Does it follow that $\pi(p_k\#) > \frac{p_{k-1}\#}{3}$ for $k \ge 7$

Question

Let:

• $p_k$ be the $k$th prime
• $p\#$ be the primorial for $p$
• $\pi(x)$ be the prime counting function

From Rosser & Schoenfeld, 1962, for $x \ge 17$:

$$\pi(x) > \frac{x}{\ln x}$$

It seems counterintuitive to me that:

$\pi(p_k\#) > \dfrac{p_{k-1}\#}{3}$

Is the following reasoning valid?

(1) From Rosser 1941 and Dusart 1999 (references here), for $k > 6$:

$$k(\ln(k\ln k)-1)<p_{k}<k{\ln(k\ln k)}\!$$

(2) From $p_k + k > k\ln(k\ln k)$,

$e^{\left(\frac{p_k+k}{k}\right)} > k\ln k$

(3) $2e^{\left(\frac{p_k+k}{k}\right)} > p_k$ since:

$2k\ln k > k\ln k + k\ln(\ln k) = k\ln(k\ln k) > p_k$

(4) For $k \ge 5, 3p_k > \ln p_k\#$ since:

• For $k \ge 7, p_k \ge 2k$ since $p_5 = 11 > 2\times 5$; Assume $p_k > 2k$; $p_{k+1} \ge p_k + 2 \ge 2k+2 = 2(k+1)$

• $3p_k > 2p_k + 2k > k\ln p_k > \ln (p_k)^k > \ln(p_k\#)$

(5) Thus, for $k \ge 7$, $\pi(p_k\#) > \dfrac{p_k\#}{\ln(p_k\#)} > \dfrac{p_{k-1}\#}{3}$

Answer 1

Your proof is valid, and it is true that $\pi(p_k\#)> p_{k-1}\#/3$. I'm not sure why you think this is counterintuitive?

Asymptotically $p_k\#\approx e^{k\ln(k)}$ and so $\pi(p_k\#)\approx e^{k\ln(k)}/\ln(k\ln k)$. Therefore

$$ \frac{\pi(p_k\#)}{p_{k-1}\#} \approx \frac{e^{k\ln(k/k-1)+\ln(k-1)}}{\ln(k\ln(k))}\gg \frac{k}{\ln(k)} $$

In particular the ratio will $\to \infty$ as $k\to \infty$, so one shouldn't be surprised that it is $>1/3$ even for small $k$.