The gaps on the edge are 40, as well as the gaps from the center to the edge. This would also support the pinwheel idea (uh oh).
So the mods are the mods for the (x+n), and (f-1) ?
I'm trying to see what Q was saying. I may have found something. These are each the values for AB(1,c) AB(a,b) then a bunch of records for the (e,1) cell. It seems like if you go to the same X value in the (e,1) cell, your (x+n) value differs from the goal (x+n) by 1 (or at least a small amount). More help would be appreciated. Also new trip
We want a(t) = na in (e,1).
(capital letters mean letters in the (e,1) cell. lowercase is (e,n))
n = (xx+e)/2a, so na = (xx+e)/2
D = na + x = (xx+e)/2 + x = (xx + 2x + e)/2
A = (xx+e)/2
D-d = (startN - 1)
D = startN - 1 + d
na + x = startN - 1 + d
na = startN - 1 + d-x
na = startN - 1 + a
na - a = startN - 1
(n-1)a = startN - 1
Is this a recursive solution?
This looks good, yes bs is big square, ls is little square. I think the only problem with what you have is that VQC said n is a factor of D-d, not that n equals D-d, but I'm sure we are close. smh the keys was 4 threads ago
I learned this in Number Theory. Another modular thing I'd recommend to look up is Chinese Remainder Theorem which is very powerful
I don't really like it. If D[t]-d = (n-1)a and A[t] = na, then
D[t] = na - a + d
D[t] = na - a + (a+x)
D[t] = na - x
D[t] = A[t] - x
Obvious because this cell also has correct x so I don't really get it. He said pattern of divisibility so I'm going to look into that.
It looks like the GCD of these two values is always either a or b. Moreover, if the GCD[t] = a, then GCD[t+a] = a. And if GCD[t] = b, then GCD[t+b] = b. Since they are prime we know that there must be a record (if we keep going) where the GCD is a*b. Maybe that can help us??
The last two are where that value it is visible.
Also I just want to say again that if F is a square, the correct x+n is the square root of F, so maybe we need to fuck with the F's more.
>you
PMA
Look at this. If we go to the record where (D[t]-d) is NOT divisible by (n-1) but where A[t] IS divisible by n, then the GCD of D[t]-d and A[t] is equal to a. Also, the X value for that record is equal to the correct D value for the record in this column that has the correct X value for the final record
I'm gonna try and do it out by hand tomorrow
Also just noticed that for the last record in each section, the (x+n) is equal to the B from the one above it minus (correctX + 1)
36 = 44 - 8 = 44 - (7 + 1)
63 = 74 - 11 = 74 - (10 + 1)
224 = 244 - 20 = 244 - (19 + 1)
https://pastebin.com/czpK8A4j
Here is the updated code. For scan it does every cell on the radar testing for x up to 100. + and - zoom their respective windows. (x+n) even is just a plain blue square for now, don't know how I should color them in. dGrid and xGrid show the values with the same D and X for the selected record. The inner rectangle looks more dense because thats where I generated the grid, everything else is just based on (e,n,x) values (at least the scanned ones). +X goes to the next valid X in the same (e,n). +D goes to the same C record with different D.
Yeah you might need to download PIL or tKinter for it but yes python
Its pretty jerry-rigged. I made it piece by piece as I went adding features so there isn't much organization in it and its a little convoluted because I tried to keep it all in one file.
I may have found something cool. So remember how if F is square then (x+n) is the square root of F? Well for every record you can shift D until you get to an F value. This works into how VQC asked about 2d(n-1) and nn because this sort of leads into those values.
For instance a=5, b=29, c=145, you set d=16, then your f=144 and (x+n)=12.
For a=5, b=31, c=155, set d=17, then f=169 and (x+n) = 13.
(I think this pattern may hold also)
Also you'll notice some similarity in the successive f and e values.
# f = 2d + 1 - e = f
# e = c - dd
# f = 2d + 1 - (c - dd)
# f = dd + 2d + (1-c)
# (x+n)(x+n) = dd + 2d + 1 - c
# (d+n)(d+n) - c = dd + 2d + 1 - c
# dd + 2dn + nn - c = dd + 2d + 1 - c
# 2dn + nn = 2d + 1
# 2d(n-1) + nn = 1
Yes I'm looking at d as just a way to look at the value. For any c you can set any d and all the d+n and x+n will be the same but your e will change and your f will change and so will some other stuff. Anyway if we look at it this way we can say
f(d) = 2d + 1 - e
e = c - d*d
f(d) = dd + 2d + (1-c)
if we find d such that f(d) = (x+n)(x+n)
then take the derivative at d*
f'(d) = 2d + 2
I noticed that f'(d) = 2d+2 = a+b
Also I noticed that if x = d - a then f'(x*) = b-a. Here is some output
Patterns common to every cell at n=1.
If you do e+2j, a=a+j, b=b+j, d=d+j, x=x.
Then we have for even e:
X(t) = 2t
A(t) = 2tt + e/2
B(t) = A(t+1)
D(t) = 2t(t+1) + e/2
Odd e:
X(t) = 2t+1
A(t) = 2tt + 2t + (e+1)/2
B(t) = A(t+1)
D(t) = 2tt + 4t + (e+3)/2
Also I was looking at stuff and correct me if I'm wrong (I probably am because i'm not VQC), but it looks like at (-f,1), A(t+1) = a(n-1) and A(t+n) = b(n-1), instead of what VQC said with A(t) = a(n-1) and A(t+n-1) = b(n-1).
Maybe its the way I generate them?? Could be the way he indexes X I do X(t) = 2t+1 he might have it as 2t-1. Not sure. But I basically the f or e is just the reference point through which you are at. Like you can have any d for any c, but the e just changes. Think of it this way.
21 is d=4 e=5, 4*4+5 = 16+5 = 21
for d=5, you do
21 = 5*5 + e =e = 21 - 5*5 = -4
21 = 6*6 + e =e = 21 - 36 = -11
Since when we look at the -f column we find these for a(n-1) and b(n-1) and they are n-1 apart, I'm thinking if we keep increasing d, then eventually we will close the gap until they are one entry apart. Since for D we have A(t) = na, b(t) = nb, then for column -f, (or d=d+1), we get A(t) = a(n-1), A(t+n-1) = b(n-1). I haven't looked into this pattern yet but I suspect that if we keep changing d we get this pattern:
d=d
A(t) = an
A(t+n) = bn
d=d+1
A(t) = a(n-1)
A(t+n-1) = b(n-1)
d=d+2
A(t) = a(n-2)
A(t+n-2) = b(n-2)
etc.
continue this pattern until
d=d+(n-1)
A(t) = a
A(t+n-(n-1))=A(t+1) = b
Here is my suspected t debacle. I think if we let x=2t-1 for odd e then it is fixed but i'm not sure
So the way VQC worded it was a little iffy (to me) so I'm going to re-say it. For your starting (e,1) you have an X value for which A = na. You also have at x=(previous X) + 2n. Then from this, if you go to the left with the different e's, for the A(t) = na value you increase X by one, then for the A(t) = bn you decrease X by one, like this output shows. Then they converge on a single record like I suspected. Look at this output. Here are two examples of the movement. To keep it in one spot I kept the na and nb records in different lists
Also the b value for the record converged on seems to be 2(x+n) + 2 = 2x + 2n + 2 = (b-a) + 2
Also it looks like with each successive shift for the na values, the d values decrease by a-1, and for the nb values, the d values decrease by b+1
I knew it would be a pointer-like system. It's the only way it could work
So think of the number line going horizontally
Think of a big post sticking up at 0
This is d=0
Any e value would just be whatever the normal value it is on the number line
d=0, e=5 =c = 0*0 + 5
d=8, e=5 =c = 64 + 5 = 69 (kek)
d=8, e=-5 =c = 64 - 5 = 59
d=8, e=17 =c = 64 + 17 = 81
Its 5 units bigger than d
Now think of (instead of d=0) d=2
Now think of the big post sticking up at 4 [2*2=4]
Any e value here would be e units away from this post
d*d is sort of your vantage point and e is the error from c per say.
that said, we can look at any c from any d, it would just change the e value [think like the F function]
So certain numbers can only have certain e values
for example lets do 15
d=-2 =e=11
d=-1 =e=14
d=0 =e=15 (*top of parabola)
d=1 =e=14
d=2 =e= 11
d=3 =e=6
d=4 =e=-1
d=5 =e=-10
d=6 =e=-21
So for the value c=15, we could say that the values for e are integer solutions to the function e = 12 - d*d,
which are {15,14,11,6,-1,-10,-20,…}
If we start with good ol' c=145, it's respective e values are {…45,24,1,-24,-51,-80,-111,-144,…} for d={…10,11,12,13,14,15,16,17,..} respectively.
We know that at our original e value [1], we have a certain x [7] value that corresponds to A[t] = na = 25 correct
If we go to the next e value and increase d, we get another e [-24] with another x [8] value which is one greater than the previous x value.
This record has an A[t] = (n-1)a = 20
The next one would have e [-51] with x=[8] and A[t] = (n-2)a
if we continue in this fashion, we eventually reach A[t] = (n-n)a = 0
Also, at the first e value [1], if we do A[t+n] = nb, like VQC said.
Remember with an increase of t by one, we increase x by two, so we get here x=7 + 2*n = 17
This is our A[t+n]=nb
If we then do the same thing with the e and go to e=-24, but for this one (nb) we decrease x by one and we get another A[m] = (n-1)b
We keep going and eventually we get A[t]=(n-n)b
Moreover, this is the same record that the na -(n-1)a -> (n-2)a etc converged on.
Moreover, at this record d=x, c=0, n=1 and b=2(x+1) = 2(d+1)
This is output for A[t] values for (e,1) and (-f,1). I took also factored each of them and took the gcd. The GCD for the two values is in the middle. You'll notice that if a is the gcd for a certain t, then a is the GCD for t+a. Same goes for b. And if we know anything about prime prime numbers we know that eventually these two numbers will meet up (cuz they cycle, modular arithmetic) and eventually there will be a value where the GCD is c.
I'm also thinking of a way to look at this, but I don't know much about math history. If we just assume that like fermat and other great mathematicians were using this machine to discover the stuff they did, perhaps we could look and see what was like the first great mathematical breakthrough, and see if we can find that pattern in the grid. Then go to the next historical breakthrough and find that in the grid, maybe through that lens we could like walk through it I'm not sure.
Easier to see the pattern here
Here's actual examples
I'm going to dump a lot we've found.
Here is how I look at D values:
Basic factoring:
For any of these c values, we have the equation (d+n)(d+n) - (x+n)(x+n) = c, which is a difference of two squares. Say d=5, n=1, x=2, then we would have (5+1)(5+1) - (2+1)(2+1) = 36 - 9 = 25. For any entry, we can increase or decrease any of these values and keep c the same. Lets increase x. Then we would need to decrease n to 0, and increase d to keep the squares the same. Then we would have d=6, x=3, n=0 to get 66-33=25. So if we can navigate to any n=0 row and maintain the same c, then it would be necessary that d^2 - x^2 = c, and from there we can find the factors. Lets say a*b = c. Then if we can show
((a+b)/2)^2 - ((b-a)/2)^2 = c
(aa+2ab+bb)/4 - (bb -2ab + aa)/4 = c
(aa + 2ab + bb - bb + 2ab - aa)/4 = c
(2ab + 2ab)/4 = c
ab = c
So this necessarily means that (d+n)=(b+a)/2 and (x+n)=(b-a)/2. Through this, if we have c as the difference of two squares, we can factor it.
Through this same view, if we have an e value that is a negative square for instance e=-ii, then c = dd + e = dd - ii which is a difference of squares so it is solved.
Significant Rows:
The n=0 row only has entries for e=-jj for any j. If e=-jj, every x=j in that cell. In this row, (a+b) = 2d always. You can use any d in any entry in this row.
The n=1 row has entries for every e. If we are at an even e, then we can take any even x value. If e is odd, we can use any odd x. If X[t]=x, then X[t+1]=x+2. Also in this row, if we have a record (e,1,d,x,a,b), then we know there exists another record (e+2,1,d+1,x,a+1,b+1). In addition, we can extend this to (e+2i, 1, d+i, x, a+i, b+i). Also if B[t] = m, then A[t+1] = m.
If we have a record with c, then there is an entry in (e,1) with A[t]=na, and A[t+n]=nb. In addition, if A[t] is divisible by a, then A[t+a] is divisible by a. If A[t] is divisible by b, then A[t+b] is divisible by b. If a and b are prime, then there exists A[t] that is divisible by c in (e,1). Also in this (e,1) cell [for starting d and n], if (D[t] - d) is not divisible by (n-1) and A[t] is divisible by n, the GCD of D[t]-d and A[t] is a. Also X[t] at this same t is such that X[t] = D[w] where X[w] = d-a or our correct x value. If we have A[t]=na and an X value for this, then if we go to the next lower e value (to get this we use our starting d (or the d for whatever e we are at relative to our starting c), we do e=e-2d+1), we see that X = X+1, and also A[t] = (n-1)a. this pattern continues as you keep calculating new e's (the next e would be e=e-2(d+1)+1). Eventually, this converges on a record where A[t]=0. At this record, X[t] = D[t] = (x+n) and e = -(x+n)*(x+n). Also B[t] = 2(x+n) + 2
Significant Columns:
For e=0, if we have an entry (0,n,d,x,a,b) then we can generate a new entry (0,ni, di, xi, ai, bi) for any value i. Also at the cell (0,2), every a and b is a square number. In addition, if A[t] = xx, then B[t] = (x+2)(x+2). Also if we have a cell (0,2,d,x,ii,jj), then there is another cell (-1,2, d + (j+i)/2 ,x+1, i(i+1), j(j+1)). Also at (0,2), if you have an entry (0,2,d,x,ii,jj), then there are cells in (-1,2) with b values that are (j(j+1)) and (j*(j-1)). If you multiply this by r to get (0,2r,rd,xr,rii,rjj), then there are records (-1, 2r, r +/- (j+i)/2, xr+/- 1, rii +/- i, rjj +/- j) (either all + or all -). So I guess for any d you could look in this column and get different ways to write it as differences of squares or something like that.
Grids:
So I have analyzed different values for d and x and have found grids. For any X value, there are entries at (-x^2,0). Then there are other entries at (-x^2 + 2in, n) for any i increase, (d+n) increases by 1 and (x+n) stays the same. Also a and b increase by 1 (also this is a regular e+2n shift). If you increase n for this formula, then a stays the same and b increases by 2*i, also (x+n) and (d+n) increase by 1.
For any D value, there is a cell at (-dd, -d). Then from here, you can generate other cells with the same d by this formula (-dd - m + i*i, -d + i) for any m or i. Moreover, if you change your d value, your position on this d grid (because its relative to any d) will stay the same relative to other cells with the same d. For instance if for my c I have a certain d value, and there is another cell at (e+j, n+i), then if I use any other d value, there is another cell at (newE + j, newN + i) so this is sort of a locked in position.
Looking at this pattern I've found some odd stuff that might be important. Its just very weird. Look at this I was analyzing as vqc said the A values a (-f,1) and the D values at (e,1). I also looked at the difference and sum between the two values. What is weird is the fact that sometimes the difference is constant 1337 or 579 for example and it is equal to the original D value, but sometimes it increases by 4 and never hits the D value.
Bit of a stretch?
>Basic picture.
>For all c.
>c = ab = dd+e = (d+n)(d+n)-(x+n)(x+n) = aa + 2ax + 2an
>Grid (p,q) where p and q are signed integers
>Elements in a cell are products with notation: e:n:d:x:a:b
>The first two of the notation correspond to the coordinates in the grid.
>Horizontal black line (e,1), (-f,1)
>Vertical black line (0,n)
>Vertical grey line (-1,n)
>For some SPECIFIC c = ab = dd+e = (d+n)(d+n)-(x+n)(x+n) = aa + 2ax + 2an
>Dark green line : column that contains e
>Dark maroon line : column that contains -f
>Pinkish-purple square cell in dark green line at (e,1) contains an and bn at elements t and t+n which are >elements:
>e:1:(na+x):x:na:(na+2x+2)
>and
>e:1:(nb+2x+2n):(x+2n):(nb+x+2n):(nb+3x+6n+2)
>Blue square in dark maroon line (-f,1) that contains a(n-1) and b(n-1) at t and t+n-1 elements
>Orange squares in -f line and e line : squares that contain c as a product… -f:n-1:d:x:a:b and e:n:d:x:a:b >respectively. THESE SQUARES ARE ONE LINE APART.
>Pick any odd c and this holds for all. ALL.
Also this is my new program. I'm going to add more stuff to it.
I don't know how to share a python package. I'm also having trouble exporting it to a exe or dmg so if anyone could help me with that I'd be really appreciative and we all could use this.
From this I noticed some stuff. Say we have 1.145:
1,61,12,11,1,145
Lets get to the 'next' D value D = D + 2(X + 2N) = 12 + 2(11 + 261) = 12 + 2(133) = 278
which of course is valid by the screenshot.
and we get the previous D by doing A-X to get D=-10.
Then our new number is 101 ((-10)*(-10) + 1) and the a value has to be 1 because 101 is prime. So we have (even though this is an invalid record as far as I know)
D = -10
A = 1
X = -11
Then if we do the transformation again but forward we do D = D + 2(x + 2n) to get:
-10 + 2(-11 + 2*61) to get:
-10 - 22 + 244 = 244 - 32 = 212
WHICH IS THE RECORD IN BETWEEN D = 12 and D = 278
forgot trip
Here is some more examples. Also the record in the negative IS valid
False alarm doesn't always work.