VA !!Nf9AmQNR7I ID: 3a32d9 July 11, 2018, 9:36 p.m. No.6901   🗄️.is 🔗kun

Ok. Here’s the idea I'm working on:

At RSA sized numbers, movements between elements are still dictated by the earliest elements of (e,1)

And their ascending factors

So what is the average distance between a and b and c at an RSA sized number ??

It’s relative

But based on observable patterns

Multiples reduce the infinite search to a limited number of possibilities

Which we can create at will using the known rules

We can create any element surrounding (na transform) for any c.

We can search those elements surrounding for Multiples

Such as an, bn, etc.

My theory is that they should still be relatively close together.(edited)

VA !!Nf9AmQNR7I ID: 3a32d9 July 11, 2018, 9:41 p.m. No.6902   🗄️.is 🔗kun

Searching the factor tree in (e,1) is the main idea.

Constructing only what we need

Moving up and down the factor tree



At 100000000000 numbers, how can (bn) and b(n-1) still be one t value apart ???????

WTF

Instead, Iteration by Creation.

Creation of each element we need, and nothing more.

Lean program.

Killer.

Start at c, then (1,c), then (na transform) then -f na transform.

 

The reason why the element movements surrounding (na transform) are key is because they hold many factors.

We have a clear blueprint of the pattern of factors we are looking for

Therefore we search the patterns of factors surrounding na transform.

(bn) and b(n-1) are only one x value apart.(edited)

We could base a search just on that.

But we have to start at (1,c)

Then na transform

Then (-f,1)

Then crate the elements surrounding (e,1) and (-f,1)

In an expanding quadrilateral

With bn being in the lower right corner of that shape

If no match, then expand the quadrilateral

 

We don’t even need the Grid. We just need the rules of adjacent element Creation FROM the Grid. That’s gonna be a badass algorithm!!!!!

You guys agree?? Is it possible??

SHIT!! It’s so beautiful!

Adapting the Grid rules for element Creation into a group of strings, one for each element.

Expanding out from the na transform

Looking for the lock

On an, ab, a(n-1), and b(n-1)(edited)

An expanding square with an offset of 1 in the bottom right corner

Which is bn

Shape is quadrilateral

But not a trapezoid

I love you @everyone ️

This is so FUN!! 

Fuck, I love Math(s)