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While trying to prove the inequality $$c_{p_n-m}+c_{m-n}>p_n+2$$ I tried the bounds of $c_n$ (denotes the $n$-th composite number) given in this paper to prove that the sum $c_{p_n-m}+c_{m-n}$ satisfies the following properties, $$ \begin{array}{l} c_{p_n-m}+c_{m-n}\ge 2c_{\left(\frac{p_n-n}{2}\right)}& \text{if $n$ is odd}\\ c_{p_n-m}+c_{m-n}\ge 2c_{\left(\frac{p_n-n+1}{2}\right)}& \text{if $n$ is even} \end{array} $$ but couldn't solve it. I think that to prove this problem we need some more stronger bounds than that of those given in the paper.

Are there any such stronger bounds? I have tried searching in the internet but I didn't find anything relevant. Any help will be appreciated.

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  • $\begingroup$ It would be very easy, and save people some work, if you would just include the problem in this post. $\endgroup$
    – Gerry Myerson
    Commented May 10, 2015 at 12:37
  • $\begingroup$ @GerryMyerson: But I thought that cross-posting is not welcome at MO. $\endgroup$
    – user57432
    Commented May 10, 2015 at 12:39
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    $\begingroup$ Crossposting is exactly what you have done. If you're going to crosspost anyway, you might as well make life easier for the people you're posting to. $\endgroup$
    – Gerry Myerson
    Commented May 10, 2015 at 12:40
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    $\begingroup$ I take back my comment and vote. The question is asking something weaker like $\pi(c_{x}) +\pi(c_y) \ge \pi(x+y)$, which would follow from precise enough versions of the prime number theorem. $\endgroup$
    – Lucia
    Commented May 10, 2015 at 15:03
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    $\begingroup$ I recommend a more informative title. Who would possibly guess the meaning of $c_n$ without context? $\endgroup$
    – Kimball
    Commented May 11, 2015 at 10:14

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I assume $p_n \gt m \gt n$ in the inequality. I have a feeling that this will be as challenging as $\pi(x) + \pi(y) \gt \pi(x+y)$ to solve. The essence to me is that composites are sparsest (primes are densest) near the origin, and that this sparsity is measured by satisfaction of the inequality. However, looking at admissible sets (there is Hensley and Richards classic work, and Hans Riesel's book on computer methods in factorization and primality testing) shows the possibility that the inequality will fail for some large $n$ and $m$ because composites get relatively sparse somewhere.

In spite of that, here is an idea which may give you a start. Note that every interval of six positive numbers has at most two primes; that is every interval avoiding 3. It is then easy to observe that the inequality holds when $4 \gt m-n \gt 0$, because $c_{m-n} > 2(m-n)$, while for large enough $n$, for consecutive values of $m$, values of $c_{p_n - m}$ can differ by 2 at most, and only twice in succession: the difference actually is 1 more often.

Similarly, there are only 19 composites at most 30, while there are at least 23 composites in larger intervals of length 30. It should be possible to show the inequality holds for $c_{m-n}$ between 4 and 30. You may be able to establish the inequality when $2m \lt p_n + n$ and $c_{m-n}$ is less than the $k$th primorial. The tricky part is when $c_{m-n}$ is larger than the $k$th primorial because you may not have good enough lower bounds on the number of primes in that region, and if the prime k-tuple conjecture is true, it may be true in a bad enough way to falsify the conjecture.

A related but worthy problem of study is to look at "unbalanced" admissible tuples. These are tuples which have more prime candidates at the higher end of the tuple. These might lead to a counterexample to the inequality. You could use the inequality to determine the shape and distribution of such a tuple, and perhaps show the existence of such a tuple.

Gerhard "Inequality Truth Not Looking Good" Paseman, 2015.05.13

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  • $\begingroup$ What is $k$ in your answer? $\endgroup$
    – user57432
    Commented May 14, 2015 at 5:29
  • $\begingroup$ It is an additional parameter such that the kth primorial is less than half of $c_{p_n - n}$. So if k =4, then we should be talking about composites larger than 420. I think it is not too hard to show the inequality holds for $m$ such that $c_{m-n}$ is less than the $k$th primorial, since the interval from 1 to the first primorial tends to have more primes in it than any successive interval of integers of that length. You might even extend it to cases where $c_{m-n}$ is less than twice the $k$th primorial. Gerhard "It's The Middle That's Toughest" Paseman, 2015.05.14 $\endgroup$
    – Gerhard Paseman
    Commented May 14, 2015 at 16:40

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