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arxiv: math/0112018 · v1 · submitted 2001-12-03 · 🧮 math.CO

Coloured permutations containing and avoiding certain patterns

classification 🧮 math.CO
keywords colouredpermutationsequalsnumberprodavoidingcoloursgroup
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Following Mansour, let $S_n^{(r)}$ be the set of all coloured permutations on the symbols $1,2,...,n$ with colours $1,2,...,r$, which is the analogous of the symmetric group when r=1, and the hyperoctahedral group when r=2. Let $I\subseteq\{1,2,...,r\}$ be subset of d colours; we define $T_{k,r}^m(I)$ be the set of all coloured permutations $\phi\in S_k^{(r)}$ such that $\phi_1=m^{(c)}$ where $c\in I$. We prove that, the number $T_{k,r}^m(I)$-avoiding coloured permutations in $S_n^{(r)}$ equals $(k-1)!r^{k-1}\prod_{j=k}^n h_j$ for $n\geq k$ where $h_j=(r-d)j+(k-1)d$. We then prove that for any $\phi\in T_{k,r}^1(I)$ (or any $\phi\in T_{k,r}^k(I)$), the number of coloured permutations in $S_n^{(r)}$ which avoid all patterns in $T_{k,r}^1(I)$ (or in $T_{k,r}^k(I)$) except for $\phi$ and contain $\phi$ exactly once equals $\prod_{j=k}^n h_j\cdot \sum_{j=k}^n \frac{1}{h_j}$ for $n\geq k$. Finally, for any $\phi\in T_{k,r}^m(I)$, $2\leq m\leq k-1$, this number equals $\prod_{j=k+1}^n h_j$ for $n\geq k+1$. These results generalize recent results due to Mansour, and due to Simion.

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