Iterating through the jump discontinuities of $f(x)$

Question

What I am attempting to do is the following. I want to find for some given function $f$ the unique step function $c$ (up to an arbitrary constant) such that $f(x) - c(x)$ is continuous assuming it exists. I technically have a formula based on a summation. This summation has to know which points have jump discontinuities. Therefore, I seek a function/sequence that returns the $n$-th jump discontinuity from $x = 0$ such that negative values of $n$ give jump discontinuities $a_n$ such that $a_n < 0$ unless such an $a_n$ does not exist in which case $a_n = a_0$ for all $n < 0$.

Is there such a discrete input function that allows me to "iterate" through those points like in a sequence and if so, how would I go about finding it?

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Originally this question was specifically about functions of the form $\lfloor f(x) \rfloor$'s discontinuities. This is in fact equivalent to what I am editing this question into. Any function with such discontinuities can be rewritten into the form $f(x) = g(x) - j(\lfloor h(x) \rfloor)$ which means that this question and the one in the previous edit are equivalent. This version is simply easier to understand. That's all.

Answer 1

It is extremely unlikely that such a function exists, and even if it does, it's extremely likely that nobody knows it, and even if someone does, they're not going to tell you about it.

Why? Warning: This gets technical, but if you've taken Calculus I and you have some patience, you have the tools to get through it.

We denote Euler's constant by $e$, and the Euler-Mascheroni constant by $\gamma$: $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n \qquad\qquad \gamma = \lim_{n\to\infty} \left(\sum_{k=1}^n \frac1k ~-\int_1^n\!\frac1x~ dx\right)$$

Moreover, let $\sigma$ be the sum-of-divisors function: that is, $\sigma(n)$ is the result of adding all of the divisors of $n$. As an example, $\sigma(12)=1+2+3+4+6+12=28$.

Now, let $f$ be a function defined as follows: $$f(n)=\frac{\sigma(n+5041) \cdot e^{-\gamma}}{(n+5041)\ln(\ln(n+5041))}$$

If you're anything like me, you have no intuition at all about such a function. So you do some computations with your favorite online calculator, and you get some numbers like

\begin{align*} f(40) &= 0.26193 \\ f(959) &= 0.83677 \\ f(10001) &= 0.52237 \\ \end{align*}

You might reasonably draw the conclusion that these numbers probably never get bigger than $1$, and therefore your $g$ would be identically zero. So your desired function is supposed to tell you, that $g$ has no jump points. Equivalently, $g=0$, equivalently $0\leq f<1$, and so your function has to be strong enough to answer this question:

Question (TheGreatDuck, 2016): Is $\lfloor f(n) \rfloor=0$ for all $n\geq 0$?

Now for the punchline:

Answer (Guy Robin, 1984): This question is equivalent to the Riemann Hypothesis.

In other words, a function which answers your question is either (1) extremely useless, since it's so complicated you can't even use it to tell if $g$ is constant, or (2) worth a million dollars :P

[If it bothers you that $\sigma$ is not defined for all real numbers, you can still complete this argument by taking a piecewise-linear extension. I am sweeping the exact construction under the rug for simplicity, but I'm not doing anything illegal.]