We're the Agartha conspiracy theory recruits for another go round of Cicada 3301. So they haven't solved it yet ?? I like the VQC way, it's fun and enjoyable to work on.
>For odd (x+n) where:
>nn + 2d(n-1) + f - 1
>This is eight triangles surrounding a single unit square.
>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)
>This might help visualise and I hinted at this in an earlier diagram.
>https://en.wikipedia.org/wiki/Polite_number
>This will REALLY help if you visualise the smaller (x+n) squares and throw in a few much larger (x+n) squares. Focus on the patterns in the way f is distributed in each of the 8 triangles. Look at symmetries and how to construct the triangles when you know you need the two different portion sizes of f. This affects the values of 2d(n-1)
>There is a pattern to these that matches the grid which is why later you'll look back at the grid (The End) and understand all the puns and analogies of using that term.
for c6107, f = 134, f-1= 133
(f-1)/8 = 16
(f-1) mod 8 = 5
sqrt 16 = 4
T(4) = 10
T(3) = 6
T(4)+T(3) = 16
8 * (T(4)+T(3)) + 5 = 133
So, the polite triangle numbers are 4+3+2+1=10 and 3+2+1=6
I think the mod = 5 needs to be accounted for as well. So for each time we fill each of the 8 triangles with 16, we also add 5. This is the "One unit asymmetry" VQC was referring to.
Thoughts, Anons?