Let $S$ be the graded polynomial ring $k[x_1,x_2]$ such that $x_i$ has degree $i$. Then it's pretty easy to show that $H_S(n)=\text{floor}(n/2)+1$. Now I'm trying to show that $\sum_{n>0}H_S(n)t^n$ is a rational function. The floor is really throwing me off... it's pretty easy to see that $H(2n)=n+1$ and $H(2n+1)=n+1$, but I'm not sure how to compute power series of floor functions...
The hint given to me is that this rational function has denominator $(1-t)(1-t^2)$
Any suggestions?
Let $a_n = H_S(n)$. Then \begin{align*} \sum_{n \geq 0} a_n t^n &= \sum_{m \geq 0} a_{2m} t^{2m} + \sum_{m \geq 0} a_{2m+1} t^{2m+1} = \sum_{m \geq 0} (m+1) t^{2m} + \sum_{m \geq 0} (m+1) t^{2m+1}\\ &= (t+1) \sum_{m \geq 0} (m+1) t^{2m} = (t+1) \left(\sum_{m \geq 0} m t^{2m} + \sum_{m \geq 0} t^{2m}\right) \, . \end{align*} Now \begin{align*} \sum_{m \geq 0} m t^{2m} &= \frac{t}{2} \frac{d}{dt} \sum_{m \geq 0} t^{2m} = \frac{t}{2} \frac{d}{dt} \frac{1}{1 - t^2} = \frac{t}{2} \frac{1}{(1-t^2)^2} \, . \end{align*} Thus \begin{align*} \sum_{n \geq 0} a_n t^n = (t+1)\left(\frac{t}{2(1-t^2)^2} + \frac{1}{1 - t^2}\right) \, . \end{align*}