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In this question, we needed to compute the Hardy-Littlewood constant for the Bateman-Horn conjecture. Is there a simple argument to show that the infinite product actually converges? Also, is there a reasonable (for your favorite value of "reasonable") estimate of the convergence speed?

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    $\begingroup$ I really wish that the answer to the first question turns out to be a "no". I would love to see a proof! (Seriously, one probably cannot avoid Chebotarev's density theorem, for if the average of $n_p$ is not $1$, the product would diverge.) $\endgroup$ – Boris Bukh Aug 16 '15 at 12:26
  • $\begingroup$ Wiki articles claims that this product is always positive (without reference), I guess it is also always finite. This is a statement of Chebotarev Density Theorem type, though quite strong. $\endgroup$ – Fedor Petrov Aug 16 '15 at 12:28
  • $\begingroup$ @BorisBukh The average of $n_p$ is $1$ when the Galois group is the whole $S_d,$ but it's far from clear for other Galois groups (at least to me). That's part of the reason why I asked the question: from the Chebotarev argument, it is far from clear that the product is always finite. $\endgroup$ – Igor Rivin Aug 16 '15 at 13:50
  • $\begingroup$ @FedorPetrov See my answer, though I don't claim to understand what is going on. $\endgroup$ – Igor Rivin Aug 16 '15 at 14:12
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    $\begingroup$ @BorisBukh: That $n_p = 1$ on average is nothing but Landau's prime ideal theorem for non-normal fields, which is very much simpler than Chebotarev. And here, for the convergence at hand, this is only needed in logarithmic form (Dirichlet rather than natural density), which by a straightforward arrangement amounts to establishing that the Dedekind zeta function has a simple pole at $s = 1$ (and can even be arranged in simple terms not mentioning the zeta function). $\endgroup$ – Vesselin Dimitrov Aug 27 '15 at 14:35
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Well, the good news is that one can find an argument in the original paper of Bateman and Horn, which also gives the convergence rate, and which does not use Chebotarev. The bad news is that I don't understand why this works if the Galois group of $f$ is such that the average number of fixed points is not equal to $1.$

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    $\begingroup$ The Galois group acts transitively on the roots, and Burnside's lemma gives that the average number of fixed points is the size of the orbit which is one. $\endgroup$ – Lucia Aug 16 '15 at 14:33
  • $\begingroup$ @Lucia I wasn't thinking :( Thanks! $\endgroup$ – Igor Rivin Aug 16 '15 at 14:37

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