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For a regular polyhedron $\{p, q\}$ the familiar set of equations,

$N_0 = \frac{4p}{4 - (p-2)(q-2)}, N_1 = \frac{2pq}{4 - (p-2)(q-2)}, N_2 = \frac{4q}{4 - (p-2)(q-2)}$

determines the number of vertices, edges and faces respectively.

Are there in general corresponding formulae for polyhedra of the form $r^n \{p, q\}$, wherein $\{p, q\}$ has been rectified $n$ times? (For instance, $r\{5, 3\}$ is the icosidodecahedron, $r\{4, 3\}$ the cuboctahedron, $rr\{5, 3\}$ the rhombicosidodecahedron, etc.)

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    Could you recall the meaning of $p$ and $q$ and/or give an example among the 5 regular Platonic polyhedra ? – 2017-02-23
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    Okay. $\{p, q\}$ has $q$ $p$-polygons around every vertex, $\{3, 3\}$ is a tetrahedron, $\{3, 4\}$ the octohedron, $\{4, 3\}$ the cube, $\{3, 5\}$ the icosahedron, $\{5, 3\}$ the dodecahedron. – 2017-02-23
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    Sorry, but 2 questions again1) by "rectifying" do you mean a "chopping" of all vertices in order to "install" a face at their place ? 2) You say the equations you give are "familiar". Not so much... Could you give a reference ? – 2017-02-24
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    Late reply incoming... Yes, that's what I mean, in other words connect the midpoints of each edge, which is equivalent to taking the reciprocal (dual) intersect the starting polyhedron. Reference for the equations: Regular Polytopes, Coxeter, page 13. This is a result of three lemmas: 1. $qN_0 = 2N_1 = p_2$ , 2. For $j \neq k$, $N_j N_{jk} = N_k N_{kj} $, 3.$ N_0 - N_1 + N_2 = 2 $ (the Euler characteristic). – 2017-03-06
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    Sorry, 1. $qN_0 = 2N_1 = pN_2$. And for 2., by $N_{jk}$ is meant how many $j$-facets are incident with each of the $k$-facets. – 2017-03-06

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