Super-Collider

When you draw it's the Sierpinski triangle! And it makes a lot of sense:

By graphing the particles against time it sorts out that this "duplications" are very similar to the construction of Pascal's Triangle. It is known that Pascal's Triangle modulo 2 is the Sierpinski Triangle. After that it's easy: these "big empty triangles" begin in every 2^n row.

8193 = 8192 +1 = 2^13 +1

So t=8193 is the first duplication of two separated particles (with no annihilation, obviously)

2*2 = 4

This solution is under construction and for now it will only serve as a hint,bookmark it if you want a full solution

They best way to visualize this is to play this game with a few checkerboard pieces or anything of that sort.

You will notice that the number of particles will always be 2 whenever $t = 2^{k}$ for any value of $k$ .

Here $8193 = 2^{13}+1$ ,SO after that many time steps, the number of particles will be $\boxed{4}$ .

We just apply condition up to time=16. We can see that, every time when time is 2^n(n should be integer and not for n=0) it will have 2 particles only. And it should be when time is 2^n+1(n not equal to 0) will have four particles. So, at time 2^13=8192 it have 2 particles. That's why at time 8193(2^13+1) there will be 4 particles...Thank you.

I think that after t=3 , the number of particles will always be 4.So that was the solution! (IDK if it is the right way to do it )