TITLE: Smallest k-term AP of primes
QUESTION [6 upvotes]: Let $S(k)$ denote the smallest integer such that there exists a k-term arithmetic progression of primes among the integers $[1,S(k)]$. Green and Tao have an unpublished note that gives a very large upper bound for $S(k)$. Conversely, an elementary argument (sketched below) gives the lower bound $S(k) > (k-1)e^{k}$. My question is, can this lower bound be improved at all?
By considering how the AP reduces mod $p$, it is easy to see that if $q$ is the step of a $k$-term AP of primes then every prime less than $k$ must divide $q$.  Thus we have that $S(k) > (k-1) \prod_{p_{i} \leq k } p_{i}$. By taking logarithms and applying the prime number theorem we have that $\prod_{p_{i} \leq k } p_{i} ~ e^{k}$. This gives $S(k) > (k-1)e^{k}$.
Update: As Thomas Bloom points out, the inequality I record isn't true for small $k$. I should have written something more along the lines of $S(k) >(k-1)e^{k-o(k)}$ or $S(k) = \Omega((k-1)e^{k})$ since we know that $psi(x) > x$ infinitely often. The spirit of my question is, are there any arguments (presumably using more sophisticated machinery) that produces a better lower bound (here, I'll accept either bounds that hold for all k, or just infinitely many k).

REPLY [3 votes]: I doubt that you can do substantially better than $\text{log}~S(k) > k$; actually, Granville believes that $\text{log}~S(k) = O(k~\text{log}~k)$.
"Combinatorially speaking," the given lower bound seems asymptotically tight (i.e., if we only use local information and PNT). But heuristically the combinatorial arguments seem to work really well when it comes to questions about additive patterns -- I'm sure there are counterexamples, but I don't know them off the top of my head.
Of course it may not be hard to improve on the $(k-1)$ factor.