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](http://www.emis.de/journals/JIPAM/images/027_01_JIPAM/027_01_www.pdf) 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. 



A question on the bounds of the $n$-th composite $c_n$
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


