Isee !kIkD/SqZ4s ID: 91c020 May 15, 2018, 9:47 a.m. No.6009   🗄️.is 🔗kun   >>6011 >>6014

>>5995

 

Another "fun" thing regarding this.

 

Compute a cell (cell_0) with (x+n) odd and n even and d even.

 

Then create a new cell with a = n, b = 2*d + n, with n and d from cell_0 and call this cell_1.

 

x+n from cell_1 is equal to d from cell_0. And the difference between d/2 - x is the same as n - d/2.

 

Sorry for derailing a bit, I know this stuff belongs back in thread #8-9, but I haven't noticed this pattern before.

Isee !kIkD/SqZ4s ID: 91c020 May 15, 2018, 10:12 a.m. No.6014   🗄️.is 🔗kun   >>6015

>>6009

 

It's quite nice the way these cells are connected.

 

Take cell_0 and add (x + n). This is cell_1's d.

Take cell_1 and add (x + n). This is cell_0's d.

 

>>6011

Let's hope so! My head is currently stuck on this, so I'll keep playing a bit with it before I go back to the triangles.

Isee !kIkD/SqZ4s ID: 91c020 May 15, 2018, 10:25 a.m. No.6015   🗄️.is 🔗kun   >>6029

>>6014

The grid is kind of hilarious in a devious sense.

 

We know that for some numbers, the d of c is equal to the (x+n) of (x+n) of c, but as far as I know, we don't know anything but e and (x+n). It's like the numbers we want is always out of reach. But just by inches.