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Let $p_n$ denote the $n$-th prime. Is it true that $p_k+p_l+1\le p_{k+l+1}$ for all $k,l\in\mathbb{N}$?

hjhjhj57
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tong_nor
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    Slightly related: http://math.stackexchange.com/questions/530722 – Bart Michels Feb 23 '15 at 20:16
  • Is it true for small $k,l?$ – Will Jagy Feb 23 '15 at 20:20
  • tong_nor, you need both upper and lower bounds if there is any hope of doing this. Well-known upper bounds are available for $k \geq 6,$ with an improvement for $k \geq 20.$ So, again, it is up to you to find out whether this thing you made up is true for small numbers. Probably involves computer, but mostly effort on your part. – Will Jagy Feb 23 '15 at 20:24
  • It is true for small $k,l$, I don't know any but $k=l=1$ example for equality. – tong_nor Feb 23 '15 at 20:35
  • Perhaps use $n\ln n+n\ln\ln n-n<p_n<n\ln n+n\ln\ln n$, see Wikipedia. – Bart Michels Feb 24 '15 at 08:00
  • It's too weak (I checked and made a graph of the difference as a function of $l$ for some fixed $k$) – tong_nor Feb 24 '15 at 20:52
  • Any thoughts/questions about my answer? – Gerry Myerson Mar 01 '15 at 23:28
  • Earth to tong_nor: Come in, please. – Gerry Myerson Mar 02 '15 at 23:44
  • Thanks for your answer. So my question is hard and undecidable because it is almost equivalent to an open number-theory problem? – tong_nor Mar 04 '15 at 12:25
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    "Undecidable" is a technical term coming from math logic, so I didn't and wouldn't use it. But, yes, your question is more-or-less equivalent to a notorious open problem. – Gerry Myerson Mar 05 '15 at 10:45
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    I'm not familiar with higher mathematics, and my english is not perfect, so sorry for unfortunate words. I meant "undecidable with current knowledge" (with the problem still open), so I understood correctly. So I mark the answer as accepted and have to wait until somebody solves the main problem, then my conjecture will also be solved :] – tong_nor Mar 06 '15 at 00:44
  • Given your choice of notation it seems likely that Dusart's 1998 paper inspired the question. It seems worth citing. http://www.unilim.fr/laco/theses/1998/T1998_01.pdf – daniel Mar 17 '15 at 21:48

1 Answers1

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Let $\pi(n)$ be the number of primes not exceeding $n$, so $\pi(p_r)=r$ for all $r$. Then $\pi(p_k+p_l)\le\pi(p_k+p_l+1)\le1+\pi(p_k+p_l)$. If $p_k+p_l+1\le p_{k+l+1}$, then $\pi(p_k+p_l+1)\le k+l+1$, so $\pi(p_k+p_l)\le k+l+1=\pi(p_k)+\pi(p_l)+1$.

Now there is a conjecture that $\pi(x+y)\le\pi(x)+\pi(y)$ for all $x\ge2$, $y\ge2$, but this conjecture is generally believed to be false (although if there are counterexamples they involve very large numbers) (see Second Hardy–Littlewood conjecture). I suspect that a close inspection of the reasons for believing it false will also cast doubt on the veracity of $\pi(p_k+p_l)\le\pi(p_k)+\pi(p_l)+1$

Gerry Myerson
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