The famous Green-Tao theorem says that there exist arbitrarily long sequences of primes in arithmetic progression.
I am wondering: How dense can a subset $S \subset \mathbb{N}$ be and still avoid
arbitrarily long sequences of elements of $S$ in arithmetic progression?
To make this more precise (following a comment by Robert Israel), 



> 
> ***Q***. What is the cardinality of the largest subset $S\_n$ of $[1,n]=\{1,2,3,\ldots,n\}$
>  that avoids $k$-term arithmetic progressions of elements in $S\_n$,
>  as a function of $n$ and $k$?
> 
> 
> 


As $n \to \infty$, can the density be significantly more dense than the primes density, $n / \log\_e n$? 


I suspect this is a well-studied question, in which case quotes and/or pointers would suffice. Thanks! 



Most dense subset of numbers that avoids arbitrarily long arithmetic progressions
You are essentially asking for quantitative estimates on Szemerédi's theorem, which states that the largest subset of $[1,n]$ without a k-term arithmetic progression has size $o(n)$. To be precise, let us define $r\_k(n)$ to be the largest subset of [1,n] with no k-term arithmetic progression. Then a construction due to Behrend (essentially projecting a high-dimensional sphere onto the integers) shows that 
$$
r\_3(n) = \Omega\left(n e^{-c \sqrt{\log n}}\right),
$$
while a result of Bloom (moderately improving on a result of Sanders), shows that
$$
r\_3(n) = O\left(n \frac{(\log \log n)^4}{\log n}\right). 
$$
For general $k$, the best known upper bound is due to Gowers and says that
$$
r\_k(n) = O\left(\frac{n}{(\log \log n)^{c\_k}}\right)
$$
for an appropriate $c\_k$. Behrend's construction clearly provides a lower bound in this case as well, but may be improved a little by projecting a collection of concentric spheres. There is some evidence (see, for example, <http://arxiv.org/pdf/1408.2568.pdf>) to believe that the lower bound is closer to the truth.


