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SOME PROPERTIES OF THE SERIES OF COMPOSED NUMBERS, p2 gives the bounds

$$l(x)=\frac{x}{\log(x)-28/29}<\pi(x) < u(x)=\frac{x}{\log(x)-1.12} \qquad (1)$$

for $x \geq 3299$.

TWO GENERALIZATIONS OF LANDAU’S INEQUALITY, p6 gives

$$x\pi(x) < 2 \pi(x^2) $$ for $x \geq 67$.

To show $\pi(x^2) > x\pi(x)/2 > \pi((x-1)^2)$ tried the bounds in (1) leading to:

$$xl(x) /2 > u((x-1)^2) \iff $$ $$ \frac{29\, x^{2}}{2 \, {\left(29 \, \log\left(x\right) - 28\right)}} - \frac{{\left(x - 1\right)}^{2}}{\log\left({\left(x - 1\right)}^{2}\right) - 1.12} >0 \qquad (2)$$

Experimentally the LHS of (2) is increasing up to $10^{120}$ and is quite large.

  1. Is this reasoning correct?
  2. Can (2) be proved?

All CAS I tried failed to solve (2) though the limit at infinity is $+\infty$.

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1 Answer 1

up vote 5 down vote accepted

Note that, in the second paper, the inequality $$x\pi(x)<2\pi(x^2)$$ is derived from the inequality $$xy\pi(x)\pi(y)>4\pi^2(xy)$$ by setting $x=y$. The proof for theorem 3 seems correct to me, so the sign flip is definitely a typo :)

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Thanks. You mean the first formula is wrong in the paper? –  joro May 5 '12 at 12:24
    
Yes, it should be $x\pi(x)>2\pi(x^2)$. –  Gjergji Zaimi May 5 '12 at 12:36
    
Thanks. The corrected formula is much worse bound than the first paper, the sign flip explains this :) –  joro May 5 '12 at 13:07
1  
Also, pi(100)=25 and pi(10,000)=1229, if memory serves. The good practice of mathematics involves verification as well as extension and refactoring. Gerhard "Let This Be A Lesson" Paseman, 2012.05.05 –  Gerhard Paseman May 5 '12 at 18:26

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