Kevin, an easier trick is to note that the $p_n$ needn't be prime; all they need is to be pairwise coprime. Also $(2x+1)(3y+1)$ does work. It does not represent $0$ but represents all other integers: write $N=2^r a$ where $a$ is odd. Then the $3y+1$ is either $2^r$ or $-2^r$.

Let's try to orient this surface. Suppose that $abc$ "goes clockwise". Then so do $acd$, $ade$, $bed$, $bce$ and $acb$. So $abc$ goes both clockwise and anticlockwise. :-)

Sorry Kevin, you're right. As an alternative, you could let $C_n=p_n^3$ where $p_n$ is the $n$-th prime congruent to $2$ mod $5$ (hmm, this depends on Dirichlet's theorem but I can see how to get round that!)

Kevin, wouldn't $(2x+1)(3y+1)$ be easier? Let $C_n=p_n$ where $p_n$ is the $n$-th prime congruent to 2 mod 3, and proceed in roughly the same manner.

Wadim, I corrected the typo you pointed out.

Wadim, you need to show that either every block of $C$ consecutive integers has an element not in the range of $f$, or that some particular block has all its elements represented (as Kevin does).

There's no $I_{0.5}$ there :-(

Could you please give us a reference to the definition of $I$?

I presume $n$ is the number of vertices, and $\Delta(G)$ and $\delta(G)$ are respectively the maximum and minimum vertex-degrees of the graph $G$?

Again what do you mean by "simultaneously diagonalizable"? Is your Lie algebra meant to be a subalgebra of some $\mathfrak{gl}_n$?