chromatic polynomial for helm graph I was just looking at Helm Graph here and looked at the chromatic polynomial here, and noticed what I think may be a discrepancy. Plugging in 3 colors and 9 vertices of a helm graph chromatic polynomial (9 vertices is the correct vertex count for the helm graph with a 4-cycle subgraph, see link 1) gives the result 0. i.e. it is not possible to color a helm graph with 3 colors. But it is plainly possible to color a helm graph with a 4-cycle subgraph, with 3 colors, although the helm graph with a 3-cycle subgraph cannot be colored with 3 colors. So it seems the chromatic polynomial is wrong. Is this a discrepancy, or am I doing something wrong.

My guess is that the OP (incorrectly) used $n=7$ or $n=9$ in the calculation. In fact the correct value is $n=4$:

$$z\left((1-z)^4(z-2)+(z-2)^4(z-1)^4\right)=\\\=z^9-12 z^8+62 z^7-179 z^6+315 z^5-346 z^4+232 z^3-87 z^2+14 z$$

Note the result is monic of degree $9$, corresponding to $9$ vertices. (or note that $n=4$ corresponding to $H_4$)

Substituting $z=3$ we get $$3\left((1-3)^4(3-2)+(3-2)^4(3-1)^4\right)=3((-2)^4+2^4)=96$$