# Two-log-convexity of the Catalan-Larcombe-French sequence

## Abstract

The Catalan-Larcombe-French sequence $$\{P_{n}\}_{n\geq0}$$ arises in a series expansion of the complete elliptic integral of the first kind. It has been proved that the sequence is log-balanced. In the paper, by exploring a criterion due to Chen and Xia for testing 2-log-convexity of a sequence satisfying three-term recurrence relation, we prove that the new sequence $$\{P^{2}_{n}-P_{n-1}P_{n+1}\}_{n\geq1}$$ are strictly log-convex and hence the Catalan-Larcombe-French sequence is strictly 2-log-convex.

## 1 Introduction

This paper is concerned with the log-behavior of the Catalan-Larcombe-French sequence. To begin with, let us recall that a sequence $$\{z_{n}\}_{n\geq0}$$ is said to be log-concave if

$$z_{n}^{2}\geq z_{n+1}z_{n-1}, \quad\mbox{for } n\geq1,$$
(1.1)

and it is log-convex if

$$z_{n}^{2}\leq z_{n+1}z_{n-1}, \quad\mbox{for } n\geq1.$$
(1.2)

Meanwhile, the sequence $$\{z_{n}\}_{n\geq0}$$ is called strictly log-concave (resp. log-convex) if the inequality in (1.1) (resp. (1.2)) is strict for all $$n\geq1$$. We call $$\{z_{n}\}_{n\geq0}$$ log-balanced if the sequence itself is log-convex while $$\{\frac{z_{n}}{n!}\}_{n\geq0}$$ is log-concave.

Given a sequence $$A=\{z_{n}\}_{n\geq0}$$, define the operator $$\mathcal {L}$$ by

$$\mathcal{L}(A)=\{s_{n}\}_{n\geq0},$$

where $$s_{n}=z_{n-1}z_{n+1}-z_{n}^{2}$$ for $$n\geq1$$. We say that $$\{z_{n}\} _{n\geq0}$$ is k-log-convex (resp. k-log-concave) if $$\mathcal{L}^{j}(A)$$ is log-convex (resp. log-concave) for all $$j=0,1,\ldots,k-1$$, and that $$A=\{z_{n}\}_{n\geq0}$$ is âˆž-log-convex (resp. âˆž-log-concave) if $$\mathcal{L}^{k}(A)$$ is log-convex (resp. log-concave) for any $$k\geq0$$. Similarly, we can define strict k-log-concavity or strict k-log-convexity of a sequence.

It is worthy to mention that besides that they are fertile sources of inequalities, log-convexity and log-concavity have many applications in some different mathematical disciplines, such as geometry, probability theory, combinatorics, and so on. See the surveys due to Brenti [1] and Stanley [2] for more details. Additionally, it is clear that the log-balancedness implies the log-convexity and a sequence $$\{z_{n}\}_{n\geq0}$$ is log-convex (resp. log-concave) if and only if its quotient sequence $$\{\frac {z_{n}}{z_{n-1}}\}_{n\geq1}$$ is nondecreasing (resp. nonincreasing). It is also known that the quotient sequence of a log-balanced sequence does not grow too fast. Therefore, log-behavior are important properties of combinatorial sequences and they are instrumental in obtaining the growth rate of a sequence. Hence the log-behaviors of a sequence deserves to be investigated.

In this paper, we investigate the 2-log-behavior of the Catalan-Larcombe-French sequence, denoted by $$\{P_{n}\}_{n\geq0}$$, which arises in connection with series expansions of the complete elliptic integrals of the first kind [3, 4]. To be precise, for $$0<|c|<1$$,

$$\int_{0}^{\pi/2}\frac{1}{\sqrt{1-c^{2}\sin^{2}{\theta}}}\,d\theta= \frac{\pi}{2} \sum_{n=0}^{\infty}\biggl( \frac{1-\sqrt{1-c^{2}}}{16} \biggr)^{n}P_{n}.$$

Furthermore, the numbers $$P_{n}$$ can be written as the following sum:

$$P_{n}=2^{n}\sum_{k=0}^{n} (-4)^{i}\binom{n-k}{k}\binom{2n-2k}{n-k}^{2},$$

see [5], A05317. Besides, the number $$P_{n}$$ satisfies three-term recurrence relations [4] as follows:

$$(n+1)^{2} P_{n+1}=8\bigl(3n^{2}+3n+1 \bigr)P_{n}-128n^{2}P_{n-1}, \quad\mbox{for } n\geq1,$$
(1.3)

with the initial values $$P_{0}=1$$ and $$P_{1}=8$$.

Recently, Zhao [4] studied the log-behavior of the Catalan-Larcombe-French sequence and proved that the sequence $$\{P_{n}\} _{n\geq0}$$ is log-balanced. What is more, the Catalan-Larcombe-French sequence has many interesting properties and the reader can refer [3, 4, 6]. In the sequel, we study the 2-log-behavior of the sequences and obtain the following result.

### Theorem 1.1

The Catalan-Larcombe-French sequence $$\{P_{n}\}_{n\geq0}$$ is strictly 2-log-convex, that is,

$$\mathcal{P}_{n}^{2}< \mathcal{P}_{n-1} \mathcal{P}_{n+1},$$
(1.4)

where $$\mathcal{P}_{n}=P_{n}^{2}-P_{n-1}P_{n+1}$$.

We will give our proof of Theorem 1.1 in the third section by utilizing a testing criterion, which is proposed by Chen and Xia [7].

To make this paper self-contained, let us recall their criterion.

### Theorem 1.2

(Chen and XiaÂ [7])

Suppose $$\{z_{n}\}_{n\geq0}$$ is a positive log-convex sequence that satisfies the following three-term recurrence relation:

$$z_{n}=a(n)z_{n-1}+b(n)z_{n-2}, \quad\textit{for } n\geq2.$$
(1.5)

Let

\begin{aligned}& c_{0}(n)= -b^{2}(n+1)\bigl[a^{2}(n+2)+b(n+1)-a(n+2)a(n+3)-b(n+3) \bigr]; \\& c_{1}(n)= b(n+1)\bigl[2a(n+2)b(n+1)+2a(n+3)a(n+2)a(n+1) \\& \hphantom{c_{1}(n)=}{} +a(n+3)b(n+2)+2a(n+1)b(n+3)-2a^{2}(n+2)a(n+1) \\& \hphantom{c_{1}(n)=}{} -2a(n+2)b(n+2)-3a(n+1)b(n+1)\bigr]; \\& c_{2}(n)= 4a(n+1)a(n+2)b(n+1)+2b(n+1)b(n+2)+a^{2}(n+1)a(n+2)a(n+3) \\& \hphantom{c_{1}(n)=}{} +a(n+1)a(n+3)b(n+2)+a^{2}(n+1)b(n+3)-3a^{2}(n+1)b(n+1) \\& \hphantom{c_{1}(n)=}{} -a(n+3)a(n+2)b(n+1)-a^{2}(n+2)a^{2}(n+1)-b(n+3)b(n+1) \\& \hphantom{c_{1}(n)=}{} -2a(n+2)a(n+1)b(n+2)-b^{2}(n+2); \\& c_{3}(n)= 2a^{2}(n+1)a(n+2)+2a(n+1)b(n+2)-a(n+1)b(n+3)-a^{3}(n+1) \\& \hphantom{c_{1}(n)=}{} -a(n+1)a(n+2)a(n+3)-a(n+3)b(n+2); \end{aligned}

and

$$\Delta(n)=4c_{2}^{2}(n)-12c_{1}(n)c_{3}(n).$$

Assume that $$c_{3}(n)<0$$ and $$\Delta(n)\geq0$$ for all $$n\geq N$$, where N is a positive integer. If there exist $$f_{n}$$ and $$g_{n}$$ such that, for all $$n\geq N$$,

1. (I)

$$f_{n}\leq\frac{z_{n}}{z_{n-1}}\leq g_{n}$$;

2. (II)

$$f_{n}\geq\frac{-2c_{2}(n)-\sqrt{\Delta(n)}}{6c_{3}(n)}$$;

3. (III)

$$c_{3}(n)g_{n}^{3}+c_{2}(n)g_{n}^{2}+c_{1}(n)g_{n}+c_{0}(n)\geq0$$,

then we see that $$\{z_{n}\}_{n\geq N}$$ is 2-log-convex, that is, for $$n\geq N$$,

$$\bigl(z_{n-1}z_{n+1}-z_{n}^{2}\bigr) \bigl(z_{n+1}z_{n+3}-z_{n+2}^{2}\bigr)> \bigl(z_{n}z_{n+2}-z_{n+1}^{2} \bigr)^{2}.$$

With respect to the theory in this field, it should be mentioned that the log-behavior of a sequence which satisfies a three-term recurrence has been extensively studied; see Liu and Wang [8], Chen et al. [9, 10], Liggett [11], DoÅ¡liÄ‡ [12], etc.

## 2 Bounds for $$\frac{P_{n}}{P_{n-1}}$$

Before proving Theorem 1.1, we need the following two lemmas.

### Lemma 2.1

Let

$$f_{n}=\frac{232n}{15(n+2)},$$

and $$P_{n}$$ be the sequence defined by the recurrence relation (1.3). Then we have, for all $$n\geq1$$,

$$\frac{P_{n}}{P_{n-1}}> f_{n}.$$
(2.1)

### Proof

We proceed the proof by induction. First note that, for $$n=1$$ and $$n=2$$, we have $$\frac{P_{1}}{P_{0}}=8>\frac{232}{45}$$ and $$\frac{P_{2}}{P_{1}}=10>\frac{464}{60}$$. Assume that the inequality (2.1) is valid for $$n\leq k$$. We will show that

$$\frac{P_{k+1}}{P_{k}}> f_{k+1}.$$

By the recurrence (1.3), we have

\begin{aligned} \frac{P_{k+1}}{P_{k}} =&\frac{8(3k^{2}+3k+1)}{(k+1)^{2}}- \frac{128k^{2}}{(k+1)^{2}}\frac{P_{k-1}}{P_{k}} >\frac{8(3k^{2}+3k+1)}{(k+1)^{2}}-\frac{128k^{2}}{(k+1)^{2}}\frac{1}{f_{k}} \\ =&\frac{8(57k^{2}+27k+29)}{29(k+1)^{2}} \\ >&f_{k+1}, \end{aligned}

in which the last inequality follows by

\begin{aligned} \frac{8(57k^{2}+27k+29)}{29(k+1)^{2}}-f_{k+1} =&\frac {8(14k^{3}+447k^{2}-873k+464)}{435(k+1)^{2}(k+3)}\\ >&0, \end{aligned}

for all $$k\geq1$$. This completes the proof.â€ƒâ–¡

### Lemma 2.2

Let

$$g_{n}=16-\frac{16}{n}-\frac{16}{n^{3}},$$

and $$P_{n}$$ be the sequence defined by the recurrence relation (1.3). Then we have, for all $$n\geq6$$,

$$\frac{P_{n}}{P_{n-1}}\leq g_{n}.$$
(2.2)

### Proof

First note that, for $$n=6$$, we have $$\frac{P_{6}}{P_{5}}=\frac{3\text{,}562}{269}< g_{6}=\frac{358}{27}$$. Assume that, for $$k\geq6$$, the inequality (2.2) is valid for $$n\leq k$$. We will show that

$$\frac{P_{k+1}}{P_{k}}< g_{k+1}.$$

By the recurrence (1.3), we have

\begin{aligned} \frac{P_{k+1}}{P_{k}} =&\frac{8(3k^{2}+3k+1)}{(k+1)^{2}}- \frac{128k^{2}}{(k+1)^{2}}\frac{P_{k-1}}{P_{k}} < \frac{8(3k^{2}+3k+1)}{(k+1)^{2}}-\frac{128k^{2}}{(k+1)^{2}}\frac{1}{g_{k}} \\ =&\frac{8(2k^{5}-2k^{3}-4k^{2}-3k-1)}{(k+1)^{2}(k^{3}-k^{2}-1)}. \end{aligned}
(2.3)

Consider

$$\frac{8(2k^{5}-2k^{3}-4k^{2}-3k-1)}{(k+1)^{2}(k^{3}-k^{2}-1)}-g_{k+1} = -\frac{8(5k^{2}+2k+3)}{(k+1)^{3}(k^{3}-k^{2}-1)}< 0,$$
(2.4)

for all $$k\geq2$$. So we see that, for all $$n\geq6$$, the inequality (2.2) holds by induction.â€ƒâ–¡

With the above lemmas in hand, we are now in a position to prove our main result in the next section.

## 3 Proof of Theorem 1.1

In this section, by using the criterion of Theorem 1.2, we can show that the Catalan-Larcombe-French sequence is strictly 2-log-convex.

To begin with, the following lemma, which is obtained by Zhao [4], is indispensable for us.

### Lemma 3.1

(Zhao [4])

The Catalan-Larcombe-French sequence is log-balanced.

By the definition of log-balanced sequence, we know that $$\{P_{n}\} _{n\geq0}$$ is log-convex.

### Proof of Theorem 1.1

By Lemma 3.1, it suffices for us to show that

$$\bigl(P_{n-1}P_{n+1}-P_{n}^{2}\bigr) \bigl(P_{n+1}P_{n+3}-P_{n+2}^{2}\bigr)- \bigl(P_{n}P_{n+2}-P_{n+1}^{2} \bigr)^{2}>0.$$

According to the recurrence relation (1.3), we see that

\begin{aligned}& a(n)= \frac{8(3n^{2}-3n+1)}{n^{2}}; \\& b(n)= -\frac{128(n-1)^{2}}{n^{2}}. \end{aligned}

By taking $$a(n)$$, $$b(n)$$ in $$c_{0},\ldots,c_{3}$$, we can obtain

\begin{aligned} c_{3}(n) =&-\frac{512}{(n+1)^{6}(n+2)^{2}(n+3)^{2}} \\ &{}\times \bigl(3n^{8}+5n^{7}-27n^{6}-32n^{5}+112n^{4}+234n^{3}+177n^{2}+63n+9\bigr) \\ < &0, \end{aligned}

for all $$n\geq1$$. Besides, we have to verify that, for some positive integer N, the conditions (II) and (III) in Theorem 1.2 hold for all $$n\geq N$$. That is,

\begin{aligned}& f_{n}\geq\frac{-2c_{2}(n)-\sqrt{\Delta(n)}}{6c_{3}(n)}; \end{aligned}
(3.1)
\begin{aligned}& c_{3}(n)g_{n}^{3}+c_{2}(n)g_{n}^{2}+c_{1}(n)g_{n}+c_{0}(n) \geq0. \end{aligned}
(3.2)

Let

$$\delta(n)=-6c_{3}(n)f_{n}-2c_{2}(n)$$

and

$$f(g_{n})=c_{3}(n)g_{n}^{3}+c_{2}(n)g_{n}^{2}+c_{1}(n)g_{n}+c_{0}(n).$$

To show (3.1), it is equivalent to show that, for some positive integers N, $$\delta(n)\geq0$$ and $$\delta^{2}(n)\geq\Delta(n)$$. By calculating, we easily find that, for all $$n\geq1$$,

\begin{aligned} \delta(n) =&\frac{8\text{,}192}{5(n+1)^{6}(n+2)^{4}(n+3)^{2}} \bigl(32n^{10}+129n^{9}+472n^{8}+3\mbox{,}556n^{7}+12\mbox{,}157n^{6} \\ &{} +17\mbox{,}632n^{5}+10\mbox{,}550n^{4}+1\mbox{,}293n^{3}-1\mbox{,}500n^{2}-798n-135 \bigr) \\ \geq&0, \end{aligned}

and for all $$n\geq3$$,

\begin{aligned} \delta^{2}(n)-\Delta(n) =&\frac{6\text{,}7108\text{,}864n}{25(n+3)^{4}(n+2)^{7}(n+1)^{12}} \bigl(699n^{18}+2\mbox{,}158n^{17}+6\mbox{,}983n^{16} \\ &{}+97\mbox{,}994n^{15}+155\mbox{,}517n^{14}-1\mbox{,}256\mbox{,}916n^{13}-3\mbox{,}302\mbox{,}168n^{12} \\ &{}+5\mbox{,}191\mbox{,}280n^{11}+25\mbox{,}505\mbox{,}142 n^{10}+14\mbox{,}486\mbox{,}584 n^{9}-63\mbox{,}005\mbox{,}002 n^{8} \\ &{}-153\mbox{,}766\mbox{,}236 n^{7}-178\mbox{,}037\mbox{,}517 n^{6}-131\mbox{,}841\mbox{,}558n^{5}-68\mbox{,}012\mbox{,}397n^{4} \\ &{} -24\mbox{,}910\mbox{,}146n^{3}-6\mbox{,}269\mbox{,}211n^{2}-975\mbox{,}888n-70\mbox{,}470\bigr) \\ \geq&0. \end{aligned}

Thus, take $$N=3$$ and, for all $$n\geq N$$, we have $$\delta(n)\geq0$$, $$\delta^{2}(n)\geq\Delta(n)$$, which follows from the inequality (3.1). We show the inequality (3.2) for some positive integer M. Note that, by Lemma 2.2 and some calculations, we have

\begin{aligned} f(g_{n}) =&c_{3}(n)g_{n}^{3}+c_{2}(n)g_{n}^{2}+c_{1}(n)g_{n}+c_{0}(n) \\ =&\frac{1\text{,}048\text{,}576}{n^{9}(n+1)^{6}(n+2)^{4}(n+3)^{2}} \bigl(54n^{15}+378 n^{14}+916 n^{13}+644 n^{12}-1\mbox{,}529 n^{11} \\ &{}-5\mbox{,}340 n^{10}-8\mbox{,}383 n^{9}-7\mbox{,}416n^{8}-2\mbox{,}284n^{7}+4\mbox{,}156n^{6}+7\mbox{,}969n^{5}+7\mbox{,}688n^{4} \\ &{} +4\mbox{,}953n^{3}+2\mbox{,}154n^{2}+576n+72\bigr). \end{aligned}

Take $$M=6$$, it is not difficult to verify that, for all $$n\geq M$$,

$$f(g_{n})>0.$$

Let $$N_{0}=\max\{N,M\}=6$$, then for all $$n\geq6$$, all of the above inequalities hold. By Lemma 3.1 and Theorem 1.2, the Catalan-Larcombe-French sequence $$\{P_{n}\}_{n\geq6}$$ is strictly 2-log-convex for all $$n\geq6$$. What is more, one can easily test that these numbers $$\{P_{n}\}_{0\leq n\leq8}$$ also satisfy the property of 2-log-convexity by simple calculations. Therefore, the whole sequence $$\{P_{n}\}_{n\geq0}$$ is strictly 2-log-convex. This completes the proof.â€ƒâ–¡

It deserves to be mentioned that by considerable calculations and plenty of verifications, the following conjectures should be true.

### Conjecture 3.2

The Catalan-Larcombe-French sequence is âˆž-log-convex.

### Conjecture 3.3

The quotient sequence $$\{\frac{P_{n}}{P_{n-1}}\}_{n\geq1}$$ of the Catalan-Larcombe-French sequence is log-concave, equivalently, for all $$n\geq2$$,

$$P_{n-2}P_{n}^{3}\geq P_{n+1}P_{n-1}^{3}.$$

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## Acknowledgements

Research supported by NSFC (No. 11161046) and by the Xingjiang Talent Youth Project (No. 2013721012).

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Correspondence to Brian Y Sun.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Sun, B.Y., Wu, B. Two-log-convexity of the Catalan-Larcombe-French sequence. J Inequal Appl 2015, 404 (2015). https://doi.org/10.1186/s13660-015-0920-0