I am trying to find the inverse of:
$$f(n) = \prod_{x=1}^{n}x^{0^{\left|\left(\operatorname{mod}\left(x,k\right)-\operatorname{mod}\left(n,k\right)\right)\right|}}$$
for $k, n \in \mathbb{Z}^+$ (as that is where it is monotonically increasing and thus has an inverse, see this proof). Alternative definitions can be found in this question, which are identical for the aforementioned condition.
These functions are used to calculate multifactorials.
So far, I have used the properties of logarithms to remove the product:
$$\exp\left(\sum_{x=1}^{n}\ln\left(x^{0^{\left|\left(\operatorname{mod}\left(x,k\right)-\operatorname{mod}\left(n,k\right)\right)\right|}}\right)\right)$$
which is the same as
$$\exp\left(\sum_{x=1}^{n}0^{\left|\left(\operatorname{mod}\left(x,k\right)-\operatorname{mod}\left(n,k\right)\right)\right|}\ln\left(x\right)\right)$$
and because I am solving for the inverse it is actually
$$\ln(f(n)) = \sum_{x=1}^{n}0^{\left|\left(\operatorname{mod}\left(x,k\right)-\operatorname{mod}\left(n,k\right)\right)\right|}\ln\left(x\right)$$
And I now need to solve for $n$ in terms of $\ln(f(n))$ (approximately), but am unsure how to proceed. It's possible that I'm going about this all wrong too, and that another form of the equation is easier to find the approximate inverse.
An example of something that I want to solve with this is:
$$n!!! = 880$$
We are given $k = 3$, and by some function we should be able to at least approximate the answer here, which is $11$ (because $11!!! = 11 * 8 * 5 * 2 = 880$).
The equations of what I have so far are here on Desmos.