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Alright. I've revised this, ended up accidentally turning it into something that closely resembles Aarex's Graham Generator, and cut it off before I got too weird with dots. I believe, based on what I've observed with other functions, that the basic White-Hex Level has a maximum growth rate of ω+3 in the fast-growing hierarchy, but the black hexagon might take it up to ω2.

White-Hex Level:

1. ab = a↑↑↑↑b

2. anb = aan-1bb

3/3b. a•(n)•b = a•(n-1)a↑↑↑↑bb

5. anb = an-1((a↑↑↑↑b))•b

Explanation:

1. ab = a↑↑↑↑b, in Knuth up-arrow notation (hexation, hence the notation's defining shape)

2. If, at the end of an expression, there is a number after a single hexagon, decrement the number by one and use the new expression as the number of up-arrows. (When that number reaches 1, remove it.)

a2•b = aabb = aa↑↑↑↑bb

a3•b = aa2•bb = aaabbb = aaa↑↑↑↑bbb

(in this way 3•64•3 = Graham's number)

3. If, at the end of an expression, there are multiple hexagons but no number, remove the last hexagon and replace it with (a↑↑↑↑b).

4. A string of n hexagons can be written (n) – parentheses mandatory, to avoid confusion. In this way you can replace rule 3 with:

3b. If, at the end of an expression, there is a number before a single hexagon, both of which are enclosed in a set of parentheses, decrement the number by one and place a↑↑↑↑b after the parentheses but before the second dot. (When the parenthesized number reaches 1, remove it.)

5. If there is a hexagon with a numerical subscript, decrement the subscript's number by one and place a↑↑↑↑b hexagons directly after the subscripted hexagon but before the second dot. (When the subscripted number reaches 1, remove it.)


So I've just set up a series:

Stage 1. the number after a hexagon determines the height of the "up-arrow tower"

Stage 2. each hexagon (n) stacks the "up-arrow tower" a↑↑↑↑b times (iterating on its height)

Stage 3. determines the number of iterations of stacking

Stage a↑↑↑↑b...?

This is where the black hexagon comes in. I'll need to rewrite the old notation so everything makes a bit more sense:

anb = a•[n]•b

a•(n)•b = a•[2n]•b

an•b = a•[3n]•b

a•b = a•[a↑↑↑↑bn]•b

I'll need to work on explaining the rules though. Consider this part a work in progress.

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