3
$\begingroup$

Define an integer $n$ to be a $k$-almost prime if it has at most $k$ distinct primes factors. A detecting function for the set of such numbers is the generalized von Mangoldt function given by $\Lambda_k(n) := \sum_{ab = n} \mu(a)\log^k b$. Let $\psi_k(t) := \sum_{n \leq t} \Lambda_k(n)$. This is a generalization of the Chebyshev function $\psi(t)$ (i.e., $\psi = \psi_1$).

In regards to $\psi$, there are theorems of so-called Hoheisel-type that essentially say that there exists $\theta_0 \in (0,1)$ such that if $y = x^{\theta}$ with $\theta \in (0,1)$, $\theta \geq \theta_0$ then $\psi(x)-\psi(x-y) =(1+o(1)) y$.

My question is: are there results in the literature for Hoheisel-type estimates for $\psi_k(x)$ with $k$ fixed (in terms of $x$), but arbitrarily large?

Bombieri's sieve (as in Chapter 3 of Opera de Cribro) allows one to take $y = xe^{-(\log x)^{\frac{k-1}{k+1}-\epsilon}}$ (see Proposition 3.3 and its application in the proof of Theorem 3.5), but this is not of Hoheisel type (I would like $y$ a fractional power of $x$). I would expect better results than this to exist given that Bombieri's sieve is designed to handle more general sequences of integers than simply $\mathcal{A} := (x-y,x] \cap \mathbb{N}$.

As far as I know, Hoheisel-type estimates emerge from the explicit formula \begin{equation*} \psi(x) = x - \sum_{\rho} \frac{x^{\rho}}{\rho} - \log(2\pi) -\frac{1}{2}\log(1-x^2), \end{equation*} via a careful analysis of the distribution of the zeroes $\rho$ of $\zeta$. Since $\sum_{n \geq 1} \frac{\Lambda_k(n)}{n^s} = (-1)^k \frac{\zeta^{(k)}(s)}{\zeta(s)}$, perhaps an explicit formula for $\zeta^{(k)}$ could be exploited in a similar way, as distributional results (namely, counts of zeroes in boxes $\{s = \sigma + i\tau : \sigma \in (a,b], \tau \in [T_1,T_2]\}$ for $T_2 > T_1 \geq 0$ and zero-free regions) for derivatives of zeta exist (for instance, a paper by Levinson and Montgomery in Acta Math., 1974). It seems like the details involved in computing such formulae would be ugly.

$\endgroup$
  • $\begingroup$ In the future, don't post identical questions on both MathOverflow and Math.Stackexchange within a 24 hour period. math.stackexchange.com/questions/1558632/… It is best to wait a couple of days to see if someone will answer on MSE before reposting it here. $\endgroup$ – Eric Naslund Dec 4 '15 at 17:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.