Does there exist a a general counting function related to the prime counting function? Does there exist a a general counting function related to the prime counting function?
Say for example I wanted all the positive integer multiples of three less than or equal to N, is there a multiples of 3 less than or equal to N counting function?
Sometimes, I might want all positive integer multiples of a product of primes, for example 15, does there exist a positive multiples of 3 and 5 less than or equal to N counting function?
 A: Just to expand on my comment: If you can imagine it, it exists. It should be well defined if it simply counts whether integers have a property or not. You can see any counting function as the cumulative sum of an indicator function which is either 1 or 0 for each integer.
Consider an indicator function for your property $A$, $\chi_A(n)$. As you have indicated you could define a counting function $\pi_A(n)$ as [assuming we are interested in strictly positive integers]
$$
\pi_A(n) = \sum_{k=1}^n \chi_A(k)
$$
interesting results related to this include a difference generating function. For primes $p_k$, and prime counting function $\pi_p(n)$
$$
\sum_{k=p_1}^\infty x^{\pi_p(k)} = \sum_{k=1}^\infty (p_{k+1}-p_k)x^k
$$
in general for any counting function, for numbers $a_k$, relating to condition $A$
$$
\sum_{k=a_1}^\infty x^{\pi_A(k)} = \sum_{k=1}^\infty (a_{k+1}-a_k)x^k
$$
for example, the number counting function $\chi_n(n)=1$, gives $\pi_n(n)=n$ and
$$
\sum_{k=1}^\infty x^{\pi_n(k)} = \sum_{k=1}^\infty (k+1-k)x^k=\frac{x}{1-x}
$$
in this case $\pi_n(n)$ is the fastest-growing counting function and it is linearly increasing.
Consider the square indicator function $\chi_{\square}(n)$, with condition $\square(n):\sqrt{n}\in \mathbb{N}$, then we have
$$
\pi_\square(n) = \sum_{k=1}^n \chi_\square(k)
$$
then
$$
\sum_{k=1}^\infty x^{\pi_\square(k)} = \sum_{k=1}^\infty ((k+1)^2-k^2)x^k =  \frac{3x-x^2 }{(x-1)^2}
$$
one of my favourites is $\pi_p(\pi_p(n))$ i.e. the nested prime counting function, which relates to the sequence A073131 as
$$
\sum_{k=3}^\infty x^{\pi_p(\pi_p(k))} = \sum_{k=1}^\infty (p_{p_{k+1}} - p_{p_k})x^k
$$
so we can see $\pi_p(\pi_p(n))$ counts prime indexed primes, such as 3, 5, 11, 17, 31, 41, 59, 67, 83, 109,... or A006450. We can see that any strictly increasing integer sequence will have an indicator function, and therefore a counting function, and also a difference generating function. We can nest different kinds of counting functions, for example
$$
\sum_{k=3}^\infty x^{\pi_\square(\pi_p(k))} = \sum_{k=1}^\infty (p_{(k+1)^2} - p_{k^2})x^k
$$
tells us $\pi_\square(\pi_p(k))$ counts primes whose indices are square, and this teaches us that a general chain of compositions gives
$$
\sum_{k=x}^\infty x^{\pi_{A} \circ \pi_{B} \circ \cdots \pi_{Z} \circ k} = \sum_{k=1}^\infty (z \circ \cdots \circ b \circ a \circ (k+1) - z \circ \cdots \circ b \circ a \circ (k))x^k
$$
and that a composition of counting functions is also a counting function, because this the right hand side is just a difference of terms.
Your examples: Positive integer multiples of three go like $3,6,9,12,...$, the indicator function could potentially be written as
$$
\chi_{m3}(n) = \bmod(1+2n^2,3)
$$
the counting function associated with this is
$$
\pi_{m3}(n) = \left\lfloor \frac{n}{3} \right\rfloor
$$
which uses notation for the floor function. We can use the concept of nesting for the multiples of 15, we have
$$
\pi_{m15} = \pi_{m3}(\pi_{m5}(n)) = \pi_{m5}(p_{m3}(n)) = \left\lfloor \frac{\left\lfloor \frac{n}{5} \right\rfloor}{3} \right\rfloor = \left\lfloor \frac{\left\lfloor \frac{n}{3} \right\rfloor}{5} \right\rfloor
$$
this should count numbers divisible by $5$ whose indices are divisible by $3$ or vice versa.
