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i don't know how to write math in Latex so i will try to explain it simply,

if we multiply $$\frac{p(i)^2}{p(i)^2-1}\prod_{j=1}^5\frac{p(i+j)^2-1}{p(i+j)^2} ,$$ where $p(i)$ denote the $i$-th prime number, is this product always less or equal to 1 when $i\geq3$ (meaning $p(i)\geq5$)?

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  • $\begingroup$ for example when i=3 -> 5^2/(5^2-1)*(7^2-1)/7^2 * (11^2-1)/11^2 *(13^2-1)/13^2 *(17^2-1)/17^2*(19^2-1)/19^2 =0.9997... which is less than 1 $\endgroup$ – Ahmad Sep 8 '16 at 19:30
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We have explicit bounds

$$ \log n + \log \log n - 1 < \dfrac{p(n)}{n} < \log n + \log \log n \ \text{for}\ n \ge 6 $$

Let $b(n) = n \log n + n \log \log n$. Thus for $n \ge 6$, your expression is less than

$$ B(i) = \dfrac{1}{1-1/(b(i)-i)^2} \prod_{j=1}^5 \left(1 - \frac{1}{b(i+j)^2}\right) $$

It appears that $B(t) < 1$ for $t \ge 9$ (though rigourous bounds will be messy). So (given that your expression is less than $1$ for $i=4,5,6,7,8$) the answer is yes.

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  • $\begingroup$ could you post the steps to the proof please $\endgroup$ – Ahmad Sep 8 '16 at 20:40
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I checked with the Maple software that for every $i\geq 3$, the mentioned formula is less than $1$, when $j=1 \cdots m$ where $m\geq2$. Please see the following picture:

enter image description here

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Let $f(a)= 1+ 1/(a^2-1)$. $f(a)$ decreases slowly to 1 as $a$ increases. If $p_j$ is a slowly increasing sequence of positive real numbers that increases slowly enough, then $\prod_{i=1}^5 f(p_{j+i}) \gt f(p_j)$. How slowly? It suffices that $p_{j+5} \lt \sqrt{5}p_j$, which holds for the primes starting with $p_7=17$. As a result, your inequality also holds for other sequences which grow faster than primes.

Gerhard "Generalization Is Good, Generally Speaking" Paseman, 2016.09.08

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    $\begingroup$ Note that this problem is more about growth rate than about primes. This question is better suited for math.stackexchange. Gerhard "Nice Question For Another Forum" Paseman, 2016.09.08. $\endgroup$ – Gerhard Paseman Sep 9 '16 at 3:40
  • $\begingroup$ I'm surprised someone down-voted this. No accounting for taste on MO! $\endgroup$ – Lucia Sep 11 '16 at 21:48
  • $\begingroup$ Indeed. Maybe it's the way I say it. Gerhard "Thanks For The Moral Support" Paseman, 2016.09.11. $\endgroup$ – Gerhard Paseman Sep 11 '16 at 22:16

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