I feel like this is a stupid question but does a formula exist that gives a continuous function of x such that its only solutions are at integer multiples of m up to n? The domain of the function is any real value x and the range is any real value y for the sake of argument, but could be a wave function, polynomial, etc or any continuous function. This is the working graph I have at m = 3 except that solutions should lie under each bell curve at y = 0 instead of at y = 1, I for some reason I lost the formula that does the former. I have an interesting zeta form of the equations that yield actual solutions at t = m but I'm still playing around with it.
I started playing around with a set of formulas one day after finding a strange relationship with Young modulus, Ramanujan congruencies, Collatz function, and logistic chaos equations I was working with at the time--as far as I could tell what I was looking at was a kernel for the zeta function for the given s expression I had, but I'm no mathematician so don't crucify me, ha. I would share the formulas but I don't know math jax or whatever that is, I only know latex. But other than what my question suggests I don't think the answer will be surprising, it's just that I've seen some weird stuff with this math which is simple so why would it do such as prime factorization and prime maps I didn't learn in school, so I've toyed with it for over a year on occasions, but mostly can't make sense of it.
Assuming that $n$ is itself a multiple of $m$, the function $$ \frac{\Gamma(1+(n-x)/m)}{\Gamma(1-x-m)} $$ (where $\Gamma$ is the Euler Gamma function) has zeros precisely at $x=m,2m,\dots,n$ and is continuous (indeed, infinitely differentiable) everywhere.
Note that this is simply a polynomial $$ \frac{\Gamma(1+(n-x)/m)}{\Gamma(1-x-m)} = \frac{(x-m)(x-2m)\cdots(x-n)}{m^{n/m}}, $$ but perhaps writing it in this fancy way makes it clearer that it's a well-behaved function.