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In the paper http://arxiv.org/pdf/1311.5319v1.pdf the author states the following theorem, which he attributes to Maynard and Tao.

For any integer $m > 2$, there exists an integer $k = k(m)$ such that the following holds. If $g_1x+h_1, . . . , g_kx+h_k$ is an admissible $k$-tuple, then for infinitely many integers $n$, there exist $m$ or more primes among $g_1n + h_1, . . . , g_kn + h_k$.

Now, Maynard's paper and Tao's blog posts only deal with the case $g_1 = \ldots = g_k = 1$. The author cites the stronger statement from Granville's article http://www.dms.umontreal.ca/~andrew/CEBBrochureFinal.pdf, where a result is stated for "every admissible $k$-tuple of linear forms" (Theorem 6.2). It is unclear to me if Granville really means to allow tuples of the form $m_i n + h_i$, or if he means only $n+ h_i$. The proof he gives is certainly in the latter case only, but I was thinking that it might be possible to extend it without much difficulty - I'll come back to this later. My reason for confusion is that on the bottom of page 5, Granville states the theorem for forms like $n+h_i$ only.

I've studied Maynard's paper a bit and it does not seem to me that the arguments go through in a straightforward way at this level of generality. The step that seems to break down to me is in Maynard's Lemma 5.2, where one is estimating

$$ S_2^{(m)} = \sum_{\substack{d_1, \ldots, d_k \\ e_1, \ldots, e_k}} \lambda_{d_1, \ldots, d_k} \lambda_{e_1, \ldots, e_k} \sum_{\substack{N \leq n < 2N \\ n \equiv \nu_0 \pmod{W} \\ [d_i, e_i \mid n+h_i \forall i}} \chi(n+h_m)$$.

The natural way to generalize this is to replace the inner sum with

$$\sum_{\substack{N \leq n < 2N \\ n \equiv \nu_0 \pmod{W} \\ [d_i, e_i \mid g_in+h_i \forall i}} \chi(g_m n+h_m)$$

This sum essentially counts the number of primes in the range $[g_mN+h_m, 2g_m N + h_m]$ satisfying a congruence condition modulo $g_m W \prod [d_i, e_i]$. So if we try to apply Bombieri-Vinogradov, we would get roughly the same main term but the error term would be a sum over moduli of the form $g_m q$, like

$$\sum_{q = 1}^{ N} E( g_m N, g_m q). $$

By Bombieri-Vinogradov we can say that this is $\ll \frac{g_m N}{(\log g_m N)^A}$, but if we are allowing the $g_m$ to become arbitrarily large, this error term could swamp the main term. I was thinking that perhaps there is a stronger form of Bombieri-Vinogradov saying that we could divided by $g_m$ in the estimate above, since we are only interested in $\frac{1}{g_m}$ of the moduli, but unfortunately this question seems to already have been asked and answered in the negative at The Bombieri Vinogradov Theorem restricted to moduli divisible by $k$.

All I can salvage for now is a statement like the quoted theorem when one has an a priori bound $|g_i | < B$ (for any $B$), and I am wondering if there is a way to get the full strength of it.

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    $\begingroup$ I have nothing to contribute here, but perhaps people would enjoy knowing that on page 2 of the arxiv paper referenced in the 1st paragraph above it says, $$\rm One\ might\ refer\ to\ strings\ of\ consecutive,\ congruent\ primes\ as\ ``Shiu\ strings''.$$ $\endgroup$ Apr 7, 2014 at 23:01
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    $\begingroup$ Andrew's survey states explicitly that the arguments extend to arbitrary linear forms (see bottom of page 6). In your notation, the $g_i$ stay fixed as $N \to \infty$, so the error terms from Bombieri-Vinogradov etc. are still acceptable. (Of course, as the $g_i$ get larger, one expects $N$ and hence $n$ to get larger also, but the theorem of James and myself does not specify any bound on these $n$, only that they appear infinitely often.) $\endgroup$
    – Terry Tao
    Apr 8, 2014 at 1:56

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Note: This question was answered in the comments, and I am posting to remove the question from the unanswered list.

The result of Maynard and Tao applies to all admissible $k$-tuples of linear forms. The argument follows in the way that you describe, however since the $g_i$ are fixed, they do not affect the asymptotic error terms arising from Bombieri-Vinogradov, as mentioned by Terence Tao in the comments above.

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