I found a fun function who's roots are only all prime numbers if anyone wants to play around with it.

Question

empty · Answer

$$f(x) = -2+ \sum_{n=1}^\infty \left\lfloor \cos^2 \left( \frac{\pi x}{n} \right) \right\rfloor$$

$$f(x)= 0 \iff \text{ x is prime}$$

empty · Answer

The square is arbitrary, it could have been absolute value sign on the cosine, oh well! https://www.desmos.com/calculator/i8hgrllcfu Really the thing I found is an infinite series for the divisor counting function, and prime numbers have 2 divisors, so that's how they become roots with that -2 heh.

imqwerty · Answer

greatest prime known till date= $2^{57,885,161}-1$ 
roots are prime numbers 
keep scrolling the graph of this function and someday we will get the new greatest prime and become rich

empty · Answer

Lolol we can do it :P

empty · Answer

Ok ikram mentioned to me something about a recurrence relation and I kinda found a sorta-kinda one... haha

ok so I define this function where we have the upper bound as an argument:

$$t\left(x, N\right)=\sum _{n=1}^{N}f\left(\frac{x}{n}\right)^m$$

So for a range, we have kinda a recurrence relation thingy:

$$\sqrt{N} < x \le N$$
$$ t(x,N)=2*t(x,\sqrt{N}) $$

empty · Answer

Maybe I wrote that wrong, but basically the idea is that after you count all the factors of a number less than the square root of it, there is an equal number above that, so I doubled it. But maybe this doesn't quite work for some reason I don't know haha.