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I just spent a full day on the brutish and thankless task of proving that the Brun-Titchmarsh bound holds for prime powers (including primes), and not just for primes, in the following senses:

(a) the number of prime powers $p^k$, $k\geq 1$, in an interval $(x,y]$ of length $y-x>1$ is at most $$\frac{2 (y-x)}{\log(y-x)},$$ with one exception: $1<x<2$, $32\leq y<33$;

(b) let $\Pi(x)$ be Riemann's prime-counting function: $\sum_{\substack{p^k\leq x\\ k\geq 1}} \frac{1}{k}$. Then, for any $x$, $y$ with $y-x>1$, $$\Pi(y)-\Pi(x) < \frac{2 (y-x)}{\log(y-x)},$$ with no exceptions.

(Obviously (b) follows from (a).)

What it takes to prove this is nothing special: the proof of Brun-Titchmarsh in Montgomery-Vaughan (1973) has some room beneath the hood, and several of its lemmas can be improved. So that's what you do. A small computational check takes care of what remains.

Now, I cannot be the first person to need this or prove this. Reference?

(Related question, with no answers: Upper bound on sum of Lambda(n) over short interval . Statement (b) here is stronger than what was asked for in the previous question.)

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  • $\begingroup$ Olivier Ramaré points out Lemma 14 in ramare-olivier.github.io/Maths/… to me: there, he and his coauthor point out that the Montgomery-Vaughan argument implies that Brun-Titchmarsh is true for $\Pi(y)-\Pi(x)$ provided that $y\leq 2 x$ (in greater generality than the above: one can work over arithmetic progressions). I started with the same observation, and then gave a supplementary argument for $y>2 x$. $\endgroup$
    – H A Helfgott
    Commented Jan 8 at 4:13
  • $\begingroup$ Out of curiosity, what problem are you considering where the classical Brun-Titchmarsh inequality is insufficient and you need the full force of what you've written here? $\endgroup$
    – Joshua Stucky
    Commented Jan 8 at 13:59
  • $\begingroup$ @JoshuaStucky I am considering mean-value theorems for Dirichlet series whose coefficients have prime-power support (such as $\zeta'(s)/\zeta(s) = \sum_n \Lambda(n) n^{-s}$). Ramaré was considering something else. $\endgroup$
    – H A Helfgott
    Commented Jan 8 at 14:21

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