Euler, Grinberg,... who's next? Given a cubic planar hamiltonian graph with $F$ faces. Let $a_k$ be the number of face of degree $k$ inside and $b_k$ outside the Hamilton cycle. We have the following:


*

*$\sum \limits_k  \left(a_k+b_k\right)k = 6(F-2)$ (due to Euler; see $(3^\ast)$  here)

*$\sum \limits_k \left(a_k-b_k\right)(k-2)=0$  (Grinberg's Theorem)


Are these two all of this type or are there more?
 A: Let's restrict it further to bipartite, cubic, hamiltonian Graphs $G$ only made up of $4$- or $6$-faces. These graphs can build out of $6$ faces of degree $4$ and the rest of degree $6$. Let $F$ be the number of faces, $V$ the number of Vertices, $E$ the number of edges and $a_k$ be the number of face of degree $k$ inside and $b_k$ outside the Hamilton cycle. Starting from Euler, we get:
$$
F+V=E+2-2g\\
F=\sum_{k\in \{4,6\}} a_k+b_k = E-V+2
$$
Here $g=0$ for graph embedded in the plane. For $3$-regular graphs, we have $2E=3V$ and $a_4+b_4=6$ so
$$
6+a_6+b_6= V\left(\frac{3}2-1\right)+2 
$$
$$
2(a_6+b_6)= V-8\tag{1}
$$
By combining $(1)$ and
$$a_4+b_4=6\tag{2}$$ with Grinberg's formula (divided by $2$) in the given case
$$
(a_4-b_4)+2(a_6-b_6)=0\tag{3}
$$
we arrive at


*$$ 
a_4+2a_6=\frac12V-1 \tag{$\frac12\left[(1)+(2)+(3)\right]$}
$$
I checked some examples and it seems to work. Tell me if there is anything wrong or unclear.

Depending on $\frac V2$ being even or odd there are $1,3,5$ or $(0,)2,4,6$ potential $4$-faces inside the HC, where I'm not sure whether $0$ or $6$ is possible at all...

UPDATE:
My interest changed to graphs on double tori $g=2$, build up from faces with $6$ and $8$ edges. There are (again!) six octagons necessary, to build these.
I focus on hamiltonian graphs that cut the double torus into two tori first, as mentioned here.
For $3$-regular graphs, we have $2E=3V$ and $a_8+b_8=6$ so
$$
6+a_6+b_6= V\left(\frac{3}2-1\right)-2\\
2(a_6+b_6)= V-16\tag{1'}
$$
By combining $(1')$ and
$$3(a_8+b_8)=18\tag{2'}$$ with Grinberg's formula (divided by $2$) in the given case
$$
2(a_6-b_6)+3(a_8-b_8)=0\tag{3}
$$
we arrive at


*$$ 
2a_6+3a_8=V+2
$$
since $V$ is even, so be $a_8$. This results in four classes of graphs where the hamiltonian cycle contains $0,2,4$ or $6$ octagons...
