1
$\begingroup$

For any real number $x$, let's define $Om_{k}(x)$ as the number of positive integers $m$ below $x$ such that $\Omega(m)-\omega(m)=k$, where $\omega(n)$ is the number of distinct primes dividing $n$, and $\Omega(n)$ the total number of prime factors of $n$ counted with multiplicity. Obviously $Om_{0}(x)$ is just the number of squarefree integers below $x$.
Do we know asymptotics and (maybe conditional) error terms for $Om_{k}(x)$? What would be the consequences of RH on such error terms?
Thanks in advance.

$\endgroup$
1
  • 1
    $\begingroup$ There is a nice discussion of this question in Kac, Statistical Independence in Probability, Analysis, and Number Theory in the MAA Carus Monograph series, pages 64-71. $\endgroup$ Jul 22, 2014 at 0:21

2 Answers 2

10
$\begingroup$

This problem was studied by Renyi, in

On the density of certain sequences of integers
Publ. Inst. Math. Belgrade 8 (1955) 157-162.
http://elib.mi.sanu.ac.rs/files/journals/publ/14/13.pdf

Let $d_k = \lim_{x\to\infty} Om_k(x)/x$. Then the $d_k$ are the coefficients in the following beautiful identity, valid for $|z| < 2$: $$ \sum_{k=0}^{\infty} d_k z^k = \prod_{p} \left(1-\frac{1}{p}\right)\left(1+\frac{1}{p-z}\right). $$

There have been several papers concerned with error estimates; a recent specimen is the following article by J. Wu:

Sur un problème de Rényi
Monatsh. Math. 117 (1994), no. 3-4, 303–322.

$\endgroup$
4
$\begingroup$

For the case $k=1$, I get $$\dfrac{Om_1(x)}{x} \to \dfrac{6}{\pi^2} \sum_p \dfrac{1}{p(p+1)} \approx 0.20076$$ the sum being over all primes. For $k = 2$, I get $$ \dfrac{Om_2(x)}{x} \to \dfrac{6}{\pi^2} \left( \sum_p \dfrac{1}{p^2(p+1)} + \sum_{p,q} \dfrac{1}{p(p+1)q(q+1)}\right) $$ where the second sum is over all unordered pairs $(p,q)$ of distinct primes.

Similarly, for each positive integer $k$, $Om_k(x)/x$ should approach a constant as $x \to \infty$, and I expect that constant to be expressible in terms of sums of rational functions over $j$-tuples of primes for $j \le k$, the expressions getting complicated as $k$ increases.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.