cases—no relations, relations only from IBP, and relations only from use of the EOM—our aim being to show results which are intuitively easy to understand, as well as simple to obtain from the framework of the previous section. In this section we also emphasize the combinatorial interpretations of our results, which have a natural and well-studied connection with Hilbert series.

No relations {#no-relations .unnumbered}
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In counting operators, the easiest place to begin is to not impose any relations. How many operators can be formed from derivatives acting on the $\ph_i$, $i=1,\dots,N$? In this case, the independent operators are monomials in $\ph_i, \pd \ph_i, \pd^2 \ph_i, \pd^3 \ph_i, \dots$, *i.e.* every operator is obtained in the expansion of $$\begin{aligned}
\prod_{i=1}^{N}\big(1+\ph_i + \ph_i^2 + \dots\big)\big(1+\pd\ph_i &+ (\pd\ph_i)^2 + \dots\big)\big(1+\pd^2\ph_i + (\pd^2\ph_i)^2 +\dots \big)\dots \\
&=\frac{1}{\prod_{i=1}^N(1-\ph_i)(1-\pd\ph_i)(1-\pd^2\ph_i)\dots} .\end{aligned}$$ We say that the operator basis is freely generated by the set of operators $\{\pd^k \ph_i\}$, $k = 0,\dots, \infty$.

The generating set of operators $\{\pd^k\ph_i\}$ have corresponding weights $\{t^ku_i\}$. Since there are no non-trivial relations among these generators, the Hilbert series is $$H_{N,\text{free}}(t,\ui) = \frac{1}{\prod_{i=1}^N (1-u_i)(1-tu_i)(1-t^2u_i)(1-t^3u_i)\dots} = \frac{1}{\prod_{i=1}^{N}(u_i;t)_{\infty}},
\label{eqn:H_free}$$ where $(u;t)_{\infty} = \prod_{k=0}^{\infty}(1-t^ku)$ is the $q$-Pochhammer symbol and the subscript “free” denotes that we 