0
$\begingroup$

I am seeing if it is possible to generalize lemma 1 in Nagura's proof that there is always a prime between $x$ and $\frac{6x}{5}$.

In a previous question, I asked whether the following inequality is valid:

If $\left\{\frac{x}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x\right\}}{2}$, $n > 2$, $x \ge n$, then:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+1}{2}\right) - \log\Gamma\left(\frac{x+1}{n}\right)$$

As I've thought about it, it seems that I can extend the argument with the following:

If $\{\frac{x}{2}\} < \frac{1}{2} + \frac{\{x\}}{2}$, $n > 2$, $x \ge n$, $\nexists{a}\in\mathbb{Z}: x < an \le x+1$, then:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+2}{2}\right) - \log\Gamma\left(\frac{x+2}{n}\right)$$

Here's the argument:

Since $\nexists{a}\in\mathbb{Z}: x < an \le x+1$, it follows that:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\left(\left\lfloor\frac{x+1}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x+1}{n}\right\rfloor!\right)$$

From the answer here, if $\left\{\frac{x}{2}\right\} < \frac{1}{2} + \frac{\left\{x\right\}}{2}$, then:

$$\left\{\frac{x+1}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x+1\right\}}{2}$$

Now, assuming that my previous question is correct, this gives us:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x+1}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x+1}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+2}{2}\right) - \log\Gamma\left(\frac{x+2}{n}\right)$$

I believe that it is possible to also handle this third condition:

If $\left\{\frac{x}{2}\}\right < \frac{1}{2} + \frac{\left\{x\right\}}{2}$, $n > 2$, $x \ge n$, $\exists{a}\in\mathbb{Z}: x < an \le x+1$, then:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+4}{2}\right) - \log\Gamma\left(\frac{x+4}{n}\right)$$

Here's the argument:

From the answer here, we know that:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\left(\left\lfloor\frac{x}{2}+1\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}+1\right\rfloor!\right) \le \frac{2}{n}\log\left(\left\lfloor\frac{x}{2}+2\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}+1\right\rfloor!\right)$$

So, it follows from $\exists{a}\in\mathbb{Z}: x < an \le x+1$:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x}{n}\right\rfloor!\right) \le \frac{2}{n}\log\left(\left\lfloor\frac{x+3}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x+3}{n}\right\rfloor!\right)$$

Since from the earlier argument, we have $\left\{\frac{x+1}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x+1\right\}}{2}$, so it follows that:

$$\left\{\frac{x+3}{2}\right\} =\left\{\frac{x+1+2}{2}\right\} = \left\{\frac{x+1}{2}\right\} \ge \frac{1}{2} + \frac{\left\{x+3\right\}}{2}$$

Again, assuming that my previous question is correct, this gives us:

$$\frac{2}{n}\log\left(\left\lfloor\frac{x+3}{2}\right\rfloor!\right) - \log\left(\left\lfloor\frac{x+3}{n}\right\rfloor!\right) \le \frac{2}{n}\log\Gamma\left(\frac{x+4}{2}\right) - \log\Gamma\left(\frac{x+4}{n}\right)$$

Does this work? Have I made a mistake?

Thanks,

-Larry

$\endgroup$
  • 1
    $\begingroup$ Is $x$ an integer? if not what does it mean $n \vert (x+1)$? $\endgroup$ – Esteban Crespi May 4 '13 at 10:09
  • $\begingroup$ Very good point. I should say that there exists $a$ such that $x < an \le x+1$. I'll update the question to address this. $\endgroup$ – Larry Freeman May 4 '13 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.