constraint on the indices in Definition \[DefEpsInd\](ii) yields that neighbouring indices must be different. See also [@SpW Prop. 3.2].

-   Let ${\varepsilon}\in M_{n+m}(\{0,1\})$ be the matrix given by: $${\varepsilon}=\begin{pmatrix}{\varepsilon}_{\operatorname{comm}} &0\\0&0\end{pmatrix},\qquad {\varepsilon}_{ij}=\begin{cases} 1 &\textnormal{ if } i\leq n\textnormal{ and } j\leq n\textnormal{ and } i\neq j\\0&\textnormal{ otherwise}\end{cases}$$ If variables $x_1,\ldots, x_n,x_{n+1},\ldots,x_{n+m}$ are ${\varepsilon}$-independent with respect to this matrix, then

    -   $x_1,\ldots,x_n$ are classically independent,

    -   $x_{n+1},\ldots,x_{n+m}$ are freely independent,

    -   and $\{x_1,\ldots,x_n\}$ and $\{x_{n+1},\ldots,x_{n+m}\}$ are free.

-   The iterated grouping of variables

    -   $x_1$ and $x_2$ are independent,

    -   $x_3$ and $x_4$ are independent,

    -   and $\{x_1,x_2\}$ is free from $\{x_3,x_4\}$

    is represented by the following matrix ${\varepsilon}\in M_4(\{0,1\})$: $${\varepsilon}=\begin{pmatrix}0&1&0&0\\1&0&0&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$

-   Swapping the terms “free” and “independent” in (d), we obtain: $${\varepsilon}=\begin{pmatrix}0&0&1&1\\0&0&1&1\\1&1&0&0\\1&1&0&0\end{pmatrix}$$

-   A constellation that cannot be obtained from iterated grouping is the following (the motivating example in [@SpW]):

    -   $x_i$ and $x_{i+1}$ are free, for $i=1,2,3,4$, as well as $x_5$ and $x_1$,

    -   but all other pairs are independent.

    The matrix ${\varepsilon}\in M_5(\{0,1\})$ in this situation is: $${\varepsilon}=\begin{pmatrix}0&0&1&1&0\\0&0&0&1&1\\1&0&0&0&1\\1&1&0&0&0\\0&1&1&0&0\end{pmatrix}$$

Like in the classical and the free case, we have a moment-cumulant formula for ${\varepsilon}$-independence. Let us first describe its combinatorics.

For $i=(i(1),\ldots,i(k))\in\{1,\ldots,n\}^k$ we define $NC^{\varepsilon}[i]$ as the set of all 