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.
1. a•⬡•b = a↑↑↑↑b
2. a•⬡n•b = a↑a•⬡n-1•bb
3/3b. a•(n⬡)•b = a•(n-1⬡)a↑↑↑↑b•b
5. a•⬡n•b = a•⬡n-1((a↑↑↑↑b)⬡)•b
1. a•⬡•b = 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.)
a•⬡2•b = a↑a•⬡•bb = a↑a↑↑↑↑bb
a•⬡3•b = a↑a•⬡2•bb = a↑a↑a•⬡•bbb = a↑a↑a↑↑↑↑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. ⬡n determines the number of iterations of stacking
This is where the black hexagon comes in. I'll need to rewrite the old notation so everything makes a bit more sense:
a•⬡n•b = a•[⬡n]•b
a•(n⬡)•b = a•[⬡2n]•b
a•⬡n•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.