It looks like the solution set is invariant under permutations of the indices. It looks like it may be enough to consider finding l such that d(lhat) <e < d((l+1)hat), where hat of a number l has the first l coordinates 1 and the rest 0. With l in hand, you can now consider linear combinations of the right number of vectors and do it up to permutation of coordinates. Gerhard "Use Symmetry To Your Advantage" Paseman, 2015.05.20

It is my hope that things like "that" are encouraged to happen (authoritative answers from such sources) when we keep this forum's quality level high. Unceasing vigilance and compassion, I say. Gerhard "Expect Quality To Follow Quality" Paseman, 2015.05.19

Indeed, such a computation has been outlined by GH from MO, who has even checked smaller values and found that $c=13/6$ does not quite cut it. Gerhard "Missed It By That Much" Paseman, 2015.05.15

You can probably use older inequalities from Rosser/Schoenfeld which apply for all $n \gt 286$ to do a proof by hand. Just check that $\epsilon$ is small enough to pull the result through. Gerhard "Rewriting Terms Really Helps Here" Paseman, 2015.05.15

No reason to be embarrassed. If it turns out that you had a thought like "Hey, maybe every even number $n$ can be a sum of two primes both at least $n/c$ in size by this easy counting argument...", you could reveal that and we could say "Yes, but ..." or "No, because ...", and answer that speculation for future readers. If the motivation is "mature enough" (it doesn't have to be "technical enough"), it's OK for MathOverflow. Just don't use too many words for it. Gerhard "Like To Know Your Thoughts" Paseman, 2015.05.15

Iterated log and a messed up version of $(\log n)^2$. It would be nice for you to edit the question not only to clarify but to include some motivation. Gerhard "If You Would, Pretty Please" Paseman, 2015.05.15

Thanks. Will fix. Apologies to both Kolyvagin and Konyagin Gerhard "Do Not Call Me Passman" Paseman, 2015.05.15

I can think of three interpretations for $\log_2 n$. Which one do you have in mind? Gerhard "Perhaps You Mean Base Two?" Paseman, 2015.05.15

Another way to look at this is to consider "generalized power sequences", where one considers a sequence such that each kth difference is positive, and then one asks how bounded such a sequence is (or what range the kth difference has). I see this as not far from asking if any "positive squarish" subsequence contained in the primes must have unbounded coefficient. Note that if the second difference were constant for long, one would have a successful prime-producing quadratic polynomial. I have a related question on squares here. Gerhard "Giving Another Avenue Of Research" Paseman, 2015.05.15

You might note that they removed a factor of logloglogn from the denominator. That may inspire more click throughs to the abstract. Gerhard "Think Of It As Teaser" Paseman, 2015.05.14

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

I ask because $m$ is not restricted in the inequality above. Even taking $c_0=1$, I get $c_{p_n -n} +c_0=p_n +2$ for $m=n$ or $m=p_n$. When $m$ is not restricted, $c_{m-n}$ can have a negative index, and it makes sense to ask if there are other failures of the inequality. Gerhard "Trying To Get It Clear" Paseman, 2015.05.13

Does $c_n$ have a meaning for nonpositive $n$? I imagine $c_0$ would be 1, but maybe not. I can also see $c_0=0$ and $c_{-n}=-c_n$. Gerhard "Saving Fractional Indices For Later" Paseman, 2015.05.12