User blog:Ikosarakt1/Unusual number naming scheme

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 = 101 = $$\uparrow\downarrow\uparrow$$. 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
 * 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 isn't always greater than n-arrowal.


 * Ninarrowal = $$3 \uparrow\downarrow\downarrow\uparrow 3$$


 * Tenarrowal = $$3 \uparrow\downarrow\uparrow\downarrow 3$$

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