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Inequality

What values of $n$ satisfy the following inequality?

$$2(n-2) < Ap_n\prod_{i=3}^n \left(\frac{p_i-2}{p_i}\right)$$

$p$ are prime numbers and the notation $p_i$ indicates the $i$th prime number.

A comes from the relationship $p_n = 6A + r : 0 \leq r < 6$

(This is a slight variation of The values of $n$ which satisfy an inequality about prime numbers)

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  • $\begingroup$ I can't find any way to approximate $\prod_{i=3}^n(\frac{p_i -2}{p_i})$ adequately to get anywhere with this. $\endgroup$ Mar 30, 2016 at 15:37
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    $\begingroup$ Except that A is Omega (p_n), so the right hand side is more like O((p_n)^2/log(p_n)), so as in the other problem the inequality is satisfied for all sufficiently large n ( I guess n at least 10 , but I haven't worked it out ). Gerhard "Don't Need Mertens For This" Paseman, 2016.03.30. $\endgroup$ Mar 30, 2016 at 16:13
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    $\begingroup$ Once you find the smallest two consecutive values of n that satisfy the inequality, you can compare $(A+1)/A$ with $(p_{n+1}-2)(p_{n+2}-2)/p_{n+1}p_{n+2}$ to see that the RHS is greater than $p_{n+2}$. Gerhard "Prefers Algebraic Reasoning Over Analytic" Paseman, 2016.03.30. $\endgroup$ Mar 30, 2016 at 16:21
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    $\begingroup$ The other question was very marginally research-level mathematics already. This tiny variation, in my opinion, definitely doesn't qualify. Much better would have been to take the answer to the other question and truly understand it, well enough to modify it for this variant (or at least to ask a specific question about the attempted modification), rather than just ask the variant (with no context or visible attempts to solve) a mere five hours later. $\endgroup$ Mar 30, 2016 at 23:58

1 Answer 1

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We already have $p_n \gt 2(n-2)$ for $n \gt 0$ by elementary (non-analytic) methods. Similar methods can establish the inequality for sufficiently large n.

Suppose the inequality $1 \lt A \prod_{i=3}^n \frac{p_i - 2}{p_i}$ holds for positive integers $n=j$ and $j+1$. I will show the inequality holds for $j+2$, leaving the inductive conclusion to the reader. Note that the term for $j+2$ is at least the term for $j$ times $B=\frac{(A+1)(p_{j+1}-2)(p_{j+2}-2)}{Ap_{j+1}p_{j+2}}$. This is because $p_{j+2} \geq 6 + p_j$, so the value of $A$ for the two terms increases by at least 1. But $\frac{A+1}{A}\geq 1 + 6/p_j$ while $(1 - 2/p_{j+1})(1- 2/p_{j+2})\gt 1 - 4/p_j)$ giving $B \gt 1$ and thus the inequality holds for $j+2$.

One can generalize this replacing $A$ by smaller values like $cp_n$ for some positive real $c$ and replacing $p_i-2$ by $p_i-k$ for a larger fixed value of $k$. The rewritten inequality will fail for small $n$ but eventually hold, as analytic methods will have the right hand side grow like a constant times $p_n^2/\log p_n$.

Gerhard "Should Anticipate The Next Questions" Paseman, 2016.03.30.

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  • $\begingroup$ I used Wolfram to find the smallest values, but I didn't know how to prove inductively how you have shown. I was forced to use analytic techniques. BTW (math.stackexchange.com/questions/1720837/…) Brad "Got the next one" Graham $\endgroup$ Mar 30, 2016 at 20:46
  • $\begingroup$ Thank you. Although I understand the intent behind your post regarding necklace colorings and the twin prime conjecture, I am not understanding your coloring. From what I do understand, I suspect that the hypothesis does not hold, and that the presentation falls short of a proof. I do like the start however, and hope you will add pictures and clarity to the stackexchang post. Gerhard "Prefers Properly Posted Picture Proofs" Paseman, 2016.03.30. $\endgroup$ Mar 30, 2016 at 21:20
  • $\begingroup$ Will take that advice thank you, and I will try and get some pictures and improve the clarity! $\endgroup$ Mar 30, 2016 at 21:30

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