>>7494

Welcome back old friend.

>>7504

Great work!

>>7510

I'm not forgetting about you either, you've been a champ the past weeks.

>>7504

So I was looking into patterns in (0, 1) and found that our n(n-1)aa will exist in (0, 1) along a multiplier. I was looking in to (0, n) where a = n(n-1)aa and noticed the pattern. It appears that our n(n-1)aa will always exist in column 0 at the row of triangle of n, that is (0, n(n-1)/2).

This also means that in (0, 1) there is an a[t] = (n(n-1) * n(n-1) * a * a)/2

>>7512

To clarify:

> I was looking in to (0, n) where a = n(n-1)aa

I was just looking for patterns in (0, i) (for some i 0) where a[t] = n(n-1)aa.

We're running out of single-letter variables.

>>7504

>>7512

Since we can multiply and a[t] in (0, 1) with any number and it will exist in some row, we can multiply our a[t] = (n(n-1) * n(n-1) * a * a)/2 with 2 and it will occur in (0, 2) as a[t] = 2 * (n(n-1) * n(n-1) * a * a) / 2.

In (0, 3) it is a[t] = 3 * n(n-1) * n(n-1) * a * a / 2

In (0, 4) a[t] = 4 * n(n-1) * n(n-1) * a * a / 2 etc.

And again in (0, 5), (0, 6) etc.. It should exist in every single row in column 0.

Wow, that is kind of mindblowing. I think I'm starting to see why VQC thinks column 0 is so special.

>>7514

> Since we can multiply and a[t] in (0, 1) with …

Since we can multiply any, it was a typo

>>7515

And the t in (0, 1) for a[t] = n(n - 1)^2 * a^2 / 2 is equal to: n(n-1) * a / 2. Meaning the t in (0, 1) is the triangle of n multiplied by the factor of a. So we "scale" the triangle of n by a?

>>7516

To clarify a bit:

This means in (0, 1):

a[ n(n-1)a/2 ] = (n(n - 1))^2 * a^2 / 2