Equidistribution of returns and height of first peak of Dyck paths - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T18:04:01Z http://mathoverflow.net/feeds/question/65363 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/65363/equidistribution-of-returns-and-height-of-first-peak-of-dyck-paths Equidistribution of returns and height of first peak of Dyck paths Martin Rubey 2011-05-18T19:53:17Z 2011-05-18T20:57:32Z <p>I believe that it is "well known" that the following two statistics on Dyck paths have symmetric joint distribution:</p> <ol> <li>number of returns to the axis $RET(D)$</li> <li>height of the first peak (or length of the last descent) $HFP(D)$</li> </ol> <p>That is: $\sum_{D} x^{RET(D)}y^{HFP(D)} = \sum_{D} x^{HFP(D)}y^{RET(D)}$</p> <p>However, I could not find a reference for that. Might it be due to Kreweras?</p> http://mathoverflow.net/questions/65363/equidistribution-of-returns-and-height-of-first-peak-of-dyck-paths/65368#65368 Answer by Gjergji Zaimi for Equidistribution of returns and height of first peak of Dyck paths Gjergji Zaimi 2011-05-18T20:44:15Z 2011-05-18T20:57:32Z <p>You can use the article <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&amp;_udi=B6V00-3SYS1V9-T&amp;_user=10&amp;_coverDate=01%2F15%2F1998&amp;_rdoc=1&amp;_fmt=high&amp;_orig=gateway&amp;_origin=gateway&amp;_sort=d&amp;_docanchor=&amp;view=c&amp;_searchStrId=1756643324&amp;_rerunOrigin=google&amp;_acct=C000050221&amp;_version=1&amp;_urlVersion=0&amp;_userid=10&amp;md5=6aa8974610b7a2e55e444388541bf632&amp;searchtype=a" rel="nofollow">"A bijection on Dyck paths and its consequences"</a> by E. Deutsch. The author has several papers on enumerative problems on Dyck/Motzkin paths. (See also <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&amp;_udi=B6V00-3X05MHG-C&amp;_user=10&amp;_coverDate=06%2F06%2F1999&amp;_rdoc=1&amp;_fmt=high&amp;_orig=gateway&amp;_origin=gateway&amp;_sort=d&amp;_docanchor=&amp;view=c&amp;_rerunOrigin=scholar.google&amp;_acct=C000050221&amp;_version=1&amp;_urlVersion=0&amp;_userid=10&amp;md5=f5864614460a2354539756923f71ea47&amp;searchtype=a" rel="nofollow">here</a>)</p>