New Sufficient Conditions for Log-Balancedness, With Applications to

Combinatorial Sequences

Rui-Li Liu and Feng-Zhen Zhao Department of Mathematics Shanghai University Shanghai 200444 P. R. China

                     liuruili@i.shu.edu.cn
                    fengzhenzhao@shu.edu.cn

Abstract In this paper, we mainly study the log-balancedness of combinatorial sequences. We first give some new sufficient conditions for log-balancedness of some kinds of sequences. Then we use these results to derive the log-balancedness of a number of log-convex sequences related to derangement numbers, Domb numbers, numbers of tree-like polyhexes, numbers of walks on the cubic lattice, and so on.

1 Introduction

A sequence of positive real numbers {z_{n}}_{n≥0} is said to be log-convex (or log-concave) if z_{n}^{2} ≤ z_{n−1}z_{n+1} (or z_{n}^{2} ≥ z_{n−1}z_{n+1}) for each n ≥ 1. A log-convex sequence {z_{n}}_{n≥0} is said to be log-balanced if \{\frac{z_{n}}{n!} \}_{n≥0} is log-concave. See Došlić [4] for more details about log-balanced sequences. It is well known that {z_{n}}_{n≥0} is log-convex (or log-concave) if and only if its quotient sequence \{\frac{z_{n+1}}{z_{n}} \}_{n≥0} is nondecreasing (or nonincreasing) and a log-convex sequence \{z_{n}\}_{n≥0} is log-balanced if and only if \frac{(n+1)z_{n}}{z_{n−1}} ≥ \frac{nz_{n+1}}{z_{n}} for each n ≥ 1. It is clear that the quotient sequence of a log-balanced sequence does not grow too quickly.

In combinatorics, log-convexity and log-concavity are not only instrumental in obtaining the growth rate of a combinatorial sequence, but also important sources of inequalities. Logconvexity and log-concavity have applications in many fields such as quantum physics, white noise theory, probability, economics and mathematical biology. See, for instance [1, 2, 5, 6, 7, 12, 14, 16]. Since log-balancedness is related to log-convexity and log-concavity, it can help us to find new inequalities. Hence, the log-balancedness of various sequences deserves to be studied.

In this paper, we are interested in the log-balancedness of some combinatorial sequences. In fact, there are many log-balanced sequences in combinatorics and number theory. Došlić [4] presented some sufficient conditions for the log-balancedness of sequences satisfying threeterm linear recurrences. As consequences, a number of sequences such as the Motzkin numbers, the Fine numbers, the Franel numbers of orders 3 and 4, the Apéry numbers, the large and little Schröder numbers, and the central Delannoy numbers, are log-balanced (see Došlić [4]). Recently, Zhao [20] gave a sufficient condition for the log-balancedness of the product of a log-balanced sequence and a log-concave sequence and she also proved that the binomial transformation preserves the log-balancedness. Zhao [21, 20] showed that the sequences of the exponential numbers and the Catalan-Larcombe-French numbers are respectively logbalanced. Zhang and Zhao [18] gave some sufficient conditions for the log-balancedness of combinatorial sequences. In addition, for a log-balanced sequence {z_{n}}_{n≥0}, Zhang and Zhao [18] proved that \{\sqrt{z_{n}}\}_{n≥0} is still log-balanced.

This paper is devoted to the study of log-balancedness of some combinatorial sequences and is organized as follows. In Section 2, we give some new sufficient conditions for logbalancedness. In Section 3, using these new results, we investigate the log-balancedness of a series of log-convex sequences.

2 Sufficient conditions for log-balancedness

Zhang and Zhao [18] proved that the sequence of the arithmetic square root of a logbalanced sequence is still log-balanced. For a log-convex sequence {z_{n}}_{n≥0}, here we prove that \{\sqrt[r]{z_{n}}\}_{n≥0} is log-balanced under some conditions, where r is a fixed positive real number.

Theorem 1. Let {z_{n}}_{n≥0} be a log-convex sequence and r be a fixed positive real number. For n ≥ 0, let x_{n} = \frac{z_{n+1}}{z_{n}}. If there exists a nonnegative integer N_{r} such that

(n + 2)^{r}x_{n} − (n + 1)^{r}x_{n+1} ≥ 0, n ≥ N_{r},

the sequence \{\sqrt[r]{z_{n}}\}_{n≥N_{r}} is log-balanced.

Proof. Since the sequence \{z_{n}\}_{n≥0} is log-convex, \{\sqrt[r]{z_{n}}\}_{n≥0} is also log-convex. In order to prove the log-balancedness of \{\sqrt[r]{z_{n}}\}_{n≥N_{r}}, it is sufficient to show that the sequence \{ \frac{\sqrt[r]{z_{n}}}{n!} \}_{n≥N_{r}} is log-concave. In fact, it is clear that \{ \frac{\sqrt[r]{z_{n}}}{n!} \}_{n≥N_{r}} is log-concave if and only if \frac{\sqrt[r]{x_{n}}}{n+1} ≥ \frac{\sqrt[r]{x_{n+1}}}{n+2} for every n ≥ N_{r}. It follows from (n + 2)^{r}x_{n} − (n + 1)^{r}x_{n+1} ≥ 0 that \frac{\sqrt[r]{x_{n}}}{n+1} ≥ \frac{\sqrt[r]{x_{n+1}}}{n+2} . Hence the sequence \{ \frac{\sqrt[r]{z_{n}}}{n!} \}_{n≥N_{r}} is log-concave. Therefore, \{\sqrt[r]{z_{n}}\}_{n≥N_{r}} is log-balanced.

Theorem 2. Suppose that a and b are positive real numbers with b < a and {z_{n}}_{n≥0} is a log-convex sequence. If the sequence {z_{n}^{a}}_{n≥0} is log-balanced, then so is the sequence {z_{n}^{b}}_{n≥0}.

Proof. Since the sequence {z_{n}^{a}}_{n≥0} is log-balanced, we have

\frac{n}{n + 1}z_{n−1}^{a}z_{n+1}^{a} ≤ z_{n}^{2a} ≤ z_{n−1}^{a}z_{n+1}^{a}.

Then we derive

(\frac{n}{n + 1})^{\frac{1}{a}} z_{n−1}z_{n+1} ≤ z_{n}^{2} ≤ z_{n−1}z_{n+1},

(\frac{n}{n + 1})^{\frac{b}{a}} z_{n−1}^{b}z_{n+1}^{b} ≤ z_{n}^{2b} ≤ z_{n−1}^{b}z_{n+1}^{b}.

Since 0 < \frac{b}{a} < 1 and 0 < \frac{n}{n+1} < 1, we have (\frac{n}{n+1})^{\frac{b}{a}} ≥ \frac{n}{n+1} and hence

\frac{n}{n + 1}z_{n−1}^{b}z_{n+1}^{b} ≤ z_{n}^{2b} ≤ z_{n−1}^{b}z_{n+1}^{b}.

It follows from the definition of log-balancedness that the sequence {z_{n}^{b}}_{n≥0} is log-balanced.

In Theorem 2, if the condition “b < a” is replaced by “b > a”, the conclusion is not valid in general. For example, the sequence {nn!}_{n≥2} is log-balanced, but {(nn!)^{2}}_{n≥2} is not log-balanced.

In the next section, we will use the results of Theorems 1–2 to derive log-balancedness of a series of sequences.

Theorem 3. Let {z_{n}}_{n≥0} be a log-concave sequence. If the sequence {n!z_{n}}_{n≥0} is logbalanced, then so is the sequence \{n!\sqrt{z_{n}}\}_{n≥0}.

Proof. Since the sequence \{z_{n}\}_{n≥0} is log-concave, then so is the sequence \{\sqrt{z_{n}}\}_{n≥0}. In order to prove the log-balancedness of \{n!\sqrt{z_{n}}\}_{n≥0}, we only need to show that \{n!\sqrt{z_{n}}\}_{n≥0} is log-convex. Since {n!z_{n}}_{n≥0} is log-balanced, we get

nz_{n}^{2} − (n + 1)z_{n−1}z_{n+1} ≤ 0,

z_{n} ≤ \sqrt{\frac{n + 1}{n}z_{n−1}z_{n+1}} < \frac{n + 1}{n}\sqrt{z_{n−1}z_{n+1}},

(n!\sqrt{z_{n}})^{2} ≤ (n − 1)!(n + 1)!\sqrt{z_{n−1}z_{n+1}}.

Hence \{n!\sqrt{z_{n}}\}_{n≥0} is log-convex.

3 Log-balancedness of some sequences

In this section, we discuss the log-balancedness of a number of log-convex sequences involving many combinatorial numbers.

3.1 The derangement numbers

The derangement numbers d_{n} (sequence \underline{A000166} in the OEIS) count the number of permutations of n elements with no fixed points. The sequence {d_{n}}_{n≥0} satisfies the recurrence

d_{n+1} = n(d_{n} + d_{n−1}), n ≥ 1,  (1)

with d_{0} = 1, d_{1} = 0, d_{2} = 1, d_{3} = 2 and d_{4} = 9; see Table 1 for some information about it. In particular, Liu and Wang [11] proved that {d_{n}}_{n≥2} is log-convex.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ d_{n} & 1 & 0 & 1 & 2 & 9 & 44 & 265 & 1854 & 14833 \end{matrix}

Table 1: Some initial values of {d_{n}}_{n≥0}

Theorem 4. For r ≥ 2, the sequence \{\sqrt[r]{d_{n}}\}_{n≥3} is log-balanced.

Proof. We first prove that the sequence \{\sqrt{d_{n}}\}_{n≥3} is log-balanced.

For n ≥ 0, let x_{n} = \frac{d_{n+1}}{d_{n}}. We prove by induction that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 3,

where λ_{n} = \frac{2n+1}{2}. It follows from (1) that

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

It is clear that λ_{3} ≤ x_{3} ≤ λ_{4}. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 3. By applying (2), we get

x_{k+1} − λ_{k+1} = \frac{k + 1}{x_{k}} − \frac{1}{2} and x_{k+1} − λ_{k+2} = \frac{k + 1}{x_{k}} − \frac{3}{2}.

Due to \frac{1}{λ_{k+1}} ≤ \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}} (k ≥ 3), we have

x_{k+1} − λ_{k+1} ≥ 0 and x_{k+1} − λ_{k+2} ≤ 0.

Then we derive λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 3.

By means of (2), we obtain

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

For any x ∈ (−∞, +∞), define a function

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

Then we have

f′(x) = 2(n + 2)^{2}x − (n + 1)^{3}.

Since f′(x) ≥ 0 for x ≥ \frac{(n+1)^{3}}{2(n+2)^{2}} , f is increasing on [\frac{(n+1)^{3}}{2(n+2)^{2}} , +∞). We can verify that

λ_{n} > \frac{(n+1)^{3}}{2(n+2)^{2}} . Hence, f is increasing on [λ_{n}, +∞). Note that

f(λ_{n}) = (n + 2)^{2}λ_{n}^{2} − (n + 1)^{3}λ_{n} − (n + 1)^{3}
= \frac{2n^{3} + 3n^{2} − 2n − 2}{4} > 0 (n ≥ 1).

Then we have f(x_{n}) > 0 for n ≥ 3. This implies that (n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} ≥ 0 for n ≥ 3. It follows from Theorem 1 that the sequence \{\sqrt{d_{n}}\}_{n≥3} is log-balanced. For r > 2, it follows from Theorem 2 that \{\sqrt[r]{d_{n}}\}_{n≥3} is log-balanced.

3.2 Numbers counting tree-like polyhexes

Let h_{n} denote the number of tree-like polyhexes with n + 1 hexagons (Harary and Read [10]); it is sequence \underline{A002212} in the OEIS. It is well known that h_{n} is equal to the number of lattice paths, from (0, 0) to (2n, 0) with steps (1, 1), (1, −1) and (2, 0), never falling below the x-axis and with no peaks at odd level. The sequence {h_{n}}_{n≥0} satisfies the recurrence

(n + 1)h_{n} = 3(2n − 1)h_{n−1} − 5(n − 2)h_{n−2}, n ≥ 2,  (3)

with h_{0} = h_{1} = 1, h_{2} = 3 and h_{3} = 10; see Table 2 for some information about it. In particular, Liu and Wang [11] showed that the sequence {h_{n}}_{n≥0} is log-convex.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ h_{n} & 1 & 1 & 3 & 10 & 36 & 137 & 543 & 2219 \end{matrix}

Table 2: Some initial values of {h_{n}}_{n≥0}

Theorem 5. For r ≥ 1, the sequence \{\sqrt[r]{h_{n}}\}_{n≥1} is log-balanced.

Proof. In order to prove that \{\sqrt[r]{h_{n}}\}_{n≥1} is log-balanced for r ≥ 1, we only need to show that {h_{n}}_{n≥1} is log-balanced by Theorem 2.

For n ≥ 0, put x_{n} = \frac{h_{n+1}}{h_{n}}. We next prove by induction that

λ_{n} ≤ x_{n} ≤ μ_{n}, n ≥ 0,

where λ_{n} = \frac{10n+3}{2n+4} and μ_{n} = \frac{5n+4}{n+1}. It follows from (3) that

x_{n} = \frac{3(2n + 1)}{n + 2} − \frac{5(n − 1)}{(n + 2)x_{n−1}} , n ≥ 1,  (4)

It is easy to find that λ_{0} ≤ x_{0} ≤ μ_{0}. Assume that λ_{k} ≤ x_{k} ≤ μ_{k} for k ≥ 0. By using (4), we derive

x_{k+1} − λ_{k+1} = \frac{3(2k + 3)}{k + 3} − \frac{10k + 13}{2k + 6} − \frac{5k}{(k + 3)x_{k}}

and

x_{k+1} − μ_{k+1} = \frac{3(2k + 3)}{k + 3} − \frac{5k + 9}{k + 2} − \frac{5k}{(k + 3)x_{k}} .

Since \frac{1}{μ_{k}} ≤ \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}} (k ≥ 0), we have

x_{k+1} − λ_{k+1} ≥ \frac{2k + 5}{2(k + 3)} − \frac{5k}{(k + 3)λ_{k}}
= \frac{16k + 15}{2(k + 3)(10k + 3)} ≥ 0

and

x_{k+1} − μ_{k+1} ≤ \frac{k^{2} − 3k − 9}{(k + 2)(k + 3)} − \frac{5k}{(k + 3)μ_{k}}
= \frac{−26k^{2} − 67k − 36}{(k + 2)(k + 3)(5k + 4)} ≤ 0.

Then we have λ_{n} ≤ x_{n} ≤ μ_{n} for n ≥ 0.

By means of (4), we obtain

(n + 2)x_{n} − (n + 1)x_{n+1} = \frac{(n + 2)(n + 3)x_{n}^{2} − 3(n + 1)(2n + 3)x_{n} + 5n(n + 1)}{(n + 3)x_{n}}.

For any x ∈ (−∞, +∞), define a function

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

It is clear that

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

We note that f is increasing on [\frac{3(n+1)(2n+3)}{2(n+2)(n+3)}, +∞]. We find that λ_{n} > \frac{3(n+1)(2n+3)}{2(n+2)(n+3)} and

f(λ_{n}) = \frac{84n^{2} − 41n − 27}{4(n + 2)} > 0 (n ≥ 1).

Then we have (n + 2)x_{n} − (n + 1)x_{n+1} > 0 for n ≥ 1. Hence {h_{n}}_{n≥1} is log-balanced.

3.3 Numbers counting walks on the cubic lattice

Consider the sequence {w_{n}}_{n≥0} counting the number of walks on the cubic lattice with n steps, starting and finishing on the xy plane and never going below it (Guy [9]); it is sequence \underline{A005572} in the OEIS. The sequence \{w_{n}\}_{n≥0} satisfies the recurrence

(n + 2)w_{n} = 4(2n + 1)w_{n−1} − 12(n − 1)w_{n−2}, n ≥ 2,  (5)

where w_{0} = 1, w_{1} = 4 and w_{2} = 17; see Table 3 for some information about it. In particular, Liu and Wang [11] showed that {w_{n}}_{n≥0} is log-convex.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ w_{n} & 1 & 4 & 17 & 76 & 354 & 1704 & 8421 \end{matrix}

Table 3: Some initial values of {w_{n}}_{n≥0}

Theorem 6. Let r be a positive real number. For r ≥ 1, the sequence \{\sqrt[r]{w_{n}}\}_{n≥0} is logbalanced. For \frac{5}{6} < r < 1, there exists a positive integer N_{r} such that \{\sqrt[r]{w_{n}}\}_{n≥N_{r}} is logbalanced.

Proof. For n ≥ 0, let x_{n} = \frac{w_{n+1}}{w_{n}}. Now we prove by induction that

λ_{n} ≤ x_{n} ≤ μ_{n}, n ≥ 0,  (6)

where λ_{n} = \frac{6n+13}{n+4} and μ_{n} = \frac{6(n+3)}{n+4}. It follows from (5) that

x_{n} = \frac{4(2n + 3)}{n + 3} − \frac{12n}{(n + 3)x_{n−1}} , n ≥ 1,  (7)

We observe that λ_{k} ≤ x_{k} ≤ μ_{k} for k = 0, 1, 2. Assume that λ_{k} ≤ x_{k} ≤ μ_{k} for k ≥ 2. By using (7), we have

x_{k+1} − λ_{k+1} = \frac{4(2k + 5)}{k + 4} − \frac{12(k + 1)}{(k + 4)x_{k}} − \frac{6k + 19}{k + 5}

and

x_{k+1} − μ_{k+1} = \frac{4(2k + 5)}{k + 4} − \frac{12(k + 1)}{(k + 4)x_{k}} − \frac{6(k + 4)}{k + 5}.

Since \frac{1}{μ_{k}} ≤ \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}}, we derive

x_{k+1} − λ_{k+1} > \frac{4(2k + 5)}{k + 4} − \frac{12(k + 1)}{(k + 4)λ_{k}} − \frac{6k + 19}{k + 5}
= \frac{8k^{2} + 17k + 72}{(k + 4)(k + 5)(6k + 13)} > 0

and

x_{k+1} − μ_{k+1} < \frac{4(2k + 5)}{k + 4} − \frac{12(k + 1)}{(k + 4)μ_{k}} − \frac{6(k + 4)}{k + 5}
= −\frac{2k^{2} + 18k + 28}{(k + 3)(k + 4)(k + 5)} < 0.

Hence we have λ_{n} ≤ x_{n} ≤ μ_{n} for n ≥ 0.

By applying (6), we obtain

(n + 2)x_{n} − (n + 1)x_{n+1} ≥ (n + 2)λ_{n} − (n + 1)μ_{n+1}
= \frac{n^{2} + 7n + 34}{(n + 4)(n + 5)} > 0 (n ≥ 0).

Then \{w_{n}\}_{n≥0} is log-balanced. For r > 1, it follows from Theorem 2 that \{\sqrt[r]{w_{n}}\}_{n≥0} is log-balanced.

For \frac{5}{6} < r < 1, by using (6), we get

(n + 2)^{r}x_{n} − (n + 1)^{r}x_{n+1} ≥ \frac{(n + 2)^{r}(n + 5)(6n + 13) − 6(n + 1)^{r}(n + 4)^{2}}{(n + 4)(n + 5)} .

It is obvious that (n + 2)^{r}(n + 5)(6n + 13) ≥ 6(n + 1)^{r}(n + 4)^{2} if and only if

r ln(n + 2) − r ln(n + 1) + ln(6n^{2} + 43n + 65) − ln(6n^{2} + 48n + 96) ≥ 0.

We note that

r ln(n + 2) − r ln(n + 1) + ln(6n^{2} + 43n + 65) − ln(6n^{2} + 48n + 96)

= r ln (1 + \frac{1}{n + 1}) − ln (1 + \frac{5n + 31}{6n^{2} + 43n + 65}).

Due to \frac{x}{1+x} < ln(1 + x) < x for x > 0, we have

r ln(n + 2) − r ln(n + 1) + ln(6n^{2} + 43n + 65) − ln(6n^{2} + 48n + 96)

> \frac{(6r − 5)n^{2} + (43r − 41)n + 65r − 62}{(n + 2)(6n^{2} + 43n + 65)}.

Since

lim_{n→+∞}[(6r − 5)n^{2} + (43r − 41)n + 65r − 62] = +∞,

there exists a positive integer N_{r} such that (6r − 5)n^{2} + (43r − 41)n + 65r − 62 > 0 for n ≥ N_{r}. Then the sequence \{\sqrt[r]{w_{n}}\}_{n≥N_{r}} is log-balanced for \frac{5}{6} < r < 1.

3.4 Numbers counting a class of arrays

For an integer r ≥ 0, let Q(n, r) denote the number of arrays (or matrices) of integers a_{i,j} ≥ 0 (1 ≤ i, j ≤ n) such that

∑_{i=1}^{n} a_{i,j} = ∑_{j=1}^{n} a_{i,j} = r

holds for all i and j. Consider the sequence {A_{n}}_{n≥0}, where A_{n} = Q(n, 2). The sequence {A_{n}}_{n≥0} satisfies the recurrence

A_{n+1} = (n + 1)^{2}A_{n} − n\binom{n + 1}{2}A_{n−1}, n ≥ 1,  (8)

where A_{0} = A_{1} = 1 and A_{2} = 3; see Table 4 for some information about it. It is sequence \underline{A000681} in the OEIS. In particular, Zhao [19] proved that the sequence \{A_{n}\}_{n≥1} is logconvex (it is clear that {A_{n}}_{n≥0} is also log-convex). See [3] for more properties of {A_{n}}_{n≥0}.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ A_{n} & 1 & 1 & 3 & 21 & 282 & 6210 & 202410 \end{matrix}

Table 4: Some initial values of {A_{n}}_{n≥0}

Theorem 7. For r ≥ 5, the sequence \{\sqrt[r]{A_{n}}\}_{n≥0} is log-balanced.

Proof. For n ≥ 0, set x_{n} = \frac{A_{n+1}}{A_{n}}. We next prove by induction that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 0,  (9)

where λ_{n} = n^{2}. It follows from (8) that

x_{n} = (n + 1)^{2} − \frac{n^{2}(n + 1)}{2x_{n−1}}, n ≥ 1,  (10)

It is clear that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k = 0, 1, 2. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 2. By applying (10), we get

x_{k+1} − λ_{k+1} = 2k + 3 − \frac{(k + 1)^{2}(k + 2)}{2x_{k}},
x_{k+1} − λ_{k+2} = −\frac{(k + 1)^{2}(k + 2)}{2x_{k}}.

Due to \frac{1}{λ_{k+1}} ≤ \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}} (k ≥ 2), we have

x_{k+1} − λ_{k+1} ≥ 2k + 3 − \frac{(k + 1)^{2}(k + 2)}{2λ_{k}}
= \frac{3k^{3} + 2k^{2} − 5k − 2}{2k^{2}} ≥ 0 (k ≥ 2)

and

x_{k+1} − λ_{k+2} ≤ −\frac{(k + 1)^{2}(k + 2)}{2λ_{k+1}}
= −\frac{k + 2}{2} ≤ 0.

Then we derive λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 0.

By using (9), we have

(n + 2)^{5}x_{n} − (n + 1)^{5}x_{n+1} ≥ (n + 2)^{5}λ_{n} − (n + 1)^{5}λ_{n+2}
= (n + 2)^{2}(n^{4} + 2n^{3} − 2n^{2} − 5n − 1) > 0 (n ≥ 2).

On the other hand, we note that (k + 2)^{5}x_{k} − (k + 1)^{5}x_{k+1} for k = 0, 1. Then (n + 2)^{5}x_{n} − (n + 1)^{5}x_{n+1} > 0 holds for n ≥ 0. It follows from Theorem 1 that the sequence \{\sqrt[5]{A_{n}}\}_{n≥0} is log-balanced. For r > 5, it follows from Theorem 2 that \{\sqrt[r]{A_{n}}\}_{n≥0} is log-balanced.

3.5 Numbers satisfying a three-term recurrence

Let t_{n} counting the number of integer sequences (f_{j}, …, f_{2}, f_{1}, 1, 1, g_{1}, g_{2}, …, g_{k}) with j + k + 2 = n in which every f_{i} is the sum of one or more contiguous terms immediately to its right, and g_{i} is likewise the sum of one or more contiguous terms immediately to its left; see Odlyzko [13]. Fishburn et al. [8] proved that the sequence {t_{n}}_{n≥1} satisfies the recurrence

t_{n+1} = 2nt_{n} − (n − 1)^{2}t_{n−1}, n ≥ 2,  (11)

where t_{1} = t_{2} = 1 and t_{3} = 3; see Table 5 for some information about it. It is sequence \underline{A005189} in the OEIS. Zhao [19] showed that the sequence \{t_{n}\}_{n≥1} is log-convex.

Theorem 8. For r ≥ 3, the sequence \{\sqrt[r]{t_{n}}\}_{n≥3} is log-balanced.

\begin{matrix}n & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ t_{n} & 1 & 1 & 3 & 14 & 85 & 626 & 5387 \end{matrix}

Table 5: Some initial values of {t_{n}}_{n≥0}

Proof. For n ≥ 0, put x_{n} = \frac{t_{n+1}}{t_{n}}. We first prove by induction that

λ_{k} ≤ x_{k} ≤ μ_{k}, k ≥ 3,  (12)

where λ_{k} = k + \sqrt{k} − \frac{1}{4} and μ_{k} = k + 1 + \sqrt{k + 1}. It follows from (11) that

x_{n} = 2n − \frac{(n − 1)^{2}}{x_{n−1}} , n ≥ 2,  (13)

It is clear that λ_{k} ≤ x_{k} ≤ μ_{k} for k = 3, 4. Assume that λ_{k} ≤ x_{k} ≤ μ_{k} for k ≥ 4. By using (13), we have

x_{k+1} − λ_{k+1} = k − \sqrt{k + 1} + \frac{5}{4} − \frac{k^{2}}{x_{k}} and x_{k+1} − μ_{k+1} = k − \sqrt{k + 2} − \frac{k^{2}}{x_{k}} .

Since \frac{1}{μ_{k}} ≤ \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}} for k ≥ 4, we get

x_{k+1} − λ_{k+1} ≥ k + \frac{5}{4} − \sqrt{k + 1} − \frac{k^{2}}{λ_{k}}
= \frac{1}{λ_{k}} (k\sqrt{k} + k + \frac{5\sqrt{k}}{4} + \frac{\sqrt{k + 1}}{4} − \frac{5}{16} − k\sqrt{k + 1} − \sqrt{k(k + 1)})
> \frac{1}{λ_{k}} (\sqrt{k} + k − \sqrt{k(k + 1)} − \frac{5}{16})
> \frac{k + 1 − \sqrt{k(k + 1)}}{λ_{k}} > 0

and

x_{k+1} − μ_{k+1} ≤ k − \sqrt{k + 2} − \frac{k^{2}}{μ_{k}}
= \frac{k + k\sqrt{k + 1} − (k + 1)\sqrt{k + 2} − \sqrt{(k + 1)(k + 2)}}{k + 1 + \sqrt{k + 1}}
≤ −\frac{\sqrt{k + 2}}{k + 1 + \sqrt{k + 1}} ≤ 0.

Then we derive λ_{k} ≤ x_{k} ≤ μ_{k} for k ≥ 3.

It follows from (12) that

(n + 2)^{3}x_{n} − (n + 1)^{3}x_{n+1} ≥ (n + 2)^{3}(n + \sqrt{n} − \frac{1}{4}) − (n + 1)^{3}(n + 2 + \sqrt{n + 2})
= (\frac{3}{4} − \frac{2}{\sqrt{n} + \sqrt{n + 2}})n^{3} + \frac{3n^{2} − 4n − 8}{2}
+3n(2(n + 2)\sqrt{n} − (n + 1)\sqrt{n + 2}) + 8\sqrt{n} − \sqrt{n + 2}
> 0 (n ≥ 3).

On the other hand, we can verify that (n + 2)^{3}x_{n} − (n + 1)^{3}x_{n+1} > 0 for 1 ≤ n ≤ 2. Hence the sequence \{\sqrt[3]{t_{n}}\}_{n≥1} is log-balanced. For r > 3, it follows from Theorem 2 that \{\sqrt[r]{t_{n}}\}_{n≥1} is log-balanced.

3.6 Numbers counting bipermutations

For a given nonnegative integer k, a relation ℜ is called a k-permutation of [n] = {1, 2, …, n} if all vertical sections and all horizontal sections have k elements. The k-permutation ℜ is called a bipermutation when k = 2. Let P (n, k) denote the number of these relations. Let P_{n} = P (n, 2). The sequence {P_{n}} satisfies the recurrence

P_{n+1} = \binom{n + 1}{2}(2P_{n} + nP_{n−1}), n ≥ 1,  (14)

where P_{0} = 1 and P_{1} = 0; see Table 6 for some information about it. It is sequence \underline{A001499} in the OEIS. In particular, Zhao [19] showed that the sequence {P_{n}}_{n≥2} is log-convex.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ P_{n} & 1 & 0 & 1 & 6 & 90 & 2040 & 67950 \end{matrix}

Table 6: Some initial values of {P_{n}}_{n≥0}

Theorem 9. For r ≥ 3, the sequence \{\sqrt[r]{P_{n}}\}_{n≥3} is log-balanced.

Proof. For n ≥ 2, put x_{n} = \frac{P_{n+1}}{P_{n}}. It follows from (14) that

x_{k} = k(k + 1) + k\binom{k + 1}{2}\frac{1}{x_{k−1}} , k ≥ 3.  (15)

We prove that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 2,  (16)

where λ_{n} = n(n + 1). It is evident that λ_{k} < x_{k} < λ_{k+1} for k = 2, 3. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 3. By applying (15), we get

x_{k+1} − λ_{k+1} = (k + 1)\binom{k + 2}{2}\frac{1}{x_{k}} > 0

and

x_{k+1} − λ_{k+2} = −2(k + 2) + \frac{(k + 1)^{2}(k + 2)}{2x_{k}}.

Since \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}}, we have

x_{k+1} − λ_{k+2} ≤ −\frac{3k^{2} + 5k − 2}{2k} < 0.

Then we have λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 2.

It follows from (15) and (16) that

(n + 2)^{3}x_{n} − (n + 1)^{3}x_{n+1} = (n + 2)^{3}x_{n} − (n + 1)^{4}(n + 2) − \frac{(n + 1)^{5}(n + 2)}{2x_{n}}
≥ (n + 2)^{3}λ_{n} − (n + 1)^{4}(n + 2) − \frac{(n + 1)^{5}(n + 2)}{2λ_{n}}
= \frac{(n + 1)(n + 2)(n^{3} − n^{2} − 5n − 1)}{2n} > 0 (n ≥ 3).

We have from Theorem 1 that the sequence \{\sqrt[3]{P_{n}}\}_{n≥3} is log-balanced. For r > 3, it follows from Theorem 2 that \{\sqrt[r]{P_{n}}\}_{n≥3} is log-balanced.

3.7 Numbers satisfying a four-term recurrence (“minus” case)

Let G_{n} stand for the number of graphs on the vertex set [n] = {1, 2, …, n}, whose every component is a cycle, and put G_{0} = 1. The sequence {G_{n}} satisfies the recurrence

G_{n+1} = (n + 1)G_{n} − \binom{n}{2}G_{n−2}, n ≥ 2,  (17)

where G_{1} = 1, G_{2} = 2, and G_{3} = 5; see Table 7 for some information about it. It is sequence \underline{A002135} in the OEIS. This example is Exercise 5.22 of Stanley [15], and one can find its combinatorial proof in Stanley [15, p. 121]. In addition, Došlić [7] showed that the sequence {G_{n}}_{n≥0} is log-convex.

Theorem 10. For r ≥ 2, the sequence \{\sqrt[r]{G_{n}}\}_{n≥0} is log-balanced.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ G_{n} & 1 & 1 & 2 & 5 & 17 & 73 & 388 \end{matrix}

Table 7: Some initial values of {G_{n}}_{n≥0}

Proof. For n ≥ 0, let x_{n} = \frac{G_{n+1}}{G_{n}}. We next prove by induction that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 0,  (18)

where λ_{n} = n. It follows from (17) that

x_{n} = n + 1 − \binom{n}{2}\frac{1}{x_{n−1}x_{n−2}} , n ≥ 2.  (19)

Firstly, we have λ_{k} ≤ x_{k} ≤ λ_{k+1} for 0 ≤ k ≤ 4. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 4. By using (19), we have

x_{k+1} − λ_{k+2} = −\binom{k + 1}{2}\frac{1}{x_{k}x_{k−1}} < 0

and

x_{k+1} − λ_{k+1} = 1 − \binom{k + 1}{2}\frac{1}{x_{k}x_{k−1}}
≥ \frac{2k(k − 1) − k(k + 1)}{2x_{k}x_{k−1}} > 0.

Then we derive λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 0.

It follows from (17) and (18) that

(n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} = (n + 2)^{2}x_{n} − (n + 1)^{2}(n + 2) + \frac{(n + 1)^{2}\binom{n+1}{2}}{x_{n}x_{n−1}}
≥ (n + 2)^{2}λ_{n} − (n + 1)^{2}(n + 2) + \frac{(n + 1)^{2}\binom{n+1}{2}}{λ_{n+1}λ_{n}}
= \frac{n^{2} − 3}{2} > 0 (n ≥ 2).

On the other hand, we note that (k +2)^{2}x_{k} −(k +1)^{2}x_{k+1} > 0 for k = 0, 1. Then (n+2)^{2}x_{n} − (n + 1)^{2}x_{n+1} > 0 holds for n ≥ 0. We have from Theorem 1 that the sequence \{\sqrt{G_{n}}\}_{n≥0} is log-balanced. For r > 2, it follows from Theorem 2 that \{\sqrt[r]{G_{n}}\}_{n≥0} is log-balanced.

3.8 Numbers satisfying a four-term recurrence (“plus” case)

Let be given a set of Δ of n straight lines in the plane, δ_{1}, δ_{2}, …, δ_{n}, lying in general position (no two among them are parallel, and no three among are concurrent). Let P be the set of their points of intersection, |P | = \binom{n}{2}. We call any set of n points from P such that any three different points are not collinear, a cloud. Let 𝒢 (Δ) stand for the set of clouds of Δ and g_{n} = |𝒢 (Δ)|. The sequence {g_{n}}_{n≥0} satisfies the recurrence

g_{n+1} = ng_{n} + \binom{n}{2}g_{n−2}, n ≥ 2,  (20)

where g_{0} = 1, g_{1} = g_{2} = 0, g_{3} = 1, g_{4} = 3 and g_{5} = 12; see Table 8 for some information about it. It is sequence \underline{A001205} in the OEIS. In particular, Zhao [19] proved that the sequence {g_{n}}_{n≥3} is log-convex. For more properties of {g_{n}}_{n≥0}, see Comtet [3].

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ g_{n} & 1 & 0 & 0 & 1 & 3 & 12 & 70 & 465 & 3507 \end{matrix}

Table 8: Some initial values of {g_{n}}_{n≥0}

Theorem 11. For r ≥ 2, the sequence \{\sqrt[r]{g_{n}}\}_{n≥5} is log-balanced.

Proof. For n ≥ 3, let x_{n} = \frac{g_{n+1}}{g_{n}}. It follows from (20) that

x_{n} = n + \binom{n}{2}\frac{1}{x_{n−1}x_{n−2}} , n ≥ 5.  (21)

Now we show that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 3,  (22)

where λ_{n} = n. We can verify that λ_{k} ≤ x_{k} ≤ λ_{k+1} for 3 ≤ k ≤ 5. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 5. By applying (21), we get

x_{k+1} − λ_{k+1} = \binom{k + 1}{2}\frac{1}{x_{k}x_{k−1}} > 0

and

x_{k+1} − λ_{k+2} = \frac{k(k + 1)}{2x_{k}x_{k−1}} − 1
≤ −\frac{k(k − 3)}{2x_{k}x_{k−1}} < 0.

Then we have λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 3.

It follows from (21) and (22) that

(n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} = (n + 2)^{2}x_{n} − (n + 1)^{3} − \frac{(n + 1)^{2}\binom{n+1}{2}}{x_{n}x_{n−1}}
≥ (n + 2)^{2}λ_{n} − (n + 1)^{3} − \frac{(n + 1)^{2}\binom{n+1}{2}}{λ_{n}λ_{n−1}}
= \frac{n^{3} − 3n^{2} − 7n + 1}{2(n − 1)} > 0 (n ≥ 5).

We have from Theorem 1 that the sequence \{\sqrt{g_{n}}\}_{n≥5} is log-balanced. For r > 2, it follows from Theorem 2 that \{\sqrt[r]{g_{n}}\}_{n≥5} is log-balanced.

3.9 Numbers counting permutation with ordered orbits

Consider the sequence {T_{n}}_{n≥2} defined by

T_{n+1} = (n − 1)T_{n} + \frac{n!}{2}, n ≥ 2,  (23)

where T_{2} = 1; see Table 9 for some information about it. The value of T_{n} is related to the number of permutations with ordered orbits. In particular, Zhao [19] proved that the sequence \{T_{n}\}_{n≥2} is log-convex. It is sequence \underline{A006595} in the OEIS. For more properties of {T_{n}}_{n≥2}, see Comtet [3].

\begin{matrix}n & 2 & 3 & 4 & 5 & 6 & 7 \\ T_{n} & 1 & 2 & 7 & 33 & 192 & 1320 \end{matrix}

Table 9: Some initial values of {T_{n}}_{n≥0}

Theorem 12. For r ≥ 2, the sequence \{\sqrt[r]{T_{n}}\}_{n≥2} is log-balanced.

Proof. For n ≥ 2, let x_{n} = \frac{T_{n+1}}{T_{n}}. It is easy to verify that

T_{n+1} = (2n − 1)T_{n} − (n − 2)nT_{n−1}, n ≥ 3.  (24)

It follows from (24) that

x_{n} = 2n − 1 − \frac{(n − 2)n}{x_{n−1}}, n ≥ 3.  (25)

Now we prove by induction that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 2,

where λ_{n} = n. It is not difficult to verify that λ_{k} ≤ x_{k} ≤ λ_{k+1} for 2 ≤ k ≤ 4. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 4. Using (25), we have

x_{k+1} − λ_{k+1} = k − \frac{(k + 1)(k − 1)}{x_{k}}
≥ k − \frac{(k + 1)(k − 1)}{k} > 0

and

x_{k+1} − λ_{k+2} ≤ k − 1 − \frac{(k + 1)(k − 1)}{λ_{k+1}} = 0.

Then we derive that λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 2. By means of (26), we obtain

(n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} = \frac{(n^{2} + 4n + 4)x_{n}^{2} − (2n^{3} + 5n^{2} + 4n + 1)x_{n} + n^{4} + 2n^{3} − 2n − 1}{x_{n}}

For any x ∈ (−∞, +∞), define a function

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

We can prove that f is increasing on [σ_{n}, +∞), where σ_{n} = \frac{2n^{3}+5n^{2}+4n+1}{2(n^{2}+4n+4)}, f is increasing on [σ_{n}, +∞). We can verify that λ_{n} > σ_{n}. Hence f is increasing on [λ_{n}, +∞). We note that

f(λ_{n}) = (n^{2} + 4n + 4)λ_{n}^{2} − (2n^{3} + 5n^{2} + 4n + 1)λ_{n} + n^{4} + 2n^{3} − 2n − 1
= n^{3} − 3n − 1
> 0 (n ≥ 2).

Then we have f(x_{n}) > 0 for n ≥ 2. This implies that (n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} ≥ 0 for n ≥ 2. It follows from Theorem 1 that the sequence \{\sqrt{T_{n}}\}_{n≥2} is log-balanced. For r > 2, it follows from Theorem 2 that \{\sqrt[r]{T_{n}}\}_{n≥2} is log-balanced.

3.10 The Domb numbers

Let {D_{n}}_{n≥0} be the sequence of the Domb numbers. The value of D_{n} is the number of 2n-step polygons on the diamond lattice. The sequence {D_{n}}_{n≥0} satisfies the recurrence

n^{3}D_{n} = 2(2n − 1)(5n^{2} − 5n + 2)D_{n−1} − 64(n − 1)^{3}D_{n−2}, n ≥ 2,  (26)

where D_{0} = 1 and D_{1} = 4; see Table 10 for some information about it. It is sequence \underline{A002895} in the OEIS. In particular, Wang and Zhu [17] proved that the sequence \{D_{n}\}_{n≥0} is log-convex.

Theorem 13. For r ≥ 2, the sequence \{\sqrt[r]{D_{n}}\}_{n≥1} is log-balanced.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ D_{n} & 1 & 4 & 28 & 256 & 2716 & 31504 & 387136 & 4951552 \end{matrix}

Table 10: Some initial values of {D_{n}}_{n≥0}

Proof. For n ≥ 0, let x_{n} = \frac{D_{n+1}}{D_{n}}. It follows from (26) that

x_{n} = \frac{2(2n + 1)(5n^{2} + 5n + 2)}{(n + 1)^{3}} − \frac{64n^{3}}{(n + 1)^{3}x_{n−1}} , n ≥ 1.  (27)

We first show that

λ_{n} ≤ x_{n} ≤ μ_{n}, n ≥ 1,  (28)

where λ_{n} = \frac{16(n−1)}{n+1} and μ_{n} = \frac{16n}{n+1}. It is obvious that λ_{k} < x_{k} < μ_{k} for 1 ≤ k ≤ 3. Assume that λ_{k} ≤ x_{k} ≤ μ_{k} for k ≥ 3. By means of (27), we have

x_{k+1} − λ_{k+1} ≥ \frac{2(2k + 3)(5k^{2} + 15k + 12)}{(k + 2)^{3}} − \frac{16k}{k + 2} − \frac{64(k + 1)^{3}}{(k + 2)^{3}λ_{k}}
= \frac{2(3k^{3} + 12k^{2} − 9k − 38)}{(k − 1)(k + 2)^{3}} > 0 (k ≥ 3)

and

x_{k+1} − μ_{k+1} ≤ \frac{2(2k + 3)(5k^{2} + 15k + 12)}{(k + 2)^{3}} − \frac{16(k + 1)}{k + 2} − \frac{64(k + 1)^{3}}{(k + 2)^{3}μ_{k}}
= −\frac{2(3k^{3} + 7k^{2} + 4k + 2)}{k(k + 2)^{3}} < 0.

Then we have λ_{n} ≤ x_{n} ≤ μ_{n} for n ≥ 1.

It follows from (28) that

(n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} ≥ \frac{16[(n − 1)(n + 2)^{3} − (n + 1)^{4}]}{(n + 1)(n + 2)}
= \frac{16(n^{3} − 8n − 9)}{(n + 1)(n + 2)}
> 0 (n ≥ 4).

On the other hand, we observe that (n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} > 0 for 0 ≤ n ≤ 3. We have from Theorem 1 that the sequence \{\sqrt{D_{n}}\}_{n≥0} is log-balanced. For r > 2, it follows from Theorem 2 that \{\sqrt[r]{D_{n}}\}_{n≥0} is log-balanced.

3.11 Numbers counting a class of n × n symmetric matrices

Let τ_{n} denote the number of n × n symmetric ℕ_{0}-matrices with every row(and hence every column) sum equals to 2 with trace zero (i.e., all main diagonal entries are zero). The sequence {τ_{n}}_{n≥0} satisfies the recurrence

τ_{n} = (n − 1)τ_{n−1} + (n − 1)τ_{n−2} − \binom{n − 1}{2}τ_{n−3},  (29)

where τ_{0} = 1, τ_{1} = 0, τ_{2} = τ_{3} = 1; see Table 11 for some information about it. It is sequence \underline{A002137} in the OEIS. In particular, Došlić [7] showed that the sequence \{τ_{n}\}_{n≥6} is log-convex.

\begin{matrix}n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ τ_{n} & 1 & 0 & 1 & 1 & 6 & 22 & 130 & 822 & 6202 \end{matrix}

Table 11: Some initial values of {τ_{n}}_{n≥0}

Theorem 14. For r ≥ 2, the sequence \{\sqrt[r]{τ_{n}}\}_{n≥6} is log-balanced.

Proof. For n ≥ 2, set x_{n} = \frac{τ_{n+1}}{τ_{n}}. We now prove by induction that

λ_{n} ≤ x_{n} ≤ λ_{n+1}, n ≥ 6,  (30)

where λ_{n} = n. It is clear that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k = 6, 7. Assume that λ_{k} ≤ x_{k} ≤ λ_{k+1} for k ≥ 7. It follows from (29) that

x_{n} = n + \frac{n}{x_{n−1}} − \frac{n(n − 1)}{2x_{n−2}x_{n−1}} , n ≥ 4,  (31)

By applying (31), we get

x_{k+1} − λ_{k+1} = \frac{k + 1}{x_{k}} − \frac{(k + 1)k}{2x_{k−1}x_{k}} and x_{k+1} − λ_{k+2} = −1 + \frac{k + 1}{x_{k}} − \frac{(k + 1)k}{2x_{k−1}x_{k}} .

Due to \frac{1}{λ_{k+1}} ≤ \frac{1}{x_{k}} ≤ \frac{1}{λ_{k}} (k ≥ 7), we have

x_{k+1} − λ_{k+1} ≥ 1 − \frac{k + 1}{2(k − 1)}
= \frac{k − 3}{2(k − 1)} ≥ 0

and

x_{k+1} − λ_{k+2} ≤ \frac{1}{k} − \frac{1}{2}
= −\frac{k − 2}{2k} ≤ 0.

Then we derive λ_{n} ≤ x_{n} ≤ λ_{n+1} for n ≥ 6.

By using (30) and (31), we have

(n + 2)^{2}x_{n} − (n + 1)^{2}x_{n+1} = (n + 2)^{2}x_{n} − (n + 1)^{3} − \frac{(n + 1)^{3}}{x_{n}} + \frac{n(n + 1)^{3}}{2x_{n}x_{n−1}}
≥ (n + 2)^{2}λ_{n} − (n + 1)^{3} − \frac{(n + 1)^{3}}{λ_{n}} + \frac{n(n + 1)^{3}}{2λ_{n}λ_{n+1}}
≥ \frac{3n^{2} − 6n − 22}{6} > 0 (n ≥ 6).

It follows from Theorem 1 that the sequence \{\sqrt{τ_{n}}\}_{n≥6} is log-balanced. For r > 2, it follows from Theorem 2 that \{\sqrt[r]{τ_{n}}\}_{n≥6} is log-balanced.

In the rest of this section, we discuss log-balancedness of some sequences by mens of Theorem 3.

3.12 The harmonic numbers

Let {H_{n}}_{n≥1} be the sequence of harmonic numbers. It is well known that

H_{n} = ∑_{k=1}^{n} \frac{1}{k}, n ≥ 1.

Theorem 15. The sequence \{n!\sqrt{H_{n}}\}_{n≥1} is log-balanced.

Proof. Using the definition of log-concavity, one can immediately prove that {H_{n}}_{n≥1} is logconcave. Moreover, from Zhao [20], {n!H_{n}}_{n≥1} is log-balanced. It follows from Theorem 3 that the sequence \{n!\sqrt{H_{n}}\}_{n≥1} is log-balanced.

3.13 The Fibonacci and Lucas numbers

Let {F_{n}}_{n≥0} and {L_{n}}_{n≥0} denote the Fibonacci and Lucas sequence, respectively. The Binet’s forms of F_{n} and L_{n} respectively are

F_{n} = \frac{α^{n} − (−1)^{n}α^{−n}}{\sqrt{5}} , L_{n} = α^{n} + (−1)^{n}α^{−n}, n ≥ 0,

where α = \frac{1+\sqrt{5}}{2}.

Theorem 16. The sequences \{n!\sqrt{F_{2n}}\}_{n≥1} and \{n!\sqrt{L_{2n+1}}\}_{n≥1} are both log-balanced.

Proof. Using the definition of log-concavity, we can prove that the sequences {F_{2n}}_{n≥1} and {L_{2n−1}}_{n≥1} are both log-concave. Zhao [20] showed that {n!F_{2n}}_{n≥1} and {n!L_{2n+1}}_{n≥1} are log-balanced. It follows from Theorem 3 that the sequences \{n!\sqrt{F_{2n}}\}_{n≥1} and \{n!\sqrt{L_{2n+1}}\}_{n≥1} are both log-balanced.

4 Conclusions

For a log-convex sequence \{z_{n}\}_{n≥0}, we have shown that the arithmetic root sequence \{\sqrt[r]{z_{n}}\}_{n≥0} is log-balanced under suitable conditions. We have also derived the log-balancedness of a number of log-convex sequences related to many famous combinatorial numbers. However, we cannot give the minimum value of r such that \{\sqrt[r]{z_{n}}\}_{n≥0} is log-balanced. We hope to solve this question in the future work. In addition, we also hope to find more functions f defined in (−∞, +∞) such that {f(z_{n})}_{n≥0} is log-balanced.

5 Acknowledgment

The authors would like to thank the anonymous referee for his (her) many 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] F. Brenti, Log-concave and unimodal sequences in algebra, combinatorics, and geometry: An update, Contemp. Math. 178 (1994), 71–89.

[3] L. Comtet, Advanced Combinatorics, Reidel, 1974.

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

[5] T. Došlić, Seven (lattice) paths to log-convexity, Acta Appl. Math. 110 (2010), 1373– 1392.

[6] T. Došlić, D. Svrtan, and D. Veljan, Enumerative aspects of secondary structures, Discrete Math. 285 (2004), 67–82.

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

[8] P. C. Fishburn, A. M. Odlyzko, and F. S. Roberts, Two-sided generalized Fibonacci sequences, Fibonacci Quart. 27 (1989), 352–361.

[9] R. K. Guy, Catwalks, sandsteps and Pascal pyramids, J. Integer Sequences 3 (2000), Article 00.1.6.

[10] F. Harary, R. C. Read, The enumeration of tree-like polyhexes, Proc. Edinb. Math. Soc. (2) 17 (1970), 1–13.

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

[12] O. Milenkovic and K. J. Compton, Probabilistic transforms for combinatorial urn models, Combin. Probab. Comput. 13 (2004), 645–675.

[13] A. M. Odlyzko, Asymptotic enumeration methods, in Handbook of Combinatorics, Vols. 1–2, Elsevier, 1995, pp. 1063–1229,

[14] D. Prelec, Decreasing impatience: A criterion for non-stationary time preference and “hyperbolic” discounting, Scand. J. Econ. 106 (2004), 511–532.

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

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

[17] Y. Wang and B. X. Zhu, Proofs of some conjectures on monotonicity of number-theoretic and combinatorial sequences, Sci. China Math. 57 (2014), 2429–2435.

[18] T. T. Zhang and F. Z. Zhao, Some sufficient conditions for log-balancedness of combinatorial sequences, J. Integer Sequences 19 (2016), Article 16.6.8.

[19] F. Z. Zhao, Log-convexity of some recurrence sequences, J. Indian Math. Soc. (N.S.) 82 (2015), 207–224.

[20] F. Z. Zhao, The log-balancedness of combinatorial sequences, Sarajevo J. Math. 11 (2015), 141–154.

[21] F. Z. Zhao, The log-behavior of the Catalan-Larcombe-French sequence, Int. J. Number Theory. 10 (2014), 177–182.

2010 Mathematics Subject Classification: Primary 05A20; Secondary 11B37, 11B83, 11B39. Keywords: log-convexity, log-concavity, log-balancedness.

(Concerned with sequences \underline{A000166}, \underline{A000681}, \underline{A001205}, \underline{A001499}, \underline{A002135}, \underline{A002137}, \underline{A002212}, \underline{A002895}, \underline{A005189}, \underline{A005572}, and \underline{A006595}.)

Received December 9 2017; revised versions received March 16 2018; June 3 2018; June 27 2018. Published in Journal of Integer Sequences, June 29 2018.

Return to Journal of Integer Sequences home page.