>>6834

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

>>6840

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

>>6862

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

>>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?

>>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

>>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)?

>>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.

>>6936

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

>>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: 66 + 216*5 + 30 - 1

While for (-30, 5) it will be: 55 + 217*4 + 65 - 1

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

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

>>6941

Just to build a bit on this.

I was reading over some older crumbs by our beloved mentor, also known as VQC.

Attached are two triangles based on the record a=7, b=37.

First one is based on (3, 6) the second one is based on (-30, 5). I used the equations in my previous post for this stuff.

I opted for the original nn + 2d(n-1) + f - 1 as I'm starting to suspect that it is irrelevant how we structure it. I don't think it's about how we structure it as much as it is about us understanding the triangle. Although I realized now that I did add f before 2d(n-1), but oh well.

Legend is:

nn = red (darker shade of red is remainder)

f-2 = yellow (darker shade of yellow is remainder)

2d(n-1) = blue (had no remainder)

The center square doesn't exists since I'm removing it (nn + 2d(n-1) + f - 1 = 8Tu + 1, moving the one over and we're left with 8 triangles).

I've also balanced the remainders on the opposite sides as a way of balancing out the distribution of the elements.

>>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.

>>6954

Added the same triangles, but in a "proper" way. These now follow the same pattern, ie nn first, 2d(n-1) second and finishes off with f-2.

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 = 2319.

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.

>>6967

One thing I've seen, which catches my fancy:

Take c, generate a[t] = Nc in (e, 1). Now generate record for a[c - t - 1]. This a[c - t - 1] = kc for some k. If we divide kc/c we get k. If we then go to -f, and repeat we will end up with (k+1)c.

Make record (1, c)

Generate record for a[t] = Nc in (e, 1)

Generate record for a[c - t - 1] in (e, 1)

a[c - t - 1] = kc for some value of k.

kc/c = k

Now make the -f record for (1, c).

Find the a[t] = (N - 1)c

Here, in -f, we have a[c - t - 1] = (k+1)c

For some reason I would assume it would be (k-1). So do we have another layer of asymmetry?

Attached shows the case of k, k+1 for (e, 1) and (-f, 1)

>>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.

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

>>6987

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

>>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.

>>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) =4u(u+1) => 2(2u*(u+1)).

2u(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 4T(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)) - ddef 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 2a, 2b. 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.

>>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

>>7019

Isn't this this just the asymmetry we've talked about before with regards to a(n-1), an, b(n-1), bn combined with a[p + t]?

Or are you referring to the parity effecting it?

I also see a difference in how you estimate t for even numbers. This is something we've gone over a few times, it seems like we're all uneven on how to estimate t from x. PMA and I had a conversation about this a few days ago as I used a different equation (indexing t=0 instead of t=1).

I believe the "correct" calculations are:

even x: x/2 + 1 = t

odd x: (x+1)/2 = t

So for your examples with even e the t should be 269 and 265 respectively. Assuming I'm not messing up with the t-calculations again.

>>7037

I've been looking into x and negative x, but I haven't found much of use yet.

What I've seen is that the number of chains / sequences in (e, n) appears to depend on the number of factors in n, but it doesn't always seem to hold. For some cases, we'll have k^2 number of chains / sequences where k is the number of factors in n, but again it doesn't always hold (For a prime number it appears they usually have only 2 chains, but c=15 which has two factors, 3 and 5, also only has two chains).

It appears that these sequences / chains come in pairs, and despite not appearing to be connected, they are at -x. Take for example c=3367 (a=37, b=91). It has 8 chains (and 3 factors):

e:

0 (3, 3367, 207, 201, 6, 7142)

1 (3, 3367, 847, 761, 86, 8342)

2 (3, 3367, 3165, 2347, 818, 12246)

3 (3, 3367, 4935, 3309, 1626, 14978)

4 (3, 3367, 5167, 3425, 1742, 15326)

5 (3, 3367, 7245, 4387, 2858, 18366)

6 (3, 3367, 11271, 5973, 5298, 23978)

7 (3, 3367, 12871, 6533, 6338, 26138)

-f:

0 (-114, 3367, 333, 318, 15, 7385)

1 (-114, 3367, 1837, 1502, 335, 10073)

2 (-114, 3367, 1989, 1606, 383, 10329)

3 (-114, 3367, 4933, 3308, 1625, 14975)

4 (-114, 3367, 5169, 3426, 1743, 15329)

5 (-114, 3367, 9033, 5128, 3905, 20895)

6 (-114, 3367, 9297, 5232, 4065, 21263)

7 (-114, 3367, 12529, 6416, 6113, 25679)

The indentation shows the connection between them. Move into negative x for any of these pairs and you'll find the chain / sequence with the same indentation level.

For example, take e and chain #3. (3, 3367, 4935, 3309, 1626, 14978)

Create the cell for -3309 and you will get: (3, 3367, -1683, -3309, 1626, 1742). Then we'll use the x + 2n and create the cell for e=3, n=3367, x=-3309 + 2*3367 and you will get the cell: (3, 3367, 5167, 3425, 1742, 15326) which is equal to chain #4 (meaning they are a connected pair).

The connected pairs here is (0, 7), (1, 6), (2, 5), (3, 4) each have the sum of 7 (indexing with 0 and 9 indexing with 1) and the sum of x in these connected pairs appear to always be 2*n (n=3367 =2*3367 = 6754).

The difference of x between the connected chains also maps to a d-value, meaning each chain is the result of a*b where 3367 is a factor of these. The chains in the middle (#3, #4) seems to always represent (1, c).

To show an example of this:

The difference between x in chain #0 and it's connected pair chain #7 is 6533 - 201 = 6332. 6332/2 = 3166.

3166^2 + e = 10023559 and 10023559/3367 =2977.

Another way of getting to the connected pairs is through a[p + 1 - t]. Take c and create the cell for Nc in (e, 1). In this case, a[p + 1 - t] = kc where k is the first a in the connected chain. For example: c = 259

It has the cell (3, 114, 16, 15, 1, 259) which gives us the Nc cell: (3, 1, 29769, 243, 29526, 30014) which is at t=122. If we now make the cell at 259 + 1 - 122 =(3, 1, 38089 275, 37814, 38366), we get a = 37814 => 37814/259 = 146.

The sequences in (3, 259) is:

1 (3, 259, 279, 201, 78, 998)

2 (3, 259, 357, 243, 114, 1118)

3 (3, 259, 421, 275, 146, 1214)

4 (3, 259, 511, 317, 194, 1346)

This means that the a, b we're after is also connected to another sequence, maybe we're going for a little hunt? Like go from our c to the alternate sequence and from that go to our proper sequence?

>>7038

To repeat, the chains that correspond to our (1, c) appears in the center/middle of these chains. Moving outwards and you'll find multiples of c.

>>7039

Column 0 is special. I haven't quite grasped it yet, but there are several interesting parts going on here.

Take c=259, in (0, 259) there is only one chain / sequence.

1 (0, 259, 1036, 518, 518, 2072)

But by multiplying c with odd multiples of 2, we increase the number of sequences:

(0, 2*259):

1 (0, 518, 777, 518, 259, 2331)

2 (0, 518, 2072, 1036, 1036, 4144)

(0, 2**3 * 259):

1 (0, 2072, 1295, 1036, 259, 6475)

2 (0, 2072, 3108, 2072, 1036, 9324)

3 (0, 2072, 5439, 3108, 2331, 12691)

4 (0, 2072, 8288, 4144, 4144, 16576)

(0, 2**5 * 259):

1 (0, 8288, 2331, 2072, 259, 20979)

2 (0, 8288, 5180, 4144, 1036, 25900)

3 (0, 8288, 8547, 6216, 2331, 31339)

4 (0, 8288, 12432, 8288, 4144, 37296)

5 (0, 8288, 16835, 10360, 6475, 43771)

6 (0, 8288, 21756, 12432, 9324, 50764)

7 (0, 8288, 27195, 14504, 12691, 58275)

8 (0, 8288, 33152, 16576, 16576, 66304)

(0, 2**7 * 259):

1 (0, 33152, 4403, 4144, 259, 74851)

2 (0, 33152, 9324, 8288, 1036, 83916)

3 (0, 33152, 14763, 12432, 2331, 93499)

4 (0, 33152, 20720, 16576, 4144, 103600)

5 (0, 33152, 27195, 20720, 6475, 114219)

6 (0, 33152, 34188, 24864, 9324, 125356)

7 (0, 33152, 41699, 29008, 12691, 137011)

8 (0, 33152, 49728, 33152, 16576, 149184)

9 (0, 33152, 58275, 37296, 20979, 161875)

10 (0, 33152, 67340, 41440, 25900, 175084)

11 (0, 33152, 76923, 45584, 31339, 188811)

12 (0, 33152, 87024, 49728, 37296, 203056)

13 (0, 33152, 97643, 53872, 43771, 217819)

14 (0, 33152, 108780, 58016, 50764, 233100)

15 (0, 33152, 120435, 62160, 58275, 248899)

16 (0, 33152, 132608, 66304, 66304, 265216)

(0, 2**9 * 259):

1 (0, 132608, 8547, 8288, 259, 282051)

2 (0, 132608, 17612, 16576, 1036, 299404)

3 (0, 132608, 27195, 24864, 2331, 317275)

4 (0, 132608, 37296, 33152, 4144, 335664)

5 (0, 132608, 47915, 41440, 6475, 354571)

6 (0, 132608, 59052, 49728, 9324, 373996)

7 (0, 132608, 70707, 58016, 12691, 393939)

8 (0, 132608, 82880, 66304, 16576, 414400)

9 (0, 132608, 95571, 74592, 20979, 435379)

10 (0, 132608, 108780, 82880, 25900, 456876)

11 (0, 132608, 122507, 91168, 31339, 478891)

12 (0, 132608, 136752, 99456, 37296, 501424)

13 (0, 132608, 151515, 107744, 43771, 524475)

14 (0, 132608, 166796, 116032, 50764, 548044)

15 (0, 132608, 182595, 124320, 58275, 572131)

16 (0, 132608, 198912, 132608, 66304, 596736)

17 (0, 132608, 215747, 140896, 74851, 621859)

18 (0, 132608, 233100, 149184, 83916, 647500)

19 (0, 132608, 250971, 157472, 93499, 673659)

20 (0, 132608, 269360, 165760, 103600, 700336)

21 (0, 132608, 288267, 174048, 114219, 727531)

22 (0, 132608, 307692, 182336, 125356, 755244)

23 (0, 132608, 327635, 190624, 137011, 783475)

24 (0, 132608, 348096, 198912, 149184, 812224)

25 (0, 132608, 369075, 207200, 161875, 841491)

26 (0, 132608, 390572, 215488, 175084, 871276)

27 (0, 132608, 412587, 223776, 188811, 901579)

28 (0, 132608, 435120, 232064, 203056, 932400)

29 (0, 132608, 458171, 240352, 217819, 963739)

30 (0, 132608, 481740, 248640, 233100, 995596)

31 (0, 132608, 505827, 256928, 248899, 1027971)

32 (0, 132608, 530432, 265216, 265216, 1060864)

Say we take 2^17, then we should have 2^((17+1)/2) number of chains (which checks out). As for doing it an even number, then it appears to have 2^(n/2) number of chains.

>>7057

>>7056

Well he is about to be kicked out of his house/embassy.

>>7092

I can relate to this. I feel pretty stuck my self and with the latest "hints", the lines of code and comments I'm kind of stuck. The square root things just popped seamingly (Not the d, though. That we've speculated on for a while, not that we have any clue on how to use it) out of nowhere and got me a bit confused.

Maybe it is being postponed, maybe VQC really wants us to figure it out ourselves, maybe something else.

I've been dividing my time between trying to figure out what square roots of things are used for, x=f-1 and just looking for more patterns in the grid. Mostly looking for more patterns and writing down my thoughts to cement them.

Regarding the square root of d, f and e I was thinking in the terms of pythagorean theorem, that this big thing could just boil down to a few triangles that we need to triangulate and that with those values, we have a start position and need to figure out the rest. But that is just a wild guess, like anything else.

The recursive tree is still lost on me, I never understood it when it was first introduced and I still don't see how we're going to use it with regards to x=f-1. All in all, I agree and I'm feeling lost in all of these hints, patterns and triangles.

I was thinking about the asymmetry regarding -f, e and an, a(n-1), bn and b(n-1).

I haven't figured out how to determine which row is off by one, but it does make sense and what I find actually more strange is that an and a(n-1) occurs at the same t.

So if you think about it, bn and b(n-1) being off by 1 makes sense. Since we know that when a[t] = an, a[t + n] = bn then for b(n - 1) it would be a[t + n - 1] = b(n-1). If it didn't then -f wouldn't abide by the grid / column rules.

Another thing which seems interesting (but I'm not sure if it will hold for all cases) is that if you were to increase n by d (essentially setting d=0, giving you the column=c) then at the correct a[t] where a[t] = a(n + d), x[t] = a.

(1/3)

I hit a doozy of a fun pattern here. Since it contains a lot of different elements to it, I'm going to split it up into a few posts describing the different patterns in it.

I have been looking into jumping up and the consequences of this. As you jump up (increasing n), d and x decreases (I've posted about this before >>6648) and e increases. After d jumps up, e = c, as d = 0. If you keep jumping up, e will start to decrease as d becomes negative. After a total of 2d jumps, you'll be back to the original e, however, our n value will now be different. It will be 2d + n, which also refers to a valid record in e, which is at (e, 2d + n). Here both d and x is negative, but our a and b value will remain normal. Keep jumping from here on out and it will grow in the negative space forever, but the interesting part is when it bounces back to column e after 2d jumps.

I made another set of posts a while ago where I described how sequences / chains in (e, n) are connected. If the number of sequences is greater than 1, then it will be paired with another set of sequences which will meet at negative x. After jumping 2d times, we end up in the paired n.

A few examples showing how this jumping works:

For this example I'll use a=7, b=37, c=259. The record for (1, c) is:

{3:114:16:15:1:259}

If we look at the column/row for (e, 259) we will see a total of four sequences, one is 114 and the other is 146. I showed in >>7038 that these are connected in negative x. With the example record above, when we jump 2*16 times, we end up with the record:

(3:146:-16:-17:1:259}

So the paired sequences meet, and our c is now in the negative x-space of the connected pair.

(2/3)

This is a neat pattern, but does it bring us any value? Yes it does. Once I saw this I started playing with other records, so let's use the record for our actual a=7,b=37:

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

Now we jump 2*16 times and we end up with:

{3:38:-16:-23:7:37}

38 is a connected pair of 6 and also exists at (3, 7) in the record a=6, b=38. Now here comes the fun part.

This means we have actually TWO different an/bn values. We have 76 and we have 738 and we also have 376 and now 3738.

What makes this pattern even more fun is the following property. Let's find the record at t where a[t] = 7*38 in (3, 1).

{3:1:289:23:266:314}

And now for bn:

{3:1:1459:53:1406:1514}

So you might think, oh cool more numbers to work with, but this isn't something THAT awesome. Except it is. The difference in t, between an and bn is not 38, but it's 15. Why does 15 make it FUN? Because 15 is equal to x + n. That's right, for these records in negative x-space, with the connected n, the difference between a[t] = an and a[t] = bn is x + n.

(3 / 3)

To move on a bit. Let's look at -f, because here is also fun time. Our records in -f usually has n = n - 1, but for these they exists in n = n + 1, and also in negative x-space. There also isn't any asymmetry between them. They exists at a[t] = an, a[t] = a(n + 1) in both e and -f. Same for a[t] = bn and a[t] = b(n - 1).

Now all of this made me think more about >>6645

> When a appears as "an" it appears another time

Take the record for a[t] = 7x6:

{3:1:51:9:42:62}

After jumping 2d times, it exists again in:

{3:103:-51:-93:42:62}

Same holds true for bn:

{3:1:243:21:222:266}

{3:487:-243:-465:222:266}

To summarize a bit:

By jumping a record up by 2d you will end up with a new n-value. This means we have 2 an-values, but with different n-values. Connected through negative x. The difference between this a[t] = an' and a[t] = bn' is x + n.

>>7135

I was THINKING about this and I realized I haven't been thinking that much. Looking back I feel like I've been more brute forcing, more rushed and not enough stepping back and thinking about it.

So to give some intuition about this, what we're actually doing is to remember that the square root of any given positive integer is NOT just another positive integer. The square root of any positive integer has two results, sqrt(d) = k AND sqrt(d) = -k.

The reasoning is simple, the square root of a positive integer is equal to k and -k. Any negative number squared will "lose" it's negativity and become a positive integer.

What I'm doing above is looking at the negative part of the square root of c, which does have a different set of values for n, d, and x. This in turn gives us a different set of an and bn to work with that is different from the record where a=a, b=b and d = the positive result from sqrt(c).

I was thinking about our numbers in (e, 1), and for odd e's they consist of 4 triangles. What does this really mean?

I was playing with it and to build some intuition about it I drew up four of these triangles. Instead of multiplication, let's say we want to simply add them together. In the image above I used t=8 for (3, 1) which is the two N's that belong to c=259.

So for an odd e, we have the equation:

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

That is 4 triangles plus half ( + 1) of the remainder. So if we want to add the two values of this cell we would get:

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

This is the same as:

2t(t+1) + 2t(t-1) + e + 1. This means adding this rows will give us a total of 8 triangles, half of which are one unit longer than the other half. Image should clear this a bit up.

I am contending now that take a number c, if this number has an odd remainder, then adding 1 to it will give you a number who's sum can be expressed as the sum of a and b from a single record in (e, 1).

So for 259, add 1 and you get 260. This number is equal to the sum of it's two big N values (Remember turns out each number has 2 n values as one belongs to a positive d and the other to the negative d).

Note: In the image I added 114 + 146, both are 4 triangles, one a unit longer than the other (114 = orange, 146 = green), but I omitted e + 1 =3 + 1 = 4. That is, there are 4 left over.

You can also see how it consists of 4 squares each of which has the square root of 8, because it belongs on t=8.

I don't think this is anything new, but it is just to help people get a better grasp of the values in (e, 1). For even e's every a and b consists of two squares added together, so a + b from any cell[t] in (e, 1) where e is even is the same as 4 squares added together. Quite similar to what we have in the image.

>>7155

I'm not too familiar with the General Number Field Sieve, but smooth numbers appear to be a big part of it.

Now first simplify our calculations for our two N's (one that is 2d "jumps" away, or negative d)

Our normal 'n' is defined as:

n1 = (a + b)/2 - d

But our paired n is:

n2 = (a + b)/2 + d

These two n's exist in (e, a) as a and b for some t.

But, over to the smooth numbers.

We know our formula for n and big N, big N is:

bigN1 = (1 + a*b)/2 - d

bigN2 = (1 + a*b)/2 + d

If we subtract bigN1 - n1 and bigN2 - n2 we have the same number, which appear to always be a smooth number.

When we do bigN1 - n1 we actually do:

((1 + ab)/2 - d) - ((a + b)/2 - d)

Here the d's cancel each other out and we're left with

(1 + ab)/2 - (a + b)/2 = k (for some smooth number k)

1 + ab - a - b = 2k

ab - a - b + 1 = 2k

(ab - a - b + 1)/2 = k

This can also be rewritten to:

(a - 1)(b - 1)/2 = k

or

(a - 1)(b - 1) = 2k

Also note that (a + b)/2 is equal to (d + n). That means if we remove (d + n) from our big (d + n) we get the smooth number we want(?)(Unless we want to remove smooth numbers from our big (d + n) to get our smaller (d+n)).

>>7215

Keep in mind that:

d = (a + b)/2 - n

2d = a + b - 2n

sqrt(2d) = sqrt(a + b - 2n)

Make the record for (1, c). 2d here could also point to the record for 4c (or a=2, b=2c) as the d there is 2d.

dd + e = c

dd = c - e

dd = ((a + b)/2 - n)*((a + b)/2 - n)

dd = ((a + b - 2n)^2)/4

dd = ((2d)^2)/4

((2d)^2)/4 = c - e

>>7223

I think the point is that big N - n is a smooth number, and they're used in number field sieving.

I've been reading a bit about GNFS and from what I can understand is that it goes through several steps to setup the algorithm before actually running, which is why it isn't used for smaller numbers. One of the steps in the algorithm is that it sieves for smooth numbers. I wonder if we're either going to find the smooth number (Big n - n) through calculations, or if this is just a hint towards finding the speed up algorithm which will lead us to solving the calculation.

We know that big N - n (column e) is the same as (big N - 1) - (n - 1) (column f). I wonder if that means we're going to find this number in two calculations using different columns. The question is if this was always about finding u, the triangle base created by x + n. I assumed it was, but now it seems we're stopping by GNFS, so I'm not sure anymore. Maybe we're stopping by GNFS before finally arriving at the calculation of our triangle base?

>>7216

I haven't had too much time to work on this lately, but I have some more patterns related to our two n-values.

We have two n-values for any given record. One is for positive d, the other is for negative d. The difference between the n-values is 2d.

For positive d, we have n. For negative d we have n' (n MARK).

For n we have the known pattern of:

a[t] = an, a[t + n] = bn

For n' we have:

a[t] = an', a[t + x + n] = bn' (Note t + x + n, as in NOT n')

We can create an f for the cell for n', this will be negative (while our normal f is positive).

The difference between these two, f and f' is e + 1.

Now some fun patterns crossing these two sets:

a[t] = an

a[t + d + n] = bn' (Again, note the [t + d + n], it is NOT n')

a[t] = bn

a[t + d] = bn'

a[t] = an'

a[t + x + n] = bn'

The difference in t between a[t] = an and a[t] = an' is b.

The difference in t between a[t] = bn and a[t] = bn' is d.

Would love it if someone verified.

>>7238

Minor correction:

a[t] = an

a[t + a] = an'

That means the difference in t between a[t] = an and a[t] = an' is not b, it is a.

a[t] = bn'

a[t - b] = bn (Might be in the negative part of the grid =-x)

>>7239

Just adding some more patterns:

a[t] = an'

a[a + 1 - t] = an (Might be negative x)

a[t] = bn'

a[b + 1 - t] = bn (Might be negative x)

We know we have sequences in (e, n), sometimes we have 1, other times there is more than 1 sequence. But our record is always tied to a sequence within (e, n).

a[t] = an

a[a + 1 - t] = ak

This time k belongs to a different sequence and points to a record where our 'a' exists as b and k is some value of from that sequence.

Example:

Take a=7,b=37

a[5] = 42 (7x6 = an for record above)

a[7 + 1 - 5] = 14 (7x2, this record exists in (3, 2) where a=1, b=7)

Another example:

Take a=37, b=91

a[11] = 222 (37x6 = an for record above)

a[37 + 1 - 22] = 1406 (37x38, this record exists in (3, 38) where a=7, b=37. Note this is also the second n (where d=-16) for the record a=7,b=37).

So a[p + 1 - t] allows us to move to a record where our a becomes b, notice also how they swap back again, 37 + 1 - 22 = 16, 37 + 1 - 16 = 22. So we can think of a[p + 1 - t] to move between records, as if they were a linked list.

Since a[p + 1 - t] points to ak, where k is some value of (e, k), we should have enough data to make that record. We start with the record: {3:6:124:33:91:169}

a[t] = an =a[17] = 546

a[91 + 1 - 17] = a[75] = 11102. 11102 / 91 = 122. We then create a record using ENB for 3, 122, 91.

{3:122:-58:-95:37:91}

a[t] = an =a[48] = 4514

a[37 + 1 - 48] = 222. 222/37 = 6

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

a[t] = an =a[5] = 42

a[7 + 1 - 5] = 14, 14/7 = 2

{3:2:2:1:1:7}

I've been thinking more about the number of records that we are interested in that exists ONLY in column e, for some c.

I've compiled a list:

The structure of the list is as follows:

(a, b) denotes the record where a=a, b=b.

(1, c) denotes the record where a=1, b=c

(e, 1) denotes the records that exists in column e at row 1

n = The n we know

n' = The shadow n

Column e:

(a, b)

• (e, n)

• (e, n')

(1, c)

• (e, n)

• (e, n')

(e, 1)

• an

• bn

• an'

• bn'

• N

• N'

• cN

• cN'

This gives us a total of 12 records for any given a, b that are of interest to us. Out of these 12 we can already generate 6 of them. Below is the same list only filtered based on what we can generate from c alone:

(1, c)

• (e, n)

• (e, n')

(e, 1)

• N

• N'

• cN

• cN'

This is not including records from -f, so we're really dealing with a lot more cells here. We should be dealing with another 12 records for -f, knowing 6 of them. So using e and -f, we should have a total of 12 records that we can generate based on c alone, out of a total of 24 records.

>>7246

Since the board is drying up a bit and people are stuck, I'm throwing out a few ideas I've been working on.

I've been trying to figure out a way of using the recursive tree by generating x=f or x=f-1 based on the columns we get from recursing on d and e. However, I've been using too small a numbers and I'm getting a lot of false positiives (think gcd on values in these records where x=f, x=f-1).

Looking at patterns of t from the 12 records we can create. Use bigger numbers to remove irrelevant patterns (of course when looking for patterns you can generate the other 12 records).

Defining a problem statement, I'm as guilty as you guys, but we should definitely do this as it's been a big hint. We could try to formulate one together here and I'm pretty convinced now that we're missing a variable that will help us.

Look for more patterns. Since I've felt a bit stuck I decided to go back to the grid and look for more, which is how I found the shadow n and the patterns within it. Those are damn cool and if you're stuck and out of ideas, go back and verify those patterns. Maybe I missed something, maybe you can find more and deeper patterns within them.

I'm not going to sit around and wait for VQC to reveal it all, I'd rather try and fail until he does.

>>6996

For the shadow n, we have the very neat pattern of:

a[t] = an' (shadow n)

a[t + x + n] = bn' (x, n from solution record where a=a, b=b and n' is shadow n).

The difference here is x + n, which means the difference in x between these two records is 2(x+n), which reminds me a lot about VQC's hint regarding 4c, or the record for a=2, b=2*c.

Example:

a=37, b=91

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

shadow n = 122

a[48] = 37x122

a[75] = 91x122

75 - 48 = 21 + 6 = 27.

The records for t=48 and t=75 is:

{3:1:4609:95:4514:4706}

{3:1:11251:149:11102:11402}

149 - 95 = 54 = 2(21 + 6).

Not sure if these are connected in a clever way, as in a solution-wise clever way.

>>7254

This also means that:

>>7238

> a[t] = an

> a[t + d + n] = bn'

The difference in x between the record for a[t] = an and a[t + d + n] = bn' is 2(d + n).

>>7255

Just looking a bit more into this.

The x-values for the record a=2, b=2*c for both the normal big N and shadow big N is equal to the two f's we have in (e, 1).

The value of x for normal n is equal to f with regards to a positive d and the value of x for the shadow n is equal to f with regards to a negative d.

Thinking more about this one too >>6469

I haven't "gotten" anything out of it yet, but I decided to look into it.

We know d = (a + x)

We know f = 2d + 1 - e

We know f = 2(a + x) + 1 - e

We know how to compute elements in -f.

a[t] in even negative f is calculated:

2tt - e/2

In this case e = f

2tt - f/2

2tt - (2d + 1 - e)/2

2tt - (2(a + x) + 1 - e)/2

2tt - (a + x + (1 - e)/2)

2tt - a - x - (1 - e)/2

Not sure where I am going with this, though.

>>7271

Correction >>6470

Just an interesting observation, not entirely sure what it means yet or if it is a pattern of use.

Take a record with an odd e.

a[t] = bn

Then we now know that d[t - 1] - d = b(n - 1). I was thinking about d[t] - d[t - 1] and noticed the pattern. For odd e's the d in (e, 1) is calculated as 2tt + (e + 1)/2. But when we are subtracting d[t] - d[t - 1] the (e + 1)/2 is removed and we're left with 2tt - 2(t-1)(t-1). Essentially, the number of integers between the two squares. What I noticed is that this is actually related to the d of our record and of the next record in the sequence in (e, n) where a and b exists.

Say we have a record in (e, n) for a, b where e is odd. Let's say we use the first chain, so we choose the record at t=1 within (e, n). We then have a second record in that chain and we know the method to calculate this (VQC also posted this in >>6736). Let's call the record at (e, n, 1) for r_1 and the next record in the sequence for r_2.

There is a t in (e, 1) where a[t] = bn from r_1. We know the difference between d[t] - d[t - 1] = 2 times the number of integers between the squares. If you add r_1.d + (d[t] - d[t - 1]) you will get the d from r_2. This means the difference between the d's in a sequence within (e, n) is two times the number of integers between the squares for t and t-1 where a[t] = bn.

This also means that b + x + 2n = 2 times the number of integers between two squares (paraphrasing d' = a' + x') >>6736).

To repeat, we have two records r_1 and r_2. They belong to the same sequence in (e, n) (for odd e) with a t-difference of 1. If we were given r_1 we can construct r_2 (>>6736). The difference in d between these two records is equal to 2tt - 2(t-1)(t-1) where t is the t-value belonging to r_2 (That is t = (x + 1)/2 for an odd x, t = x/2 + 1 for even x).

Example:

r_1 = {627:6:1310:117:1193:1439}

r_2 = {627:6:1568:129:1439:1709}

In (627, 1) we have:

a[65] = 1439*6, which gives the record: {627:1:8763:129:8634:8894}

The previous record is at t 64: {627:1:8505:127:8378:8634}

The difference in d[t] - d[t - 1] = 8763 - 8505 = 258.

1310 + 258 = 1568.

For even e's this holds true for a-values (In an even e, d's are 4 triangles with a base of t).

>>7316

I think this is actually quite amazing, but not in the sense that it is going to help us solve our factorization problem, but rather as insight. It essentially allows us to get a better understanding of how the values in (e, n) relate.

We know that dd + e = c. What this does is show a relationship between this c and the next one in (e, n). This is important to get a better understanding of the grid, which I think is the actual main goal here. I'm not here anymore to factorize, but rather to understand the grid. This is more about learning how numbers truly are related and less about breaking cryptography.

This shows us a big part of the relationship between the records in a sequence. We can take a record and observe how the differences in d is related to the integers between two perfect consecutive squares. It is quite beautiful.

What we now know is that given a record, the difference between d^2 + e = ab and (d + k)^2 + e = bi (where i is the next b' in the sequence) is equal to the difference in squares multiplied by two, where the squares are defined to be the t^2 and (t-1)^2 of the next sequence.

>>7319

This is a neat pattern. What we're actually seeing is:

dd - d + e = 2t - 1 (for odd e)

dd - d + e = 2t (for even d)

When solving for t we get:

Odd e:

dd - d + e + 1 = 2t

(dd - d)/2 + (e + 1)/2 = t

d(d-1)/2 + (e + 1)/2 = t

If we treat d as the base of a triangle and add half of e (rounded up) we get the t for Nc (big N transformation for nb for a=1, b=c).

We also know the equation a[c + 1 - t] = n'c (shadow big N for c). This gives us:

c + 1 - d(d-1)/2 - (e + 1)/2

dd + e + 1 - d(d-1)/2 - (e+1)/2

dd - d(d-1)/2 + (e+1)/2

(For an even e it just turns into e/2)

What is neat here is how the formula is so similar to our a's and b's in (e, 1). And what it "means" is that if we calculate the T_d (the dth triangular number) and add half of e (rounded up) we get the t for our Nc (bigN, nb for a=1, b=c record) in (e, 1).

If we remove the triangle with a base of d, from the d^2 and add half of e, we get the counter part N'b (shadow big N, nb transformation for the record a=1, b=c).

(Trailing off a bit, but I hope this makes sense. Looking over the jargon I can't imagine anyone not familiar with this grid would be able to make sense of anything).

>>7320

This pattern is interesting, because the t for the nb transformation for c is located at d(d-1)/2 + (e + 1)/2, but there are values in (e, 1) that does not fit into this equation.

I wonder if that means this "separates" the values in (e, 1), since this does not hold true for all values in (e, 1). As in, this pattern will hold for all c, meaning a number that doesn't fit this pattern is not a c, but something else.

>>7324

Generate from 1 to k:

2(i(i-1)/2 + (e + 1)/2) - 1

Each of those will be an x-value in (e, 1) which also has a corresponding record with a=1, b=i^2 + e

Each of these records exist in an (e, n) as a=1, b=i^2 + e. Which means they should all exist in unique n's. They will only exist in n's that have a record with a=1, meaning there are n's that is not covered by this (ones that doesn't contain a=1).

That should also mean it includes every a, b in column e that has a=1, b=ab, but not necessarily the n where a and b exists (I think).

This makes me wonder what's so interesting about the numbers that doesn't fit in these patterns in column e.

>>7344

Yeah, this is a different way of looking at what I posted about regarding shadow n.

In your first example you have 17864/203 = 88. Here 88 is the big N of 203. Then you move t + d (or in your case x + d + 1) and you get 23548/203 = 116.

116 is the shadow big N of 203. And those patterns are the same as the ones I posted in >>7238 except with a perspective of x rather than t.

This is what bugs me a lot. I post tons of stuff, but no one is verifying and checking to see if it actually makes sense. It's like there is no team working here, just a bunch of people doing whatever they're doing and ignoring what everyone else is doing.

The patterns I've posted about the shadow n seems to be ignored and I don't get it. It seems like a pretty damn big step. There's lots of very cool patterns in it.

>>7346

So the first pattern:

x = c - d

It's the same as

t = d(d - 1)/2 + (e + 1)/2

You can see it being described in >>7320

And x = c + d is the same as

t = d(d + 1)/2 + (e + 1)/2

Meaning we get big N and shadow big N by treating d as triangles. One with Triangle(d) and the other at Triangle(d + 1).

>>7344

To build a bit on it.

The patterns I found for n and shadow n are with regards to the a[t], but if you look at them with regards of x you'll have:

x = d - a (x for an)

x = d + a (x for an')

x = b - d (x for bn)

x = b + d (x for bn')

Where n is normal n and n' is shadow n.

>>7529

Well if you remember a bit back, I ranted a lot about the shadow n. Every record has a "main" record and the shadow record. The shadow record is a valid record, but with values shifted by 2d, this is also true for the x. In the shadow record, x is negative and is equal to the "original" x - 2d.

Not sure if this is what we're looking for, but if we find the "shadow" x, then we could use that to find the "correct" x.

>>7520

Yup it was me.

Not sure I'm heading down the correct path (or if I know where I am heading), but I'm trying to think about the aan(n-1) and what it means.

We know n(n-1) is two triangles and a singular triangle is defined as n(n-1)/2. This is also the sum from 1 to n.

So two triangles is the same as:

2*(1 + 2 + .. + n)

or

2 + 4 + 6 + .. + 2n

Essentially the sum of even numbers from 2 to 2n.

Let's ignore that for now and pretend our number is really aan(n-1)/2, that is, we're only dealing with a single triangle.

Then what we're really doing is:

aa(1 + 2 + 3 + .. + n)

Or

aa + 2aa + 3aa + … + naa)

Which is summing the square up to n times. For our case, since we're dealing with two triangles, we instead have:

2(aa + 2aa + 3aa + 4aa + … + naa)

or

2aa + 4aa + 6aa + … + 2naa

Not sure where I'm heading with it, though. I was thinking that maybe we could coherence information from this.

>>7535

Note though, that bn and b(n-1) occur in different indexes of t (a[t] = b(n - 1) in f, a[t + 1] = bn in e). So they won't appear when we do the naive multiplication.

That might be the "key", but I haven't seen any obvious patterns there yet.

>>7545

If you use an offset, that is, you multiply a[t] from f with a[t + 1] from e, you'll get the bbn(n-1).

So I've been looking a bit into our grid, thinking more about what we actually are looking at and I was reading up on arithmetic progression, geometric progression and series in general.

Our grid is just a bunch of series with different offsets and triangles and squares, right.

So I started to play a bit with sequences and did some analysis using that way of thinking about our (e, n) and I found what I believe is the general equation for generating the nth record in (e, n).

It's remarkably simple, in fact so simple I'm surprised we haven't noticed it before. First I want to clarify on our n's. We have two types of n, one that occurs as a in (e, 1) and one that occurs as a factor of a in (e, 1). The first one I refer to as primary n, or root n, while the second one is a sub-primary n, or factor n. The first one will always have a=1 in (e, n), thus following the rules of the grid (a[t] = an in (e, 1), where a[t] = 1n for some t). The second one will not, it will have a "start" record with a=k (for some k, factor of a[t] = kn).

We know the equations for (e, 1). They are either the sum of half of e + 1 if e is odd added with either a product of triangles or squares.

odd: a[t] = (e/2 + 1) + 4t(t-1)/2

even: a[t] = e/2 + 2tt

or, if we want to limit our selves to purely triangles:

a[t] = e/2 + 2(t(t-1)/2 + t(t+1)/2)

The pattern for primary n's, or root n's is:

a = 1

b = 1 + 2(n + x)t + 4n(t(t-1)/2)

d = x + 1

while for sub-primary n's it is:

a = k

b = k + 2(n + x)t + 4n(t(t-1)/2)

d = k + x

In the first pattern, the x refers to the x in (e, 1) where it first occurs (which is also the same the cell in (e, n) where a = 1).

For the second pattern, the k is an offset, some factor where a[t] = kn in (e, 1). Here x is also the same as for when the n first occurs as a factor in (e, 1).

Now this also makes me think about VQC's hints. We know using -f and e, we're supposed to find an offset (in the second pattern you can think of k as an offset) and we know by multiplying -f and e we will end up with aan(n-1). That means, if we multiply the product of -f and e with 2, we will have one part of our equation for calculating records in (e, aa).

That is, k + 2(n + x)n + 2aan(n-1). The latter part will be known (although only in the sense that it exists as a product of a[t] in e and -f, but not which product).

I've been playing a bit with the idea from VQC about multiplying -f and e, but I trailed of a bit and found something funny.

If you imagine a number line with d's as poles (GA talked about this before) then our c exists between dd and (d + 1)(d + 1). It will exist at dd + e and at (d+1)(d+1) - (2d + 1 - e).

That is the variable we call f, I decided to take a look at c from "below" our d, that is d-1. I have called it g, simply because I needed a name. In g we will have n+1 instead of n-1. This means if we were to multiply e and g, there is a result that is equal to aan(n+1) instead of aan(n-1).

It's just a different point of view, but I decided to just play around with g and f out curiosity and I decided to ignore e and just play with g and f. The idea was that we know that in -f we have a[t] = a(n-1) and at g we will have a[t] = a(n+1). Multiplying these two rows will give us aa(n-1)(n+1). But I trailed off again and decided to play with addition. This means at some t we will add a(n-1) + a(n+1). This should result in 2an. However, I did a mistake in my function and instead of adding a[t] from -f with a[t] from g, I ended up skewing it and added a[t] from f with a[t + 1] from g. The result surprised me, as it generated the list that exists in (4e, 2) instead. The funny thing is that in (4e, 2) we have (at least) two sequences. One sequence will contain c and the other is the result of this addition.

I'm sure it can be explained easily using algebra, I haven't looked into it yet, but I was just surprised and found it very interesting. Not sure if this is a blind alley or not, but I figured I should share.

>>7586

Meh, I think it's a blind alley. I looked into the algebra and it boils down to the difference of g and f. The series created from it won't always match 4e either, although the pattern inside of it should still be valid.

As in we have two sequences, one with a[t]=c and the other where an should appear at some t.

>>7588

> How do you stop releases being scrubbed from the internet?

Only way to do that while minimizing the risk of scrubbing would be to expose it to as many people as possible in one big go, or to the right people who could then help spreading it. If enough people are spreading it faster than they can scrub it, it's out there.

> Can you write it to the blockchain?

Yup, might take multiple transactions depending on what you're releasing. Releasing the whole thing vs just enough equations to solve the problem.

> Can you write it to many?

Yes, should be able to write it to any blockchain that supports extra data. But you would still be limited to the amount of data pr. transaction you could store.

> Would they take out all crypto to scrub?

Most definitely. At least try.

>>7632

Almost looks like Sierpinski triangle.

>>7633

I want to believe!

I'm not sure what I'm actually doing. I've been looking into the VQC thinking "lookup" table and I was playing around with x=d in column 0, row 1 for even d's. I noticed a pattern for BigN - n, but I don't know if it's anything usable. It doesn't solve anything, just a pattern.

Anyway, if you take an (e, n) and create the cell in (0, 1, x=d). The a will relate to the bigN - n, but not in any obvious way. What I've looked at is creating the smooth number BigN - n, then subtracting it from (0, 1, x=d).a, which gives another number. If you do this for a record, it will generate a new sequence of numbers that correspond to another column where a=(0, 1, x=d).a - (BigN - n).

Example:

a=7, b=37, c = 259, d = 16

BigN = 114,

n = 6

114 - 6 = 108

Column 0, row 1 x=d: {0:1:144:16:128:162}

128-108 = 20.

The next record in the sequence of (3, 6) is a=37, b=91, c=3367, d=58. This has:

BigN = 1626

n = 6

1626 - 6 = 1620

{0:1:1740:58:1682:1800}

1682 - 1620 = 62.

These two numbers will exist in (15, 6): {15:6:35:15:20:62}.

This appears to hold true for all records.

At first glance it could look like it is (e + 2n, n), but it isn't true for all the records. If you look at the other sequence in (3, 6) you will have one record that has a=19, b=61. So it could also look like 19 + 1, 61 + 1, but again, that doesn't work for all the records. Sometimes it is an increase of something else.

Another interesting thing regarding n's and shadow n's and smooth numbers.

Create both Big N's for a record (Normal and shadow). Subtract the smooth number from these two. It will give you a new set of numbers that will exist in (e, k) (for some k) where x + n = d.

Example:

a=7, b=37, c=259, d=16

BigN = 114, shadow BigN = 146

Smooth number (114 - 6) = 108.

114 - 108 = 6, 146 - 108 = 38

{3:7:15:9:6:38}

7+9 = 16

Another example:

a=61, b=131, n=7, d=89, e=70

BigN = 3907, shadow BigN=4085

Smooth number (3907-7) = 3900

3907 - 3900 = 7

4085 - 3900 = 185

{70:61:35:28:7:185}

61 + 28 = 89.

Meaning there is a record in column e, where d = x+n that is related to our own record. In this record the d is equal to x+n.

>>7663

> Meaning there is a record in column e, where d = x+n that is related to our own record. In this record the d is equal to x+n.

Just to clarify a bit

Here a is the n for the record a=61, b=131 while 185 is the shadow record. That means a=n, b=shadow n exists in (e, a).

These two records are the "normal" record and the shadow record:

{70:7:89:28:61:131}

{70:185:-89:-150:61:131}

This is the record that is created when you remove the smooth number from BigN and shadow BigN, it is also the record where a=n and b=shadow n.

{70:61:35:28:7:185}

>>7664

Just one last quick note, a=7, b=185 will also exist in n=131 (b-value), but here it will be negative.

{70:131:-35:-42:7:185}

In some cases the a's in column e will match the parity of x in column -f. If you highlight, or "mark" these columns in -f you will see that they are spaced out by 2n - 1 (The difference between square n^2 - (n-1)^2 for some n). For smaller values, if you highlight the center between these marked cells (since 2n-1 is odd there will be a center piece) it tends to contain our factors. Might just be a side-effect of using smaller numbers, though.

>>7666

The product of the two n's (normal n and shadow n) for our cell (a, b) is equal to (x + n)^2 + e.

Given the cell in column e for (a, b) for some a and b. This cell will have n, but there will also exist a shadow n' for it. The product of n and n' is equal to the (a, b)s cells (x + n)^2 + e.

The cell in column e for (n, n') will have x + n = d from the cell of (a, b).

This gives us: n * n' = (x + n)^2 + e

>>7670

So knowing nn' = (x + n)^2 + e

We get

nn' = xx + 2xn + nn + e

We know xx + e = 2an

nn' = 2xn + nn + 2an

nn' = n(2x + 2a + n)

n' = 2x + 2a + n

>>7671

This also means:

n' = 2(a + x) + n

n' = 2d + n

>>7670

To build a bit on this.

We have the cell: {3:6:58:21:37:91}

This has another cell with negative d (shadow cell?): {3:122:-58:-95:37:91}

The records for where a=6, b=122 is:

{3:37:27:21:6:122}

{3:91:-27:-33:6:122}

In these cells, x + n = 58 (d). The product of these records are 732 (6 * 122) which is (x + n)^2 + e (6 + 21)^2 + 3. Meanwhile those records will have (x + n) as d. (6 + 21 = 27).

This means we can find the triangular base for the shadow cells, but we'll only know e and x+n.

I'm not entirely sure what to call this, but I noticed something when I was looking at 4c. In my head I've been thinking of it as a kind of record transformation. Take a record and multiply it by 4 (you can think of it as by dividing the 4 into 2 by 2, multiplying a by 2 and b by 2.).

The reason I'm calling it a transformation is because we skew or transform the patterns in (e, 1). Take a column e and row 1. Multiply it by 4. In (4e, 4) you will find the same values, but this time the patterns are skewed, as in, the sequences / chains within this record no longer matches the ones we have in (e, 1).

Example, look at (3, 1). Multiply it by 4 and we get (12, 4).

{3:1:3:1:2:6} {12:4:4:2:2:14}

{3:1:9:3:6:14} {12:4:12:6:6:26}

{3:1:19:5:14:26} {12:4:24:10:14:42}

{3:1:33:7:26:42} {12:4:40:14:26:62}

{3:1:51:9:42:62} {12:4:60:18:42:86}

{3:1:73:11:62:86} {12:4:84:22:62:114}

{3:1:99:13:86:114} {12:4:112:26:86:146}

{3:1:129:15:114:146} {12:4:144:30:114:182}

{3:1:163:17:146:182} {12:4:180:34:146:222}

{3:1:201:19:182:222} {12:4:220:38:182:266}

{3:1:243:21:222:266} {12:4:264:42:222:314}

Here we can see the sequences have changed. In (3, 1, 1) we have a=2, b=6 followed by (3, 1, 2) where a=6, b=14. This forms a single chain / sequence within (3, 1). In (12, 4) however, we can see that we now have transformed (3, 1) into two sequences. If we were to multiply (12, 4) again by 4 the pattern would repeat (transforming yet again).

If we were to multiply (e, 1) by n^2 (For example n=6 where a=7, b=37 exists) we would end up with a record that connects / pairs all the an, bn transformations in (e, 6).

{3:6:16:9:7:37} =an = 42, bn = 222 => {108:36:96:54:42:222}

{3:6:58:21:37:91} =an = 222, bn = 546 => {108:36:348:126:222:546}

{3:6:124:33:91:169} an = 546, bn = 1014 ={108:36:744:198:546:1014}

I haven't looked enough into this yet, there's so many damn patterns to try and understand, but I suspect that multiplying any (e, 1) with a square will result in (e * k^2, k^2) records to "skew" the original record, generating multiple sequences.

>>7674

Forgot to mention, 108 is also the smooth number belonging to record a=7, b=37, c=259. We have n = 6 and BigN = 114 with 114 - 6 = 108. I believe I've seen patterns before, but forgot to write up about the smooth numbers within a column running in a pattern similar to the other patterns we're seeing.

Also, note that there are two sequences in (3, 6) and both of them are "paired" in (108, 36). Example a=1, b=19, n=6. This gives an = 6, bn = 114, which is at (108, 36, 2).