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 $x$?