Any number of the form n(n-1) or multiple of it, has distinguishing properties and can be pick out of a crowd, especially and more easily if the multiple of n(n-1) is a square.

What if you multiplied each corresponding element in (-f,1) and (e,1)?

Would that series be totally predictable?

Could you quickly lookup or figure out any values that are a square multiples by two consecutive numbers like n(n-1)?

If you do this, you will quickly spot the value of t or x that is a.a.n.(n-1) at x.

This is the solution.

Multiply each entry a[t] at (e,1) by the corresponding entry at (-f,1). These pairs form a series of products.

You will quickly start to c.

Try it on some lower numbers.

Then you will be able to do it for all odd products.

Prime numbers will apear to be special.

Godspeed anons.

>>7492

If you define n-1 as k, then n(n-1) becomes k(k+1) or twice a triangular number.

So, you're looking for when a series becomes two triangles where every unit of the triangle is a square.

There is a known shortcut to do this between two limiting numbers or range.

Importantly, the search is now a calculation with a known shortcut that is at most O(log t). Think about the values in 0,1 and 0,1.

Everything that has been discussed points at this solution.

Euler? Euler? (Mispronounced)