= Solution
The <single-vertex scalar two-point graph> at first order in $g_{2k}$ has one $2k$-valent <interaction vertex>. Attach the two external lines to it and pair its remaining $2k-2$ legs into $k-1$ <tadpole diagrams>. The graph is connected, has $L=k-1$ independent <loop momenta>, and is momentum independent after amputation apart from the overall <Dirac delta function>. In particular, first order in the coupling need not mean one loop.
The number of contractions with fixed external labels is
$$
(2k)(2k-1)\frac{(2k-2)!}{2^{k-1}(k-1)!}.
$$
Dividing by the interaction factorial $(2k)!$ gives the inverse <Feynman-diagram symmetry factor> $1/[2^{k-1}(k-1)!]$. Use $m>0$ for the mass in the lowest-order propagator and define the <tadpole integral>
$$
T_D(m)=\int\frac{d^Dq}{(2\pi)^D}\frac1{q^2+m^2}.
$$
The linear-order correction to the amputated function is consequently
$$
\boxed{\Gamma^{(2)}_{\mathrm{lin}}(p_1,p_2)
=-(2\pi)^D\delta^{(D)}(p_1+p_2)
\frac{\mu^{d_k}g_{2k}}{2^{k-1}(k-1)!}\,T_D(m)^{k-1}.}
$$
This is the correction to the paper's amputated connected correlator, not a definition of an inverse-propagator self-energy with an unspecified sign. At this order using $m_0$ instead of $m$ gives the same lowest-order integral.
For completeness, <Schwinger parameterization> and a <Gaussian integral> give
$$
\begin{aligned}
T_D(m)&=\frac1{(4\pi)^{D/2}}\int_0^\infty d\alpha\,\alpha^{-D/2}e^{-\alpha m^2}\\
&=\frac{(m^2)^{D/2-1}}{(4\pi)^{D/2}}\Gamma(1-D/2).
\end{aligned}
$$
This first converges for $0<D<2$ and then defines its <dimensional regularization> continuation. At $D=4-2\epsilon$, the <Gamma function> expansion yields
$$
\mu^{2\epsilon}T_D(m)
=-\frac{m^2}{16\pi^2}\left[\frac1\epsilon+1-\gamma_E+\log\frac{4\pi\mu^2}{m^2}+O(\epsilon)\right].
$$
Thus its highest <ultraviolet pole> is
$$
\boxed{\Gamma^{(2)}_{\mathrm{lin}}
=(2\pi)^D\delta^{(D)}(p_1+p_2)
\frac{(-1)^k g_{2k}\mu^{4-2k}(m^2)^{k-1}}
{2^{k-1}(k-1)!(16\pi^2)^{k-1}\epsilon^{k-1}}
+O(\epsilon^{-(k-2)}).}
$$
For $k=2$ the remainder means finite terms. For $k>2$ the unrenormalized graph has a higher-order pole from its independent <ultraviolet subdivergences>, rather than only a simple pole. A massless <tadpole integral> is scaleless and vanishes in <dimensional regularization>; that does not justify discarding the massive <ultraviolet divergence> by setting $m=0$ before separating ultraviolet and infrared effects.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2018-multitadpole-and-bubble-diagrams.png]
{title=A single higher-valence vertex with three tadpole loops, and the three one-loop quartic bubble channels}
{description=The first panel illustrates k=4; in general there are k minus one loops. The three remaining panels have identical topology and different labelled external attachments.}
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