I have no specific reason to ask this question other than seeing that it comes up quite often when I'm playing around with prime numbers. Let $$f(n,k)=\prod_{i \leq n\vee p_n>k}(p_i-k)$$ Where $p_i$ is the $i$'th prime. I was wondering whether there are good lower and upper bounds for this function. From here, I know that: $$f(n,0)<n(1+\dfrac{1}{2\log n})$$ After some testing it looks like $f(n,k)\sim\log p_n\cdot f(n,0)$ But I can't prove anything.
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... where $p_i$ is the $i$'th prime? – Robert Israel Jun 17 '16 at 18:35
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Yes, it is. I'll add it for clarification – Mastrem Jun 17 '16 at 18:38
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Let $m = \pi(k)$. Then for $n > m$, $$\log f(n,k) - \log f(n,0) = \sum_{i=m+1}^n \log(1 - k/p_i) - \sum_{i=1}^m \log(p_i) $$ As $i \to \infty$, $\log(1 - k/p_i) \sim \dfrac{-k}{p_i} \sim \dfrac{-k}{i \log i}$, and $\int_{2}^n 1/(t \log t)\; dt \sim \log \log n$, so we should get $$ \log f(n,k) - \log f(n,0) \sim -k \log \log n$$ and thus $$\dfrac{f(n,k)}{f(n,0)} = (\log n)^{-k + o(1)}$$ I think that $o(1)$ can be improved to $O(1/\log n)$.
Robert Israel
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Does this also mean that for every $k$ there is a constant $C$ such that: $\dfrac{f(n,k)}{f(n,0)}<(\log n)^{-k+C}$? – Mastrem Jun 17 '16 at 18:55
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Oh, one more thing. Do you happen to know how one can find the value of such a $C$? I'm specifically trying to find a $C$ fo $k=2$. – Mastrem Jun 17 '16 at 19:36
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You'd need explicit bounds that reduce it to computations on finitely many $n$. The bounds $$ \log n + \log \log n - 1 < \dfrac{p_n}{n} < \log n + \log \log n$$ for $n \ge 6$ may help. – Robert Israel Jun 17 '16 at 21:14
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Numerical calculations appear to indicate that $$\dfrac{f(n,2)}{f(n,0)} < (\log n)^{-2}\ \text{for}\ n \ge 3$$ while $$\dfrac{f(2,2)}{f(2,0)} =\frac{ 1}{6} \approx (\log 2)^{-2+6.8866}$$ Of course your inequality doesn't make sense for $n=1$. And of course $C=2$ works for $n \ge 3$. – Robert Israel Jun 17 '16 at 21:56