Higher Order Log-Concavity in Euler's Difference Table

TL;DR

This paper investigates the higher order log-concavity properties of a permutation-related array derived from Euler's difference table, revealing new concavity characteristics in combinatorial sequences.

Contribution

It establishes that the array formed by scaled entries of Euler's difference table exhibits 2-log-concavity and reverse ultra log-concavity for all n, a novel insight into their combinatorial structure.

Findings

The sequence is 2-log-concave for all n.

The sequence is reverse ultra log-concave for all n.

Provides new combinatorial properties of Euler's difference table entries.

Abstract

Let $e_{n}^k$ be the entries in the classical Euler's difference table. We consider the array $d_{n}^{k}=e_n^k/k!$ for $0\leq k \leq n$, where $d_n^k$ can be interpreted as the number of k-fixed-points-permutations of [n]. We show that the sequence $\{d_n^k\}_{0\leq k\leq n}$ is 2-log-concave and reverse ultra log-concave for any given n.

Equations

$$e_n^{k-1}=e_n^k-e_{n-1}^{k-1},$$

$$d_n^k$$

$$d_n^k=d_{n-1}^{k-1}+(n-k)d_{n-1}^k.$$

$$(d_n^k{)}^2\ge d_n^{k-1}{d}_n^{k+1},$$

$$b_k=a_k^2-a_{k-1}{a}_{k+1}.$$

$${\left((d_n^k{)}^2-d_n^{k-1}{d}_n^{k+1}\right)}^2-\left((d_n^{k-1}{)}^2-d_n^{k-2}{d}_n^k\right)\left((d_n^{k+1}{)}^2-d_n^k{d}_n^{k+2}\right)\ge 0.$$

$$I=[n+\frac{n-k}{n},n+\frac{n-k}{n}+\frac{n-k}{n^2}],$$

$$\frac{n-k}{n}\le \frac{d_{n+1}^k}{d_n^k}\le n+\frac{n-k}{n}+\frac{n-k}{n^2}.$$

$$f\left(n+\frac{n-k}{n}+\frac{n-k}{n^2}\right)\ge 0.$$

$$d_n^{k-1}=(k+1)(n-k)d_n^{k+1}-(n-2k+1)d_n^k.$$

$$d_n^{k-1}=k{d}_n^k-d_{n-1}^{k-1}.$$

$$d_n^{k-1}$$

$$d_n^k$$

$$d_n^{k-1}=(k+1)(n-k)d_n^{k+1}-(n-2k+1)d_n^k,$$

$$\frac{d_{n+1}^k}{d_n^k}\ge n+\frac{n-k}{n}.$$

$$\frac{d_{n+1}^k}{d_n^k}$$

$$=n+(n-k)\frac{d_{n-1}^k}{d_n^k}$$

$$=n+(n-k)\frac{d_{n-1}^k}{(n-1)d_{n-1}^k+(n-k-1)d_{n-2}^k}.$$

$$(n-1)+(n-k-1)\frac{d_{n-2}^k}{d_{n-1}^k}\le n.$$

$$\frac{d_{n-1}^k}{d_{n-2}^k}\ge n-k-1.$$

$$\frac{d_{n-1}^k}{d_{n-2}^k}$$

$$\frac{d_{n+1}^k}{d_n^k}\le n+\frac{n-k}{n}+\frac{n-k}{n^2}.$$

$$\frac{d_{n+1}^k}{d_n^k}$$

$$=n+(n-k)\frac{d_{n-1}^k}{(n-1)d_{n-1}^k+(n-k-1)d_{n-2}^k}.$$

$$(n-1)+(n-k-1)\frac{d_{n-2}^k}{d_{n-1}^k}\ge \frac{n^2}{n+1},$$

$$\frac{d_{n-1}^k}{d_{n-2}^k}\le (n+1)(n-k-1).$$

$$\frac{d_{n-1}^k}{d_{n-2}^k}\le n-1,$$

$$d_n^{k-2}$$

$$d_n^{k-1}$$

$$d_n^{k+1}$$

$$d_n^{k+2}$$

$$A\cdot \left(C_3(n,k){\left(d_{n+1}^k\right)}^3+C_2(n,k){\left(d_{n+1}^k\right)}^2\left(d_n^k\right)+C_1(n,k)\left(d_{n+1}^k\right){\left(d_n^k\right)}^2+C_0(n,k){\left(d_n^k\right)}^3\right)\ge 0,$$

$$A$$

Full text

Higher Order Log-Concavity in Euler’s Difference Table

William Y.C. Chen1, Cindy C.Y. Gu2, Kevin J. Ma3 and Larry X.W. Wang4

Center for Combinatorics, LPMC-TJKLC

Nankai University, Tianjin 300071, P. R. China

1 chen@nankai.edu.cn, 2 guchunyan@cfc.nankai.edu.cn,

3 majun@cfc nankai.edu.cn, 4 wxw@cfc.nankai.edu.cn

Abstract. Let $e_{n}^{k}$ be the entries in the classical Euler’s difference table. We consider the array $d_{n}^{k}=e_{n}^{k}/k!$ for $0\leq k\leq n$, where $d_{n}^{k}$ can be interpreted as the number of $k$-fixed-points-permutations of $[n]$. We show that the sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ is $2$-log-concave and reverse ultra log-concave for any given $n$.

Keywords: log-concavity, 2-log-concavity, reverse ultra log-concavity, Euler’s difference table

Classification: 05A20; 33F10

1 Introduction

Euler introduced the difference table $(e_{n}^{k})_{0\leq k\leq n}$ defined by $e_{n}^{n}=n!$ and

[TABLE]

for $1\leq k\leq n$; see [5]. The combinatorial interpretation of the numbers $e_{n}^{k}$ was found by Dumont and Randrianarivony [6]. Clarke, Han and Zeng [5] further gave a combinatorial interpretation of the $q$-analogue of Euler’s difference table, and this interpretation has been extended by Faliharimalala and Zeng [8, 9] to the wreath product $C_{\ell}\wr S_{n}$ of the cyclic group with the symmetric group.

It is easily seen from the recurrence (1.1) that $k!$ divides $e_{n}^{k}$. Thus we can define the integers $d_{n}^{k}=e_{n}^{k}/k!$. Rakotondrajao [14] has shown that $d_{n}^{k}$ counts the number of $k$-fixed-points-permutations of $[n]$, where a permutation $\pi\in\mathfrak{S}_{n}$ is called $k$-fixed-points-permutation if there are no fixed points in the last $n-k$ positions and the first $k$ elements are in different cycles. Based on this combinatorial interpretation, Rakotondrajao [15] has found bijective proofs for the following two recurrence relations for $0\leq k\leq n-1$,

[TABLE]

Notice that $d_{k}^{k}=1$. Recently, Eriksen, Freij and Wästlund [7] have generalized these formulas to fixed point $\lambda$-colored permutations. Employing (1.2) and (1.3), we can easily derive the following recurrence relation for $0\leq k\leq n-1$,

[TABLE]

Using the above recurrence relations (1.2) (1.3) and (1.4), we shall prove that the sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ has higher order log-concave properties. To be more specific, we shall show that this sequence is $2$-log-concave and reverse ultra log-concave for any $n\geq 1$.

2 $2$-log-concavity

In this section, we shall show that the sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ is $2$-log-concave for any $n\geq 1$. Recall that a sequence $\{a_{k}\}_{k\geq 0}$ of real numbers is said to be log-concave if $a_{k}^{2}\geq a_{k+1}a_{k-1}$ for all $k\geq 1$; see Stanley [16] and Brenti [2]. From the recurrence relation (1.4), it is easy to prove by induction that the sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ is log-concave.

Theorem 2.1

For $1\leq k\leq n$, we have

[TABLE]

that is, the sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ is log-concave.

The notion of high order log-concavity was introduced by Moll [13]; see also, [10]. Given a sequence $\{a_{k}\}_{k\geq 0}$, define the operator $\mathfrak{L}$ as $\mathfrak{L}\{a_{k}\}=\{b_{k}\}$, where

[TABLE]

The log-concavity of $\{a_{k}\}$ becomes the positivity of $\mathfrak{L}\{a_{k}\}$. If the sequence $\mathfrak{L}\{a_{k}\}$ is not only positive but also log-concave, then we say that $\{a_{k}\}$ is $2$-log-concave. In general, we say that $\{a_{k}\}$ is $l$-log-concave if $\mathfrak{L}^{l}\{a_{k}\}$ is positive, and that $\{a_{k}\}$ is infinite log-concave if $\mathfrak{L}^{l}\{a_{k}\}$ is positive for any $l\geq 1$. From numerical evidence, we pose the following conjecture.

Conjecture 2.2

The sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ is infinitely log-concave.

Recently, Brändén [1] and Cardon [3] have independently proved that if a polynomial has only real and nonpositive zeros, then its Taylor coefficients form an infinite log-concave sequence. However, this is not the case of the polynomials $\sum d_{n}^{k}x^{k}$. For example, for $n=2$, the polynomial $x^{2}+x+1$ does not have real roots. Nevertheless, we shall show that the sequence $\{d_{n}^{k}\}$ is $2$-log concave in support of the general conjecture.

Theorem 2.3

The sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ is $2$-log-concave. In other words, for $n\geq 4$ and $2\leq k\leq n-2$, we have

[TABLE]

The idea to prove Theorem 2.3 may be described as follows. As the first step, we reformulate the left hand side of the above inequality (2.1) a cubic function $f$ on $\frac{d_{n+1}^{k}}{d_{n}^{k}}$ by applying the recurrence relations (1.2), (1.3), (1.4) and the recurrence relation presented in the following Lemma 2.4. Then Theorem 2.3 is equivalent to the assertion that $f\geq 0$ on the interval

[TABLE]

since it can be verified that for $n\geq 4$ and $2\leq k\leq n-2$,

[TABLE]

Moreover, when $f(x)$ is considered as a continuous function on $x$, we will be able to show that $f^{\prime}(x)<0$ for $x\in I$ and

[TABLE]

Hence we deduce that $f>0$ on the interval $I$ so that Theorem 2.3 is immediate.

As mentioned above, the following recurrence relation will be needed in the proof of Theorem 2.3.

Lemma 2.4

For $1\leq k\leq n$, we have

[TABLE]

*Proof.  *First, it is easy to establish the following recurrence relation for $1\leq k\leq n$,

[TABLE]

By (1.2) and (1.4), we have

[TABLE]

as claimed. By (1.4), (2.4), for $1\leq k\leq n$, we find

[TABLE]

Consequently,

[TABLE]

as desired.

In order to prove (2.2), we first give a lower bound for $d_{n+1}^{k}/d_{n}^{k}$.

Lemma 2.5

For $n\geq 1$ and $1\leq k\leq n-1$, we have

[TABLE]

*Proof.  *We proceed by induction on $n$. It is clear that (2.5) holds for $n=1$ and $n=2$. We now assume that (2.5) holds for positive integers less than $n$. By the recurrence (1.2), we have

[TABLE]

Thus (2.5) can be recast as

[TABLE]

So it suffices to check that

[TABLE]

Since $n\geq 3$, by the inductive hypothesis, we have

[TABLE]

as required.

Next we give an upper bound for $d_{n+1}^{k}/d_{n}^{k}$.

Lemma 2.6

For $n\geq 4$ and $2\leq k\leq n-2$, we have

[TABLE]

*Proof.  *It follows from the recurrence (1.2) that

[TABLE]

Thus (2.6) can be rewritten as

[TABLE]

that is,

[TABLE]

By recurrence (1.3) for $2\leq k\leq n-2$, we see that

[TABLE]

which implies (2.7). This completes the proof.

We are now ready to give the proof of Theorem 2.3.

Proof. It is easy to check that that the theorem holds for $n=4,5,6$ and $2\leq k\leq n-2$. So we may assume that $n\geq 7$.

We claim that the left hand side of (2.1) can be expressed as a cubic function $f$ on $\frac{d_{n+1}^{k}}{d_{n}^{k}}$. By the recurrences (1.2), (1.3), (1.4) and (2.3), we can derive the following relations,

[TABLE]

It follows that (2.1) can be rewritten as

[TABLE]

where

[TABLE]

Since $d_{n}^{k}$ are positive integers, it suffices to show that

[TABLE]

We now consider the function

[TABLE]

with

[TABLE]

We are going to show that $f^{\prime}(x)<0$, for $2\leq x\leq n-1$. As will be seen, the quadratic function $f^{\prime}(x)$ has a zero in the interval $[-1,k]$ and a zero in the interval $[k,n]$. At the point $x=1$, we have

[TABLE]

Since for $n\geq 7$ and $2\leq k\leq n-2$, we find

[TABLE]

This yields that $f^{\prime}(-1)<0$. Similarly, for $x=k$, we obtain that

[TABLE]

Moreover, for $x=n$, we have

[TABLE]

To prove $f^{\prime}(n)<0$, it is sufficient to show that for $2\leq k\leq n-2$,

[TABLE]

We have two cases for the ranges of $k$. For $2\leq k\leq n-3$, we have

[TABLE]

Meanwhile, for $k=n-2$,

[TABLE]

Thus $f^{\prime}(n)<0$ is valid for $2\leq k\leq n-2$. Then we reach the conclusion that $f^{\prime}(x)$ has a zero in the interval $[-1,k]$ and a zero in the interval$[k,n]$.

We continue to demonstrate that $f^{\prime}(x)<0$ in the interval $I$. By Lemma 2.5, for $k\geq 2$ we have

[TABLE]

which means that $f^{\prime}(x)$ has no zero on the interval $I$. Since $n\geq k+2$, it is easily seen that

[TABLE]

Since $f^{\prime}(n)<0$, we see that $f^{\prime}(x)<0$ in the interval $I$, as expected. In other words, $f(x)$ is strictly decreasing on this interval.

Up to now, we have shown that $f(x)$ is strictly decreasing on the interval $I=[n+\frac{n-k}{n},n+\frac{n-k}{n}+\frac{n-k}{n^{2}}]$. So it remains to prove that

[TABLE]

Since

[TABLE]

where

[TABLE]

Clearly, the proof will be complete as long as we can show that $h(k)\geq 0$ for $n\geq 7$ and $2\leq k\leq n-2$.

Regard $h(x)$ as a continuous function on $x$, that is,

[TABLE]

Since the leading coefficient $-10n^{4}-26n^{3}-28n^{2}-18n-6$ of $h(x)$ is negative, we only need to prove that $h(2)>0$ and $h(n-1)>0$. For $n\geq 7$, we have

[TABLE]

and

[TABLE]

In summary, we have confirmed that $h(k)>0$ for $n\geq 7$ and $2\leq k\leq n-2$. This completes the proof.

3 The reverse ultra log-concavity

This section is concerned with the reverse ultra log-concavity of $d_{n}^{k}$. Recall that sequence $\{a_{k}\}_{0\leq k\leq n}$ is called ultra log-concave if \left\{a_{k}\big{/}{n\choose k}\right\} is log-concave; see Liggett [12]. This condition can be restated as

[TABLE]

It is well known that if a polynomial has only real zeros, then its coefficients form an ultra log-concave sequence. As noticed by Liggett [12], if a sequence $\{a_{k}\}_{0\leq k\leq n}$ is ultra log-concave, then the sequence $\{k!a_{k}\}_{0\leq k\leq n}$ is log-concave.

In comparison with ultra log-concavity, a sequence is said to be reverse ultra log-concave if it satisfies the reverse relation of (3.1), that is,

[TABLE]

Chen and Gu [4] have shown the Boros-Moll polynomials have this reverse ultra log-concave property. We shall show that the sequence $\{d_{n}^{k}\}_{0\leq k\leq n}$ also possesses this property.

Theorem 3.1

For $1\leq k\leq n-1$, we have

[TABLE]

or equivalently,

[TABLE]

*Proof.  *According to the recurrence relations (1.4) and (2.3), we find that (3.3) can be reformulated as

[TABLE]

The discriminant of the quadratic polynomial of the left side of (3.4) in $d_{n+1}^{k}/d_{n}^{k}$ equals

[TABLE]

We claim that $\Delta>0$ for $1\leq k\leq n-1$. Put

[TABLE]

Since $n\geq k+1$, we have

[TABLE]

which implies that $f(k)$ is monotone decreasing for $1\leq k\leq n-1$. Furthermore,

[TABLE]

Thus, $\Delta>0$ for $1\leq k\leq n-1$. Consequently, the quadratic function has two distinct real zeros. If we can show that for $1\leq k\leq n-1$, $d_{n+1}^{k}/d_{n}^{k}$ is larger than the maximal zero, then (3.4) holds since $n-k+1>0$. Thus we still have to show that

[TABLE]

In view of (2.5), we see that (3.5) can be deduced from the following inequality

[TABLE]

which is equivalent to

[TABLE]

Since both sides are positive, we can transform the above relation into the following form

[TABLE]

Evidently,

[TABLE]

for $1\leq k\leq n-1$. This completes the proof.

Acknowledgments. This work was supported by the 973 Project, the PCSIRT Project of the Ministry of Education, and the National Science Foundation of China.