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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

The log-concavity of the q-derangement numbers of type B

Eric H. Liu
  • Corresponding author
  • School of Statistics and Information, Shanghai University of International Business and Economics, Shanghai, 201620, China
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  • De Gruyter OnlineGoogle Scholar
/ Wenjing Du
Published Online: 2018-02-23 | DOI: https://doi.org/10.1515/math-2018-0009

Abstract

Recently, Chen and Xia proved that for n ≥ 6, the q-derangement numbers Dn(q) are log-concave except for the last term when n is even. In this paper, employing a recurrence relation for DnB(q) discovered by Chow, we show that for n ≥ 4, the q-derangement numbers of type B DnB(q) are also log-concave.

Keywords: The q-derangement numbers of type B; Unimodality; Log-concavity

MSC 2010: 05A15; 05A19; 05A20

1 Introduction

Let 𝓓n denote the set of derangements on {1, 2, …, n} and let D(π) := {i|1 ≤ in – 1, π(i) > π(i + 1)} denote the descent set of a permutation π. Define the major index of π by

maj(π):=iD(π)i.(1)

The q-derangement number Dn(q) is defined by

Dn(q):=πDnqmaj(π).(2)

Gessel [1] (see also [2]) discovered the following formula

Dn(q):=[n]!k=0n(1)kqk21[k]!,(3)

where [n] = 1 + q + q2 + … + qn – 1 and [n]! = [1][2] … [n]. Combinatorial proofs of (3) have been found by Wachs [3] and Chen and Xu [4]. Chen and Rota [5] showed that the q-derangement numbers are unimodal, and conjectured that the maximum coefficient appears in the middle. Zhang [6] confirmed this conjecture by showing that the q-derangement numbers satisfy the spiral property. Recently, Chen and Xia [7] introduced the notion of ratio monotonicity for polynomials with nonnegative coefficients, and they proved that, for n ≥ 6, the q-derangement numbers Dn(q) are strictly ratio monotone except for the last term when n is even. The ratio monotonicity implies the spiral property and log-concavity.

Let Bn denote the hyperoctahedral group of rank n, consisting of the signed permutations of {1, 2, …, n}. Let DnB denote the set of derangements on Bn, which is defined as

DnB:={π|πBn,π(i)ifor alli{1,2,,n}}.(4)

Let N(π) := #{i|1 ≤ in, π(i) < 0} be the number of negative letters of π and let maj(π) be defined as before. In [8], Chow considered the q-derangement number of type B DnB(q) which is defined as

DnB(q):=πDnBqfmaj(π),(5)

where fmaj(π) := 2 maj(π) + N(π). Chow [8] (see also [9]) established the following formula

DnB(q):=[2][4][2n]k=0n(1)kq2k2[2][4][2k],(6)

where [n] is defined as before. Furthermore, Chow [8] discovered that for all integers n ≥ 1,

Dn+1B(q)=(1+q++q2n+1)DnB(q)+(1)n+1qn2+n.(7)

Chen and Wang [10] proved the normality of the limiting distribution of the coefficients of the usual q-derangement numbers of type B.

Recall that a positive sequence a0, a1, …, an or the polynomial a0 + a1x + … + anxn is called log-concave if the ratios

a0a1,a1a2,,an1an(8)

form an increasing sequence. Clearly, if a positive sequence is log-concavity, then it is unimodality. In this paper, we prove that for n ≥ 4, the q-derangement numbers of type B DnB(q) are log-concave.

Suppose that n is given. It is easy to prove that the degree of DnB(q) is n2 and the coefficient of qn2 is 1. Set

DnB(q)=Bn(1)q+Bn(2)q2++Bn(n2)qn2.(9)

The log-concavity of DnB(q) can be stated as the following theorem.

Theorem 1.1

For all integers n ≥ 4, the q-derangement numbers of type B DnB(q) are log-concave, namely,

Bn(1)Bn(2)<Bn(2)Bn(3)<<Bn(n22)B<(n21)<Bn(n21)Bn(n2).(10)

For example, by (6), we have

D4B(q)=q+4q2+8q3+13q4+18q5+22q6+26q7+28q8+28q9+25q10+21q11+17q12+11q13+7q14+3q15+q16.

It is easy to check that

14<48<813<1318<1822<2226<2628<2828<2825<2521<2117<1711<117<73<31.

2 Some lemmas

To prove Theorem 1.1, we first present some lemmas. By (7), it is easy to check that

Lemma 2.1

For n ≥ 4,

Bn+1(k)=i=1kBn(i),1k2n+2,i=k2n1kBn(i),2n+2<kn2,i=n2n1n2Bn(i)+(1)n+1,k=n2+n,i=k2n1n2Bn(i),n2k(n+1)2andkn2+n.(11)

Based on recurrence relation (11), it is easy to verify the following lemma.

Lemma 2.2

Let n4 be an integer. Then Bn(i) are positive integers for 1in2 and

Bn(n2)=1,Bn(n21)=n1,(12)

Bn(n22)=n2n+22,Bn(n23)=n3+5n186.(13)

To prove Theorem 1.1, we require the following lemma.

Lemma 2.3

For positive integers a1, a2, …, ak+1, ak+2 (k ≥ 1) satisfying

aiai+1<ai+1ai+2,1ik,(14)

i=1kaii=1k+1ai<i=1k+1aii=1k+2ai,(15)

i=1kaii=1k+1ai<i=1k+1aii=2k+2ai,(16)

i=1kaii=2k+1ai<i=2k+1aii=3k+2ai,(17)

i=1kaii=2k+1ai<i=2k+1aii=3k+1ai,(18)

i=1kaii=2kai<i=2kaii=3kai.(19)

Proof

We only prove (15). The rest can be proved similarly and the details are omitted. Based on (14),

aiak+2<ai+1ak+1(1ik),

and

ak+2(a1+a2++ak)<ak+1(a2+a3++ak+1).

Therefore,

(a1+a2++ak)(a1+a2++ak+ak+1+ak+1)=(a1+a2++ak)2+ak+1(a1+a2++ak)+ak+2(a1+a2++ak)<(a1+a2++ak)2+ak+1(a1+a2++ak)+ak+1(a2+a3++ak+1)<(a1+a2++ak)2+ak+1(a1+a2++ak)+ak+1(a2+a3++ak+1)+a1ak+1=(a1+a2++ak+ak+1)2,

which yields (15). This completes the proof of this lemma.

3 Proof of Theorem 1.1

We prove Theorem 1.1 by induction on n. It is easy to check that Theorem 1.1 holds for 4 ≤ n ≤ 12. Thus, we always assume that n ≥ 13 in the following proof. Suppose that Theorem 1.1 holds for n = m, namely,

Bm(i)Bm(i+1)<Bm(i+1)Bm(i+2),1im22.(20)

We proceed to show that Theorem 1.1 holds for n = m + 1, that is,

Bm+1(k)Bm+1(k+1)<Bm+1(k+1)Bm+1(k+2),1k(m+1)22.(21)

Employing (11), (15) and (20), we see that (21) holds for 1 ≤ k ≤ 2m. It follows from (11), (16) and (20) that (21) is true for the case k = 2m + 1. In view of (11), (17) and (20), we find that (21) holds for 2m + 2 ≤ km2 – 2. From (11), (18) and (20), we deduce that (21) is true for the case k = m2 – 1. By (11), (19) and (20), we can prove that (21) holds for m2km2 + m – 3 and m2 + m + 1 ≤ k ≤ (m + 1)2 – 2.

Now, special attentions should be paid to three cases k = m2 + m – 2, k = m2 + m – 1 and k = m2 + m.

By (12) and (13), it is easy to check that for m ≥ 4,

Bm(m22)+Bm(m21)+Bm(m2)m3Bm(m21)+Bm(m2)+1Bm(m23)Bm(m22)=m413m2+72m1326(m3)(m2m+2)>0.(22)

In view of (20) and (22),

Bm(m2m3)Bm(m2m2)<Bm(m23)Bm(m22)<Bm(m22)+Bm(m21)+Bm(m2)m3Bm(m21)+Bm(m2)+1.(23)

From (11), it is easy to prove that for m ≥ 4,

Bm(m2m2)>Bm(m2m1)>>Bm(m21)>Bm(m2).(24)

Thanks to (23) and (24),

Bm(m2m3)(Bm(m21)+Bm(m2)+(1)m+1)+(1)m+1i=0m+2Bm(m2i)<Bm(m2m3)(Bm(m21)+Bm(m2)+1)+(m+3)Bm(m2m2)<Bm(m2m2)(Bm(m22)+Bm(m21)+Bm(m2)).(25)

By (20),

Bm(m2m3)i=2m+1Bm(m2i)<Bm(m2m2)i=3m+2Bm(m2i).(26)

Combining (25) and (26) yields

i=m2m3m2Bm(i)i=m2m2m2Bm(i)<i=m2m2m2Bm(i)i=m2m1m2Bm(i)+(1)m+1,(27)

which can be rewritten as

Bm+1(m2+m2)Bm+1(m2+m1)<Bm+1(m2+m1)Bm+1(m2+m).(28)

Therefore, (21) holds for the case k = m2 + m – 2.

Based on (12) and (13), we deduce that for m ≥ 13,

Bm(m22)+Bm(m21)+Bm(m2)2(m+2)Bm(m21)+Bm(m2)Bm(m23)Bm(m22)=m412m313m2+36m366m(m2m+2)>0.(29)

By (20) and (29),

Bm(m2m2)Bm(m2m1)<Bm(m23)Bm(m22)<Bm(m22)+Bm(m21)+Bm(m2)2(m+2)Bm(m21)+Bm(m2).(30)

It follows from (24) and (30) that

Bm(m2m2)(Bm(m21)+Bm(m2))<Bm(m2m1)(Bm(m22)+Bm(m21)+Bm(m2)2(m+2))<Bm(m2m1)(Bm(m22)+Bm(m21)+Bm(m2))2i=0m+1Bm(m2i)Bm(m2m1)(Bm(m22)+Bm(m21)+Bm(m2))+2×(1)m+1i=0m+1Bm(m2i).(31)

In view of (20),

Bm(m2m2)i=2mBm(m2i)<Bm(m2m1)i=3m+1Bm(m2i).(32)

It follows from (31) and (32) that

i=m2m2m2Bm(i)i=m2m1m2Bm(i)+(1)m+1<i=m2m1m2Bm(i)+(1)m+1i=m2mm2Bm(i).(33)

By (11), we can rewrite (33) as follows

Bm+1(m2+m1)Bm+1(m2+m)<Bm+1(m2+m)Bm+1(m2+m+1),(34)

which implies that (21) holds for the case k = m2 + m – 1.

In view of (12) and (13), we see that for m ≥ 4,

Bm(m22)+Bm(m21)+Bm(m2)mBm(m21)+Bm(m2)Bm(m23)Bm(m22)=(m+1)(m37m2+12m+12)6m(m2m+2)>0.(35)

By (20) and (35), we find that for m ≥ 4,

Bm(m2m1)Bm(m2m)<Bm(m23)Bm(m22)<Bm(m22)+Bm(m21)+Bm(m2)mBm(m21)+Bm(m2).(36)

It follows from (24) and (36) that

Bm(m2m1)(Bm(m21)+Bm(m2))+(1)m+1i=0m1Bm(m2i)<Bm(m2m1)(Bm(m21)+Bm(m2))+mBm(m2m)Bm(m2m)(Bm(m22)+Bm(m21)+Bm(m2)).(37)

By (20),

Bm(m2m1)i=2m1Bm(m2i)<Bm(m2m)i=3mBm(m2i).(38)

In view of (37) and (38), we can prove that

i=m2m1m2Bm(i)+(1)m+1i=m2mm2Bm(i)<i=m2mm2Bm(i)i=m2m+1m2Bm(i).(39)

By (11), we can rewrite (39) as follows

Bm+1(m2+m)Bm+1(m2+m+1)<Bm+1(m2+m+1)Bm+1(m2+m+2),(40)

which implies that (21) holds for the case k = m2 + m. This completes the proof.

Acknowledgement

The authors would like to thank the anonymous referee for valuable corrections and comments. This work was supported by the National Science Foundation of China (11701362).

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About the article

Received: 2017-09-22

Accepted: 2018-01-11

Published Online: 2018-02-23


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 127–132, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0009.

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© 2018 Liu and Du, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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