Consider the ring $S=k[x_1,\ldots,x_r]$, graded by $\mathbb N$ with each variable in degree $1$. Let $M$ be finitely generated graded $S$-module. The Hilbert function is defined by $H_M(d)=\dim_k M_d$. It is a well-known theorem of Hilbert that if $M$ is a finitely generated graded $S$-module, then $H_M(d)$ agrees with a polynomial for large $d$.
Now, consider the graded ring $S=k[z_1,z_2]$, with $\deg z_1=2,\deg z_2=3$. I have to show that the Hilbert function $H_S(d)$ is not eventually equal to a polynomial function of $d$.
The homogeneous component $S_d$ is the $k$-submodule of $S$ generated by the monomials of degree $d$. Since $\deg z_1=2$ and $\deg z_2=3$, a monomial $z_1^{\alpha_1}z_2^{\alpha_2}$ has degree $2\alpha_1+3\alpha_2$ (I'm not quite sure about this statement. Can someone verify please?). So given $d$, I have to find the total number of non-negative integer solutions to $2\alpha_1+3\alpha_2=d$. So this reduces to a problem in combinatorics. The generating function associated with the term $2\alpha_1$ is $\frac1{1-t^2}$ and with $3\alpha_2$ is $\frac1{1-t^3}$. After some tedious calculations(which I'm not including here), I got the final answer as $$H_S(d)=\begin{cases} \dfrac{(-1)^d}{4}+\dfrac14+\dfrac{d+1}6+\dfrac13,& d\equiv 0\pmod 3\\ \dfrac{(-1)^d}{4}+\dfrac14+\dfrac{d+1}6-\dfrac13,& d\equiv 1\pmod 3\\ \dfrac{(-1)^d}{4}+\dfrac14+\dfrac{d+1}6,& d\equiv 2\pmod 3 \end{cases},$$ and it seems to agree with the coefficients given here. But even after all these calculations, I'm struggling to find a solid argument to conclude the proof.
The generating function of $H_S(d)$ is $$f(t)=\frac1{(1-t^2)(1-t^3)}.$$ The generating function of a sequence $(a_n)_{n=0}^\infty$ of a function which is eventually polynomial always has the form $$g(t)=\frac{G(t)}{(1-t)^N}$$ where $G$ is a polynomial. In particular, $g(t)$ has poles in the complex plane only at $t=1$. But $f(t)$ has a pole at $t=-1$.