Anonymous ID: decc88 June 24, 2018, 7:19 a.m. No.6464   🗄️.is 🔗kun

https://files.catbox.moe/elspzb.ods

I have at the very least made a start. I have a horizontal block for each cell with each calculation (not all of them are there yet I don’t think) and each different cell related to (e, n) is shown vertically (also probably missing something). All values are calculated from the (e, n, t) block at the top. I’ve started implementing the calculations for each of the other related cells. I think something’s weird about the t calculation, though. I just did what it said in the image at the top of this thread. The numbers I used in the example in this spreadsheet give a t value of 3 with this calculation, but the real t value in the grid is 4. Not sure why. I’m also calculating other values based on t (such as x in some of the (e, 1) and (-f, 0) cells), so that might be a bit of a problem. But if anyone would like something to work on, or something to critique (I’m not holding my breath considering the slow pace lately), here’s some progress.

 

>>6420

I haven't implemented the math in the spreadsheet above yet, but I had a brief look at the positive version of this (but +2n/4n/6n, because it seems to be the same pattern but this worked better for my example). I picked {7:4:27:11:16:46}, t=7, f=-48. As follows,

{15:4:28:11:17:47}, t=7, f=-42

{23:4:29:11:18:48}, t=7, f=-36

{31:4:30:11:19:49}, t=7, f=-30

{39:4:31:11:20:50}, t=7, f=-24

Then they stop.

As e increases by 2n, |f| decreases by 6 (in this case 2(n-1)), d increases by 1, x stays the same, a increases by 1 and b increases by 1. This is the exact same pattern I saw when I figured out that for all n=1, f=(x+n)(x+n) as an infinite set of increasing a, b and d values with those screenshots. Pic related is an example for (x+n)=9. Relating it to our triangles, we are meant to be finding 2d(n-1). Maybe it's related. I would do more examples but I need to sleep.

AA !dTGY7OMD/g ID: decc88 June 24, 2018, 7:23 a.m. No.6465   🗄️.is 🔗kun

>>6462

>>6463

Forgot to mention, it turns out I had figured out that between (a, b) = (1, c) and the (e, 1) of (a, b) = (1, c), xx+e=2na is the same, because you're basically just swapping the a and n values.