The formula for the number of order-preserving selfmappings of a fence

Abstract

LetY be a fence of sizem andr=⌊m−1/2⌊. The numberb(m) of order-preserving selfmappings ofY is equal toA _r-B_r-C_r-D_r, where, ifm is odd,

$$\begin{gathered} A_r = 2(r + 1)\sum\limits_{s = 0}^r {\left( {\begin{array}{*{20}c} {r + s} \\ {2s} \\ \end{array} } \right)} 4^s , B_r = 2r\sum\limits_{s = 1}^r {\left( {\begin{array}{*{20}c} {r + s} \\ s \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {r - 1} \\ {s - 1} \\ \end{array} } \right),} \hfill \\ C_r = 4r\sum\limits_{s = 0}^{r - 1} {\left( {\begin{array}{*{20}c} {r + s} \\ s \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {r - 1} \\ s \\ \end{array} } \right), D_r = \sum\limits_{s = 0}^{r - 1} {(2s + 1)} \left( {\begin{array}{*{20}c} {r + s - 1} \\ s \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {r - 1} \\ s \\ \end{array} } \right)} \hfill \\ \end{gathered} $$

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Ifm is even, a similar formula forb(m) is true. The key trick in the proof is a one-to-one correspondence between order-preserving selfmappings ofY and pairs consisted of a partition ofY and a strictly increasing mapping of a subfence ofY toY.