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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).

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