-1
$\begingroup$

Let $\Lambda$ be the von Mangoldt function. I think the following is known to hold under GRH: Given any $q \geq 1$, $(a,q)=1$, and $X \geq 1$, we have $$ \sum_{\substack{1 \leq n \leq X \\ n \equiv a\,(\text{mod }q) }} \Lambda(n) = \frac{X}{\phi(q)} + O(X^{1/2} \log (qX)). $$ I was wondering if someone could give me a reference for this. I keep finding a slight variant of this without the von Mangoldt, but I just wanted this version so that I can directly reference in a paper. Thank you very much!

Improve this question
$\endgroup$
3
  • $\begingroup$ What's the relation between $a, X$ and $q, X$? $\endgroup$
    – T. Amdeberhan
    Commented Dec 26, 2016 at 22:33
  • $\begingroup$ I suppose the sum is over $n\equiv a\pmod q$ within this bound? $\endgroup$
    – Wojowu
    Commented Dec 26, 2016 at 22:39
  • $\begingroup$ Your answer to Wojowu is enough for my question. $\endgroup$
    – T. Amdeberhan
    Commented Dec 26, 2016 at 22:43

3 Answers 3

Reset to default
3
$\begingroup$

The usual error term in this problem is not $O(X^{1/2} \log (qX))$ but $O(X^{1/2} \log^2(X))$. This version of the result is part of Corollary 13.8 in Montgomery-Vaughan: Multiplicative number theory I. For a more general result (also featuring the slightly weaker error term above) see Theorem 5.15 in Iwaniec-Kowalski: Analytic number theory.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Is there a (conjectured) $O$ term uniformly on $q,a$ (or $\chi$) ? $\endgroup$
    – reuns
    Commented Dec 27, 2016 at 22:08
  • $\begingroup$ @user1952009: GRH implies the uniform error term $O(X^{1/2}\log^2(X))$. The implied constant is independent of $q$ and $a$. $\endgroup$
    – GH from MO
    Commented Dec 27, 2016 at 22:32
  • $\begingroup$ I see. I'd say it is a consequence of an uniform estimate $A T \ln T+B T+O(\ln T)$ (or $O(\ln T/\ln \ln T)$) for the number of non-trivial poles of $\sum_{n \equiv a \bmod q} n^{-s} \Lambda(n)$ ? $\endgroup$
    – reuns
    Commented Dec 27, 2016 at 22:52
  • $\begingroup$ @user1952009: I gave two references for this estimate (Montgomery-Vaughan and Iwaniec-Kowalski). You can study the proof in those books. $\endgroup$
    – GH from MO
    Commented Dec 27, 2016 at 23:27
3
$\begingroup$

The inequality you state is not a known consequence of GRH, not even in the case $q=1$. In this case von Koch proved 1901 the error term $\mathcal{O}(X^{1/2}\log^2 X)$. Gallagher and Mueller showed that $\mathcal{O}$ can be replaced by $o$ if we assume Montgomery's pair correlation conjecture, and Heath-Brown proved the same under some weakening of the pair correlation conjecture, but no unconditional improvement is known. For an overview of these results see http://www.math.sjsu.edu/~goldston/article35.pdf . The general case runs parallel to the case $q=1$, although the work of Ozluk (J. Number Theory 59, 319-351) gives a little hope that something non-trivial can be proven, at least on average over $q$.

Improve this answer
$\endgroup$
2
$\begingroup$

You may like to look into

C. Hooley, The distribution of sequences in arithmetic progression, Proceedings of the International Congress of Mathematicians (Vancouver, 1974), 357-364. Canad. Math. Congress, Montreal 1975.

Or, for a quick reference, see this paper and its references.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .