VA !!Nf9AmQNR7I ID: ffd9d7 April 29, 2018, 4:09 p.m. No.5703   🗄️.is 🔗kun   >>5708

>>5690

This part is cool! It also matches an earlier PMA diagram!

>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)

VA !!Nf9AmQNR7I ID: ffd9d7 April 29, 2018, 5:57 p.m. No.5708   🗄️.is 🔗kun   >>5710 >>5762

>>5703

Here's a quick draw. Each inner rectangle has dimensions of u by u+1. This is just for visualizing the idea, and for small n it would be different. But I think for large n the pieces of the squares are ordered correctly. Thoughts, Anons?

VA !!Nf9AmQNR7I ID: ffd9d7 April 30, 2018, 11:34 p.m. No.5727   🗄️.is 🔗kun   >>5728

>>5716

Alright guys, I think I just found something really cool. Need your help to verify Pls!

 

So, for c = 6107

f = 134

f-2 = 132

(f-2)/8 = 16

(f-2) mod 8 = 6

 

Based on VQC's latest hints, there are two staircase numbers that make up (f-2)/8 which give the correct n0 inside each one of the 8Tu. I think I found out how to find them:

 

4+3+2+1 = 10

3+2+1 = 6

= 16

The correct f shape inside each of the 8Tu could be a triangle?

 

Key idea is that the overlay of (1+2+3) and (1+2+3+4) = 16

Overlay doesn't work for filling the square.

But making a triangle along the straight side of each Tu works. Thoughts, Anons?

VA !!Nf9AmQNR7I ID: ffd9d7 May 1, 2018, 12:16 a.m. No.5729   🗄️.is 🔗kun

>>5728

Thanks PMA! More verification needed, but hopefully this turns out to be a good step forward. Can we find the two triangular numbers that create (f-2)/8 ? Is that the formula you're talking about?

VA !!Nf9AmQNR7I ID: ffd9d7 May 1, 2018, 10:19 p.m. No.5741   🗄️.is 🔗kun

>>5740

CA, check out row 1!

Sqrt(f) is ALWAYS (x+n)

f is also (x+n)^2 always.

That's why (e,1) row 1 is our reference to the correct solution.

Somehow row 1 gives us a clue to the (prime) solution.

Thoughts, Anons?

VA !!Nf9AmQNR7I ID: ffd9d7 May 5, 2018, 10:50 a.m. No.5764   🗄️.is 🔗kun   >>5765

>>5763

Hello VQC, nice to wake up on a Saturday and see your posts! Studying now. The adjacent polite triangle numbers are the key. We're working on how to make the remainders and "leftovers" fit. I think that's what Topol's new drawings are hinting at. Thanks for the new crumbs, back to work!

VA !!Nf9AmQNR7I ID: ffd9d7 May 5, 2018, 5:22 p.m. No.5775   🗄️.is 🔗kun   >>5776 >>5777 >>5779

>>5773

Trying to work this out in a real example.

c=6107

f=134

(f-2)=132

(f-2)/8=16

(f-2) mod 8 = 4

(x+n) = 83

(d+n) = 114

u1 = 41

u2 (u+1) =42

correct n value = 36

Here's the formula: nn + 2d(n-1) + f - 1

f-1 = 134-1 = 133

2d(n-1) = 2*78(36-1) = 5460

nn = 1296

1296 + 5460 + 133 = 6889

SQRT(6889) = 83 = correct (x+n)

Thoughts, Anons?

VA !!Nf9AmQNR7I ID: ffd9d7 May 5, 2018, 5:47 p.m. No.5778   🗄️.is 🔗kun

>>5777

Nice digits, AA! I'm just building out the c=6107 example to try and complete a medium sized problem IRL. I'll post updates, trying to fit everything together like a block puzzle.

VA !!Nf9AmQNR7I ID: ffd9d7 May 6, 2018, 11:41 a.m. No.5787   🗄️.is 🔗kun   >>5788

>>5781

Here's layer 4.

c=6107

f=134

(f-2)=132

(f-2)/8 = 16 (green triangles = 6 + yellow triangles =10)

(f-2) mod 8 = 4 (blue squares)

 

Current Values for layer 4:

(f-2)/8 value = 128 18 = 2304 = 24 * 24 4

(f-2) mod = 4 * 18 = 72

2304 + 72 = 2376

Check 132 * 18 = 2376,

mods + (f-2) values currently matching.

 

I think the mods are going to form a new border to create the +1 differential need for u and u+1.

VA !!Nf9AmQNR7I ID: ffd9d7 May 7, 2018, 6:44 p.m. No.5810   🗄️.is 🔗kun   >>5813

Ok, lads. I have a match. Still don't quite understand how it all works. But additional iterations of 16 filled the gap. Take a look, can't confirm it's 100% correct, but the symmetry / asymmetry is pretty cool.

 

The border needed corners and midpoints of 40/8=5. So each corner and midpoint has and odd number. The 8 remaining gaps to be filled were all 36 units. Found a way to make them all fit.

Yellow = 6

Green = 10

They all fit.

But shouldn't there be extra mod units to add in? I'm stoked but at the same time confused. Any ideas welcome. Posting a work in progress for group discussion.

VA !!Nf9AmQNR7I ID: ffd9d7 May 8, 2018, 7:07 p.m. No.5846   🗄️.is 🔗kun

>>5844

>5844

>>5844

>5844

FRESH BREAD

 

Let's fill this one up!

 

I kept it clean and very simple for my first bake ever. No time for fixing funky dough. We can work on getting all the code links up later.

VA !!Nf9AmQNR7I ID: ffd9d7 May 9, 2018, 1:40 a.m. No.5858   🗄️.is 🔗kun

>>5844

>5844

>>5844

>5844

>>5844

>5844

>>5844

>5844

FRESH BREAD HERE

Simple loaf, but delicious!

My first Bake.

Please post your favorite diagrams, or links to code. Batter was funky. I chose to err on the side of simplicity. Let's set up a pastebin for batter.

VA !!Nf9AmQNR7I ID: ffd9d7 May 9, 2018, 1:49 a.m. No.5859   🗄️.is 🔗kun

Also, every Anon archive everything offline. I've got most of it, but need to save the maps first. Baker did a great job with that.