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Consider the QCD beta function. Its expansion in powers of the coupling is

$$\beta(\mu)=-(\beta_0a(\mu)+\beta_1a^2(\mu)+\ldots)$$

where $a=\alpha/4\pi$. For simplicity let's neglect everything but the one loop term $\beta_0$. This term is given by

$$\beta_0=11-\frac{2}{3}N_f$$

where $N_f$ is the number of fermion flavors. Notice that if the number of quark flavors is higher than $16$ the beta coefficient changes sign.

While performing the running of the coupling constant, when some energy thresholds are crossed we must increase the number of active fermions up to a total of 6. Now, for the fun of it, assume that more quark flavors are lurking in even higher energies. If this were to be true, and we reached a number higher than $16$, the sign of $\beta_0$ would change and the coupling would become large at high energies deconfining QCD. Is this picture right or I am I assuming something I shouldn't?

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In fact, a strong coupling regime in the IR sector, which is provided by the RG flow for $N_{f} < 16$, is a necessary condition for confinement: without strong coupling, you won't get out from the perturbative regime, so you always may treat quarks as free in/out-states in most cases. Bound states, of course, may exist, but their existence is possible only due to changes in the effective constant of the expansion. Think of the electron-proton system which forms a hydrogen atom in QED.

Note that the strong coupling regime is necessary, but not sufficient for confinement, since a coupling constant of order of one only means that the perturbative treatment isn't valid. There are theories with a strong coupling regime, but without the confinement - nonperturbative dynamics doesn't lead to it automatically. You may understand by the example of QCD: the QCD coupling constant isn't really a free parameter, since in the result of the RG flow, it is absorbed into defining the quarks mass-scale through dimensional transmutation; these masses are small, so we can just set them to zero at some scales.

To summarize, the answer on your question is yes.

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