Hello VQC! Thanks for stopping by, glad you were over on /qr/ to see us fishing for ya! :)
For c=6107
f-2/8 = (134-2)8 = 16
f-2 mod 8 = 4
Interesting that the correct n=36 is a multiple of our mod, which is 4
Right, AA. CA, we can't know a and b starting from only c.
From c we only have d, e, and f.
VQC is hinting that we can solve the (x+n) square using f.
CA, check out these recent VQC posts.
>Going from a search to a calculation is an almost inconceivable step.
>Think of a Rubik's cube.
>Trying to solve it without a plan.
>Then remembers those books on how to solve it?
>You have to know which side to start with and which steps to take.
>All we will end up doing with the next step is figuring out a pretty pattern with odd squares with a single unit hole in the middle.
>The symmetry is beautiful in the solution.
>When that symmetry cannot exist in more than one way, you have a neat way of spotting prime numbers.
>Recap and exercise.
>Starting with odd squares.
>(x+n) mod 2 = 1
>Starting with the smallest odd squares.
>(x+n) = 1,3,5,..,
>For each odd square what are the permutations of possible values of f,d and (n-1)?
>What are the patterns?
>What do these look like?
>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.
Oops! Here's a shot from Topol's VQC Fishing Expedition last night. Highly relevant.
Here's 2(f-2) + 2(mod 4).
(f-2)+(mod 4) is a few posts above. Don't want to keep referencing myself. Baker is gonna shit his pants. Native American symbols everywhere. They must have had a few mathematicians in the tribe!
Here's layer 6:
(f-2) div 8 = 128 * 50 = 6400
(f-2) mod 8 = 4 * 50 = 200
Check Math: 6400 + 200 + 1 = 6601
Distance remaining to correct (x+n)^2 = 6889 - 6601 = 288
PMA, is that close enough for rm 2d(n-1) to lock on it?
ALSO: The remaining 288 can be divided by our polite (f-2) div 8 triangle of 16 ( 6 green + 10 yellow)
288 / 16 = 18
This could just be chance, but still pretty cool!!
Well, that was fun! I didn't find a perfect match, but got very close. Main point is the the BASES of the polite triangle number are 3+4=7 (our correct potential n0), and our correct u1 and u2 values are 41 + 42 = 83. So we could iterate n0 = 7 * 6 = 42, and then go -1 and +1 to test which one is our other u value.
For this last diagram, I added back in the (f-2), and we get another chunk.
2 * 50 = 100
6601 + 100 = 6701
correct (x+n) = 6889
6889 - 6701 = 188 remaining
Close enough to lock with PMA's formulas???
I think so.
Thoughts, Anons?
I know what Baker is thinking ;)
Wow! Great job on these diagrams, CA. Studying now.
For all you nerdy fags ;)
Update:
Properly filled borders.
10 * 18 = 180
6 * 18 = 108
= 288, our remaining units to be filled.
Still confused about additional mods related to the 288, but here is the diagram for your consideration and analysis.
Thoughts, Anons?
new baker on standby
Novice Baker here, but have my cookbook from /qr/ handy. Topol haz Dough? I'll bake in 2-3 hours if nobody more qualified steps up.
Damn. Nice work PMA! Studying in detail now.