Number theoretical function with logarithmic properties

Question

Introduction

This is all about the function $v_a(x)$, what i could find about it and what questions i still have.

First of all, I define $v_a(x)$ as the function that counts how often $x$ is divisible by $a$ with $x>0,a>1$ and $x,a \in \mathbb{N}$. For example $v_2(24)=3$ because $24$ is $3$ times divisible by $2$, whereas $v_2(3)=0$ because $3$ is not divisible by $2$.

To make $v_a(x)$ a little more general, I introduced the rule, that

$v_a(a^n) = n$ iff $a^n \in \mathbb{N}$ and $n\in\mathbb{Q}$

Now things as $v_4(2) = \frac 12$ are possible (Which, if you think about it, makes kind of sense).

I first thought about it when i was playing a little bit with the Collatz-conjecture (here). I noticed that Collatz function can be redefined involving $v(x)$ as $3\frac x{2^{v(x)}}+1$ ($v(x)$ without a base always refers to $v_2(x)$).

Logarithmic properties

One of the most important properties of the $v$ function is that $$v_a(x*y)=v_a(x)+v_a(y)$$ and its trivial to see why. If you look at the properties of $v_a(x)$ so far, they are very similar to the logarithmic function. And we can even prove that $$\max(v_a(x))=\log_a(x)$$ Proof:

Assume that there exist $n\in\mathbb{N}$ so that $$v_a(n) > log_a(n)$$ Note that we can write every natural number as $ka^b$ with $k,a,b\in\mathbb{N}; a,k>0; \gcd(a,k) =1$

Therefore $$v_a(ka^b)>\log_a(ka^b)$$ $$v_a(k)+v_a(a^b)>\log_a(k)+\log_a(a^b)$$ $$0+b>\log_a(k)+b$$ $$0>log_a(k)$$ $\unicode{x21af}$

This is a contradiction simply because $k\in\mathbb{N}$

Other results

Here are all my resuts about the $v$ function:

By definition, the equality $$\prod_{p\in\mathbb{P}} p^{v_p(x)}=n$$ holds for every $n\in\mathbb{N}$ $$\implies \sum_{p\in\mathbb{P}} \frac {v_p(x)}{\log_p(x)}=1$$ Other things that I managed to prove are $$\sum_{n=1}^x v_n(x)=\sum_{n=1}^\infty v_n(x)$$ or that $$\sum_{n=1}^\infty \frac1{v_a(n)+1}$$ diverges (not very hard to prove results). But the more interesting things are $$\sum_{n=1}^\infty \frac {z^{v_a(n)}}{n^s} = \zeta(s)\frac {a^s-1}{a^s-z}$$ for $s>1;a\neq1;z\neq0;a^s\neq z$ which is also quite easy to prove using the Euler product (here)

My most recent result is $$-\sum_{n=1}^\infty \frac {v_a(n)}{n^s}=\zeta(s)^2\sum_{n=1}^\infty \frac {v_a(n)}{n^s}\mu(n)$$ where $\mu(x)$ is the Möbius function (here) (Which is equal to -$\frac {\zeta(s)}{a^s-1}$ see Edit 4). This equality can be proved using Möbius inversion (here) altough the proof is not that easy. If anyone wants to prove it, they are welcome to try and share. I would love to see someone find a simple proof.

Last note and open question

Even though it's not really much I found and im not a real mathematician, I think it's quite interesting. I dont know, if this function has already been studied, I couldn't find it anywhere. But if someone knows where to search, I would be happy about references and known results. Or maybe you can come up with other interesting or related stuff. If I´ll find anything else, I´ll make an edit and write it in here.

The only thing I couldnt figure out was a closed form expression of $v_a(n)$ or a formula to calculate it's value. And I really tried. That was frustrating. So if you find a formula, please share it. Share any of your thoughts about $v_a(n)$ if you like.

Edits

Edit 1

if you plot $v(x)$ for natural number

it gives a fractal-like structur. I think maybe $v(x)$ could be calculated iteratively. Also, here is the python code:

def v(a, x):
    if a > 1 and x > 0:
        count = 0
        while x % a == 0:
            x /= a
            count += 1
        return count
    else:
        return False

Edit 3

Here is the wikipedia page to this function

Edit 4

See @reuns comment