>>5851

I tok a stab at this using a=7, b=37 in which case the (n - 1) center piece surrounded by (f -2) + remainder (resulting in unbalanced triangles) still left room from the base (base is 3 for f2/8).

However, this could be filled with one d. Leaving the outside ready to be filled by nn + (2d - 1)*(n - 1).

So I tried again with a = 19, b = 61 and this time, with base = 8 (f2/8 and no remainder) it left 219 free cells, after filling in for n - 1.

However, if I filled in (n - 1) four times in the center the remaining slots could be filled by d.

It's been a while since I checked in so I'm a bit outdated, but I'm reading up on the stuff you guys have been doing and so far I'm blown away! Great work guys!

And, as always, great to see you to VQC!

>>5866

I'm going to try and make some sense here using the cell for a = 19, b = 61.

This cell is equal to: (3, 6, 34, 15, 19, 61)

We can see that 6 + 15 = 21, thus x+n is odd.

e = 3

d = 34

f = 2 * d + 1 =66

f2 = 64

f2 / 8 = 8 (with no remainder as f2 is 8**2).

We now know that the base of our f2-triangles is 8 and we calculate the total amount of cells that is contained in a single f2 triangle: n*(n + 1)/2 = 36.

Multiply this by 8 and we have 8 triangles that fill inn 288 cells.

We then remove f2 from this leaving:

288 - f2 =288 - 64 = 224.

Now we know that in the center (n - 1) exists, but we don't know enough yet. We also assume that n < 8 (our base) and since we don't have any more of f2 to fill in the triangle we assume we have to use some 2*d's.

So we compute 224 % (2*d) =20. Thus we will have a remainder of 20 if we remove some amount of 2*d.

204 / (2d) = 3. So we have to remove 2d3 from our triangle which leaves a total of 20 cells left for us to fill. Our d is 34, so we can't fill it in with a complete d, instead we know that we have to have AT LEAST one (n - 1) in here. Since we already removed 3 2d's, we know n - 1 has to be at least 3. Assuming n = 3, we check 20 % 3, but this leaves us with 2, so we bump it up a notch and say n = 4. This gives us 20 / 4 = 5 (which happens to be our (n - 1)). This means we have to add (n - 1) four times to the center piece to fill it up.

I know this is a wall of text, but I'm just trying to think out loud. It might also be gibberish.

Oh dang, I see you guys have learned more. So if I get this straight, we now use f-1 (can anyone explain why?) and (f-1)/8 doesn't represent the triangle, but rather the number of cells we fill our triangle with.

>>5872

>(f-1)/8 doesn't represent the triangle

• Represent the BASE of the triangle

>>5875

Dangit, yes, but the numbers are still okay. I just typed it in, but the code was using the proper f = 2*d + 1 - e. But it looks like we've changed our perception of the base of the triangle

>>5887

In a previous post (RSA100 from #10 I think) you noticed that the f-2 was divisible by 5, so you divided it by 5*8 (40) and then referred to the blue base in the triangle as consisting of a base of five units. I'm spitballing a bit here, but that implies that when we divide (f-1)/8, we're looking for a base with a single unit (mind the remainder of course) and as we divide by multiples of 8 we will look for wider units?

Am I on the right track with trying to wrap my head around that part?

>>5891

And if I get it, if we make a base (f-1)/8 which is smaller than n, we would have to increase it (by for example making a new base (f-1)/(8*2)). We continue to do so for some time (which I guess will become more apparent once we understand more) which will result in either a solution or we determine that the number is a prime number.

I'm looking into this part

> This restricts the multiples of 2d that can be used because they must always be in a number that supplies the gap in f and satisfy another property. That property can be broken down.

It sounds like it means that there is a gap between f and the capstone (n-1) which is to be filled with multiples of 2d, but is "guarded" by so-far an unknown property. I gave it a go and it could most easily be a fluke, but the remaining parts, which are filled in with parts of d is equal to (n - 1) * 3, leaving a total of (n - 1)4 inside the capstone.

I haven't looked at any other numbers yet, but I'll take a look at another cell to see if it's a fluke or not.

>>5893

Green represents nn, red represents (n - 1), yellow is f-2/8 and shades of blue inside of f-2 represents different multiples of 2d.

>>5491

I'm trying another approach, before this I would divide (f-1)/8 and then just randomly place the remainder of (f-1)/8 against the different triangles. What if it's supposed to be more like this (image).

In this case I used f/8 and not (f-1)/8 and instead removed the extra square from (2*d - 1)/8.

So I end up with two types of triangles, with regards to f (yellow), two which is 3 base wide and 6 which are 4 base wide. The inside of this fits (n - 1) (red) and nn (green) with 2 to spare. The 2d (blue) expands out and finishes the triangle (x+n) with 2 to spare, which will fit inside of the minor f-triangle.

I'm replying two VQC's post from #11:

>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.

I think this approach fits in with what he says as the outer layer of the triangles of f/8 consists of 2 * (4 + 5 + 6 + 7) and 6 * (5 + 6 + 7) leaving two of the triangles being one unit longer than the other 6.

>>5965

VQC has hinted (or flat out said) that we have four sub-types for when (x+n) is odd and n is even, could that be related to the possible dissymmetry between the triangles bases?

In the image I posted there are two triangles that has a base of 3 units and 6 with a base of 4. If we iterate the possible patterns we get:

1 triangle base x, 7 triangles base y

2 triangles base x, 6 triangles base y

3 triangles base x, 5 triangles base y

4 triangles base x, 4 triangles base y

Where x != y.

This excludes the possibility of when all triangles have the same base (meaning it's divisible by 8 exactly). But in that case we wouldn't have two triangle sets to solve (based on the assumption that what I'm referring to is in fact the triangle sets VQC was talking about back in >>5491).

>>5968

I suspect I might on to something, in >>5690 VQC refers to

>Using an arbitrary divisor for f, then each of the eight triangles will have one OR one of two (the latter when c is large and the product of two different prime numbers) configurations in each triangle. The difference between those two configurations of a portion of f are that they are both staircase numbers where the base of one is a unit longer than the other. E.g. (3,4,5) and (4,5,6)

If I get this correctly, when you divide f-1 by 8, you have only a single unit. This works fine for smaller examples as the base of this triangle will be smaller than (x+n)*(x+n)/8's base (VQC has mentioned (x+n)/8, but I'm guessing it was supposed to be (x+n)(x+n)/8?), but if f-1/8 is bigger than (x+n)(x+n)/8 base, then it won't work.

Since our goal is to find a base for f, which is (x+n)(x+n)/8 < f-1/x n/8 we will have to divide f-1 by some multiple of 8. For each multiple of 8 we increase our unit width (vertically).

So if I divide f-1 by 16 I should have a unit of the type (2, 3) or (5, 6) or (11,12). Since it's (f-1)/(28). If I did (f-1)/(38) it would be three units wide (vertically), for example (2, 3, 4) or (5, 6, 7) etc.

I could be completely wrong, though. But I think I've managed to convince myself that it makes sense.