# Upper bound for prime product $\frac{a \sigma(b)}{b\phi(a)}$

Let numbers $a = \prod_{p \leq \sqrt{x}} p$ and $b = \prod_{\sqrt{x} <p \leq x} p$ be products of consecutive primes. Now I wonder is it possible to prove that $$r = \frac{a \sigma(b)}{b\phi(a)} < e^\gamma \log(\log(a^2b))$$ when $x$ is sufficiently large? Here $\phi$ is the Euler's totient function and $\sigma$ is the divisor function. I did some numerical experiements and I found out that the inequality is true when $2557 \leq x \leq 10 ^ 8$ assuming there where no numerical errors in my computations.

Using some simple manipulations the ratio $r$ can be written as

$$r = \prod_{p \leq x}\left(1 - p^{-1} \right)^{-1} \prod_{\sqrt{x} < p \leq x}\left(1 - p^{-2} \right).$$

For the left product, I know that $$\prod_{p \leq x}\left(1 - p^{-1} \right)^{-1} < e^\gamma \log(x) \left(1 +\frac{1}{2 \log^2{x}}\right)$$ when $286 \leq x$. I know hardly anything about the right product other than $$\frac{1}{\zeta(2)} <\prod_{\sqrt{x} < p \leq x}\left(1 - p^{-2} \right) < 1.$$ At the moment it does seem like I do not have enough knowledge to prove the inequality.

• $-\log \prod_{p \le x} (1-p^{-2}) = -\sum_p \log (1-p^{-2})- \mathcal{O}(1/x)$ so $$\frac{1}{\zeta(2)} \ge \prod_{p \le x} (1-p^{-2}) = \frac{1}{\zeta(2)} e^{\mathcal{O}(1/x)} = \frac{1}{\zeta(2)} (1+\mathcal{O}(1/x))$$ With the PNT we can improve it to $\mathcal{O}(\frac{1}{x \log x})$ and $\int_x^\infty \frac{dt}{t^2 \log t}+ \mathcal{O}(x^{-3/2+\epsilon})$ with the Riemann hypothesis. Estimating the other product is harder and needs the PNT or the Mertens theorem – reuns Aug 21 '17 at 19:39

A simple equivalent to your question is $\prod \limits_{p \leq \sqrt{x}} \frac{p}{p-1} \prod \limits_{ \sqrt{x} < p \leq x} \frac{p+1}{p} < e^\gamma \ln(\theta(x)+\theta(\sqrt{x}))$, So we are back to the old Robin's inequality, and since he proved its true for square full,square free numbers the above is what left to be proven true, which will imply the famous RH, so i don't think you would get any solution in the near future, nonetheless hope it will be answered one day!!!

For your last line, it looks like no other mathematician also have enough knowledge to prove the inequality.

Edit : Elaborate ,

$a = \prod \limits_{p \leq \sqrt{x}} p$ and $\phi(a) = \prod \limits_{p \leq \sqrt{x}} p-1$ so $\frac{a}{\phi(a)} = \prod \limits_{p \leq \sqrt{x}} \frac{p}{p-1}$

$b = \prod \limits_{\sqrt{x} <p \leq x} p$ and $\sigma(b) =\prod \limits_{\sqrt{x} <p \leq x} p+1$ so $\frac{\sigma(b)}{b}= \prod \limits_{\sqrt{x} <p \leq x} \frac{p+1}{p}$

Then $\frac{a \sigma(b)}{b \phi(a)} =\prod \limits_{p \leq \sqrt{x}} \frac{p}{p-1} \prod \limits_{\sqrt{x} <p \leq x} \frac{p+1}{p}$

On the other hand $a = \prod \limits_{p \leq \sqrt{x}} p = e^{\sum \limits_{p \leq \sqrt{x}} \ln p}$

And since $\theta(x) = \sum \limits_{p \leq x} \ln p$ by definition

We get that $a = e^{\theta(\sqrt{x})}$, in the same way we get that $b=e^{\theta(x)-\theta(\sqrt{x})}$

So $e^\gamma \ln \ln a^2 b = e^\gamma \ln \ln (e^{\theta(\sqrt{x})})^2 e^{\theta(x)-\theta(\sqrt{x})} =e^\gamma \ln \ln e^{2\theta(\sqrt{x})}e^{\theta(x)-\theta(\sqrt{x})}$

$=e^\gamma \ln \ln e^{2\theta(\sqrt{x})+\theta(x)-\theta(\sqrt{x})} = e^\gamma \ln (\theta(x)+\theta(\sqrt{x}))$.

• Your answer is unclear. – reuns Aug 21 '17 at 21:28
• @reuns the Edited part did not help ? – Ahmad Aug 21 '17 at 21:28
• The OP said he doesn't know how to estimate $\prod_{\sqrt{x} < p \leq x}\left(1 - p^{-2} \right)$ so Robin's inequality isn't the topic. Now if you can show OP's observed inequality implies the Robin's inequality (for primorials) and hence the RH, simply state it in a clear way. – reuns Aug 21 '17 at 21:33
• @reuns Thanks, will do – Ahmad Aug 21 '17 at 21:36