I was recently learning about renormalization in quantum field theory (in particular I was looking at the renormalization of phi to the fourth theory). The superficial degrees of divergence of a Feynman diagram for a theory involving one scalar field with a self interaction term that goes like $gϕ^n$ is given by the following formula: $D = 4 - [g]V - E$ where $[g]$ is the mass dimension of the coupling constant $g$ (which in our case is equal to $4-n$, where $n$f is the exponent of the interaction term), $V$ is the number of vertices in the diagram, and $E$ is the number of external lines in the Feynman diagram. This works out to be $D = 4 -V(4-n) - E$.
Now, with $ϕ^3$ theory the superficial degree of divergence $D = 4 - V - E$. Any diagrams where $D$ is greater than or equal to $0$ have UV divergences that need to absorbed into some bare parameter in the Lagrangian. There are only a few diagrams that I can find in $ϕ^3$ theory with $D$ greater than or equal to zero (to one loop order). Naively I'd expect that $ϕ^3$ theory should be 'more' renormalizable than $ϕ^4$ theory, since for $ϕ^4$ theory $D = 4 - E$, whereas for $ϕ^3$ theory $D = 4 - V - E$ (it get's lower for each additional vertex). But this is probably not the case since I can't seem to find anyone anywhere talking about the renormalization of $ϕ^3$ theory in 3+1 dimensions (in 5+1 dimensions I believe I've seen someone tackle it). The only divergent diagrams that I can find to one loop order are some vacuum diagrams with no external legs and the one loop correction to the $ϕ$-propagator ($D = 4 - 2 - 2 = 0$, so a log divergence). So am I just missing something here? Can $ϕ^3$ theory not be re-normalized in 3+1 dimensions?