= Solution
In the normalization $c(\mathbf8)=1$ of <2/c>, the <Adjoint representation> of $D_4$ has <highest weight> $\lambda_{\mathrm{adj}}=(0,1,0,0)$, the <highest root>. The supplied <Gram matrix> has second row $(1,2,1,1)$, whose sum is five. Hence
$$
C_2(\mathrm{adj})=\frac12(2+2\cdot5)=6.
$$
The <Adjoint representation> has the same dimension as the <Lie algebra>, so its <Dynkin index> is also $c(\mathrm{adj})=6$. The $k$ <Weyl fermions> form a <direct sum> of $k$ vector <representations>; additivity of the <trace> gives $c(r_W)=k$.
Substitution into the given one-loop <renormalization-group beta function> yields
$$
\boxed{\beta(g)=-\frac{g^3}{16\pi^2}\left(22-\frac{2k}{3}\right)
=-\frac{33-k}{24\pi^2}g^3.}
$$
For positive $g$, <asymptotic freedom> at this order requires a negative <beta function>, equivalently
$$
\boxed{k<33;\quad\text{for a nonnegative integer number of fermions, }k=0,1,\ldots,32.}
$$
Indeed, if $b_0=22-2k/3>0$, integrating $\mu\,dg/d\mu=-b_0g^3/(16\pi^2)$ gives
$$
\frac1{g^2(\mu)}=\frac1{g^2(\mu_0)}+\frac{b_0}{8\pi^2}\log\frac\mu{\mu_0},
$$
so the <gauge coupling> decreases towards zero as the energy scale increases. At $k=33$ the one-loop coefficient vanishes and the given equation cannot decide <asymptotic freedom>; $k>33$ has the opposite one-loop running.
With the alternative convention $C(\lambda)=\lambda^TG(\lambda+2\mathbf1)$ described in <2/c>, $c(\mathbf8)=2$, $c(\mathrm{adj})=12$, and the same source formula reads $\beta(g)=-(44-4k/3)g^3/(16\pi^2)$. Rescaling the <gauge coupling> together with the <Lie algebra generators> relates the conventions; the bound $k<33$ is identical.
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