Let:
- $p_k$ be the $k$th prime.
- $p\#$ be the primorial for $p$
- gcd$(a,b)$ the the greatest common divisor of $a$ and $b$
- $\mu(x)$ be the möbius function
- $x$ be an integer such that $p_k < x < p_k\#$ and gcd$(x,p_k\#)=1$
- $X$ be the set: $x, x+p_k\#, x+2p_k\#, \dots, x+(p_k\#-1)p_k\#$
Does it follow that if $p_k > 300$:
There are less than $p_k\# - \left(\dfrac{1.99p_k}{\ln 2p_k}\right)\left(\dfrac{p_{k-1}\#}{2}\right)$ primes in $X$.
Here's my thinking:
(1) If $p_k > 300$, there are at least $\dfrac{1.99p_k}{\ln 2p_k}$ primes between $p_k$ and $2p_k$
From Rosser & Schoenfeld (1962), if $x \ge 17$: $$\dfrac{x}{\ln x} < x < \dfrac{1.25506x}{\ln x}$$
$\pi(2x) - \pi(x) > \dfrac{2x(\ln x) - 1.25506(\ln 2x)}{(\ln x)(\ln 2x)} > \dfrac{(2x-1.25506)\ln x - 1.25506\ln 2}{(\ln x)(\ln 2x)} > \dfrac{2x - 1.25506}{\ln 2x} - 0.0871$
Since for $x \ge 300$, $\dfrac{x}{\ln 2x}$ is increasing: $$\dfrac{2x - 1.25506}{\ln 2x} - 0.0871 > \dfrac{2x - 0.01x}{\ln 2x} = \dfrac{1.99x}{\ln 2x}$$
(2) Since $x, x+p_k\#, x+2p_k\#, \dots, x+(p_k\#-1)p_k\#$ forms a set of complete residue systems modulo each prime, it follows that the number of primes is equal to:
$$p_k\# - \sum\limits_{\text{gcd}(i,p_k\#)=1\text{ and }i < p_k\#}\mu(i)\left\lfloor\frac{p_k\#}{i}\right\rfloor$$
(3) Then, the number of primes is less than:
$$p_k\# - \sum\limits_{p_k < s < 2p_k}\left\lfloor\frac{p_k\#}{s}\right\rfloor + \sum\limits_{p_k < s,t < 2p_k\text{ and }s \ne t}\left\lfloor\frac{p_k\#}{st}\right\rfloor$$
(4) Which is less than:
$$p_k\# - \left(\frac{1.99p_k}{\ln 2p_k}\right)\left(\frac{p_{k-1}\#}{2}\right)$$
Is this correct? Is there a more straight forward way to show the same conclusion? Is there are any holes in my argument that need to be demonstrated?