Okay I think I have something.

Any cell generates an (e,n,d,x,a,b). From this if we have the entry (e+d*d, n+d) we have our position on the D grid. The D grid can be generated in two different ways.

Start at (0,0) and move to the left by square amounts (ie (0,0) -(-1,0) -> (-4,0) etc) and from here generating parabolas up and down (or just down for positive a and b) to get

(0,0): (1,1), (4,2), (9,3), (1,-1),(4,-2) etc

(-1,0): (0,1), (3,2), (8,3), etc

This way you have a bunch of parabolas that are basically parallel.

Or you can generate it another way by starting at (0,0) and making the parabola that goes (w^2, w). Then for any cell (w^2, w) you can generate lines that go (w^2+2wt, w+t) so you get a bunch of rays going out from a parabola. Lets focus on this way to generate it and talk about the rays. We can identify these Rays from the bottom up by 1,2,3, etc. This is because one is generated after w=1, then w=2 etc.

If you generate values for (a,b) = (1,c) and place them on this grid, they are always on the bottom-most line. Then if you want any other factors of C they are on the same vertical line on the D grid. (ie they have the same E value). If you notice where these are placed on the D grid, these entries are always either ON the D grid, or they are halfway between two points on the D grid and have the same N value as the first value on the ray (or the one with the smaller N). Also from our D value, we know that any solution cell is on a ray (or exactly between 2 points on a ray) and the ray must be of a value less than D. So basically we need to get our value and check if there is a solution to one of our lines (or halfway between one) with the same E. Then that N value is the correct N value

>>2889

this is the D grid and the two lines are the coordinates of the D value of the cells

>>2890

Also this is a cool pic I forgot exactly how I made it. I think it is the C values with their respective values in each D grid (up until a certain amount, then I did the F function on each of them. This looks pretty neat but I think its worthless.

>>2889

Basically for this type of algorithm you would get your number and get the E value for the record and then transform it into a D coordinate by doing E + D*D (which is actually just C, neato!)

So actually from the D grid we know that the E coordinate is just the C value. If we check the lines we can only be on the d lines from 1 to D, starting from the bottom line. Then we need to check for each d (from 1 to D) if (C - d^2) is divisible by d [d^2 is the origin of the line which is tangential to the parabola]. If we get this IS divisible, then we have the correct N and it is at whatever value that would calculate to. I don't know it off the top of my head. This would involve factoring a bunch of numbers, but we could probably do it recursively which could be simple.

>>2893

>>2893

damn this is worthless

>>2895

if you're checking to see if C-d^2 is divisible by d, you already know d^2 is, so checking if C is would be presumably worthless because you could just check C anyway. There would be no point in going past D for the little d's because thats just common sense

>>2897

Yeah I think the Gif and the coordinates will help. Just the specific algorithm is bad. My idea is that if we can transform into the D grid and somehow navigate that grid to find the right coordinate and then transform it back. I'm also thinking it might be worth it to (idk exactly how this would be done) but to basically stretch everything into parallel lines instead of the parabolas, then we would have a normal looking grid which could potentially be even easier to navigate. Maybe then the pattern on that grid would be worth noting? I'll try and render an image of that now. I also have some other stuff I was working on and I think that the parabolas on the left which start at (-d^2,0) each produce all products of d. Also I reflected the image over the y axis to to do negative factors, but idk if that is useful yet.

>>2899

The last paragraph is the important part. Download the gif and look at the lines when the red dots shift. The red dots are either on a line or between two lines

>>2898

This is what happens when you transform it into the grid where the parabolas are actually just straight lines and the vertical is based on the T value which is (b-a)/2. This T value also corresponds to which entry it is from the origin of the ray. Basically if you look at the D grid, I ripped out the prabola on the right and turned all the lines into vertical lines and closed the holes. idk if that helps. Also I generated the same for negative factors. For this, we have a hyperbola. If we can find the two lines for the parabola, then we can solve integer solutions for it and those would all be valid entries for C.

>>2904

Just an input variable.

Here is all the values colored in a gradient

>>2901

>>2906

lmao could we then transform this into another, more linear graph? We'll see next time. Also on a more serious note, I am posting this stuff because I find doing this stuff fun and I figure that it can't hurt to have more info. If this has already been discovered or if you guys think this is irrelevant I won't post it anymore.

>>2907

Here it is although this isn't exactly what I was looking for

>>2913

Not sure what you're getting at. Here is the grid with all the multiples of any give number. Notice how there is a little box that is the same just repeated over and over again.

Here are the primes. Notice they have a little zig zag guy in every square.

>>2915

These are actually squares. I accidently saved over the prime image with the square image. I'll remake the prime one. With this square image you can see that there is one cell that kind of follows it's way through the charts. Sort of near the origin. I'd guess that is the entry for d*d.

Here are the prime entries

>>2917

This is the biggest version without the skew. You can really notice the patterns for this one. Notice there are squares for each thing and for odd bois you have a little zigzag guy going horizontal through the middle and for even bois you have a straight line going through the middle of a square.

Here is the grid for products of 2 primes. Notice that on each of these they have the same spicy boi zig zag grid that the odd entries had. These are also the only ones that matter for RSA.

>>2964

>using boiling point of vinegar transform

Ah yes, the classic boiling point of vinegar transform

>>2965

My previous pictures were fucked up. Let t = (b-a)/2. Then if we generate cells by (n-t,t) we have their entries on the rays which is the first pic.

If we generate cells by (t,n) (which corresponds to their parabolas) we get the second pic.

>>2967

Could plain sight be looking at this again?

n = (x^2+e)/2a

so then a new record would be

(e,n,d,x,a,b) -(e,((x-1)^2+e)/2),x,x-1,1,1+2(newX) + 2(newN))

because that would generate another on the same vertical (same e) (which is the goal) but with a different N. Then maybe you could generate higher or lesser values in the cell until you can get to the next cell back. Then go back another cell, shift it to see if D = D, then go to another cell and repeat.

>>2972

Yeah but first I'd need to figure out the constant for Dc for a (e,n) cell such that D(t) = D(t-1) + 4tn + Dc

I haven't thoroughly investigated this equation but it might work. I'm working with my grids above right now