## FANDOM

10,825 Pages

First, what is an "usual" number-naming scheme? We take a certain hierarchy, $h_\alpha(n)$, and map a word for it up to certain n and $\alpha$ (not necessarily do it for each $\beta < \alpha$). The restriction is only that the rule $h_\alpha(n) = h_{\alpha[n]}(n)$ always keeps. So, under that, Bowers' scheme is usual:

$h_1(100)$ = googol

$h_2(100)$ = giggol

$h_3(100)$ = gaggol

...

$h_\omega(100)$ = boogol

$h_{\omega*2}(100)$ = biggol (even though we have no word for $h_{\omega+1}(100)$, it still works)

$h_{\omega*3}(100)$ = baggol

...

$h_{\omega^2}(100)$ = troogol

## Mixed-arrow notation

Let's define so-called mixed-arrow notation as follows:

1. is an expression composed by $\uparrow$'s and $\downarrow$'s.

Rule 1. (# is empty, no arrows at all)

$a\#b = ab = a*b$

Rule 2. (b=1, regardless of #)

$a\#1 = a$

Rule 3. (# ends at up-arrow)

$a\#\uparrow (b+1) = a\# (a\#\uparrow b)$

Rule 4. (# ends at down-arrow)

$a\#\downarrow (b+1) = (a\#\downarrow b) \# a$

Examples:

$3 \uparrow\downarrow 3 = (3 \uparrow 3) \uparrow 3 = 27 \uparrow 3 = 19683$

$3 \downarrow\uparrow 3 = 3 \uparrow (3 \uparrow 3) = 3 \uparrow 27 = 7625597484987$

$3 \uparrow\uparrow\downarrow 3 = (3 \uparrow\uparrow 3) \uparrow\uparrow 3 = (3 \uparrow\uparrow 3)^{{(3 \uparrow\uparrow 3)}^{(3 \uparrow\uparrow 3)}}$

$3 \uparrow\downarrow\downarrow 3 = (3 \downarrow\downarrow 3) \downarrow\downarrow 3 = 19683 \downarrow\downarrow 3 = 19683^{19683^2}$

Note that $a \uparrow \# b = a \downarrow \# b$ because they both drop to multiplication which is associative.

## Unusual naming scheme

Now we want to form names for this unique notation.

Let $\downarrow$ be 0 and $\uparrow$ be 1. Then we can map a binary expansion of each number to the arrow-expression. For example, 5 = 1012 = $\uparrow\downarrow\uparrow$. So we can form the prefix for our names, adopting English numbers. For the suffix, I propose "-arrowal". For a and b we choose "3" as the smallest non-trivial case.

So it forms the names for the numbers as follows:

• Zerarrowal = $3 \downarrow 3$ = 27
• Onarrowal = $3 \uparrow 3$ = 27 (not greater than previous term)
• Twarrowal = $3 \uparrow\downarrow 3$ = 19683
• Thrarrowal = $3 \uparrow\uparrow 3$ = 7625597484987
• Fourarrowal = $3 \uparrow\downarrow\downarrow 3 = 19683^{19683^2}$
• Fivarrowal = $3 \uparrow\downarrow\uparrow 3 = 3 \uparrow\downarrow (3 \uparrow\downarrow 3) = 3 \uparrow\downarrow 19683 = 3^{3^{19682}}$
• Sixarrowal = $3 \uparrow\uparrow\downarrow 3 = (3 \uparrow\uparrow 3) \uparrow\uparrow 3 = 7625597484987 \uparrow\uparrow 3 = 7625597484987^{7625597484987^{7625597484987}}$
• Sevarrowal = $3 \uparrow\uparrow\uparrow 3$ (exactly tritri in Bowers' system)
• Eigarrowal = $3 \uparrow\downarrow\downarrow\downarrow 3 = (3 \uparrow\downarrow\downarrow 3) \uparrow\downarrow\downarrow 3 = 19683^{19683^2} \uparrow\downarrow\downarrow 3 = 19683^{19683^2} \downarrow\downarrow\downarrow 3$

To estimate eigarrowal we have to determine how fast $f(n) = n \downarrow\downarrow\downarrow 3$ grows. Consider it:

$n \downarrow\downarrow\downarrow 3 = (n \downarrow\downarrow n) \downarrow\downarrow 3 \approx (n \uparrow\uparrow 3) \uparrow\uparrow 3 \approx n \uparrow\uparrow 6$. So eigarrowal is about $19683^{19683^2} \uparrow\uparrow 6$, which is, as we know, much less than sevarrowal = $3 \uparrow\uparrow 7625597484987$. It turned out that in our naming scheme (n+1)-arrowal can be even lesser than n-arrowal.

• Ninarrowal = $3 \uparrow\downarrow\downarrow\uparrow 3$
• Tenarrowal = $3 \uparrow\downarrow\uparrow\downarrow 3$

...

• Twentarrowal $3 \uparrow\downarrow\uparrow\downarrow\downarrow 3$
• Hundarrowal $3 \uparrow\uparrow\downarrow\downarrow\uparrow\downarrow\downarrow 3$
• Thousarrowal $3 \uparrow\uparrow\uparrow\uparrow\uparrow\downarrow\uparrow\downarrow\downarrow\downarrow 3 = 3 \uparrow^5 \downarrow\uparrow \downarrow^3$ (superscripts still can be used).

I see no way to pin down this sequence to a standard ordinal hierarchy $h_\alpha(n)$.