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Some months ago, or about 1 year ago? I calculated Moser's number, using up-arrow notation. Later I posted movie about Moser, to Nico-nico-douga:

However, those movies are in Japanese, so I want to write down the process of calculation, in English.

It needs some knowledges of approximation between up-arrow notation and Steinhaus-Moser notation.

## Approximation section 1

we need some preparations to calculate moser.

• $(\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n}) ^{(\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n})}$ when a and n is not small.

This can be calculated as follows:

• $=a^{a^{\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2} +\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}}$

Here $\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2}$is so big, but $\ll\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}$

• Therefore, $a^{a^{\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n-2} +\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}} \approx a^{a^{\underbrace{a^{a^{.^{.^{.^{a^{a^b}}}}}}}_{n-1}}} =\underbrace{a^{a^{.^{.^{.^{a^b}}}}}}_{n+1}$

If b=a, this can be described as follows, using tetration operator,

$(a\uparrow\uparrow n)\uparrow\uparrow2\approx a\uparrow\uparrow(n+1)$

using that step,

• $(a\uparrow\uparrow b)\uparrow\uparrow3\approx a\uparrow\uparrow(b+2)$
• $(a\uparrow\uparrow b)\uparrow\uparrow4\approx a\uparrow\uparrow(b+3)$
• $(a\uparrow\uparrow b)\uparrow\uparrow c\approx a\uparrow\uparrow(b+c-1)$

and,

• $(a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow b)\approx a\uparrow\uparrow(b+(a\uparrow\uparrow b)-1)\approx a\uparrow\uparrow a\uparrow\uparrow b$

Because of this approximation, the next can be also approximated as follows,

• $(a\uparrow\uparrow a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow a\uparrow\uparrow b)\approx a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b$
• $(a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b)\uparrow\uparrow(a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b)\approx a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow b$

so, using pentation operator,

• $(a\uparrow^3 n)\uparrow^3 2\approx a\uparrow^3(n+1)$

and also,

• $(a\uparrow^3 b)\uparrow^3 3\approx a\uparrow^3(b+2)$
• $(a\uparrow^3 b)\uparrow^3 4\approx a\uparrow^3(b+3)$
• $(a\uparrow^3 b)\uparrow^3 c\approx a\uparrow^3(b+c-1)$
• $(a\uparrow^3 b)\uparrow^3 (a\uparrow^3 b)\approx a\uparrow^3 a\uparrow^3 b$

Similarly and generally,

• $(a\uparrow^m b)\uparrow^m c\approx a\uparrow^m(b+c-1)$
• $(a\uparrow^m b)\uparrow^m(a\uparrow^m b)\approx a\uparrow^m a\uparrow^m b$

And of course,

• $(a\uparrow^m b)\uparrow^{m+1}(a\uparrow^m b)\approx a\uparrow^{m+1} a\uparrow^m b$
• $(a\uparrow^m b)\uparrow^{m+c}(a\uparrow^m b)\approx a\uparrow^{m+c} a\uparrow^m b$

## Approximation section 2

For example, if n is not small,

• $3\uparrow\uparrow n\approx(10\uparrow)^{n-1}1.1\approx10\uparrow\uparrow(n-1)$
• $10^{12}\uparrow\uparrow n\approx(10\uparrow)^{n+1}1.1\approx10\uparrow\uparrow(n+1)$

The differences of exponential heights between tetration of 10, are only 1 or 2.

So if N>>0, for any a,

• $a\uparrow\uparrow N\approx10\uparrow\uparrow N$

Nextly, for example,

• $a\uparrow\uparrow\uparrow3=a\uparrow\uparrow a\uparrow\uparrow a\approx10\uparrow\uparrow 10\uparrow\uparrow a'$ when a' is a, or a number near a.
• $a\uparrow\uparrow\uparrow4 =a\uparrow\uparrow a\uparrow\uparrow a\uparrow\uparrow a \approx10\uparrow\uparrow10\uparrow\uparrow10\uparrow\uparrow a'$

So, if N>>0,

• $a\uparrow\uparrow\uparrow N\approx10\uparrow\uparrow\uparrow N$

Similarly,

• $a\uparrow^m n\approx(10\uparrow^{m-1})^{n-1}a'$
• $a\uparrow^m N\approx10\uparrow^m N$

Those dispositions are necessary to calculate polygon notation.

## Polygon notation

Now let's consider about n in p-gon, as n[p]. n is not small (at least 3 or more).

• $n[3]=n^n$
• $n[3]_2=(n[3])[3]=(n^n)^{n^n}=n^{n\times n^n}=n^{n^{n+1}}$
• $n[3]_3=(n[3]_2)[3]=(n^{n^{n+1}})^{n^{n^{n+1}}}=n^{n^{n+1}\times{n^{n^{n+1}}}}=n^{n^{n+1+{n^{n+1}}}}\approx{n^{n^{n^{n+1}}}}$ if n is not small
• $n[3]_4\approx(n^{n^{n^{n+1}}})^{n^{n^{n^{n+1}}}} =n^{n^{n^{n+1}+n^{n^{n+1}}}}\approx{n^{n^{n^{n^{n+1}}}}}$

Here we are passing the first steps of approximation.

So when it reach n in square,

• $n[4]=n[3]_n\approx (n\uparrow)^n(n+1)\lesssim(n+1)\uparrow\uparrow(n+1)$
• and if N>>0, $N[4]\approx N\uparrow\uparrow N$

if n' =: n+1,

• $n[4]\approx n'\uparrow\uparrow n'$
• $n[4]_2=(n[4])[3]_{(n[4])} \approx(n'\uparrow\uparrow n')\uparrow\uparrow(n'\uparrow\uparrow n') \approx n'\uparrow\uparrow n'\uparrow\uparrow n'$ (using second steps of approximation)
• $n[4]_3\approx n'\uparrow\uparrow n'\uparrow\uparrow n'\uparrow\uparrow n'$
• $n[5]=n[4]_n\approx n'\uparrow^3(n+1)=n'\uparrow^3 n'$
• $n[5]_2=(n[5])[4]_{(n[5])} \approx (n'\uparrow^3 n')\uparrow^3(n'\uparrow^3 n') \approx n'\uparrow^3 n'\uparrow^3 n'$
• $n[6]=n[5]_n\approx n'\uparrow^4(n+1)=n'\uparrow^4 n'$

...Finally, in general,

• $n[p]\approx(n+1)\uparrow^{p-2}(n+1)$

## Calculation of 2[p]

2 is small, and so week, among up-arrow notation, so it's hard to apply $n[p]\approx(n+1)\uparrow^{p-2}(n+1)$. However, 2[p]=(2[p-1])[p-1], and if p is not small, 2[p-1] is large number. So (2[p-1])[p-1] can be applied approximation of n[p].

• $2[3]=2^2=4$
• $2[4]=(2[3])[3]=4[3]=4^4=256$
• $2[5]=(2[4])[4]=256[4]\approx(256\uparrow)^{256} 257\lesssim257\uparrow\uparrow257$ = Mega
• $2[6]=(2[5])[5]\approx(257\uparrow\uparrow257)[5] \approx(257\uparrow^2 257)\uparrow^3(257\uparrow^2 257)$

$\approx257\uparrow^3 257\uparrow^2 257 \approx10\uparrow^3 257\uparrow^2 257$ = A-ooga

• $2[7]=(2[6])[6]\approx(10\uparrow^3 257\uparrow^2 257)[6]$

$\approx(10\uparrow^3 257\uparrow^2 257)\uparrow^4(10\uparrow^3 257\uparrow^2 257) \approx10\uparrow^4 10\uparrow^3 257\uparrow^2 257$

• $2[8]=(2[7])[7]\approx(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)[7]$

$\approx(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)\uparrow^5(10\uparrow^4 10\uparrow^3 257\uparrow^2 257)$

$\approx10\uparrow^5 10\uparrow^4 10\uparrow^3 257\uparrow^2 257$

## Another approach to 2[p]

$2[p]=2[p-1][p-1]=2[p-2][p-2][p-1]=2[p-3][p-3][p-2][p-1]$

$\dots=2^2[3][4][5]\dots[p-3][p-2][p-1]$

now [3] is exponentiation, and [4] is approximately tetration, similarly [5] is pentation, [6] is hexation, and [p] is, approximately $\uparrow^{p-2}$.

So concidering pattern of previous section,

• $2[p]=2^2[3][4][5]\dots[p-3][p-2][p-1]$

$\approx10\uparrow^{p-3}10\uparrow^{p-4}10\uparrow^{p-5}\dots\uparrow^4 10\uparrow^3 257\uparrow^2 257$

## And Moser's number is,

• $2[2[5]]\approx10\uparrow^{\text{Mega}-3}10\uparrow^{\text{Mega}-4}10\uparrow^{\text{Mega}-5}\dots\uparrow^4 10\uparrow^3 257\uparrow^2 257$

Can we approximate this more simply? Yes, there is two way to describe this structure.

• $3\uparrow^{\text{Mega}-2}3$

$3\uparrow^n 3$ can be written down as follows,

$=3\uparrow^{n-1}3\uparrow^{n-2}\dots\uparrow^3 3\uparrow^2 3\uparrow(3\times(3+3+3))$ It is similar to 2[p]

• $2\uparrow^{\text{Mega}-2}4$

$2\uparrow^n 4(=2\uparrow^{n+1} 3)$

$=2\uparrow^{n-1}2\uparrow^{n-2}\dots\uparrow^3 2\uparrow^2 2\uparrow(2\times(2+2+2+2))$

Compareing these and Moser, in fact,

• $3\uparrow^{\text{Mega}-2}3\gtrsim2\uparrow^{\text{Mega}-2}4\gtrsim\text{Moser}$

Because,

• $3\uparrow^{\text{Mega}-2}3=\dots\uparrow^4 3\uparrow^3 3\uparrow^2 7625597484987$

$\approx\dots\uparrow^4 10\uparrow^3(10\uparrow)^{7625597484986}1.1$

• $2\uparrow^{\text{Mega}-2}4=\dots\uparrow^4 2\uparrow^3 2\uparrow^2 65536$

$\approx\dots\uparrow^4 10\uparrow^3(10\uparrow)^{65533}4.3$

• $\text{Moser}\approx\dots\uparrow^4 10\uparrow^3(10\uparrow)^{257}2.8$

After all, However, Mega is already quite large number.

• $\text{Moser}\approx3\uparrow^{257\uparrow\uparrow257}3$

This might be the most simple approximation, using up-arrow notation...