Logarithmic behavior of some combinatorial sequences

arXiv:math/0603379v1 [math.CO] 15 Mar 2006 Logarithmic behavior of some combinatorial sequences Tomislav Došlić⋆† and Darko Veljan⋆‡ February 2, 2008 † Department of Informatics and Mathematics, Faculty of Agriculture, University of Zagreb, Svetošimunska c. 25, Zagreb, CROATIA ‡ Department of Mathematics, University of Zagreb, Bijenička 30, Zagreb, CROATIA ⋆ To whom correspondence should be addressed, e-mail : [email protected] Proposed running head: Logarithmic behavior of combinatorial sequences 1 Abstract Two general methods for establishing the logarithmic behavior of recursively defined sequences of real numbers are presented. One is the interlacing method, and the other one is based on calculus. Both methods are used to prove logarithmic behavior of some combinatorially relevant sequences, such as Motzkin and Schröder numbers, sequences of values of some classic orthogonal polynomials, and many others. The calculus method extends also to two- (or more- ) indexed sequences. Keywords: log-concavity, log-convexity, special combinatorial numbers, calculus, integer sequences AMS subject classifications: 05A20, 05A16, 05E35, 11B83, 11B37, 11B39, 11B68 2 1 Introduction Let a(n), n ≥ 0, be a sequence of positive real numbers. We want to examine the rate of growth of this sequence, i.e. to examine whether the quotient a(n) a(n−1) decreases, increases or remains constant. In other words, we want to see whether the sequence is log-concave, i.e. a(n)2 ≥ a(n − 1)a(n + 1), log-convex, i.e. a(n)2 ≤ a(n − 1)a(n + 1), or log-straight (or geometric), i.e. a(n)2 = a(n − 1)a(n + 1) for all n ≥ 1. Under log-behavior we also sometimes include log-Fibonacci behavior, meaning sign[a(n)2 − a(n − 1)a(n + 1)] = sign(−1)n (or (−1)n+1 ). It is of great interest, especially in combinatorics, as it can be seen from many examples in [25], to know the log-behavior of a given sequence. It is, in fact, just one instance of the whole paradigm of “positivity questions” ([28]). If a(n) has a combinatorial meaning it would be ideally to provide a combinatorial proof of its logbehavior. For example, if we want to prove that a(n) is log-convex and if we know that a(n) = |S(n)|, where S(n) is a certain finite set, then we would like to find an injection S(n) × S(n) → S(n − 1) × S(n + 1), or a surjection S(n − 1) × S(n + 1) → S(n) × S(n), and similarly for log-concavity. It is usually a hard task to find such a (natural) injection or surjection. Still, examples of this type include binomial coefficients, Motzkin numbers ([8]) and permutations with a prescribed number of runs ([6]). Of course, the explicit formulae give another possibility to prove results of this type, but they are rarely on disposal. Instead, other methods for proving such inequalities have been developed, e.g. see [22], [25], [7] or [5]. In this paper besides using old methods to prove some new results on log-behavior, we shall also introduce some new methods and use them to prove log-behavior of certain interesting combinatorial sequences, and apply this method to other sequences, the most prominent example being values of classical orthogonal polynomials. 3 2 Log-behavior of some sequences using known results Let us quote some known results and apply them to examine the log-behavior of certain combinatorial and other sequences. Lemma 2.1 (Newton’s lemma) Let P (x) = Pn k k=0 ak x be a real polynomial whose all roots are real numbers. Then the coefficients of P (x) are log-concave, i.e. a2k ≥ ak−1 ak+1 , k = 1, . . . , n − 1. Moreover, the (finite) sequence ak ( nk ) is log-concave in k. Let us briefly recall how to apply this lemma to binomial coefficients and Stirling numbers c(n, k) of the first kind (the number of permutations on the set [n] = {1, 2, . . . , n} with exactly k cycles) and Stirling numbers S(n, k) of the second kind (the number of partitions of [n] into exactly k blocks). The following formulae are well known: (x + 1)n = n   X n k=0 xn̄ = n X k xk , (2.1) c(n, k)xk , (2.2) S(n, k)xn , (2.3) k=0 xn = n X k=0 where xk = x(x − 1) . . . (x − k + 1) is the k-th falling power, and xk = x(x + 1) . . . (x + k − 1) the k-th rising power of x. From (2.1) and (2.2) we see that (x + 1)n and xn̄ have only real roots. So, by Newton’s lemma we conclude that the sequences sequence S(n, k) is a bit more involved. Let Pn (x) = n X k=0 4 n
k and c(n, k) are log-concave. The case of the S(n, k)xk . From P0 (x) = 1 and from the basic recursion S(n, k) = S(n − 1, k − 1) + kS(n − 1, k), it follows at once that Pn (x) = x[Pn′ (x) + Pn−1 (x)]. The function Qn (x) = Pn (x)ex has the same roots as Pn (x) and it is easy to verify Qn (x) = xQ′n (x). By induction on n and using Rolle’s theorem it follows easily that Qn , and hence Pn , have only real and non-positive roots. So, we conclude: Theorem 2.2 The sequences n
k k≥0 , (c(n, k))k≥0 and (S(n, k))k≥0 are log-concave. Hence, they are unimodal. An inductive proof of Theorem 2.2 is given in [22]. The next easy lemma is sometimes useful in proving log-convexity results. Lemma 2.3 Let f : [a, b] → R be a positive, continuous function, and In = Z b f (x)n dx, a n ≥ 1. Then (In )n≥2 is a log-convex sequence. Proof By Cauchy-Schwarz inequality, we have In2 = Z a b f (x)n dx 2 = Z b f (x) n−1 2 f (x) n+1 2 dx a 5 2 ≤ Z b a f (x)n−1 dx Z b a f (x)n+1 dx = In−1 In+1 . As an example, we apply this lemma to Legendre polynomials Pn (x). It is well known (e.g. [31]) that the following Laplace formula holds: Pn (x) = 1 π Z π (x + 0 p x2 − 1cosϕ)n dϕ. (2.4) Hence, from Lemma 2.3 and (2.4) we obtain Theorem 2.4 The values Pn (x), n ≥ 0, for x ≥ 1 are log-convex. Another proof of this fact will be presented in Section 4. We say that a sequence (an )n≥0 has no internal zeros if there do not exist integers 0 ≤ i < j < k such that ai 6= 0, aj = 0, ak 6= 0. Theorem 2.5 (Bender-Canfield, see [5]) Let 1, a1 , a2 , . . . be a log-concave sequence of nonnegative real numbers with no internal zeros and let (bn )n≥0 be the sequence defined by X n≥0 bn xn n! Then the sequence (bn )n≥0 is log-convex and  = exp 
bn n! n≥0 X ak k≥1 xk k!  . (2.5) is log-concave. As our first application of this theorem consider the Bell numbers (Bn )n≥0 . B0 = 1, and Bn is the number of partitions of an n-set. It is well known that the exponential generating function for (Bn )n≥0 is given by X n≥0 By taking ak = 1 (k−1)! , Bn xn n!  = exp(ex − 1) = exp  X xk k≥1 k!  . k ≥ 1, and checking that the sequence 1, a1 , a2 , . . . is log-concave, we conclude from Theorem 2.5 that (Bn )n≥0 is log-convex and 6 Bn n! n≥0
is log-concave sequence. More generally, for an integer l ≥ 2, define expl to be l times iterated exponential function, i.e. expl (x) = exp(exp(. . . (exp(x)) . . .),   (l) exp written l times. Define the sequence bn X bn(l) n≥0 n≥0 by xn = expl (x), n! and Bell numbers of order l by (l) Bn(l) = bn . expl (0) (2.6) (2) So, Bn = Bn are ordinary Bell numbers. Now it is not hard to prove by induction on l the following result. Theorem 2.6   (l) is log-convex, and the For any fixed l ≥ 2, the Bell numbers of order l, i.e. the sequence Bn n≥0  (l)  sequence Bn!n is log-concave. n≥0 The following lemma is an easy consequence of the definition of log-convexity and log-concavity. Lemma 2.7 Suppose (an )n≥0 is a positive, log-convex sequence and a0 = 1. Then an am ≤ an+m . If, in addition, an n! n≥0
is log-concave, then an+m ≤ m+n
an am . n Now Theorem 2.6 and Lemma 2.7 imply semi-additivity inequalities for Bell numbers of order l. Corollary 2.8 (l) Bn(l) Bm ≤ (l) Bm+n ≤  m+n n 7  (l) Bn(l) Bm , m, n ≥ 0. The next application of the Bender-Canfield theorem concerns the number of certain permutations, i.e. elements π ∈ Σn , the symmetric group of permutations on n letters. For a fixed integer k ≥ 1 let ck (n) be the number of all permutations from Sn that have cycles of length not exceeding k. By definition, ck (0) = 1. Theorem 2.9 The sequence (ck (n))n≥0 is log-convex and the sequence   ck (n) n! n≥0 is log-concave. Proof It is well known (see, e.g. [27]) that the exponential generating function Ck (x) of the sequence (ck (n))n≥0 is given by   k j X x . Ck (x) = exp  j j=1 Hence the sequence (ak )k≥0 from the Bender-Canfield theorem in (2.5) is as follows: a0 = 1, a1 = 1, . . . , ak = 1, ak+1 = 0, ak+2 = 0, . . ., which has no internal zeros and is obviously a log-concave sequence. Hence the claim follows from Theorem 2.6. The requirement that the sequence (an ) from Bender-Canfield theorem does not have internal zeros is essential, and in general can not be weakened. As an illustration, let us consider a class of sequences related to ck (n). Again, for a fixed integer k ≥ 1 let ek (n) be the number of all permutations π from Σn such that π k = id, for n ≥ 1 and ek (0) = 1. Then the exponential generating function Ek (x) of this sequence is given by (see, e.g. [27])  and the corresponding recurrence is Ek (x) = exp  ek (n) = X xj j|k j  , X (n − 1)j−1 ek (n − j), j|k 8 (2.7) with appropriate initial conditions. The sequence (an ) from (2.5) is given by a0 = a1 = 1, aj = 1 if j divides k and aj = 0 otherwise. This binary sequence 1, 1, a2 , . . . , ak , 0, 0, . . . is log-concave if and only if it does not contain 1, 0, 1 as a subsequence, but it contains internal zeros for all k > 2. Taking, for example, k = 5, the first few terms of e5 (n) being 1, 1, 1, 1, 1, 25, 145, 505, 1345, . . ., we can easily see that this sequence does not exhibit any logarithmically definite behavior, although the sequence 1, 1, 0, 0, 0, 1, 0, . . . is log-concave. For higher values of k, sequences ek (n) log-behave even more chaotically. 3 The interlacing (or “sandwich”) method. Secondary structures. Let a(n), n ≥ 0, be a sequence of positive numbers defined by a homogeneous linear recurrence, say of second order: a(n) = R(n)a(n − 1) + S(n)a(n − 2), (3.1) where R and S are known functions, together with given initial values a(0) = a0 , and a(1) = a1 . Let our task be to examine the rate of growth of a(n). We define the sequence of consecutive quotients q(n) = a(n) , a(n − 1) n ≥ 1. Dividing (3.1) by a(n − 1) we obtain the recurrence q(n) = R(n) + with initial condition q(1) = a1 a0 S(n) , q(n − 1) (3.2) := b1 . The log-concavity or log-convexity of (a(n)) is equivalent, respectively, to q(n) ≥ q(n + 1) or q(n) ≤ q(n + 1), for all n ≥ 1. So, what we want to see is whether the sequence (q(n))n≥1 decreases or increases. To prove that (q(n)) increases, it is enough to find an 9 increasing sequence (b(n)) such that b(n) ≤ q(n) ≤ b(n + 1) (3.3) holds for all n ≥ 1, or at least for all n ≥ n0 for some n0 . Then we can conclude that (a(n)) behaves log-convex at least from some place on. Analogously for log-concavity. This “sandwich method” or “interlacing method” works in some simple cases, but often it is very hard to hit the right sequence (b(n)) which is simple enough. In the rest of this section we show how this method works for some combinatorially important sequences. We also show some consequences of the obtained results. Example 3.1 (Derangements) Let Dn be the number of derangements on n objects, i.e. the number of permutations π ∈ Σn without fixed points, for n ≥ 1, and D0 := 1. It is well known (and easy to prove) that the following recurrence holds: Dn = (n − 1)[Dn−1 + Dn−2 ], (3.4) with initial conditions D0 = 1, D1 = 0. Then D2 = 1, D3 = 2, D4 = 9, D5 = 44, D6 = 265 etc., and we expect from these initial values that Dn2 ≤ Dn−1 Dn+1 for n ≥ 3. Indeed, divide (3.4) by Dn−1 and denote q(n) = Dn Dn−1 . Then  q(n) = (n − 1) 1 +  1 , q(n − 1) (3.5) with q(3) = 2, q(4) = 9/2. Let b(n) = n − 1/2. It is easy to check by induction on n and using (3.5) that b(n) ≤ q(n) ≤ b(n + 1), for all n ≥ 4. Since (b(n)) is clearly increasing and q(3) ≤ q(4), we conclude that (Dn ) is log-convex for n ≥ 3. 10 Example 3.2 Let T2 (n) denotes the number of n×n symmetric N0 -matrices with every row (and hence every column) sum equal to 2 with trace zero (i.e. all main-diagonal entries are zero) (Example 5.2.8 in [27]). The exponential generating function of T2 (n) is given by 1 exp T (x) = √ 1−x  x2 x − 4 2  . The numbers T2 (n) satisfy the recurrence T2 (n) = (n − 1)T2 (n − 1) + (n − 1)T2 (n − 2) −  n−1 2  T2 (n − 3) with the initial conditions T2 (0) = 1, T2 (1) = 0, T2 (2) = T2 (3) = 1. The corresponding recurrence for successive quotients q2 (n) = T2 (n)/T2 (n − 1) is given by n−1 q2 (n) = (n − 1) + − q2 (n − 1)  n−1 2  1 , q2 (n − 1)q2 (n − 2) n ≥ 5, with the initial conditions q2 (3) = 1, q2 (4) = 6. Tabulating the first few values of q2 (n), we see that, after some initial fluctuations, this sequence seems to behave like n − 21 . So, we guess that n − 1 ≤ q2 (n) ≤ n, and indeed, this follows easily by induction on n for n ≥ 6. Hence, the sequence (T2 (n))n≥6 is log-convex. For our next application we need some preparations. A Motzkin path of length n is a lattice path in (x, y)-plane from (0, 0) to (n, 0) with steps (1, 1) (or U p), (1, −1) (or Down) and (1, 0) (or Level), never falling below the x-axis. Denote by M(n) the set of all Motzkin paths of length n. The number Mn = |M(n)| is the n-th Motzkin number. By definition, M0 = 1. A handful of other combinatorial interpretations of Mn are listed in Ex. 6.38 in [27]. A typical member of the Motzkin family M(20) is shown in Fig. 1 as a “landscape path”. A peak of a Motzkin path is a place where an U p step is immediately followed by a Down step. A plateau of length l is a 11 peak terrace plateau trench valley plain (0,0) (20,0) Figure 1: A Motzkin path of length 20 sequence of l consecutive Level steps, immediately preceded by an U p step and immediately followed by a Down step. Similarly we define a terrace, trench, valley and plain of a Motzkin path. A Motzkin path without any Level steps is called a Dyck path or a “mountain path”. It is well known that the set D(n) of all Dyck paths of length 2n is enumerated by Catalan numbers Cn , i.e. |D(n)| = Cn = 1 n+1 2n
n . From the explicit formula it follows immediately that Cn2 ≤ Cn−1 Cn+1 , i.e. Catalan numbers are log-convex. It is also easy to find a simple combinatorial proof of this fact. By counting the number of Level steps on a Motzkin path, it can be easily shown that Catalan and Motzkin numbers are related as follows: Mn = X n  Ck , 2k Cn+1 = k≥0 X n k≥0 k Mk . Proposition 3.2 a) The Motzkin numbers satisfy the following convolutive recursion: Mn+1 = Mn + n−1 X Mk Mn−k−1 . (3.6) k=0 b) The generating function of (Mn ) is given by M (x) = X n Mn x = 1−x− n≥0 12 √ 1 − 2x − 3x2 . 2x2 (3.7) c) Mn ’s satisfy the short recursion (n + 2)Mn = (2n + 1)Mn−1 + 3(n − 1)Mn−2 . (3.8) Proof a) M0 = 1 and M1 = 1. A Motzkin path of length n + 1 either starts by a Level step and then can proceed in Mn ways, or starts by an U p step and returns for the first time to the x-axis after k + 1 steps (to the point (k + 2, 0)). The number of latter is equal to the number of pairs (P1 , P2 ), where P1 is a (translated) Motzkin path from (1, 1) to (k + 1, 1) which is not below the line y = 1, and P2 is a (translated) Motzkin path from (k + 2, 0) to (n + 1, 0). The number of paths P1 is equal to Mk , while the number of paths P2 is equal to the number of Motzkin paths on n + 1 − (k + 2) = n − k − 1 steps, and this is Mn−k−1 . Thus a) follows. b) If we multiply (3.6) by xn+1 and sum over n ≥ 0, we get the functional equation x2 M 2 (x) + (x − 1)M (x) + 1 = 0, and since M (0) = M0 = 1, we obtain (3.7). c) The generating function M (x) is algebraic, hence D-finite. Therefore (Mn ) is P -recursive. Now from Eq(6.38) in [27] for the polynomial A(x) = 1 − 2x − 3x2 and r = 2, a0 = 1, a1 = −2, a2 = −3, d = 2, we get the claim. Theorem 3.3 a) The sequence Mn of Motzkin numbers is log-convex. b) Mn /Mn−1 < 3, for all n ≥ 1. c) The sequence Mn /Mn−1 is convergent and limn→∞ Mn /Mn−1 = 3. 13 Proof a) Divide (3.8) by (n + 2)Mn−1 and let q(n) := Mn /Mn−1 . Then we obtain   1 3(n − 1) q(n) = 2n + 1 + , n+2 q(n − 1) n ≥ 2, (3.9) 2n . This with the initial condition q(1) = 1. Then q(2) = 2. Now define the sequence bn = 3 2n+3 sequence is obviously increasing. We claim that bn ≤ q(n) ≤ bn+1 , for all n ≥ 3. This is obviously true for n = 3. Let n ≥ 3. By inductive hypothesis, we have q(n) = 2n + 1 3(n − 1) 1 2n + 1 3(n − 1) 1 + ≥ + . n+2 n + 2 q(n − 1) n+2 n + 2 bn But n−3 3 2n + 1 3(n − 1) 1 + ≥ 0, − bn = n+2 n + 2 bn 2 n(n + 2)(2n + 3) for all n ≥ 3, and the inequality q(n) ≥ bn follows. On the other hand, by the inductive hypothesis bn−1 ≤ q(n − 1), we have q(n) ≤ 2n + 1 3(n − 1) 1 + . n+2 n + 2 bn−1 Subtracting bn+1 from the right hand side, we get 2n + 1 3(n − 1) 1 1 1 + − bn+1 = − ≤ 0, n+2 n + 2 bn−1 2 (n + 2)(2n + 5) and the inequality q(n) ≤ bn+1 follows. So, the claim is proved by induction. Hence, the sequence Mn /Mn−1 is increasing, and the Motzkin sequence is log-convex. 2n b) We have proved in a) that Mn /Mn−1 ≤ bn = 3 2n+3 < 3. c) By a) and b) it follows that q(n) = Mn /Mn−1 is an increasing and bounded sequence: 2 ≤ q(n) < 3, hence convergent. Passing to limit in (3.9), or passing to limit in the sandwich inequality bn ≤ q(n) ≤ bn+1 above, the last claim in c) follows. 14 Corollary 3.4 a) The sequence (Mn /n!) is log-concave. 1 n b) Mn2 ≤ Mn−1 Mn+1 ≤ 1 + c) Mm Mn ≤ Mm+n ≤ m+n n

Mn2 , for all n ≥ 1. Mm Mn , for all m, n ≥ 0. Proof a) The log-concavity of Mn /n! is equivalent to n+1 n q(n). Mn+1 Mn ≤ n+1 Mn n Mn−1 . So, we have to prove that q(n + 1) ≤ We know that the sequence q(n) is increasing. Starting from the short recursion (3.9) for q(n + 1), we have q(n + 1) = = 2n + 3 1 1 3n 2n + 3 3n + ≤ + n+3 n + 3 q(n) n+3 n + 3 q(n − 1)   1 n+2 n (2n + 3)(n − 1) 2n + 1 3(n − 1) + . n+3n−1 2n(n + 1) n+2 n + 2 q(n − 1) The claim now follows by noting that the term that the inequality n+2 n n+3 n−1 < n+1 n (2n+3)(n−1) 2n(n+1) is clearly less than one for all n ≥ 1, and is valid for all n ≥ 3. The validity of our claim can easily be checked for 1 ≤ n ≤ 3. b) This follows from log-convexity of Motzkin numbers and a). c) A simple combinatorial proof of the left inequality follows from the fact that a concatenation of two Motzkin paths of lengths m and n, respectively, is again a valid Motzkin path of length m + n. To prove the right inequality, start from q(n) ≥ n n+1 q(n + 1). Using this inequality repeatedly, we get Mm+n 1 M2 1 M3 1 M1 ≥ ≥ ≥ ... ≥ , M0 2 M1 3 M2 m + n Mm+n−1 for all n ≥ 0, m ≥ 1. 15 Hence, for any 0 ≤ j ≤ m − 1, we have Mj+1 j + 1 Mm+n ≥ . Mj m + n Mm+n−1 From this we get M1 M2 Mm ... ≥ M0 M1 Mm−1  1 Mn+1 n + 1 Mn  2 Mn+2 n + 2 Mn+1  ...  Mm+n m m + n Mm+n−1  , and after the cancellation: m!n! Mm+n Mm ≥ . M0 (m + n)! Mn Since M0 = 1, we get the claim. The case m = 0, n ≥ 0 is trivial. Remark 3.5 An algebraic proof of log-convexity of Motzkin numbers was given in [1], and a combinatorial proof in [8]. We shall give yet another (“calculus”) proof later in Section 4. Now we proceed in applying the “sandwich method” to combinatorial structures that generalize both Dyck and Motzkin structures in a sense that they are counted by Catalan and Motzkin numbers in some special cases. They are called secondary structures and come from molecular biology (see, e.g. [29], [33], [34], [17], [23], [19]). More details are in [12] and [11]. Let n and l are integers, n ≥ 1, l ≥ 0. A secondary structure of size n and rank l is a labeled graph S on the vertex set V (S) = [n] = {1, 2, . . . , n} whose edge set E(S) consists of two disjoint subsets, P (S) and H(S), satisfying the following conditions: (a) {i, i + 1} ∈ P (S), for all 1 ≤ i ≤ n − 1; (b) {i, j} ∈ H(S) and {i, k} ∈ H(S) =⇒ j = k; (c) {i, j} ∈ H(S) =⇒ |i − j| > l; (d) {i, j} ∈ H(S), {p, q} ∈ H(S) and i < p < j =⇒ i < q < j. 16 The vertices of a secondary structure are (in biology) usually called bases, the edges in P (S) are called p-bonds and those in H(S) h-bonds. A secondary structure S with H(S) = ∅ is called trivial. The number of h-bonds of S is called its order. An example of a secondary structure of size 12, rank 2 and order 3 is shown below in Fig. 2. Note that every h-bond “leaps” over at least two bases. We p−bonds 1 2 3 h−bonds 4 5 6 7 8 9 10 11 12 Figure 2: An example of a secondary structure denote the set of all secondary structures of size n and rank l by S (l) (n), and the set of such structures (l) (l) of order k bySk (n). The respective cardinalities we denote by S (l) (n) and Sk (n). By definition, we put S (l) (0) = 1, for all l. Another interpretation of secondary structures is as follows. Denote by M(l) (n) the set of all Motzkin paths in M(n) whose every plateau is at least l steps long. For l = 0, M(0) (n) = M(n), i.e. all Motzkin paths on n steps. Proposition 3.6 There is a bijection between S (l) (n) and M(l) (n) for all n ≥ 1, l ≥ 0. Proof We shall only briefly describe the correspondence S (l) (n) → M(l) (n). Take a secondary structure S ∈ S (l) (n) and scan it from left to right. To each base, starting from 1, we assign a step in a lattice path as follows. To any unpaired base (i.e. base in which no h-bond starts or ends) we assign a Level step. To a base in which a new h-bond starts assign an U p step, and to each base in which an already 17 encountered h-bond terminates, assign a Down step. The obtained path P is in M(l) (n) and it is not hard to check that S 7−→ P is a bijection. An example of the correspondence S (1) (10) ←→ M(1) (10) is shown in Fig. 3. Therefore, the rank 0 1 10 (0,0) (10,0) Figure 3: A secondary structure and the corresponding Motzkin path secondary structures, i.e the border case, are just all Motzkin paths, hence S (0) (n) = Mn . Note that for l ≥ 1 a secondary structure of rank l is a simple graph. If we allow rank to degenerate to the value of l = −1, then an h-bond can terminate in the very vertex it starts from, i.e. we allow loops. Proposition 3.7 There is a bijection between the set S (−1) (n) of all secondary structures of size n and rank −1 and the set D(n + 1) of all Dyck paths on 2(n + 1) steps. Hence, S (−1) (n) = Cn+1 , for all n ≥ 0. Proof Again, we shall only briefly describe how to assign a member of S (−1) (n) to a Dyck path P ∈ D(n + 1). So, take a Dyck path on 2(n + 1) steps. Discard the first and the last step and divide the remaining path in groups of two consecutive steps. Assign to each group a base in a secondary structure according to the following rule. To a group of two U p steps assign a base in which an h-bond starts, to a group 18 of two Down steps assign a base in which the last started h-bond terminates. To a group (U p, Down) assign a base with a loop attached to it, and, finally, to a group (Down, U p) assign an unpaired base. The obtained graph S is in S (−1) (n) and P 7−→ S is a bijection. An example of the correspondence D(8) ←→ S (−1) (7) is shown in Fig. 4. The third combinatorial 1 6 (0,0) (14,0) Figure 4: The correspondence between secondary structures of rank -1 and Dyck paths interpretation of secondary structures is related to pattern avoidance in permutations. A pattern is a permutation σ ∈ Σk , and a permutation π ∈ Σn avoids σ if there is no subsequence in π whose members are in the same relative order as the members of σ. It is well known ([18]) that the number of permutations from Σn avoiding σ ∈ Σ3 is equal to the Catalan number Cn , for all patterns σ ∈ Σ3 . The concept of pattern avoidance was generalized in [3], by allowing the requirement that two letters adjacent in a pattern must be adjacent in the permutation. An example of a generalized pattern is 1 − 32, where an 1 − 32 sub-word of a permutation π = a1 a2 . . . an is any sub-word ai aj aj+1 such that i < j and ai < aj+1 < aj . Generalized pattern avoidance is treated in more detail in [9], where it is shown that the permutations from Σn that avoid both 1 − 23 and 13 − 2 are enumerated by the Motzkin numbers. Proposition 3.8 For any l ≥ 0, there is a bijection between S (l) (n) and the set of all permutations from Σn that avoid 19 {1 − 23, 13 − 2, ij}, where j ≥ i + l. Proof This bijection is just the bijection given in the proof of Proposition 24 in [9]. It is easy to see that any pattern i, i + k in a permutation σ ∈ Σn avoiding {1 − 23, 13 − 2} generates a plateau of length k − 1 in the corresponding Motzkin path. So, the claim follows from Proposition 3.6. Let us find now the recurrences and generating function for the general secondary structure numbers analogous to those for the Motzkin numbers (the border case) as was shown in Proposition 3.2. Proposition 3.9 a) For any fixed integer l ≥ −1, the numbers S (l) (n) of secondary structures of rank l satisfy the following convolutive recurrence (l) (l) S (n + 1) = S (n) + n−1 X m=l S (l) (m)S (l) (n − m − 1), n ≥ l + 1, (3.10) together with the initial conditions S (l) (0) = S (l) (1) = . . . = S (l) (l + 1) = 1. b) The generating function S (l) (x) = P n≥0 S Sl (x) = (l) (n)xn −ωl (x) − (3.11) is given by q ωl2 (x) − 4x2 2x2 , (3.12) where ωl (x) = x − 1 − x2 − . . . − xl+1 = 2x − (1 + x + x2 + . . . + xl+1 ) = 1 − x2 1 − xl . 1−x (3.13)
c) The sequence S (l) (n) n≥0 satisfies the following “short” recursion (l) (n + 2)S (n) = 2l+2 X k=1 A(l) (n, k)S (l) (n − k), 20 (3.14) where A(l) (n, k) =            − 12 (k − 3)(2n + 4 − 3k) , 1≤k ≤l+1 − 12 (l − 3)(2n − 3l − 2) , k =l+2 − 12 (2l + 3 − k)(2n + 4 − 3k) , l + 3 ≤ k ≤ 2l + 2. Proof a) Clearly, for n ≤ l + 1 there is only one (the trivial) secondary structure of rank l and size n, and hence initial conditions (3.11) hold. Let n ≥ l + 1. A secondary structure on n + 1 bases either does not contain an h-bond starting at the base 1, in which case there are S (l) (n) such structures, or it has an h-bond from the base 1 to some base m + 2, at least l bases apart. In this case, there are S (l) (m)S (l) (n − m − 1) such structures, and (3.10) follows. b) Multiplying (3.10) by xn+1 , summing over n ≥ 0 and taking into account the initial conditions (3.11), we obtain the functional equation x2 [Sl (x)]2 + ωl (x)Sl (x) + 1 = 0, and this, in turn, implies (3.12). c) As in the proof of Proposition 3.2 c), we use D-finiteness of the above generating function and again formula (6.38) in [27] with d = 2, r = 2l + 2 to obtain the claim. More details are in [11] and [12]. Let us write down explicitly the recurrences (3.14) in cases l = 1, 2, 3 (of course, for l = 0, it coincides with (3.8)): S (1) (n) = 2n + 1 (1) n − 1 (1) 2n − 5 (1) n − 4 (1) S (n − 1) + S (n − 2) + S (n − 3) − S (n − 4), n+2 n+2 n+2 n+2 S (1) (0) = S (1) (1) = S (1) (2) = 1, S (1) (3) = 2; 21 (3.15) S (2) (n) = 2n + 1 (2) n − 1 (2) n − 4 (2) 2n − 11 (2) n − 7 (2) S (n−1)+ S (n−2)+ S (n−4)− S (n−5)− S (n−6), n+2 n+2 n+2 n+2 n+2 (3.16) S (2) (0) = S (2) (1) = S (2) (2) = S (2) (3) = 1, S (2) (4) = 2, S (2) (5) = 4; n − 1 (3) n − 4 (3) 2n + 1 (3) S (n − 1) + S (n − 2) − S (n − 4) n+2 n+2 n+2 3n − 21 (3) 2n − 17 (3) n − 10 (3) − S (n − 6) − S (n − 7) − S (n − 8), n+2 n+2 n+2 S (3) (n) = (3.17) S (3) (0) = S (3) (1) = S (3) (2) = S (3) (3) = S (3) (4) = 1, S (3) (5) = 2, S (3) (6) = 4, S (3) (7) = 8. Remark 3.10 The recurrences (3.14) do not have any polynomial solutions or solutions in hypergeometric terms for l ≥ 0. This fact follows by applying the algorithm Hyper, described in [21]. Finally, we now return to our main theme: the log-convexity of secondary structure numbers. Theorem 3.11
The sequence S (1) (n) n≥0 is log-convex. Proof We start from the short recursion (3.15). Dividing (3.15) by S (1) (n−1) and denoting S (1) (n)/S (1) (n−1) by qn , we obtain the following recursion for the numbers qn :   1 2n − 5 n−4 n−1 qn = + − , 2n + 1 + n+2 qn−1 qn−1 qn−2 qn−1 qn−2 qn−3 n ≥ 4, with the initial conditions q1 = q2 = 1, q3 = 2. It is easy to check that q1 ≤ q2 ≤ . . . ≤ q6 . 22 (3.18) Assume, for the moment (and we shall prove it later on) that the sequence (qn ) is convergent with limit q when n → ∞. By passing to limit in (3.18), we obtain the equation q 4 − 2q 3 − q 2 − 2q + 1 = 0, whose maximal positive solution is q = ϕ2 = ϕ + 1, where ϕ = Define now the sequence an = 2n 2 2n+3 ϕ . √ 1+ 5 2 is the golden ratio. It is clearly an increasing sequence and its limit is ϕ2 . We claim that (an ) is interlaced with our sequence (qn ). More precisely, we shall prove by induction that an ≤ qn ≤ an+1 , (3.19) for n ≥ 6. First we check directly the cases n = 6, 7, 8 and 9. Now take n ≥ 9. From the induction hypothesis and (3.18), we have n−1 n−4 n−4 n−1 + + − qn−1 qn−1 qn−2 qn−1 qn−2 qn−1 qn−2 qn−3 n−1 n−1 qn−3 − 1 = 2n + 1 + + + (n − 4) qn−1 qn−1 qn−2 qn−1 qn−2 qn−3 n−1 an−3 − 1 n−1 + + (n − 4) . ≥ 2n + 1 + an an an−1 an an−1 an−2 (n + 2)qn = 2n + 1 + We would like the right hand side to be at least (n + 2)an . But this is equivalent to (2n + 1)an an−1 an−2 + (n − 1)an−1 (an−2 + 1) + (n − 4)(an−3 + 1) − (n + 2)a2n an−1 an−2 ≥ 0. Inserting the formulae for an ’s, we get 12(5 + √ √ √ √ √ 5)n4 − 2(241 + 121 5)n3 + 2(847 + 382 5)n2 − 3(341 + 146 5)n + 126 + 54 5 ≥ 0. (2n − 3)(2n − 1)(2n + 1)(2n + 3)2 The denominator is positive for all integers n ≥ 2. Denote the numerator by L(n) and shift its argument for 6. The polynomial L(n + 6) has only positive coefficients, so it can not have a positive root. It then follows that L(n) can not have a root γ ≥ 6. So the left inequality is valid for all n ≥ 6, and hence qn ≥ an . 23 To prove the other inequality, note that the induction hypothesis implies (n + 2)qn ≤ 2n + 1 + n−1 an−2 − 1 n−1 n−1 + + + (n − 4) . an−1 an−1 an−1 an−2 an−1 an−2 an−3 The condition that the right hand side of this inequality does not exceed (n + 2)an+1 is equivalent to (2n + 1)an−1 an−2 an−3 + (n − 1)an−3 (an−2 + 1) + (n − 4)(an−2 − 1) − (n + 2)an+1 an−1 an−2 an−3 ≤ 0. Substituting the formulae for an ’s, we get √ √ √ √ (82 + 42 5)n3 − (572 + 248 5)n2 + (1103 + 474 5)n − (529 + 247 5) −3 ≤0 (2n − 3)(2n − 1)(2n + 1)(2n + 5) If we put n + 5 instead of n in the numerator, we get a polynomial with all the coefficients positive, and from this we conclude that the numerator does not change the sign for n ≥ 6. So, we have proved the inequality qn ≤ an+1 , and thus completed the induction step. This proves the theorem. Corollary 3.12 S (1) (n) S (1) (n−1) √ 3+ 5 2 . The sequence qn = limn→∞ qn = ϕ2 = is strictly increasing for all n ≥ 5, bounded from above by ϕ2 and Remark 3.13 It is proved in [12] that the asymptotic behavior of S (l) (n), n ≥ 0, is given by S (l) (n) ∼ Kl αnl n−3/2 , (l) where Kl and αl are constants depending only on the rank l. Denote qn = S (l) (n)/S (l) (n − 1). Then it follows qn(l)   1 3/2 ∼ αl 1 − ր αl . n (l) In other words qn asymptotically behaves as an increasing sequence tending to αl as n → ∞. It can be shown that αl ∈ [2, 3]. The exact values of αl are known for l ≤ 6. So, α0 = 3, α1 = (3 + α2 = 1 + √ 2, etc., and αl ց 2 as l → ∞. 24 √ 5)/2, Theorem 3.14 The sequences S (l) (n) are log-convex, for l = 2, 3 and 4. Outline of the proof We present only the case l = 2. Dividing the short recursion for S (2) (n) by S (2) (n − 1) and denoting the quotient qn(2) S (2) (n) S (2) (n−1) (2) (2) by qn , we obtain the recursion for the sequence (qn ) " # n−1 n−4 2n − 11 n−7 1 2n + 1 + (2) + (2) (2) (2) − (2) (2) (2) (2) − (2) (2) (2) (2) (2) = n+2 qn−1 qn−1 qn−2 qn−3 qn−1 qn−2 qn−3 qn−4 qn−1 qn−2 qn−3 qn−4 qn−5 (2) (2) (2) (2) (2) with the initial conditions q1 = q2 = q3 = 1, q4 = q5 = 2. We want to prove that the sequence (2) (qn ) is increasing. √ (2) From Remark 3.13 we conclude that the sequence (qn ) behaves asymptotically as (1 + 2)(1 − n1 )3/2 . Denote this quantity by bn , i.e. bn = (1 + expansion of bn in powers of 1 n+2 . an = α2 where α2 = 1 + √  √ 2)(1 − n1 )3/2 . Now take the first three terms of the series Define 1 3 3 1 + 1− 2 n + 2 8 (n + 2)2  = α2 8n2 + 20n + 11 , 8(n + 2)2 2. The sequence (an ) tends increasingly toward α2 . We shall show now that the (2) sequences (qn ) and (an ) are interlaced, i.e. an−1 ≤ qn(2) ≤ an , for sufficiently large n. (2) Suppose inductively that an−i−1 ≤ qn−1 ≤ an−i for i = 1, 2, 3, 4, 5. Then qn(2)   1 n−4 2n − 11 n−7 n−1 ≤ + − − . 2n + 1 + n+2 an−2 an−2 an−3 an−4 an−1 an−2 an−3 an−4 an−1 an−2 an−3 an−4 an−5 (2) If we prove that the right hand side of this inequality does not exceed an , the right inequality, qn ≤ an , 25 will follow. But this is equivalent to the condition P10 (n) ≥ 0, Q12 (n) where P10 (n) and Q12 (n) are certain polynomials in n of degree 10 and 12, respectively. (Using Mathematica, the polynomials P10 (n) and Q12 (n) can easily be computed explicitly.) Their leading √ √ coefficients are 262144(1+ 2) and 86 (1+ 2)5 , respectively, and we can conclude that this quotient is positive for all n big enough. Again, the biggest real roots of the polynomials P10 (n) and Q12 (n) can be easily found using Mathematica. It turns out that their quotient becomes (and remains) positive for n ≥ 39. (2) On the other hand, from the induction hypothesis and the recursion for qn it follows that qn(2)   1 n−4 2n − 11 n−7 n−1 ≥ + − − . 2n + 1 + n+2 an−1 an−1 an−2 an−3 an−2 an−3 an−4 an−5 an−2 an−3 an−4 an−5 an6 That the right hand side of this inequality is ≥ an−1 is equivalent to P13 (n) Q15 (n) ≥ 0, where P13 and Q15 are certain polynomials in n of degree 13 and 15, respectively, with the positive leading coefficients. Their quotient is positive for n big enough (n ≥ 6). The claim now follows by checking that (q (2) (n)) is increasing for n ≤ 44. A similar proof works for l = 3 and l = 4. We omit the details. 4 Calculus method Let us start again as in (3.1) with a linear homogeneous recursion for positive numbers a(n), and consider again the corresponding recurrence (3.2) for the quotients q(n) = a(n)/a(n − 1). Suppose 26 again we want to prove that a(n) is log-convex, i.e. that (q(n)) is an increasing sequence (at least from some place n0 on). This time we do the following. Define a continuous function f : [1, ∞) → R (or f : [n0 , ∞) → R) starting with the appropriate linear function on the segment [1, 2] (or [n0 , n0 + 1], for some n0 ∈ N) determined by the initial and the next value of q(n), and then continue to the next segment by the same rule as (3.2). In other words, by replacing q → f , n → x, (3.2) becomes f (x) = R(x) + S(x) , f (x − 1) (4.1) defined so for x ≥ 2 (or x ≥ n0 + 1). In a sense, f is a dynamical system patching the discrete values f (n) = q(n) if R and S are “good” enough functions. For example, if R and S are rational functions (and in combinatorics this is mostly the case) without poles on the positive axis, then f is a piecewise rational function, i.e. rational on every open interval (n, n + 1) for any integer n ≥ 1 (or n ≥ n0 ). If we can prove that f is smooth on such open intervals (usually by proving that f is bounded from above and below by some well-behaved functions), we can consider the derivative f ′ (x), for any x ∈ (n, n+1). The idea is to show inductively on n that f ′ (x) ≥ 0 (or f ′ (x) ≤ 0 if we want to prove log-concavity of a(n)). This will imply that f is an increasing function (or a decreasing function in the log-concave case) on any open interval (n, n + 1), and then, by continuity of f it will follow that f is increasing (or decreasing) on its whole domain, and hence q(n) = f (n), n ∈ N, will increase (or decrease), too. Now, in general, if we want to prove that a sequence q(n) defined by (3.2) is increasing, we form the corresponding functional equation (4.1) with the appropriate start, i.e. we define f (x) on some starting segment [n0 , n0 + 1] to be an increasing function and then prove inductively on n that f ′ (x) ≥ 0 for x ∈ (n, n + 1). It is always necessary to have some a priori bounds for f . So, suppose we know 0 < m(x) ≤ f (x) ≤ M (x), for all x ≥ n0 . Let us find some sufficient conditions which ensure that f ′ (x) ≥ 0. Of course, we assume that R and S are smooth on all open intervals (n, n + 1), n ≥ n0 . 27 Fix an x ∈ (n, n + 1) and write f = f (x), f1 = f (x − 1), R = R(x), S = S(x). Then (4.1) can be written as f =R+ S , f1 or equivalently f f1 = Rf1 + S. (4.2) Taking the derivative d/dx of both sides of (4.2), we get f ′ f1 + f f1′ = R′ f1 + Rf1′ + S ′ , implying f ′ f1 = R′ f1 + S ′ + (R − f )f1′ . From (4.2), we obtain S ′ f . f1 1 (4.3) 0 < m(x) ≤ f (x) ≤ M (x), (4.4) f ′ f1 = R ′ f1 + S ′ − Assume that for all x ≥ n0 , and suppose inductively that f is increasing on [n0 , n]. This means that f1′ ≥ 0. Then, with our notation convention m1 = m(x − 1) etc., we have: Theorem 4.1 If R′ ≥ 0, R′ m1 + S ′ ≥ 0 and S ≤ 0, then f ′ ≥ 0. Proof Obvious from (4.3), (4.4) and the above discussion. Theorem 4.2 Suppose R′ ≥ 0, S ′ ≥ 0, S ≥ 0 and m1 m2 (R′ m1 + S ′ ) ≥ S(R1′ M2 + S1′ ). Then f ′ ≥ 0. 28 Proof Divide (4.3) by f1 and substitute in by the same rule f1′ . WE obtain   R ′ f1 + S ′ R ′ f1 + S ′ S ′ S R1′ f2 + S1′ S1 ′ f = − 2 f1 = − 2 − 2 f2 , f1 f1 f2 f1 f1 f1 ′ and this implies f′ = S R′ f2 + S1′ SS1 ′ R ′ f1 + S ′ − 2 1 + f . f1 f2 (f1 f2 )2 2 f1 (4.5) By the inductive hypothesis, f2′ = f ′ (x − 2) ≥ 0, and since SS1 ≥ 0, it follows that the last term in (4.5) is non-negative. On the other hand, R ′ f1 + S ′ S R′ f2 + S1′ − 2 1 ≥ 0 ⇐⇒ f1 f2 (R′ f1 + S ′ ) ≥ S(R1′ f2 + S1′ ). f1 f2 f1 (4.6) Since R′ , S ′ , S ≥ 0 and m1 m2 (R′ m1 + S ′ ) ≥ S(R1′ M2 + S1′ ), by our assumptions, it follows that (4.6) holds, and we are done. Now let us show how to apply the above method to some combinatorially relevant numbers. Recall that the n-th big Schröder number rn is the number of lattice paths from (0, 0) to (n, n) with steps (1, 0), (0, 1) and (1, 1) that never rise above the line y = x. Equivalently, rn is the number of lattice paths from (0, 0) to (2n, 0) with steps (1, 1), (1, −1) and (2, 0) that never fall below the x-axis. For n = 0 we put r0 := 1. So, rn is the number of Motzkin paths whose every plateau, every valley, every terrace and every plain is of even length. Hence, via Proposition 3.6, big Schröder numbers count secondary structures whose unpaired bases form contiguous blocks of even length. The n-th little Schröder number is sn = 21 rn , s0 = 1. Theorem 4.3 √ The Schröder numbers are log-convex. The sequences rn /rn−1 and sn /sn−1 increasingly tend to 3+2 2 as n → ∞. 29 Proof Similarly to the proof of Proposition 3.2, it can be shown (and it is well known) that (rn ) satisfies the following (convolutional) recurrence: rn+1 = rn + n X rj rn−j . j=0 This implies that the generating function for (rn ) is given by r(x) = 1−x− √ 1 − 6x + x2 . 2x This, in turn, implies (as in Proposition 3.2) that rn satisfy the following short recursion rn = 1 [3(2n − 1)rn−1 − (n − 2)rn−2 ], n+1 (4.7) together with the initial conditions r0 = 1, r1 = 2. (There are also combinatorial proofs of (4.7), see [15] or [24].) Divide (4.7) by rn−1 and denote rn /rn−1 by q(n). We get q(n) = 3(2n − 1) n − 2 1 − , n+1 n + 1 q(n − 1) n ≥ 2, (4.8) with q(1) = 2. Then q(2) = 3 and q(3) = 11/3. Define the function f : [2, ∞) → R by  1  3 (2x + 5) , x ∈ [2, 3] f (x) =  1 x−2 x ≥ 3. x+1 [6x − 3 − f (x−1) ] , It is easy to show by induction on n that f is bounded on [2, n], for n ≥ 2. More precisely, 3 ≤ f (x) ≤ 6, for all x ≥ 2. Also, f is continuous everywhere and differentiable on open intervals (n, n + 1), n ≥ 2. In the notations of Theorem 4.1, we have R(x) = 3(2x − 1) , x+1 S(x) = − x−2 , x+1 m(x) = 3, M (x) = 6. Then R′ (x) = 9 ≥ 0, (x + 1)2 30 S ′ (x) = − 3 . (x + 1)2 Clearly, S(x) ≤ 0 for x ≥ 2, and let us check that R′ m1 + S ′ ≥ 0. But this is obvious, since 3 24 9 ·3− = ≥ 0. 2 2 (x + 1) (x + 1) (x + 1)2 Since f ′ (x) = 2/3 ≥ 0 for x ∈ (2, 3), it follows from Theorem 4.1 that f ′ (x) ≥ 0 for all x ∈ (n, n + 1), n ≥ 2. Hence, by continuity, f is increasing, so (q(n)) is increasing and therefore (rn ) is log-convex. By passing to limit in (4.9) we get the last claim. Now that we are more familiar with this method, let us show once again (but this time almost automatically) the log-convexity of Motzkin numbers. Recalling the short recursion (3.9), we define the function f : [2, ∞) → R by  2 , x ∈ [2, 3]  f (x) =  1 3(x−1) x ≥ 3. x+2 [2x + 1 + f (x−1) ] , Here we have R(x) = 2x + 1 , x+2 S(x) = 3(x − 1) , x+2 so R′ (x) = 3 ≥ 0, (x + 2)2 S ′ (x) = 9 ≥ 0. (x + 2)2 Further,it is easy to check that 2 ≤ f (x) ≤ 7/2 for all x ≥ 2. So, we may take m(x) = 2, M (x) = 7/2. Let us check the inequality m1 m2 (R′ m1 +S ′ ) ≥ S(R1′ M2 +S1′ ) from Theorem 4.2. But this is equivalent to     3 3 9 9 3x − 3 7 4 2· + + ≥ , (x + 2)2 (x + 2)2 x + 2 2 (x + 2)2 (x + 2)2 and this is equivalent to 1.5x2 + 61.5x + 117 ≥ 0. This last inequality is certainly true for all x ≥ 1. Hence, by Theorem 4.2, f ′ ≥ 0, and this implies the log-convexity of Motzkin numbers. 31 A directed animal of size n is a subset S ⊆ (N ∪ {0})2 of cardinality n with the following property: if p ∈ S, then there is a lattice path from (0, 0) to p with steps (1, 0), (0, 1), all of whose vertices lie in S. Let an be the number of directed animals of size n. It is well known that the generating function for (an )n≥1 is given by A(x) = X n≥1 1 an x = 2 n r ! 1+x − 1 = x + 2x2 + 5x3 + 13x4 + 35x5 + . . . 1 − 3x These same numbers appear also as the row sums of Motzkin triangle (see [10] and [4]). Theorem 4.4 The sequence (an )n≥1 of numbers of directed animals is log-convex. Hence the sequence an /an−1 increasingly tends to 3. Proof Taking the derivative A′ (x) of the generating function A(x) we get (1 + x)(1 − 3x)A′ (x) = 2A(x) + 1, and equating coefficients of xn−1 above yields the recurrence nan = 2nan−1 + 3(n − 2)an−2 , n ≥ 3, with a1 = 1, a2 = 2. The corresponding function for the successive quotients is given by f (x) = 2 + and f (x) = x+2 2 3(x − 2) 1 , x f (x − 1) x ≥ 3, on [2, 3]. It is easy to check that 2 ≤ f (x) ≤ 72 . Here R(x) = 2, S(x) = 3(x−2) x ≥ 0 and R′ (x) = 0, S ′ (x) = follows from Theorem 4.2. 32 6 x2 ≥ 0, and m = 2, M = 7/2, so the claim We aplly now our approach to Franel’s sequences. Recall that the n-th Franel number of order r is defined by Sn(r) = n   X n r k=0 (r) k . The numbers Sn satisfy a linear homogenous recurrence of order ⌊ r+1 2 ⌋ with polynomial coefficients. For r = 3, the recurrence reads as follows ([27]): (3) (3) n2 Sn(3) = (7n2 − 7n + 2)Sn−1 + 8(n − 1)2 Sn−2 , (3) (4.9) (3) with the initial conditions S0 = 1, S1 = 2. Theorem 4.5   (3) The sequence Sn n≥0   (4) is log-convex. The same is true for the sequence Sn n≥0 . Proof (3) (3) Again, taking the quotients q(n) = Sn /Sn−1 in (4.9), we get q(n) = 7n2 − 7n + 2 8(n − 1)2 1 + , n2 n2 q(n − 1) n ≥ 2, (4.10) with q(1) = 2. Then q(2) = 5. Define the function f : [1, ∞) → R by  3x − 1 , x ∈ [1, 2]  f (x) =  7x2 −7x+2 8(x−1)2 1 + x2 f (x−1) ] , x ≥ 2. x2 The function f is bounded. More precisely, 5 ≤ f (x) ≤ 9 for all x ≥ 2. Indeed, f is clearly bounded on [1, 2]. On [2, 3] f is given by f (x) = 21x2 − 41x + 18 , 3x2 − 4x and one can check easily that 5 ≤ f (x) ≤ 9 on this interval. Now, by induction on n ≥ 3, it is not hard to verify that if 5 ≤ f (x) ≤ 9 for x ∈ [2, n], then also 5 ≤ f (x) ≤ 9 for x ∈ [n, n + 1]. Also, f is a continuous function. So, here we have, in the notations of Theorem 4.1: R(x) = 7x2 − 7x + 2 , x2 33 S(x) = 8(x − 1)2 x2 and we may take m(x) = 5, M (x) = 9 for x ≥ 2. Since R′ (x) = 7x−2 x3 ≥ 0, S ′ (x) = 16(x−1) x3 ≥ 0 and S(x) ≥ 0 for x ≥ 2, it remains only to check the last condition in Theorem 4.2, i.e.    16(x − 2) 8(x − 1)2 7x − 9 7x − 2 16(x − 1) + ·9+ ≥ . 25 5 · x3 x3 x2 (x − 1)3 (x − 1)3  This is equivalent to 643x2 − 1021x + 650 ≥ 0, and this last inequality is certainly true for x ≥ 1. So, as before, we conclude that f increases, hence (q(n))n≥0 is an increasing sequence and therefore   (3) is log-convex. Sn n≥0 Note that (q(n)) converges, being increasing and bounded, and so, passing to limit in (4.10) we obtain that (3) Sn lim (3) n→∞   (4) A similar proof applies to the case of Sn Sn−1 n≥0 = 8. , starting with the recurrence (4) (4) n3 Sn(4) = 2[6n3 − 9n2 + 5n − 10]Sn−1 + (4n − 3)(4n − 4)(4n − 5)Sn−2 , (4) (4) S0 = 1, S1 = 2. It turns out that (4) lim n→∞ Sn (4) Sn−1 = 16. We leave the details to the reader. As a bit more simple case, let us show the log-behavior of sequences defined by two-term recurrences with constant coefficients. 34 Proposition 4.6 Let (an )n≥0 be a positive sequence defined by an = C1 an−1 − C2 an−2 , where C1 , C2 > 0 are constants, and with some initial conditions a0 > 0, a1 > 0. Then the logbehavior of (an ) is completely determined by the log-behavior of its first three terms. In other words, the sequence (an ) is log-convex if a21 ≤ a0 a2 , and log-concave if a21 ≥ a0 a2 . Proof Let q(n) = an /an−1 . The recursion for q(n) is given by q(n) = C1 − C2 , q(n − 1) and the first two values are q(1) = a1 /a0 , q(2) = a2 /a1 . Set ∆ = a0 a2 − a21 and define the function f : [1, ∞) → R by f (x) =  ∆  a0 a1 (x − 1) +  C1 − C2 f (x−1) a1 a0 , x ∈ [1, 2] , x ≥ 2. f is a linear fractional mapping on any interval (n, n + 1), n ∈ N, and f is continuous. Suppose first that ∆ ≥ 0. Then f is obviously increasing on [1, 2] and bounded from below by f (1) = q(1) = a1 /a0 . Namely, suppose that f (x) ≥ f (1) on [1, n]. For x ∈ [n, n + 1] we have f (x) = C1 − C2 C2 ≥ C1 − = f (2) ≥ f (1), f (x − 1) f (1) and so f (x) is bounded from below on the whole interval [1, n + 1]. Hence f is bounded from below by f (1). This also shows that f is differentiable on any open interval (n, n + 1), n ∈ N. Here we have R(x) = C1 , S(x) = −C2 (so R′ (x) = S ′ (x) = 0 for x 6∈ N) and we may take m(x) = a1 /a0 . It follows by Theorem 4.1 that f ′ (x) ≥ 0 for all x 6∈ N. By continuity, f is an increasing function, and hence f (n) = q(n) is an increasing sequence. 35 If ∆ ≤ 0, then f decreases on [1, 2] and similarly as above f is, in this case, bounded from above by f (1). Since f ′ (x) = C2 f ′ (x − 1) , f (x − 1) we see that the sign of the derivative is propagated to the right and is determined by its sign on (1, 2). Hence, in this case, f ′ (x) < 0. Corollary 4.7 The sequence (F2n+1 )n≥0 of odd-indexed Fibonacci numbers is log-convex, while the sequence of evenindexed Fibonacci numbers (F2n )n≥1 is log-concave. Proof Let an = F2n+1 . The numbers an satisfy the recursion an = 3an−1 −an−2 , and a21 = 4 < 5 = a0 a2 . The numbers bn = F2n satisfy the same recursion bn = 3bn−1 − bn−2 , but this time with initial conditions b1 = 1, b2 = 3. However, b22 = 9 > 8 = b1 b3 . So, the claim follows from Proposition 4.6. (Of course, this Corollary is an immediate consequence of the Cassini identity Fn2 − Fn−1 Fn+1 = (−1)n .) As our next application of the method, we consider some classical orthogonal polynomials. Let ν > (ν) −1/2 be a parameter. The Gegenbauer (or ultraspherical) polynomials Cn (t) are defined by ([31]) (ν) (ν) nCn(ν) (t) = 2t(ν + n − 1)Cn−1 (t) − (2ν + n − 2)Cn−2 (t), (ν) n ≥ 2, (4.11) (ν) together with the initial conditions C0 (t) = 1, C1 (t) = 2νt. (1/2) For ν = 1/2, Cn (1) (t) = Pn (t) is the Legendre polynomial and for ν = 1, Cn (t) = Un (t) is the Chebyshev polynomial of the second kind. 36 Theorem 4.8   (ν) Cn (t) is log-concave for t ≥ 1, ν ≥ 1, and log-convex for 0 < ν < 1, t ≥ n≥0  1 . The sequence  max √1 , √ 2ν 2(1−ν) Proof Let qt (n) = (ν) Cn (t) (ν) Cn−1 (t) (ν) . Dividing the recursion (4.11) by nCn−1 (t) we obtain the recursion qt (n) = 2t 1 ν + n − 1 2ν + n − 2 − ,n ≥ 1 n n qt (n − 1) (4.12) 1 with initial condition qt (1) = 2νt. Then qt (2) = t(ν + 1) − 2t . Now define the function ft : [1, ∞) → R by formula ft (x) =  2 2t (1−ν)−1  x+  2t   2t(ν+x−1) − x 2t2 (3ν−1)+1 2t 2ν+x−2 1 x ft (x−1) , if x ∈ [1, 2], (4.13) , if x ≥ 2. The function ft (x) is piecewise rational, i.e. it is rational on all segments [n, n + 1], n ∈ N. It is easy to show, by induction on n, that ft (x) is continuous at the points x = n, for all n ∈ N, and that ft (n) = qt (n). Let us first consider the case ν ≥ 1 and t ≥ 1. We claim that ft (x) is smooth on all intervals (n, n + 1), n ∈ N. To prove this, it is enough to show that ft (x) is bounded from below by some positive quantity. Let us check that ft (x) ≥ 1 t for all x ≥ 1. Since ν ≥ 1, t ≥ 1, the function ft (x) is non-increasing on the interval [1, 2], so it is enough to show that ft (2) ≥ 1t . But this is equivalent to t(ν + 1) > 3 2t , or 2t2 (ν + 1) ≥ 3. Since ν ≥ 1, we have 4t2 ≥ 3, and this is true for all t ≥ 1. Suppose now inductively that ft (x) ≥ 1 t for all x ∈ [1, n], and take an x ∈ [n, n + 1], n ≥ 2. From (4.13) we have ft (x) ≥ 2t(ν + x − 1) 2ν + x − 2 1 − 1 = t, x x t 37 and t ≥ 1 t for all t ≥ 1. So, the function ft (x) is nowhere zero on [1, ∞), hence it has no poles on [1, ∞) and hence is smooth on all intervals (n, n + 1), n ∈ N. The function ft (x) is obviously decreasing on [1, 2]. Suppose now that it is decreasing on [1, n]. Let x ∈ (n, n + 1). Then ft′ (x)   ν − 1 ft′ (x − 1) 2(1 − ν) [tft (x − 1) − 1] + 1 + 2 . = 2 x ft (x − 1) x ft2 (x − 1) (4.14) The term in square brackets is positive, since ft (x) ≥ 1t , so the whole first term in (4.14) is negative. The second term is negative by the induction hypothesis, hence ft′ (x) ≤ 0. This completes the step of induction, and we can conclude that the function ft (x) is decreasing on [1, ∞). Hence, the sequence   (ν) Cn (t) is log-concave for all ν ≥ 1, t ≥ 1. n≥0 Let us now consider the case 0 < ν < 1, t ≥ max  √1 , 2ν √ 1 2(1−ν)  . Obviously, for such values of t, the function ft (x) is increasing on [1, 2], and ft (1) ≥ 1t , hence ft (x) ≥ induction on n that ft (x) ≥ 1 t 1 t on [1, 2]. It follows easily by on [1, ∞), hence ft (x) is bounded from below by a positive quantity, hence ft (x) is smooth on all open intervals (n, n + 1), n ∈ N. Suppose now that ft (x) is increasing on [1, n] and let x ∈ (n, n + 1), n ≥ 2. Take a look at (4.14) again. The first term is positive since ft (x) ≥ 1 t, and the second term is positive by the induction hypothesis for all x ≥ 2. Hence ft′ (x) ≥ 0 and ft (x) is increasing on [1, n + 1]. The increasing behavior   (ν) of ft (x) implies the log-convexity of the sequence Cn (t) . n≥0 In Fig. 5 we can see the areas of log-convexity (denoted by Λ) and log-concavity (Ξ) of sequences   (ν) Cn (t) . Our analysis is not sufficient to determine the logarithmic behavior of sequences n≥0   (ν) Cn (t) whose parameters (ν, t) fall into the areas denoted by I and II. n≥0 38 t Λ 1 I Ξ II ν −1/2 0 1/2 1 Figure 5: Areas of log-convexity (Λ) and log-concavity (Ξ) for sequences of values of Gegenbauer polynomials Remark 4.9 So far we have restricted our definitions of log-convexity and log-concavity to positive (or perhaps non-negative) sequences, since most combinatorial sequences have only non-negative terms. The definitions, however, make sense even if we include negative numbers. Corollary 4.10   (ν) The sequence Cn (t) is log-concave for |t| ≥ 1, ν ≥ 1, and log-convex for 0 < ν < 1, |t| ≥   n≥0 max √12ν , √ 1 . 2(1−ν) 39 Proof It follows from Theorem 4.7 and the relation Cn(ν) (−t) = (−1)n Cn(ν) (t), valid for all ν 6= 0 and all t. Corollary 4.11 The sequence (Un (t))n≥0 of the values of Chebyshev polynomials of the second kind is log-concave for |t| ≥ 1. In particular, F2n = Un (3/2) is log-concave. Proof (1) Follows from Corollary 4.10, since Un (t) = Cn (t). Corollary 4.12 The sequence (Pn (t))n≥0 of the values of Legendre polynomials is log-convex for |t| ≥ 1 and log-concave for |t| ≤ 1. Proof (1/2) Pn (t) = Cn (t) and log-convexity follows from Corollary 4.10. The log-concavity follows from the well known fact that |Pn (t)| ≤ 1 for |t| ≤ 1 and from the inequality ([2], [30]) Pn2 (t) − Pn−1 (t)Pn+1 (t) ≥ 1 − Pn2 (t) , (2n − 1)(n + 1) valid for |t| ≤ 1. Our results on Legendre polynomials complement and generalize in some respects those of Turán and Szegö ([32]), who considered determinants of the form Pn (x) Pn−1 (x) Pn+1 (x) Pn (x) . Now, a combinatorial consequence is in order. The n-th central Delannoy number D(n) counts 40 the lattice paths in (x, y) coordinate plane from (0, 0) to (n, n) with steps (1, 0), (0, 1) and (1, 1). (Such paths are also known as “king’s paths”.) The generating function of (D(n))n≥0 is given by D(x) = √ 1 1−6x+x2 ([27]). This implies easily that D(n)’s satisfy the recurrence (4.11) for ν = 1/2 and t = 3 and D(0) = 1, D(1) = 3. Hence D(n) = Pn (3), the value of the n-th Legendre polynomial Pn at t = 3. Corollary 4.13. The sequence D(n) of Delannoy numbers is log-convex and limn→∞ D(n) D(n−1) √ = 3 + 2 2. Proof The log-convexity follows from Corollary 4.12. Hence q3 (n) = D(n)/D(n − 1) is increasing. From (4.12) for ν = 1/2 and t = 3 it follows easily by induction that q3 (n) ≤ 6. Passing to limit, we obtain the second claim. Remark 4.14 No wonder that the above limit coincides with the limit for Schröder numbers from Theorem 4.3. This can be explained by showing that rn = Pn′ (3) . Pn′ (1) (For a proof, see e.g. [11].) The derivatives of ultraspherical polynomials behave log-concavely. More precisely, the following is valid. Theorem 4.15   (ν)′ The sequence Cn (t) n≥0 is log-concave, for all ν > 0, t ≥ 1. 41 Proof Deriving the recurrence (4.11) with respect to t and using the relation (ν) (ν)′ ′ 2(ν + n − 1)Cn−1 (t) = Cn(ν) (t) − Cn−2 (t), we get the two-term recursion for derivatives: ′ Cn(ν) (t) = 2t (ν)′ which starts with C0 2ν + n − 1 (ν)′ ν + n − 1 (ν)′ Cn−1 (t) − Cn−2 (t), n−1 n−1 (ν)′ (ν)′ (t) = 0, C1 (t) = 2ν, C2 (ν)′ (t) = 4ν(ν + 1)t and C3 (ν)′ (t) = 4ν(ν + 1)(ν + 2)t2 − (ν)′ 2ν(ν + 1). Hence, the successive quotients qt (n) = Cn (t)/Cn−1 (t) satisfy the recurrence qt (n) = 2t 1 ν + n − 1 2ν + n − 1 − , n ≥ 3, n−1 n − 1 qt (n − 1) starting with qt (2) = 2(ν + 1)t, qt (3) = (ν + 2)t − defined by ft (x) = 1 2t . The appropriate function ft : [2, ∞) → R is  1 2  − 2t (2νt + 1)x + (4ν + 1)t +  2t(ν+x−1) − x−1 2ν+x−1 1 x−1 ft (x−1) 1 t , if x ∈ [2, 3], , if x ≥ 3. The function ft (x) is obviously decreasing on [2, 3] for all ν > 0, t ≥ 1. Also, ft (x) is continuous in all points x = n, for n ∈ N, and ft (n) = qt (n). As before, it is not hard to check that ft (x) ≥ 1t , for all x ≥ 2. Hence, the derivative ft′ (x) exists for any x ∈ (n, n + 1), n ≥ 2. ft′ (x)   ′ ft (x − 1) 2ν 2ν . [1 − tft (x − 1)] + 1 + = 2 (x − 1) ft (x − 1) x − 1 ft2 (x − 1) This derivative is obviously non-positive on (2, 3). Suppose inductively that ft′ (x − 1) ≤ 0. The first term above is negative since ft (x) ≥ 1t , and the second term is negative by the induction hypothesis. Hence, ft (x) is decreasing on (n, n + 1), and then, by continuity, on [1, n + 1]. This completes the inductive step and hence qt (n) = ft (n) is also decreasing. We conclude our review of orthogonal polynomials with Laguerre polynomials. The (ordinary) Laguerre polynomials satisfy the recursion nLn (t) = (2n − 1 − t)Ln−1 (t) − (n − 1)Ln−2 (t) 42 with the initial conditions L0 (t) = 1, L1 (t) = 1 − t. Theorem 4.16 The sequence (Ln (t))n≥0 is log-concave for all t ≤ 0. The ratio Ln (t)/Ln−1 (t) is greater than one for all t < 0 and it tends to 1 decreasingly as n → ∞. Proof Dividing the above recursion by nLn−1 (t) and denoting qt (n) = Ln (t) Ln−1 (t) by qt (n), we get a recursion for qt (n) 2n − 1 − t n − 1 1 − n n qt (n − 1) with qt (1) = 1 − t. By computing qt (2) = t2 −4t+2 2(1−t) , (4.15) we see that qt (1) ≥ qt (2), for all t ≤ 0. Define ft : [1, ∞) → R by formula ft (x) =  t2   − 2(t−1) x +   2x−1−t − x 3t2 −4t+2 2(t−1) x−1 1 x ft (x−1) , if x ∈ [1, 2], , if x ≥ 2. This function is, again, piecewise rational, continuous for x = n ∈ N and ft (n) = qt (n) for all n ∈ N. By induction on n ≥ 2, one can easily check that, for x ∈ [1, n], we have 1 ≤ ft (x) ≤ 1 − t for all t ≤ 0. Hence, 1 ≤ ft (x) ≤ 1 − t, x ≥ 1, t ≤ 0. (4.16) By computing the derivative, we see that ft′ (x) ≤ 0 for x ∈ (1, 2), and for x ∈ (n, n + 1), n ≥ 2 we find ft′ (x)   1+t 1 1 ft′ (x − 1) 1 = 2 − 2 . + 1− x x ft (x − 1) x ft2 (x − 1) Suppose ft′ (x − 1) ≤ 0. Then for t ≤ −1, all three terms on the right hand side are negative, hence ft′ (x) ≤ 0. For −1 < t ≤ 0, the second and third terms are still negative, but the first term is positive. The claim will follow if we prove 1+t 1 1 − 2 ≤ 0. 2 x x ft (x − 1) 43 But this is equivalent to (1 + t)ft (x − 1) − 1 ≤ 0, and this reduces to ft (x − 1) ≤ and for −1 < t ≤ 0 we have 1 − t ≤ 1 1+t . 1 1+t . But ft (x) ≤ 1 − t, Hence, the sum of the first two terms is negative, and ft′ (x) ≤ 0. By continuity, ft is decreasing, and so qt (n) is decreasing. Since by (4.16) this sequence is also bounded, it is convergent and passing to limit in (4.15) we get that this limit is 1 as claimed. Some of our results on orthogonal polynomials, in particular those concerning Legendre and Laguerre polynomials, have been already known (see, e.g. [14]), but here we derived them in a simple and unified manner, almost automatically. So far we have been considering only the two-term recurrences. The “calculus” method, however, works as well for higher order recurrences, as the following example shows. Let L(n) be the number of graphs on the vertex set [n], whose every component is a cycle, and define L(0) := 1. Then L(1) = 1, L(2) = 2, L(3) = 5, L(4) = 17 etc. The following recurrence holds ([27], Ex. 5.22): L(n) = nL(n − 1) −  n−1 2  L(n − 3), n ≥ 3. (4.17) Theorem 4.17 The sequence (L(n))n≥0 is log-convex. Proof Let q(n) = L(n)/L(n − 1). From (4.17), dividing with L(n − 1), we obtain q(n) = n −  n−1 2  1 , q(n − 1)q(n − 2) n ≥ 3, (4.18) with the initial conditions q(1) = 1, q(2) = 2. The value q(3) is equal to 5/2. We claim that (q(n))n≥1 44 is an increasing sequence. To this end, define the function f : [1, ∞) → R by  x , if x ∈ [1, 2],      x+2 , if x ∈ [2, 3], f (x) = 2      x − (x−1)(x−2) 1 , if x ≥ 3. 2 f (x−1)f (x−2) (4.19) The function f is obviously continuous, rational on any [n, n + 1], n ∈ N, without poles on [n, n + 1] and clearly increasing on [1, 3]. Let us first prove an a priori bound f (x) ≥ √ x + 1, x ≥ 2. (4.20) This is clearly true for x ∈ [2, 3]. Suppose that it is true for x ∈ [2, n], and let x ∈ [n, n + 1]. Then from (4.20) we have √ 1 (x − 1)(x − 2) (x − 2) x − 1 √ , f (x) ≥ x − =x− √ √ 2 2 x x x−1 and it is easy to check that √ (x − 2) x − 1 √ √ ≥ x + 1, x− 2 x x ≥ 2. The function f is differentiable for all x ≥ 1, x 6∈ N. Let x ∈ (n, n + 1) for some n ≥ 2 and denote f1 = f (x − 1), f2 = f (x − 2). Suppose inductively that f1′ ≥ 0, f2′ ≥ 0. Then f ′ (x) = 1 − (x − 1)(x − 2) f1′ f2 + f1 f2′ 2x − 3 1 + . 2 f1 f2 2 (f1 f2 )2 The last term is positive, and from (4.21) it follows that 1− p √ √ 2x − 3 1 3 ≥ 0 ⇐⇒ f1 f2 ≥ x x − 1 = x2 − x ≥ x − , 2 f1 f2 2 for x ≥ 2. Hence, f ′ (x) ≥ 0 and theorem is proved. Consider now a general three-term recurrence of the form a(n) = R(n)a(n − 1) + S(n)a(n − 2) + T (n)a(n − 3), 45 n ≥ 4. (4.21) Dividing this recurrence by a(n − 1) and denoting q(n) = a(n)/a(n − 1), we obtain the recurrence for q(n): q(n) = R(n) + T (n) S(n) + , q(n − 1) q(n − 1)q(n − 2) n ≥ 3, (4.22) with some given initial conditions, say, q(1) = b1 , q(2) = b2 , b2 > b1 > 0. Suppose we want to prove that (q(n))n≥1 is an increasing sequence. Again, form the function f : [1, ∞) → R mimicking the rule (4.22) and starting with the linear functions on [1, 2] and [2, 3], connecting the points (1, b1 ) and (2, b2 ) and the points (2, b2 ) and (3, b3 ), respectively, where b3 = q(3) = R(3) + S(3) b2 + T (3) b1 b2 (supposing, of course, b3 ≥ b2 ). For x ≥ 3, the function f is defined by replacing q by f and n by x in (4.22). f (x) = R(x) + S(x) T (x) + , f (x − 1) f (x − 1)f (x − 2) (4.23) for x ≥ 3 (or x ≥ n0 + 1, for some n0 ∈ N). For a fixed x we write simply f (x) = f , R(x) = R, S(x) = S, T (x) = T , f (x − 1) = f1 , f (x − 2) = f2 , etc. So, we write (4.23) simply as f =R+ S T + ⇐⇒ f1 f2 f = Rf1 f2 + Sf1 + T. f1 f1 f2 (4.24) We assume that R, S and T are “good enough” functions, in the sense that f is differentiable on open intervals (n, n + 1) for integers n ≥ 2 (or n ≥ n0 ). This is usually a consequence of an a priori bound of the type 0 < m(x) ≤ f (x) ≤ M (x), (4.25) for some well-behaved functions m(x) and M (x). Taking the derivative d/dx of both sides of (4.24) gives us   T ′ Sf2 + T ′ 1 f1 − f2 , F− f = f1 f2 f1 f2 ′ 46 (4.26) where F = R ′ f1 f2 + S ′ f2 + T ′ . (4.27) Suppose that R, S, T ≥ 0 and R′ , S ′ , T ′ ≥ 0. By substituting the analogous expressions for f1′ and f2′ in (4.26) and supposing inductively that f2′ , f3′ and f4′ ≥ 0 we obtain (denoting F1 = F (x − 1), R1 = R(x − 1), etc.) that: f1 f2 f ′ = F − Sf2 + T T F1 − F2 + Bf2′ + Cf3′ + Df4′ , f1 f2 f3 f2 f3 f4 where B, C and D are non-negative. To prove that f ′ ≥ 0, assuming inductively that f2′ , f3′ and f4′ ≥ 0, it is enough to prove the following: F− Sf2 + T T F1 − F2 ≥ 0, f1 f2 f3 f2 f3 f4 or equivalently f1 f2 f3 f4 F ≥ (Sf2 + T )f4 F1 + T f1 F2 . (4.28) Taking into account (4.25), it suffices to prove a stronger inequality m4 F ≥ (M S + T )M F1 + M T F2 (4.29) for all x ≥ n0 , for some n0 ∈ N, and to check that f ′ (x) ≥ 0 for x ∈ (k, k + 1), k < n0 . Let us now consider a concrete example. A Baxter permutation is defined in [27]. The numbers B(n) of Baxter permutations in Σn satisfy the recurrence (n + 1)(n + 2)(n + 3)(3n − 2)B(n) = 2(n + 1)(9n3 + 3n2 − 4n + 4)B(n − 1) +(3n − 1)(n − 2)(15n2 − 5n − 14)B(n − 2) +8(3n + 1)(n − 2)2 (n − 3)B(n − 3), n ≥ 4, together with the initial conditions B(0) = 1, B(1) = 1, B(2) = 2, B(3) = 6. 47 (4.30) Theorem 4.18 The numbers B(n) of Baxter permutations are log-convex. The limit limn→∞ B(n)/B(n − 1) exists and is equal to 8. Proof Let q(n) = B(n)/B(n − 1). From (4.30) we form the recurrence for q(n)’s, and then according to the initial values q(1) = 1, q(2) = 2, q(3) = 3, we form the function f : [1, ∞) → R defined by f (x) = x on [1, 3] and for x ≥ 3 by the rule (x + 1)(x + 2)(x + 3)(3x − 2)f (x) = 2(x + 1)(9x3 + 3x2 − 4x + 4) (3x − 1)(x − 2)(15x2 − 5x − 14) f (x − 1) 8(3x − 1)(x − 2)2 (x − 3) + . f (x − 1)f (x − 2) + (4.31) This, written in the form of (4.23), yields to conclude that R, S, T are positive rational functions with no poles on [1, ∞). We also see that R(x) ր 6, S(x) ր 15, T (x) ր 8 as x → ∞. It is apparent from (4.31) that f (x) is a piecewise rational function, i.e. rational on intervals [n, n + 1], n ∈ N. For example, for x ∈ [3, 4], f (x) = 18x5 + 51x4 − 122x3 − 87x2 + 200x + 12 . (x − 1)(x + 1)(x + 2)(x + 3)(3x − 2) Clearly, f (x) is continuous everywhere. It can be checked, by induction on n, that the function f is bounded. More precisely, 7 ≤ f (x) ≤ 9, x ≥ 47. (4.32) We want to prove that f is an increasing function. From the above a priori bound, it follows that f is differentiable on all open intervals (n, n + 1), n ∈ N. The non-negativity of f ′ (x) can be checked for x ≤ 49, using e.g. Mathematica, and for x ≥ 49, it follows from the stronger inequality, 74 F ≥ (9 · 15 + 8) · 9F1 + 9 · 8F2 , 48 obtained by substituting appropriate values for m, M in (4.29). As this inequality is true for all x ≥ 9, the first claim follows. The second claim follows passing to the limit in (4.30). Many other three-or-higher-term recurrences can be investigated by this method. Let us only mention that the four-term recurrence (3.15) for the numbers S (1) (n) of secondary structures of rank 1 can also be shown to be log-convex by this “calculus” method. The details (rather tedious) are given in [11], [13]. Another example is the number Sn of n × n symmetric matrices with entries 0, 1, 2, whose sums of all rows and all columns are equal to 2. These numbers satisfy ([27]) the following recurrence 1 Sn = (2n − 1)Sn−1 − (n − 1)(n − 2)Sn−2 − (n − 1)(n − 2)Sn−3 + (n − 1)(n − 2)(n − 3)Sn−4 , 2 starting with S0 = 1, S1 = 1, S2 = 3, S3 = 11. The sequence (Sn )n≥0 is also log-convex. Finally, as it should be clear by now, this method applies to any P -recursive sequence (a(n)), satisfying a recurrence of the form Q(n)a(n) = Pd (n)a(n − 1) + Pd−1 (n)a(n − 2) + . . . + P0 (n)a(n − d − 1), n ≥ d + 1, (4.33) where d ≥ 0 is an integer and P0 , . . . , Pd , Q real polynomials, Q > 0. The corresponding function for the successive quotients q(n) = a(n) a(n−1) Q(x)f (x) = Pd (x) + is given by the functional equation P0 (x) Pd−1 (x) + ... + , f (x − 1) f (x − 1) . . . f (x − d) x ≥ d + 1. (4.34) As we have seen, the most important thing in this approach is to express the derivative f ′ (x) in terms of previous derivatives. So, fix a point x ∈ (n, n + 1), n > d, and write as before for short f = f (x), fj = f (x − j), j = 1, 2, . . ., Pi = Pi (x), Qi = Qi (x), i = 0, 1, . . . , d. Then (4.34) can be written as Qf = d X i=0 Pi , f1 f2 . . . fd−i or by denoting the product of all values by Π, i.e. Π = f1 f2 . . . fd , and by Πj = f1 f2 . . . fj the partial 49 products (so Πd = Π and Π0 = 1), as ΠQf = d X (fd−i+1 . . . fd )Pi . (4.35) i=o Taking the derivative d/dx of both sides in (4.35), after some manipulations, we obtain the following formula (in terms of Wronskians): d 1 X Q f = 2 Q Q′ ′ i=0 d Pi 1 1 X Pi Π′d−i − . Pi′ Πd−i Q Π2d−i (4.36) i=0 In particular, for d = 1 this reduces to f′ = Q 1 Q2 f 1 Q′ P0 1 Q + 2 P0′ Q Q′ P1 P0 ′ − f . P1′ Qf12 1 (4.37) With a priori bounds 0 < m(x) ≤ f (x) ≤ M (x) and by substituting f1′ in (4.37), one gets almost instant proofs of log-behavior. For example, if P0 ≤ 0 and if we want to prove the log-convexity, hence assuming f1′ ≥ 0, then if the first Wronskian W0 in (4.37) is positive, we only have to check 1 M W0 + W1 ≥ 0, and check that f increases at the beginning. Of course, not every (combinatorially relevant) sequence satisfying a recurrence of this type can be expected to have a reasonable log-behavior; it is enough to recall here the sequences ek (n) from Section 2, whose log-behavior is rather chaotic for k ≥ 3. Let us only mention here that the log-convexity of secondary structure numbers of general rank l can also be proved by calculus method, using the explicit formulae from Proposition 3.9 and formula (4.36). The details will appear elsewhere. As a finall remark, note that our approach applies also to linear nonhomogeneous recurrences for positive numbers. So, for example, let (a(n)) be given by the linear recurrence of the first order a(n) = R(n)a(n − 1) + S(n). 50 (4.38) Consider the quotients q(n) = a(n) a(n−1) and note that a(n) = q(n)q(n − 1) . . . q(2)a(1), n ≥ 2. (4.39) Then, dividing (4.38) by a(n − 1) we obtain a (long) recurrence for q(n)’s: q(n) = R(n) + S(n) . q(n − 1)q(n − 2) . . . q(2)a(1) (4.40) To get a short recurrence for q(n)’s, substitute for a(n) and a(n − 1) the corresponding products (4.39) in (4.38) (n ≥ 3): q(n)q(n − 1) . . . q(2)a(1) = R(n)q(n − 1) . . . q(2)a(1) + S(n) = R(n) q(n)q(n − 1) . . . q(2)a(1) q(n)q(n − 1) . . . q(2)a(1) + S(n) . q(n) q(n)q(n − 1) . . . q(2)a(1) From there we get   1 1 R(n) = 1− , q(n)q(n − 1) . . . q(2)a(1) S(n) q(n) and then   1 1 R(n − 1) = 1− . q(n − 1) . . . q(2)a(1) S(n − 1) q(n − 1) Substituting (4.41) in (4.40) yields a short recursion for q(n)’s: q(n) = R(n) + 1 R(n − 1)S(n) S(n) − . S(n − 1) S(n − 1) q(n − 1) Similarly, for a second order linear recurrence a(n) = R(n)a(n − 1) + S(n)a(n − 2) + T (n), we obtain   S(n) T (n) S(n − 1) R(n − 1) q(n) = R(n) + + − 1− . q(n − 1) T (n − 1) q(n − 1) q(n − 1)q(n − 2) Then we can proceed as before. 51 (4.41) 5 Calculus method in two variables We shall outline our method for non-negative sequences a(n, k) in two integer variables n, k ≥ 0 (or n ≥ n0 , k ≥ k0 ). Suppose (as often in combinatorics) that the numbers a(n, k) satisfy a two-term recurrence of the form a(n, k) = R(n, k)a(n − 1, k − 1) + S(n, k)a(n − 1, k), (5.1) with some known functions R and S, together with some initial values, usually of the type a(0, 0) = a, a(1, 0) = b, a(1, 1) = c. Suppose we want to prove that the sequence a(n, k) is log-concave in k, i.e. that a(n, k)2 ≥ a(n, k −1)a(n, k +1), for all n, k. Here is what we do. Write down (5.1) with k replaced by k − 1: a(n, k − 1) = R(n, k − 1)a(n − 1, k − 2) + S(n, k − 1)a(n − 1, k − 1). (5.2) Denote q(n, k) = a(n, k) , a(n, k − 1) (5.3) and divide (5.1) by (5.2) (always assuming we do not divide by zero). q(n, k) = R(n, k)a(n − 1, k − 1) + S(n, k)a(n − 1, k) R(n, k) + S(n, k)q(n − 1, k) = R(n,k−1) R(n, k − 1)a(n − 1, k − 2) + S(n, k − 1)a(n − 1, k − 1) + S(n, k − 1) q(n−1,k−1) = q(n − 1, k − 1) R(n, k) + S(n, k)q(n − 1, k) . R(n, k − 1) + S(n, k − 1)q(n − 1, k − 1) Equivalently, q(n, k)[R(n, k − 1) + S(n, k − 1)q(n − 1, k − 1)] = q(n − 1, k − 1)[R(n, k) + S(n, k)q(n − 1, k)]. (5.4) The log-concavity of a(n, k)’s is equivalent to q(n, k) ≥ q(n, k + 1), for any fixed n and all k. The idea is again to pass to a “continuation” of (5.4) by letting n → x, k → y, q → f and obtaining the functional equation f (x, y)[R(x, y − 1) + S(x, y − 1)f (x − 1, y − 1)] = f (x − 1, y − 1)[R(x, y) + S(x, y)f (x − 1, y)]. (5.5) 52 We assume that R and S are “good enough” functions, in the sense that f is continuous everywhere and smooth on open cells (n, n + 1) × (m, m + 1), for all m, n. What we want to prove is that f is decreasing in y for any fixed x. Fix a point (x, y) in an open cell Q = (n, n + 1) × (m, m + 1), and prove inductively that ∂f (x, y) ≤ 0. ∂y (5.6) For the fixed pair (x, y) write for short fij = f (x − i, y − j), for i, j = 0, 1, 2, . . ., and similarly for R and S. So, f00 = f (x, y), R01 = R(x, y − 1) etc. In this notation, (5.5) can be written as f [R01 + S01 f11 ] = f11 [R + Sf10 ]. (5.7) Now take the partial derivative ∂/∂y of both sides in (5.7). We have     ∂f ∂R10 ∂f11 ∂S01 ∂f11 ∂S ∂f10 ∂R [R01 + S01 f11 ] + f + f11 + S01 [R + Sf10 ] + f11 + f10 +S = . ∂y ∂y ∂y ∂y ∂y ∂y ∂y ∂y     ∂R ∂S ∂S01 ∂f11 ∂R10 ∂f10 ∂f [R01 + S01 f11 ] = f11 + f10 + f11 f11 S + [R + Sf10 − S01 f ] . − f+ ∂y ∂y ∂y ∂y ∂y ∂y ∂y Substituting here f from (5.7), we get ∂f [R01 + S01 f11 ] = ∂y   f11 ∂S ∂R01 ∂S01 ∂R + f10 ) − (R + Sf10 )( + f11 (R01 + S01 f11 )( R01 + S01 f11 ∂y ∂y ∂y ∂y ∂f10 (R + Sf1,0 )R01 ∂f11 +f11 S + . (5.8) ∂y R01 + S01 f11 ∂y Assume that R and S are positive. Hence if f starts with some positive values, then f can be considered positive, too. Suppose inductively that ∂f10 ∂f11 , ≤ 0. ∂y ∂y Then the last two terms in (5.8) are negative and to prove (5.6), it is enough to prove that the first term is negative, too. In other words, to conclude (inductively) that (5.6) holds, it is enough to prove that the “free” term is non-positive, i.e. that F = (R01 + S01 f11 )  ∂R ∂S + f10 ∂y ∂y  − (R + Sf10 ) 53  ∂S01 ∂R01 + f11 ∂y ∂y  ≤ 0. (5.9) So, if we can check that f begins decreasingly in y and assuming inductively that f is decreasing in y, then by (5.9) we can conclude that f is decreasing in y at the point (x, y) and then, by continuity, that f is decreasing in y everywhere. Note that the following inequalities imply (5.9) (simply by comparing similar terms): R01 ∂R01 ∂R ∂S01 ∂S ∂R01 ∂S ∂S01 ∂R ≤R , S01 ≤R , R01 ≤S , S01 ≤S . ∂y ∂y ∂y ∂y ∂y ∂y ∂y ∂y In terms of Wronskians, writing G′ for R01 ′ R01 ∂G ∂y , R S01 ≤ 0, ′ R′ S01 these inequalities can be written in the form: R R01 ≤ 0, ′ R′ R01 S01 S ≤ 0, ′ S′ S01 S ≤ 0. S′ (5.10) Instead of formalizing everything (which can be done with a little care), let us take an example. Example 5.1 The Eulerian number E(n, k), is the number of permutations π from Σn with exactly k ascents, i.e. with exactly k places where πj < πj+1 . We know (see, e.g. [16]) that these numbers satisfy the recurrence E(n, k) = (n − k)E(n − 1, k − 1) + (k + 1)E(n − 1, k), with the initial conditions E(0, k) = δ0k , E(n, 0) = 1, n, k ≥ 0. Then (5.4) becomes q(n, k)[n − k + 1 + kq(n − 1, k − 1)] = q(n − 1, k − 1)[n − k + (k + 1)q(n − 1, k)] with q(n, k) = E(n, k)/E(n, k − 1), for 1 ≤ k ≤ n. The initial conditions are q(2, 1) = 1, q(3, 1) = 4, q(3, 2) = 1/4 and (we extra define) q(2, k) = 0, for k ≥ 2. We want to prove that q(n, k) ≥ q(n, k + 1), for any fixed n and all k. In the sense of the above discussion and notations, here we have R(n, k) = n − k, S(n, k) = k + 1. Passing to the natural “continuation”, (5.7) becomes f [x − y + 1 + yf11 ] = f11 [x − y + (y + 1)f10 ]. 54 (5.11) In fact, we define the function f : [2, ∞) × [1, ∞) → R first on two shaded strips in Fig. 6 below and then continue by the rule (5.11). In Fig. 6 we indicated the values f (n, k) = q(n, k) for n ≥ 2, k ≥ 1 in the lattice nodes. On the vertical walls of the (square) cells Q2 , Q3 , Q4 , . . ., as well as on their lower horizontal walls, we define f to be appropriate linear functions. The upper walls of Q3 , Q4 , . . ., are determined by (5.11). The right walls of Q′3 , Q′4 , . . . are also determined by (5.11). We fill in f on cells Qn , Q′n by appropriate homotopies connecting (possibly nonlinear) functions on the walls. 6 0 0 Q 0 5 0 4 0 3 0 0 0 , Q f1,0 6 , Q5 0 0 0 1/57 0 0 1/26 57/302 0 1/11 26/66 1 1/4 1 66/26 302/57 f f1,1 , Q4 , Q3 2 0 1 1 Q2 Q3 Q4 4 Q5 11 Q6 26 Q7 57 Q8 120 247 0 1 2 3 4 5 6 7 8 9 Figure 6: The boundary values of function f So, let f (x, 2) = 2n (x − n) + 2n − n − 1, for x ∈ [n, n + 1], n ≥ 2, f (2, y) = 2 − y for y ∈ [1, 2] and f (2, y) = 0 for y ≥ 2. Further, let f (3, y) = 31−15y 4 for y ∈ [1, 2], and, in general, let f (n, y), y ∈ [1, 2], be the linear function in y between the points (n, q(n, 1)) and (n, q(n, 2)). The cell Q2 is surrounded by linear functions a(x) = 3x − 5, x ∈ [2, 3], b(y) = 55 31−15y , 4 y ∈ [1, 2], c(x) = x−2 4 , x ∈ [2, 3] and d(y) = 2 − y, y ∈ [1, 2]. Extend f to Q2 by the homotopy f (x, y) = (2 − y)a(x) + (y − 1)c(x) = The cell Q′3 is surrounded by functions a′3 (x) = 1 (23x + 18y − 11xy − 38), 4 x−2 4 , x ∈ [2, 3], b′3 (y) = (x, y) ∈ Q2 . (3−y)2 4−y+y(3−y) , y ∈ [2, 3], c′3 (x) = 0, x ∈ [2, 3] and d′3 (y) = 0, y ∈ [2, 3]. Extend f to Q′3 by the homotopy f (x, y) = (3 − x)d′3 (y) + (x − 2)b′3 (y) = (x − 2)(3 − y)2 , 4 − y + y(3 − y) (x, y) ∈ Q′3 . Extend f to Q′4 , Q′5 , . . . to be zero. Next, extend f to Q3 , Q4 , . . . by the homotopies connecting the lower walls of Qn , given by an (x) = 2n (x − n) + 2n − n − 1, for x ∈ [n, n + 1], n ≥ 2, and the upper walls, given by rational functions cn (x) determined inductively on n by (5.11), thus obtaining f f (x, y) = (2 − y)an (x) + (y − 1)cn (x), For example, since a2 (x) = 3x − 5, a3 (x) = 7x − 17, c2 (x) = c3 (x) = and hence f Q3 Qn by (x, y) ∈ Qn . x−2 4 , then 3x − 8 a2 (x − 1)[x − y + (y + 1)c2 (x − 1)] = , x − y + 1 + ya2 (x − 1) 4 is given by f (x, y) = (2 − y)a3 (x) + (y − 1)c3 (x) = 53x + 60y − 25xy − 128 . 4 In this way, f is well defined on the shaded strips on Fig. 6 and extended to [2, ∞) × [1, ∞) by the rule (5.11). It is easy to check that f is continuous and nonnegative and that it is a rational function with no poles on any open cell, and hence smooth on any open cell. It is also easy to check inductively on n that f is decreasing in y on Qn and Q′n for n ≥ 2, i.e. ∂f ∂y (x, y) ≤ 0 for (x, y) ∈ int(Qn ) ∪ int(Q′n ). Now that we have elaborated carefully the “beginning” of f , the rest is more-or-less automatic. The inequality (5.9) reduces to (x − y + 1 + yf11 )(−1 + f10 ) ≤ (x − y + yf10 + f10 )(−1 + f11 ). This is 56 equivalent to −1 + xf10 + f10 ≤ −f10 + xf11 + f10 f11 , or, after some rearrangement, f11 f10 − 2f10 + 1 + x(f11 − f10 ) ≥ 0. The second term is non-negative by the induction hypothesis, and the rest is non-negative since 2 f11 f10 − 2f10 + 1 ≥ f10 − 2f10 + 1 = (f10 − 1)2 ≥ 0. Hence, ∂f ∂y (x, y) ≤ 0 for (x, y) ∈ int(Q), for all Q. So, f is decreasing in y on every open cell, and hence by continuity, f is decreasing in y everywhere. In particular, q(n, k) = f (n, k) ≥ f (n, k+1) = q(n, k+1) and we are done. In the same manner we can prove that ∂f ∂x (x − y + 1) ≥ f , hence ∂f ∂x ≥ 0, for any (x, y) ∈ int(Q), for any Q, and this implies q(n + 1, k) ≥ q(n, k), for any fixed k and all n. Hence, E(n + 1, k) E(n, k) E(n + 1, k) ≥ ⇐⇒ E(n + 1, k − 1) E(n, k − 1) E(n, k) E(n + 1, k − 1) ≥ 0. E(n, k − 1) Apart from settling the “beginning” of f , we can (almost automatically now) prove the well-known log-concave behavior in the second variable of the binomial coefficients (and in general find the logconcave behavior when R and S in (5.1) are constants), q-binomial coefficients, Stirling numbers of the first and second kind, Schläfli numbers, cover many particular results (e.g [20]) and so on. Of course, the method can be extended in a few ways; for example, to three-or-more term recurrences, recurrences for three or more variables a(n, k, l) etc. but we shall not consider it here. In a word, a general idea of this method is as follows. Combinatorics gives a recurrence. 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