# Can the Brun-Titchmarsh theorem be improved when the modulus is smooth?

For $q,a$ relatively prime, let $\pi(x,q,a)$ denote the number of primes less than $x$ which are congruent to $a$ modulo $q$. The Brun-Titchmarsh theorem states that $$\pi(x,q,a)\leq \frac{(2+o(1))x}{\phi(q)\log(x/q)}$$ for all $q<x$. Letting $\theta=\frac{\log q}{\log x}$, the Brun-Titchmarsh theorem may be rewritten as $$\pi(x,q,a)\leq(C+o(1)) \frac{x}{\phi(q)\log x},$$ where $C=\frac{2}{1-\theta}$. There have been some improvements to this theorem, depending on the range of $\theta$. The optimal results based on the range are:

• $\begin{array}{ccc} \frac{2}{3}\leq\theta & \Rightarrow & C=\frac{2-((1-\theta)/4)^{6}}{1-\theta}\end{array}.$ (Friedlander and Iwaniec)
• $\begin{array}{ccc} \frac{9}{20}\leq\theta\leq\frac{2}{3} & \Rightarrow & C=\frac{8}{6-7\theta}\end{array}.$ (Iwaniec)
• $\begin{array}{ccc} \frac{1}{8}\leq\theta\leq\frac{9}{20} & \Rightarrow & C=\frac{16}{8-3\theta}\end{array}.$ (Motohashi)
• $\begin{array}{ccc} \theta\leq\frac{1}{8} & \Rightarrow & C=2\end{array}.$ (Maynard)

(For $\theta$ very close to $1$, Bourgain and Garaev recently showed that we may take $C=\frac{2-c_{1}(1-\theta)^{2}}{1-\theta}$ for an absolute constant $c_1$.)

Question: Can the Brun-Titchmarsh theorem be improved further if the modulus $q$ is $y$ smooth for $y=x^\delta$ a small power of $x$? That is, if $q$ is composed of prime factors less that $x^\delta$, can we show that $$\pi(x,q,a)\leq (2+o(1)) \frac{x}{\phi(q)\log x}$$ for all $q$ up to $x$?

• Doesn't Goldmakher give an improvement to character sum estimates when the modulus is friable? – Greg Martin Dec 17 '13 at 4:10
• You're asking for a very strong bound when $q$ is close to $x$; one that tends to a constant rather than infinity. I doubt that one can get such a strong result. In general, the situation for progressions is similar to (but usually worse than) the situation for short intervals. For a very short interval also, we do not have such a strong Brun-Titchmarsh. The work of Graham-Ringrose, Iwaniec-Postnikov etc puts our knowledge for certain smooth progressions on the same footing as zeros of zeta (=short intervals; but never better). The best test case for your problem would be to take $q=3^N$. – Lucia Dec 17 '13 at 5:08

(Credit goes also to James Maynard for this answer.) If you assume that $x\ge q^{12/5+\epsilon}$ for some fixed $\epsilon>0$ and you take $q$ to be $x^\theta$-smooth, then you do get that $$\pi(x;q,a)\le (2+\delta) \frac{x}{\phi(q)\log x},$$ with $\delta$ tending to 0 as $\theta\to0$ and $x\to\infty$. This follows by Theorem 10 in this paper of Chang http://math.ucr.edu/~mcc/paper/143%20Char060113.pdf and the zero-density estimates of Jutila (On Linnik's constant, Math. Scand. 41 (1977), no. 1, 45–62) and Huxley (Large values of Dirichlet polynomials. III. Acta Arith. 26 (1974/75), no. 4, 435–444). See also the derivation of Theorem 2 in Iwaniec's paper "On zeros of Dirichlet's L series", Invent. Math. 23 (1974), 97–104. The constant 2 here corresponds directly to a potential exceptional character, whose contribution is bounded trivially. As Lucia also mentions, we cannot answer this question if $x$ is very close to $q$.