>>1672

Since this is exactly what we don't want to be doing, I created a seperate thread for it. Also I don't have the batter for the other one. I'll probably jump back into the new thread once it is created.

I have devised a way to generate the (e,n,d) coordinates for any d value. This requires new rules. For now, I'm ignoring the x,a,b,c values because they aren't required for this pattern. That said, if we analyze steps of this pattern we may have patterns in x,a,b,c, I just haven't checked yet. I'm still trying to validate if this is the case. Also, for this, you generate (e,n,d,x,a,b,c) values through END(e,n,d) which is posted in my other code dump

(E,N,D,O,T)

Pic related is for D = 6.

On this graph, the +'s indicate a value that was generated by theend(1500) code. The 's indicate a suspected d at that value. Whether or not the 's actually correspond to a cell we can overlook for now, because the pattern seems to hold.

O is the origin cell of a line. Indicated by the integers in the chart (except for the 6,7,8,9 on the line). 0 is the zero origin, 1 is the 1 origin etc.

For each origin cell, the value of T is 0. For the cell on the N+1 row relative to an origin line, we have T = T+1. The T's for O=1 are written for T=6,7,8,9

Now that you can read the values, here are the rules I've discovered:

(-D^2, -D, D, 0, 0) is a valid start record for any D

(e,n,d,o,t) -(e+t^2, n, d, o+t, 0) (to generate origin cell for same row given any cell)

(e,n,d,o,t) -(e+2o, n+1, d, o, t+1) (to get next t) *

(e,n,d,o,t) -(e-2o, n-1, d, o, t-1) (to get prevoius t) *

(e,n,d,o,0) -(e+2ot, n+t, d, o, t) (any t from origin) (basically the same as the two * above)

(e,n,d,o,0) -(e-t^2, n, d, o-t, t) (to generate other cells in row n from t=0) (t<o)

(e,n,d,o,t) -(e+ (2t-1), n, d, o+1, t-1) (to shift from o to o+1)

(e,n,d,o,t) -(e- (2t+1), n, d, o-1, t+1) (shift from o to o-1)

BIG (e,n,d,o,t) -(e+2d-1, n-1, d-1, o,t)

This last equation comes from noticing that the layout of all the D grids is the EXACT SAME, just with a different 0 origin cell.

To factor C, our record (e,n,d,x,1,c) has the correct D already. I suspect that we can generate the correct O and T values from some other method to zip right to our correct cell.

>>1762

FUCK the T values indicated on the graph are all wrong by one 6,7,8,9, they should be 5,6,7,8. SORRY

>>1763

Also the origin O cell for D is at the following

(-D^2 + O^2, -D + O, D, O, 0)

Thanks for setting this up CA! I'll check in sometimes. You've got me schooled in the programming department, but I really enjoy following your ideas!

>>1762

Did you ever succeed with this or was it a restatement of our problem?

>>1762

I'm gonna update this thread a bit. I'm not going to talk in those coordinates in OP either because I never really did much with them it's more of a way to generate where they'd be. That pattern in OP is the 'D grid' for every d. No matter what number you have, the d grid is the same but in a different spot.

First pic is the entry of 121. What is highlighted in red is the D grid. The second pic is me on the center of the d grid. If you would like to know how it is constructed, basically start from the center point (which is at (e,n) = (-dd, -d) ) and go to the left square intervals (-dd-1, -dd-4, -dd-9 etc,) and your n values are going to stay the same at -d. Then from each of these points you draw parabolas out to the right. This creates the d grid that looks sort of like a web. The first parabola at -dd that goes to the right kind of spawns off straight lines. The first tangent line for this has the slope that goes over 2 and down 1. The next tangent goes over 4 and down one. This pattern continues. It's cool because these are linear patterns that come from parabolic origins. Third pic is on the second 'tangent' line

If we check each of these 'tangent' lines, then we notice a few things.

Pic one (for d=12) shows that if we are on the first tangent, which starts at (-d*d + 1, -d+1) and goes down 1 over 2 from then on, the d+n and x+n values differ by 1. Then if you go down to the next n, (d+n) and (x+n) both increase by 1.

>>7593

If we start at something on the first tangent (this is also where our starting c is because a = 1 = d - x =(d+n) - (x+n) = 1) and shift to the second tangent, this corresponds to decreasing (x+n) by one. This also means that the difference between (x+n) and (d+n) is now two on the second tangent. This pattern extrapolates to every tangent. The third pic, compared with the second, shows that if you go down one n level on the second tangent, x+n and d+n increase by one as well. This happens on every tangent.

A way to look at factoring from this perspective is that you need to find a n value in the same e column on this d grid. For instance here is a picture of c=65. Notice the only other red part in the column in pic 1. If I click on that entry it brings me to the correctly factored cell. This makes sense because if you have an entry that has the same d value (which would put in on the same d grid) then if there were any other selection on the same e column it would have to be another factorization of the same number. Last pic is for c = 75 = 355. Since this has 3 prime factors it is located in three entries in the same e column.

Also, if we follow the column up further, it brings us into negative n values. If you look at pic 3, that is the negative value in this grid which was the like the ghost cell in the grid. If you look at it, you'll notice that compared to its mirrored partner below, the mirrored A value is -b and the mirrored B value is -a. Because -a*-b = c still this is valid.

Here is the starting entry for c=21. If we follow to the left, we get to the zero column, which is located at e = -d. This column has every c=0. If we go to the left one more time we get to this entry, which has the same (d+n) as the starting but increased x+n by 2. This makes sense because every row in the d grid would obviously have the same n, and so it would obviously have the same d+n values. Also since this works like this every shift to the left increases the x value by one and every shift to the right does the opposite.

From this horizontally mirrored cell on the left, if we go up the -1 tangent one entry, it brings us to our 'f' record we know and love

>>7597

>>7597

nvm that isn't the f record

>>7597

This would be the f record for c=21. Similar to before, there is one entry above it in the same column, which is the factored entry

Now lets take a look at the X grid. In this first picture, we are located in the x=7 grid. Similar to the D grid, from pics 2 and 3, if you go down one of these legs, then x+n and d+n increase by one. If you shift to the left in the x grid, then your d+n gets smaller by one and your x+n stays the same. If you go to the right x+n remains and d+n gets bigger by one obviously. This also corresponds to an e+2n shift, so you can repeat this to keep getting more numbers.

>>7600

If you look at the center of the X grid, it is located at (e,n) = (-xx, 0) [sorry about the pic a shortcoming of this little app I made is that it doesn't work for n=0 records]. Then from that origin of the x grid, you can go to any cell (-xx + 2hi, 0 + i). This leads to a lot of cells with x values.

Then if you remember that the vertical column for the d grid has c=0, this is true for the vertical part of the x grid also. This is always where d and x are the same. The yellow is where d and x grids are meeting up.

This is a pretty zoomed out picture, but it shows what a factored entry looks like. The d grids and the x grids are off set by a certain amount.

Ok I've been working on this for a while. Basically I've wanted to analyze these grids but I could never find a way to do it, so I decided to write a console program to do it. The GUI with buttons never really worked well so I decided to make a command line terminal app. The code is here

https://pastebin.com/btdwEpuk

just make sure you open it in a big enough terminal window.

>>7752

Pic_0.

So here we have the C grids. These are simple to look at but hard to navigate.

Notice how the points (%, 0, 1, 2) are in the same column. They are all also different factorings for c

% = 1, 45

0 = 3, 15

1 = 5, 9

2 = -5, -9

Also notice that the d's are all 0. This is also why the e is equal to 45.

The MAX value that e could possibly be for a number c is c, because e = c - d*d,

if d < 0 or d 0, dd > 0, but if d=0, dd=0, and since d*d is a negative in that function, we

minimize the subtraction at d=0. Therefore the highest e = c. This is an upper limit for e for given c.

I showed #2 because I wanted to illustrate that there are negative values for factors. This also shows that

in the grid, for a given e, the different factoring for them with the same d value, leaves them in the same

e value, which makes sense. The change in the grid comes from the different n value. The change in an n value

with the same d would correspond to a change in (d+n) value. This makes sense, because (d+n) = (b+a)/2, which

would obviously change because these different records have different a's and b's. That basically explains

the movement in the columns.

So now if you look at this, it goes off to the left with a decreasing n value. Also e changes by a square amount,

45

44 = 45 - 1

41 = 45 - 4

36 = 45 - 9

29 = 45 - 16

That is because we are increasing the d value, and e = c - d*d, and d is incrementing so we are taking away squares.

• Why does the n decrease by one? *

For a given (a,b) pair, the d+n and x+n are always the same. Notice how all the records in the parabola

have the same a and b and d+n and x+n. We are increasing our d value as we go off to the left, BUT since

a and b are the same, d+n must be the same and for that to happen n must decrease by one if d increases to

keep the squares the same size. Then, since n decreased by one, x must increase by one as we go down the parabola,

because (x+n) must also stay the same size. This explains the e, d, n and x movements as we go along these parabolas.

[ sorry for skipping the x=0 one. For some reason you can't track those records its a bug ]

So we know that the c value can exist on any of these parabolas and the origin of the parabola is at e=c

In addition, since it is all the way to the right and e=c, then d=0, which woudl mean that n=(d+n) for

any of these factorizations.

Therefore,

(e, n) = (c - d*d, (d+n) - d)

which is sort of obvious algebraically, but from the visual approach I think it brings it into a new light.

Here is the x grid for x=3.

Similar to the d grid, the origin of this grid has an e value of -x*x. But differently, it has n=0 for every x.

Moreover, the origin of x grids are the only place in the row n=0 where there are any records. Moreover you can

have any factor a and any factor b in any of the x origin cells.

You'll notice that stemming from this origin cell, are lines going out in a (e+2i, n+i) fashion for any i.

Here I made it easier to spot the lines.

So if you stay on one of these x lines, then your d value will stay the same.

Points % - 3 are all d=4. Their a stays the same (a=1=4-3) when you go down it and the b increases by 2.

With these points, your f is decreasing by 2 each time. d+n increments and x+n decrements

>>7756

Points 4 - 8 are d=10. Here the a value is 7 (10 - d). When you go down this also the b value increases by two

d+n increments and x+n decrements. Here your f value decreases by 14 = 2*7 each time.

Here shows if you start at an entry in the x grid and go to the right 2*n, you have the same x. This is

equivalent to an e+2n shift that vqc said. Notice here the (x+n) stays the same while the (d+n) increases by one.

Also when you do that you are shifting d by one.

Since you increase by 2n for each row there is a periodicity to the x grid that keeps it constant relative

to the origin of the x grid, much like the d grid.

That said, if you are at n=15 for a certain x, you know that 15 is divisible by 3, so you can bounce down

to that record where n=5, 3, or 1, because you know that when it's repeating it would hit 15*2. For this

I'm speaking in relative terms to the origin. If we are at x=3 and I'm at the '15' relative to the origin,

then I would be at #5. So I guess in a way if you can navigate the X grid also and get an entry in the same

column, you factored it also.

Here the vertical and horizontal lines are the e=0 and n=1 lines just for reference.

Our main record '%', has a d value of 6 and the factors are 3 and 15.

the record at '0', (the arrow might help) has factors 5,9, so they are both for c=45.

If the map was a little bigger, towards the bottom would be the 1,45 record in the same

column as well. The amount is the same as as many factorizations there are for a number.

What the 'D' cells are are every record (e,n,d) where d=6.

These have one of the best patterns because you can look at it in so many ways.

Here I have highlighted the different factorizations for c=45 on the d grids where d = 6 (green) and d = 4 (cyan)

You'll notice that relative to the records 5 and 6, the d grids are exactly the same no matter which d you have, it just

shifts the position of the origin of the d grid.

Also, you may notice that while the (e,n) coordinates are different for the records on the d grids, their position

relative to the origins is the same (e is noted by the striking yello '45's). So, you can shift d's for a record which would

change the e,n values, but you will always be in the same spot in the d grid. I hope that makes sense.

In addition, at the origin of the d grid, your c=0. Then if you increase e by one c=1 etc.

ex: if d=7, origin (e,n) = (-d*d, -d)

c at (-d*d, -d) would be 0

c at (-d*d+1, -d+1) would be 1

etc.

Also, you'll notice that the origin of each d grid is located at (e, n) = (-d*d, -d)

Now we're going to look into how to d grid is structured before we look into the patterns.

Highlighted on this grid, are some parabolas. These are interesting because in a sense,

the d grid is created from all of these. From the origin of the d grid, there are points that

extend to the left at square amounts. From these points, stem other parabolas that are, in

a sense, parallel to the original parabola.

If you just look at the cells that start these parabolas, you'll see that from the origin, they

each go to the left a square amount. In addition, the c values are equal to this negative square.

At the origin if you go to the left, you can either go UP x by one, or go DOWN x by one. After that,

you kind of have to stay going up or down. You could obviously go back to the origin then go up so

its a trivial calculation.

If you look at the records from %,0,..,5, you'll see that as you move around the parabola in the

positive n direction, your x decreases. This makes sense, because we are on all the points

of a parabola going to the right, so we would have square values and so x+n=0

So as your n increases for these points around the rightmost parabola, your n decreases.

Then from any of these points on the rightmost parabola, you can go to the left in square increments.

If you do this, similar to the last picture, you can either increase or decrease x by one. I chose to decrease

it in this case. When you go to the left here and decrease x, a values decrease by one and b values increase by one.

Here is an example of looking at a different parabola.

If you count out from the origin of the d grid, you'll notice that the parabola I've selected is the 7th one.

If we start at %, you'll notice that x decreases by one as we go down. Also, a is increasing by one as we go around.

The changes in f increase by increasing odd numbers. The most interesting thing here is that at the origin of this

parabola that is highlighted, the a value is equal to 7. This is cool and it follows for every parabola. The 10th

one out has a=10 at the origin.

This picture shows how a values work on the d grid. For any a value, you can shift over e by 2*a and increase n by one

and you'll reach another value with the same d and a values. As you do this, your b value increases by 2 each time.

Because of this, your c value increases by 2*a each time you go to the right and down one.

Also notice that at poin #6, (x+n) is equal to 0. Also this line looks like it is the derivative of the

origin parabola at point #6. Similarly the other lines are derivatives at other points.

Here illustrates that each little ray that is tangent to the parabola has one factor only and they are as labeled.

As you go to the right and increase a, b decreases by one. Also the (d+n) is the same for the entire row while x and x+n

decrease by one. (again a glitch is there so I skipped one record between #6 and #7)

Thats basically it for the d grid unless I forgot something.

Next will be the A grid.

Here is the a grid for a = 3. Notice that at (e,n)=(0,0) cell there is a=5. In fact, if you look at the

columns on the right, you'll notice that every single a value is in this row. This is the same for the

entire n=0 row. You can have any a in it. From this a, if you go over 2a and down one (e+2a, n+1), then you will

still be at the same a, you will just have b increase by two. The d value also stays the same. This makes

sense because it is analogous to moving along one d grid for a certain factor. A better way to look at the d grid row

for given a is that the line is just a piece of this much bigger a grid.

Also you'll notice that similar to before, if we go to the left from one of these top right entries, we can increase or

decrease x by one and inc/dec b by two each time. e changes predictably (because you're moving in squares). This was

expected from prior behavior in the d grid.

Also in the a grid, there are tons of parabolas. For this a=3 grid, the parabolas sort of stretch out as you

go down. In this case, when you transition from the smaller parabola to the bigger parabola, if you're going

to start at the origin of the smaller parabola, you are already on position 3 of the bigger parabola.

If you go to the origin of that parabola, then you are on position 3 of the next parabola.

This illustrates that if you have a=5, your little parabola starts at the point when you're on your 5th entry from

the origin.

If we fully flesh this pattern out, it looks like fish scales. Just because I drew these lines, doesn't mean we can't

go to the other a values either, they are all in this same grid as well. In fact, this grid repeats itsself 5 times

Count the numbers of A values inside of these little 'scales'. For 5 it looks like 16.

Here it is for a=4. Notice that if we slice around these parabolas, we can get a*a amount of a entries.

heres an example of two different paths we could go down. The one on the left would be if we went around the origin

and then went on the next parabola once you're at e=(origin - a*a). The one on the right would be if at every origin

you reach, you go down to the right, and once you do that you're at e=(next_origin - a*a).

You can keep doing this and the n skip amount just increases by 1 each time.

This is if you take left every time or right every time. Could this be a ?decision tree?

Isn't also worth noting that these two paths, that represent similar decision making choices

sort of represent a parabola as a whole, even though they aren't one?

Here is the (d+n) grid. Notice that everything is inside of this parabola.

In this case, d+n = 7, so the parabola's origin is at (e,n) = (7*7, 7),

kind of like the opposite of how the d grid origin is.

Then, no suprise here, from this spot, you can go to the left in square increments

and still be on the same (d+n) grid. From each of these next cells, you can go out

in parabolas to the left, kind of like the opposite of the d grid.

[maybe we could link the two]

Then, if you look at the values along these parabolas, you'll see that as you go down,

the x values decrease by one each time. Also the d value decreases by one (necessity with n-1 and d+n being constant)

Since this is the sixth parabola, the (x+n) stays at the value 6. Also for this motion our a and b values stay constant.

This means that the (x+n) motion is parabolic to the left. For every (x+n) this grid is constant relative to the origin

of the (x+n) grid, which is acutally the same as the origin of the x grid.

Here is the x+n grid. It looks similar to the x grid, but it's a bit different because we can find parabolas in

this as well. For each ray, instead of shifting over 2a and down 1 as in the x grid, these go over 2a and down 2.

You'll also notice that for these parabolas, they have a mirrored side. For this picture, highlighted in purple

is a parabola. What's interesting here, is that parabolas seem to have a constnat c, a, b and d+n value.

It also looks like that at the origin of the parabola, you are always at d = 0.

Also every parabola here intersects the origin of the (x+n) cell. The first yellow one is the second parabola,

and it intersects the origin at the second entry.

The purple parabola is the 5th parabola from the center, and it intersects the origin at the 5th entry from the

origin.

Also notice that around this parabolic motion there is a decreasing f value. For this, the purple grid

has c = -200, and the f value at the origin of that parabola is 200, and then from there the f values increase by

square amounts as we go to the left.

Now if you look at this picture, we are going down the rays fo the x+n grid.

As we go down, our x values decrease by decrease by 2 (makes sense because it is x+n and n is increasing by 2)

Then our f value is decreasing by twice whatever ray we are on. I have labeled the rays.

Also, unlicke the a grid, we are decreasing our a value by one and increasing our b value by one.

This makes sense because (x+n) = (b-a)/2 so 2x+2n = b-a, so if we take from a we must give to b.

Also notice that the (x+n) grid is mirrored over to the left so unlike other grids, the parabolas

go both ways so you could theoretically generate another record with the same x+n by flipping over the axis of the

x and (x+n) origin cell.

Here is the f grid. This is by far the most confusing grid. Here is f=7.

f is a value that represents the distance from c to the next biggest square.

No this is the one for f=7. Other f grids make more sense but I want to illustrate what I got going on here.

You'll notice that the origin of this grid is located at (e,n) = (-8, 1)

You'll notice that when we flip over across this origin of the f grid, our (d+n) value turns negative.

Also with this transformation, the a and b values switch and invert signs.

Other than that, this is a tough to understand grid. Unlike the other grids, this one is NOT the same

but shifted for different f's.

Here is the f=5 grid. See it is different. For this one, the origin of the f grid is at (-6, 1) so I think

it is safe to say that for any f value, the origin will be at (e,n) = (-f-1, 1).

You can still flip over that point and generate a value with the same f value and an inverted (d+n) value.

BUT here, notice the cell I'm highlighted on. Look at the f values for that cell.

They look awful familiar don't they??

>>7778

Here are the values at (e,n) = (1,1). Notice how for this, the A[t] values are the same as all of the F[t] values for (-10, 3).

Bizarre. I think this should get some further investigation in the future.

>>7779

This is also a f = 9 grid. Notice how here it actually makes a little bit of sense.

There are two vertical lines at n=0 and x=-4, x=-16, so where f=mm, the origin is at e=-mm

and there are vertical lines at e=(m-1)(m-1) and e=(m+1)(m+1).

You'll notice that from here, we can go off to the right on these little rays again.

From %-3, you see that a values increase by 1 and b values increase by 9

You'll notice by followint records 1,4-6, that the b values increase by amounts of 8, which is one less than 9

obviously. Those are really the only consistent rays for these f values. Everything else is rather chaotic other than the

vertical and horizontal lines.

>>7780

Here we follow the line that comes down and the horizontal axis.

The one that comes down is from records % - 5.

As we go down this line, we see that e and n both decrease by 2.

Also x is increasing by 2 every time we go down this path.

Also as we go down a and b increase by one.

Also the d decreases by one and d+n increases by 1.

(x+n) remains constant.

Then on the path 3, 6,7,8,9, we are in the n=1 row.

For this, t is always the same, and a and b increase by 1 which makes

d+n increase by 1 also (d+n) = (b+a)/2 = (b+a+1+1)/2 = (b+a)/2 + 1 = (d+n) + 1

>>7781

This grid, f=16, look similar to the last f grid, which was f=9, but it isn't the same. You'll notice that the two columns

are seperated by a little bit more. That is because where the line coming from the top intersects with n=0 is equal

to -16 here. Therefore the columns highlighted are e=-9 and e=-25, which are more seperated.

Looking at I, you can see that see that going down this path, we still

have e and n decrease by two, so that is constant across all f = m*m grids.

Moreover, a and b both increase by one as you move down, so that is constant across all f=m*m grids.

Then, what is really weird is how in part II, the b value increases by 9 again.

It made sense for f=9, but also f=16? This is bizarre but It holds true.

Then if you look at III, it increases by 9 again and and a still increases by 1

If you look at IV, you'll see that here, the a value is increasing by 2, and

the b value increases by 8. This also always is true for any f grid.

Also, there are always only these two movements that are reliable.

Here is f=25

Notice here from 0-2 (messed up selecting #3) a increases by one and b increases by 9.

Yet another proof that this pattern is always 9 for some reason.

The other one is always 8

https://www.youtube.com/watch?v=D0CfhdkrEmU

Then I wanted to explain these columns because I have yet to do that.

First off, why are these columns here?

They are here because these represent 0 for different values of b.

As you go up this line, b decreases by two and x+n and d+n decrease by one

Also here d=x

Heres another thing to ponder. Look at this grid for f=9. If we flip to f=-9, we get something neat.

It looks completely different (pic 2). f is the distance from your c to the (d+1)(d+1). If you had a square

value for f, then you would have that c + f*f = (d+1)(d+1) === c = (d+1)(d+1) - f*f,

which is a difference of two squares, which exist in so many ways, so this f could almost represent an (x+n)

value if you want.

Here is the grid for f = -40. Notice how the points make out a hyperbola that has the origin at (e,n)=(39,1).

Also, you can flip over the origin like other f grids. This involves inverting your a and b values and flipping the t value.

This is very neat because unlike the positive f grid, when it is negative it is always in this hyperbola shape.

You'll also notice that inside this sort of chaotic hyperbola are parabolas again.

Still f = -40

from % - 5 we can see one larger parabola.

The origin of this parabola is at (e,n) = (43, 11)

Then e shifts by 4 each time you go down the parabola and n shifts in square amounts.

As you go around these parabolas, the a values all stay the same, and the b values change

in amounts of 2*(2k+1) but they are centered at e=43 the origin line and aren't related

to their position the the parabola relative to n.

As you zoom around the parabola, d and x both increase by 2 and a stays the same.

The inner parabola is centered at (e,n) = (41, 21)

You can see from this parabola, that the increase x by two pattern stays,

Also the t values increase by one. a and b and d+n and x+n all kind of are the same from 7 to 8

and I think that is because they go around the origin axis, whereas the last parabola hit the axis.

It might be hard to tell from this picture, but the origin of this bigger hyperbola for f=-60

is at (e,n) = (59, 1). It always seems to be at (-f-1,1).

From here, you can see the larger inner parabola has its origin at (e,n) = (63, 16)

and it has a inner parabola with the origin at (e,n) = (61, 31)

So how do these coordinates work? The bigger parabola starts at e = origin_e + 4, n = origin_n + f/4

The smaller parabola is at e = origin_e + 2, n = origin_n + f/2

Here is f = -42. Notice how unlike the last two this one does not have the bigger parabola, just the smaller one.

That is because f is not divisible by 4, so we couldn't do (n= origin_n + f/4), but it is divisible by 2.

Of course, since this is an f grid, the origin is located at (42-1,1) = 41,1.

Then there is obviously, near the bottom, the parabola which represents it's divisibility by 2.

It's origin is at (e,n) = (43, 22) which is (origin_e + 2, origin_n + 42/2)

This has values that co around similar to the last one

BUT now look at #6, where (e,n) = (47, 8). This is also (origin_e + 6, + origin_n + 42/6)

Crazy enough this works for every single factor.

For each parabola, the amount with which you shift is scaled also. For instance the #6 to #7

transition, relative to a normal parabola would be a (e+1,n+1) shift, but here it is being shifted

like (e+12, n+6), which is just (e+2m, and n+m) where m is the factor of f.

So in theory every factorization of every number would have cells organized like this in the f grid.

I think we could use our f value and factor it to find similar entries along this f grid.

Once you notice that this works for every factor its easer to see that this hyperbola is just made of a lot

of stretched out parabolas.

Thats all I got for now. Peace.

>>7788

https://pastebin.com/ZKcX61s5

BETTER CODE

Some Grids analysis I've done