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\begin{array}{ll}
\sum\limits_{i,j,k,l=0}^{N-1}a_{ijk}^{l}a_{jil}=1, & l\in [N]\cap (\mathbb{Z}_N), \\
a_{ijk}^{\overline{l}}=a_{ijk}^{l}, & l\in [N]\cap (\mathbb{Z}_N),
\end{array} where $a_{ijk}^{l}$ is the coefficient of $X^iY^jZ^k$ in $(Y+Z)(X+Z)^{N-1}$, and $a_{ijk}^l=0$ if $i,j,k,l$ are not all distinct or if they are not all in $\mathbb{Z}_N$. Using an algorithm similar to that in the proof of Theorem $thm:polynomial$, we find that the functional equation in ($eqn:functional\_eqn\_for\_catalan$) holds for any fixed $N$.

The rest of the proof goes similarly to that of Theorem $thm:polynomial$.

It is clear that both $\ell_\infty^{N^2}$ and ${\ell_1}^N$ are free modules over their respective algebras. We can also easily show that $\mathcal{F}$ is free over its algebra. We will skip the details and only point out that the method of proving the functional equation in Theorem $thm:functional\_eqn$ is still applicable here.

Finally, we note that the formula ($eqn:general\_formula$) can be generalized to the case of ${\ell_2}^N$ with $N$ even. The proof is almost the same as that of Theorem $thm:polynomial$, but note that we only have to deal with the case that $i+j+k=N/2$ in the proof of the functional equation.

{} M. Bożejko and R. Speicher, *Completely positive maps
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