Anonymous ID: 64f02a Aug. 31, 2018, 8:37 p.m. No.7440   🗄️.is 🔗kun   >>7441 >>7442 >>7444 >>7445 >>7503

Hey all. I stopped keeping up with this research a while ago since I couldn't keep up with the programming aspect, but I've been browsing sporadically. So, I broke off on my own efforts autistically looking for patterns that may be helpful. So far I've identified a couple of patterns within fibonacci sequences that might be relevant. Bare with me, this is extremely autistic…

 

Im working on an infograph at the moment, but for now here's a rundown of it.

 

The first (ones) digits in the numbers of a fib sequence make up a pattern of 60 numbers which repeats ad infinitum. Within this pattern there are subpatterns: every 5th and 10th number is a 5, every 15th a 0. Every 3rd number is even. The second half (#'s 31-60) is opposite the first half, i.e. the 1st and 31st add up to 10.

 

If you break the pattern of 60 into sets of 15 more patterns emerge, 1, 7, 9, 3 and 2, 4, 8, 6. As well as a helical sort of distribution throughout the set of 60 numbers.

 

For the time being that is all that needs to be said about the first digit pattern.

 

All digit columns subsequent of the first digit pattern also have patterns which they follow. Each pattern being 5 times larger than the pattern before it. The first digit pattern is 60 numbers long, the second digit pattern is 300. Every digit in a fibonacci sequence always has a repetitious pattern. And, these patterns appear to follow rules established by the first digit pattern.

 

If you break up any of the subsequent patterns into subsets Nx numbers long where x = the length of the previous pattern, you can establish at least one that I've proven thus far rule which dictates that pattern. The second (10's) digit pattern being 300 numbers long, and the previous pattern being 60, break this pattern up into 5 columns of 60 and the rule for this pattern becomes obvious. At first glance the numbers appear highly chaotic, but surprisingly follows a very simple rule. It follows the rule of "2, 4, 8, 6" in reverse. i.e. {6,8,4,2} the 1st + 6 = the 61st, the 4th + 2 = the 64th, ad infinitum.

 

The next (100's) digit follows the same rule, but this time the sequence is shifted two spaces so that it is {4, 2, 6, 8}

 

I'm going to cut myself off for now. If you guys find this interesting and relevant let me know and I will keep going.

Anonymous ID: 64f02a Aug. 31, 2018, 10:01 p.m. No.7443   🗄️.is 🔗kun   >>7448

>>7441

>why only the first digits

Because the first digits establish a set of solid patterns and I am operating under the hypothesis that all subsequent digits amount to an abstraction of the first.

 

>>7442

Check out pics related. Like I said I am not programmer and I started running into problems with floating integers fucking up the numbers so I've been calculating everything manually. I have the first 7500 iterations of a fibonacci sequence starting with 1 calculated.

 

First pic is the infograph that Im working on. I just slapped it together so apologies if its hard to follow but this should help to visualize what I am talking about. It gets a bit difficult (as you can see with the 3rd pattern getting chopped off on the bottom cuz its huge.

 

I also made a gif to show the juxtaposition of the numbers in the first pattern. In the gif the pattern is arranged in order but it is playing through to highlight the opposite numbers and their relationship, i.e. rather than sequential order it is 1 followed by 9, 2 followed by 8, and then these numbers paired together.

 

I'll have to continue this either in a little bit or tomorrow I pushed my computer a little to hard tonight with this shit and its not playing ball right now.

Anonymous ID: 64f02a Sept. 1, 2018, 4:32 p.m. No.7447   🗄️.is 🔗kun

>>7444

>>7445

 

Actually, at the very beginning I started off looking at it in base 9 because way back when we were still trying to figure out what vqc was talking about several anons were wondering if 0 was an actual number. There were patterns but it didnt start 'making sense' until I went back to base 10.

 

>Maybe we could use the fact that binary is doubling…

That is the assumption that I am running on, that if all the rules of this pattern can be identified we would be able to use it as an abstract means of simplifying very large numbers.

 

I've ran into a problem with my theory. The 2nd pattern is 5 times the first and the third 5 times the 2nd but the 4th is not 5 times the 3rd. I made the assumption that the patterns always being 5 times larger without paying attention to the pattern of 5, 5, 0 in the first pattern. So, I have a feeling that the 4th pattern will be 10 times the 3rd. Good news is Im already at 7500, bad news is I need to get to 1500. My computer might explode.

Anonymous ID: 64f02a Sept. 2, 2018, 4:07 p.m. No.7449   🗄️.is 🔗kun   >>7450

>>7448

Interesting. Im giving it some attention right now.

>e is x, n is y

Like I said I haven't been keeping tabs here in a while which thread should I browse to get a refresher on the vqc formulas?

Anonymous ID: 64f02a Sept. 2, 2018, 6:39 p.m. No.7451   🗄️.is 🔗kun   >>7452 >>7453 >>7455

>>7450

>You understand that?

insert us navy seal copypasta

 

I both do and don't understand it. I'd need to be able to play around with it to really get an appreciation for it but aside from getting numbers to print in console I have no idea how to code to get any sort of graphic output. I've been doing all my work in excel and mspaint, jej Im pretty decent with python but all documentation I've found on it is in redditese so I've never been able to quite figure out how to open a window and draw to screen.

 

Anyway, as for >>7448 it appears all odd y have a pattern x=y long and all even have a pattern x=y*2 long

Anonymous ID: 64f02a Sept. 3, 2018, 7:13 a.m. No.7454   🗄️.is 🔗kun

>>7452

It pretty much is isnt it . The odd y are growing by 2 and the even y by 4. I remember way back in like december anons were associating Lorentz transforms with vqc

Anonymous ID: 64f02a Sept. 3, 2018, 3:55 p.m. No.7459   🗄️.is 🔗kun   >>7460

>>7458

Is the missing data the result of floating integer rounding? If that is the case I got around that with my fibonacci computations by taking a couple extra steps to create variables and booleans in order to add each digit in an equation individually and carry over the one if needed (like you would doing the math in your head)

 

>can you do it with missing data

Maybe. I can certainly try. Hopefully I will have time tonight to sit down and go over the equations, until I understand what I am looking at Im just playing hyper autistic connect the dots.

Anonymous ID: 64f02a Sept. 3, 2018, 6:44 p.m. No.7461   🗄️.is 🔗kun

>>7460

>for example

I knew that.. But for some reason I needed you to say that exactly in that way. Give me a day or two and I should be able to contribute much better.

Anonymous ID: 64f02a Sept. 4, 2018, 7:38 a.m. No.7463   🗄️.is 🔗kun   >>7464

>>7460

I got two questions if you don't mind. For finding c are we limiting to the squares of integers? And for a,b is a the smallest factor or any factor smaller than b and vice verse? I can work with flash so Im writing something up.

Anonymous ID: 64f02a Sept. 4, 2018, 8:04 a.m. No.7465   🗄️.is 🔗kun

>>7464

>no fractions

Is there a reason for this or just how its being done for the moment for simplicities sake?

>floor(sqrt(c))

wouldnt d=floor(sqrt(c)) and c = (xx)-(yy) ?

Anonymous ID: 64f02a Sept. 22, 2018, 7:25 a.m. No.7658   🗄️.is 🔗kun   >>7659 >>7660 >>7661

>>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.

Anonymous ID: 64f02a Sept. 22, 2018, 7:41 a.m. No.7659   🗄️.is 🔗kun

>>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.

Anonymous ID: 64f02a Sept. 28, 2018, 8:02 p.m. No.7688   🗄️.is 🔗kun   >>7689

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.