\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \usepackage{xspace} \usepackage{multirow} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf Enumeration of Two Particular Sets of \\ \vskip .12in Minimal Permutations } \vskip 1cm \large Stefano Bilotta, Elisabetta Grazzini,\footnote{ Corresponding author.} and Elisa Pergola\\ Dipartimento di Matematica e Informatica ``Ulisse Dini''\\ Universit\`a di Firenze\\ Viale G. B. Morgagni 65\\ 50134 Firenze \\ Italy\\ \href{mailto:egrazzini@unifi.it}{\tt egrazzini@unifi.it} \end{center} \newcommand{\mini}{ minimal permutations with $d$ descents and size $d+2$ } \vskip .2in \begin{abstract} Minimal permutations with $d$ descents and size $d+2$ have a unique ascent between two sequences of descents. Our aim is the enumeration of two particular sets of these permutations. The first set contains the permutations having $d+2$ as the top element of the ascent. The permutations in the latter set have $1$ as the last element of the first sequence of descents and are the reverse-complement of those in the other set. The main result is that these sets are enumerated by the second-order Eulerian numbers. \end{abstract} \section{Introduction} \subsection{Preliminary definitions} A permutation of size $n$ is a bijective map from $\{1..n\}$ to itself. We denote by $S_n$ the set of permutations of size $n$. We consider a permutation $\sigma \in S_n$ as the word $\sigma_1 \sigma_2 \cdots \sigma_n$ of $n$ letters on the alphabet $\{1,2,\ldots,n\}$, containing each letter exactly once (we often use the word \emph{element} or \emph{entry} instead of letter). For example, $6\,2\,4\,3\,5\,1$ represents the permutation $\sigma \in S_6$ such that $\sigma_1 = 6, \sigma_2 = 2, \ldots, \sigma_6 = 1$. \begin{definition} Let $\sigma$ be a permutation in $S_n$. We say that $\sigma$ has a \emph{descent} in position $i$ whenever $\sigma_i>\sigma_{i+1}$. In the same way, we say that $\sigma$ has an \emph{ascent} in position $i$ whenever $\sigma_i<\sigma_{i+1}$. \end{definition} \begin{example} The permutation $\sigma = 6\,9\,8\,4\,1\,3\,7\,2\,5 \in S_9$ has $4$ descents, namely in positions $2$, $3$, $4$, $7$, and $4$ ascents in positions $1$, $5$, $6$ and $8$. \label{ex:descent} \end{example} \begin{definition}Let $\sigma$ be a permutation in $S_n$. The \emph{reverse} of $\sigma$ is the permutation $\sigma^r=\sigma_n \sigma_{n-1} \cdots \sigma_1$. The \emph{complement} of $\sigma$ is the permutation $\sigma^c= (n+1-\sigma_1)(n+1-\sigma_2) \cdots (n+1-\sigma_n)$.\label{def:revcompl} \end{definition} \begin{example} If $\sigma = 4\,2\,6\,5\,3\,1$, then $\sigma^r = 1\,3\,5\,6\,2\,4$, $\sigma^c= 3\,5\,1\,2\,4\,6$ and $\sigma^{rc} = \sigma^{cr} = 6\,4\,2\,1\,5\,3$. \label{ex:revcompl} \end{example} \begin{definition} A permutation $\pi \in S_k$ is a \emph{pattern} of a permutation $\sigma \in S_n$ if there is a subsequence of $\sigma$ which is order-isomorphic to $\pi$, i.e., if there is a subsequence $\sigma_{i_1} \sigma_{i_2} \cdots \sigma_{i_k}$ of $\sigma$, $1 \leq i_1 < i_2 <\cdots \sigma^{rc}_2 > \cdots >\sigma^{rc}_{d+2-i-1} >\sigma^{rc}_{d+2-i}=1 <\sigma^{rc}_{d+2-i+1}<\sigma^{rc}_{d+2-i+2}< \cdots <\sigma^{rc}_{d+2}$$ $\sigma^{rc}$ is a permutation of size $(d+2)$ with $d$ descents where the unique ascent is in position $d+2-i$ and the first sequence of descents ends with $1$. Moreover, since $\sigma^{rc}_{d+2-i-1}\sigma^{rc}_{d+2-i}\sigma^{rc}_{d+2-i+1}\sigma^{rc}_{d+2-i+2}$ forms an occurrence of either the pattern $2143$ or the pattern $3142$, (depending on $\sigma$), $\sigma^{rc}$ is a minimal permutation of size $d+2$ with $d$ descents, (see Theorem~\ref{thm:characterization}). Therefore the permutations in $\mathcal{M}_2$ are the reverse-complement of those in $\mathcal{M}_1$ and the following theorem holds. \begin{theorem} \label{thm:m2} The\mini having $1$ as the bottom element of the diamond are enumerated by the sequence $(m_2)_d$ defined as follows: \begin{equation}\label{eq:m2}(m_2)_d= 2^{d+1}-2(d+1).\end{equation} \end{theorem} Since in a minimal permutation $\sigma$ with $d$ descents and size $d+2$ the entry $1$ is at the end of the first sequence of descents or it is the last element of $\sigma$, the proof of the following corollary is straightforward. \begin{corollary}\label{cor:last}The\mini whose last entry is $1$ are enumerated by the sequence $(f)_d$ defined in Corollary~\ref{cor:first}. \end{corollary} \begin{corollary}\label{cor1d}The\mini whose unique ascent is $1\,(d+2)$ are enumerated by the sequence $(g)_d$ defined as follows: \begin{equation}\label{eq:tot1d} (g)_d=2^{d}-2.\end{equation} \end{corollary} \begin{proof} By the definition of the bijection $\Phi^2$, the \emph{unique} minimal permutation with $d$ descents and size $d+2$ of type \texttt{A} in which the bottom element of the diamond is $1$ is the the permutation $2\,1\,(d+2)\,(d+1)\,d \cdots 3$. Similarly, if a permutation of type \texttt{B} has 1 as the bottom element of the diamond then the associated non-interval subset has cardinality $2$. Therefore, there is an \emph{unique} minimal permutation with $d$ descents and size $d+2$ of type \texttt{B} whose first sequence of descents ends with 1, and it is the permutation $\Phi^2(s)$ where $s=\{1,3\}$ and $w=\{2,4,\ldots,d+1\}$, that is the permutation $d\,(d-1)\cdots 2\,1\,(d+2)\,(d+1)$. The\mini of type \texttt{C} in $\mathcal{M}_2$ are those in which the segment $r_2$ of the second sequence of descent is empty. Recall that the first sequence of descent contains at least three elements. By the definition of $\Phi^2(s)$ for permutations of type\texttt{C}, $r_2$ is empty if $n-p-1=0$, that is $w_1=2$, as $p=n+1-w_1$. Therefore, for each value of the cardinality $n$ of $s$ there is only one permutation of type \texttt{C} in which 1 is the bottom element of the diamond. Since $n$ ranges from $3$ to $d-1$, the number of\mini of type \texttt{C} in $\mathcal{M}_2$ is $d-3$. To sum up, the total number $M_{ABC}$ of\mini of type \texttt{A}, \texttt{B}, and \texttt{C} with unique ascent $1\,(d+2)$ is \begin{equation}\label{eq:totABC} M_{ABC} = 1+1+ (d-3) = d-1. \end{equation} Now we have just to count the permutations in $S^1$ having $1$ at the end of the first sequence of descents. The first sequence of descents in a permutations $\Phi^1(s)$ contains the elements of $s$ in descending order. Therefore, it is sufficient to count the non-interval sets $s$ containing the entry $1$. Given the cardinality $n$, if $s$ contains $1$ then the other $n-1$ elements are \begin{itemize} \item a non-interval set of cardinality $n-1$. As we have seen before, the intervals of length $n-1$ are $d+1-(n-1)$, so the non-interval sets of cardinality $n-1$ are ${d \choose n-1}- (d+1)+(n-1)$, \item or an interval of length $n-1$ without the entry $2$. As before, it is simple to see that the intervals of length $n-1$ starting with $3$ are $d+1-n$. \end{itemize} Thus, the non-interval sets $s$ of cardinality $n$ containing the entry $1$ are \begin{equation}\label{eq:totsn} {d \choose n-1}- (d+1)+(n-1)+d+1-n = {d \choose n-1}-1. \end{equation} Hence, the permutations in $S^1$ having $1$ at the end of the first sequence of descents are \begin{eqnarray}\label{eq:totS1} M_{S^1}&=& \sum_{n=2}^d[{d \choose n-1}-1]\nonumber\\ &=& \sum_{n=2}^d{d \choose n-1} - \sum_{n=2}^d 1\nonumber\\ &=& 2^d- {d \choose 0}-{d \choose d}-(d-1)\nonumber\\ &=&2^d-d-1. \end{eqnarray} The number of\mini having the pair $1\,(d+2)$ as unique ascent is $$M_{S^1}+M_{ABC}= 2^d-d-1+d-1 = 2^d-2.$$ \end{proof} The first terms of the sequence~\eqref{eq:tot1d} are $2,6,14,30,62,126,254 \ldots$, for $d \geq 2$. They correspond to the sequence \seqnum{A000918} in the On-line Encyclopedia of Integer Sequence \cite{OEIS}. This sequence is the first differences of \seqnum{A005803}, as noted in \cite{OEIS}. Then the\mini whose unique ascent is $1\,(d+2)$ are a combinatorial interpretation of the first differences of \seqnum{A005803}. \begin{corollary}\label{cord1}The\mini having the first or the second sequence of descents starting with $d+2$ and ending with $1$ are enumerated by the sequence $(h)_d$ defined as follows: \begin{equation}\label{eq:totd1} (h)_d=2^{d}-2d.\end{equation} \end{corollary} \begin{proof}The number of\mini whose first sequence of descents is $(d+2) \cdots 1$ is given by the difference between the number of\mini having $1$ as the bottom element of the diamond (see Theorem~\ref{thm:m2}) and the number of those having the pair $1\,(d+2)$ as unique ascent (see Corollary~\ref{cor1d}), that is $$2^{d+1}-2(d+1)-(2^{d}-2) = 2^d-2d.$$ The number of\mini whose second sequence of descents is $(d+2) \cdots 1$ is obtained in a similar way from Theorem~\ref{thm:m1} and Corollary~\ref{cor1d}. \end{proof} \begin{corollary}\label{cord21} The\mini having $d+2$ as the first entry and $1$ as the last one are enumerated by the sequence $(k)_d$ defined as follows: \begin{equation}\label{eq:totd21} (k)_d=2^{d}-d(d-1)-2.\end{equation} \end{corollary} \begin{proof}The number of\mini having $1$ as the last entry is $2^{d+1}-d(d+1)-2$ (see Corollary~\ref{cor:last}). If from this set we cancel those permutation having $d+2$ as the top element of the diamond (see Corollary~\ref{cord1}) we obtain the set of\mini having $d+2$ as the first entry and $1$ as the last one, whose cardinality is given by $$2^{d+1}-d(d+1)-2-(2^d-2d)= 2^d-d(d-1)-2.$$ \end{proof} The first terms of the sequence~\eqref{eq:totd21} are $0,0,2,10,32,84,198 \ldots$, for $d \geq 2$. \begin{table}[H] \begin{center} \resizebox{\columnwidth}{!}{% \begin{tabular}{|l|l|l|l|l|} \hline \multicolumn{1}{|c|}{\small{Shape of the permutation}}&\multicolumn{1}{|c|}{\small{Number of}} &\multicolumn{1}{|c|}{\small{Formula}}&\multicolumn{1}{|c|}{\small{OEIS}}&\multicolumn{1}{|c|}{\small{Reference}}\\ &\multicolumn{1}{|c|}{\small{permutations}}&&&\\ \hline \hline \multirow{7}*{\includegraphics[height=3cm]{riga1.eps}}&&&&\\ &&&&\\ &\small{$2,8,22,52,$}&\small{$2^{d+1}-2(d+1)$}&\small{\seqnum{A005803}}&\small{Theorems \ref{thm:m1}, \ref{thm:m2} }\\ &\small{$114,240,494, \ldots$}&&&\\ &&&&\\ &&&&\\ &&&&\\ \hline \multirow{7}*{\includegraphics[height=3cm]{riga2.eps}}&&&&\\ &&&&\\ &\small{$0,2,10,32,$} &\small{$2^{d+1}-d(d+1)-2$}&&\small{Corollaries \ref{cor:first}, \ref{cor:last}}\\ &\small{$84, 198, 438, \ldots$}&&&\\ &&&&\\ &&&&\\ &&&&\\ \hline \multirow{7}*{\includegraphics[height=3cm]{riga3.eps}}&&&&\\ &&&&\\ &\small{$2,6,14,30,$} &\small{$2^{d}-2$}&\small{\seqnum{A000918}}&\small{Corollary \ref{cor1d}}\\ &\small{$62, 126, 254, \ldots$}&&&\\ &&&&\\ &&&&\\ &&&&\\ \hline \multirow{7}*{\includegraphics[height=3cm]{riga4.eps}}&&&&\\ &&&&\\ &\small{$0,2,8,22,$}&\small{$2^{d}-2d$}&\small{\seqnum{A005803}}&\small{Corollary \ref{cord1}}\\ &\small{$52,114,240, \ldots$}&&&\\ &&&&\\ &&&&\\ &&&&\\ \hline \multirow{7}*{\includegraphics[height=3cm]{riga5.eps}}&&&&\\ &&&&\\ &\small{$0,0,2,10,$} &\small{$2^{d}-d(d-1)-2$}&&\small{Corollary \ref{cord21}}\\ &\small{$32,84, 198,\ldots$}&&&\\ &&&&\\ &&&&\\ &&&&\\ \hline \end{tabular} } \end{center} \caption{Number sequences for some subsets of minimal permutations with $d$ descents and size $d+2$, $d \geq 2$. OEIS refers to entry in \cite{OEIS}.} \label{tab:listcount} \end{table} \begin{thebibliography}{9} \bibitem{BV10} J.-L. Baril and R. Vernay, Whole mirror duplication-random loss model and pattern avoiding permutations, \textit{Inform. Process. Lett.} \textbf{110} (2010), 474--480. \bibitem{math.CO} M. Bouvel and E. Pergola, Posets and permutations in the duplication-loss model: Minimal permutations with $d$ descents, \textit{Theoret. Comput. Sci.} \textbf{411} (2010), 2487--2501. \bibitem{BR09} M. Bouvel and D. Rossin, A variant of the tandem duplication-random loss model of genome rearrangement, \textit{Theoret. Comput. Sci.}, \textbf{410} (2009), 847--858. \bibitem{SODA06} K. Chaudhuri, K. Chen, R. Mihaescu, and S. Rao, On the tandem duplication-random loss model of genome rearrangement, in \textit{Proc. 17th Ann. ACM-SIAM Symposium on Discrete Algorithms (SODA)}, ACM, 2006, pp.~564--570. \bibitem{MY10} T. Mansour and S. H. F. Yan, Minimal permutations with $d$ descents, \textit{European J. Combin.} \textbf{31} (2010), 1445--1460. \bibitem{OEIS} N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, \url{http://oeis.org}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 05A15; Secondary 05A05. \noindent \emph{Keywords: } minimal permutation, enumeration. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000918} and \seqnum{A005803}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received April 29 2015; revised version received August 25 2015. Published in {\it Journal of Integer Sequences}, September 8 2015. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .