Estimating the maximum number of primes in the sequence $x, x+p\#, x + 2p\#, \dots, x+ (p\#-1)p\#$ where $p$ is prime and $p\#$ is the primorial Let:


*

*$p$ be a prime

*$p\#$ be the primorial for $p$

*gcd$(a,b)$ be the greatest common divisor of $a$ and $b$.

*$x$ be an integer such that $p < x < p\#$ and gcd$(x,p\#)=1$

*$X$ be the set of of integers $x, x+p\#, \dots, x+(p\#-1)p\#$ 

*$\mu(x)$ be the möbius function
Observations:


*

*There are $p\#$ elements in $X$ 

*For each element $i \in X$, gcd$(i,p\#)=1$

*For each element $i \in X$, there exists an integer $0 \le j < p\#$ where $i = jp\# + x$
Question:
What is the maximmum number of elments in $X$ that are prime?
My Thoughts:


*

*Since $x, x+p\#, x+2p\#, \dots$ forms a set of complete residue systems modulo each prime less than $p\#$, the following sum is an upper bound:



$p\# + \sum\limits_{\text{gcd}(t,p\#)=1 \text{ , } t < p\#}\mu(t)\left\lfloor\dfrac{p\#}{t}\right\rfloor$

I am stuck on how to find a slightly worse upper bound that does not include the möbius function or an alternative approach that would be more fruitful than the trivial answer of $p\#$
Would the mmöbius inversion formula help here?
 A: Let $\pi(x;q,a)$ denote the number of primes $p\le x$ such that $p\equiv a\pmod q$, so that $\pi(x+y;q,a)-\pi(x;q,a)$ counts the number of such primes $p\in(x,x+y]$.
The Brun–Titchmarsh theorem tells us that
$$
\pi(x+y;q,a)-\pi(x;q,a) \le \frac{2y}{\phi(q)\log(y/q)},
$$
where $\phi$ is the Euler phi-function. (The Wikipedia page shows this theorem for the special case $\pi(x;q,a)$ instead of the fully general $\pi(x+y;q,a)-\pi(x;q,a)$, Montgomery and Vaughan's Multiplicative Number Theory. I gives the general case but with an additional term on the right-hand side; the strong version appears in their paper "The large sieve" in Mathematika 20, 1973.)
Applying this with $q=p\#$ and $a=x$, and with $x$ replaced by $x-p\#$ and y replaced by $(p\#)^2$, yields
\begin{align*}
\pi(x+(p\#-1)p\#;p\#,x)-\pi(x-p\#;p\#,x) &\le \frac{2(p\#)^2}{\phi(p\#)\log(p\#)} \\
&= \frac{2p\#}{\log(p\#)} \prod_{q\le p} \bigg( 1-\frac1q \bigg)^{-1},
\end{align*}
where the product runs over all primes $q$ up to and including $p$. The left-hand side (once we decipher the notation) is exactly what you're trying to count.
This bound is probably close to best possible; it's definitely not possible to improve the fraction on the right-hand side to anything less than $p\#/\log x$, by the prime number theorem for arithmetic progressions (if we average over lots of nearby $x$).
