Anonymous ID: d032c1 June 21, 2018, 7:47 a.m. No.6424   🗄️.is 🔗kun   >>6427

>>6395

 

I've been thinking about

> Any product of 2 primes will be divisible by 3 if you add either 2 or 4, or if you subtract 2 or 4.

> Any product of 2 primes will be divisible by 5 if you…

 

And I can't see any specific patterns here. Adding or subtracting 2, 4 from a product of 2 primes for mod 3 is fine. This makes sense. Mod 3 will either be 0, 1 or 2. If it is 1 then adding 2 will "remove" the remainder, and subtracting 4 will also "remove" the remainder.

If it is 2 then adding 4 or subtracting 2 will also "remove" the remainder.

 

But as for 5, 7, 9(are we limited to primes only?), 11, … I can't see any reasonable pattern as the number grows.

 

If you multiply two, any given, primes and then compute the modulus on a specific value (ex. 5) then it should cover all the values of 0, 1,…, (5 - 1). In which you will always have to cover these numbers.

 

I was thinking in the line that this might, somehow, be related to parts of the grid, but nevertheless I can't say I see anything straight forward.

Anonymous ID: d032c1 June 23, 2018, 4:56 a.m. No.6447   🗄️.is 🔗kun   >>6449 >>6450 >>6452

>>6439

Great list AA.

 

>big_N’s D minus the real d to our semiprime c contains (n-1) as a factor

 

Is this true though?

 

Take a look at a=7,b=37. That gives us the record:

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

 

For (1, c) the record is:

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

 

Big Ns record in (3, 1) is

{3:1:129:15:114:146}

 

129 - 16 = 113. 113 Does not have (n - 1) as a factor, which would be 5.

Anonymous ID: d032c1 June 23, 2018, 5:13 a.m. No.6452   🗄️.is 🔗kun   >>6453

>>6450

Ugh I'm an idiot. Yeah big N's D - d thing doesn't hold.

 

As for the d[t] - d that is a reference to the d[t] at a[t] where a[t] = an.

 

So based on the example in >>6447

The record for 7*6 is:

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

 

Here 51 - 16 = 35 (contains (n-1) as factor).

 

I think that one holds, I was just a dummie before and though it also held for bn and I guess big N. But that means it doesn't hold as we have a counter example above.

Anonymous ID: d032c1 June 23, 2018, 5:42 a.m. No.6454   🗄️.is 🔗kun

>>6445

Are we now going to back the first few threads where you talked about controlling our rows by multiplying primes with c? Feels like we're going back now, but this time we know so much more.

 

>>6453

No problem, I haven't checked it all, some parts I recognize others I've checked just to see if it makes sense to me. I've been though like half of those at least, but I barely remember half of the properties of the grid.

Anonymous ID: d032c1 June 23, 2018, 6:41 a.m. No.6455   🗄️.is 🔗kun   >>6457

Trying to interpret. Comment if you interpret it different, want to supplement or for any other reason.

 

>>6445

> Given two columns f+e or 2d+1 apart.

> How can we tell or construct the list of factors in a[t] that are 1 unit apart?

We know how to generate records for any given [t] in (e, 1) and (-f, 1). They are on the shape of:

 

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

even e: 2tt + (e/2)

 

odd f: 2t(t + 1) - (e - 1)/2

even f: 2(t + 1)*2 - e/2

 

How to construct a list of factors for those numbers I don't know. We could start with t = 1 and then count upwards and add to our list. Some will be products of numbers, but will we always have a factor for those? I think we do, but I'm not sure.

 

>Those factors represent the cells with elements in each column.

 

>The values of a[t] in each column are all the possible values of x squared plus e (or -f).

We know this already, but I don't think we've though too much about it. It's been known for many threads and used in a lot of our methods for calculating records based on different values: a = (xx + e) / 2n.

 

We divide a by 2n, because in those context we already know n, x and e and we know a[t] = 2an. So we remove 2n from 2an to get a.

 

>The key to using the grid is at hand.

Somewhere in this thread we have the key. Triangulation, jumping from column to column? Solving our triangles?

 

>The difference in the element position of b(n-1) and bn by ONE element at (-f,1) and (e,1).

The difference in t between b(n-1) and bn is 1. In (e, 1) a[t] = bn and in (-f, 1) a[t-1] = b(n-1).

 

> What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.

Anonymous ID: d032c1 June 23, 2018, 9:13 a.m. No.6457   🗄️.is 🔗kun   >>6458

>>6455

For our example a=5, b=29, c = 145 we have the following:

 

For f: a(n - 1) = 2tt - 24/2

For e: an = 2t(t-1) + (1+1)/2

 

an - a = 2tt - 12

an = 2tt - 12 + a

 

an = 2tt - 12 + a

an = 2t(t-1) + 1

 

That means also that

2tt - 12 + a = 2t(t-1) + 1

a = 2t(t-1) + 1 - 2tt + 12

a = 2tt - 2t -2tt + 13

a = -2t + 13

 

This means our a can be expressed as -2t + 13 for our specific t. This gives us some limitations in the possible t values of a.

 

For a bigger example:

Take a =1217, b =1979. This has the record:

(e, 1) = {2842:47:1551:334:1217:1979}

 

Here f = 261 so we know e is even and f is odd.

For an odd f, the equation for a is:

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

 

and for even e it is:

a[t] = 2tt + e/2

 

e/2 = 2842/2 = 1421

(f-1)/2 = 260/2 = 130

 

So we now now:

 

2tt + 1421 = an

2t(t+1) - 130 = a(n-1)

2t(t+1) - 130 = an-a

 

2t(t+1) - 130 + a = an

2tt + 1421 = an

 

2t(t+1) - 130 + a = 2tt + 1421

a = 2tt + 1421 - 2tt - 2t + 130

a = -2t + 1551

 

Meaning the t for the a in (e, 1) and (-f, 1) is somewhere between 0 and (1551/2). Not entirely sure where I'm going with this, but if I am correct then the maximum t value for our a[t] is f/2 + e/2.

Anonymous ID: d032c1 June 23, 2018, 9:32 a.m. No.6458   🗄️.is 🔗kun

>>6457

I now wonder what happens if you keep this up. I don't think this is a rabbit hole worth following, but we know also that you can jump with 2(2d + 2) - e which represents the "next" f.

 

So for our 145 example we can also express the set of equations as:

an = 2t(t+1) - 25 + 2a

an = 2tt - 12 + a

an = 2t(t-1) + 1

 

Note the 25 comes from our "next" jump which is -51 (2(2d + 2)) in which case a = 2t(t+1) - (e - 1)/2.

 

I'm not sure if this would help us, though. I'm not sure how many jumps like this we can make. If we do it n times a*(n-n) would be 0. In which case both a(n-n) and b(n-n) would be equal to 0.

Anonymous ID: d032c1 June 24, 2018, 9:55 a.m. No.6466   🗄️.is 🔗kun

>>6445

>What happens when you compare the -f and e columns in the grid for 4c? The square for x+n has now sides of 2(x+n) compared to c.

 

The interesting thing about this approach is that you can control the parity of the n, and thus the parity of n-1. Take our beloved example of 145. If we multiply it by 4, we can think of it as 25 * 229. This record exists in e=4, n=10.

 

Exactly how to use it, I'm not sure yet. It should allow us though to use the t + p, t + 2*p and p + t - 1 to search through the records of that column.

 

Since we don't know any of the factors of a, b we will introduce the factor of 2 into the mix. This gives us knowledge about how to "search" in our new column (for 4*145 that is e=4).

 

For example, using p + 1 - t on column 4 we can find 100, which is an (a = 25 and n=10) in 4 iterations.

 

This would require iterative search, which isn't what I think we're after. I wonder if we could multiply it with a large prime number instead to allow us to "limit" the iterative search in whatever column it ends up in.

 

I think it also requires the prime we multiply in to exists in that column for us to use the t-part as the t is supposed to be the first t where that primes is (think a[t] = <large prime>).

Anonymous ID: d032c1 June 24, 2018, 10:22 a.m. No.6467   🗄️.is 🔗kun

>>6445

Interestingly enough. I compared (1, 1) and (4, 1) with regards to our c=145 and c=4*145.

 

It seems like the a[t] in (4, 1) is equal to the a[t] - x[t-1] from (1, 1) and the d[t] + x[t] in (4, 1) is equal to the d[t] in (1, 1).

Anonymous ID: d032c1 June 27, 2018, 9:34 a.m. No.6516   🗄️.is 🔗kun

>>6499

I'm assuming by t we are referring to t at a[t] = an in (3, 1).

 

I haven't checked them all, but in the first row I think you got the t wrong. According to my program it's 14 and not 15. I'm not sure if we count the first record in an (e, n)/(e, 1) as 0 or 1. If it's 0 then I believe it's 13, and not 14. But I don't know if we've developed a strict enough consensus around it.

 

As for 87*177 I get 19 and not 20.

Anonymous ID: d032c1 June 27, 2018, 9:49 a.m. No.6517   🗄️.is 🔗kun   >>6518

>>6499

But I think you noticed some unknown pattern. I haven't checked the math to see if there is anything obvious behind it yet, but I've attached some output data showing it.

Anonymous ID: d032c1 June 27, 2018, 10:08 a.m. No.6518   🗄️.is 🔗kun   >>6519 >>6520

>>6517

It almost appears as if when x is even the t, at a[t] = an, is equal to (x/2 - 1). Is this something we already knew?

 

I added another image showing it for (1348, 44), but it holds for the other test cases I posted from the image before.

Anonymous ID: d032c1 June 27, 2018, 10:16 a.m. No.6520   🗄️.is 🔗kun

>>6518

And of course I just realized that I'm running in circles. The x at a[t] = an is the SAME x as for our record of (a, b). Which explains what I'm running in circles for.

 

As for the base, that's just (x+n-1)/2 for odd x+n. Since x is a result of t then that would explain why the base - T is so consistent. Since n doesn't change and the x depends on the t.

Anonymous ID: d032c1 June 30, 2018, 2:41 p.m. No.6572   🗄️.is 🔗kun   >>6575

>>6569

PMA, you've done a pretty good job looking into the f % 8 / f-1 % 8. I was looking at that for a, b's in a specific (e, n) (example (3, 6)) and I noticed that it alternates between 1 and 5 (in (3, 6)).

 

I was looking at a = 7, b = 37 then the next one where a = 37, b = 91, and again at a = 91, b = 169 etc.

 

I was wondering about what kind of patterns you might have seen in this. It appears to hold for other (e, n)s' (but not necessarily in the 1, 5 pattern).

Anonymous ID: d032c1 July 1, 2018, 2:51 a.m. No.6588   🗄️.is 🔗kun   >>6589 >>6593 >>6606 >>6609 >>6615

>>6586

Hello VQC! Great to see you again.

 

>>6574

I'll try and go over this a bit later.

 

I was looking into the d[t] - d with regards to patterns of (n - 1) and I noticed something I haven't noticed before. Not sure if someone else has seen it either, but either way here it is:

 

If you take the cell for bn in (e, 1) we have a[t] = bn. What I noticed was that d[t - 1] - d = b(n-1).

 

If we look at a[t] = an we will have d[t] - d = a(n-1). So the d[t] for an and d[t-1] for bn appears to match up with a and b in (-f, 1).

 

In the image I posted I show some test cases for the records in (3, 6).>>6574

Anonymous ID: d032c1 July 1, 2018, 4:26 a.m. No.6600   🗄️.is 🔗kun   >>6601 >>6605

>>6599

Given an e and a d you can find the specific t using a quadratic solution (there is a similar one for a as well):

 

For d:

if e % 2 == 0:

t = d - (e / 2)

t = t / 2

t = t * 4 + 1

t = t / 4

t = math.sqrt(t) + 1

return -1/2 + t

else:

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

t = t / 2

t = t + 1

t = math.sqrt(t)

t = t

return t

 

For a:

if e == 0:

t = a * 2 - e

t = math.sqrt(t) + 2

t = t / 2

return t - 1

if e % 2 == 0:

t = a * 2 - e

t = math.sqrt(t) + 2

t = t / 2

return t

else:

return math.sqrt((a - (e / 2))/2) + 0.5

 

But for clarity: I haven't actually solved anything yet. All I've done is understand more of the pattern behind d[t] - d.

Anonymous ID: d032c1 July 1, 2018, 5:08 a.m. No.6602   🗄️.is 🔗kun   >>6603 >>6604

>>6601

No, not at all.

 

All I've seen is that at the t where a[t] = nb, the d[t - 1] - d = b(n-1). That is the previous records d[t] - our d will be equal to b * (n - 1). So far I got no good solution.

 

I've tried a few things, like we now know that in (-f, 1) there is an a such that d[t] - a[t] = d, but this seems to hold for all cases, which makes sense since the formula for d's in -f, 1 is equal to the formula for a's in e, 1.

 

If we had b(n-1) we could simply do gcd of c and b(n-1) and throw away the n-1 since we now have b. This would yield c/b=a. Now I think the important thing is to figure out what VQC is referring to when he talks about the offset and how to use this efficiently to let (e, 1) do all the work for us.

Anonymous ID: d032c1 July 1, 2018, 6:10 a.m. No.6609   🗄️.is 🔗kun   >>6612

>>6588

I haven't had time to properly test it, but it looks as if the d[t-1] for nb in (-f, 1) is equal to (2d + 1)(n - 1).

 

It also seems like d[t-1] for an in (-f, 1) is divisible by n, but I haven't had a good time to see if this is a pattern that holds.

 

If the first part holds, then it looks similar (2d(n-1) vs (2d+1)(n-1)) to part of our triangle equation.

Anonymous ID: d032c1 July 1, 2018, 11 p.m. No.6620   🗄️.is 🔗kun   >>6626

>>6615

Yeah either that or gcd(c, 201 - 16).

 

What we see with this pattern is that the asymmetry for b(n-1) from (-f, 1) is also reflected in (e, 1).

 

>>6605

> The assymetry of a(n-1) for -f and an for e or using b(n-1) and bn (which are one position different) hold the key.

 

I haven't wrapped my head around this yet, but I think this means for some values of C we will use the asymmetry of a(n-1) for -f and an for e, and for other values of C we will use the asymmetry of b(n-1) and bn.

Anonymous ID: d032c1 July 3, 2018, 8:25 a.m. No.6646   🗄️.is 🔗kun   >>6647

>>6626

Sorry, I missed this.

 

No, I base it of the correct record. I start with correct a, b and then generate that record, the (1, c) record and then transform by making n*b, finding the t from (e, 1) and then generating the cell for t in (e, 1).

Anonymous ID: d032c1 July 3, 2018, 9:52 a.m. No.6648   🗄️.is 🔗kun   >>6649

I'm not sure how much we've looked into it, but I was thinking about the jumps left and right. What we essentially do is increase, or decrease d for each jump. But we're also modifying a.

 

d = a + x and when we do our jumps, the x is locked. So I was wondering how it would look if we locked a instead.

 

It doesn't jump up or down in e, but instead it jumps up and down between n. So to jump one n up, you decrease d and x by 1. To jump one n down, you increase d and x by one.

 

Moving up would work like this:

d = d - 1

e = c - d^2

x = x - 1

 

Moving down would work like this:

d = d + 1

e = c - d^2

x = x + 1

 

From those values you can create a new record that holds a=a, b=b. I'm not sure if it's of any direct use yet. d - (n - 1) will, of course, put a, b at n=1, but if you have to iterate over these values, then it is of little use. Maybe someone else finds it interesting / useful. Maybe there are some shortcuts available too.

 

The part that intrigues me is that when we jump up or down like this, both (1, c) and (a, b) will end up on the same e's. Unlike our jumping left and right since then our n values will modify the length we jump.

 

So think of n as our marker/center. Jumping left and right moves left of that n, right moves right of that n. Jumping up moves the n up and jumping down moves the n down.

 

Jumping left decreases a, b but locks n. Jump right increases a, b but locks n.

Jumping up locks a, b but increases n. Jumping down locks a, b but decreases n.

 

Example time:

a = 7, b = 37, c = 259

 

The cell is: {3:6:16:9:7:37}

 

Say we want to move up. We then do

d = 16 - 1 = 15,

e = 259 - 15*15 = 34

x = 9 - 1 = 8

 

This will result in the cell:

{34:7:15:8:7:37}

 

Now with (1, c) the cell is: {3:114:16:15:1:259}

Then we do:

d = 16 - 1 = 15

e = 259 - 15*15 = 34

x = 15 - 1 = 14

 

This will result in the cell:

{34:115:15:14:1:259}

 

Now say we want to move down (decrease n):

 

The cell is: {3:6:16:9:7:37}

 

Say we want to move up. We then do

d = 16 + 1 = 17,

e = 259 - 17*17 = -30

x = 9 + 1 = 10

 

This will result in the cell:

{-30:5:17:10:7:37}

 

Now with (1, c) the cell is: {3:114:16:15:1:259}

Then we do:

d = 16 + 1 = 17

e = 259 - 17*17 = -30

x = 15 + 1 = 16

 

This will result in the cell:

{-30:113:17:16:1:259}

 

Note:

n will change (increase or decrease) as to balance out, since x is either increasing or decreasing. The easiest way to rationalize this is the b equation:

b = a + 2x + 2n

 

In order to keep b = 37 while decreasing x, n has to increase That is for moving up. For moving down, x increases and n decreases.

Anonymous ID: d032c1 July 3, 2018, 10:46 a.m. No.6651   🗄️.is 🔗kun

>>6650

Finishing this up. We have f = 2d + 1 - e, this is the same as moving one n down. To move 1 n up, we can do 2d + e - 1. Instead of increasing d and x we decrease. That should simplify the moving up as well.

Anonymous ID: d032c1 July 4, 2018, 9:40 a.m. No.6667   🗄️.is 🔗kun

>>6654

I've been looking into this:

>When is the first time a squared appears?

 

I've been looking into this with the reasoning that this applies to both perfect squares and a perfect square * j for some other factor of j. Example 75 = 553. Here 3 is the j and 5*5 is the square.

 

I've noticed that they come in pairs of two so I've been thinking along out t + p, t + 2p, t + 3p.. and p + 1 - t, 2p + 1 - t, …

 

Once you know the t where a[t] = kkj or a[t] = kk the next one can be found at a[t + kk].

 

And for (1, 1) I got this:

a(t-1)(t-1) - t + 1

aa + t - a(t-1)(t-1)

 

But it doesn't seem to hold for anything else.

Anonymous ID: d032c1 July 4, 2018, 12:53 p.m. No.6674   🗄️.is 🔗kun   >>6675 >>6678

>>6670

Long time no see :-)

 

I'm just posting a lot, but I have a lot on my mind. I don't know exactly how I can use it, if I can (might need some of those sweet short cuts), but we have our a, b in (e, n) and we have our n in (e, c). By creating a record for ENA you can make the cell for (e, c) with a=n. Essentially flipping it. But like I said. I don't know how to actually get there without knowing the n so it's not a solution.

Anonymous ID: d032c1 July 5, 2018, 8:58 a.m. No.6700   🗄️.is 🔗kun

>>6697

Just a quick note.

 

I'm also interpreting it like that, but it should be noted that:

 

At (e, 1) with a[t] = k*(a^2) we have the following records:

 

(e, 1) a[t] = k(a^2)

(e, k) a[t] = a^2

(e, a) a[t] = ka

(e, ak) a[t] = a

(e, k(a^2) a[t] = 1

 

Essentially we have (e, i) where i is some permutation of k(a^2). They all have the same x.