Some Sufficient Conditions for the Log-Balancedness of Combinatorial Sequences

Ting-Ting Zhang Department of Mathematics Shanghai University Shanghai 200444 P. R. China

                ztingting@i.shu.edu.cn

Feng-Zhen Zhao Department of Mathematics Shanghai University Shanghai 200444 P. R. China

                fengzhenzhao@shu.edu.cn

Abstract

In this paper, we give some new sufficient conditions for log-balancedness of combinatorial sequences. In particular, we show that the product of two log-convex sequences is log-balanced under a mild condition. Then, we apply this result to a series of special combinatorial sequences. In addition, we show some results by using the definition of log-balancedness directly.

1 Introduction

For convenience, we first recall some concepts that will be used later on. The following definition is well known in combinatorics.

Definition 1. (i) For a sequence of real numbers {z_{n}}_{n≥0}, we say that {z_{n}}_{n≥0} is concave (resp., convex) if 2z_{n} ≥ z_{n−1} + z_{n+1} (resp., 2z_{n} ≤ z_{n−1} + z_{n+1}) for all n ≥ 1. (ii) For a sequence of positive numbers {z_{n}}_{n≥0}, we say that {z_{n}}_{n≥0} is log-concave (resp., log-convex) if z_{n}^{2} ≥ z_{n−1}z_{n+1} (resp., z_{n}^{2} ≤ z_{n−1}z_{n+1}) for all n ≥ 1.

Došlić [2] gave the following definition.

Definition 2. Let {z_{n}}_{n≥0} be a log-convex sequence. We say that {z_{n}}_{n≥0} is log-balanced if \{\frac{z_{n}}{n!} \}_{n≥0} is log-concave.

Log-concavity and log-convexity play important roles in many subjects. For example, in combinatorics, they are not only instrumental in obtaining the growth rate of a combinatorial sequence, but also fertile sources of inequalities. See, e.g., [1, 6] for more applications of logconcavity and log-convexity.

For a sequence of positive numbers, it is easy to see from the arithmetic-geometric mean inequality that its concavity implies its log-concavity and its log-convexity implies its convexity. Obviously, a sequence {z_{n}}_{n≥0} is log-convex (resp., log-concave) if and only if its quotient sequence \{\frac{z_{n+1}}{z_{n}} \}_{n≥0} is nondecreasing (resp., nonincreasing). A log-balanced sequence is naturally log-convex, but its quotient sequence does not grow too fast. Moreover, a sequence {z_{n}}_{n≥0} is log-balanced if and only if z_{n}^{2} ≤ z_{n−1}z_{n+1} and (n + 1)z_{n}^{2} ≥ nz_{n−1}z_{n+1} for every n ≥ 1. Došlić [2] showed that many combinatorial sequences, including the Motzkin numbers, the Fine numbers, the Franel numbers of order 3 and 4, the Apéry numbers, the large Schröder numbers, and the central Delannoy numbers, are log-balanced. Zhao [7, 8] proved that the sequences of the exponential numbers and the Catalan-Larcombe-French numbers are respectively log-balanced.

The main purpose of this paper is to discuss log-balancedness of some combinatorial sequences. In the next section, we present some new sufficient conditions for log-balancedness of combinatorial sequences. In particular, we provide a sufficient condition for log-balancedness of the product of two log-convex sequences. Then, based on this result, we obtain some similar results for a series of special combinatorial sequences.

2 Main results

Zhao [7] gave a sufficient condition for log-balancedness of the product of a log-balanced sequence and a log-concave sequence. Here, we consider log-balancedness of the product of two log-convex sequences.

Theorem 3. Suppose that the sequences {x_{n}}_{n≥0} and {y_{n}}_{n≥0} are both log-convex. Let s_{n} = \frac{x_{n+1}y_{n+1}}{(n+1)x_{n}y_{n}} for n ≥ 0. If \{s_{n}\}_{n≥0} is decreasing, then \{x_{n}y_{n}\}_{n≥0} is log-balanced.

Proof. By the log-convexity of the sequences {x_{n}}_{n≥0} and {y_{n}}_{n≥0}, we know that {x_{n}y_{n}}_{n≥0} is log-convex. Note that \{s_{n}\}_{n≥0} is the quotient sequence of \{\frac{x_{n}y_{n}}{n!} \}_{n≥0}. Since \{s_{n}\}_{n≥0} is decreasing, \{\frac{x_{n}y_{n}}{n!} \}_{n≥0} is log-concave. Hence, the sequence \{x_{n}y_{n}\}_{n≥0} is log-balanced.

Next, we apply Theorem 3 to deduce log-balancedness of some combinatorial sequences.

Corollary 4. For the sequence {C_{n}}_{n≥1} of the Catalan numbers, we have that {C_{n}^{2}}_{n≥3} is log-balanced.

Proof. Since \{C_{n}\}_{n≥1} is log-convex, \{C_{n}^{2}\}_{n≥1} is log-convex. For n ≥ 1, let s_{n} = \frac{C_{n+1}^{2}}{(n+1)C_{n}^{2}} . It is well known that

C_{n} = \frac{1}{n}\binom{2n − 2}{n − 1}, n ≥ 1.

Then we have

s_{n} = \frac{4(2n − 1)^{2}}{(n + 1)^{3}} .

It is not difficult to verify that {s_{n}}_{n≥3} is decreasing. By Theorem 3, the sequence {C_{n}^{2}}_{n≥3} is log-balanced.

Corollary 5. For the sequence {M_{n}}_{n≥0} of the Motzkin numbers, we have that {M_{n}^{2}}_{n≥1} is log-balanced.

Proof. The Motzkin numbers satisfy the recurrence

(n + 3)M_{n+1} = (2n + 3)M_{n} + 3nM_{n−1}, M_{0} = M_{1} = 1.  (1)

For n ≥ 0, let t_{n} = \frac{M_{n+1}}{M_{n}} and s_{n} = \frac{t_{n}^{2}}{n+1}. It follows from (1) that

t_{n} = \frac{2n + 3}{n + 3} + \frac{3n}{(n + 3)t_{n−1}} .  (2)

Then we have

s_{n} − s_{n+1} = \frac{(n + 2)t_{n}^{2} − (n + 1)t_{n+1}^{2}}{(n + 1)(n + 2)} .

It follows from (2) that

(n + 2)t_{n}^{2} − (n + 1)t_{n+1}^{2}

= \frac{(n + 2)(n + 4)^{2}t_{n}^{4} − (n + 1)(2n + 5)^{2}t_{n}^{2} − 6(n + 1)^{2}(2n + 5)t_{n} − 9(n + 1)^{3}}{(n + 4)^{2}t_{n}^{2}} .

For any real number x, we let

f(x) = (n + 2)(n + 4)^{2}x^{4} − (n + 1)(2n + 5)^{2}x^{2} − 6(n + 1)^{2}(2n + 5)x − 9(n + 1)^{3}.

Then we have

f′(x) = 4(n + 2)(n + 4)^{2}x^{3} − 2(n + 1)(2n + 5)^{2}x − 6(n + 1)^{2}(2n + 5)

and

f′′(x) = 12(n + 2)(n + 4)^{2}x^{2} − 2(n + 1)(2n + 5)^{2}.

Since f′′(x) > 0 when x ≥ 1, we know that f′ is increasing over [1, ∞). Došlić and Veljan [3] showed that

t_{n} ≥ q_{n},

where q_{n} = \frac{6(n+1)}{2n+5}. Since

f′(q_{n}) = \frac{18(n + 1)^{2}[48(n + 1)(n + 2)(n + 4)^{2} − (2n + 5)^{4}]}{(2n + 5)^{3}} > 0,

the function f is increasing over [q_{n}, ∞). Note that

f(q_{n}) = \frac{81(n + 1)^{3}(16n^{3} + 72n^{2} + 24n − 113)}{(2n + 5)^{4}} > 0.

By the definition of f, we have

(n + 2)t_{n}^{2} − (n + 1)t_{n+1}^{2} = \frac{f(t_{n})}{(n + 4)^{2}t_{n}^{2}} > 0

for each n. This means that {s_{n}}_{n≥0} is decreasing. On the other hand, {M_{n}}_{n≥1} is logbalanced. It follows from Theorem 3 that the sequence {M_{n}^{2}}_{n≥1} is log-balanced.

Denote by A_{n} the number of directed animals of size n (see [5, Exercise 6.46]), which satisfies the recurrence

(n + 1)A_{n+1} = 2(n + 1)A_{n} + 3(n − 1)A_{n−1}  (3)

with A_{0} = 1, A_{1} = 1, and A_{2} = 2.

Corollary 6. Both \{A_{n}^{2}\}_{n≥2} and \{\frac{A_{n}}{n} \}_{n≥2} are log-balanced.

Proof. It is clear that the sequence \{\frac{1}{n}\}_{n≥1} is log-convex. Liu and Wang [4] proved that the sequence \{A_{n}\}_{n≥0} is log-convex. For n ≥ 0, let t_{n} = \frac{A_{n+1}}{A_{n}} and s_{n} = \frac{t_{n}^{2}}{n+1}. By (3), we have

t_{n} = 2 + \frac{3(n − 1)}{(n + 1)t_{n−1}} , n ≥ 1.  (4)

It follows from (4) that

(n + 2)t_{n}^{2} − (n + 1)t_{n+1}^{2}

= \frac{(n + 2)^{3}t_{n}^{4} − 4(n + 1)(n + 2)^{2}t_{n}^{2} − 12n(n + 1)(n + 2)t_{n} − 9n^{2}(n + 1)}{(n + 2)^{2}t_{n}^{2}}

and

n(n + 2)^{2}t_{n} − (n + 1)^{3}t_{n+1}

= \frac{n(n + 2)^{3}t_{n}^{2} − 2(n + 2)(n + 1)^{3}t_{n} − 3n(n + 1)^{3}}{(n + 2)t_{n}} .

For any real number x, let

f(x) = (n + 2)^{3}x^{4} − 4(n + 1)(n + 2)^{2}x^{2} − 12n(n + 1)(n + 2)x − 9n^{2}(n + 1),
g(x) = n(n + 2)^{3}x^{2} − 2(n + 2)(n + 1)^{3}x − 3n(n + 1)^{3}.

Then we have

f′(x) = 4(n + 2)^{3}x^{3} − 8(n + 1)(n + 2)^{2}x − 12n(n + 1)(n + 2),
f′′(x) = 12(n + 2)^{3}x^{2} − 8(n + 1)(n + 2)^{2},
g′(x) = 2n(n + 2)^{3}x − 2(n + 2)(n + 1)^{3}.

It is obvious that f′′(x) > 0 when x ≥ 1 and hence f′ is increasing over [1, +∞). Noting that f′(2) > 0, we have f′(x) > 0 when x ≥ 2.

Liu and Wang [4] showed that

t_{n} ≥ μ_{n},

where μ_{n} = \frac{6n}{2n+1}. Since f′(μ_{n}) > 0, f is increasing over [μ_{n}, ∞). It is evident that g′(x) > 0 for x ≥ 1 and hence the function g is also increasing over [1, ∞).

Note that

f(μ_{n}) = \frac{9n^{2}}{(2n + 1)^{4}} [144n^{2}(n + 2)^{3} − 16(n + 1)(n + 2)^{2}(2n + 1)^{2}
−8(n + 1)(n + 2)(2n + 1)^{3} − (n + 1)92n + 1)^{4}]
= \frac{9n^{2}}{(2n + 1)^{4}} (144n^{4} + 370n^{2} − 72n^{2} − 513n − 81)

and

g(μ_{n}) = \frac{3n}{(2n + 1)^{2}} [12n^{3}(n + 2)^{3} − 4(n + 2)(2n + 1)(n + 1)^{3} − (2n + 1)^{2}(n + 1)^{3}]
= \frac{3n}{(2n + 1)^{2}} (12n^{4} + 27n^{3} − 15n^{2} − 51n − 9).

Clearly, f(μ_{n}) > 0 and g(μ_{n}) > 0 for n ≥ 2. This implies that

(n + 2)t_{n}^{2} − (n + 1)t_{n+1}^{2} > 0

and

n(n + 2)^{2}t_{n} − (n + 1)^{3}t_{n+1} > 0

for n ≥ 2. Then \{s_{n}\}_{n≥2} and \{\frac{nt_{n}}{(n+1)^{2}} \}_{n≥2} are both decreasing. It follows from Theorem 3 that the sequences \{A_{n}^{2}\}_{n≥2} and \{\frac{A_{n}}{n} \}_{n≥2} are both log-balanced. Corollary 7. For the sequence \{B_{n}\}_{n≥0} of the Fine numbers, we have that \{\frac{B_{n}}{n} \}_{n≥2} is log-balanced. Proof. The Fine numbers satisfy the recurrence

B_{n+1} = \frac{7n + 2}{2(n + 2)}B_{n} + \frac{2n + 1}{n + 2}B_{n−1}, B_{0} = 1, B_{1} = 0.  (5)

For n ≥ 2, let t_{n} = \frac{B_{n+1}}{B_{n}}. Došlić [2] showed that the sequence \{B_{n}\}_{n≥2} is log-balanced. We next prove that \{\frac{nt_{n}}{(n+1)^{2}} \}_{n≥2} is decreasing.

By (5), we have

t_{n} = \frac{7n + 2}{2(n + 2)} + \frac{2n + 1}{(n + 2)t_{n−1}} .

Then we have

n(n + 2)^{2}t_{n} − (n + 1)^{3}t_{n+1}

= \frac{2n(n + 3)(n + 2)^{2}t_{n}^{2} − (7n + 9)(n + 1)^{3}t_{n} − 2(n + 1)^{3}(2n + 3)}{2(n + 3)t_{n}}.

For any real number x, let

f(x) = 2n(n + 3)(n + 2)^{2}x^{2} − (7n + 9)(n + 1)^{3}x − 2(n + 1)^{3}(2n + 3).

Then we obtain

f′(x) = 4n(n + 3)(n + 2)^{2}x − (7n + 9)(n + 1)^{3}.

It is obvious that f′(x) > 0 for x ≥ 3. Then f is increasing over [3, ∞).

Liu and Wang [4] proved that

t_{n} ≥ λ_{n},

where λ_{n} = \frac{2(2n+5)}{n+4}. Since

f(λ_{n}) = \frac{1}{(n + 4)^{2}} [8n(n + 3)(n + 2)^{2}(2n + 5)^{2}
−(7n + 9)(2n + 5)(n + 4)(n + 1)^{3} − 2(2n + 3)(n + 4)^{2}(n + 1)^{3}]
= \frac{14n^{6} + 185n^{5} + 968n^{4} + 458n^{3} + 4314n^{2} + 1023n − 596}{(n + 4)^{2}}
> 0,

we have

2n(n + 3)(n + 2)^{2}t_{n}^{2} − (7n + 9)(n + 1)^{3}t_{n} − 2(n + 1)^{3}(2n + 3) > 0.

Then n(n + 2)^{2}t_{n} − (n + 1)^{3}t_{n+1} > 0, and \{\frac{nt_{n}}{(n+1)^{2}} \}_{n≥2} is decreasing. It follows from Theorem 3 that the sequence \{\frac{B_{n}}{n} \}_{n≥2} is log-balanced.

Theorem 8. For a given sequence \{z_{n}\}_{n≥0}, if it is log-balanced, then \{\sqrt{z_{n}}\}_{n≥0} is also log-balanced.

Proof. Suppose that {z_{n}}_{n≥0} is log-balanced, that is,

z_{n}^{2} ≤ z_{n−1}z_{n+1}, (n + 1)z_{n}^{2} ≥ nz_{n−1}z_{n+1}, n ≥ 1.

For n ≥ 1, we immediately derive

z_{n} ≤ \sqrt{z_{n−1}z_{n+1}}

and

z_{n} ≥ \sqrt{\frac{n}{n + 1}z_{n−1}z_{n+1}} > \frac{n}{n + 1}\sqrt{z_{n−1}z_{n+1}}.

This means that the sequence \{\sqrt{z_{n}}\}_{n≥0} is log-convex and the sequence \{\frac{\sqrt{z_{n}}}{n!} \}_{n≥0} is logconcave. As a result, the sequence \{\sqrt{z_{n}}\}_{n≥0} is log-balanced.

Theorem 9. Suppose that the sequences {x_{n}}_{n≥0} and {y_{n}}_{n≥0} are both log-convex. If both \{\frac{x_{n}}{n!} \}_{n≥0} and \{\frac{y_{n}}{n!} \}_{n≥0} are concave, then \{x_{n} + y_{n}\}_{n≥0} is log-balanced.

Proof. Since {x_{n}}_{n≥0} and {y_{n}}_{n≥0} are both log-convex, the sequence {x_{n} + y_{n}}_{n≥0} is logconvex. We next prove that \{\frac{x_{n}+y_{n}}{n!} \}_{n≥0} is log-concave.

It is well known that {x_{n}}_{n≥0} is concave if and only if its difference sequence {x_{n+1} − x_{n}\}_{n≥0} is decreasing. Therefore, by the concavity of \{\frac{x_{n}}{n!} \}_{n≥0} and \{\frac{y_{n}}{n!} \}_{n≥0}, the sequence \{\frac{x_{n+1}+y_{n+1}}{n!} − \frac{x_{n}+y_{n}}{n!} \}_{n≥0} is decreasing. Then the sequence \{\frac{x_{n}+y_{n}}{n!} \}_{n≥0} is concave and it is also log-concave. Hence, the sequence {x_{n} + y_{n}}_{n≥0} is log-balanced.

It follows from Theorem 9 that the sequence {n! + (n + 1)!}_{n≥0} is log-balanced.

In the rest of this section, we devote to discuss the log-balancedness of some sequences by means of Definition 2 directly. Our first example is to consider some sequences related to harmonic numbers. Let H_{n} denote the n^{th} harmonic number. Then we have the following result.

Proposition 10. Both \{\frac{H_{n}}{n} \}_{n≥1} and \{\frac{H_{n}}{n^{2}} \}_{n≥1} are log-balanced.

Proof. In order to prove the log-balancedness of \{\frac{H_{n}}{n} \}_{n≥1}, it is sufficient to show that \{\frac{H_{n}}{n} \}_{n≥1} is log-convex and the sequence \{\frac{H_{n}}{nn!}\}_{n≥1} is log-concave. In fact, for n ≥ 2, we have

\frac{H_{n}^{2}}{n^{2}} − \frac{H_{n−1}H_{n+1}}{n^{2} − 1} = \frac{1}{n^{2}(n^{2} − 1)}[(n^{2} − 1)H_{n}^{2} − n^{2}(H_{n} − \frac{1}{n})(H_{n} + \frac{1}{n + 1})]
= −\frac{n(H_{n}^{2} − H_{n} − 1) + H_{n}^{2}}{n^{2}(n + 1)^{2}(n − 1)} .

Note that

2(H_{2}^{2} − H_{2} − 1) + H_{2}^{2} > 0, H_{n} > H_{2} > 2 (n ≥ 3).

Now we prove that n(H_{n}^{2} − H_{n} − 1) + H_{n}^{2} > 0 for n ≥ 3. For any real number x, let

f(x) = x^{2} − x − 1.

It is clear that f′(x) = 2x − 1 > 0 for x ≥ 2. Then f is increasing over [2, ∞) and f(H_{n}) > f(H_{3}) = \frac{19}{36} > 0 for n ≥ 3. Hence, the sequence \{\frac{H_{n}}{n} \}_{n≥1} is log-convex. On the other hand, for n ≥ 2, we have

\frac{H_{n}^{2}}{nn!} − \frac{H_{n−1}H_{n+1}}{(n^{2} − 1)(n − 1)!(n + 1)(n + 1)!}

= \frac{1}{n^{2}(n^{2} − 1)n!(n + 1)!}[(n^{2} − n − 1)H_{n}^{2} + n^{2}(\frac{H_{n}}{n + 1} + \frac{1}{n + 1})]

> 0.

Hence the sequence \{\frac{H_{n}}{nn!}\}_{n≥1} is log-concave. It follows from Definition 2 that the sequence \{\frac{H_{n}}{n} \}_{n≥1} is log-balanced.

Now we consider the sequence \{\frac{H_{n}}{n^{2}} \}_{n≥1}. Since both \{\frac{H_{n}}{n} \}_{n≥1} and \{\frac{1}{n}\}_{n≥1} are log-convex, \{\frac{H_{n}}{n^{2}} \}_{n≥1} is log-convex. On the other hand, for n ≥ 2, we get

(\frac{H_{n}}{n^{2}n!})^{2} − \frac{H_{n−1}H_{n+1}}{(n − 1)^{2}(n − 1)!(n + 1)^{2}(n + 1)!}

= \frac{1}{n^{2}(n − 1)^{2}(n + 1)^{2}n!(n + 1)!}[(n^{4} − 2n^{3} − 2n^{2} + 3n + 1)H_{n}^{2}

+n^{4}(\frac{H_{n}}{n + 1} + \frac{1}{n + 1})].

For n = 2,

(\frac{H_{n}}{n^{2}n!})^{2} − \frac{H_{n−1}H_{n+1}}{(n − 1)^{2}(n − 1)!(n + 1)^{2}(n + 1)!} = \frac{25}{20736}.

We find that n^{4} − 2n^{3} − 2n^{2} + 3n + 1 > 0 for n ≥ 3. Thus the sequence \{\frac{H_{n}}{n^{2}n!}\}_{n≥1} is log-concave. It follows from Definition 2 that the sequence \{\frac{H_{n}}{n^{2}} \}_{n≥1} is log-balanced.

Our second example is to consider some sequences related to the Fibonacci (Lucas) sequence. The Binet form of the Fibonacci sequence {F_{n}}_{n≥0} and the Lucas sequence {L_{n}}_{n≥0} are

F_{n} = \frac{α^{n} − (−1)^{n}α^{−n}}{\sqrt{5}} , L_{n} = α^{n} + (−1)^{n}α^{−n},

where α = \frac{1+\sqrt{5}}{2}. It is well known that log-convexity and log-concavity of \{F_{n}\}_{n≥0} and {L_{n}}_{n≥0} depend on the parity of n. In fact, by using the definition of log-convexity, we can easily prove that both {F_{2n+1}}_{n≥0} and {L_{2n}}_{n≥2} are log-convex. Now we discuss the log-balancedness of some sequences related to F_{n} and L_{n}. We first give a lemma.

Lemma 11. For n ≥ 1, we have

F_{2n+1} ≥ 2n  (6)

and

L_{2n} ≥ 3n.  (7)

Proof. It is well known that {F_{n}}_{n≥0} and {L_{n}}_{n≥0} satisfy the recurrence relation

W_{n+1} = W_{n} + W_{n−1}, n ≥ 1.  (8)

We can prove (6)–(7) by induction. We only give a proof of (6) and (7) can be shown in a similar way. In fact, it is clear that F_{2n+1} ≥ 2n for 1 ≤ n ≤ 5. Assume that F_{2n+1} ≥ 2n for n ≥ 5. By (8), we have

F_{2n+3} = F_{2n+1} + F_{2n+2}.

Then we have F_{2n+3} ≥ F_{2n+1} + 2 ≥ 2n + 2. By mathematical induction, (6) holds for each n ≥ 1.

Proposition 12. The sequences \{\frac{F_{2n+1}}{n} \}_{n≥1} and \{\frac{L_{2n}}{n} \}_{n≥2} are log-balanced.

Proof. Because \{F_{2n+1}\}_{n≥0}, \{L_{2n}\}_{n≥2} and \{\frac{1}{n}\}_{n≥1} are log-convex, the sequences \{\frac{F_{2n+1}}{n} \}_{n≥1} and \{\frac{L_{2n}}{n} \}_{n≥2} are log-convex. Next we show that \{\frac{F_{2n+1}}{nn!} \}_{n≥1} and \{\frac{L_{2n}}{nn!} \}_{n≥2} are log-concave.

For n ≥ 2, we obtain

(\frac{F_{2n+1}}{nn!} )^{2} − \frac{F_{2n−1}F_{2n+3}}{(n^{2} − 1)(n − 1)!(n + 1)!} = \frac{(n + 1)(n^{2} − 1)F_{2n+1}^{2} − n^{3}F_{2n−1}F_{2n+3}}{n^{2}(n^{2} − 1)n!(n + 1)!}
= \frac{n^{3}(F_{2n+1}^{2} − F_{2n−1}F_{2n+3}) + (n^{2} − n − 1)F_{2n+1}^{2}}{n^{2}(n^{2} − 1)n!(n + 1)!}

and

(\frac{L_{2n}}{nn!})^{2} − \frac{L_{2n−2}L_{2n+2}}{(n^{2} − 1)(n − 1)!(n + 1)!} = \frac{(n + 1)(n^{2} − 1)L_{2n}^{2} − n^{3}L_{2n−2}L_{2n+2}}{n^{2}(n^{2} − 1)n!(n + 1)!}
= \frac{n^{3}(L_{2n}^{2} − L_{2n−2}L_{2n+2}) + (n^{2} − n − 1)L_{2n}^{2}}{n^{2}(n^{2} − 1)n!(n + 1)!} .

By means of the equalities

F_{2n+1}^{2} − F_{2n−1}F_{2n+3} = −1 and L_{2n}^{2} − L_{2n−2}L_{2n+2} = −5,

we have

(\frac{F_{2n+1}}{nn!} )^{2} − \frac{F_{2n−1}F_{2n+3}}{(n^{2} − 1)(n − 1)!(n + 1)!} = \frac{−n^{3} + (n^{2} − n − 1)F_{2n+1}^{2}}{n^{2}(n^{2} − 1)n!(n + 1)!} ,
(\frac{L_{2n}}{nn!})^{2} − \frac{L_{2n−2}L_{2n+2}}{(n^{2} − 1)(n − 1)!(n + 1)!} = \frac{−5n^{3} + (n^{2} − n − 1)L_{2n}^{2}}{n^{2}(n^{2} − 1)n!(n + 1)!} .

For n ≥ 2, put

R(n) = −n^{3} + (n^{2} − n − 1)F_{2n+1}^{2} and S(n) = −5n^{3} + (n^{2} − n − 1)L_{2n}^{2}.

It follows from Lemma 11 that

R(n) ≥ n^{2}(4n^{2} − 5n − 4), S(n) ≥ n^{2}(9n^{2} − 14n − 9).

Note that

R(n) > 0 (n ≥ 2), S(n) > 0 (n ≥ 3).

This implies that \{\frac{F_{2n+1}}{nn!} \}_{n≥1} and \{\frac{L_{2n}}{nn!} \}_{n≥2} are both log-concave. By Definition 2, \{\frac{F_{2n+1}}{n} \}_{n≥1} and \{\frac{L_{2n}}{n} \}_{n≥2} are both log-balanced. This completes the proof.

3 Conclusions

We have derived some new sufficient conditions for log-balancedness of combinatorial sequences. We have further applied these results to show log-balancedness of some special sequences. One future work is to study log-balancedness of the partial sums for log-balanced sequences.

4 Acknowledgment

This work was supported in part by a grant of the First-Class Discipline of Universities in Shanghai and the Innovation Fund of Shanghai University. The authors would like to thank an anonymous referee for his or her helpful comments and suggestions.

References

[1] N. Asai, I. Kubo, and H. H. Kubo, Roles of log-concavity, log-convexity, and growth order in white noise analysis, Infin. Dimens. Anal. Quantum Probab. Relat. Top. 4 (2001), 59–84.

[2] T. Došlić, Log-balanced combinatorial sequences, Int. J. Math. Math. Sci. 4 (2005), 507–522.

[3] T. Došlić and D. Veljan, Logarithmic behavior of some combinatorial sequences, Discrete Math. 308 (2008), 2182–2212.

[4] L. L. Liu and Y. Wang, On the log-convexity of combinatorial sequences, Adv. in Appl. Math. 39 (2007), 453–476.

[5] R. P. Stanley, Enumerative Combinatorics, Vol. 2, Cambridge University Press, 1999.

[6] R. P. Stanley, Log-concave and unimodal sequences in algebra, combinatorics and geometry, Ann. N. Y. Acad. Sci. 576 (1989), 500–535.

[7] F. Z. Zhao, The log-balancedness of combinatorial sequences, Sarajevo J. Math. 11 (2015), 14–154.

[8] F. Z. Zhao, The log-behavior of the Catalan-Larcombe-French sequence, Int. J. Number Theory. 110 (2014), 177–182.

2010 Mathematics Subject Classification: Primary 05A20; Secondary 11B37, 11B39. Keywords: log-convexity, log-concavity, log-balancedness.

(Concerned with sequences \underline{A000032}, \underline{A000045}, \underline{A000108}, \underline{A000957}, \underline{A001006}, and \underline{A005773}.)

Received May 9 2016; revised versions received May 11 2016; June 22 2016; July 7 2016. Published in Journal of Integer Sequences, August 29 2016.

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