Counting and recounting: The aftermath

This is a recount of the letters that I received from readers who continued where I left off by offering solutions to the problem. The problem is not that new. P. Erd6s, on reading the article, was quick to point out to me that Hungarian mathemat ic ians tackled it in the thirties: P. Veress proposing and G. Hajos solving it. In his letter to me I. Gessel (M.I.T.) has given a survey of the more recent history of the problem. Proofs were published by D. Kleitman (Studies in Applied Mathematics 54. (1975), also by his student D. J. Kwiatowski (Ph.D. Thesis, MIT, 1975). It also found its way into texts (Feller, Mohanty). In addition, I received solutions by A. Bondesen (Royal Danish School of Educational Studies, Copenhagen), K. Gr~nbaum (Roskilde Universitetscenter, Denmark), J. Hofbauer, jointly with N. Fulwick (Universit/it, Wien, Austria), D. Zeilberger (Drexel University, Philadelphia), and verbally from C. Pearce (Adelaide University), directly after reading the article. All solutions are based, with some variations, on the count of lattice paths, or equivalently (1,0) sequences. Figure 1 is used to illustrate the simplest version. It represents a two-dimensional coordinate lattice, or a network of streets running East and North. We consider paths of length 2n, beginning at O, proceeding in unit steps, heading East or North. It is clear that there are 22n ways in which a lattice point on the boundary AB can be reached. This gives the left-hand side of the identity. Counting in a different way, assume that the last crossing of a path with the NE line (OM on the diagram) is at K(k,k), which of course may coincide with O or M. It is easy to see that there are (2~) possible paths from O to K. Assuming for the moment that the number of ways the remaining 2n 2k steps, (avoiding OM) may be taken, is similarly (2n_-k2k), we obtain the desired right-hand side: ~=ot2k) (2n-~k) 9