/vqc/ - Virtual Quantum Computer

I figured we could go with this one since it has the pastebin link dump and all the other introductory information, and because >>6746. Hopefully you all found your way here.

>>6746 (pb&j)

(That's Chris sans trips in case anyone didn't notice.)

ISP here, but no trip for now. I'm stocked with this stuff, but I'm also traveling for a few days so I have no means of participation. For safe keeping if someone archives or as new stuff and shares it (bunker? discord?) In the event that we get shut down I'll be able to review in a couple of days. Either way godspeed anons and as always Hello VQC! Always a pleasure to see you.

Currently working on proper verification of the latest VQC post. Just reminding myself of how BigInteger works.

>>6746

I'm happy for you to wait for someone else to confirm as well (not to mention I'm currently in my bedroom while the other people who live here are having a party I'll have to show my face at eventually), but this seems far more important than not drinking around a bunch of drunk people.

>Next, we show how to calculate BigN of an odd number.

>All odd numbers are the difference of two squares.

>Every odd number is the difference of two consecutive squares.

1 = 1-0, 3 = 4-1, 5 = 9-4, etc.

>The product of two primes, is the difference two sets of squares.

Semiprimes are all odd since odd*odd, so they fit that pattern.

>The value of n for the product of 1 and c is defined now as BigN for odd numbers.

>Since d-a=x and x+n is the smaller square in the difference of two squares, then for odd numbers, BigN is ((c-1)/2)-x

>x is d-1, since a=1

This is fairly straight forward (for me, anyway). Algebra.

>Anyone want to show:

>RSA 2048

>c=RSA 2048

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

>d=

I copied the C# code to Java so hopefully I did it all accurately. Obviously you'll be able to confirm that this is all correct.

floor(sqrt(above big number)) = 158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

>e=

c - dd = 149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

>BigN=

x is d-1 since a=1, so x = 158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807843

n = ((c-1)/2)-x = 12597954237828946747013591620024199285714641063102016013888568918021831010353797778132009262940392203459145320624757541094649279574588092251404244560036422496343696403643888367985709173635130948187507485912345582538806689929547850048665229874404214200898714550321229345908597559373060757586327316141108434993615859020166427425368088424069958780819853230147577508164643339364712441831390650677628862077837124697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335

>Remember BigN is the row in column e, where we always find (e,n,d,x,1,c)

So (1,c) for RSA2048 is at cell (149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021, 12597954237828946747013591620024199285714641063102016013888568918021831010353797778132009262940392203459145320624757541094649279574588092251404244560036422496343696403643888367985709173635130948187507485912345582538806689929547850048665229874404214200898714550321229345908597559373060757586327316141108434993615859020166427425368088424069958780819853230147577508164643339364712441831390650677628862077837124697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335).

>>6754

Oh for sure. Although I'm not sure that I'd tell a bunch of normalfags that an anonymous stranger on the internet has been trying to teach me and some others integer factorization for the last 7 months. I think I recognize that picture. Is it from a painting of a UFO? Or is it something completely different?

>>6753

Thanks AA.

I can verify once I'm at a desktop.

Now, quick reminder on triangular numbers and the construction of the grid.

Column 0.

The values of a[t] at 0,0 are twice the square numbers.

The values of d[t] at 0,0 are 4 multiples by the triangular numbers. All other cells at row 1 can be constructed from these values by adding to them or subtracting from them.

Confirming trip.

>>6760

I understand the last post you had. It's the same as PMA's calculation for the next record but worded a bit differently. Also if you go into negative x then back into positive you can get the record in between c and c'. Could be irrelevant though

>>6763

New trip also why not new thread

>>6764

>>6753

Wrote some python for this before noticing the full bread or reading your post. Got the same results as you:

c =

25195908475657893494027183240048398571429282126204032027777137836043662020707595556264018525880784406918290641249515082189298559149176184502808489120072844992687392807287776735971418347270261896375014971824691165077613379859095700097330459748808428401797429100642458691817195118746121515172654632282216869987549182422433637259085141865462043576798423387184774447920739934236584823824281198163815010674810451660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

d =

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

e =

149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

BigN =

12597954237828946747013591620024199285714641063102016013888568918021831010353797778132009262940392203459145320624757541094649279574588092251404244560036422496343696403643888367985709173635130948187507485912345582538806689929547850048665229874404214200898714550321229345908597559373060757586327316141108434993615859020166427425368088424069958780819853230147577508164643339364712441831390650677628862077837124697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335

Now I need to go get some coffee…

>>6759

Guess I'll keep doing the deconstruction thing if nobody else is.

>Column 0.

>The values of a[t] at 0,0 are twice the square numbers.

Not sure I understand this one, actually. So here's the first 7 records of (0,0).

{0:0:1:0:1:1}, t=1, f=-3

{0:0:2:0:2:2}, t=1, f=-5

{0:0:3:0:3:3}, t=1, f=-7

{0:0:4:0:4:4}, t=1, f=-9

{0:0:5:0:5:5}, t=1, f=-11

{0:0:6:0:6:6}, t=1, f=-13

{0:0:7:0:7:7}, t=1, f=-15

>The values of a[t] at 0,0

t=1 for all of these records. That means there aren't multiple t values, so a[t] is always a[1], even though there are multiple values of it.

>are twice the square numbers

a and b are equal for all of these, but they just increment: 1, 2, 3, 4, 5, 6, 7. So, I mean, if you're referring to the fact that every square multiplied by 2 will come up in this infinite set of a values, then sure, but I'm not sure if that's what you're actually talking about.

>The values of d[t] at 0,0 are 4 multiples by the triangular numbers.

d also increments by 1 upwards from 1 like a and b. It would also contain every instance of a number being equal to a triangle number multiplied by 4. Is that what you mean? I can't tell.

>All other cells at row 1 can be constructed from these values by adding to them or subtracting from them.

I do remember that being the case but I don't remember the specific rules. Does this have to do with the a[t] and d[t] values?

You might have to verify this one before moving on maybe.

>>6766

Thanks for double checking. I did it in a bit of a hurry so I did wonder if it was completely accurate.

>>6757

Man, that is not what I expected. Not just the picture too; I mean, I know you all as text on a screen so seeing actual pictures with one of you on the other end of the camera and a weird story related to it is always strange.

See you all in about 8 hours.

>>6695

Thanks baker!

>>6736

Morning VQC.

The formulas posted work to construct infinite elements in a chain and where n=1 will generate all records.

For n>1, there will be gaps.

This is equivalent to creating elements by ent where t=t+n.

>>6759

>>6767

These d[t] and a[t] rules are for (0,1).

>>6726

Thanks! Its great to see you posting more often, and I really appreciate the guidance.

I'm personally very interested in your path VQC. I have so many questions - may I ask a few?

Based on the date of the youtube post, you obviously discovered this years ago - what happened when you first found this solution?

Were you working in university research? NZ, I assume, at the time? How did your peers and superiors respond?

I imagine many people would go very far to get their hands on this. Were you safe, or at risk?

I'm really very curious about your story.

I understand that many of these questions you may not be able to answer. But thank you!

>>6728

I've read your post GA, still trying to process it into what it means. I'll definitely continue to think on it.

Sorry that I have nothing to add on this at the moment, great post though :)

>>6723

I couldn't imagine this community without you Topol.

Thanks for defending all of us.

>>6730

Thank you for your thoughts. I've been here in spirit! But I too missed being part of this math family, I missed you all!

>>6771

Honestly, if I knew someone who did this…

I'd hole 'em up in an embassy with 24/7 security for like… 7 years.

>>6772

Thats where I went with it too.

I actually share a homeland with someone who has been in exactly that situation…

I guess my next thought is, of all the people in the world, who knows this algorithm? Just 1 person? Any governments? Or NSA/military?

What do you think Topol?

Tomorrow we will go through where the product of BigN and c are found in (e,1) and where the other value of c is in (e,1) and what information that gives us.

Then we will look at the key that is made by column -f with the locations of c in (-f,1) and how to find x for an and a(n-1).

Then we build the algorithm.

Then one of you anons can edit the Wikipedia page with the remaining RSA numbers and their new factors.

>>6773

Well… people in business… government… Intel…

But… Intel… or Cicada? ;)

>>6771

This was all done alone.

From scratch.

I realised that if I learned to build my own software for multiplying large numbers together and trying new views of number properties, I could see something where information was lost.

The first truly new idea came out of reading Revelation one afternoon in 2011.

The red text in the first chapter looked like description of odd (prime) numbers in binary, since "I" is the first and last number of all binary odd numbers.

I think binary is not new.

It looked like a coded algorithm using allegories to keep it secret.

The first version of the grid, looks very different, was in binary.

>>6774

1 2 3 NOT IT!

>>6769

Sorry, yes, they are derived from 0,0 but are in 0,1.

Good catch.

A good reason to have checks and balances.

>>6774

I can't wait. Thank you again for sharing your knowledge with us.

>>6777

Don't ever change Top :)

>>6779

In the scheme of things, given the "coincidences", I think this was always meant for you guys.

Just appreciate being given the chance to share.

Thank you.

>>6774

Appreciate you doing this walk through, although a bit disappointed that we haven’t yet solved it.

So much still to discover.

>>6783

I agree, but I have a feeling that we're just getting started. Integer factorization doesn't prove P = NP. It will prove P = BQP and thus prove that quantum computers can be simulated on a classical machine (self verifying the VQC). So we still have work to do.

I mean, we got Mandelbrot set as phase 2, right?

>>6785

Definitely just getting started.

>>6769

Hey PMA! The d[t] pattern is found in (0,1), but the a[t] pattern repeats!

Each n = a perfect square ((0,1), (0,4), (0,9)…), a[t] is 2 * a square.

>>6774

Looking forward to the process. Thank you.

>>6775

I've thought about this so much. If other people know, how many and how is it kept secret? Military organizations are structured in a way that embeds secrecy, so I can understand those groups, as I can secret organizations. But in the business sector, I really have no idea.

But if John knew (>>6776), then it's likely others have known over time.

>>6777

I wonder if we can anonymously update the wikipedia page via vpn's or tor or a public wifi access point. A starbucks in a major city would be pretty anonymous.

>>6776

Thanks for sharing this.

I'm really curious about the religious connections, but I'm very ignorant about religion unfortunately. Your quotes from Revelations took me down a research path, one that I intend to continue. One of the things I found interesting was the etymology of apocalypse - "to uncover".

Do you have any info you could share on following your interpretation of binary in Revelations?

>>6787

Thanks, Teach.

That repeating a[t] pattern for perfect squares is interesting.

Also appears that difference in x values between those records is 2*(sqrt(n)).

>>6787

Hey Teach, do you use Discord by any chance?

>>6787

Also, the a[t] pattern repeats where n=a[t] from (e,1).

(e,1) a[t] = 2, 8, 18, 32, 50, etc.

((0,2), (0,8), (0,18), (0,32)…), a[t] are the perfect squares 1,4,9,16…

>>6789

I love it.

>>6791

I don't at the moment, should I? Am I missing much over there?

>>6790

Yep, good find.

I think this has to do with another link, I'll try my best to describe it:

* in (0, 1) we know a[t] = t*t*2.

* we also know how to "move" up and down (0,1) whereby t=p+t (I think thats the equation… please correct me if i'm off-by-one)

* This means that if we try to find all the a[t] that are divisible by 3, we can simply list:

t = 3, 6, 9, 12, etc.

* This means that at (0,9) for example, we'll see the x[t] values are equal to the x[t] values in (0,1) at t = 3, 6, 9, etc.

>>6792

I'm trying to follow this one, but i'm missing something, can you give an example or something PMA?

>>6793

It's more of a speed thing.

We're able to spitball over there and that's how we started making progress again after the lull.

Also, there's less pressure to come up with a post that goes far enough to be "worth adding to the bread yet".

Super useful for getting on the same page and I make sure everyone remembers to post findings on the board.

Simple lock and key mechanism to get into the server:

Post something (UNIQUE) in two parts.

Post one half with your trip here, and then send the other half to Ta.io in a private message (server owner):

https://discord.gg/pQDaMkU

>>6795

What type of unique thing?

keys? signature?

>>6796

Any type of iFagFactor screenshot that wasn't posted here before will do. Or a whiteboard drawing!

>>6796

Could be a obscure ass picture, an output style that only you do, specific something where the reference doesn't make sense until you send it to me… is whatever.

I'm just making sure you hit the correct points along a curve.

>>6797

lol, ifagfactor ;) I'm going to change the name.

>>6798

gotcha, thanks guys!

>>6801

>interchangeable in certain circumstances.

*all circumstances.

>>6803

O_o coulda just cut that one in half… lol

>>6800

I'd been wondering who BO 'round here was…

>>6804

>O_o coulda just cut that one in half… lol

oops

Ok, so I was investigating d[t] at (0,n) and its relationship to triangle numbers…

so, lets say we're at n = 15, what are the set of d[t]?

which rows are in (0,1) have 15 as a factor?

start with t = 16:

t=16: a = 450 = 2*15*15 x = 30 = 2*15 d = ( 15 +1)( 15)*2

t=31: a = 1800 = 4*2*15*15 x = 60 = 2*2*15 d = (2*15 +1)(2*15)*2

t=46: a = 4050 = 9*2*15*15 x = 90 = 3*2*15 d = (3*15 +1)(3*15)*2

t=51: a = 7200 = 16*2*15*15 x = 120 = 4*2*15 d = (4*15 +1)(4*15)*2

move to (0,15):

t= 1: a = 30 = 2*15 x = 30 = 2*15 d = 2*2*15 *1

t= 2: a = 120 = 4*2*15 x = 60 = 2*2*15 d = 2*2*15 *3

t= 3: a = 270 = 9*2*15 x = 90 = 3*2*15 d = 2*2*15 *6

t= 4: a = 480 = 16*2*15 x = 120 = 4*2*15 d = 2*2*15 *10

What this is all saying is, for odd n, d[t] = n*4*T(t)

1:8 - "I am Alpha and Omega", says the Lord God, "who is and who was and who is coming, the Almighty."

Imagination is the Beginning and the End.

Er… let me rephrase that.

The Beginning, the Infinite Now, and the Continuance.

Imagine that.

<3

~The Phither

I'm thinking this may be where the recursive algorithm starts to come in? With an 'e' over 300 digits in length, it seems like a few more rounds of squaring and storing results might be called for. It's so satisfying to watch that number get chopped down to something comprehendable, even if it's just play-acting.

>>6812

Did you see >>6774 ? VQC is taking us through an interactive example of how to use the grid. Today is the day.

>>6812

Forgot trip.

VQC, I don't think I can edit the article as I'm basically barebacking. But don't you want credit for this? And if so, how should you be given it? Twitter handle?

I'm sure it will all come in time. As magnanimous as you are, though, do you think it might be a good idea to claim credit?

Consider how many brilliant inventors there are that most people have never heard of–maybe you're comfortable knowing that God sees your good acts, and J know this to be true and reward enough. But there is something to be said about the hijacking of the message; we have no idea how Doug Engelbart saw things, but we know exactly what the men who pirated his ideas think–they're still venerated to this day.

Then again, maybe that has more to fo with corruption than anything. I can't imagine there will be much joy from the bad actors in Europe once we fly their dirty laundry.

I hope this is not what you mean when you speak of sacrificing yourself to this. I really hope you're just referencing the money you stand to lose by sharing the idea rathing than capitalizing on it…

>>6814

Dearest Fagwit:

STFU.

I'm putting this where it needs to be.

I'll deal with the media.

I'm not looking to famefag but I got this becuase…

Torture me all they like, I don't know aaaanything of value.

I just say Trangles funny.

;)

I still want Tucker, but that's quite the request.

Especially since his Hair could do the whole interview.

>>6814

His whole point has been to make this about the information being public. It's the message, not the messenger. You really should read through all the old threads if you haven't.

>>6814

Hi, thanks for the question. No I don't want any personal credit for this. I have a family, a career and mild Aspergers. I'm also a recovering alcoholic/addict. Fame, infamy or money would only end in a quick death pursuing women, drugs and alcohol into the gates of death. It wouldnt be my intention, but that is exactly where it would lead after just one drink. That's all it would take. But it is valuable to demonstrate that there is a whole world of maths and engineering that is waiting and using mathematical objects to solve problems in the virtual and real world in this way could open up a new dimension, so to speak.

Other anons can make the billions but I don't have any needs except the thrill of finding no problems to solve and helping the group effort to make the world better where I can.

The only way forward for me is to be helpful and share ideas.

If it sounds altruistic, it's closer to self preservation.

Maybe with a few years of recovery under my belt?

Prolly not.

Now, my computer is frozen (the new one) as soon as it's up and running, I'll write the e column definitions with RSA 100 and 2048, to show how easily everything scales up and to give examples for knowns and unknowns.

>>6817

*new problems

Please also anons, choose caution as the temptation to use this maths for hacking is huge.

Bitcoin has been talked about before.

Bear in mind that this maths can lead to shortcutting the mining process before it can be used even to do anything else.

Second laptop is not functioning.

Before I attribute it to malice, I will let it run to apply background updates.

Hasn't been switched on in a while.

Do not worry.

I have this week off work too.

I will keep you updated.

>>6819

Some thoughts on all this. Of course everyone is thinking about crypto-currencies with this process. They are a HUGE target and will be destroyed but, IMO they are a small portion of where this will go. Of primary importance is the idea that this creation cannot be lost. I have been backing up the breads here to the best of my ability. So many technologies have been suppressed and buried over the years. Much of the troubles we have in the world are related to that fact. It is frustrating and difficult to accept but I am pretty much convinced this is true. I fear that going after BTC as an early example of this will cause a colossal shit-storm. I haven’t mentioned much about cryptos since coming here but I have been following them since around 2010. I managed to mine a tiny bit in the early days. Certainly not enough to be called wealth but it was something. Most importantly I learned a LOT and found an accepting community. Everyone will want to go for the Satoshi wallets… or simply mine the rest of the block chain in a single day or something like that. I caution against this action for the following reason.

Those will be high-profile events. Especially the mining. If anyone here managed to crank out the code to pre-mine BTC someone will notice. Instead, simply identify and crack the private keys on “lost” wallets. What I mean about lost wallets is this. There are tons of early BTC wallets whose private keys have been lost. Many of these wallets are vast by today’s standards. Not a lot of people are looking at them. We could all do this for YEARS with no one noticing. There are block chain forensic firms that WILL notice but it will be nothing but a curiosity to them probably for YEARS because of the un-believability of the cracking of this encryption. They will write it off as a super-hodler finally opening that wallet. The reason I say this is because I hope a lot of good will come from this as opposed to it getting buried like all those other technologies. Topol, I think you would be a good face to speak to the media about this but, honestly, they won’t care. They certainly won’t want to hear a REAL story like the one coming from this board and report it as such. It won’t happen. Instead they will try HARD to bury this and bury it DEEP (probably all of you along with it) and probably in the worst way.

I do not have the skills to actually use this tool in its current form but if I did or (if one of you helped me to do it) I would use much of what I gathered cracking lost wallets to try to release buried technologies and to push for Full Disclosure. I would support a man named Steven Grier and do so in a huge way. I want to live in a world with advanced energy technologies at the very least. This is a noble goal in my opinion and one that is worth fighting for. It is the one I will fight for if I have the means to do so. I encourage you to all find something similar to actually help build a new world with this tool. I, again, caution you all deeply to consider carefully before approaching ANY mainstream news or government organization with this in any form at all. Always remember that their goal will be to use the vast (and terrible) resources at their disposal to destroy and hide this VQC tool. If it is given to them FIRST the THEY WILL CONTROL IT. Instead I propose first, each of us set ourselves up so none of us have to waste time at menial jobs and such and can live in some degree of comfort then BUILD STUFF OF VALUE with this thing and give that away for free, essentially under a GNU license or some such thing. Let the world figure it out that way as opposed to pursuing an easily suppressible fame-fest where this tool will be captured, hidden, and then used for nefarious purposes. The media and the government are not our friends, they are the greatest enemy. They WILL use this tool to do harm as opposed to good.

>>6820

Thanks VQC.

>>6819

>>6821

Well said Hobo. I agree with you.

I'm glad we're talking about this, I'd love to continue the conversation further about how to use this for good, and not cause total destruction in the process.

I've been following the NIST post-quantum crypto RFP: https://csrc.nist.gov/Projects/Post-Quantum-Cryptography/Round-1-Submissions

and: https://en.wikipedia.org/wiki/Post-Quantum_Cryptography_Standardization

My fear is that all the world's data will be hackable until we have better crypto, and I'm not sure we have anything that is proven to be secure yet.

>>6821

>I caution against this action for the following reason.

>Those will be high-profile events.

>their goal will be to use the vast (and terrible) resources at their disposal to destroy and hide this VQC tool

Now, I actually think these are a little contradictory. It depends on our goals (more importantly, in my opinion, VQC's goals), but if we aim to make the VQC public knowledge, the perfect way would be through a massive shitstorm. If it's the goal to make the VQC public knowledge, cracking into the Satoshi wallets and mining the rest of the blockchain in one day would do exactly that, and there would be no way to stop it from becoming public. That would be exactly the kind of event that would guarantee the public would learn of the existence of the VQC without the possibility of diversion by the media.

Of course, I don't think it's a good idea for other reasons. Firstly, I would think the only way to do it would be to take all of this huge mass of bitcoins and then not do anything with them. If anyone tried to cash them in for money, obviously they would be mobbed with publicity. Not to mention, that kind of money very easily corrupts people. Just leave this ridiculous scenario of a bitcoin wallet overflowing with bitcoins without actually doing anything with them as a public testament to the ridiculousness of it all. Secondly, and also in regards to your lost wallet idea, while that actually sounds like a brilliant way to fund the rest of this, keep in mind, there are anons lurking. While you can go through the threads and get a list of regular contributors together, and then distribute large sums of cash around to the people who still contribute, there are going to be people who feel entitled to this money. This is where fighting, manipulation and sabotage come into play. It's a dangerous thing. If we try to use this with lost wallets for a long period of time, like you said, and someone isn't getting a cut, what's stopping them from ruining our funding plans and getting the lost Satoshi wallets just to fuck it all up? Not to mention, it's going to be public by then, so anyone else could do the same thing.

Personally, while I think that idea could be very useful, and I can't predict the future, I'm holding out for this thing VQC briefly mentioned a couple months ago. He said he wanted to start a company making futuristic tech. I don't know how he plans on funding it if he isn't going after this money, but if it'll provide a viable and stable source of income, I would be part of that. I couldn't agree more about your motivations for the lost wallet idea, and if there's one main thing keeping me here it's the fact that we've all been given an extremely unique opportunity to change the world for the better, so I feel like it's a responsibility at this point to stick with it since there are so few of us. Not to mention, it would give me the funding and the free time to pursue music full time. That's the only selfish reason I have for being here, as far as I can tell.

>Steven Grier

This is just speculation, but if you've seen the end of Undisclosed, he had an interview with John Podesta, talking as if voting for Hillary would have led to disclosure (yeah fucking right). I definitely don't think he's a bullshit artist, but I don't know how helpful sending him a bunch of money would actually be if he has the ability to get someone as heavily cabal as John Podesta to be in his movie talking about UFO disclosure, you know?

Computer that can be connected to the internet is now functioning.

Wife is binge watching The Handmaid's Tale right now with it, which rewards her patience this very hot afternoon and allows me to check it's not spiking on CPU or network traffic as this will pause the show.

Once she is done (still recovering from jetlag) I'll install VS C# and it will be a number storm for a couple of post.

We'll be looking at values in a[t] a lot with examples.

As an exercise, can you find BigN-1 in column -f in terms of t? This is unfair as I said I would walkthrough but need to balance a wife who has gone the extra mile and patient anons.

Meanwhile, Q has been very quiet.

Something big happening?

>>6824

>As an exercise, can you find BigN-1 in column -f in terms of t? This is unfair as I said I would walkthrough but need to balance a wife who has gone the extra mile and patient anons.

Family first.

even d:

t = d/2 + 1

odd d:

t = (d+1)/2

I think this is right?

>>6825

To clarify, this is the t in (-f, 1) at which a[t] = N-1… I think thats what you meant?

>>6824

If we're waiting around then, could you possibly elaborate on that idea you posted a while ago about starting a company making futuristic tech?

>BigN-1 in column -f in terms of t

Did you mean like Teach did above, or do you mean the cell in which we would find BigN-1 in column -f for RSA2048?

>>6823

"there would be no way to stop it from becoming public"

Well, AA I hadn't thought of it that way. It would be Sooo easy for them to say something like "Russian Hackers… blah blah" though then just shoot us all and use the tool to end what Trump is doing. It will have to be done in a way that the tool is released to TONS of people with a really easy way to use it. It will simply destroy the cryptos, then be reworked by smart people and will wipe all cryptography that is not air-gapped or pen and paper.

Yeah I see your point on the wallet thing. Boy, what a fascinating situation this is. How would we do mass distribution. We will need a normie-tier tool to get this done. I mean a 1 click crypto cracker that is easy to use by anyone. Something like "input public wallet address " Click button. Output private key for that wallet. Then distribute it to people who do crypto for a living or something like that, They will spread it automatically. The tool will have to have a back-end description of what is happening behind the scenes so it is spread.

As for Grier, Yeah, I have no idea if he really is legit. What I DO know is he is far more legit than other personalities out there (Gaia/etc). Maybe Catherine Austin Fitts would be able to help? She has a pretty negative view of BTC and has been attacked relentlessly by the big guys. I could easily see her having some ideas of where to go from here? Yeah, I just urge caution. I do like the idea of leaking it as far and as wide as possible as fast as possible but the opportunity for manipulation of WHAT it is is sooooo easy. Soooo many normies. It has to be super-easy to use/understand. Perhaps a new thread for this conversation since this is not really research-related?

While we wait, just to go over the solution.

The Grid called The End is a superposition of all products where those products are the difference of two squares. This includes all odd numbers and even numbers that are zero mod 4.

Row 1, n=1 is special, since all factors of all products are present in it and by knowing d and e, and calculating f, this superposition collapses into a method to use cells e and -f in the first row, which are unique to d and e.

If there are more than two prime factors, d and e are the same, you can just find the largest two first, one will just be the product of further primes, divide and go again.

The problem we are trying to solve appears to be one where we don't have enough variables known in the equations we have.

The different factors (n-1) in -f, and n in e, coupled with BigN in e AND BigN-1 in -f as well as their 'shadow' partners where they exist at a[c+1-t] give us enough information to determine 1. Is the number prime 2. The value of x.

Remember, x is related to t.

So, for this problem, the 'superposition' collapses on two variables, d and e.

That was my assumption back in 2010 as it happens, that these two properties of c would be enough to factorise.

Integers are not a number line.

They are a family that grows like a fractal tree.

This work can probably be optimised.

There are more ways to do this, as implied, one just uses column zero. The other uses the diagonals but all are related. Like turning a higher dimensional object in 3D. It can look different.

>>6829

>>6831

I would love to buy physical copies of your work Tops.

Elliptic curves are DEFINITELY related.

The universality of the extended Riemann zeta function, the Mandelbrot Set. Related.

Whether you like to get lost in the beauty of colourful representations or found in the ruthless logic of it all, there is something for everyone.

>>6833

Nice.

Thanks for waiting just a bit longer.

Looking at morning UK time to get started as I am still limited to a phone.

A few extra reminders…

t is the order of members in a cell

In the first row where n=1, t gives a values of x that alternates between odd and even.

The value of x in e and -f will never both be even of odd since the gap between the column is always odd or 2d+1

a[t] is half the square of x plus e for the column

When e is larger than x by more than 2x+1, the value at a[t] will be found in a previous column in the first row.

For odd e, x=2t-1; even e, x=2(t-1)

a[t]=(((2t-1)(2t-1)+e)/2, (((2(t-1)2(t-1))+e)/2 for odd, even e

d[t]=a[t]+x[t]

In the rows below the first, where n>1, have you noticed that there is sometimes more than one set of products in the cell?

If there are two, both are growing by 2n added to x in that cell. The first set you can spot the same starting value of x from the value of x in the first cell at n=1. The second begins at -x until adding 2n makes it positive. I'll show this in the other thread after the walkthrough, if anyone is still interested at that point!

>>6834

We're not here just for integer factorization, we're here for the virtual quantum machines :-)

>>6834

Noticed and still interested. But are there only 2 sequences at n > 1?

So I've been looking into the frequency of repeating A[t] values and B[t] values in (e,1), (-f,1) and other cells with same c but different d. Look at the following records images. Pay attention to the block on the right. The center dot is (e,1,t) for whatever record we have at the moment with the same x or t. Column to the left would be -f, then the next would be for d=d+2, etc. Then if you go to the right each column is d=d-1. Going down you have higher t values, and going vertical up you have lower t values. If there is a red dot that means it shares a as factor of A[t], green dot means it shares b as factor of A[t] and cyan is c as factor. Notice that for any entry in with the t=t+n shift, the location of that c row (which is origianlly the cyan one going down to the right) turns into a B line for the next record at t+n, then it is always a B line for the rest of those records. This is pretty neato I think.

>>6834

Could you elaborate more on Revelation? Where do you see roots and remainders in it, and the other variables? I've already gotten the idea that it has variables corresponding to the churches and found 5461 in there, which is 7 0's and 1's

From discord convo:

https://en.wikipedia.org/wiki/Diffie–Hellman_key_exchange

>>6840

If you look at (0, 8) you will also see there are multiple sequences.

>>6816

That was not my point. My point was this: when a message is received, if the messenger isn't there to interpret, pretenders will certainly be more than willing to make the interpretation. This is trivially easy to prove: the New Testament is rife with Jesus condemning the Pharisees and Sadducees for twisting the Word of God to their benefit; similarly, we have the Catholic Church's storied history, Buddhist monks in designer fashions, hundreds of nuclear bombs for every nuclear reactor, etc.

There is a reason why the membership of the Council on Foreign Relations is structured as it is: those in media, social media and higher ed twist our perception of right and wrong, while bankers, international corporations and government officials do wrong. By placing their members at the critical distribution points of our shared reality, they've managed to put rings in our collective noses and bring us to heel.

Rest assured—in the absence of an intervention of Biblical proportions, this knowledge will be used (in large part) to deliver porn before it is used to save lives. Maybe it really wouldn't matter if Chris revealed himself–Doug Engelbart wasn't hiding; it's just that nobody cared to solicit his views.

>>6817

I understand, in a very personal way. I went down that path and ended up just short of homelessness. I got your "three rivers"(?) reference awhile back, but didn't want to bring it up–shitty way to introduce oneself.

It's your message to give–God gave you the understanding for a reason, and this student will respect the teacher's wishes. At this point, I have no idea what I will (or even can) do with the knowledge, but I pledge to you that, so long as I make money off of it, and I have a way of contacting you (or vice-versa), neither you or any member of your family will want for their share.

>>6821

Thank you for sharing your insight, Hobo–these are wise considerations. I know nothing about cryptocurrencies; I would be lying if I didn't hope to get a few bitcoins out of this, but I never expected there to be any left by the time I caught up. It sucks, given that I've probably invested over 2k hours toward the Q movement and this by now; that could've translated into enough scholarships that I wouldn't have to work while going to school. But money can't buy this (unless Chris is giving private lessons somewhere).

I honestly think it is inevitable that cryptocurrencies go down, once this knowledge is out–if not from one of us, then from the next ones who draw the connection. We have no idea who is monitoring this thread; my guess is that some are prepared just as you have described. Once the news spreads that RSA has been broken, there will be a rush to sell. It is Chris' wish to let this news out, none here will deny it; it would be best to cash out asap, imho.

Also, as you have mentioned—I think there's a lot to be said for doing the right thing. From the outset, I felt that the real test here was not whether we could be one of the first to understand the language of Creation, but what we did with that knowledge. This point in time is precarious–it feels, for lack of a better word, "pregnant." If we are reaching a pivotal point in our existence–one in which the Veil is lifted and we come face-to-face with our Creator, I wouldn't want to have to explain why $2 Billion worth of bitcoin was spent developing dildos that could read one's mind and be several places at once.

Morning anons, have woken in the night with the kind of sore throat I haven't had since I was little. Horrible timing. This should only push things to tomorrow or hopefully this afternoon. Feels like swallowing broken glass.

Thanks for the questions since last night, I will get to them once I can.

Anyone have recommendations besides Manuka honey?

>>6846

The best cure for that kind of thing is to overthrow the cabal and release the cure for the common cold.

>>6846

Finally.

I was about to start fishing again.

Feel better.

>>6846

I always enjoy a nice cup of ginger tea. Relaxing under a warm blanket is also highly important.

Take care, VQC!

>>6846

Also, try some dayquil mixed with SUPER MALE VITALITY and BRAINFORCE!

>>6846

This kung fu movie is suggesting chrysanthemum tea for sore throats…

Brilliant ideas.

My wife made a manuka honey drink with a berocca, plus coffee, plus paracetamol, plus this throat numbing thing.

Just doing the install of VS so can use some proper numbers.

I cannot quite believe the amount of the Spyware that comes these days on a cheap laptop. They don't even try and hide it any more.

Customer experience is second to full on data gibs.

Important to get the trip correct.

Brain at 70%

When we start, we'll build some functions.

We want to get a[t] (some na) for a column. We supply the column e and the t. The return is a[t].

Then we want to show that if p is a factor of a[t], p is also a factor of a[p+1-t] for some column e.

>>6858

How far do you think we'll get today? I was about to go to sleep so if you think we'll get to the final step in the next 8 hours I might stay up for it.

>>6857

Also this

>>6857

Good question.

>>6859

Another good question.

Should be ok until the moment the correct factorisation of RSA 2048 appears or just before.

This isn't about hacking or breaking encryption, that's a consequence. This is solving a math problem.

Nothing illegal is occuring here.

Also, nothing attracts attention like having linux pop up from a USB drive all over the place. That's got hacker written all over it, right?

Hackers don't use £200 laptops with Micro$oft on them.

>>6862

Get some rest, we have some big days a head of us :-)

>>6865

Can you guys let me back in the discord? I closed the tab about a week ago and I miss it

New "coincidence" I found that I think we haven't spoken about:

In (e,1) at a[t] = nb

x[t] = x+2n

d+x[t] = a+2x+2n = b

>>6866

Let you back in?

Can you not log into your account or something?

I believe I still see your screen name in the list but if yer not, just follow the instructions I gave teach.

>>6846

I think you'd be very good at writing suspense novels with how on the edge of our seats you're making us.

>>6866

You have to log back into your account. If you didn't complete making the account you have to make another. The discord URL is on one of the older threads, just DM ta.io and ask for the vqc server.

Update from last night's shadow boxing:

Their, the nerds, mental kung fu seems to have leveled up after meditation.

Which you'd kinda think they'd be better at considering sci-fi, neckbeards, and tabletop/MasterRacePC gaming…

Maybe I'll help show them how to larPLAY during a lull as a continuance of the visualization walkthrough I started yesterday.

Basic gist was (Kindergarten Cop Arnie) "Which are your Trangulata's and what do zey dooo?"

Now… The Alpha∞End Grid…

From the side it'd look like a line with a point on it.

Maybe it's that thing we desire. Buuuut…

The Grid. Let's call that our base"line" even though we just jumped to 2D, because puns.

There's a Trangle on it and a mid point but you don't know derp about it cuz you ain't got no references.

So let's make the grid an array of 4 grids by defining (0,0) {ooooor (0,1)? moving on.}

Seems like we have an idea of what's going on a lil' bit better now.

though… where did we place the common "midpoint" of the trangle?

At (0,0) or (0,1)? I guess if it's a mid"point" it's have to have a value or something, like binary, maybe. It's me, moving on.

So let's say we put it at (0,0)…

Wait… did you just assume my midpoint?

What if we go all Neo and push the grid back into space… (0,0) should turn into (0,0,0), with a newer, "bigger yet same" 0 in this new… Matrix.

Since we can imagine The Grid as a Square, we get a Cube, so the midpoint of that.

Now… that's fine and dandy, but if our Trangulated Common-Midpoint is (0,1) and we bust into the Third dimension, something happens to the Trangle.

The first time would have had it going dimensionally batshit trying to maintain the common 0 and its relationship to the three other points in this "new" third dimension.

With the TCM at (0,1), and assume this just plays out like you're opening a translucent accordion with a pop-up book style Trangle within it that locks upon full expansion of the Mathcordion.

And while we're here, you pick 3 notes/chords/combinations, or however that works, and upon expansion you can see where they reside in 3space with the revealed position of (0, 1, z-value).

Maybe it's that the chords have (x,y,z) values (cuz chords) and we're decrypting the third note for (0, 1, z) out of that. MOVING ON!

Toooo waveforms and frequencies…

Take our 3D The Griddles and turn it into a hypercube. (0, 1, z, ∆)

Which makes me wonder if locking our (Trangle) gives us the ability to see in 4space (or wherever I am)… because without it we'd lose (0,0,0,0) in the permutations (however they "operate) of the hypercube.

From the Chris-fu lessons last night:

"in a static equilibrium yes but the center is moving".

That way, no matter how the Trangle mutates with the permutations of the cube, we aways know that our TCM is always (0,1,,) away from it.

Aaaaaand that might kinda be how you move through "time/4D". Maybe.

-wanders off-

>>6869

>>6869

Yeah I never made an account I was only temporary and I updated my computer and I thought chrome would keep the tabs open but it didn't.

>>6872

I'll find the url. Thanks guise.

I don't remember where we were putting Mandelbrot and Pi stuff, but here's a bunch of mandelbrot stuff that someone out in the open waters was aksin' aboot.

Had to do with mutating/permutating/transforming a Flower of Life into a Mandelbrot or vice a verso.

Something like that.

3d Mandelbrot

https://www.youtube.com/watch?v=kaGcXUIg2qQ

Hidden areas in Julia and Mandelbrot sets (3d)

https://www.youtube.com/watch?v=vfteiiTfE0c

Times Table Modulo 200 (vid related)

https://www.youtube.com/watch?v=qhbuKbxJsk8

>>6654

>When does c first appear at a[t]?

>When is the second time it appears?

Not sure about the first time yet, but the second time is always in the same (e,n) at t = t + n.

Stumbled upon the third time c appears at a[t].

Borrowing from QVC's variable style >>6736, the following formulas represent the move to the third occurrence for odd e only, starting from the 1,c entry record:

e' = e

n' = 2x + n + 2

x' = 3x + 2n + 2

a' = b

d' = a' + x'

b' = a' + 2x' + 2n'

t' = (c+1)/2 + t

Examples:

entry

(1,61,6) = {1:61:12:11:1:145} = 145

2nd a[t]=c

(1,61,67) = {1:61:278:133:145:533}

3rd a[t]=c

(1,85,79) = {1:85:302:157:145:629}

n' = 2*11 + 61 + 2 = 85

entry

(31,128,8) = {31:128:16:15:1:287} = 287

2nd a[t]=c

(31,128,136) = {31:128:558:271:287:1085} = 311395 (second occurrence)

3rd a[t]=c

(31,160,152) = {31:160:590:303:287:1213} = 348131 (third occurrence)

n' = 2*15 + 128 + 2 = 160

Not sure if this is useful other than to show an example of how n can be calculated in terms of x and n.

>>6880

Thats an awesome meme man. I like it.

>>6881

I'm looking a bit into this as well, and while my results aren't great. I did find something that appears as a minor pattern, maybe?

I've only tested this on a few cells, but from what I've seen it is interesting.

Take a=7, b=37, c=259. The first occurrence of 259 in (e, 1) is at t=101. a = 20202, but note a % 4 = 2. We need to divide 20202 by 2, which gives us 10101.

So why is this interesting?

10101/259 = 39. And the cell for a=39, b=259 is:

{101:49:100:61:39:259}

Let's try this again with another cell like a=19, b=61, c=1159.

The first occurrence in (3, 1) appears at t = 502 => (3, 1, 504009, 1003, 503006, 505014)

503006 % 4 = 2 => We need to divide by two => 251503 which mod 4 = 3.

Then we do 251503 / 1159 = 217. We get the cell for a=217, b=1159 and lo and behold, we get:

(502, 187, 501, 284, 217, 1159).

What I've seen so far is that the e will always be equal to the t where c first occurs. The d will always be equal to e - 1 (or e is equal to d + 1?).

I've only run this through a few of the (3, 6) numbers, but so far it checks out.

The first occurrence of c in (e, 1) will have a corresponding cell with b=c at (t, k) for some k. I haven't locked down any patterns relating the cells, except for the e and b. So far I'm unable to "predict" where it would be given just a c.

And it also appears that the a at a[t] where a is the first occurrence of c, a % 4 == 2. Again, I've only tested cells from (3, 6), but so far it looks promising.

>>6887

I might have jumped the gun a bit fast. I is an interesting pattern, but does not hold for (1, 5) or (27, 6). That is, the e=t does not hold. There might be some differences between odd n and even n? Maybe the parity of e also affects such a pattern, I'm not entirely sure yet.

I'm still looking into it though. I'll post any new observations.

Also >>6735

>Because the RoD approach has Big Oh the same as the square root function, it doesn't matter how big they are, they would be insecure and too impractical to use.

RoD? Recursion on D?

So obviously he's not doing what he said he would again. I've been having a look at the RoD thing. Over the next few posts, I'm going to post every cell containing a and b values that multiply together to produce a given c value for the sequence of c values starting from RSA2048 replacing c with d recursively until reaching the lowest possible cell (obviously excluding RSA2048, since we don't know its a and b). It's a lot of numbers, so give me a bit. He said look at big semiprimes to see the pattern, so why not use the biggest semiprime we’ve dealt with so far?

This is the sequence of c values where each time we replace c with d (excluding RSA2048 itself):

158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844, 12598896421924866518220704506279920262503827873668266660431014021495031828185862096976477460760574959532030199266536614252274764167453116999041838483272614, 112244805768128382102848738141332806509123753930790945631219843869937264197591, 335029559543823509028205969685172744250, 18303812705112110303, 4278295537, 65408, 255, 15, 3, 1, 1

Let's start at the bottom because I'm waiting for this webtool to factor the second d (I somehow doubt I'll get to the first d, but who knows). To begin with, I accidentally added an extra 1. The lowest cell we find is c = 1. Above that, c = 3, then above that c = 15, then c = 255, then c = 65408, then c = 4278295537, then c = 18303812705112110303, then c = 335029559543823509028205969685172744250, then c = 112244805768128382102848738141332806509123753930790945631219843869937264197591, then c = 12598896421924866518220704506279920262503827873668266660431014021495031828185862096976477460760574959532030199266536614252274764167453116999041838483272614 (and I'm probably not going to do the original d since it's quite large, and then the original c we want to factor is obviously too large itself). I'll do 65408 etc in the next post because that one is even so it has a lot of factors.

c = 1, factors are 1

(0, 0, 1) = {0:0:1:0:1:1} f = -3

c = 3, factors are 1 and 3

(3, 2, 0) = {3:2:0:-1:1:3}, f = 4 (which has a negative x value and t = 0, so it doesn't exist)

c = 15, factors are 1, 3, 5 and 15

(6, 5, 2) = {6:5:3:2:1:15}, f = -1

(6, 1, 1) = {6:1:3:0:3:5}, f = -1

c = 255, factors are 1, 3, 5, 15, 17, 51, 85 and 255

(30, 113, 8) = {30:113:15:14:1:255}, f = -1

(30, 29, 7) = {30:29:15:12:3:85}, f = -1

(30, 13, 6) = {30:13:15:10:5:51}, f = -1

(30, 1, 1) = {30:1:15:0:15:17}, f = 1

c = 65408, factors are 1, 2, 4, 7, 8, 14, 16, 28, 32, 56, 64, 73, 112, 128, 146, 224, 292, 448, 511, 584, 896, 1022, 1168, 2044, 2336, 4088, 4672, 8176, 9344, 16352, 32704 and 65408

(383, 32449, 127) = {383:32449:255:254:1:65408}, f = -126

(383, 16098, 127) = {383:16098:255:253:2:32704}, f = -126

(383, 7923, 126) = {383:7923:255:251:4:16352}, f = -126

(383, 4420, 124) = {383:4420:255:248:7:9344}, f = -126

(383, 3837, 124) = {383:3837:255:247:8:8176}, f = -126

(383, 2088, 121) = {383:2088:255:241:14:4672}, f = -126

(383, 1797, 120) = {383:1797:255:239:16:4088}, f = -126

(383, 927, 114) = {383:927:255:227:28:2336}, f = -126

(383, 783, 112) = {383:783:255:223:32:2044}, f = -126

(383, 357, 100) = {383:357:255:199:56:1168}, f = -126

(383, 288, 96) = {383:288:255:191:64:1022}, f = -126

(383, 229, 91) = {383:229:255:182:73:896}, f = -126

(383, 93, 72) = {383:93:255:143:112:584}, f = -126

(383, 64, 64) = {383:64:255:127:128:511}, f = -126

(383, 42, 55) = {383:42:255:109:146:448}, f = -126

(383, 3, 16) = {383:3:255:31:224:292}, f = -126

c = 4278295537, factors are 1, 4561, 938017 and 4278295537

(89073, 2139082361, 32704) = {89073:2139082361:65408:65407:1:4278295537}, f = -41742

(89073, 405881, 30424) = {89073:405881:65408:60847:4561:938017}, f = -41742

c = 18303812705112110303, factors are 1, 73, 229, 16717, 1094922097572059, 79929313122760307, 250737160344001511 and 18303812705112110303

(3197991934, 9151906348277759615, 2139147769) = {3197991934:9151906348277759615:4278295537:4278295536:1:18303812705112110303}, f = -5358599139

(3197991934, 125368575893705255, 2139147733) = {3197991934:125368575893705255:4278295537:4278295464:73:250737160344001511}, f = -5358599139

(3197991934, 39964652283084731, 2139147655) = {3197991934:39964652283084731:4278295537:4278295308:229:79929313122760307}, f = -5358599139

(3197991934, 547456770498851, 2139139411) = {3197991934:547456770498851:4278295537:4278278820:16717:1094922097572059}, f = -5358599139

>>6893

Fuck it I'm way too tired, I'll finish this tomorrow.

>>6894

You are my hero AA. Get some rest man!

So I have been thinking about this distribution problem and I see it like this.

Assumptions:

1. Chris is not working with/for Q. He is independent but still on the side of the "good guys".

2. He does not know the best way to distribute this thing without hurting (and helping) a hell of a lot of people.

3. He is certain that the release will do more good than bad overall. (I deeply agree) and wants us to figure the details out.

The worse-case scenario is that the "Bad guys" (Think Hillary Clinton/Podesta/Pizzagate-tier bad guys) capture this tool and use it to crush what Trump is doing to fix the US and the world. If this thing is in the hands of just a few bad guys this is very possible, likely even.

Best case scenario is that we drop this thing in the right way. In a positive way that helps more than hurts people and the good guys use it to clear many of the bad things out of the system.

My guess is we will be somewhere in-between. See the movie “Sneakers” if you want a fun example of this exact problem. My voice is my passport, Verify Me!

The obvious target here is BTC. When I mentioned a one-click tool the reason I did was because this is easy to distribute to many people and would he hard hitting and attention getting. Most importantly though it could easily include the raw source code as well as the source files from this place compiled in text format so someone with a large enough brain could reproduce the work all in one package. It would be OUT. Think about the old keygen’s that were common in the software piracy days of the late 90s and early 0’s except you input a public key, hit generate and a private key pops out.

This would protect all of us from the men with the dead eyes, silenced pistols and blue gloves, which is very real risk going forward. Also it would make sure that no one could lock it up, which if you look into so many of these amazing inventions that have never come out is a very likely possibility that could happen. In fact, I think that is still the most likely possibility at this point. There will soon come a critical time where we must distribute far and wide or die trying. To simply squash BTC like a bug and post the code around may wipe cryptos from the face of the earth but that is just a tiny fraction of what is possible. A gnat buzzing around the ear of the elephant. What will be built and discovered with this tech is what really matters. THAT is what we are fighting for here weather we know it or not. Cryptos are simply a way to get there.

AA mentioned just making a script that rips all the bitcoin wallets to zero and sends the coins all to one wallet, essentially destroying the chain. Chris mentioned updating Wikipedia. I think the Wikipidia idea is VERY dangerous to alone. Definitely a way to get wacked in a hot-minute. It solves nothing at all related to distribution and puts up a gigantic bullseye on us all. I think that is the right move AFTER WE HAVE DISTRIBUTED THE CODE FAR AND WIDE. Wikipedia would be a good step-3 right there with Topol's Media release. Step 1 and step 2 though need to be more subtle and broader in scale. Step 1: This thing needs to be sent to people who UNDERSTAND what the hell we are doing here. This is tough for me because I barely understand what we are doing here but that is what needs to happen first. This means crypto-anarchists and netsecurity guys/math nerds at the academic level. Very shortly thereafter we need to use other broad distribution methods such as the BTC cracker I mentioned. Something that will be shared and shared far and wide. Something everyone would want and everyone would want to share. Bit Torrent as well as stuff like 'chan boards and social media. Lastly we drop the hammer and update Wikipedia/post all over social media.

I think a source of “People who understand this stuff” would be in the Net Security community as well is the crypto currency community. This could be done with a white paper. Then the BTC cracker, then the academic-tier Wikipedia release and official release to the media as Topol suggested. That is what I think is the best direction to go. What do you guys think?

>>6894

There are honestly probably over a hundred factors of c = 335029559543823509028205969685172744250. This is not a useful way for me to spend my time.

>>6891

>>6892

My point in all of this was to show that if we are meant to use d as c recursively (which got the thumbs up from VQC, so we probably are meant to), we can create a path from the lowest possible c in this sequence upwards, but there are differing numbers of (e,n,t) cells for each one. I think it would be very useful to find where d[t]-d = b(n-1), where BigN * c pops up in (e,1), and the triangle bases for each of the cells I've posted, and see if there's some way in which we can use one of the cells one lower to generate the next one up. That is the point of recursion, after all.

>>6896

I have a feeling he's holding it back because of the Wikileaks cables. There was someone on 4chan last year at some point claiming to be from Wikileaks saying that the huge US-wide internet blackout was caused by international powers trying to stop Julian Assange's dead man's switch. This person was on the run trying to get to a place where they could release it themselves, and they said Wikileaks was now compromised. This is around when weird shit started happening to do with Wikileaks. They didn't say this, but putting it together in hindsight with what we've learned from Q, it would have possibly meant releasing information that would have given everyone what we now refer to as the 40,000ft view. It would have crashed economies and started global conflicts, apparently. If DJT and Q really know what they're doing, this giant international campaign to stop JA's dead man's switch could be considered a good thing. Q et al are trying to get this information out gradually, because if it all came out at once without the media changing its tune and without key players having their power taken away, the world would turn to chaos. So think about it: is the stage set? Because once we figure this out, we can read those Wikileaks cables.

That's why I think VQC is stalling for a good reason without saying it out loud, rather than just being a super unreliable person (if you take everything he says completely literally). VQC did say he would give us time but keep up with the happenings and release it when the time is right. Once we solve the VQC puzzle, depending on the contents of these cables, the major worldwide events we've been waiting for are unstoppable. So the world needs to be ready. I think that might be fairly soon, personally, considering the huge growth in the Q thing publicly, as well as that stuff with the server.

Ok. Here’s the idea I'm working on:

At RSA sized numbers, movements between elements are still dictated by the earliest elements of (e,1)

And their ascending factors

So what is the average distance between a and b and c at an RSA sized number ??

It’s relative

But based on observable patterns

Multiples reduce the infinite search to a limited number of possibilities

Which we can create at will using the known rules

We can create any element surrounding (na transform) for any c.

We can search those elements surrounding for Multiples

Such as an, bn, etc.

My theory is that they should still be relatively close together.(edited)

Searching the factor tree in (e,1) is the main idea.

Constructing only what we need

Moving up and down the factor tree



At 100000000000 numbers, how can (bn) and b(n-1) still be one t value apart ???????

WTF

Instead, Iteration by Creation.

Creation of each element we need, and nothing more.

Lean program.

Killer.

Start at c, then (1,c), then (na transform) then -f na transform.

The reason why the element movements surrounding (na transform) are key is because they hold many factors.

We have a clear blueprint of the pattern of factors we are looking for

Therefore we search the patterns of factors surrounding na transform.

(bn) and b(n-1) are only one x value apart.(edited)

We could base a search just on that.

But we have to start at (1,c)

Then na transform

Then (-f,1)

Then crate the elements surrounding (e,1) and (-f,1)

In an expanding quadrilateral

With bn being in the lower right corner of that shape

If no match, then expand the quadrilateral

We don’t even need the Grid. We just need the rules of adjacent element Creation FROM the Grid. That’s gonna be a badass algorithm!!!!!

You guys agree?? Is it possible??

SHIT!! It’s so beautiful!

Adapting the Grid rules for element Creation into a group of strings, one for each element.

Expanding out from the na transform

Looking for the lock

On an, ab, a(n-1), and b(n-1)(edited)

An expanding square with an offset of 1 in the bottom right corner

Which is bn

Shape is quadrilateral

But not a trapezoid

I love you @everyone ️

This is so FUN!! 

Fuck, I love Math(s)

>>6903

I was honestly kind of surprised as to how many people didn't notice that until it was so overtly pointed out.

Then again, part of my journey was r/T_D, so I rarely miss a Ben Garrison.

Has anyone found anything to do with this clue?

>where the product of BigN and c are found in (e,1)

I'm using a=7, b=29, c=203 in this example. I'll do another example soon.

(7,4,4) = {7:4:14:7:7:29} f = -22

(7,88,7) = {7:88:14:13:1:203} f = -22

Product of BigN and c = 88*203 = 17864

Here’s (e,1) for this cell:

(7,1,1) = {7:1:5:1:4:8} f = -4

(7,1,2) = {7:1:11:3:8:16} f = -16

(7,1,3) = {7:1:21:5:16:28} f = -36

(7,1,4) = {7:1:35:7:28:44} f = -64

(7,1,5) = {7:1:53:9:44:64} f = -100

(7,1,6) = {7:1:75:11:64:88} f = -144

(7,1,7) = {7:1:101:13:88:116} f = -196

(7,1,8) = {7:1:131:15:116:148} f = -256

(7,1,9) = {7:1:165:17:148:184} f = -324

(7,1,10) = {7:1:203:19:184:224} f = -400

(7,1,11) = {7:1:245:21:224:268} f = -484

(7,1,12) = {7:1:291:23:268:316} f = -576

88 appears as b in (7,1,6) and as a in (7,1,7). 203 appears as d in (7,1,10). I don't know if 17864 itself appears though.

>>6905

a=13, b=43, c=559

(30,5,6) = {30:5:23:10:13:43} f = -17

(30,257,12) = {30:257:23:22:1:559} f = -17

Product of BigN and c = 257*559 = 143663

Here’s (e,1) for this cell:

(30,1,1) = {30:1:15:0:15:17} f = -1

(30,1,2) = {30:1:19:2:17:23} f = -9

(30,1,3) = {30:1:27:4:23:33} f = -25

(30,1,4) = {30:1:39:6:33:47} f = -49

(30,1,5) = {30:1:55:8:47:65} f = -81

(30,1,6) = {30:1:75:10:65:87} f = -121

(30,1,7) = {30:1:99:12:87:113} f = -169

(30,1,8) = {30:1:127:14:113:143} f = -225

(30,1,9) = {30:1:159:16:143:177} f = -289

(30,1,10) = {30:1:195:18:177:215} f = -361

(30,1,11) = {30:1:235:20:215:257} f = -441

(30,1,12) = {30:1:279:22:257:303} f = -529

…

(30,1,17) = {30:1:559:32:527:593} f = -1089

257 appears as b in (30,1,11) and as a in (30,1,12). 559 appears as d in (30,1,17). This does appear to be a pattern. I’m going to quickly put together a spreadsheet to extend these into further ts and see if I can actually find the product of BigN and c as the product, rather than the two separate variables.

>>6907

People have been asking on /qr/ and Discord, so just FYI, this image wasn't anything important. When you paste something from the clipboard, sometimes 8ch will create an image containing the thing you pasted. It was just an unformatted version of one of those cells.

//Function. RoD key pairs

//given two columns, which values of n are separated by 1, where e(n)>-f(n)

//Returns one pair for primes, two for product of different primes, etc.

List<BigInteger[2]> ColumnKeys(BigInteger minus_f, BigInteger e);

>>6910

Ayy lmao

>>6910

IS A FISHY YAY! ^_^

LEWDING=LOOTING

THOSE BASTARDS!

>>6913

Booty.

>>6910

RoD = Roots of D = Roots of David?

Okay, I'll bite.

>>6742 >>6739

Checks out. It will generate the next cell in the the sequence (or chain as I've referred to them before).

>>6746

INFO[0000] Starting [Big]VQC program

INFO[0000] E: 149730827186590819409161975355863102499674499386960841184873248110419525705142414412560340457818822465695973580080183089801154577153055685206470161899420076208154943348665238647007714095089342669905666884517354400122833485897064098153958317993833609635063079613328500485188782423277886515290863800949716792021

INFO[0000] N: 1259795423782894674701359162002419928571464106310201601388856891802183101035379777813200926294039220345914532062475754109464927957458809225140424456003642249634369640364388836798570917363513094818750748591234558253880668992954785004866522987440421420089871455032122934590859755937306075758632731614110843499361585902016642742536808842406995878081985323014757750816464333936471244183139065067762886207783712

4697517250971910788373374647441598985076306038901094587160385923033013611772476591818466822702222519806172328865872316642926190459725502720591581076567956723729060778728578126316270798047266909911419313225552335

INFO[0000] D: 158732191050391204174482508661063007579358463444809715795726627753579970080749948404278643259568101132671402056190021464753419480472816840646168575222628934671405739213477439533870489791038973166834068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807844

INFO[0000] X: 1587321910503912041744825086610630075793584634448097157957266277535799700807499484042786432595681011326714020561900214647534194804728168406461685752226289346714057392134774395338704897910389731668

34068736234020361664820266987726919453356824138007381985796493621233035112849373047484148339095287142097834807843

INFO[0000] A: 1

INFO[0000] B: 2519590847565789349402718324004839857142928212620403202777713783604366202070759555626401852588078440691829064124951508218929855914917618450280848912007284499268739280728777673597141834727026189637501497182469116507761337985909570009733045974880842840179742910064245869181719511874612151517265463228221686998754918242243363725908514186546204357679842338718477444792073993423658482382428119816381501067481045

1660377306056201619676256133844143603833904414952634432190114657544454178424020924616515723350778707749817125772467962926386356373289912154831438167899885040445364023527381951378636564391212010397122822120720357

>>6753

Appears to check out.

Adding again, but this time I'm adding the following:

(1, c)

(1, cN)

(-f, c)

(-f, c(N-1))

If anyone can double-check them, that would be appreciated.

Added in pastebin due to size of text: https://pastebin.com/6jn9Dmni

>>6916

Nice work!

Here's my working idea for order of operations.

Start at c (of course!)

Create (1,c)

Create na transform records for (e,1) and (-f,1)

Then begin expanding outward from the na transform records.

We already have a starting t value from the na transforms

We know an and a(n-1) have a lower t value

We know bn and b(n-1) have a larger t value

We also know that at the correct t values, we can move from an to bn, and they are [t+n] elements apart

We also know that at the correct t value, a(n-1) and b(n-1) are [t+n-1] elements apart

So we “iterate” (not really, because we’re moving by multiples and factors), looking for the 4 elements that match and give us our an, bn, a(n-1), and b(n-1) values.

It’s definitely an o(log t) way to search

Thoughts, Anons?

>>6918

How would we you expand though?

Say we have a start t, which is the t for the nc(na, but for c) transformation.

How would I move (or iterate)?

>>6920

3*5 looks particularly dank

black = divisible by a

red = divisible by b

yello = divisible by c

white = divisible by nothing

>>6920

Ima smoke a toke to that! Dank. Nice work GA!

>>6919

By creating only the needed records around each na transform record. Surrounding (-f,1) and (e,1)

>>6919

and looking for the an, ab, a(n-1), and b(n-1) records. They occurs in a semi predictable pattern surrounding (na transform).

We're searching the factor tree now, niggaz.

All that "iterate" stuff pales in comparison.

Here's a tired thought off the top of my head that I've applied no context to: e and -f in n=1 are 2d+1 apart. This is equivalent to 2d(n-0) + nn from the (x+n)(x+n), since n=1. We're finding (e,1) and (-f,1) based on a set d, e and f value with an assumed n. So what if we instead assumed n=2, and we found (e,2) and (-f,2) and so on upwards? Would this relationship hold (2d(n-0) + nn gap between (e,2) and (-f,2))?

What does the asymmetry in (-f, 1) and (e, 1) mean?

That (e, 1) has one cell more? That (-f, 1) has one cell less?

>>6931

I've been looking more into this, but I can't help but feel a bit stuck.

I feel a bit bad. I'm sure VQC is sitting here watching us while ripping his hair out because the solution is right in front of us.

>>6931

>>6933

To clarify, this is for c145

In (-f,1) at x=8, t= (8)/2, a value is a(n-1) = 20 = 5(5-1) = 20

In (e,1) at x = 7, t = (7+1)/2 = 4, a(n-1) value is 5(5) = 25

>>6934

> That n-1 is a factor of col -f and n is a factor of e will show you a shortcut to the triangle solution.

We're still doing triangular numbers, but now we're looking for a shortcut with -f and e.

When e is odd and -f is even, then every a-value in -f is a square multiplied by two and subtracted f/2. Every e is a triangle multiplied by two and added e/2.

Equation for even -f: 2tt - f/2

Equation for odd e: 2t(t+1) + (e/2)

But how this is leading us to a shortcut I'm still not sure.

We know we have n and n-1. We know for example that n*(n-1)/2 is a triangle and is also the sum up to n-1. I was thinking that maybe we can find this n*(n-1)/2 triangle somewhere either in e or maybe column 0, which should allow us to do some kind of shortcut. Not sure yet though.

I wonder if the person who said "you have until July 14th" on /qresearch/ a while back will pop up today.

>>6936

Was that just a blank statement or a response to a fishing attempt?

>>6937

It was a response to a fishing attempt by Topol.

>>6938

I guess we'll see. Worst case we will just keep at it. By the sounds of VQC we've gotten close (although close doesn't necessarily mean close to a solution, but rather closer to a solution).

Either way I'm not go anytime soon.

>>6935

f_d[t] - e_d[t] < 0 until we reach a t, then

f_d[t] - e_d[t] > 0

This t is at a[t] = N.

>>6940

We have our equation nn + 2d(n-1) + f - 1 which is nice and dandy.

For -f when our a, b is in (-f, n-1) the equation still holds, since we are having -30 it turns into +30

(n-1)(n-1) + 2d'(n-1) + f - 1 where f is now defined as 2*d + 1 -(-f)) turning it into a + f.

For example, take a=7, b=37

For (3, 6) the equation will be: 6*6 + 2*16*5 + 30 - 1

While for (-30, 5) it will be: 5*5 + 2*17*4 + 65 - 1

We get 65 by doing 2*17 + 1 - (-30).

Essentially by moving to f, we move another part of 2*d*(n-1) into f.

Working on what VQC said he was going to explain to us seemed to be enough for him to taunt us with his presence yesterday, so I'll continue.

>>6774

>Then we will look at the key that is made by column -f with the locations of c in (-f,1)

I'll start with the same two examples from my last posts, c=203 (7*29) and c=559 (13*43). I'll just do the first in this post and probably do the other in the next post.

c=203, d=14, e=7, f=-22

Relevant n=1 records are (7,1) and (-22,1).

559 doesn't seem to come up as a, b or d in (-22,1), nor does it appear to come from some kind of equation between different variables. So I thought the only way I would see it explicitly come up was by extending the records until x=203 or t=203. x is always even for (-22, 1), so c only appears in relation to x as x+n or x-n (since n=1).

(-22, 1, 101) = {-22:1:20593:202:20391:20797} f=41163

(-22, 1, 102) = {-22:1:21001:204:20797:21207} f=41979

Blindly looking for c here, the difference between d values here is 408, which is (203+1)*2, the difference between a values is 406, which is just 203*2, and the difference in f values here is 816, which is (203+1)*4. So I guess you can technically find c here, when x is adjacent to c.

The only other way you can explicitly find c in (-f,1) is when t=c.

(-22, 1, 203) = {-22:1:82813:406:82407:83221} f=165603

Here also, we have 406 pop up again, which is 203*2.

So, yeah, c does kinda come up in (-f,1), as 2c when you make t=c. I'll have to see about the other one. I also don't see how this could be useful at all, but he mentioned it, so I guess I'll keep looking.

>>6942

So then for c=559 (13*43), d=23, e=30, f=-17.

Again, 559 doesn't appear as an a, b or d value, nor as an equation of the variables (that I can find). a, b and d stay the same parity, too, so considering 559 is odd, which through addition or subtraction would change the parity of an incrementing number, it isn't going to appear as the gap between successive values either. So, I'll do the same thing: x=c, t=c. x is odd this time, so 559 does appear as an x value.

x=c

(-17, 1, 280) = {-17:1:156791:559:156232:157352} f=313564

The gap between a and b here is 1120, which is (559+1)*2. I can't really see anywhere else that I could get 559 from this record.

t=c

(-17, 1, 559) = {-17:1:624953:1117:623638:626072} f=1249888

The only thing close to being c here is 1117, which is actually one less than 559*2. But that isn't consistent with the other example in my previous post.

In conclusion, I have no idea what VQC was talking about when he said

>Then we will look at the key that is made by column -f with the locations of c in (-f,1)

because the only definite location of c in (-f,1) is when t=c, and the relationship between the variables at that record and our c is inconsistent.

>>6946

VQC explicitly said c would be in (-f,1) though. In your example, it's (-f,6), (-f,60) and (-f,86).

>>6954

Note I've also only divided by 8 as I haven't yet understood what VQC was talking about when he increased the part we divide for. I still think about >>5965 but VQC hasn't acknowledged those and has afterwards stated that we should keep the diagrams we have. Reason for that is that it aligns well with his first posts about triangles from thread #10 >>4342.

>>6957

also (prime) d = 78, and SQRT d = 8 remainder 14 = 8^2 +14

I'm gonna predict that at least of one the key values will be [t] +/- 8 from the na transform at [t]=39

so what's going on at [t]=31 ??

Do we find a(n-1) and (an) ??

>>6960

Thanks, GA! Ok, I'll start studying now!

VQC, hoping for a correction from you here.

I was looking into:

> If an integer p is a factor of a[t], then p will be a factor of a[p+1-t] for ALL cells in row n=1.

From the thread Grid patterns, you posted that in >>6561

I'm thinking that we generate Nc in (e, 1). We then know that N and c are integer factors of Nc.

That should mean that N is a factor in a[N + 1 - t] and that c is a factor in a[c + 1 - t]. However I'm struggling to actually make it work.

Take the example a=7, b=37. c=259. big N = 114.

We generate Nc in (e, 1) which is {3:1:29769:243:29526:30014} which is at t = 121.

Then 114 should be a factor in a[114 + 1 - 121] and 259 again in a[259 + 1 - 121]. Neither are.

The cell for a[114 + 1 - 121] is {3:1:51:-11:62:42} and for a[259 + 1 - 121] has cell {3:1:39201:279:38922:39482}.

This lead me to think it only applies to prime factors of a[t] which lead me to factorize 114 to test. 114 = 2*3*19.

So this means a[2 + 1 - 121], a[3 + 1 - 121] and a[19 + 1 - 121] should have 2, 3 and 19 as factors.

It holds true for a[2 + 1 - 121], but not for a[3 + 1 - 121] or a[19 + 1 - 121].

What I'm wondering is if I've misunderstood it or I'm applying it wrongly. My initial though when using it was that when p is less than t, we enter negative x space, but you've stated earlier that we should be afraid of that.

Of course if anyone else knows I'd love a reply. I can't quite grasp how this is supposed to work.

>>6964

Unless you made a mistake and in reality it is a[p - (t + 1)] in which case it appears to hold.

>>6965

I'm chalking this one up to one of VQCs' happy mistakes. It appears to hold fine for a[p - (t + 1)] (or a[p - t - 1] depending on how you want to look at it).

>>6774

Product of Big N and c are important, as it combined with the "other" value of c from (e, 1) will give us some extra information. This combined with -f and the locations of c will somehow lead us to x for an, a(n-1). Does this mean we want to find x for either of them, both?

>>6858

We want to make a function to get a[t] from some column (Presumably a[t] = Nc). Then we want to take advantage of a[p - (t+1)] = kp for some k, p.

I'm speculating that we want to get Nc, use a[c - (t+1)] to get another location for c and from here move over to -f and then find the value of x for either/or an or a(n-1). But if we're actually going to find x or t I'm not sure.

Hell I don't know the step to go from knowing different locations of c in (e, 1) and (-f, 1) to solve for x, but given that these are the last posts he made I'm willing to speculate that they are somehow related.

>>6966

>>6965

>>6967

>>6968

Chalk this one up to Anon's happy mistake and not VQC. I used the wrong calculation on x to t and t to x, indexing the first record in (e, 1) as t=0. As a result, a[p + 1 - t] wouldn't work but a[p - 1 - t] worked. I've corrected my function and a[p + 1 - t] works fine now.

Remember 2na = xx + e. If n=1 then xx + e = 2a and so a = (xx+e)/2. Since we know that we can reverse engineer this to put our c value in the a position, so we can find where c is in A[t] in n=1.

c = (xx + e)/2 -> 2c = xx + e

So we do the same calculation the square root then the remainder with the first thing. So for 145 we would get 2c = 290 = xx + e = 17*17 + 1.

Also 145 = 12*12 + 1. Also we know that (of course) 2c - c = c. This gives us the equation 17*17 + 1 - (12*12 + 1) = 145, which simplifies to 17*17 - 12*12 = 145, the correct answer to our problem, because d+n = 17, x+n = 12, d-x = 5 (I think this is why VQC liked this number). This got me to look into more equations and they aren't always that simple. Here is another:

17*41 = 697

697 = 26 * 26 + 21

697*2 = 37 * 37 + 25

37*37 + 25 - 26*26 - 21 = 697

37*37 - 26*26 + 4 = 697

So if anything this may just give us a close approximation of a difference of two squares. But we may be able to notice that when you shift d up one by either of these e gets shifted by 2d+1 or 2d-1 depending on the direction. Maybe we could use that and algebra to match up the columns to whatever makes e values equal then when e is equal we have the solution.

>>6975

Also there are always 2 columns that match up with the values

I've been looking into a[p + 1 - t] and I wanted to know what was happening under the hood. So I wrote it up using algebra and I got the following:

For odd e:

2(p + 1 - t)(p + 1 - t - 1) + ((e +1) / 2)

This can be rewritten as:

2p(p - 2t + 1) + 2t(t-1) + (e+1)/2

We can immediately see that 2t(t-1) + (e+1)/2 is the value of a[t] (the t we're doing a[p + 1 - t] from). Meaning we're doing 2p(p - 2t + 1) + a[t].

Then I was wondering what (p - 2t + 1) and 2p(p - 2t + 1) would look like for a few values and I wrote up a test. Attached is an image showing the test cases with values from (e, 1) and (-f, 1).

For these tests I did the na, nb and nc transformations on records from (3, 6) starting in the sequence a=7, b=37 and the 9 following records.

Note, aT = t where a[t] = na, bT = t where a[t] = nb and cT = t where a[t] = Nc. (Same for the negative records)

What sticks out is that b - 2bT + 1 = d (the d for our record) and c - 2cT + 1 = d (same d as our record d) while a - 2aT + 1 = d'

Note I marked it d' because it refers to the previous record in the sequence of (7, 37)s' d. The first record in the list is also the first "valid" record in the sequence which is probably why it shows d = -2, but you can see in the following records that a - 2aT + 1 will match the previous record in the sequences d.

>>6978

Thank you for uploading, it's the correct image. 8ch was being a dick so someone on discord uploaded for me.

>>6980

I was trying to replicate with a=7, b=37.

The t-values for the first occurrences of a^2 in (e, 1) was 19 and 31. However 19 + 31 - 1 == 49. Same for b^2: First occurrences 582 and second at 788, 582 + 788 = 37^ + 1.

I suspect then that there is more to it than just odd/even e, maybe odd/even n's as well?

In your second image where e=odd and n=even you can see what I'm talking about.

>>6981

Correction (missing 2 after ^): 582 + 788 = 37^2 + 1

>>6980

If you look at the record a=91, b=169 you'll also see that 91^2 appears first at t=2175, second at t=3020 and the third time at t=5262

2175 + 3020 = 5195 which is not correct, while 3020 + 5262 = 8282 = 91^2 + 1

Thinking this through again:

>>6629

>As stated at the start, the square root of (D)avid and what remains are the key to unsealing the The End, the grid.

I'm thinking it isn't that we treat d as c recursively (even though VQC's reply to my post saying that implied it was), but maybe it's literally as he said, and it the square root of d and the difference between d and its square root.

I tried seeing if this stuff PMA has been posting examples of with the first three times a[t]=c shows any relationship with these root(d) and remainder values but it's quite hard actually finding examples without just making a brute force algorithm. I do have part of another example if anyone's interested.

7*29=203

d=14, root(d)=3, 14-3=11 (before you get any ideas, I realize this is equal to e+n but this is definitely not the case universally)

The true cells for this are {7:4:14:7:7:29:4} -22} and {-22:3:15:8:7:29:4} 7}

-f — a[t] = c

First: {-22:63:363:160:203:649} t=80 e=703

Second: {-22:87:391:188:203:753} t=94 e=759

Third: {-22:117:421:218:203:873} t=109 e=819

Fourth: {-22:149:449:246:203:993} t=123 e=875

e — a[t] = c

First: {7:88:392:189:203:757} t=95 f=-778

Second: {7:116:420:217:203:869} t=109 f= -834

Third: I couldn't find

I have no idea what any of this has to do with 3 or 11. Possibly coincidentally, x[t] = c at {7:64:525:203:322:856} t=102 f=-1044, which is the cell in the sequence that exists in the negative space but was missed in the positive space. I also did this with 13*43=559 and found x[t]=c at (30,1) when the first cell where a[t]=c in the negative space was (-17,4), so I don't know if this is actually relevant or a pattern.

{

List<BigInteger> result = new List<BigInteger>();//list to return

BigInteger 2d_plus_1 = e - minus_f;

BigInteger _2d = BigInteger.Subtract(2d_plus_1,BigInteger.One);

BigInteger sqrt_2d = Lib.Sqrt(_2d);

>>6987

Oh come on man, hahaha

>>6987

Okay so this is a snippet of the code and it isn't necessarily the beginning, the opening bracket indicates that it is the start of a function or a loop or an if statement or something, it seems like there have been a few things defined already, namely e and minus_f and they have been defined before the '{'. Again this could be after an 'if' in a loop or something where e and minus_f change in the loop. Then you take the sqrt of 2d which could mean to recursively do the function on 2d [145 = 12*12 + 1, 24 = 4*4 + 8, 8 = 2*2 + 4, 4 = 2*2, 4 = 2*2, etc]. under some circumstance. I'm trying to read between the lines here

>>6990

But also he creates the return value so its probably the start of the function

>>6980

>>6983

Thanks for double checking.

In reviewing further, I think I may have just restated how the p+t formulas work in a different way.

Was looking for a way to jump sequences, but don't believe these formulas enable that kind of movement.

>>6987

Great to see you even if you're in your cryptic corner.

>>6994

Ignore the 11^2, that came from when I had 12 blue instead of 11.

I feel like this was skewed on purpose to obfuscate something.

Also, the floating blue block in the top left of the small square on the original is offset by half a block, for some reason.

Something feels "on purpose, to obscure a thing" here.

>>6987

2d is the same as multiplying 4 to d (and should also be the d for 4c) which is related to a crumb he dropped back in thread #12 >>6445

I don't remember if anyone actually did anything with this crumb. I know I didn't.

>>6996

I typed this out too fast. Let me rephrase it.

If we multiply c by 4, which was a crumb by VQC in >>6445 then the d off that record will equal the 2*d from c.

Example:

{595:170:4698:1105:3593:6143}

Here d = 4698, 2*d = 9396. The record for 4 * 3595 * 6143 = {2380:11026505:9396:9392:4:22071799}

So what he is doing is taking the square root of the d for 4c. Why I don't know yet, but since this is related to the 4c crumb I suspect there is more to that crumb than we figured out.

>>6997

>>6996

I'm sure you're already aware, but since we're using floors of square roots, 2d doesn't equal d of 4c in every case. Here's an example:

c=145, d=12, e=1, 2d=24

4c=580, d=24, e=4, 2d=48

4c=2320, d=48, e=16, 2d=96

4c=9280, d=96, e=64, 2d=192

4c=37120, d=192, e=256, 2d=384

This is where it stops following the pattern.

4c=148480, d=385, e=255, 2d=770

4c=593920, d=770, e=1020, 2d=1540

It then also resets here too.

4c=2375680, d=1541, e=999, 2d=3082

4c=9502720, d=3082, e=3996, 2d=6164

And so on, every time e becomes greater than d.

>>6999

Also trips

>>7000

Double trips!

Also, I checked and this is true. It appears that when e > d the difference is 1.

So d in (1, 4c) = 2*d' + 1 for d' in (1, c).

>>6973

Yeah, I think we were doing those back in thread #3-#4? I remember a lot of talks about Chirstmas trees (as a joke since it looks like a christmas tree).

I think you'd be better off if you started skimming all the threads.

I think I realized a crumb from way back in thread #11 that deals with this 4c.

So some thoughts regarding c and 4c. Right now we're focusing on odd squares (x+n)^2. By multiplying c with 4 we're creating a new record and distributing the 4 (2x2) to both a and b, giving us 2a, 2b. This also gives us an even square to work with 2(x+n)^2.

The crumb from post >>5491

>"The 'heart' of the problem is that breaking the problem down involves two sets of triangle solving (otherwise it is prime), with one set of triangles (say half of the eight) working on a triangular problem that is one unit longer than the other set. This is one reason why no current approach I have seen works."

Using 4c we now have another 2(x+n)^2 square However, since this is even we can't divide it into 8 perfect triangles like we can with our odd (x+n)^2. Luckily for us, PMA has done some great work on even squares at >>6272 and >>6273 and showed that an even (x+n)^2 can be expressed as 4*(Tu + T(u-1)).

This gives us two equations for our x+n squares:

Odd (x+n)^2 = 8(Tu) + 1

Even (x+n)^2 = 4(Tu + T(u-1))

We now have two triangle problems to solve. For an even square, half of the triangles is one unit longer than the other half.

Now for some insight into our equations (This is just for insight, not sure if it can be used for solving):

8Tu = 8(u*(u+1)/2) => 4*u*(u+1) => 2*(2*u*(u+1)).

2*u*(u+1) should be something we all recognize as it's the same method we calculate a's or d's depending on parity of e (Missing + (e + 1)/2 or e/2). This should mean (I might be wrong) that we are using values from (0, 1) as the 4*T(u), as in 4T(u) should appear in (0, 1) as d. Just to re-iterate T(u) is the triangle with a base of u, has the equation u*(u+1)/2. But by multiplying 4 we get 4(u(u+1)/2) => 2u(u+1). This should mean that our odd (x+n)^2 - 1 = 2(d[u + 1]) from (0,1).

For even squares we have the equation:

4(Tu + T(u-1)) which is also equal to 4Tu + 4T(u-1) => 2u(u+1)) + 2(u)(u-1). Here we have two d's from (0, 1) with a difference of t by 1. Essentially d[u+1] + d[u].

I think this might be why VQC has said that column (0, 1) is so special. Our x+n squares, regardless of parity, can be expressed using d's from (0, 1).

>>7003

Assuming we're trying to figure out the triangles using a similar method Get_N_From_… we need two new functions to work with even triangles.

We have GetNFromOddTriangleBase and we now need GetNFromEvenTriangleBase and a way to get the remainder needed for even squares:

def GetNFromEvenTriangleBase(bs, c, d):
  big_triangle = triangleN(bs)
  small_triangle = triangleN(bs-1)
  eight_base = big_triangle * 4 + small_triangle * 4
  XPN = eight_base # (x + n)(x + n)
  DPN = XPN + c # (d + n)(d + n)
  return math.floor(math.sqrt(DPN)) - d

def GetRemainderEven2dnm1(bs, d, n, f):
  big_triangle = triangleN(bs)
  small_triangle = triangleN(bs - 1)
  eight_base = big_triangle * 4 + small_triangle * 4
  XPN = eight_base # (x + n)(x + n)
  two_d = d + d
  nm1 = n - 1
  two_d_m1 = two_d * nm1
  XPN_mfp1 = XPN - f - 1
  resultpN = XPN_mfp1 - two_d_m1
  result = resultpN - (n * n)
  return result

>>7003

Just a quick unverified thing:

When we multiply our c with 4 we do 2*a, 2*b. This gives us a new square (x+n)^2 that is even.

If we divide this by 4 we end up with our original x+n square, however that means that our original x+n square, which is defined to by (x+n)^2 = 8Tu + 1, can also be expressed as the sum of two triangle numbers, with a difference of a unit of one. Specifically the triangle numbers used in the even square.

The one we're after

(x+n)^2 = 8Tu + 1

The one we get after multiplying 4

(x+n)^2 = 4(Tu + T(u-1))

Dividing the new square after multiplying 4 gives us the (x+n)^2 from our a, b. This also means that our (x+n)^2 = Tu + T(u-1).

So we have two equations for out x+n.

(x+n)^2 = 8(Tu) + 1

(x+n)^2 = Tu' + T(u' - 1).

Note the two u's are different, hence the u' vs u.

I haven't had time to properly verify it, but I *think* it holds.

Dissecting these code snippets a little bit.

>>6910

>//Function. RoD key pairs

>returns one pair for primes two for product of different primes, etc

So this function is only returning the n values. It doesn't seem to be doing anything with t of the correct (e,n). That means there's something inherent to the root of d that is also inherent to n, right? Because that seems to be the point of this function (given e, -f and d, return valid n values).

>given two columns, which values of n are separated by 1, where e(n)>-f(n)

It also means that something about e, -f and d tells us the number of factors that a number has. Every n value in the entire positive e space has a corresponding n-1 somewhere in the negative f space, but obviously all the ns in one e column aren't going to all be in the same -f column. So it returns every instance of an n in a given e having an n-1 in the given -f of our c. That means it returns every possible n value for our c. That means it knows how many n values there are for our c (and thus how many factors it has).

>>6987

I don't think these are the same function. This starts with an open parenthesis, but the last piece of code he posted was comments followed by a variable. That variable from the last one was the list of n and n-1 values for a given e and -f. That's definitely a local variable rather than a global variable, so those comments were inside its function. That means this has to be the beginning of a separate function. So we have one function to find all of the n values for our c, and another to do whatever this one's doing (which seems to be creating a list of BigIntegers (as opposed to a list of BigInteger arrays with 2 values each) and doing something to them that has something to do with sqrt(2d)).

>>7003

Just to add some more insight into even (x+n)^2.

Even (x+n)^2 = 4(Tu + T(u-1))

4(Tu + T(u-1)) = 4(u(u+1)/2 + u(u-1)/2)

4(u(u+1)/2 + u(u-1)/2) = 2(u(u+1) + u(u-1))

2(u(u+1) + u(u-1)) = 2(uu + u + uu - u) = 2(2uu)

This means our even (x+n)^2 is 2(a[u]) from (0, 1) and 2(d[u]) from (1, 1).

In this case, based on some testing u = x+n.

>>7005

Also, just to point out (x+n)^2 = Tu' + T(u' - 1) makes sense. I think I was a bit tired or fatigued. Two triangles (one being a unit longer than the other) added together makes a square, in this case the sum of Tu' and T(u' - 1) is equal to our x+n.

I'm probably re-iterating something we already know, but the nn + 2d(n - 1) + f - 1 = (x+n)^2 comes from (d+n)^2 - d*d.

(d+n)^2 - d^2 = (d+n)(d+n) - dd

(d+n)(d+n) = dd + dn + nd + nn

dd + 2dn +nn - dd = (x+n) + e

2dn + nn = (x+n) + e

nn + 2dn - e = (x+n)

Then we "steal" from 2dn to create f, since f = 2d + 1 - e, we steal a 2d

nn + 2d(n-1) + 2d - e = (x+n)^2

Then we add 1 to both sides

nn + 2d(n-1) + 2d - e + 1 = (x+n)^2 + 1

nn + 2d(n-1) + f = (x+n)^2 + 1

nn + 2d(n-1) + f -1 = (x+n)^2

I've made a brute-force version of this algorithm

>>6910

It doesn't use the root of d (since obviously we don't know how you're meant to use that to do this), but it's useful for considering the context in which this function fits within the entire algorithm. It doesn't do (1,c) for even cs since n would have .5 at the end with an odd a+b (so those cells aren't in the grid).

import java.util.*;

public class brute{
	public static void main(String[] args){
		int e = Integer.parseInt(args[0]);
		int f = Integer.parseInt(args[1]);
		int d = (e-f-1)/2;
		int c = (d*d)+e;
		int twod = d*2;
		int roottwod = (int)Math.floor((int)Math.sqrt(twod));
		System.out.println("c="+c+", d="+d+", e="+e+", f="+f+", 2d="+twod+", sqrt(2d)="+roottwod);
		for(int i=0; i<((((1+c)/2)-d)+1); i++){
			int aplusb = (i+d)*2;
			int x = ((int)Math.floor((int)Math.sqrt(((d+i)*(d+i))-c)))-i;
			int testa = d-x;
			int testb = aplusb-testa;
			if(testa*testb==c){
				System.out.println(i + " is a valid n, with a=" + (int)testa + " and b=" + (int)testb);
			}
		}
	}
}

So you input e and f values (f has to be negative) and it outputs every valid n value for that pairing

BigInteger sqrt_f = Lib.Sqrt(Lib.Abs(minus_f));

BigInteger sqrt_e = Lib.Sqrt(e);

//Section: determine rows (-f,n-1) and (e,n)

>>7013