/vqc/ - Virtual Quantum Computer

The fact that we're almost a year into this and we still don't have an easily accessible pastebin for making new threads with is a bit aggravating. https://pastebin.com/VkaTJcX1

I apologize for necesitating an emergency bake. I was busy.

In a clearer format (the one I keep) (also, fixed some typos):

post 1, glossary - https://pastebin.com/gWV7vySY

post 2, rules - https://pastebin.com/cE195ptk

post 3, code - https://pastebin.com/heweAtWs

>>7695

Don't sweat it.

The Dimension equals 8?

Q=D5=Minecraft Avalanche

VQC=8D=Mathcraft Avalanche, starting with a single particle of water.

Difference of 3 Dimensions.

E8 H2O

>>7702

>7702

Also theres definitely room for optimization. I tried to generate the records in a dictionary and then load them so I didn't have to iterate through x values for every pixel in the map but I couldn't get it for some reason and I wasn't concerned about it really so I just generate every pixel. Someone could easily fix that. I plan on making improvements and adding more features I just think this is at a pretty solid state so I want to send it out. Also don't fuck with pressing j or k on a record in the n=0 row. It will crash the thing because I index with x and x is all the same with that so it never finds the next x and I dont feel like adding a whole nother indexing set for n=0 so just understand the patterns there.

I've been spending some more time looking into the smooth numbers. I've noticed that given (e, n) the smooth numbers for those records (bigN - n) exists in (e - 2*n, 1). If e is odd, then the smooth numbers will exist as d's. If e is even then smooth number will appear as a. If you jump to (e + 2n, n) the smooth values of those records will match the a's or d's in (e, 1) at t for bigN. Essentially, given a record a, b we get bigN. If e is odd then the d at a[t] = aBigN (e, 1) will be a "smooth" number for the jump (or a + 1, b + 1) in (e + 2n, n).

The smooth numbers of (0, 1) exist in (-2, 1) at t = 1, 4, 9, 16…

The smooth numbers of (1, 1) exist in (-1, 1) at t = 1, 4, 9, 16…

It also appears that the t's for the smooth numbers in (e, 1) exist in (e - 1, 2). Although that is just a preliminary result. I haven't had much time to look into that result.

>>7704

The patterns aren't always [1, 4, 9, ..] but offsets.

Take (7, 1). The first 10 smooth numbers are:

[2, 5, 10, 17, 26, 37, 50, 65, 82]

Here the relationship with squares is 2 + 2, 5 + 4, 10+6, 17 + 8.

(8, 1) has first 10 smooth numbers: [4, 8, 14, 22, 32, 44, 58, 74, 92] which has the pattern 4 + 0, 8 + 1, 14 + 2, 22 + 3, 32 + 4 …

>>7705

Take cell (3, 6). Moving to the left will put us in (3 + 2*6, 6) => (15, 6).

The first 10 rows in (3, 6) is:

and their smooth numbers are:

{3:6:4:3:1:19}

{3:6:16:9:7:37}

{3:6:34:15:19:61}

{3:6:58:21:37:91}

{3:6:88:27:61:127}

{3:6:124:33:91:169}

{3:6:166:39:127:217}

{3:6:214:45:169:271}

{3:6:268:51:217:331}

{3:6:328:57:271:397}

[0, 108, 540, 1620, 3780, 7560, 13608, 22680, 35640]

While the first 10 rows in (15, 6) are:

{15:6:17:9:8:38}

{15:6:35:15:20:62}

{15:6:59:21:38:92}

{15:6:89:27:62:128}

{15:6:125:33:92:170}

{15:6:167:39:128:218}

{15:6:215:45:170:272}

{15:6:269:51:218:332}

{15:6:329:57:272:398}

{15:6:395:63:332:470}

and their smooth numbers are:

[129, 579, 1683, 3873, 7689, 13779, 22899, 35913, 53793]

Note that (15, 6) has "shifted" one record down, meaning the first cell in (3, 6) (a=1, b=19) is gone.

Each of the smooth numbers in (15, 6) are the d's at (3, 1) where a[t] = BigN.

The big n for a=7, b=37 is 114 and the d at that location = 129 and exists at t = 8.

This means that the t's in (3, 1) where d = smooth number for (15, 6) are: [8, 17, 29, 44, 62, 83, 107, 134, 164]

>>7706

Minor typo, pasted the records along with the smooth numbers, but you'll see the difference.

The smooth numbers for (3, 6) (0, 108, …) exists in (3 - 2*6, 1) => (-9, 1) at t's: [2, 8, 17, 29, 44, 62, 83, 107, 134]

Which we can see occur at the same location in (3, 1) as the smooth numbers do for (15, 6).

This has completely destroyed my life. It didn't hit me how terrible it was until today, but fitting in enough time to study everything I've studied about it over the past 3 months or so and doing Grid Patterns completely by myself has led to my friends thinking I'm a useless, selfish cunt, my sleep schedule being ruined to the point of it probably affecting me long-term, my university grades tanking, my diet becoming very unhealthy, and in general my mood being very negative. I'll try to stop this from turning into a wall of text and just say that I'll still be around to do mod duties, but unless something big happens very soon, I'm possibly going to have to take an extended break, for my own good.

>>7709

Hey man, that's understandable. You need to take care of yourself. You've done a massive job and I'm very impressed (and a bit guilty for not participating as much in the pattern thread). I hope you'll straighten things out.

>>7709

Ive found that in my studies I always wind up in that same position after about 3 months or so. Also, think of it like a video game. Ever play a game and you just cant get past one part? But then if you put the game down and come back after a couple of days you beat that level on your first attempt? Take a break. Sometimes we need to get out and see some trees in order to get the inspiration we need to make any progress.

It's easy to want to be 'the one' to figure it all out but we are racing against them, not each other. When I figure it out while you are taking a break I'll include you in a footnote somewhere towards the back of my dissertation :^)

>>7709

Hello AA, I can relate. I was spending so much time on this that my family life and health was suffering. I'm still thinking and working every day, but I've had to dial back my involvement a bit and make sure the other parts of my life are getting the focus and attention they need. Also, I had to ask for help to quit drinking which was fueling my late night obsessive working till early hours and wrecking my sanity and health. I have 37 days sober today.

I think I can speak for everyone here when I say Thank You for all your work over the last few months. Take the break you need to put your life back in balance, my friend. Love ya!

A concept you all might find interesting, from the General Number Field Sieve (it is pretty simple if you know what a smooth number is.)

Let's say we're going to look at 3-smooth numbers, numbers which at most have 3 as a prime factor. We can represent all of the 3-smooth numbers as evaluations of:

2^x * 3^y = 3-smooth number

where the exponents determine which number it is. For example, 12 = 2^2 * 3^1

We can represent this in a different notation called exponent vectors (the premise of the notation is simple, don't let that scare you).

Since we know we're looking at 3-smooth numbers, we only need to know the values of the exponents to represent a given 3-smooth number.

2^2 * 3^1 becomes (2, 1). (I haven't confirmed a connection here, but does this look familiar?)

So, our 3-smooth numbers become sets of exponent values. E.g, (3,1) = 24. We can also define multiplication of these "exponent vectors." Since a pair of values now represents a number, we can simplify multiplying these pairs by adding their exponent values instead of calculating the actual multiplication.

(4,2) = 144

(4,1) = 48

(4, 2) times (4, 1) = (4+4, 1+2) = (8, 3) = 6912 = 144*48 (This also defines squaring a 3-smooth number, as in (2,1) times (2,1) = (4,2) = 12*12.

Incidentally (or maybe not), any exponent vector that only contains even exponent values happens to be a perfect square. As in, (4,2) = 12*12, (4,4) = 36^2, (6,2) = 24^2, and so on. (This type of notation is used in GNFS to find one of the nontrivial squares (i^2 or j^2) which, interestingly enough happen to be equal when you are operating in GF(c) where c = ab, a field modulo c. When you are operating in GF(c) (you may have seen Galois fields in my ECC explanations), i^2 = j^2 (mod c) and i^2 - j^2 = 0 (mod c) (because they are equivalent to c, when it's modulo c then it's 0 since c%c=0).

So there you have it. Smooth numbers in Shor, smooth numbers in GNFS in a relatively simple introduction, and N-n as a smooth number.

c is b-smooth.

If you write c in exponent vector notation then every c value is in (1,1) because c = a^1 * b^1.

for future reference on our variable names (we are working on formulating these papers in the language of the VQC)

d = floored sqrt(c) = sqrt(c)//1 = sqrt(c) with no decimals

g = sqrt(c)

So we'll call the actual square root of c with the decimals g

>>7712

>>7709

Its amazing that it has come this far. Been following this for a while with a friend, but I really gotta say that this whole thing here is problematic and everyone here should take a break honestly. It says a lot that Chris would let it go this far. Simple as that.

That said, this is the most fascinating experiment in human nature I have ever seen in my life. Someone could get a PHD from this place without ever even doing a math problem to be honest. Simply analyzing the people and their interactions and such would be enough. I encourage you to all just maybe come here once a week or so AT MOST. Also, I encourage Chris to move things along. Things have gone too far imo.

Some formulae for deriving Kurzweg's variable k (from solution values j and i so far), in (A-New-Method-for-Factoring-Large-Semiprimes.pdf)

if (c%6) == 1 {

k = ( ( sqrt( (2j)^2 + 4c ) - (c+1) ) / 36) * -1

}

if (c%6) == 5 {

k = ( ( sqrt( (2i)^2 - 4c ) - (c-1) ) / 36 ) * -1

}

in the program "calc," they'd be written as (in the same order as above for the cases of which to use):

define kF(c, j) = (-1*(sqrt((2*j)^2 + 4*c) - (c+1)))/36

define kG(c, i) = -1*((sqrt((2*i)^2 - 4*c) - (c-1))/36)

examples of k values (examples c's from the paper):

kF(34417, 72) = 945

kF(21428053, 594) = ~594964.444

kF(21428053, 594)*36 = 21418720

kG(106577, 369) = ~2950.889

kG(106577, 369) * 36 = 106232

kG(732010841, 29229) = 20333020

In his paper,

G = (2i)^2 = 2(d+n)^2

F = (2j)^2 = 2(x+n)^2

>>7722

So, in the second paper, Kurzweg's second factorization method went from being a search for 2i to being a more informed search for 2(d+n)^2 or 2(x+n)^2 based on the remainder of c%6.

Interestingly, (c-1)/6 or (c+1)/6 is always in the ring of Z, which is number theory speak for it's always a whole integer, and which one it is is based on whether c%6==1 or c%6==5, respectively.

Also, an idea on what the point of the remainder tree was:

We were told to repeatedly divide each term by 2 until it was no longer a multiple of two. What does that make? A list of odd numbers, which also isolates the primes.

So , hypothesis:

with smooth numbers you are looking at the primes that make up a number by multiplication and with the tree you are looking at the primes in the remainders.

The tree wasn't the solution, it was a hint.

Kurzweg's 3 papers all have confusingly similar names, so I'm going to give them distinct names:

(A-Quick-Way-to-Factor-Large-Semiprimes.pdf) = the 2i paper, the 2(d+n) paper (because that's the value he's searching for, in his terms, Nf(N))

(A-New-Method-for-Factoring-Large-Semiprimes.pdf) = the 2i^2 and 2j^2 paper, because this paper contains a description of an iterative search for the values of 2i^2 and 2j^2 in our terms

(A-New-Approach-to-Factoring-Semiprimes.pdf) = the d, e and f paper, because the variables d, e and f all show up in this paper (I'll show how in a moment).

Next I'm going to outline his paper (the d e and f paper) in terms of our equations. I'm going to use greek letters for his variables that don't exactly map to ours because we've almost taken up the entire alphabet, and he introduces a lot of different variables.

A NEW APPROACH TO FACTORING SEMI-PRIMES:

step 1:

g = sqrt(c)

g is a variable we've defined before, this one includes the decimals

step 2:

α = the closest whole integer to g (d or (d+1))

step 3 - realize that there are two unknown values which add and subtract from α to make a and b:

a = α - β

(where β = x or x+1)

b = α + γ (this isn't a y (why), this is a γ (gamma))

(where γ = (x+2n) or (x+2n-1))

step 4 - which become, if we rearrange:

α * (γ - β) - β*γ = c - α^2

In our terms,

β*γ = x * (x+2n)

or

β*γ = (x+1) * (x+2n-1)

Here's where it gets fascinating:

c - α^2 = e or f

step 5, solve for (β*γ) and (γ - β):

β*γ = δ + ε (where δ and ε are an unknown integers)

γ - β = (c - α^2 + δ + ε)/α

δ < α

step 6, eliminate γ from the equations and set δ=c - α^2 (I'm not sure of the exact path of this mathematical jump so I'll figure it out later if it is important):

β^2 + β*(ε / α) - (δ + ε) = 0

step 7, solve for β:

β = -(ε / (2*α)) + sqrt( (ε / (2*α))^2 + (δ + ε) )

step 8, since we know ε / (2*α) must be an integer (because it's defining β and a = α - β), we can get to the value that we need to search for to factorize:

let ε / (2*α) = ζ

β = -ζ + sqrt( ζ^2 + 2*α*ζ + δ)

step 9, rearrange for factors:

a = (α + ζ) - sqrt( ζ^2 + 2*α*ζ + δ ) = (α + ζ) - η

b = (α + ζ) + sqrt( ζ^2 + 2*α*ζ + δ ) = (α + ζ) + η

letting η = sqrt( ζ^2 + 2*α*ζ + δ )

step 10:

α and δ are known from c ( δ = c - α^2 and α is the closest integer to g (d or d+1)), so now it just becomes finding a value of ζ that makes η evaluate to a whole integer.

next, Kurzweg solves for ζ and defines the search as between 0 and 1 (decimals), but I don't think that is the right direction to go with this.

>>7729

I'll program an algorithm of this tomorrow.

>>7729

We don't have to calculate g to determine whether g is closer to d or (d+1), we just have to see whether e or f is smaller. If f < e then α = (d+1), if f > e then α = d

Kurwzeg has been treading on our territory since 2013 (most recent paper on factorization from what I've seen so far was in 2015).

>>7729

You're right, that is damn close. I was following his last example in A-New-Approach.. and he almost hits a few of our variables (d+1 instead of d, thus missing our e-value).

>>7732

I wouldn't consider it a miss. Using d+1 produces f.

Ulrich H Kurzweg factorization papers list (to eliminate confusion) (they are all describing different algorithms and approaches):

November 1, 2012 - (Factorizing-Semiprimes-with-the-Integer-Spiral.pdf)

August 2013 - the search for 2(x+n)^2 or 2(d+n)^2 paper (A-New-Method-for-Factoring-Large-Semiprimes.pdf)

Christmas (December?) 2013 - the search for 2(d+n) [N(f(N)] paper (A-Quick-Way-to-Factor-Large-Semiprimes.pdf)

January 2015 - d, e and f paper, (A-New-Approach-to-Factoring-Semiprimes.pdf) (hereby renamed the zeta search paper since it is a search for the value of a variable I've renamed to zeta ζ for our purposes)

August 2015 - Factorizing with Q primes (Factoring-Large-Semiprimes.pdf)

April 7, 2016 - (UNIVERSAL-CURVE.pdf)

Who knows where he's gotten by now.

ZetaSearch (UHK) —— https://pastebin.com/iUVg7q2u

Kurzweg contacted.

Location of forum and of us was not revealed, only mathematical concepts shared in order to prove we are serious and are making enough progress (even to the point of improving on current research).

Pursuit of more of his research and his expected interest in our research was made clear.

>>7741

Exciting.

>>7743

start it in a terminal and make sure the window is big enough or resize the settings or change the terminal font size. If you want to learn type !help basic

Then from there there will be a tutorial type path to go !help commands or !help map

Really cool tool

>>7743

>>7744

Great work.

It is a lookup via the x values in row 1.

It is about finding how to make the lookup with column -f and e.

Something about the product of squares and triangles. I(t) stands out. Once you c it, you cannot un-c (i)t.

In The End, it is a lookup.

No more secrets.

Sneakers.

>>7743

Outstanding.

One pattern that might be important…

Negative x values in row 1.

According to the -x value hint, maybe look at x - n values since (-x + n) = (x - n) and see the u that that creates.

Just a quick look:

c287

X = 15, N = 128

(X+N) = 143, U = 2556, (X+N)^2 = 20449

(X-N) = 113, U = 56, (X-N)^2 = 12769

f = 2, e = 31

(31, 1, 7) -> {31:1:113:13:100:128}

(31, 1, 8) -> {31:1:143:15:128:160}

(-2, 1, 8) -> {-2:1:111:14:97:127}

(-2, 1, 9) -> {-2:1:143:16:127:161}

|

(31, 1, 0) -> {31:1:15:-1:16:16}

(31, 1, 1) -> {31:1:17:1:16:20}

(-2, 1, 1) -> {-2:1:-1:0:1:-1}

(-2, 1, 2) -> {-2:1:3:2:1:7}

287 = 7 * 41, x = 9, n = 8

(x+n) = 17, u = 8, (x-n)^2 = 289

(x-n) = 1, u = 1, (x-n)^2 = 1

The patterns say, look, I, the solution am here! But can we bring it out?

>>7752

>>7753

>>7754

>>7755

>>7756

>>7757

>>7758

>>7759

>>7760

>>7761

>>7762

>>7763

>>7764

>>7765

>>7766

>>7767

>>7768

>>7769

>>7770

>>7771

>>7772

>>7773

>>7774

>>7775

>>7776

>>7777

>>7778

>>7779

>>7780

>>7781

>>7782

>>7783

>>7784

>>7785

>>7786

>>7787

>>7788

Made a big string of posts on my grid thread. I think it's interesting that the get 7777 was the introduction of the f grid. Not planned.

I hope you guys enjoy and learn!!

>>7751

correction: U value of X+N is 71

if (e, n, t) = (e, n, d) then sqrt(2d) ~= sqrt(x)

>>7793

This is explainable with algebra. If d=t, depending on the parity of e, 2t will be within 1 of 2d. The square root of a number is very close to the square root of itself plus or minus 1.

Since I couldn't find any triangle visualization tools written in Java, I've been putting one together myself. It isn't finished. If anyone did already write one I must have missed it. Nobody will probably use this so it doesn't matter. Here it is, anyway. https://pastebin.com/3nDs75dy It relies on this GridCell object https://pastebin.com/HEC87GAQ

>>7798

With colours now (that was fucking horrendous to get my head around): https://pastebin.com/4CE3tirj

I just wasted another 8 hours straight doing something that probably isn't going to solve the thing. I haven't even eaten dinner and it's already past midnight. I have a problem, and I think the only way I'm ever not going to keep doing this is either if I quit altogether or if we finally solve it. I don't know which of those is going to come sooner than the other.

>>7748

Tops, my love for the will never waiver.

Maths without Art is like POTUS without genius.

>>7749

>>7799

>>7798

>>7797

>>7795

>>7794

>>7790

The best and most pragmatic advice I can offer is to use known numbers at scale as well as what you are doing.

Reconcile an earlier post with a HUGE drop in BTC.

Everything is connected.

You are on the verge.

100%

>>7801

https://news.bitcoin.com/why-the-cryptosphere-is-losing-its-mind-over-this-18-digit-number/

Product of a VQC?

>>7801

>>7805

https://blockexplorer.com/block/00000000000000000021e800c1e8df51b22c1588e5a624bea17e9faa34b2dc4a

>>7801

We've been on the verge for almost a year but okay

An interesting relation I found while working on the new hints that could become a quicker way to calculate integer square roots:

TM1(c-1/2) = d or d-1

>>7808

Clarification:

TM1((c-1)/2) = d or d-1

>>7809

>>7808

I just realized that the TM1 function relies on the integer sqrt function, so nevermind. But still an interesting relation.

Another triangle identity:

k = an integer

k^2 = 2T(k) - k

I've been avoiding talking about this because I know I'm not going to get an answer that actually explains jack shit but what was the point of the rm2dnm1 thing from RSA #10? n0 is somewhere between f/8 and 1. We want some number between them that is greater than n and less than x+n. That divided f value is a staircase number. You put it into the rm2dnm1 function and it gives you a number that isn't 0 because it isn't going to equal the correct u value. We never seemed to do anything with that value aside from checking whether it was zero. What was the point of that entire thing? If n0 is never going to equal n, we're never actually going to use that function for the sake of checking that a number gives a remainder of 0 and is thus u. So we do want the remainder, and we do want a specific value from the staircase number, but what are we even using them for?

Now you put a and b in through terminal and you can iterate through all cells with the same x+n. For even n odd x, it takes one away from n (to make it an odd square in the center) and adds the difference to f.

Triangles - https://pastebin.com/Kjm6jhZE

GridCell - https://pastebin.com/KRg2QTH9

>>7813

Thanks for this

>The best and most pragmatic advice I can offer is to use known numbers at scale as well as what you are doing.

>Reconcile an earlier post with a HUGE drop in BTC.

>Everything is connected.

And theeen you see that we're looking for the element in (e,1) where x + (e a value) - (-f a value) = original d value.

for c145 here's the calc:

original d value = 12

(e na transform) x value = 11

at x = 7, we have the following calc: x= 7 + (25) - (20) = 12 = original d value.

which gives us d = x + a, 12 = 7 + 5

So my idea is to use the original d value as the target that we're trying to make x+(e a)-(-f a) add up to. Have we already done this lads? Just thinking the clue through again and working it through.

>>7818

The bign thing probably isn't working when you view a BigN cell for an even a even b cell, by the way..

>>7818

Take a break.

>>7820

That was all I was going to do aside from potentially turning it into a bitmap generator actually.

>>7821

I know, but that's all you have to do to handle this well. Just take breaks.

>>7822

>that's all you have to do to handle this well

I'll handle this well when something happens that justifies the amount of time I've put into this.

>>7823

What kind of VQC Proof would you like?

>>7824

>>7823

This is proof to me. "Oh, let's explain it away as randomness." Nah

https://news.bitcoin.com/why-the-cryptosphere-is-losing-its-mind-over-this-18-digit-number/

>>7825

Sure, we were thinkin' that.

Fiiiiiiiine.

But what about my valtrex medication?

My husband has YET to pick any up.

>>7825

>>7824

>>7826

"when something happens that justifies the amount of time I've put into this" != proof. If I didn't think this was real, why would I have spent all that time? That would be fucking moronic. What I mean is finding the solution. That would justify me destroying a friendship and completely neglecting university because I was spending too much time doing Grid Patterns.

>>7827

I remember friends… be glad you only have VQC to sink into.

You want the solution… but you haven't earned it and we're not at a place where we need to have it finished yet.

What are you going to school for?

Something that will be meaningless when this is successful?

Even if you're going for something that will still exist… I'm an artist.

You have way more of an advantage by doing this, hopefully.

>>7828

Super constructive, thanks Topol

I'm only as good as what I have to work with.

Want me to send you money for a whore?

I'd… I might throw some cash at that, but I need proof.

No Proof = Refund.

c6107

{23:2976:78:77:1:6107} (23, 2976, 39)

{23:1:18641629:-6107:18647736:18635524} (23, 1, -3053)

{23:1:18653843:6107:18647736:18659952} (23, 1, 3054)

{-134:1:18647657:-6108:18653765:18641551} (-134, 1, -3053)

{-134:1:18659873:6108:18653765:18665983} (-134, 1, 3055)

α = 18641607

β = 9322268

γ = 18647657

δ = 18647712

γ - α = 6050

δ - γ = 55

γ - α - c = -57

α - (-134, 1, -3053)[b] = 56

α - (-134, 1, -3053)[a] = -12158

(α - (-134, 1, -3053)[a])/2 = -6079

α - (23, 1, -3053)[b] = 6083

α - (23, 1, -3053)[a] = -6129

α - (23, 1, -3053)[b] - α - (23, 1, -3053)[a] = -46

6107[x] = 47

c34117

{261:16875:184:183:1:34117} (261, 16875, 92)

{261:1:581950858:-34117:581984975:581916743} (261, 1, -17058)

{261:1:582019092:34117:581984975:582053211} (261, 1, 17059)

{-108:1:581984790:-34118:582018908:581950674} (-108, 1, -17058)

{-108:1:582053026:34118:582018908:582087146} (-108, 1, 17060)

α = 581950598

β = 290983839

γ = 581984790

δ = 581984713

γ - α = 34192

δ - γ = -77

γ - α - c = 75

α - (-108, 1, -17058)[b] = -76

α - (-108, 1, -17058)[a] = -68310

(α - (-108, 1, -17058)[a])/2 = -34155

α - (261, 1, -17058)[b] = 33855

α - (261, 1, -17058)[a] = -34377

α - (261, 1, -17058)[b] - α - (261, 1, -17058)[a] = -522

34117[x] = 75

This hasn't been ironed out, as one can see.

I give you guys presents but I'm not sure anyone opens them.

The only one who takes me seriously is VQC.

I guess it's because I connect dots in a non-linear fashion.

And because I have the barest understanding of the math you guys do.

You guys are amazing.

Seriously. You guys are smart as hell.

But it takes all kinds to complete the puzzle.

I need you.

And you need me.

My "crumbs" will stand the test of time.

Of that I am certain.

My mathematical intuition has never failed me.

I have the box, but not the key.

I cannot open them by myself.

So here are more presents for those anons who wish to open them.

May you find some inspiration!

#1.

>https:// medium.com/@thoughttheory/lets-see-how-deep-the-rabbit-hole-goes-9af5f9896941

=

>http:// archive.is/64L2h

#2.

>https:// www.academia.edu/7489568/The_pythagorean_relationship_between_Pi_Phi_and_e

>>7834

Are you "rambles in randomly labeled, yet complexly generated math" guy?

If yes…

When did VQC take you seriously?

>>7835

nope

Shall Capwn?

If yes, chat me up on faybo.

Let's catch you up.

>>7834

>I give you guys presents but I'm not sure anyone opens them.

>I have the barest understanding of the math you guys do.

I don't mean to be patronizing, but, I mean, duhhhh. I don't know who you are because your IP hash only has three posts associated with it, but chances are if you're complaining about nobody reading your posts because they aren't the same kind of math that the rest of us are doing, it's because you haven't done the following:

(a) read all of the threads so you understand the grid (it's not difficult, it's just time consuming - I was going to work on a simple guide in June (before that too) but nobody ever wanted to help me so it never happened). VQC explained it pretty well at the beginning of the first thread. We're finding a and b, the factors of c, c being any arbitrary number of your choice. c has a bunch of variables associated with it. You can grid these sets of variables with e as the x axis and n as the y axis, and they create infinite sets. Relationships between cells of variables (different a and b pairs) can be used to factor any arbitrarily large number. Understanding the multitude of patterns will lead us to the solution. Go read the threads. You don't turn up to a calculus class with a chemistry textbook and pass the class, man.

(b) figure out just how related your math is to the grid - if VQC is taking you seriously, maybe it's important stuff, but if you don't even understand the grid, how do you actually know it's related? Intuition is a strong thing, but you have to back your shit up with facts, anon. If we're all talking about (x+n)(x+n) and you're trying to relate the golden ratio to the Mandelbrot set, we're in completely different worlds. Apply it to the concepts we're all studying or we're not going to have a use for it (at least with our current objective).

(c) now that you'll know how the math we're doing works, you'll have the terminology and the context to apply your math to it in such a way that you can explain where you're coming from. We're working explicitly on what VQC is trying to explain to us. If you post something that seems like it doesn't have anything to do with that (like weird graphs with lots of geometric shapes in seemingly arbitrary places like that one anon), obviously you're not going to get very much attention. If it is related, explain in terms that we understand (i.e. with our math). Let's say you found out that f divided by d is equal to a (that's definitely not true but it's an example). Let's say you found it because you're super into geometry and you were messing around with the grid and found this weirdly shaped line. If you didn't understand the grid (which you said you don't), you'd be telling us "hey guys, there's this weird shape here and you really need to know about it" then all you're doing is showing us a weird line. What are we meant to do with that? We've seen lots of weird lines and none of them got us anywhere. If you understood the grid, you would be able to say "hey guys, f/d=a, isn't that neat". Then we'd actually know what you're talking about.

tl;dr: read the threads

>>7832

>>7831

A thread specifically about this type of thing: >>7839

I've got a new idea and it's simpler than what I was working on earlier. It's putting the remainder tree and the grid together and it also unifies the "the solution is recursive" and "the grid does all the work for you" hints.

The grid took me from (d+N) to (d+n)

you start at the odd column (it's either f or e), where t=-(d+N-1) then you take sqrt(2d) and it equals c, then you take sqrt d and it gives you another value and then you set that value as x and it will send you down the column, and then you take sqrt(d) again and use that as x, then keep doing that, sometimes it'll have you switch columns

Also, (d+N) - (d+n) is N-n because d is the same

>>7850

Without seeing your code, it doesn't take into account

>(e,1)

>(0,n)

>triangles

>diagonals

etc so if you generate triangle diagrams and output the respective (e,1) cells, just as an example, during your process, it might display some patterns.

>>7851

Do you have an idea on what to add to it?

>>7852

Well for a start, if you're starting in whichever column is odd, chances are it's different for odd e and even e, so you might just have to add one or divide by two somewhere. Just in general, you aren't utilizing every concept. While (0,n) and (e,1) might not be used in one particular solution, VQC said the three solutions he's aware of involve either (0,n) or (e,1) or diagonals. So as much as you could potentially find another solution, I think it's far more likely that you've found something that you're meant to use in conjunction with the other concepts. So maybe if you run your algorithm and every time you calculate a new cell you also find that cell's (e,1), for example, it'll have some important value in it, or its triangle configuration will have something in common with the solution cells, or something. My point is to start analyzing this at a deeper level. One other idea is that you could work backwards. If you're taking a cell and setting the square root of its d to x, you could take xx and xx+1 as potential d values in the cell that leads to the solution cell.

Here's the code.

https://pastebin.com/sEse8cxm

try exploring setting x = f or f-1 (depending on whether the column is even and odd), and try exploring setting x to root of f and root of e. Also, we should figure out what a diagonal is.

Also, I have no idea how to mix 0,n into this.

There was a "ghost in the machine," so to speak, as to the reason why some of the records are invalid in the screenshots, but the code has been corrected to reflect the original idea of the movement from (-(odd column), 1, -(d+N)) to (col, 1, -(d+n). Also, it wasn't terminating in one step, I just forgot to print out the movement, but Chris has already said an instant prime test will be included with the solution.

Enumerate the patterns in the first row (and into negative x).

It will be something that was there that you didn't c that will trigger it.

I will come back to clarify the other by the end of the week but briefly. Using the values of RSA 100 just as a large number example is a suggestion. In column e, the values at (e,1) for each element includes a value for a[t] which is an. The value bn is at a[t+n]. At (-f,1) the value a(n-1) is at the same value x in that cell as "an" for (e,1), the value b(n-1) at cell (-f,1) is one element less than bn at (e,1). This difference is key. It is not the only key, as you are seeing. I have found three, not including yours. Those three are in row 1, column zero and the side by side diagonal cells from the origin. There may be infinite keys.

Two sets of equations that run together and then merge, lock and key.

diagonal = one thing that is one unit apart from another thing, the offset is one

>>7859

Are you paraphrasing VQC or did you figure it out? Some of the lines appear to be quotes by VQC from past threads.

>>7862

just paraphrasing vqc

//using the diagonal to find pairs that are one row apart

c1541

e=20, f = 59

sqrt(e) = 4, sqrt(f) = 7

sqrt(f) - sqrt(e) = 3

2d = e + f - 1

{20:732:39:38:1:1541} i[t] = 771; j[t] = 770; sqrt(d[t]) = 6; sqrt(2d[t]) = 8; f[t] = 59

{-59:731:40:39:1:1541} i[t] = 771; j[t] = 770; sqrt(d[t] = 6; sqrt(2d[t]) = 6; f[t] = 140

140 - 59 = 81

difference between f[t] in (e,N, t) and (e, N-1, t) = 81

{20:6:39:16:23:67} = 1541; i[t] = 45; j[t] = 22; sqrt(d[t]) = 6; sqrt(2d[t]) = 8; f[t] = 59;

{-59:5:40:17:23:67} = 1541; i[t] = 45; j[t] = 22; sqrt(d[t]) = 6; sqrt(2d[t]) = 8; f[t] = 140;

140 - 59 = 81

sqrt(140) - sqrt(59) = 4

difference between f[t] in (e,n, t) and (e, n-1, t) = 81

{20:1:770:38:732:810} i[t] = 771; j[t] = 39; sqrt(d[t]) = 12; sqrt(2d[t]) = 17; f[t] = 1521; c[t] = (d+N)^2 - d^2

{-59:1:770:39:731:811} i[t] = 771; j[t] = 40; sqrt(d[t]) = 27; sqrt(2d[t]) = 39; f[t] = 1600; c[t] = (d+N-1)^2 - d^2

1600 - 1521 = 79

sqrt(1600) - sqrt(1521) = 1

{20:1:154:16:138:172} f[t] = 289; c[t] = (d+n)^2 - d^2

{-59:1:132:17:115:151} f[t] = 324; c[t] = (d+n-1)^2 - d^2

324 - 289 = 35

sqrt(324) - sqrt(289) = 1

I'm making the most progress working backwards, seeing that it's not about going from c to n, it's about going from d and e to n and N.

"Two sets of equations running together."

since f=(x+1)^2 at (e,1), f increases in squares by t of the opposite parity of e. if e is even, f increases in odd squares, and if e is odd, f increases in even squares.

eventually you realize x+n is so special because it's a view of the diagonal, of the key

>>7812

Bump

>>7876

>The smaller staircase number, T1, is simply the T1- 1

Typo here lads. Should read:

The smaller staircase number, T1, is simply T2- 1. (the next smallest triangle number)

>>7876

>>7878

What does that picture with the mini yellow and green triangles represent? Is it c? All I've ever seen going through past threads is the picture itself and not an explanation, so I haven't been taking note of it since I have no idea what it is. It doesn't look like (x+n)(x+n) because of the whitespace. But if it's c, I haven't ever seen anyone break it down into smaller parts that aren't dd+e.

>>7879

Here's the steps for a test program:

start with c

get d, e, and f

calc (f-1)/8

calc SQRT((f-1)/8) = base T2

define base T1 as T2-1 = T1

T1 + T2 = potential u value

iterate potential u value looking for lock using rm(2d-1) formula, basically looking for a perfect square.

hopefully find a lock for the correct (x+n)^2 value.

Basically the correct u value should be a multiple of the combined bases of T2 and T1. Remainders may need to be accounted for as well, not sure yet.

>>7880

>>7881

An example of the staircase numbers:

in the even column, (x+n)^2+3 = (f-1/8) + ((f-1/8)+1)

>>7882

Where did you get that equation from?

(18:1:5) = {18:1:49:8:41:59} f=-81

(1+8)(1+8)+3=(80/8)+((80/8)+1)

84=/=21

>>7883

It was typed hastily while I was typed, but the pattern observed. Add a division by 4 on the left side and it works.

84/4 = 21

>>7884

Tired*.

rsa100u = 529865753494301776968715634328960133662340771465

>>7881

Chris dropped a different formula at the very end of RSA#12 for N that I didn't notice.

We've been using

N = (c+1)/2 - d

but

N = (c-1)/2 - x

also works

>>7884

And this is just in (e,1), yes? Because you should be specific about that. Otherwise you would have solved it for even e.

>>7889

Yes, just for (e,1). I apologize for the false positive by not clarifying that.

Also, 2 values of c at (e,1) (a different idea)

2 values of c in both columns in -x and +x

145:

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

{1:1:10368:-145:10513:10225} (1, 1, -72)

{1:1:10658:145:10513:10805} (1, 1, 73)

{-24:1:10500:-146:10646:10356} (-24, 1, -72)

{-24:1:10792:146:10646:10940} (-24, 1, 74)

34117:

{261:1:581950858:-34117:581984975:581916743} (261, 1, -17058)

{261:1:582019092:34117:581984975:582053211} (261, 1, 17059)

{-108:1:581984790:-34118:582018908:581950674} (-108, 1, -17058)

{-108:1:582053026:34118:582018908:582087146} (-108, 1, 17060)

rsa100c:

{61218444075812733697456051513875809617598014768503:761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876:39020571855401265512289573339484371018905006900194:39020571855401265512289573339484371018905006900193:1:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139} (61218444075812733697456051513875809617598014768503, 761302513961266680267809189066318714859034057480651323757098846024549192056136964455981270339102876, 19510285927700632756144786669742185509452503450097)

{61218444075812733697456051513875809617598014768503:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306382919000431748126203278565987540269446653259579647945903160561369722164795555272230598393346522221773:-1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306384441605459670659563814184365672906876371327694609326591818469864302287758814225128252393697214227912:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306381396395403825592842742947609407632016935191464686565214502652875142041832296319332944392995830215636} (61218444075812733697456051513875809617598014768503, 1, -761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003069)

{61218444075812733697456051513875809617598014768503:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306385964210487593192924349802743805544306089395809570707280476378358882410722073178025906394047906234051:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006139:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306384441605459670659563814184365672906876371327694609326591818469864302287758814225128252393697214227912:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306387486815515515726284885421121938181735807463924532087969134286853462533685332130923560394398598240192} (61218444075812733697456051513875809617598014768503, 1, 761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003070)

{-16822699634989797327123095165092932420211999031886:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306384441605459670659563814184365672906876371327694609287571246614463036775469240885643881374792207327717:-1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006140:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306385964210487593192924349802743805544306089395809570668259904522957616898432499838541535375142899333857:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306382919000431748126203278565987540269446653259579647906882588705968456652505981932746227374441515321579} (-16822699634989797327123095165092932420211999031886, 1, -761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003069)

{-16822699634989797327123095165092932420211999031886:1:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306387486815515515726284885421121938181735807463924532048948562431452197021395758791439189375493591339997:1522605027922533360535618378132637429718068114961380688657908494580122963258952897654000350692006140:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306385964210487593192924349802743805544306089395809570668259904522957616898432499838541535375142899333857:1159163035527489297252269855748911245456985761764903865191919139010631678686397433667992819640306389009420543438259645421039500070819165525532039493429637220339946777144359017744336843375844283346139} (-16822699634989797327123095165092932420211999031886, 1, 761302513961266680267809189066318714859034057480690344328954247290061481629476448827000175346003071)

Also another thing worth noting, the b value at the record in (e,1) where a=N, b=N' (shadow N)

shadow N = 2d+N

>>7893

Tempted to try but I'm not even gonna bother trying to open that.

>>7899

There's a full parity chart here >>7386

a prime number appears once (2^(1-1)) in a column.

a product of two primes appears twice (2^(2-1)) in a column.

a product of three primes appears four (2^(3-1)) times in a column.

a product of four primes appears eight (2^(4-1)) times in a column.

a product of five primes appears sixteen (2^(5-1)) times in a column.

focus on solving it for odd x+n and it will expand into all 4 solutions.

Introducing the grid and the diagonals then moving to a series of odd x+n hints was not pointless. It's crucial the final step of the journey is solved organically and not given away.

The x+n hints are one solution and will lead to the grid solution.

It will all make sense in hindsight.qGncp

>>7903

>introducing the diagonals

Other than a code comment mentioning diagonals, have they really been introduced to any significant degree?

>>7903

Single post… talks like Chris… not same id tho.

qGncp takes you Down Under tho…

https://youtu.be/QgncP-f3XVs

>>7901

I believe this can also be visualized in (e, c) where c is the product. It's tied into the number of chains in that row, although I haven't quite figured that part out yet.

>>7906

I've been suspecting for a while that VQC has actually been dropping hints for all three keys, mixed in with each other. I believe the smooth numbers is part of one key, the x+n triangle is part of another and that some of his other hints belong to the third part.

I've been busy with life for a while, but I did have some realizations the past few weeks.

I can't believe I missed the sqrt(2d), it's obvious now, but it's how you find the triangular base of d. Take a triangle n(n-1)/2 = k. To find the triangular base you do 2k = n(n-1), then the square root will be sqrt(2k) = sqrt(n(n-1)). Floor this number and you will have (n-1).

>>7907

Yeah here I suspect aan(n-1) belongs to either the smooth numbers key or the third key.

>>7893

THANK YOU!!!!

>>7912

Why did you want that so much?

Create diagrams for examples -> see the patterns.

>>7915

That's super pretty!

Who is this? One of the nerds or has a new player appeared?

>>7906

>>>7906

>This is interesting too. Look at xx + e = 2na

>If you set a = 1, then xx + e = 2c because n = c

>>7913

Sometimes this is the best way to view it. It's in a grid for a reason right? I threw out my old pc and lost that pic and I missed it sorely

>>7917

sorry about that first part forgot I was working on something..

>>7915

>d^2 + e + f = 6141 //next square in structure

d^2 + e + f = 6241

same as (d+1)^2 or 79^2

nice diagrams!

>>7915

Nice work on this diagram, Anon!

Alright, I have an idea lads.

In the linked c6107 diagram, (n-1) = 35

Let's derive the staircase numbers, as I explained here: >>7881

(f-1)/8 = 16, T2 = 4 , T1= 3, T1+T2 = 7

The 2d(n-1) rectangles are 2730 + 2730, or 2 * 78 * 35

35 = (n-1)

T1 + T2 * iterate = correct (n-1)

7 * 5 = 35

So (f-1)/8 gives us the staircase triangle number bases to add together, and then multiply out looking for a lock. It will scale properly upward, even with huge c values.

Obviously we need to test some other examples, but deriving the T2 + T1 staircase numbers from f , and then using it to iterate/solve for (n-1) works in this c6107 example.

>>7915

25921

>>7924

204646843863289

Slow your horses VQC, no need to spoon feed us! I still need some time to process it and make sure I got it right.

Transpose—

The transpose of a given matrix is formed by switching the columns and rows.

{0:1:2:3:4:5}

{1:2:3:4:5:6}

{2:3:4:5:6:7}

{3:4:5:6:7:8}

{4:5:6:7:8:9}

{5:6:7:8:9:10}

{6:7:8:9:10:11}

{7:8:9:10:11:12}

{8:9:10:11:12:13}

(9:10:11:12:13:14}

>>7930

Those aren't valid cells though

>>7932

Correction: "with no error" -> "with an extremely small margin of error about the equivalent of a calculation."

Noob here, just wondering, what c value is implied in

>na and nb for any c can be found n apart in the cell at e,1?

I'm not sure if I should be throwing gardening tools at a grid cell but GCD seems to fit really well there

Wouldn't it be amazing if dark matter didn't exist and the grid did all the work for us?

I was really shitfaced last night and I could have sworn there is a pattern of n being a factor of the a values in the cell at e,1 and that n-1 had a pattern of being a factor of the a values in the cell at -f,1

>>7938

Ja M.O.N.D. :P

Hey Chris!

Since the parade was pushed back… are you still coming stateside?

>>7936

>what c value is implied

"any c"

>>7942

If you look at the patterns of n in a[t] you find pairs of elements of the n we're looking for multiplied by different a's and b's for a different, smaller c, that are separated by n for that c's elements

If you could calculate two multiples of n where the coefficients of n are not multiples of eachother you could calculate n by doing gcd(na, nb)

Kind of like the GCDFactor algorithm, but intelligent navigation. This may be the lookup.

These particular patterns may be coincedence but this is the approach I'd like to be investigated:

c21428053:

N=10709398

n=38

So we want to find a way to find multiples of 38 in e,1 (work backwards).

In e,1 if we look for n and its multiples, we do find

{412:1:318:14:304:334} a = n*8 t=8

{412:1:470:22:448:494} a = n*13 t=13

8*13 = 104

104's n value = 5 and these elements are 5 elements apart.

So we find our n times a different set of a and b for a different c with elements that are that c's n apart at e,1.

>>7946

and gcd(n*8, n*13) = n

so if two multiples of n are found, n can be split from them efficiently (unless the two coefficients are multiples of eachother, as in 7 and 21)

HOW DOES THE GRID ENCODE INFORMATION?

And the name of the city from that day shall be: "THE LORD IS THERE."

an interesting observation: N%4=n%4 generally

>>7953

What about with 8?

Some random food for thought:

c=ab

2 inputs

infinite solutions

infinite wrong solutions

1 right solution - two primes

encode solution not as a number

but as a pattern, algorithm, approach towards the number

>>7957

Encode as a Fractal?

An Algorithm that is Self-Replicating and Binary, perchance?

>>7962

this is so sad, Alexa factor integers in logarithmic time

What do we get when we sort numbers by their distance from a square instead of as a line?

2, 5, 10, 17, 26, 37, 50, 65, 82, 101

3, 6, 11, 18, 27, 38, 51, 66, 83, 102

4, 7, 12, 19, 28, 39, 52, 67, 84, 103

5, 8, 13, 20, 29, 40, 53, 68, 85, 104

6, 9, 14, 21, 30, 41, 54, 69, 86, 105

7, 10, 15, 22, 31, 42, 55, 70, 87, 106

8, 11, 16, 23, 32, 43, 56, 71, 88, 107

9, 12, 17, 24, 33, 44, 57, 72, 89, 108

[iterating d for e=31], relevant c=287

32

35

40

47

56

67

80

95

na = 56

so e,1 in our grid analyzing every value of d in d^2 + e for us. understanding why the grid is configured the way it is is a good exercise.

We want to master the pattern of factors of a[t] in e,1.

Every n value that exists in column e is a factor of a[t] in e,1.

>>6579

>>6580

What does "position of column 0" mean?

From our even column, (f or e), distance from column 0 can be expressed in jumps of 2n (where n=1), but not for the odd column.

anyone ever notice how 145 is the sum of squares?

VQC, please read this post and respond to it

As far as I can tell. I've finished getting all of the patterns and concepts together in the Grid Patterns thread >>6506 here. I have a couple things I wanted to ask about.

For whatever range of reasons, none of the other anons have been helping, and none of them have been proofreading or checking that I'm correct and that I haven't missed anything. That means it's probably likely that I missed something or that maybe something is incorrect. I know you of all people will know if anything is right or if anything important has been missed. There's a lot of information in that thread, but would there be any chance you might have the time to go through and check everything?

Secondly, in your explanations of (x+n)(x+n)=nn+2d(n-1)+f-1, you introduced us to n0 and the guess triangle base made of f/8 or f/(8*something). As far as I could tell going through the threads, you never actually seemed to do anything with either of these values. If n0=n and guess base=u, you can construct the (x+n)(x+n) square, but you even said yourself that we know n0 won't equal n and we know the guess base won't equal u. After explaining that, you stopped talking about n0 altogether, and only mentioned the guess base a bit cryptically. So what was the point of those variables?

Thirdly, with the grid patterns thread done, do you have any newer advice as to what we should be looking into? Is there anything specific (as in looking at this concept and that concept at the same time) that we should be doing that we can now that everything's in one place?

>>7986

I'm thinking fractals. That has to be what we are missing here. It's a way you can turn that massive search space into nothing—because the same patterns that apply to the massive apply to the small, in a fractal.

What happens when equations are evaluated in higher dimensions?

c = 15105

123 = floor_sqrt(-14883 + 244i)

Re: 38! + 1

>>7979

Hmmmmmm….

8,9

8,9,10

8,9,10,11

8,9,10,11,12

etc etc

Jumps?

Why are a and b defined as i-j and i+j?

What is the geometrical meaning behind it?

c=i^2-j^2 or is it

c=i^2+ij-ji-j^2

Keep in mind j is always < i

What does a and b MEAN in this context?

>>7988

>>7990

Read the threads

Riemann Zeta + Mandelbrot

http://www.dhushara.com/DarkHeart/RH2/RH.htm

Riemann Zeta + Non-Mandelbrot Set Fractals

ttp://primepatterns.tumblr.com/tagged/Riemann-zeta-function

>>7986

Also, why do you think the grid patterns thread is done

>>7995

Because I've gone through all of the threads and put everything from each one in there. Why do you think it isn't done?

>>7998

There's plenty of other grid patterns to look at that haven't been fully enumerated.

>>7999

Then enumerate them and put them in the thread yourself.

>>8001

I'll "calm my feckin' tits" when getting a giant two-month-long set of work done doesn't get a response of "you have more work to do". I'll calm down when I'm not the only one who does that work (I'm not the only one who is meant to work on it by the way, don't know if anyone realizes that). I'll calm down when suggesting that we work on something together for once rather than hiding off in our own little research corners completely ignoring everyone else doesn't get a response (this is a direct quote) of "who cares" on Discord. This is ridiculous. Give me one good reason not to ditch this board and all of you lazy pricks and figure this out by myself, Topol. I might as well considering how little interaction there is between any of us that isn't just mindless chatting on Discord. I'm over being told to calm down when every time I do anything useful here I either get completely ignored or I get this bullshit.

>>8002

You're real entitled, you know that?

>>8003

What do you think I feel entitled to exactly?

>>8004

You're going about this as if you've earned the solution, as if you deserve more. You're too obsessed with the endpoint to the point of putting your friends down just because they don't work on what you want them to.

>>8005

>you're going about this as if I've earned the solution

How am I doing that? Topol told me to calm down. I told him why I'm not calm. Everything I'm saying is backed up with facts. I'm not just having a big emotional rant. These are genuine issues (to varying degrees).

>point 2: nobody helped me with grid patterns

Now if everyone explicitly told me "we don't think this is useful, maybe you should do it yourself", maybe I wouldn't mind so much. You actually said multiple times that you would help me with it (so did a couple of the others) and yet you didn't do anything. So this isn't me putting people down because they don't do what I want them to. This is people saying they will do something and never doing it.

>point 3: nobody is working together

I'm not the only one who has complained about this. Several other anons have. Isee was one. I think he was complaining about it on Discord. I suggested we work together on something for once and the only response I got was "who cares". That's a thing that happened. Do you think that's a useful outlook for everyone to have in regards to working together?

Now how am I acting entitled? Are any of the things I said objectively incorrect?

>>8006

When did I say I was the only one doing anything? I said I was the only one working on one thing in particular. And then when I said I was done, someone told me to do more work (that they could do themselves, might I add). Also, I can admit I did snap that one time (that was generally a shitty period) but the only snapping I've done today is calling people lazy pricks. Nothing else I'm saying isn't based purely around facts. They might be angry facts, but they're facts.

>>8009

>And yeah, when you finish one thing and there's more that needs to be done… what kind of response do you expect?

When it's something VQC told all of us to do, and it therefore isn't just my job, I don't expect a response of "there's more work to do, go do more work".

>But you're not uniquely frustrated and we don't deserve to be the butt of your anger

All I'm doing is pointing out things that I think are problems. If there's a bit of anger in there, it isn't for the sake of making people upset.

>and frankly, neither does Chris.

Would you like to point out where I got angry at Chris? Because as far as I'm aware I didn't do that.

>>8010

There you go saying you're the only one doing something again…

And you've seen my try to herd the cats into playing with the same ball of yarn.

You're dealing with maths-people.

It's complex.

We get it. You're upset. Quit lashing at us.

Yeah: >>7986 <<< there you are being pissy and making entitled commands.

>>8011

How the hell is that pissy? He misses postss addressed to him pretty often. That's why I made it redtext.

>>8012

Jesus, you sound like you're new here!

I've made bomb ass points before that he didn't acknowledge at the time.

IF he mentions something I've worked on outside of saying he likes my style… it's when it's coming into play.

For starters… let's look at another pattern.

Have you NOT noticed "someone" ip hopping and making posts?

Sometimes they respond to someone, and sometimes they seem to be in their own little world…

Haaaaave you not figured out what's going on with that, yet?

You are the only one here who can see the ip hashes, after all.

>>8013

No offense to whoever posted these if they're new and they're just firing off whatever ideas come to them, but is there anything we can actually use any of these posts for?

>>7950

>>7957

>>7988

>>7990

I don't think VQC posted any of those. They're the kinds of musings we all had at the beginning, which makes me think it's a newfag with a dynamic IP or a VPN. When VQC has crumbs, he posts more specific stuff like >>7746 this. Whatever the case, I really don't see how I was acting pissy towards VQC.

>>8016

>the one with the clue in the email

Where? I don't see one.

Okay, but if the goal is to rethink our approach to mathematics as a way of rewriting it, then why would any musing be unwelcome? None of them are things that have been completely explored

>>8019

Maybe read my post again because I never said they were unwelcome. I said I don't think VQC posted them because they're musings and VQC posts more specific hints about variables.

>>7988

>Re: 38! + 1

38! + 1 = 523022617466601111760007224100074291200000001

Is a semi-prime with factors 14029308060317546154181 × 37280713718589679646221

Interesting write-up and sample code at https://codegolf.stackexchange.com/questions/8629/fastest-semiprime-factorization

>>7988

>c = 15105

>123 = floor_sqrt(-14883 + 244i)

Relevant records:

entry c: (221,7431,61) = {221:7431:122:121:1:15105} = 15105; f=24;

na: (221,1,61) = {221:1:7552:121:7431:7675} = 57032925; f=14884