= Solution
Set $g=g_1$ and $g_2=0$. The <superficial degree of divergence> is then $\omega=6-2n$, independently of the number of cubic <interaction vertices>. After subtracting <ultraviolet subdivergences>, the divergent part of a <one-particle irreducible> <Feynman diagram> is a local <polynomial> in external <momenta> of degree at most $\omega$. <Lorentz invariance> and <momentum conservation> leave the following possibilities:
* $n=0$: a constant <vacuum energy density>.
* $n=1$: a linear <tadpole counterterm>. The single external <momentum> is zero, so derivatives cannot generate further independent terms.
* $n=2$: a <mass counterterm> and a <wave-function renormalization>, since the <polynomial> has the form $A+Bp^2$.
* $n=3$: a constant cubic <interaction vertex>.
* $n\geq4$: no overall <ultraviolet divergence>, although <ultraviolet subdivergences> still require their lower-point <counterterms>.
Thus a finite set of <counterterms> closes under <perturbation theory>. In terms of renormalized coefficients, write
$$
\boxed{\mathcal L+\mathcal L_{\mathrm{ct}}
=\frac{1+\delta_Z}{2}(\partial\phi)^2
-\frac{m^2+\delta_m}{2}\phi^2
-\frac{g+\delta_g}{3!}\phi^3-\delta_h\phi-\delta_\mathcal E.}
$$
Here $\delta_m$ is a coefficient of squared <mass>, rather than a correction to $m$ itself. The <tadpole counterterm> fixes the chosen expansion point, for example $\langle\phi\rangle=0$; it cannot be omitted merely because the original <Lagrangian density> lacks a linear term. The <vacuum energy density> term cancels out of normalized <Green functions>, but is required if the unnormalized <vacuum functional> is also to be renormalized. This establishes perturbative <renormalizability>; the unbounded <scalar potential> of a real cubic <scalar field theory> does not by itself define a stable nonperturbative <vacuum state>.
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