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Let $d_1<d_2<\dots<d_k$ be integers. Then the number of integers $n\leq x$, such that $n+d_1, n+d_2, \ldots, n+d_k$ are simultaneously prime, is bounded above by $$ \mathfrak{S}(d_1, \ldots, d_k) (Ck)^k\frac{x}{\log^k x}, $$ where $\mathfrak{S}(d_1, \ldots, d_k)$ is a singular series.

For $k$ fixed this follows without to much trouble from Selberg's sieve or the large sieve. Very precise results of this type are given e.g. in the book by Halberstam and Richert. However, for my application I need $k$ of order $\sqrt{\log\log x}$. With some effort the large sieve should still work. However, this question appears so natural that certainly someone already solved this problem.

So my first question is: If there a reference for a result as above, which is uniform in $k$ (at least up to $(\log\log x)^{1/2+\epsilon}$?

The second question is: From which point onward can one replace the factor $k^k$ by something like $k^{o(k)}$? I guess that $k^{k/2}$ would already be pretty difficult, at least for not too large $k$.

Third question: When bounding prime $k$-tuples, for which values of $k$ should one switch from Selberg's sieve to the large sieve, and for which value of $k$ from the large sieve to the larger sieve? From the work of Elsholtz it is clear that the larger sieve is best for $k$ of order $\log x$, which is probably the right order, but the change between Selberg's sieve and the large sieve appears less clear.

Jan-Christoph Schlage-Puchta

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Welcome on mathoverflow! –  Stefan Kohl Jul 22 '13 at 14:26
Have you seen Lemma 5.1 on p. 12 of this recent paper of Ford, Konyagin, and Luca: math.uiuc.edu/~ford/wwwpapers/chains.pdf ? –  so-called friend Don Aug 9 '13 at 3:12
This Lemma pretty much does what I need! Thank you! –  Jan-Christoph Schlage-Puchta Aug 9 '13 at 11:22
@so-calledfriendDon: If you turn your comment into an answer, I'd award the bounty to you. -- But please be quick -- MO doesn't leave much time (see below)! –  Stefan Kohl Aug 15 '13 at 16:48

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