^{34}$$ is a large number of Mashimo scale 3.

Here is a list of googological numbers classified with Mashimo scale.

$$n$$ Value or approximation of $$M(n)$$Numbers with Mashimo scale $$n$$
1 $$10^{10}$$ Dialogue
2 $$10^{20}$$ Avogadro's number
3 $$10^{30}$$ Belphegor's prime
4 $$10^{40}$$ Septillion (long scale)
5 $$10^{50}$$ Gougasha, Asougi
6 $$10^{60}$$ Nayuta, Fukashigi, Muryoutaisuu
7 $$10^{70}$$ Gazillion, Eddington number
8 $$10^{80}$$ Sexvigintillion (short scale)
9 $$10^{90}$$ Trigintillion (short scale)
10 $$10^{100}$$ Googol
11 $$10^{110}$$ Eleventyplex
12 $$10^{120}$$ Shannon number
13 $$10^{130}$$ Duovigintillion (long scale)
14 $$10^{140}$$ Dvajagranisamani
15 $$10^{150}$$ ³4
16 $$10^{160}$$ Septenvigintillion (long scale)
17 $$10^{170}$$ Novemvigintillion (long scale)
18 $$10^{210.801}$$ Millillion
19 $$10^{4557.17}$$ Googolgong
20 $$10^{10^{6.219}}$$ Trialogue
21 $$10^{10^{12.547}}$$ Femtillion
22 $$10^{10^{36.647}}$$ Googolplex, Tritet Jr.
23 $$10^{10^{210.439}}$$ Ecetonplex, Legion's number of the second kind
24 $$10^{10^{4556.808}}$$ Googolplexigong
25 $$10^{10^{10^{6.219}}}$$ Tetralogue
26 $$10^{10^{10^{12.547}}}$$ First Skewes number
27 $$10^{10^{10^{36.647}}}$$ Googolduplex
28 $$10^{10^{10^{210.439}}}$$ Ecetonduplex, Second Skewes number
29 $$10^{10^{10^{4556.808}}}$$ Googolduplexigong
30 $$10^{10^{10^{10^{6.219}}}}$$ Pentalogue
31 $$10^{10^{10^{10^{12.547}}}}$$ Betillion
32 $$10^{10^{10^{10^{36.647}}}}$$ Googoltriplex
33 $$10^{10^{10^{10^{210.439}}}}$$ Ecetontriplex
34 $$10^{10^{10^{10^{4556.808}}}}$$ Googoltriplexigong
35 $$10^{10^{10^{10^{10^{6.219}}}}}$$ Hexalogue
36 $$10^{10^{10^{10^{10^{12.547}}}}}$$ Duseptimevalka
37 $$10^{10^{10^{10^{10^{36.647}}}}}$$ Googolquadriplex
38 $$10^{10^{10^{10^{10^{210.439}}}}}$$ Ecetonquadriplex
39 $$10^{10^{10^{10^{10^{4556.808}}}}}$$ Googolquadriplexigong
40 $$10^{10^{10^{10^{10^{10^{6.219}}}}}}$$ Heptalogue
41 $$10^{10^{10^{10^{10^{10^{12.547}}}}}}$$ Duoctimevalka
42 $$10^{10^{10^{10^{10^{10^{36.647}}}}}}$$ Googolquinplex
43 $$10^{10^{10^{10^{10^{10^{210.439}}}}}}$$ Ecetonquintiplex
44 $$10^{10^{10^{10^{10^{10^{4556.808}}}}}}$$ Googolquintiplexigong
45 $$10^{10^{10^{10^{10^{10^{10^{6.219}}}}}}}$$ Octalogue
46 $$10^{10^{10^{10^{10^{10^{10^{12.547}}}}}}}$$ Bentley's Number, Giggol, Tritri, Googolplexidex
47 $$10↑↑10^{10^{619}}$$ Googolduplexidex
48 $$4↑↑↑4 = 4 → 4 → 3$$ Tetra-teraksys
49 $$4↑↑↑5 = 4 → 5 → 3$$ Gaggol, Tritet
50 $$4 → 5 → 4$$ Geegol
51 $$4 → 5 → 5$$ Gigol
52 $$10 \rightarrow 5 \rightarrow 6$$ Goggol
53 $$5 \rightarrow 7 \rightarrow 7$$ Tridecal, Moser's number
54 $$10 → 10 → (10 → 4 → 3)$$ Boogolplex
55 $$10 → 10 → 3 → 2$$ Boogolduplex
56 $$10 → 10 → 4 → 2$$ Boogoltriplex
57 $$10 → 10 → 6 → 2$$ Graham's number
58 $$A(1,1,8 \times 10^5)$$ Forcal
59 $$A(1,1,5↑↑↑5)$$ {3, {3, 3, 4}, 1, 2}
60 $$A(1,1,3 \rightarrow 8 \rightarrow 7)$$ Fish number 1, Fish number 2
61 $$A(3,3,3,3,4)$$ Quadreegol
62 $$A(7,7,7,7,7,7,7,7,8)$$ Okojo-stoat number
63 $$f_{\omega^\omega}(4↑↑↑3)$$
64 $$f_{\omega^\omega}(10 \rightarrow 10 \rightarrow 3 \rightarrow 2)$$ { 3, Boogolduplex (1) 2 }
65 $$f_{\omega^\omega}(A(1,1,10^{10^8}))$$ Duperdecal, Goobolplex
66 $$f_{\omega^\omega＋1}(5)$$ Gibbol
67 $$f_{\omega^{\omega}+1}(10 \rightarrow 4 \rightarrow 3 \rightarrow 2)$$ $$\lbrace 3, \lbrace 3,3,3,3 \rbrace ,2 (1) 2 \rbrace$$
68 $$f_{\omega^\omega＋1}(A(3,3,3,3,4))$$ $$\lbrace 3, \lbrace 3,3,3,3,3,3,3 \rbrace,2 (1) 2 \rbrace$$
69 $$f_{\omega^\omega＋1}(f_{\omega^\omega}(A(1,1,5 \uparrow \uparrow \uparrow 5)))$$ Latri, Boobol
70 $$f_{\omega^\omega＋\omega}(A(3,3,3,2,4))$$ {3,3,{3,3,3,3,3,3,3} (1) 2}
71 $$f_{\omega^{\omega}+\omega}(f_{\omega^{\omega}+1}(A(1,1,10^{10^8})))$$ Fish number 3, Xappol
72 $$f_{\omega^{\omega^{\omega}}}(3) = f_{\omega^{\omega^3}}(3) = f_{\epsilon_0}(3)$$ Gongulus, Dulatri
73 $$f_{\omega^{\omega^{\omega^{\omega^{\omega}}}}}(3) \approx f_{\epsilon_0}(5)$$ Graltertathol
74 $$f_{\epsilon_0}(7)$$ Goppatoth
75 $$f_{\epsilon_0}(10^{10^8})$$ Fish number 5
76 $$f_{\epsilon_0 3}(3)$$ Tethrathor
77 $$f_{\epsilon_1 3}(3)$$ Great and Terrible Tethrathoth
78 $$f_{\phi(2,0)}(3)$$ Triakulus, Fish number 6
79 $$f_{\phi(2,1)}(3)$$ $$\{3,3[1/1/3]2\}$$ (Bird's array notation)
80 $$f_{\phi(2,2)}(3)$$ Quadrunculus
81 $$f_{\Gamma_0}(3)$$ Pentacthulhum
82 $$f_{\Gamma_1}(3)$$ Pentacthulhusquare
83 $$f_{\phi(1,0,0,0)}(3)$$ Enormabixul
84 $$f_{\phi(1,0,0,0,0)}(3)$$ Bird's number, TREE(3) (lower limit)
85 $$f_{\psi(\Omega^{\Omega^{\Omega}})}(2)$$ Golapulus
86 $$f_{\psi(\Omega^{\Omega^{\Omega^\Omega}})}(3)$$ $$\{X,X,2(0,1)2\}\&3$$
87 $$f_{\vartheta(\Omega_2)}(2)$$ Golapulusplex
88 $$f_{\vartheta(\Omega_3)}(3)$$ SCG(13)
89 $$f_{\psi(\Omega_{\Omega_\omega})}(3)$$ Nucleabixul
90 $$f_{\psi(\Omega_{\Omega_{\Omega_\omega}})}(4)$$ Nucleatrixul
91 $$\lbrace L,10\rbrace_{10,10} \approx f_{\psi(\psi_I(0))}(10)$$ Big hoss
92 $$\lbrace L2,10\rbrace_{10,10}$$ Goshomity
93 $$\lbrace L3,10\rbrace_{10,10}$$ Bongo Bukuwaha
94 $$\lbrace L4,10\rbrace_{10,10}$$ Meameamealokkapoowa oompa
95 $$D^5(10)$$ loader.c
96 $$D^{10}(10)$$
97 $$D^{15}(10)$$
98 $$D^{20}(10)$$
99 $$D^{25}(10)$$
100 $$\Sigma(1000)$$ $$\Sigma(1000)$$
101 $$\Sigma_2(1000) \approx f_{\omega^\text{CK}_2}(1000)$$
102 $$\Sigma_{3}(1000) \approx f_{\omega^\text{CK}_{3}}(1000)$$
103 $$f_{\omega^\text{CK}_{\omega}}(1000)$$ $$\Xi(10^6)$$ (Xi function)
104 $$f_{\omega^\text{CK}_{\omega+1}}(1000)$$
105 $$f_{\omega^\text{CK}_{\omega 2}}(1000)$$
106 $$f_{\omega^\text{CK}_{\omega^2}}(1000)$$
107 $$f_{\omega^\text{CK}_{\omega^{\omega}}}(1000)$$ Fish number 4
108 $$f_{\omega^\text{CK}_{\omega^{\omega^{\omega}}}}(1000)$$
109 $$f_{\omega^\text{CK}_{\epsilon_0}}(1000)$$
110 $$f_{\omega^\text{CK}_{\epsilon_1}}(1000)$$
120 $$R_1(10^{10})$$ Rayo's number
121 $$R_2(10^{10})$$
122 $$R_3(10^{10})$$
123 $$R_\omega(10^{10})$$
130 $$R_{\epsilon_1}(10^{10})$$
131 $$R_{\phi(2,0)}(10^{10})$$ Fish number 7
132 $$R_{\phi(1,0,0)}(10^{10})$$
133 $$R_{\phi(1,0,0,0)}(10^{10})$$

## Calculation

When $$47 \le x \le 71$$, $$M(x) = H(^{x/20}2,2)$$, where $$H$$ is H function.

\begin{eqnarray*} H(0,2) &=& H_{0}(2) = 2 \\ H(1,2) &=& H_{1}(2) = 3 \\ H(2,2) &=& H_{\omega}(2) = 4 \\ H(3,2) &=& H_{\omega+1}(2) = 6 \\ H(4,2) &=& H_{\omega^\omega}(2) = H_{\omega+2}(2) = 8 \\ H(5,2) &=& H_{\omega^\omega+1}(2) = H_{\omega^\omega}(3) = H_{\omega^2 2 + \omega 2 + 3}(3) = f_2^2(24) \approx 10^{10^8} \\ H(6,2) &=& H_{\omega^\omega+\omega}(2) = H_{\omega^\omega+2}(2) = H_{\omega^\omega}(4) \approx 5↑↑↑5 \\ H(7,2) &=& H_{\omega^\omega+\omega+1}(2) = H_{\omega^\omega+\omega}(3) = H_{\omega^\omega}(6) \approx 7↑^{5}7 \\ H(8,2) &=& H_{\omega^{\omega+1}}(2) = H_{\omega^\omega・2}(2) = H_{\omega^\omega}^2(2) = H_{\omega^\omega}(8) \approx 9↑^{7}9 \\ H(9,2) &=& H_{\omega^{\omega+1}+1}(2) = H_{\omega^{\omega+1}}(3) = f_{\omega+1}(3) = f_{\omega}^3(3) \\ &=& f_{\omega}^2(H_{\omega^2 2 + \omega 2 +3}(3)) \approx f_{\omega}^2(10^{10^8}) > f_{\omega}^2(10^{10}) \approx f_{\omega}(10 \rightarrow 10 \rightarrow 2 \rightarrow 2)\\ &\approx& 10 \rightarrow 10 \rightarrow 3 \rightarrow 2 \\ H(10,2) &=& H_{\omega^{\omega+1}+\omega}(2) = H_{\omega^{\omega+1}}(4) = f_{\omega+1}(4) \approx 10 \rightarrow 10 \rightarrow 4 \rightarrow 2 \\ H(11,2) &=& H_{\omega^{\omega+1}+\omega+1}(2) = H_{\omega^{\omega+1}+\omega}(3) = f_{\omega+1}(6) \approx 10 \rightarrow 10 \rightarrow 6 \rightarrow 2 \\ H(12,2) &=& H_{\omega^{\omega+1}+\omega^\omega}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega}(2)) = f_{\omega+1}(8) \approx 10 \rightarrow 10 \rightarrow 8 \rightarrow 2 \\ H(13,2) &=& H_{\omega^{\omega+1}+\omega^\omega+1}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega+1}(2)) \approx A(1,1,10^{10^8}) \\ H(14,2) &=& H_{\omega^{\omega+1}+\omega^\omega+\omega}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega+\omega}(2)) \approx A(1,1,5↑↑↑5) \\ H(15,2) &=& H_{\omega^{\omega+1}+\omega^\omega+\omega+1}(2) = H_{\omega^{\omega+1}}(H_{\omega^\omega+\omega+1}(2)) \approx A(1,1,7 \uparrow^5 7) \\ H(16,2) &=& H_{\omega^{\omega^\omega}}(2) = f_{\omega^\omega}(2) = f_{\omega+2}(2) = f_{\omega+1}^2(2) \\ &=& f_{\omega+1}(f_{\omega}^2(2)) = f_{\omega+1}(f_{\omega}(8)) \\ &\approx& A(1,1,3 \rightarrow 8 \rightarrow 7) \\ H(17,2) &=& H_{\omega^{\omega^\omega}+1}(2) = f_{\omega^\omega}(3) \approx A(1,0,0,0,3) = A(2,2,3,3) \\ H(18,2) &=& H_{\omega^{\omega^\omega}+\omega}(2) = f_{\omega^\omega}(4) \approx A(1,0,0,0,0,4) = A(3,3,3,4,4) \\ H(19,2) &=& H_{\omega^{\omega^\omega}+\omega+1}(2) = f_{\omega^\omega}(6) \approx A(1,0,0,0,0,0,0,6) = A(5,5,5,5,5,6,6) \\ H(20,2) &=& H_{\omega^{\omega^\omega}+\omega^\omega}(2) = f_{\omega^\omega}(8) \approx A(1,0,0,0,0,0,0,0,0,8) = A(7,7,7,7,7,7,7,8,8) \\ H(21,2) &=& H_{\omega^{\omega^\omega}+\omega^\omega+1}(2) \approx f_{\omega^\omega}(10^{10^8}) \\ \end{eqnarray*}

Interpolation between $$H(5,2)$$ and $$H(6,2)$$ \begin{eqnarray*} H(5,2) &=& H(27,3) = H_{\omega^\omega}(3) = H_{\omega^3}(3) = H_{\omega^2 3}(3) = H_{\omega^2 2 +\omega 2 + 3}(3) \\ &=& H_{\omega^2 2 +\omega 2 + 2}(4) = H(4^2*2 + 4*2 + 2,4) = H(32+8+2,4) = H(42,4) \\ H(6,2) &=& H(28,3) = H_{\omega^\omega + 1}(3) \\ &=& H_{\omega^\omega}(4) = H(256,4) \\ \end{eqnarray*} Therefore, $$H(5+r,2) = H(42+(256-42)r,2) = H(42+214r)$$

\begin{eqnarray*} M(47) &=& H(^{2.35}2,2) = H(2^{2^{2^{0.35}}},2) = H(5.34893, 2) \\ &=& H(42+214 \times 0.34893,4) = H(116.67178, 4) \\ &>& H(116,4) = H_{\omega^3+\omega^2*3+\omega}(4) = f_3(f_2^3(f_1(4))) = f_3(f_2^3(8)) \\ &>& f_3(2^{2^{2059}}) > 10 \uparrow \uparrow 10^{10^{619}} \\ M(48) &=& H(5.64045, 2) = H(42+214 \times 0.64045,4) = H(179.05579) \\ &>& H(179, 4) = f_3^2(f_2^3(7))) > f_3^2(2^{2^{905}}) > 4 \uparrow \uparrow 4 \uparrow \uparrow 10^{10^{271}} \\ &>& 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 4 = 4 \uparrow \uparrow \uparrow 4 = \{4, 4, 3\} \\ M(49) &=& H(5.96955, 2) = H(42+214 \times 0.96955,4) = H(249.48282) \\ &>& H(249, 4) = f_3^3(f_2^3(f_1^2(5))) = f_3^3(f_2^3(20)) > f_3^3(2^{2^{20971544}}) \\ &>& 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 10^{10^{6313063}} > 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \uparrow \uparrow 4 \\ &=& 4 \uparrow \uparrow \uparrow 5 = \{4, 5, 3\} \\ \end{eqnarray*}

Interpolation between $$H(6,2)$$ and $$H(7,2)$$. Since $$H(6,2) = H(28,3)$$ and $$H(7,2) = H(30,3)$$, $$H(6.5,2) = H(29,3)$$.

\begin{eqnarray*} H(6,2) &=& H(28,3) = H(256,4) = H(5^3 \times 3 + 5^2 \times 3 + 5 \times 3 + 3,5) = H(468,5) \\ H(6.5,2) &=& H(29,3) = H(257,4) = H(3125,5) = H(6220,6) \\ H(7,2) &=& H(30,3) = H(258,4) = H(3126,5)= H(46656,6) \\ \end{eqnarray*}

Therefore, \begin{eqnarray*} M(50) &=& H(6.3429,2) = H(28.6858,3) = H(256.6858,4) \\ &=& H(468+2657 \times 0.6858, 5) > H(2290,5) \\ &=& f_4^3(f_3^3(f_2(f_1^3(5)))) = f_4^3(f_3^3(f_2(40))) > f_4^3(f_3^3(2^{45})) \\ &\approx& f_4^3(3 \uparrow \uparrow 3 \uparrow \uparrow 3 \uparrow \uparrow 3 \uparrow \uparrow 3) = f_4^3(3 \uparrow^3 5) > f_4^3(4 \uparrow^3 4) \\ &=& 4 \uparrow^3 4 \uparrow^3 4 \uparrow^3 4 \uparrow^3 4 = \{4, 5, 4\} \\ M(51) &=& H(6.76873,2) = H(29.53747,3) = H(3125.53747,5) \\ &=& H(6220+40436 \times 0.53747, 6) > H(27952,6) = f_5^3(f_4^3(f_3^3(f_2^2(f_1^2(10))))) \\ &=& f_5^3(f_4^3(f_3^3(f_2^2(40)))) > f_5^3(f_4^3(f_3^3(2^{45}))) > f_5^3(f_4^3(4)) \\ &\approx& f_5^3(4 \uparrow^4 4) = 4 \uparrow^4 4 \uparrow^4 4 \uparrow^4 4 \uparrow^4 4 = \{4, 5, 5\} \\ M(51) &<& H(27953,6) = f_5^3(f_4^3(f_3^3(f_2^2(f_1^2(11))))) = f_5^3(f_4^3(f_3^3(f_2^2(44)))) \\ &<& f_5^3(f_4^3(f_3^3(2^{50}))) < f_5^3(f_4^3(f_3^3(5 \uparrow^2 5))) \approx f_5^3(f_4^3(5 \uparrow^3 5)) \approx f_5^3(5 \uparrow^4 5) \\ &\approx& \{5, 5, 5\} \end{eqnarray*}

Interpolation between $$H(7,2)$$ and $$H(8,2)$$. Since $$H(7,2) = H(30,3) = H(258,4)$$ and $$H(8,2) = H(31,3) = H(260,4)$$, $$H(7.5,2) = H(259,4)$$.

\begin{eqnarray*} H(7,2) &=& H(258,4) = H(6^6,6) = H(98040,7) \\ H(7.5,2) &=& H(259,4) = H(7^7,7) = H(823543,7) = H(1797558,8) \\ H(8,2) &=& H(260,4) = H(8^8,8) = H(16777216,8) \\ \end{eqnarray*}

Therefore, \begin{eqnarray*} M(52) &=& H(7.25718,2) = H(258.51435,4) = H(46656.51435,6) > H(471203,7) \\ &=& f_6^4(f_3(f_2^5(f_1^2(12)))) > f_6^4(10) \\ &\approx& 10 \uparrow^5 10 \uparrow^5 10 \uparrow^5 10 \uparrow^5 10 = 10 \uparrow^6 5 = 10 \rightarrow 5 \rightarrow 6 \\ M(53) &=& H(7.8209,2) = H(259.6418,4) = H(823543.6418,7) > H(11411690,8) \\ &=& f_7^5(f_6^3(f_5^4(f_4^2(f_2^3(f_1^5(10)))))) > f_7^5(f_6^3(f_5^4(5))) \\ &>& f_7^5(f_6^3(5 \uparrow^3 5 \uparrow^3 5 \uparrow^3 5 \uparrow^3 5)) \\ &=& f_7^5(f_6^3(5 \uparrow^4 5) = f_7^5(5 \uparrow^5 5 \uparrow^5 5 \uparrow^5 5 \uparrow^5 5) \\ &=& f_7^5(5 \uparrow^6 5) = 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 \uparrow^6 5 = 5 \uparrow^7 7 = 5 \rightarrow 7 \rightarrow 7 \\ \end{eqnarray*}

Interpolation between $$H(8,2)$$ and $$H(9,2)$$. Since $$H(8,2) = H(31,3)$$ and $$H(9,2) = H(81,3)$$, $$M(54) = H(8.47588,2) = H(54.794,3)$$. As $$H(54,3) = H(298,4)$$ and $$H(55,3) = H(512,4)$$, \begin{eqnarray*} M(54) &=& H(54.794,3) = H(467.912,4) > H(467,4) \\ &=& H_{\omega^\omega + \omega^3 3 + \omega^2 + 3}(4) = f_\omega(f_3^3(f_2(7))) \\ &=& f_\omega(f_3^3(896)) \approx f_\omega(10 \uparrow^3 4) \\ &\approx& 10 → 10 → (10 → 4 → 3) \\ \end{eqnarray*}

Calculation of $$M(55)$$ to $$M(57)$$.

\begin{eqnarray*} M(55) &=& H(9.2424,2) > H(9,2) \approx 10 \rightarrow 10 \rightarrow 3 \rightarrow 2 \\ M(56) &=& H(10.146,2) > H(10,2) \approx 10 \rightarrow 10 \rightarrow 4 \rightarrow 2 \\ M(57) &=& H(11.222,2) > H(11,2) \approx 10 \rightarrow 10 \rightarrow 6 \rightarrow 2 \\ \end{eqnarray*}

Calculation of M(58).

\begin{eqnarray*} M(58) = H(12.5131,2) \approx A(1,1,H(4.5131,2)) \\ H(4.5131) = H(15.8022,3) \\ \end{eqnarray*} As $$H(15,3) = H(70,8)$$ and $$H(16,3) = H(72,8) = H(256,16)$$, $$H(15.5,3) = H(71,8) = H(238,16)$$, and \begin{eqnarray*} H(15.8022,3) &=& H(248.8791,16) \approx 2^{15} \times 24.8791 \\ &\approx& 8 \times 10^5 \\ \end{eqnarray*} Therefore, $$M(58) \approx A(1,1,8 \times 10^5)$$

Calculation of $$M(59)$$ and $$M(60)$$.

\begin{eqnarray*} M(59) &=& H(14.0793,2) > H(14,2) \approx A(1,1,5 \uparrow \uparrow \uparrow 5) \\ M(60) &=& H(16,2) \approx A(1,1,3 \rightarrow 8 \rightarrow 7) \\ \end{eqnarray*}

Approximation of $$M(61)$$ and $$M(62)$$.

\begin{eqnarray*} H(17,2) &=& H(3^{3^3},3) \approx H(4^{4^2*2+4*2+2}*2) \approx H(4 \times 10^{25},4) \\ H(18,2) &=& H(3^{3^3}+1,3) = H(4^{4^4},4) \approx H(1.3408 \times 10^{154},4) \\ A(3,3,3,3,4) &\approx& H_{\omega^{\omega^3 3 + \omega^2 3 + \omega 3 + 3}}(4) \\ &\approx& H(3.352 \times 10^{153},4) \approx H(17.25,2) \\ M(61) &=& H(17.6255,2) > H(17+1/4) \approx A(3,3,3,3,4) \\ M(62) &=& H(19.6042,2) > H(19+1/8) \approx A(7,7,7,7,7,7,7,7,8) \\ \end{eqnarray*}

Calculation of $$M(63)$$ to $$M(65)$$.

\begin{eqnarray*} M(63) &=& H(22.045,2) = f_{\omega^{\omega}}(H(6.045,2)) \approx f_{\omega^{\omega}}(H(206,5)) \approx f_{\omega^{\omega}}(4 \uparrow \uparrow \uparrow 3) \\ M(64) &=& H(25.1015,2) = f_{\omega^{\omega}}(H(9.1015,2)) \approx f_{\omega^{\omega}}(3 \rightarrow 3 \rightarrow 3 \rightarrow 2) \\ M(65) &=& H(28.9944,2) = f_{\omega^{\omega}}(H(12.9944,2)) \approx f_{\omega^{\omega}}(A(1,1,10^{10^8}) \\ \end{eqnarray*}

Calculation of $$M(66) = H(34.0487,2)$$.

\begin{eqnarray*} H(34,2) &=& f_{\omega^{\omega}+1}(4) = H(4^{4^4+1},4) \approx H(5^{5^5} \cdot 4,5) \\ H(35,2) &=& H(4^{4^4+1}+2,4) = f_{\omega^{\omega}+1}(6) \\ H(34.5,2) &=& H(4^{4^4+1}+1,4) = H(5^{5^5+1},5) = f_{\omega^{\omega}+1}(5) \\ M(66) &\approx& H(34.05,2) \approx H(5^{5^5} \cdot 4 + 5^{5^5} \cdot 0.1,5) > H(5^{5^5} \cdot 4 + 5^{5^5-2},5) \\ \end{eqnarray*}

Therefore, $$f_{\omega^{\omega}+1}(4) < M(66) < f_{\omega^{\omega}+1}(5)$$ and $$M(66)$$ is better approximated as $$f_{\omega^{\omega}+1}(5)$$.

Calculation of $$M(67)$$. Since $$H(8,2) = H(31,3)$$ and $$H(9,2) = H(81,3)$$,

\begin{eqnarray*} M(67) &=& H(40.7558,2) = f_{\omega^{\omega}+1}(H(8.7558,2)) = f_{\omega^{\omega}+1}(H(68.7901,3)) \\ &=& f_{\omega^{\omega}+1}(f_{\omega}^2(H(14.7901,3))) \approx f_{\omega^{\omega}+1}(f_{\omega}^2(H_{\omega^2 + \omega + 2}(3))) \\ &\approx& f_{\omega^{\omega}+1}(f_{\omega}^2(10^4)) \\ &\approx& f_{\omega^{\omega}+1}(10 \rightarrow 4 \rightarrow 3 \rightarrow 2) \\ \end{eqnarray*}

Calculation of $$M(68)$$ to $$M(71)$$.

\begin{eqnarray*} M(68) &=& H(49.882,2) = f_{\omega^{\omega}+1}(H(17.882,2)) \approx f_{\omega^{\omega}+1}(A(3,3,3,3,4)) \\ M(69) &=& H(62.6632,2) = f_{\omega^{\omega}+1}(f_{\omega^{\omega}}(H(14.6632,2))) \\ &\approx& f_{\omega^{\omega}+1}(f_{\omega^{\omega}}(A(1,1,5 \uparrow \uparrow \uparrow5))) \\ M(70) &=& H(81.1718,2) = f_{\omega^{\omega}+\omega}(H(17.1718,2)) \\ &\approx& f_{\omega^{\omega}+\omega}(A(3,3,3,2,4)) \\ M(71) &=& H(109.041,2) = f_{\omega^{\omega}+\omega}(H(45.041,2)) \\ &=& f_{\omega^{\omega}+\omega}(f_{\omega^{\omega}+1}(H(13.041,2)) \\ &\approx& f_{\omega^{\omega}+\omega}(f_{\omega^{\omega}+1}(A(1,1,10^{10^8}))) \\ \end{eqnarray*}