>>1984

Okay, so I have no idea if this is actually the proper method, but it appears to sometimes work (it's not like math is an exact science right/s)

We know e, d and c. Compute f (e - (2 * d + 1)).

If you take d + k where k is an a from (f, 1) you will find a * n. However, it doesn't always lie on (f, 1, 1). It can be further out.

I'm not sure if it helps.

>>1986

Note, while t isn't always 1. I haven't found a case where a from (f, 1) + d is not equal to the na we want.

>>1987

This is also why I want an method of generating negative e's, so we can test this further.

>>1993

Do you have any code you're willing to share?

One thing I just noticed,not sure if it's known or helpful.

Have you noticed that the a and b's in (e, 1) is the same values as the d's in the next (e + 1, 1)?

And for (e + 2, 2) the d's alternate between (e, 2) a's and b's. For example, (12, 2) has d's equal (8, 2) a's while (16, 2) has d's equal (12, 2) b's.

I've probably now gone back to thread 1, but it's possible to traverse (e, 1) by swapping a's, b's and d's. No idea how useful that is.

It branches depending on if you want to follow the a's or b's. But again, I think this was covered quite early on and I don't know if it of any use.

>>2126

Here:

{4:1:26:6:20:34} {5:1:34:7:27:43} {6:1:27:6:21:35}

For (4, 1) you take b as (5, 1)'s d. Then you take (5, 1)'s a as (6, 1)'s d. Then (6, 1)'s b as (7, 1)'s d. Then (7, 1)'s a as (8, 1)'s d etc.

For (e, 1) it alternates between a for d and b for d.

>>2127

It's also possible to take (6, 1)s a as d for (7, 1) (thus the branching I guess)