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This question is an extension of this question. There the asymptotic density of k-almost primes was asked.

By subsets I mean the following: Let $\lambda$ be a partition of $k$ and $P_{\lambda}=\{ \prod p_m^{\lambda_m} \; |\; p_m\neq p_k \}$. So $P_{(1,1)}$ would be all semiprimes, despite squares.

What I got are results on $k$-almost primes, being the union of all subsets $P_{\lambda}$. Here are some explicite formulas, like $$ \pi_2(n)=\sum_{i=1}^{\pi(n^{1/2})}\left[\pi\left(\frac{n}{p_i}\right)-i+1\right]. $$ A general asymptotic is given by $$ \begin{eqnarray*} \pi_k(n) &\sim& \left( \frac{n}{\log n} \right) \frac{(\log\log n)^{k-1}}{(k - 1)!}\\ \end{eqnarray*} $$ For the case of $P_{(1,1)}$ we just subtract the number of squares from $\pi_2(n)$ and get $$ \pi_{P_{(1,1)}}=\pi_2(n)-\pi(n^{1/2}), $$ but I don't see how to extend this.

So again: How do the counting function $\pi_{P_{\lambda}}(n)$ or their asymptotics look like?

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draks ...
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  • I think all the asymptotic weight goes to (1,...,1) since the sum of the squares (or cubes, etc.) of the reciprocals of the primes converge. So the k=2 case seems more general than it appears at first. – Charles Mar 30 '12 at 13:44
  • @Charles, What are referring to? (Maybe you remember this?) – draks ... Mar 30 '12 at 14:36
  • Consider the 4-almost primes. The density of all the forms is n (log log n)^3/(6 log n). I think that the density of (1,1,1,1) is (log log n)^3/(6 log n) and the density of the others is negligible in comparison. For example, the density of (2,1,1) is certainly less than 3 times the density of (1,1,1) which is at most (log log n)^2/(2 log n). – Charles Mar 30 '12 at 15:01

2 Answers2

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For a complete answer, see this paper:

Integers with a predetermined prime factorization.

There, the author defines two functions, $\sigma_{\lambda}(x)$ and $\pi_{\lambda}(x)$ where $\lambda$ is some vector in $\mathbb{N}^k$. These count the number of integers up to $x$ of the form $p_1^{\lambda_1}\cdots p_r^{\lambda_r}$ where $\pi_{\lambda}(x)$ has the added condition that $p_i\neq p_j$ when $i\neq j$. The result is general asymptotics for both.

Also see this Math Stack Exchange question and answer.

Examples: Using your notation above, we have: $$\pi_{(1,1,2)}(x)\sim \sum_p P(2) \frac{x \log \log x}{\log x}, $$ where $P(s)=\sum_p \frac{1}{p^s}$ is the prime zeta function.

Another example is: $$\pi_{(1,1,2,5)}(x)\sim C_{(1,1,2,5)} \frac{x \log \log x}{\log x},$$ where $$C_{(1,1,2,5)}=P(2)P(5)-P(10)=\sum_{p}\frac{1}{p^2}\sum_{p}\frac{1}{p^5}-\sum_{p}\frac{1}{p^{10}}.$$

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Eric Naslund
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I've been asked to submit this partial answer.

Asymptotically, almost all k-almost primes are squarefree. This follows from Landau's asymptotic $$ \pi_k(x)\sim\tau_k(x)\sim\frac{x}{\log x}\cdot\frac{(\log\log x)^{k-1}}{(k-1)!} $$ which holds both for numbers with $\omega(n)=k$ and for numbers with $\Omega(n)=k.$

First, consider $k$-almost primes which are divisible by the square of some prime. They are $\ell$-almost primes for some $\ell<k$ and so there are at most $\pi_\ell(x)$ such numbers. Repeating this for each of the other $p(k)-1$ classes of $k$-almost primes one gets their total density at most $\left(p(k)-1\right)\pi_{k-1}.$ Hence the number of squarefree divisors must make up the whole asymptotic density.

Explicitly, in your notation, $$ \pi p_{1,\ldots1}\sim\frac{x}{\log x}\cdot\frac{(\log\log x)^{k-1}}{(k-1)!} $$ where there are $k$ 1s, and $$ \pi p_\lambda\ll\frac{x}{\log x}\cdot(\log\log x)^{k-2} $$ if there are any numbers larger than 1 in $\lambda.$

Charles
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