I have recently learned that numbers with no square (prime, assumed in the following) factor are called square-free numbers. I have read that it would asymptotically grows towards
$$\#\{SquareFree\} under\ n = \frac{6n}{\pi^2}$$
I am curious and am wondering if there's a similar behaviour for people with exactly 1 square prime factor or even k?
What I am referring to here: For example, 2100 = $2^2\times3\times5^2\times7$ which has 2 square prime factors, namely $2^2$ and $5^2$. Here $1^2$ and $10^2$ doesn't count for my purpose.
Thanks for reading and thinking 🙂
Best Answer
Here I would proudly present to you the result that I and some friends have actually got after doing more math on it. Would someone read it and check if it's correct?
First, it is known that the number of square-free is
$$\sum_{i=1}^{\sqrt{n}}\mu{(i)}\left\lfloor{\frac{n}{i^2}}\right\rfloor$$ Where $\mu$ is the mobius function.
Now I would apply this to the question, counting numbers with exactly k prime square factors. If $i$ is not square-free, its coefficient is still 0. If $i$ has less than k prime factors, its coefficient is 0 as well because we don't count it at all. If it has m prime factors where m≥k, it's coefficient will be $$(-1)^{m-k}\binom{m}{k}$$ by "Generalized Inclusion-Exclusion Principle". Hence, if we denote the number of primes factors of $i$ by $p(i)$, formula is
$$C_k(n)=(-1)^k\sum_{i=1}^{\sqrt{n}}\left\lfloor{\frac{n}{i^2}}\right\rfloor\binom{p(i)}{k}$$
Would someone like to calculate the ratio of this to n? i.e. $\lim_{n\rightarrow\inf}\frac{C_k(n)}{n}$?
Much love, Gareth