aan(n-1) rings a bell with all these new posts.

Supposedly this type of integer is easy to spot.

Although we haven't discovered the identification process.

Lol.

product of triangles and squares, i think.

Best idea I can think of is to find the next largest perfect square, which should be (aann)

Subtract the difference of (aann) and aan(n-1)

For c145 it's 625-500 = 125

Then, aan(n-1) / 125 = 500 / 125 = 4 = (n-1)

>>8587

Above is a small example, but here is another key idea we have yet to solve.

aan(n-1) is supposed to be easy to spot when comparing (-f,1) and (e,1) a[t] values.

We worked for a week on this, and we need to finish it, as Saga correctly pointed out yesterday.

We fucking work our asses off on an idea, then bail on it when VQC distracts us with a new idea. That distracting faggot.

He's doing it on purpose to slow us down.

Let’s make a list of unsolved crumbs.

And then work on them independently, while sharing our results. That would work well for us as a group. We are all very independent souls but love to check out each other's ideas.

And we share a common purpose, to do our part for Q / VQC etc. and our fellow humans worldwide.

Thoughts, fellow Anons and Math Fam?

>>8567

Let's fill this bread up with some excellent posts about unsolved crumbs to work on. Then, we can move to our new bread.

>>8567

>>>8567

>>8567

>>>8567

The easiest method I can think of is this:

Get the (e na transform) and (-f na transform) elements.

Then create the list of a[t] values for both columns up to a[1].

Then simply move up subtracting e a[t] from -f a[t] and divide c by each result.

First whole integer result is prime a

It could be millions and millions of calcs, but it will run super fast, bc it’s so simple.

(e,1) gives us all factors.