868475 No.6694 [View All]
The virtual quantum computer (VQC) is a grid made of infinite yet constructable sets that follow a known pattern. Like a quantum spreadsheet.
The grid is the superposition. The collapse of that superposition will be two input parameters, d and e which can be calculated easily for all integers, c, where c is the difference of two squares. Its purpose and our goal is to learn and show the TRUTH, one of them being P=NP. Cracking RSA will be a consequence.
When the integers that are the difference of two squares are arranged into the grid and their corresponding properties are shown, a pattern emerges that shows calculation instead of searching is possible.
Glossary
Column
All cells for a given e.
Row
All cells for a given n
The grid is indexed using e, n, and t, where e is the rows, n is the columns, and t is the specific element in the cell.
Entry, record, element
one set of variables that represents one factorization for a number.
an entry = {e:n:d:x:a:b} (e, n, t)
{1:5:12:7:5:29} (1, 5, 4) is a record AKA an element AKA an entry.
ab record, nontrivial factorization, prime record
the element that contains the factorization of c that is not 1*c, hence, nontrivial.
1c record, trivial factorization
the element generated from setting a=1 and b=c
Cell
All entries for a given e,n (not to be confused with an entry itself.)
Genesis cell
e,1
Remainder Tree
The remainder tree is the result of treating d and e as c's recursively until 1 is reached, creating a tree with several to many branches.
Functions
na transform
a movement from a record in (e, n) into (e,1) where n becomes 1 and a becomes a times the n of the (e,n) record. It has also been used to refer to moving n*a records down in a cell.
T
T of number or T(input) is the triangle number function. If our input is 7, T(7) returns the 7th triangle number
T-1, inverse T
the inverse function of the triangle number function that returns the index of a given triangle number. If our input is the 7th triangle number, the function returns 7.
Variables
The map's legend is {e:n:d:x:a:b}, where c is any number that is the difference of two squares, so odd numbers are included. It is the number you want to factor. It is the number that the a and b in an entry multiply to make.
a and b are, to reiterate, the factors of c. a is the smaller factor of c, and b is the larger one.
d is the integer square root of c
e is the remainder of taking the integer square root of c. Unless c is a perfect square, a remainder will be left over.
i is the root of the large square. it is the same thing as (d+n)
j is the root of the small square. it is the same thing as (x+n). i^2 - j^2, difference of squares.
n is what you add to d to be exactly halfway between a and b, and it is the root of the large square. So it takes you from d to the large square.
x is what you add to a to make d. When added to n it makes the root of the small square.
f is what you add to c to make a square. (e is what you subtract from c to make the square below it, f adds to make the square above c.)
g is the square root of c with decimals, opposed to d, which discards decimals.
t is the third coordinate in the VQC, it is a function of x.
u is the base of a triangle that helps us calculate (x+n) for certain c values. simply put, it is a representation of (x+n). 8 times the triangle number of u plus one is x+n.
s was a variable used to demonstrate patterns in (e, 1). See "(e, 1)."
When capitalized versions of the variables are used in comparison to lowercase versions, the capitalized versions refer to the variable's value for the trivial record, and the lowercase variables refer to the values for the nontrivial record.
{e:N:d:X:A:B} (e, N, T) is the trivial element.
{e:n:d:x:a:b} (e, n, t) in this context is the nontrivial element, the prime factorization of c.
____________________________
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358135 No.7578
>>7577
possibly two close circle rotations around x and Y and all the others are diagonal at a lower scale - maybe. The first circle rotation is about 8 degrees and then is spins into 3d and warps into ellipses is what I see.
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358135 No.7579
I know this is off topic as it is about my discovery that even physics say 99 percent of mass is not matter but binding energy. But I need a better math fag than I am. I basically want to figure out the energy per meter of a proton. Here is my feeble attempt:
How big is the proton?
Because protons are not fundamental particles, they possess a physical size, though not a definite one; the root mean square charge radius of a proton is about 0.84–0.87 fm or0.84×10−15 to0.87×10−15 m.
Mass of Proton
Mass
1.672621898X. 10−27 kg[1]
938.2720813(58) MeV/c2[2]
the speed of light =
299 792 458 m / s
E=mc^2
EProton= 1. 672621898 X10^-27 Kg* (299 792 458 m / s)^2
= 1. 672621898e-27 Kg* 8.987551787e+16
=1.50327759289610516349668072e-10 Kgm/s
How big is the proton?
Because protons are not fundamental particles, they possess a physical size, though not a definite one; the root mean square charge radius of a proton is about 0.84–0.87 fm or0.84×10−15 to0.87×10−15 m.
Thus the Energy per meter area is:
The energy of Proton / the area of the proton
1.50327759289610516349668072e-10 Kgm/s / 0.85e-15m=
= 176856.18739954178394078596705882 Kg/s
Option 2:
Ratio of area of proton to c^2=
0.85e-15 m / 8.987551787e+16 m/s=
=9.4575254768431856157937707803052e-33 seconds
This is the area of energy the proton contains.
If you then divide the EProton by the ratio of area of proton to c^2=
15895041431120167484788.273385791 kgm
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358135 No.7580
>>7577
I have been looking at the figure - looks more like a spiral spun in 3 or 4 d. Do you have a higher res picture?
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08d201 No.7582
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23f1bb No.7583
Hello Anons! Working on diagrams, found a better offset layout which clearly shows how the u and u+1 dimensions form 8 equal triangles surrounding one unit. Maybe it's old news, but I had fun making it, lol ;)
for c6107, T(41) * 8 + 1 = 861 * 8 + 1 = 6889
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9a0997 No.7584
>>7576
Have taken this a bit further, and found a way to calculate all triangle portions of the aan(n-1) formula for the initial factor record for any odd n-1 in (0,n).
Pics attached for c21025 (145^2) and c104329 (323^2) show the aan(n-1) derivations for all factor records, and then all square multiples in terms of triangles for the first factor record.
The revised formulas are:
aan(n-1) = aa * 2T(n-1) = aa * 2(T(u) + remainder)
where u = m * x - 1
and remainder = m * x / 2 - m^2
m is just a multiplier, but it turns out to work inversely to each "a" square multiple.
For c21025, as an example, there are 12 possible values that represent the aa square multiple of aan(n-1). 1,2,3,4,6,8,9,12,18,24,36,72.
Using the above formulas, and using m=4 as an example, we can find an "a" factor as:
53742528 = aa * (T(4*144 - 1) + (4*144/2 - 4^2)) = aa * (T(575) + 272)
aa = 53742528 / 165872
a = 18
And for each possible value of a, a * m will have the same value of 72, hence the inverse relationship between a and m.
Therefore, to find all valid square multiples, we can simply start at the lowest m = 1, find the largest "a" value, and then iterate halfway upwards.
This isn't a solution to our general factoring problem, as it doesn't yet appear to provide a way to jump to any other factor record for our starting c.
It does, however, show that there is a way to calculate both the triangle base and remainders to enable extracting factors from larger products.
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a332a2 No.7585
So I've been looking a bit into our grid, thinking more about what we actually are looking at and I was reading up on arithmetic progression, geometric progression and series in general.
Our grid is just a bunch of series with different offsets and triangles and squares, right.
So I started to play a bit with sequences and did some analysis using that way of thinking about our (e, n) and I found what I believe is the general equation for generating the nth record in (e, n).
It's remarkably simple, in fact so simple I'm surprised we haven't noticed it before. First I want to clarify on our n's. We have two types of n, one that occurs as a in (e, 1) and one that occurs as a factor of a in (e, 1). The first one I refer to as primary n, or root n, while the second one is a sub-primary n, or factor n. The first one will always have a=1 in (e, n), thus following the rules of the grid (a[t] = an in (e, 1), where a[t] = 1n for some t). The second one will not, it will have a "start" record with a=k (for some k, factor of a[t] = kn).
We know the equations for (e, 1). They are either the sum of half of e + 1 if e is odd added with either a product of triangles or squares.
odd: a[t] = (e/2 + 1) + 4t(t-1)/2
even: a[t] = e/2 + 2tt
or, if we want to limit our selves to purely triangles:
a[t] = e/2 + 2(t(t-1)/2 + t(t+1)/2)
The pattern for primary n's, or root n's is:
a = 1
b = 1 + 2(n + x)t + 4n(t(t-1)/2)
d = x + 1
while for sub-primary n's it is:
a = k
b = k + 2(n + x)t + 4n(t(t-1)/2)
d = k + x
In the first pattern, the x refers to the x in (e, 1) where it first occurs (which is also the same the cell in (e, n) where a = 1).
For the second pattern, the k is an offset, some factor where a[t] = kn in (e, 1). Here x is also the same as for when the n first occurs as a factor in (e, 1).
Now this also makes me think about VQC's hints. We know using -f and e, we're supposed to find an offset (in the second pattern you can think of k as an offset) and we know by multiplying -f and e we will end up with aan(n-1). That means, if we multiply the product of -f and e with 2, we will have one part of our equation for calculating records in (e, aa).
That is, k + 2(n + x)n + 2aan(n-1). The latter part will be known (although only in the sense that it exists as a product of a[t] in e and -f, but not which product).
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a332a2 No.7586
I've been playing a bit with the idea from VQC about multiplying -f and e, but I trailed of a bit and found something funny.
If you imagine a number line with d's as poles (GA talked about this before) then our c exists between dd and (d + 1)(d + 1). It will exist at dd + e and at (d+1)(d+1) - (2d + 1 - e).
That is the variable we call f, I decided to take a look at c from "below" our d, that is d-1. I have called it g, simply because I needed a name. In g we will have n+1 instead of n-1. This means if we were to multiply e and g, there is a result that is equal to aan(n+1) instead of aan(n-1).
It's just a different point of view, but I decided to just play around with g and f out curiosity and I decided to ignore e and just play with g and f. The idea was that we know that in -f we have a[t] = a(n-1) and at g we will have a[t] = a(n+1). Multiplying these two rows will give us aa(n-1)(n+1). But I trailed off again and decided to play with addition. This means at some t we will add a(n-1) + a(n+1). This should result in 2an. However, I did a mistake in my function and instead of adding a[t] from -f with a[t] from g, I ended up skewing it and added a[t] from f with a[t + 1] from g. The result surprised me, as it generated the list that exists in (4e, 2) instead. The funny thing is that in (4e, 2) we have (at least) two sequences. One sequence will contain c and the other is the result of this addition.
I'm sure it can be explained easily using algebra, I haven't looked into it yet, but I was just surprised and found it very interesting. Not sure if this is a blind alley or not, but I figured I should share.
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a332a2 No.7587
>>7586
Meh, I think it's a blind alley. I looked into the algebra and it boils down to the difference of g and f. The series created from it won't always match 4e either, although the pattern inside of it should still be valid.
As in we have two sequences, one with a[t]=c and the other where an should appear at some t.
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9f8800 No.7588
How do you stop releases being scrubbed from the internet?
Can you write it to the blockchain?
Can you write it to many?
Would they take out all crypto to scrub?
Journey to the Centre of the Earth
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d4b724 No.7589
>>7588
Senpai! Nice to see you. Can you give a hint on how to correctly use the f-1 div 8 mods in combination with the f-1 div 8 multiples? I'm puzzling over the diagrams now.
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a332a2 No.7590
>>7588
> How do you stop releases being scrubbed from the internet?
Only way to do that while minimizing the risk of scrubbing would be to expose it to as many people as possible in one big go, or to the right people who could then help spreading it. If enough people are spreading it faster than they can scrub it, it's out there.
> Can you write it to the blockchain?
Yup, might take multiple transactions depending on what you're releasing. Releasing the whole thing vs just enough equations to solve the problem.
> Can you write it to many?
Yes, should be able to write it to any blockchain that supports extra data. But you would still be limited to the amount of data pr. transaction you could store.
> Would they take out all crypto to scrub?
Most definitely. At least try.
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9a0997 No.7604
>>7584
Turns out there is a way to calculate these triangle bases in terms of n-1.
Continuing with the c21025 example, attached tree shows a path to each of the triangle bases in aa2T(u) starting from u = n-1 = 10367.
The u value for each subsequent node is calculated either as (previous u - 1)/2 for blue nodes or (previous u - 2)/3 for orange nodes.
The denominator for each formula when multipled up the tree will equal the corresponding "a" square multiple.
For example, from (863-1)/2 = 431, the denominators are 2 x 3 x 2 x 2 = 24.
It is also possible to calculate this u value of 431 directly from the starting n-1 as (10367 - 23)/24 = 431. Other examples are indicated by curved lines.
The formula used in this example to calculate the triangle base u value from n-1 is:
((n-1) - (m-1)) / m
where m is any factor of aa.
A more generalized formula, that requires further testing, is:
((n-1) - (n-1 % m)) / m
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e906b3 No.7631
>>1762
I've updated some of this long overdue D-grid thread I made before. I will try to post more stuff. Maybe I'll get the A grid in there too.
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52c782 No.7632
>>7604
>A more generalized formula, that requires further testing:
Attached pic is a test of the u mod formula to verify the triangle base calculations from valid known factors.
Starting from the third factor record for c1020100, (0,9216,481) = {0:9216:1010:960:50:20402}, each known factor represents a square multiple "a" value in aan(n-1).
The formula for each cell is:
( u - mod( u, p ) ) / p
where p is a valid factor of aa shown in the header row, and u is the triangle base starting position in the "u" column.
Green cells indicate an exact match to a valid u value.
Yellow cells indicate a mismatch by 1.
For the case of yellow cells, the correct formula would be: ( u - mod( u, p ) ) / p - 1, though not yet sure when that -1 rule applies.
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736f39 No.7633
In case yall missed it. Q just said there are aliens.
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a332a2 No.7634
>>7632
Almost looks like Sierpinski triangle.
>>7633
I want to believe!
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06d37a No.7636
You can do it.
Think lookup.
Think col 0.
Think col 1.
Think -1.
Golden.
Route of Dafide.
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06d37a No.7637
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06d37a No.7638
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08d201 No.7640
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08d201 No.7641
>>7636
Route of Da Fide…
Route of "Give Faith"
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e906b3 No.7642
>>7636
>Think col 0
>Think col 1
>Think -1
Column 0 and column 1 are specified here.
Think -1. He doesn't say column -1 so I'd say they mean row -1, which as all the negative squares in it. And if you notice from all these n=-1 entries there are little projections of them that map going from (-d^2, -1) to (-d^2 + 2, 0), (-d^2 + 4, 2), …. (-d^2 + 2i, i)
So maybe we could start from these -1 entries with the right d (or d and d+1 because our f record) and check something like in between these entries in the column. Obviously since one of d or d+1 will be odd and the other will be even, one will get cast into the col 0 and one will go into col -1. (I thought it was gonna be 1 so it would make more sense with this crumb but whatever) maybe this could be a path into using the center columns.
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e906b3 No.7643
So if you draw a square and a smaller square inside of it, you can tilt the inner square a bit and then it makes four identical right triangles around the edge. One length of the triangle is the base of the big square, (d+n), then from the corner it goes into the middle corner of the tilted one like pic related. These are all the squares for products where the factors differ by 2. Then there is 4 and 10. Now I don't know if this is completely related, but it is a completely unique shape to every product. Also it looks like they start off moving then sort of set into place but approach a limit. Idk. You could also see that this is just two rectangles also and you could pull this out and fractal it up
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c01508 No.7645
I have 2 good leads that are coming to me right now.
1. I was thinking about the asymmetry thing.
And 0.
And if 0 is a number.
And symmetry.
And vqc's euler? euler?.
And e.
And how the background temperature of the universe is e.
Instead of absolute 0.
Weird.
Cosmic Microwave Background they call it.
Or maybe they call it something else now.
Which is directly connected to free energy and white noise and the casimir effect and all that fun stuff.
A force that can be tapped into for seemingly infinite energy.
It's the noise that's left over when everything else is quiet.
It's the ghost in the machine.
The noise in the lines when there shouldn't be anything there.
We're getting rid if it like it's something bad, when we could be tapping into it.
A background frequency that's always there.
And where is it coming from?
Seemingly everywhere.
Propogating in all directions.
Or maybe you could call it directionless?
You can also think of it like a temperature.
So what is that temperature?
It happens to be unfucking believably close to the value of e.
The background temp of this universe is e. Instead of 0.
So my thinking is this - What if 0 isn't the center? What if the center of all numbers is e? How sexy does that feel with all this talk of asymmetry? Because to me it feels perfect.
2. Number 2 isn't as cool, but I'm wondering what patterns would be noticed in base e. Some anons were talking about changing base to notice patterns (9 and 2 were suggested). I think base e would be perfect if there is a change from 10.
And
3. Number 3 because I just noticed it. And this is gonna really blow your minds because I'm about to casually state one of the most profound things in mathematics, physics, spirituality, religion, philosophy and life in general:
e is God.
I'm gonna let you think about that for a second.
It probably won't connect until I give you the next piece:
phi is Satan.
phi is the law of the real. phi is the lord of this physical realm. WHO DO THEY WORSHIP? Why the SPIRALS? Covered in GOLD? What's it MEAN? (kek)
To put it in different terms:
e is the SPIRIT and phi is the ANIMAL
Also - here is a .pdf (yeah I know people don't usually like pdfs) that you guys probably understand more than me.
>Relationship between irrational constants Phi and e - viXra
Also Also -
>What is the connection between euler's number, phi (golden ratio) and pi (3.14…)???
<https://www.researchgate.net/post/What_is_the_connection_between_eulers_number_phi_golden_ratio_and_pi_314
>I've been struggling with this question for a while and I'm wondering if you may help me with any insight/references? My intuition tells me that it is growth that links these irrational numbers…growth from negative to positive matter. The link between these numbers may possibly lead to an explanation of the origin of matter, or 'substratum'. Any help….?
>An elegant solution to this question is:
phi = ei*pi/5 + e-i*pi/5
I give to you the key.
I know of no other nerds that are more perfect for this task.
Godspeed.
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c01508 No.7646
Sorry for the link, the < should have broken it I had thought.
I may never get credit for this, but a rare pepe would be appropriate. Nothing good on this machine.
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64f02a No.7658
>>7643
This is interesting, what would be the dimensions of 1st pic related?
>>7645
Also interesting. Im the anon that was sperging about fibonacci sequences earlier in the thread. I managed to get to the next set of patterns and I noticed something. First off the 1st pattern is 60 numbers long, the 2nd is 300, the 3rd is 1500, I had assumed the 4th would be 7500 but that was incorrect, it is 15,000 numbers long. Im working on getting to the next pattern but it seems that it too is a 10 fold increase, as ive gotten to the 75,000th iteration of the fibonacci sequence and it didnt repeat as predicted at 75,000.
Anyway. in an attempt to make sense of such large groups of numbers I tried organizing them all into columns of as long as the 2nd prior pattern. (so with the 1500 pattern I arranged it into columns of 60 and with the 15,000 pattern I arranged it into columns of 300 numbers. I then assigned each number a color. 2nd pic related is the 15,000 number long pattern and 3rd pic related is the 1,500 number long pattern. First off, wew that is quite the image that it creates. Second off, look at how similar they are. the 1,500 number long pattern looks like a much lower resolution version of the same image.
For a refresher on where I'm getting these patterns from, the first (60) pattern is the first digit spot in a fibonacci sequence, after 60 numbers the singles digit of said fibonacci sequence will repeat forever. The 300 number long pattern is the 10's digits, 1500 the hundreds and the 15,000 pattern is the thousands.
So far my theory that they will always repeat holds true.
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64f02a No.7659
>>7658
the first picture of the pattern is kinda hard to see the way it maximizes in browser, heres it zoomed out a bit and broken into two images (couldnt get the whole image into two screencaps its missing the center portion of it) for phone lurkers.
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08d201 No.7660
>>7658
Maybe this'll help?
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e906b3 No.7661
>>7660
>>7658
So the dimensions of this boils down to a quadratic equation. If you have the inner square as x+n and the outer as d+n, then d+n would be the hypotenuse. You would also see that the little extra length (on the far right and far left) would be the same distance, lets call it w for no specific reason.
This function screenshot returns the length and angle
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a332a2 No.7662
I'm not sure what I'm actually doing. I've been looking into the VQC thinking "lookup" table and I was playing around with x=d in column 0, row 1 for even d's. I noticed a pattern for BigN - n, but I don't know if it's anything usable. It doesn't solve anything, just a pattern.
Anyway, if you take an (e, n) and create the cell in (0, 1, x=d). The a will relate to the bigN - n, but not in any obvious way. What I've looked at is creating the smooth number BigN - n, then subtracting it from (0, 1, x=d).a, which gives another number. If you do this for a record, it will generate a new sequence of numbers that correspond to another column where a=(0, 1, x=d).a - (BigN - n).
Example:
a=7, b=37, c = 259, d = 16
BigN = 114,
n = 6
114 - 6 = 108
Column 0, row 1 x=d: {0:1:144:16:128:162}
128-108 = 20.
The next record in the sequence of (3, 6) is a=37, b=91, c=3367, d=58. This has:
BigN = 1626
n = 6
1626 - 6 = 1620
{0:1:1740:58:1682:1800}
1682 - 1620 = 62.
These two numbers will exist in (15, 6): {15:6:35:15:20:62}.
This appears to hold true for all records.
At first glance it could look like it is (e + 2n, n), but it isn't true for all the records. If you look at the other sequence in (3, 6) you will have one record that has a=19, b=61. So it could also look like 19 + 1, 61 + 1, but again, that doesn't work for all the records. Sometimes it is an increase of something else.
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a332a2 No.7663
Another interesting thing regarding n's and shadow n's and smooth numbers.
Create both Big N's for a record (Normal and shadow). Subtract the smooth number from these two. It will give you a new set of numbers that will exist in (e, k) (for some k) where x + n = d.
Example:
a=7, b=37, c=259, d=16
BigN = 114, shadow BigN = 146
Smooth number (114 - 6) = 108.
114 - 108 = 6, 146 - 108 = 38
{3:7:15:9:6:38}
7+9 = 16
Another example:
a=61, b=131, n=7, d=89, e=70
BigN = 3907, shadow BigN=4085
Smooth number (3907-7) = 3900
3907 - 3900 = 7
4085 - 3900 = 185
{70:61:35:28:7:185}
61 + 28 = 89.
Meaning there is a record in column e, where d = x+n that is related to our own record. In this record the d is equal to x+n.
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a332a2 No.7664
>>7663
> Meaning there is a record in column e, where d = x+n that is related to our own record. In this record the d is equal to x+n.
Just to clarify a bit
Here a is the n for the record a=61, b=131 while 185 is the shadow record. That means a=n, b=shadow n exists in (e, a).
These two records are the "normal" record and the shadow record:
{70:7:89:28:61:131}
{70:185:-89:-150:61:131}
This is the record that is created when you remove the smooth number from BigN and shadow BigN, it is also the record where a=n and b=shadow n.
{70:61:35:28:7:185}
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a332a2 No.7665
>>7664
Just one last quick note, a=7, b=185 will also exist in n=131 (b-value), but here it will be negative.
{70:131:-35:-42:7:185}
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a332a2 No.7666
In some cases the a's in column e will match the parity of x in column -f. If you highlight, or "mark" these columns in -f you will see that they are spaced out by 2n - 1 (The difference between square n^2 - (n-1)^2 for some n). For smaller values, if you highlight the center between these marked cells (since 2n-1 is odd there will be a center piece) it tends to contain our factors. Might just be a side-effect of using smaller numbers, though.
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a332a2 No.7670
>>7666
The product of the two n's (normal n and shadow n) for our cell (a, b) is equal to (x + n)^2 + e.
Given the cell in column e for (a, b) for some a and b. This cell will have n, but there will also exist a shadow n' for it. The product of n and n' is equal to the (a, b)s cells (x + n)^2 + e.
The cell in column e for (n, n') will have x + n = d from the cell of (a, b).
This gives us: n * n' = (x + n)^2 + e
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a332a2 No.7671
>>7670
So knowing nn' = (x + n)^2 + e
We get
nn' = xx + 2xn + nn + e
We know xx + e = 2an
nn' = 2xn + nn + 2an
nn' = n(2x + 2a + n)
n' = 2x + 2a + n
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a332a2 No.7672
>>7671
This also means:
n' = 2(a + x) + n
n' = 2d + n
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a332a2 No.7673
>>7670
To build a bit on this.
We have the cell: {3:6:58:21:37:91}
This has another cell with negative d (shadow cell?): {3:122:-58:-95:37:91}
The records for where a=6, b=122 is:
{3:37:27:21:6:122}
{3:91:-27:-33:6:122}
In these cells, x + n = 58 (d). The product of these records are 732 (6 * 122) which is (x + n)^2 + e (6 + 21)^2 + 3. Meanwhile those records will have (x + n) as d. (6 + 21 = 27).
This means we can find the triangular base for the shadow cells, but we'll only know e and x+n.
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a332a2 No.7674
I'm not entirely sure what to call this, but I noticed something when I was looking at 4c. In my head I've been thinking of it as a kind of record transformation. Take a record and multiply it by 4 (you can think of it as by dividing the 4 into 2 by 2, multiplying a by 2 and b by 2.).
The reason I'm calling it a transformation is because we skew or transform the patterns in (e, 1). Take a column e and row 1. Multiply it by 4. In (4e, 4) you will find the same values, but this time the patterns are skewed, as in, the sequences / chains within this record no longer matches the ones we have in (e, 1).
Example, look at (3, 1). Multiply it by 4 and we get (12, 4).
{3:1:3:1:2:6} {12:4:4:2:2:14}
{3:1:9:3:6:14} {12:4:12:6:6:26}
{3:1:19:5:14:26} {12:4:24:10:14:42}
{3:1:33:7:26:42} {12:4:40:14:26:62}
{3:1:51:9:42:62} {12:4:60:18:42:86}
{3:1:73:11:62:86} {12:4:84:22:62:114}
{3:1:99:13:86:114} {12:4:112:26:86:146}
{3:1:129:15:114:146} {12:4:144:30:114:182}
{3:1:163:17:146:182} {12:4:180:34:146:222}
{3:1:201:19:182:222} {12:4:220:38:182:266}
{3:1:243:21:222:266} {12:4:264:42:222:314}
Here we can see the sequences have changed. In (3, 1, 1) we have a=2, b=6 followed by (3, 1, 2) where a=6, b=14. This forms a single chain / sequence within (3, 1). In (12, 4) however, we can see that we now have transformed (3, 1) into two sequences. If we were to multiply (12, 4) again by 4 the pattern would repeat (transforming yet again).
If we were to multiply (e, 1) by n^2 (For example n=6 where a=7, b=37 exists) we would end up with a record that connects / pairs all the an, bn transformations in (e, 6).
{3:6:16:9:7:37} => an = 42, bn = 222 => {108:36:96:54:42:222}
{3:6:58:21:37:91} => an = 222, bn = 546 => {108:36:348:126:222:546}
{3:6:124:33:91:169} an = 546, bn = 1014 => {108:36:744:198:546:1014}
I haven't looked enough into this yet, there's so many damn patterns to try and understand, but I suspect that multiplying any (e, 1) with a square will result in (e * k^2, k^2) records to "skew" the original record, generating multiple sequences.
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a332a2 No.7675
>>7674
Forgot to mention, 108 is also the smooth number belonging to record a=7, b=37, c=259. We have n = 6 and BigN = 114 with 114 - 6 = 108. I believe I've seen patterns before, but forgot to write up about the smooth numbers within a column running in a pattern similar to the other patterns we're seeing.
Also, note that there are two sequences in (3, 6) and both of them are "paired" in (108, 36). Example a=1, b=19, n=6. This gives an = 6, bn = 114, which is at (108, 36, 2).
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a47a6b No.7676
Did a bunch of posts get deleted?
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e906b3 No.7677
>>7676
check the grid patterns board. I thought this once too
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d4c3ed No.7678
>>7677
I think it was because of the changes to the board last night. I suddenly couldn't see all my latest posts, the last post was yours >>7661
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d4c3ed No.7684
I've been looking a bit more into movements. One thing I noticed was cells at (e + nn, n). In those records the d-values will be equal to the a-values in (e, n) and the x-values will be equal to (x + n) while the a's (and thus b's) will be equal to (d + n) values.
Since we don't know n, this isn't usable, but I've been thinking more about the fractal nature VQC has been talking about. I'm wondering if we should be trying to draw some figures that attempt to match up with what our fractal actually looks likes.
Having said that, anyone actually looked into the mandlebrot set? I was reading up on it (Only quickly between breaks), but the damn thing is screaming VQC. f_c(z) = z^2 + c. That's pretty much what we're doing, except instead of dealing with complex numbers or float numbers we're working with integers.
I've been thinking that we will either extend the grid (to support float / complex numbers) or there is a deeper connection (Is our fractal of mandlebrot nature?).
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d4c3ed No.7685
So we have (e * nn, n) and (e + nn, n), both of which extend or transform the records in some way.
(e * nn, n) will extend the sequence (Essentially increase the number of sequences in our row) while (e + nn, n) will transform / modify our row by exchanging values of d, x and a with other values. I'm going to try and think more about what this actually means and try and formulate a better description of what actually is happening. I'm struggling a bit with visualizing what I'm doing, so I would like to try and step back a bit to get a better understanding of why (e + nn, n) and (e * nn, n) works the way they do.
As for (e * nn, n) we will see repetitions of (1, 1) in every (nn, n) as 1 * nn = nn, meanwhile this explains the repetitions of (0, 1) in (0, nn).
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d4c3ed No.7686
>>7684
Regarding the fractal / mandelbrot idea I'm wondering if that's actually our fractal. As in, (0, 1) or (1, 1) is the "base" mandelbrot set, as in the grand picture and why VQC has been telling us to use bigger numbers. Because by doing so we'll start seeing the repeated numbers.
I'm just spitballing, so I'm probably way off, but I can't help but shake the fact that our equation for c (d^2 + e) is exactly the same as the one for mandelbrot (z^2 + c).
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d4c3ed No.7687
>>7686
In the mandelbrot set you have a term called "escaping" in which a point that is "escaping" means that the point is not within the boundaries of the fractal (If I grasped it correctly). I wonder if prime numbers are the ones that "escape".
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64f02a No.7688
Fibonacci autist here.. This might be really fucking important. I need to test something. I need a couple of large sub-prime numbers that I do not know the factors to. Fucking pronto.
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412edb No.7689
>>7688
>guys I totally actually figured it out this time, spoonfeed me
My ass. 22222268356688393753
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