Enumeration Via Ballot Numbers Martin Aigner
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View metadata, citation and similar papers at core.ac.uk brought to you by CORE provided by Elsevier - Publisher Connector Discrete Mathematics 308 (2008) 2544–2563 www.elsevier.com/locate/disc Enumeration via ballot numbers Martin Aigner Mathematics Institute, Freie Universität Berlin, Arnimallee 3, D-14195 Berlin, Germany Received 27 April 2004; received in revised form 19 June 2007; accepted 21 June 2007 Available online 22 October 2007 Abstract Several interesting combinatorial coefficients such as the Catalan numbers and the Bell numbers can be described either via a 3-term recurrence or as sums of (weighted) ballot numbers. This paper gives some general results connecting 3-term recurrences with ballot sequences with several applications to the enumeration of various combinatorial instances. © 2007 Elsevier B.V. All rights reserved. Keywords: Ballot numbers; Catalan-like numbers; Weighted sums; Combinatorial instances 1. Introduction 2n Our starting point is a two-fold description of the Catalan numbers Cn = (1/(n + 1)) n . First, consider the ballot number recurrence: a0,0 = 1,a0,k = 0 (k > 0), an,k = an−1,k−1 + an−1,k +···+an−1,n−1 (n1). Then a classical result says n an,k = Cn+1,an,0 = Cn for all n. k=0 The matrix A = (an,k) indexed by N0 is lower triangular with 1’s in the main diagonal. The following table shows the first rows of A with the 0’s omitted: 1 11 221 A = 5531 14 14 9 4 1 42 42 28 14 5 1 ... 2n−k The numbers an,k =[(k + 1)/(n + 1)] n are called the (ordinary) ballot numbers. E-mail address: [email protected]. 0012-365X/$ - see front matter © 2007 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2007.06.012 M. Aigner / Discrete Mathematics 308 (2008) 2544–2563 2545 The ubiquity of the Catalan numbers in enumeration problems (the book of Stanley [20] lists more than 70 instances) is partly explained by this description via the ballot number recurrence. Very often, the set Sn to be counted can be partitioned in a natural way into subsets Sn,k (0k n) such that the sizes an,k =|Sn,k| obey the ballot number recurrence, resulting in |Sn|=Cn+1. We will see several examples later on. The second description of the Catalan numbers is less well-known. Consider the 3-term recurrence: b0,0 = 1,b0,k = 0 (k > 0), bn,k = bn−1,k−1 + 2bn−1,k + bn−1,k+1 (n1). Then bn,0 = Cn+1 for all n. The matrix B = (bn,k) is again lower-triangular. The table shows the first rows: 1 21 541 B = 14 14 6 1 42 48 27 8 1 132 165 110 44 10 1 ... S = n a b We see that the sum n k=0 n,k of the ballot table equals the coefficient n,0 computed via a 3-term recurrence. It is the purpose of this paper to describe several instances where the number Sn of generalized ballot numbers is equal to bn,0 for general 3-term recurrences. In particular, this approach via ballot tables will give some indica- tion why, next to the Catalan numbers, the Motzkin numbers [1,7], Riordan numbers [7] and Fine numbers [11,12] appear so often in enumeration problems. Hence the present article may be considered as a companion paper to [7,12]. Let us consider generalized ballot numbers first. Let = (m0,m1,m2,...) be a sequence of real numbers. We associate with the lower triangular matrix A = (an,k) as follows (omitting the index ): a0,0 = 1,a0,k = 0 (k > 0), an,k = an−1,k−1 + mkan−1,k + an−1,k+1 +···+an−1,n−1 (n1). (1) n We call A the ballot table corresponding to . We are interested in the sum Sn = an,k and the element in k=0 column zero, Zn := an,0, for n0. Next we turn to 3-term recurrences. Let = (s0,s1,s2,...), = (t1,t2,...) be two sequences of real numbers, , , where ti = 0 for all i. We associate with , the matrix B = (bn,k) in the following way (again omitting the super- scripts): b0,0 = 1,b0,k = 0 (k > 0), bn,k = bn−1,k−1 + skbn−1,k + tk+1bn−1,k+1 (n1), (2) , and call Bn := bn,0 the Catalan-like numbers corresponding to and . The recurrence (2) has been considered by many authors via continued fractions, orthogonal polynomials, generating functions, etc. for special sequences and (see e.g. [2,4,13,21]). In view of later applications to enumeration problems, 2546 M. Aigner / Discrete Mathematics 308 (2008) 2544–2563 , let us note some important instances where the numbers Bn are known (see [4]): Consider = (1, 1, 1,...). The most interesting Catalan-like numbers arise in the case when is constant from the term s1 on; we then write shortly = (s0,s1): = (0, 0), B2n = Cn Catalan number,B2n+1 = 0, = (1, 1), Bn = Mn Motzkin number, = (2, 2), Bn = Cn+1, n = (1, 0), Bn = , n/2 2n + 1 = (3, 2), Bn = , n = (0, 1), Bn = Rn Riordan number, = (1, 2), Bn = Cn, = (0, 2), Bn = Fn Fine number. Other examples are 2n = (2, 2), Bn = , n = (2, 1, 1,...), = (2, 3), Bn = Schn Schröder number, = (2, 2, 2,...) and n−1 n−k = (s, s + 1), Bn = N(n, k)s , k=0 1 n n = (s,s,s,...) where N(n, k) = is the Narayana number. n k k + 1 Finally, we note that for = (1, 2, 3, 4,...), = (1, 2, 3, 4,...)we obtain the Bell number Bn = Belln, counting the number of partitions of an n-set, and for = (2, 3, 4, 5,...), = (1, 2, 3, 4,...)we obtain Bn = Belln+1 (see e.g. [3]). It was shown in [2] that the generating function , , n B (x) = Bn x for = (a,s,s,...), = (t,t,t,...) n 0 is given as follows: 2 2 , 1 − (2a − s)x − 1 − 2sx + (s − 4t)x B (x) = . (3) 2(s − a)x + 2(a2 − as + t)x2 Actually, in [2] only the case t = 1 is considered, but the more general result follows in the same way. In the next section we prove the main theorem connecting ballot numbers and Catalan-like numbers, together with some ramifications. In Section 3 we discuss several combinatorial instances, with more general weighted versions appearing in Section 4. We follow the usual terminology, for all terms not defined the reader may consult any of the standard texts [14,17,19]. M. Aigner / Discrete Mathematics 308 (2008) 2544–2563 2547 2. Ballot numbers and Catalan-like numbers S = n a A B, = b, To equate a sum n k=0 n,k of a ballot table as in (1) with a Catalan-like number n n,0 of the matrix B, as in (2), there are two approaches: , (A) Find an explicit expression of the coefficients bn,k in terms of an,k. (B) Compute the respective generating functions. We will illustrate both methods as we go along. In the sequel we use the following short notation: S , , ←→ means that Sn = Bn for all n, where Sn and Bn are defined according to (1) and (2), Z , ←→ means that Zn = Bn for all n. In this notation our starting example reads: S = (2, 2, 2,...) = (1, 1, 1,...)←→ (Sn = Bn = Cn+1), = (1, 1, 1,...) Z = (1, 2, 2,...) = (1, 1, 1,...)←→ (Zn = Bn = Cn). = (1, 1, 1,...) ←S→ (a, a, ) ←S→ (a+1,) (a, ) ←Z→ (a,) Theorem 1. If , then (1,) and (1,) . Proof. We prove the first assertion, the second is shown similarly. Consider the ballot matrix A with respect to . The columns 0, 1,...,2n determine all elements down to row 2n + 1, and hence all sums Sk (k = 0,...,2n + 1). Similarly, in the matrix B, corresponding to and , the columns 0, 1,...,n determine the Catalan-like numbers B, = b k = , ,..., n + C(x)= c xn C(x)= c xn k k,0 for 0 1 2 1. For any generating function n 0 n set n 0 n+1 , thus C(x)=(C(x)−C(0))/x. Considering the ballot recurrence for the columns 0, 1,...,2n we obtain for the generating functions Ai(x) of the ith column up to i = 2n the system, where for brevity we omit here and later often the variable x: A − 0 1 = A = m A + A +···+A x 0 0 0 1 2n A 1 = A = A + m A +···+A x 1 0 1 1 2n . A 2n = A = A + m A x 2n 2n−1 2n 2n and therefore (1 − m0x)A0 − xA1 − xA2 −···−xA2n = 1, −xA0 + (1 − m1x)A1 − xA2 −···−xA2n = 0, . −xA2n−1 + (1 − m2nx)A2n = 0. 2548 M. Aigner / Discrete Mathematics 308 (2008) 2544–2563 Denote by ⎛ ⎞ 1 − m0x −x ... −x ⎜ ⎟ ⎜ −x − m x ... −x ⎟ ⎜ 1 1 ⎟ M = ⎜ ⎟ ⎜ .. ⎟ ⎝ . ⎠ 0 −x 1 − m2nx the matrix of coefficients. By Cramer’s rule it follows that j (−1) Ej (M) Aj (x) = , det M where Ej (M) (j = 0,...,2n) is the determinant of the submatrix of M with the 0th row and jth column deleted. In particular, we obtain 2n 2n (− )j E (M) i j=0 1 j Q(x) = qix = Aj (x) = , det M i 0 j=0 and we know that qi = Si for 0i 2n + 1. Turning to the Catalan-like numbers defined by and , we find in an analogous way for the column-generating functions B0(x),...,Bn(x) up to n the coefficient matrix ⎛ ⎞ 1 − s0x −t1x 0 ...0 ⎜ ⎟ ⎜ −x − s x −t x ..