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Catalan Numbers, Their Generalization, and Their Uses Peter Hilton and Jean Pedersen

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Catalan Numbers, Their Generalization, and Their Uses Peter Hilton and Jean Pedersen Catalan Numbers, Their Generalization, and Their Uses Peter Hilton and Jean Pedersen Probably the most prominent among the special in- stepping one unit east or one unit north of P,. We say tegers that arise in combinatorial contexts are the bino- that this is a path from P to Q if Po = P, Pm= Q. It is mial coefficients (~). These have many uses and, often, now easy to count the number of paths. fascinating interpretations [9]. We would like to stress one particular interpretation in terms of paths on the THEOREM 0.1. The number of paths from (0,0) to (a,b) is integral lattice in the coordinate plane, and discuss the the binomial coefficient if+b). celebrated ballot problem using this interpretation. A path is a sequence of points Po P1 9 9 Pro, 9 m >I O, Proof: We may denote a path from (0,0) to (a,b) as a where each P, is a lattice point (that is, a point with sequence consisting of a Es (E for East) and b Ns (N for integer coordinates) and Pz+l, i 1> 0, is obtained by North) in some order. Thus the number of such paths 64 THE MATHEMATICALINTELUGENCER VOL 13, NO 2 @ 1991 Spnnger-Verlag New York is the number of such sequences. A sequence consists of (a + b) symbols and is determined when we have decided which a of the (a + b) symbols should be Es. p=2 p=3 Thus we have (o+b) choices of symbol. k=3 k=2 3 4 5 2 3 Figure 1. Left: a binary (or 2-ary) tree with 3 source-nodes (Q) and 4 end-nodes (o) and, at right, a ternary (or 3-ary) tree with 2 source-nodes (o) and 5 end-nodes (o). p=2 p=3 k=3 k=2 (Sl((S2 S3)S4)) (sl s2(ss Figure 2. Left: an expression involving 3 applications of a binary operation applied to 4 symbols and, at right, an ex- pression involving 2 applications of a ternary operation ap- plied to 5 symbols. We will present another family of special integers, called the Catalan numbers (after the nineteenth-cen- tury Belgian 1 mathematician Eugene Charles Catalan; see the historical note at the end of this ar~cle), which p=2 p=3 are closely related, both algebraically and concep- k=3 k=2 tually, to the binomial coefficients. These Catalan numbers have many combinatorial interpretations, of which we will emphasize three, in addition to their interpretation in terms of 2-good paths on the integral lattice. Because all these interpretations remain valid if one makes a conceptually obvious generalization of the original definition of Catalan numbers, we will present these interpretations in this generalized form. Figure 3. Left: a S-gun subdivided into three 3-gons by 2 Thus the definitions we are about to give depend on a diagonals and, at right, a 6-gon subdivided into two 4-gons parameter p, which is an integer greater than one. The by I diagonal. original Catalan numbers--or, rather, characteriza- tions of these numbers--are obtained by taking p = 2. ak: rao = 1, ~a k = the number of p-ary trees with k Let us then present three very natural combinatorial source-nodes, k ~> 1 (see Fi- concepts, which arise in rather different contexts, but gure 1). which turn out to be equivalent. bk: pbo = 1, ~bk = the number of ways of associating Let p be a fixed integer greater than 1. We define k applications of a given p-ary three sequences of positive integers, depending on p, operation, k I> 1 (see Figure 2). as follows: ck: fo = 1, t,ck = the number of ways of subdi- viding a convex polygon into k disjoint (p + 1)-gons by means of t He lived his life m Belgium; but, being born in 1814, he was not non-intersecting diagonals, k I> 1 born a Belgian! (see Figure 3). THE MATHEMATICALINTELLIGENCER VOL. 13, NO. 2, 1991 65 However, relating the trees of Figure 1 (or the paren- p=2 p=3 thetical expressions of Figure 2) to the corresponding k=3 k=2 (1 ((23) 4)) (~ 2 s 4 5)) dissected polygons of Figure 3 is more subtle (and the hint given in [8], in the case p = 2, is somewhat cryptic), so we will indicate the proof that bk = Ck. Thus, suppose that we are given a rule for associating k applications of a given p-ary operation to a string of (p - 1)k + 1 symbols sl, s2..... s0,_l~,+ v Label the successive sides (in the anticlockwise direction) of the convex ((p - 1)k + 2)-gon with the integers from 1 to 3 (p - 1)k + 1, leaving the top, horizontal side unla- Figure 4. The dissected polygons of Figure 3, with diag- beled. Pick the first place along the expression onals and last side labeled. (counting from the left) where a succession of p symbols is enclosed in parentheses. If the symbols en- closed run from st+ 1 to sj+p, draw a diagonal from the (1(23(4(567) 8))9) initial vertex of the (j + 1)st side to the final vertex p=3 of the (j + p)th side, and label it (j + 1..... j + p) k=4 Also imagine the part (sj+a .... sj+p) replaced by a 3(4(567) 8)) /~ single symbol. We now have effectively reduced our word to a set of (k - 1) applications and our polygon to a set of 2 polygons, one a (p + 1)-gon, the other a [(p - 1)(k - 1) + 2]-gon, so we proceed inductively to complete the rule for introducing diagonals. More- over, the eventual label for the last side will corre- spond precisely to the string of (p - 1)k + 1 symbols 'Y with the given rule for associating them, that is, to the original expression. Figure 4 illustrates the labeling of the dissections of the pentagon and hexagon that are determined by the corresponding expressions of Figure 2. The converse procedure, involving the same initial labeling of the original sides of a dissected polygon, and the successive labeling of the diagonals 5 introduced, leads to an expression that acts as label for the last (horizontal) side. Figure 5. The polygonal dissection associated with We now illustrate a slightly more complicated situa- (s~(s2s3(s,(sss~ )ss) ) ~. tion: Example O. 1. Let p = 3, k = 4 and consider the expres- (As indicated, we suppress the 'p' from the symbol if sion no ambiguity need be feared.) sl(s2s3(s,(sss6sT)sg))sg). Note that, if k t 1, (i) a p-ary tree with k source-nodes has The dissection of the convex 10-gon associated with (p - 1)k + 1 end-nodes and pk + 1 nodes in this expression, with the appropriate labeling, is all (Figure 1); shown in Figure 5. The corresponding ternary tree is (ii) the k applications of a given p-ary operation are shown in Figure 6, which also demonstrates how the applied to a sequence of (p - 1)k + 1 symbols tree may be associated directly with the dissected (Figure 2); polygon. The rule is "Having entered a (p + 1)-gon (iii) the polygon has (p - 1)k + 2 sides and is sub- through one of its sides, we may exit by any of the divided into k disjoint (p + 1)-gons by (k - 1) other p sides." diagonals (Figure 3). In the case p = 2, any of a t, bk, Ck may be taken as the A well-known and easily proved result (for the case definition of the kth Catalan number. Thus we may re- p = 2 see Sloane [8]) is the following: gard any of pat, pbk, pck as defining the generalized k th Catalan number. Moreover, the following result is THEOREM 0.2. pak = pbk = pCk" known (see [6]). The reader will probably have no trouble seeing THEOREM 0.3. how the trees of Figure I are converted into the corre- Pak= k 1 (p-1)k + l ' sponding expressions of Figure 2--and vice versa. k~>l. 66 THE MATHEMATICAL INTELLIGENCER VOL 13, NO 2, 1991 The "classical" proof of this result is rather sophisti- particular stress on the flexibility of this fourth inter- cated. One of our principal aims is to give an elemen- pretation, which we now describe. tary proof of Theorem 0.3, based on yet a fourth inter- We say that a path from P to Q is p-good if it lies pretation of the generalized Catalan numbers which, entirely below the line y = (p - 1)x. Let dk = pdk be like that of the binomial coefficients given earlier, is the number of p-good paths from (0, - 1) to (k, (p - 1)k also in terms of paths on the integral lattice. We lay - 1). (Thus, by our convention, do = 1.) We extend Theorem 0.2 to assert the following. THEOREM 0.4. pak = pbk = pCk =pd k. Proof: First we restate (with some precision) the defini- tions of b k and dk, which numbers we will prove equal. p=3 bk = the number of expressions involving k appli- k=4 cations of a p-ary operation (on a word of 1 9 k(p - 1) + 1 symbols). Let E be the set of all O•j• such expressions. dk = the number of good paths from (0,-1) to (k, (p - 1)k - 1); by following such a path by a 2 single vertical segment, we may identify such paths with certain paths (which we will also call 'good') from (0,- 1) to (k, (p - 1)k).
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