Abstract

We give an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dicyclic group 𝐵4𝑛 of order 4𝑛 by finding its generating function of multivariables.

1. Introduction

Throughout this paper, all groups are assumed to be finite. The lattice of subgroups of a given group 𝐺 is the lattice (𝐿(𝐺),) where 𝐿(𝐺) is the set of all subgroups of 𝐺 and the partial order is the set inclusion. In this lattice (𝐿(𝐺),), a chain of subgroups of 𝐺 is a subset of 𝐿(𝐺) linearly ordered by set inclusion. A chain of subgroups of 𝐺 is called 𝐺-rooted (or rooted) if it contains 𝐺. Otherwise, it is called unrooted.

The problem of counting chains of subgroups of a given group 𝐺 has received attention by researchers with related to classifying fuzzy subgroups of 𝐺 under a certain type of equivalence relation. Some works have been done on the particular families of finite abelian groups (e.g., see [14]). As a step of this problem toward non-abelian groups, the first author [5] has found an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dihedral group 𝐷2𝑛 of order 2𝑛 where 𝑛 is an arbitrary positive integer. As a continuation of this work, we give an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dicyclic group 𝐵4𝑛 of order 4𝑛 by finding its generating function of multivariables where 𝑛 is an arbitrary integer.

2. Preliminaries

Given a group 𝐺, let 𝒞(𝐺), 𝒰(𝐺), and (𝐺) be the collection of chains of subgroups of 𝐺, of unrooted chains of subgroups of 𝐺, and of 𝐺-rooted chains of subgroups of 𝐺, respectively. Let 𝐶(𝐺)=|𝒞(𝐺)|, 𝑈(𝐺)=|𝒰(𝐺)|, and 𝑅(𝐺)=|(𝐺)|.

The following simple observation is useful for enumerating chains of subgroups of a given group.

Proposition 2.1. Let 𝐺 be a finite group. Then 𝑅(𝐺)=𝑈(𝐺)+1 and 𝐶(𝐺)=𝑅(𝐺)+𝑈(𝐺)=2𝑅(𝐺)1.

For a fixed positive integer 𝑘, we define a function 𝜆 as follows: 𝜆𝑥𝑘=12𝑥𝑘,𝜆𝑥𝑘,𝑥𝑘1,,𝑥𝑗𝑥=𝜆𝑘,𝑥𝑘1,,𝑥𝑗+1𝑥1+𝜆𝑘,𝑥𝑘1,,𝑥𝑗+1𝑥𝑗(2.1) for any 𝑗=𝑘1,𝑘2,,1.

Proposition 2.2 (see [5]). Let 𝑛 be the cyclic group of order 𝑛=𝑝𝛽11𝑝𝛽22𝑝𝛽𝑘𝑘,(2.2) where 𝑝1,,𝑝𝑘 are distinct prime numbers and 𝛽1,,𝛽𝑘 are positive integers. Then the number 𝑅(𝑛) of rooted chains of subgroups in the lattice of subgroups of 𝑛 is the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=1𝜆𝑥𝑘,,𝑥1.(2.3)

Let be the set of all integer numbers. Given distinct positive integers 𝑖1,,𝑖𝑡, we define a function 𝜋𝑖1𝑖𝑡𝑘𝑘,𝑥1,,𝑥𝑘𝑦1,,𝑦𝑘,(2.4) where 𝑦=𝑥,if𝑖𝑗𝑥𝑗=1,,𝑡,1,=𝑖𝑗forsome𝑗suchthat1𝑗𝑡.(2.5)

Most of our notations are standard and for undefined group theoretical terminologies we refer the reader to [6, 7]. For a general theory of solving a recurrence relation using a generating function, we refer the reader to [8, 9].

3. The Number of Chains of Subgroups of the Dicyclic Group 𝐵4𝑛

Throughout the section, we assume that 𝑛=𝑝𝛽11𝑝𝛽22𝑝𝛽𝑘𝑘,(3.1) is a positive integer, where 𝑝1,,𝑝𝑘 are distinct prime numbers and 𝛽1,,𝛽𝑘 are nonnegative integers and the dicyclic group 𝐵4𝑛 of order 4𝑛 is defined by the following presentation: 𝐵4𝑛=𝑎,𝑏𝑎2𝑛=𝑒,𝑏2=𝑎𝑛,𝑏𝑎𝑏1=𝑎1,(3.2) where 𝑒 is the identity element.

By the elementary group theory, the following is wellknown.

Lemma 3.1. The dicyclic group 𝐵4𝑛 has an index 2 subgroup 𝑎, which is isomorphic to 2𝑛, and has 𝑝𝑖 index 𝑝𝑖 subgroups 𝑎𝑝𝑖,𝑏,𝑎𝑝𝑖𝑎,𝑎𝑏,,𝑝𝑖,𝑎𝑝𝑖1𝑏,(3.3) which are isomorphic to the dicyclic group 𝐵4𝑛/𝑝𝑖 of order 4𝑛/𝑝𝑖 where 𝑖=1,2,,𝑘.

Lemma 3.2. (1) For any 𝑖=1,2,,𝑘, 𝑎𝑝𝑖,𝑎𝑟𝑏𝑎𝑝𝑖,𝑎𝑠𝑏=𝑎𝑝𝑖2𝑛/𝑝𝑖,(3.4) where 0𝑟<𝑠𝑝𝑖1.
(2) For any distinct prime factors 𝑝𝑖1,𝑝𝑖2,,𝑝𝑖𝑡 of 𝑛, 𝑎𝑝𝑖1,𝑎𝑟1𝑏𝑎𝑝𝑖2,𝑎𝑟2𝑏𝑎𝑝𝑖𝑡,𝑎𝑟𝑡𝑏𝐵4𝑛/𝑝𝑖1𝑝𝑖𝑡,(3.5) where 𝑟1,,𝑟𝑡 are nonnegative integers.

Proof. (1) To the contrary suppose that 𝑎𝑝𝑖,𝑎𝑟𝑏𝑎𝑝𝑖,𝑎𝑠𝑏𝑎𝑝𝑖.(3.6) Then 𝑎𝑝𝑖𝑢+𝑟𝑏=𝑎𝑝𝑖𝑣+𝑠𝑏 for some integers 𝑢 and 𝑣. This implies 𝑝𝑖𝑠𝑟. Since 0𝑟<𝑠𝑝𝑖1, we have 𝑠=𝑟, a contradiction.
(2) We only give its proof when 𝑡=2. The general case can be proved by the inductive process. Let 𝐾=𝑎𝑝𝑖1,𝑎𝑟1𝑏𝑎𝑝𝑖2,𝑎𝑟2𝑏.(3.7) Clearly, 𝑎𝑝𝑖1𝑝𝑖2𝐾. Since gcd(𝑝𝑖1,𝑝𝑖2)=1, there exist integers 𝑢 and 𝑣 such that 𝑝𝑖1𝑢+𝑝𝑖2𝑣=1. Note that 𝑎𝑝𝑖1(𝑢(𝑟1𝑟2))+𝑟1𝑏=𝑎𝑝𝑖1(𝑢(𝑟1𝑟2))𝑎𝑟1𝑏𝑎𝑝𝑖1,𝑎𝑟1𝑏. On the other hand, 𝑎𝑝𝑖1(𝑢(𝑟1𝑟2))+𝑟1𝑏=𝑎𝑝𝑖1𝑢(𝑟1𝑟2)+𝑟1𝑏=𝑎𝑝𝑖2𝑣(𝑟1𝑟2)(𝑟1𝑟2)+𝑟1𝑏since𝑝𝑖1𝑢+𝑝𝑖2𝑣=1=𝑎𝑝𝑖2𝑣(𝑟1𝑟2)+𝑟2𝑏𝑎𝑝𝑖2,𝑎𝑟2𝑏.(3.8) Considering the order of 𝐾, one can see that 𝐾=𝑎𝑝𝑖1𝑝𝑖2,𝑎𝑝𝑖1(𝑢(𝑟1𝑟2))+𝑟1𝑏. Since 𝑎𝑝𝑖1𝑝𝑖24𝑛/𝑝𝑖1𝑝𝑖2𝑎=𝑒,𝑝𝑖1(𝑢(𝑟1𝑟2))+𝑟1𝑏2=𝑏2=𝑎𝑛=𝑎𝑝𝑖1𝑝𝑖2𝑛/𝑝𝑖1𝑝𝑖2,𝑎𝑝𝑖1(𝑢(𝑟1𝑟2))+𝑟1𝑏𝑎𝑝𝑖1𝑝𝑖2𝑎𝑝𝑖1(𝑢(𝑟1𝑟2))+𝑟1𝑏1=𝑎𝑝𝑖1𝑝𝑖21,(3.9) we have 𝐾𝐵4𝑛/𝑝𝑖1𝑝𝑖2.

By Lemma 3.1, we have 𝒰𝐵4𝑛=𝒞𝑎2𝑛𝑘𝑝𝑖=0𝑖1𝑗=0𝒞𝑎𝑝𝑖,𝑎𝑗𝑏𝐵4𝑛/𝑝𝑖.(3.10) Using the inclusion-exclusion principle and Lemma 3.2, one can see that the number 𝑈(𝐵4𝑛) has the following form: 𝑈𝐵4𝑛=𝐶2𝑛+1𝑖1<<𝑖𝑡𝑘,1𝑡𝑘𝑧𝑖1,,𝑖𝑡𝐶2𝑛/𝑝𝑖1𝑝𝑖𝑡+1𝑖1<<𝑖𝑡𝑘,1𝑡𝑘𝑏𝑖1,,𝑖𝑡𝐶𝐵4𝑛/𝑝𝑖1𝑝𝑖𝑡(3.11) for suitable integers 𝑧𝑖1,,𝑖𝑡 and 𝑏𝑖1,,𝑖𝑡. In the following, we determine the numbers 𝑧𝑖1,,𝑖𝑡 and 𝑏𝑖1,,𝑖𝑡 explicitly.

Lemma 3.3. (1) 𝑏𝑖1,𝑖2,,𝑖𝑡=(1)𝑡+1𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡.
(2) 𝑧𝑖1,𝑖2,,𝑖𝑡=(1)𝑡𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡.

Proof. (1) Clearly 𝑏𝑖1=(1)1+1𝑝𝑖1=𝑝𝑖1 for any 𝑖1=1,,𝑘. For any integer 𝑡2, one can see by Lemma 3.2 that among intersections of the subgroups of the right-hand side of (3.10), the group isomorphic to 𝐵4𝑛/𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡 only appears in 𝑡-intersection of the subgroups 𝑎𝑝𝑖1,𝑎𝑗1𝑏,𝑎𝑝𝑖2,𝑎𝑗2𝑏𝑎,,𝑝𝑖𝑡,𝑎𝑗𝑡𝑏,(3.12) where 0𝑗𝑟𝑝𝑖𝑟1 and 1𝑟𝑡. Since there are 𝑝𝑖11𝑝𝑖21𝑝𝑖𝑡1=𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡 such choices, we have 𝑏𝑖1,𝑖2,,𝑖𝑡=(1)𝑡+1𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡.
(2) By Lemma 3.2, one can see that among intersections of the subgroups of the right-hand side of (3.10), the group isomorphic to 2𝑛/𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡 only appears one of the following two forms: 𝑎𝑎𝑝𝑖1,𝑎𝑗1𝑏𝑎𝑝𝑖2,𝑎𝑗2𝑏𝑎𝑝𝑖𝑡,𝑎𝑗𝑡𝑏,𝑎𝑝𝑖1,𝑎𝑗1𝑏𝑎𝑝𝑖2,𝑎𝑗2𝑏𝑎𝑝𝑖𝑡,𝑎𝑗𝑡𝑏,(3.13) where 0𝑗𝑟𝑝𝑖𝑟1 and 1𝑟𝑡, and each subgroup type in the first form must appear at least once, and it can appear more than once, while each subgroup type in the second form must appear at least once, and one of the subgroup types must appear more than once. Let 𝛾 be the number of the groups isomorphic to 2𝑛/𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡 obtained from the first form, and let 𝛿 be the number of the groups isomorphic to 2𝑛/𝑝𝑖1𝑝𝑖2𝑝𝑖𝑡 obtained from the second form. Then clearly 𝑧𝑖1,𝑖2,,𝑖𝑡=𝛾+𝛿. Note that 𝛾=𝑝𝑖1++𝑝𝑖𝑡𝑡𝑘=0(1)𝑡+2+𝑘𝑗1++𝑗𝑡=𝑡+𝑘,1𝑗𝑟𝑝𝑖𝑟𝑡,1𝑟𝑘𝑟=1𝑝𝑖𝑟𝑗𝑟=𝑘0𝑗1++𝑗𝑡=𝑡+𝑘,1𝑗𝑟𝑝𝑖𝑟,1𝑟𝑡(1)𝑡𝑡+𝑘𝑟=1𝑝𝑖𝑟𝑗𝑟=1𝑗𝑟𝑝𝑖𝑟,1𝑟𝑡(1)𝑗1++𝑗𝑡𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟=𝑡𝑟=11𝑗𝑟𝑝𝑖𝑟(1)𝑗1++𝑗𝑡𝑝𝑖𝑟𝑗𝑟=(1)𝑡.(3.14) On the other hand, 𝛿=𝑝𝑖1++𝑝𝑖𝑡𝑡1𝑘=0(1)𝑡+2+𝑘𝑗1++𝑗𝑡=𝑡+1+𝑘,1𝑗𝑟𝑝𝑖𝑟𝑡,1𝑟𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟=𝑝𝑖1++𝑝𝑖𝑡𝑡1𝑘=0(1)𝑡+2+𝑘𝑗1++𝑗𝑡=𝑡+1+𝑘,1𝑗𝑟𝑝𝑖𝑟𝑡,1𝑟𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟+(1)𝑡+1𝑗1++𝑗𝑡=𝑡,1𝑗𝑟𝑝𝑖𝑟𝑡,1𝑟𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟(1)𝑡+1𝑗1++𝑗𝑡=𝑡,1𝑗𝑟𝑝𝑖𝑟𝑡,1𝑟𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟=𝑝𝑖1++𝑝𝑖𝑡𝑡𝑘=0(1)𝑡+1+𝑘𝑗1++𝑗𝑡=𝑡+𝑘,1𝑗𝑟𝑝𝑗𝑟𝑡,1𝑟𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟(1)𝑡+1𝑗1++𝑗𝑡=𝑡,1𝑗𝑟𝑝𝑖𝑟𝑡,1𝑟𝑡𝑟=1𝑝𝑖𝑟𝑗𝑟=(1)𝑡(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡.(3.15) Therefore, we have 𝑧𝑖1,𝑖2,,𝑖𝑡=(1)𝑡𝑝𝑖1𝑝𝑖𝑡.

By Proposition 2.1 and Lemma 3.3, (3.11) becomes 𝑅𝐵4𝑛=2𝑅2𝑛+21𝑖1<<𝑖𝑡𝑘,1𝑡𝑘(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑅2𝑛/𝑝𝑖1𝑝𝑖𝑡+21𝑖1<<𝑖𝑡𝑘,1𝑡𝑘(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑅𝐵4𝑛/𝑝𝑖1𝑝𝑖𝑡.(3.16) Let 𝑎𝛽1,,𝛽𝑘=𝑅(𝐵4𝑛) and let 𝑏𝛽1,,𝛽𝑘=𝑅(2𝑛). Then (3.16) becomes 𝑎𝛽1,,𝛽𝑘=2𝑏𝛽1,,𝛽𝑘+21𝑖1<<𝑖𝑡𝑘,1𝑡𝑘(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑏𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)+21𝑖1<<𝑖𝑡𝑘,1𝑡𝑘(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑎𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘).(3.17)

Throughout the remaining part of the section, we solve the recurrence relation of (3.17) by using generating function technique. From now on, we allow each 𝛽𝑖 to be zero for computational convenience.

Let 𝜓𝛽1,,𝛽𝑘𝑥𝑘,𝑥𝑘1,,𝑥𝑗=𝛽𝑗=0𝛽𝑘1=0𝛽𝑘=0𝑎𝛽1,,𝛽𝑘𝑥𝛽𝑘𝑘𝑥𝛽𝑘1𝑘1𝑥𝛽𝑗𝑗,𝜙𝛽1,,𝛽𝑘𝑥𝑘,𝑥𝑘1,,𝑥𝑗=𝛽𝑗=0𝛽𝑘1=0𝛽𝑘=0𝑏𝛽1,,𝛽𝑘𝑥𝛽𝑘𝑘𝑥𝛽𝑘1𝑘1𝑥𝛽𝑗𝑗,(3.18) where 𝑗=𝑘,𝑘1,,1.

For a fixed integer 𝑛=𝑝𝛽11𝑝𝛽22𝑝𝛽𝑘𝑘 such that 𝑝1,,𝑝𝑘 are distinct prime numbers and 𝛽1,,𝛽𝑘 are non-negative integers, we define a function 𝜇 as follows. 𝜇𝑥𝑘=12𝑝𝑘𝑥𝑘,𝜇𝑥𝑘,,𝑥𝑗𝑥=𝜇𝑘,,𝑥𝑗+1𝑥1+𝜇𝑘,,𝑥𝑗+1𝑝𝑗𝑥𝑗(3.19) for any 𝑗=𝑘1,𝑘2,,1.

Lemma 3.4. Let 𝑘 be a positive integer. If 𝑘=1, then 𝜇𝑥1𝜓𝛽1𝑥1=𝑥1+𝜇1𝜙𝛽1𝑥1.(3.20) If 𝑘2, then 𝜇𝑥𝑘,,𝑥𝑗𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥𝑗=𝑥1+𝜇𝑘,,𝑥𝑗×𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥𝑗+1𝑖1<<𝑖𝑡𝑗1,1𝑡𝑗1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗+1𝑖1<<𝑖𝑡𝑗1,1𝑡𝑗1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗(3.21) for any 𝑗=𝑘,𝑘1,,2.

Proof. Assume first that 𝑘=1. Then (3.17) with 𝑘=1 gives us that 𝑎𝛽1=2𝑏𝛽1+2𝑝1𝑎𝛽112𝑝1𝑏𝛽11.(3.22) Taking 𝛽1=1 to both sides of (3.22), we have 12𝑝1𝑥1𝜓𝛽1𝑥1=22𝑝1𝑥1𝜙𝛽1𝑥1(3.23) because 𝑎0=𝑅(𝐵4𝑝01)=𝑅(22)=22 and 𝑏0=𝑅(2𝑝01)=𝑅(2)=2 by a direct computation.
From now on, we assume that 𝑘2. We prove (3.21) by double induction on 𝑘 and 𝑗. Equation (3.17) with 𝑘=2 gives us that 𝑎𝛽1,𝛽2=2𝑏𝛽1,𝛽22𝑝1𝑏𝛽11,𝛽22𝑝2𝑏𝛽1,𝛽21+2𝑝1𝑝2𝑏𝛽11,𝛽21+2𝑝1𝑎𝛽11,𝛽2+2𝑝2𝑎𝛽1,𝛽212𝑝1𝑝2𝑎𝛽11,𝛽21.(3.24) Taking 𝛽2=1𝑥𝛽22 of both sides of (3.24), we have 12𝑝2𝑥2𝜓𝛽1,𝛽2𝑥2=22𝑝2𝑥2𝑝1𝜓𝛽11,𝛽2𝑥2+𝜙𝛽1,𝛽2𝑥2𝑝1𝜙𝛽11,𝛽2𝑥2(3.25) because 𝑎𝛽1,0=𝑎𝛽1 and 𝑏𝛽1,0=𝑏𝛽1 by the definition, and 𝑎𝛽1,02𝑏𝛽1,02𝑝1𝑎𝛽1,0+2𝑝1𝑏𝛽11,0=0(3.26) by (3.17) with 𝑘=1. That is, 𝜇𝑥2𝜓𝛽1,𝛽2𝑥2=𝑥1+𝜇2𝜙𝛽1,𝛽2𝑥2𝑝1𝜙𝛽11,𝛽2𝑥2+𝑝1𝜓𝛽11,𝛽2𝑥2.(3.27) Thus (3.21) holds for 𝑘=2.
Assume now that (3.21) holds from 2 to 𝑘1 and consider the case for 𝑘. Note that the last two terms of the right-hand side of (3.17) can be divided into three terms, respectively, as follows: 21𝑖1<<𝑖𝑡𝑘,1𝑡𝑘(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑏𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)=2𝑝𝑘𝑏𝛽1,,𝛽𝑘1,𝛽𝑘12𝑝𝑘1𝑖1<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑏𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,𝛽𝑘1)+21𝑖1<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑏𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘),21𝑖1<<𝑖𝑡𝑘,1𝑡𝑘(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑎𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)=2𝑝𝑘𝑎𝛽1,,𝛽𝑘1,𝛽𝑘12𝑝𝑘1𝑖1<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑎𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,𝛽𝑘1)+21𝑖1<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑎𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘).(3.28) Taking 𝛽𝑘=1𝑥𝛽𝑘𝑘 of both sides of (3.17) and using (3.28), one can see that 12𝑝𝑘𝑥𝑘𝜓𝛽1,,𝛽𝑘𝑥𝑘=22𝑝𝑘𝑥𝑘×𝜙𝛽1,,𝛽𝑘𝑥𝑘+1𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,𝛽𝑘)𝑥𝑘+1𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘+𝑎𝛽1,,𝛽𝑘1,02𝑏𝛽1,,𝛽𝑘1,021𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑏𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,0)21𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑎𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,0).(3.29) Further since 𝑎𝛽1,,𝛽𝑘1,02𝑏𝛽1,,𝛽𝑘1,021𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝑏𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,0)21𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝑎𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,0)=0(3.30) by (3.17), we have 12𝑝𝑘𝑥𝑘𝜓𝛽1,,𝛽𝑘𝑥𝑘=22𝑝𝑘𝑥𝑘×𝜙𝛽1,,𝛽𝑘𝑥𝑘+1𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘1,𝛽𝑘)𝑥𝑘+1𝑖<<𝑖𝑡𝑘1,1𝑡𝑘1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘.(3.31) Thus (3.21) holds for 𝑗=𝑘. Assume that (3.21) holds from 𝑘 to 𝑗 and consider the case for 𝑗1. Note that the last two terms of the right-hand side of (3.21) can be divided into three terms, respectively, as follows: 1𝑖1<<𝑖𝑡𝑗1,1𝑡𝑗1(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗=𝑝𝑗1𝜙𝛽1,,𝛽𝑗2,𝛽𝑗11,𝛽𝑗,,𝛽𝑘𝑥𝑘,,𝑥𝑗𝑝𝑗11𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑗2,𝛽𝑗11,𝛽𝑗,,𝛽𝑘)𝑥𝑘,,𝑥𝑗+1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗,(3.32)1𝑖1<<𝑖𝑡𝑗1,1𝑡𝑗1(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗=𝑝𝑗1𝜓𝛽1,,𝛽𝑗2,𝛽𝑗11,𝛽𝑗,,𝛽𝑘𝑥𝑘,,𝑥𝑗𝑝𝑗11𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑗2,𝛽𝑗11,𝛽𝑗,,𝛽𝑘)𝑥𝑘,,𝑥𝑗+1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗.(3.33) Taking 𝛽𝑗1=1𝑥𝛽𝑗1𝑗1 of both sides of (3.21), we have 𝜇𝑥𝑘,,𝑥𝑗,𝑥𝑗1𝜓𝛽1,,𝛽𝑘𝑥𝑗1,,𝑥𝑘=𝑥1+𝜇𝑘,,𝑥𝑗,𝑥𝑗1×𝜙𝛽1,,𝛽𝑘𝑥𝑗1,,𝑥𝑘+1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗,𝑥𝑗1+1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗,𝑥𝑗1𝑥+𝜇𝑘,,𝑥𝑗𝜓𝛽1,,𝛽𝑗2,0,𝛽𝑗,,𝛽𝑘𝑥𝑘,,𝑥𝑗𝑥1+𝜇𝑘,,𝑥𝑗1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑗2,0,𝛽𝑗,,𝛽𝑘)𝑥𝑘,,𝑥𝑗𝑥1+𝜇𝑘,,𝑥𝑗1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑗2,0,𝛽𝑗,,𝛽𝑘)𝑥𝑘,,𝑥𝑗.(3.34) Note that 𝜇𝑥𝑘,,𝑥𝑗𝜓𝛽1,,𝛽𝑗2,0,𝛽𝑗,,𝛽𝑘𝑥𝑘,,𝑥𝑗𝑥1+𝜇𝑘,,𝑥𝑗1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑗2,0,𝛽𝑗,,𝛽𝑘)𝑥𝑘,,𝑥𝑗𝑥1+𝜇𝑘,,𝑥𝑗1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑗2,0,𝛽𝑗,,𝛽𝑘)𝑥𝑘,,𝑥𝑗=0(3.35) by induction hypothesis. Thus 𝜇𝑥𝑘,,𝑥𝑗,𝑥𝑗1𝜓𝛽1,,𝛽𝑘𝑥𝑗1,,𝑥𝑘=𝑥1+𝜇𝑘,,𝑥𝑗,𝑥𝑗1×𝜙𝛽1,,𝛽𝑘𝑥𝑗1,,𝑥𝑘+1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡𝑝𝑖1𝑝𝑖𝑡𝜙𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗,𝑥𝑗1+1𝑖1<<𝑖𝑡𝑗2,1𝑡𝑗2(1)𝑡+1𝑝𝑖1𝑝𝑖𝑡𝜓𝜋𝑖1𝑡𝑖(𝛽1,,𝛽𝑘)𝑥𝑘,,𝑥𝑗,𝑥𝑗1.(3.36) Therefore, (3.21) holds for 𝑗1.

Equation (3.21) with 𝑗=2 gives us that 𝜇𝑥𝑘,,𝑥2𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥2=𝑥1+𝜇𝑘,,𝑥2×𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥2𝑝1𝜙𝛽11,𝛽2,,𝛽𝑘𝑥𝑘,,𝑥2+𝑝1𝜓𝛽11,𝛽2,,𝛽𝑘𝑥𝑘,,𝑥2.(3.37) Taking 𝛽1=1𝑥𝛽11 of both sides of (3.37), we get that 𝜇𝑥𝑘,,𝑥1𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥2,𝑥1=𝑥1+𝜇𝑘,,𝑥1𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥2,𝑥1𝑥+𝜇𝑘,,𝑥2𝜓0,𝛽2,,𝛽𝑘𝑥𝑘,,𝑥2𝑥1+𝜇𝑘,,𝑥2𝜙0,𝛽2,,𝛽𝑘𝑥𝑘,,𝑥2.(3.38)

Lemma 3.5. If 𝑘2, then 𝜇𝑥𝑘,,𝑥2𝜓0,𝛽2,,𝛽𝑘𝑥𝑘,,𝑥2=𝑥1+𝜇𝑘,,𝑥2𝜙0,𝛽2,,𝛽𝑘𝑥𝑘,,𝑥2.(3.39)

Proof. If 𝑘=2, then since 𝜓0,𝛽2(𝑥2)=𝜓𝛽2(𝑥2) and 𝜙0,𝛽2(𝑥2)=𝜙𝛽2(𝑥2), the equation 𝜇𝑥2𝜓0,𝛽2𝑥2=𝑥1+𝜇2𝜙0,𝛽2𝑥2(3.40) holds by (3.20). Assume now that (3.39) holds for 𝑘. Then by (3.38) we get that 𝜇𝑥𝑘,,𝑥1𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥2,𝑥1=𝑥1+𝜇𝑘,,𝑥1𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥2,𝑥1,(3.41) which implies that 𝜇𝑥𝑘+1,,𝑥2𝜓0,𝛽2,,𝛽𝑘+1𝑥𝑘+1,,𝑥2=𝑥1+𝜇𝑘+1,,𝑥2𝜙0,𝛽2,,𝛽𝑘+1𝑥𝑘+1,,𝑥2.(3.42) Thus (3.39) holds for 𝑘+1.

By Lemmas 3.4 and 3.5 and (3.38), we have 𝜇𝑥𝑘,,𝑥1𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=𝑥1+𝜇𝑘,,𝑥1𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1.(3.43) We now need to find the function 𝜙𝛽1,,𝛽𝑘(𝑥𝑘,,𝑥1) explicitly.

Lemma 3.6. If 𝑝1=2, then 𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=2𝜆𝑥11,𝑖𝑓𝑘=1,1+𝜆𝑥𝑘,,𝑥21𝜆𝑥𝑘,,𝑥1,𝑖𝑓𝑘2.(3.44) If 𝑝𝑖2 for 𝑖=1,2,,𝑘, then 𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=11+𝜆𝑥𝑘,,𝑥11𝜆𝑥𝑘,,𝑥1.(3.45)

Proof. We first assume that 𝑝1=2. Then by Proposition 2.2, 𝑏𝛽1,𝛽2,,𝛽𝑘=𝑅𝑝𝛽11+1𝑝𝛽22𝑝𝛽33𝑝𝛽𝑘𝑘(3.46) is the coefficient of 𝑥𝛽11+1𝑥𝛽22𝑥𝛽33𝑥𝛽𝑘𝑘 of 1𝜆𝑥𝑘,,𝑥1,(3.47) which implies that 𝑏𝛽1,𝛽2,,𝛽𝑘 is the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽33𝑥𝛽𝑘𝑘 of 2𝜆𝑥11if𝑘=1,1+𝜆𝑥𝑘,,𝑥21𝜆𝑥𝑘,,𝑥1if𝑘2,(3.48) and hence by the definition of 𝜙 we get that 𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=2𝜆𝑥11,if𝑘=1,1+𝜆𝑥𝑘,,𝑥21𝜆𝑥𝑘,,𝑥1,if𝑘2.(3.49)
Assume now that 𝑝𝑖2 for 𝑖=1,2,,𝑘. Since 𝑏𝛽1,,𝛽𝑘=𝑅(2𝑝𝛽11𝑝𝛽22𝑝𝛽𝑘𝑘), by Proposition 2.2𝑏𝛽1,,𝛽𝑘 is the coefficient of 𝑥11𝑥𝛽12𝑥𝛽23𝑥𝛽𝑘𝑘+1 of 1𝜆𝑥𝑘+1,,𝑥1.(3.50) Since 1𝜆𝑥𝑘+1,,𝑥1=1𝜆𝑥𝑘+1,,𝑥2𝑥1+𝜆𝑘+1,,𝑥2𝑥1=1𝜆𝑥𝑘+1,,𝑥21𝑥11+1/𝜆𝑘+1,,𝑥2𝑥1(3.51) by the definition, 𝑏𝛽1,,𝛽𝑘 is the coefficient of 𝑥𝛽12𝑥𝛽23𝑥𝛽𝑘𝑘+1 of 1𝜆𝑥𝑘+1,,𝑥211+𝜆𝑥𝑘+1,,𝑥2.(3.52) By changing the variables 𝑥2,𝑥3,,𝑥𝑘+1 by 𝑥1,𝑥2,,𝑥𝑘, respectively, we get that 𝑏𝛽1,,𝛽𝑘 is the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 1𝜆𝑥𝑘,,𝑥111+𝜆𝑥𝑘,,𝑥1.(3.53) By the definition of 𝜙𝛽1,,𝛽𝑘(𝑥𝑘,,𝑥1), we have 𝜙𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=1𝜆𝑥𝑘,,𝑥111+𝜆𝑥𝑘,,𝑥1.(3.54)

By Proposition 2.1, (3.43), and Lemma 3.6, we have the following theorem.

Theorem 3.7. Let 𝑛=𝑝𝛽11𝑝𝛽22𝑝𝛽𝑘𝑘,(3.55) be a positive integer such that 𝑝1,,𝑝𝑘 are distinct prime numbers and 𝛽1,,𝛽𝑘 are positive integers. Let 𝐵4𝑛=𝑎,𝑏𝑎2𝑛=𝑒,𝑏2=𝑎𝑛,𝑏𝑎𝑏1=𝑎1(3.56) be the dicyclic group of order 4𝑛. Let 𝑅(𝐵4𝑛) be the number of rooted chains of subgroups in the lattice of subgroups of 𝐵4𝑛. (1)If 𝑝1=2, then 𝑅(𝐵4𝑛) is the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=11+𝜇𝑥12𝜆𝑥11,𝑖𝑓𝑘=1,1+𝜇𝑥𝑘,,𝑥111+𝜆𝑥𝑘,,𝑥21𝜆𝑥𝑘,,𝑥1,𝑖𝑓𝑘2.(3.57)(2)If 𝑝𝑖2 for 𝑖=1,2,,𝑘, then 𝑅(𝐵4𝑛) is the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1=11+𝜇𝑥𝑘,,𝑥11𝜆𝑥𝑘,,𝑥111+𝜆𝑥𝑘,,𝑥1.(3.58)
Furthermore, the number 𝐶(𝐵4𝑛) of chains of subgroups in the lattice of subgroups of 𝐵4𝑛 is the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 2𝜓𝛽1,,𝛽𝑘𝑥𝑘,,𝑥1𝑘𝑖=111𝑥𝑖.(3.59)

We now want to find the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 𝜓𝛽1,,𝛽𝑘(𝑥𝑘,,𝑥1) explicitly. Since 1𝜇𝑥𝑘,,𝑥1=1𝜇𝑥𝑘,,𝑥21𝑥11+1/𝜇𝑘,,𝑥2𝑝1𝑥1,(3.60) by the definition, the coefficient of 𝑥𝛽11 of 1/𝜇(𝑥𝑘,,𝑥1) is 1𝜇𝑥𝑘,,𝑥211+𝜇𝑥𝑘,,𝑥2𝛽1𝑝𝛽11=𝑝𝛽11𝛽1𝑖1=0𝛽1𝑖11𝜇𝑥𝑘,,𝑥2𝑖1+1=𝑝𝛽11𝛽1𝑖1=0𝛽1𝑖11𝜇𝑥𝑘,,𝑥3𝑖1+11𝑥11+1/𝜇𝑘,,𝑥3𝑝2𝑥2𝑖1+1.(3.61) Thus the coefficient of 𝑥𝛽11𝑥𝛽22 of 1/𝜇(𝑥𝑘,,𝑥1) is 𝑝𝛽11𝑝𝛽22𝛽1𝑖1=0𝛽1𝑖1𝑖1+𝛽2𝛽21𝜇𝑥𝑘,,𝑥3𝑖1+111+𝜇𝑥𝑘,,𝑥3𝛽2=𝑝𝛽11𝑝𝛽22𝛽1𝑖1𝛽=02𝑖2=0𝛽1𝑖1𝛽2𝑖2𝑖1+𝛽2𝛽21𝜇𝑥𝑘,,𝑥3𝑖1+𝑖2+1.(3.62) Continuing this process, one can see that the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 1/𝜇(𝑥𝑘,,𝑥1) is 2𝛽𝑘𝑝𝛽11𝑝𝛽22𝑝𝛽𝑘𝑘𝛽1𝑖1𝛽=02𝑖2=0𝛽𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑖𝑟𝛽𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1.(3.63) Similarly one can see that the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 1/𝜆(𝑥𝑘,,𝑥1) is 2𝛽𝑘𝛽1𝑖1𝛽=02𝑖2=0𝛽𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑖𝑟𝛽𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1,(3.64) the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of [1+(1/𝜆(𝑥𝑘,,𝑥2))](1/𝜆(𝑥𝑘,,𝑥1)) is 2𝛽𝑘𝛽1+1𝑖1𝛽=02𝑖2𝛽=03𝑖3=0𝛽𝑘1𝑖𝑘1=0𝛽1𝑖+11𝛽2+𝑖1𝛽2𝑘1𝑟=2𝛽𝑟𝑖𝑟𝛽𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1(3.65) and the coefficient of 𝑥𝛽11𝑥𝛽22𝑥𝛽𝑘𝑘 of 1/𝜆(𝑥𝑘,,𝑥1)2 is 𝛽12+1𝛽𝑘𝛽1𝑖1𝛽=02𝑖2=0𝛽𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑖𝑟𝛽𝑟+1+𝑟𝑚=1𝑖𝑚𝛽+1𝑟+1.(3.66)

Therefore, one can have the following.

Corollary 3.8. Let 𝑛 and 𝐵4𝑛 be the positive integer and the dicyclic group, respectively, defined in Theorem 3.7. Let 𝑅(𝐵4𝑛) be the number of rooted chains of subgroups in the lattice of subgroups of 𝐵4𝑛. (1)If 𝑝1=2, then 𝑅𝐵4𝑛=2𝛽𝑘𝛽1+1𝑖1𝛽=02𝑖2𝛽=03𝑖3=0𝛽𝑘1𝑖𝑘1=0𝛽1𝑖+11𝛽2+𝑖1𝛽2𝑘1𝑟=2𝛽𝑟𝑖𝑟𝛽𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1+2𝛽𝑘𝛽1𝑗1𝛽=02𝑗2=0𝛽𝑘𝑗𝑘=0𝑝𝑗11𝑝𝑗22𝑝𝑗𝑘𝑘𝑗1𝑖1𝑗=02𝑖2=0𝑗𝑘1𝑖𝑘1=0𝑘1𝑟=1𝑗𝑟𝑖𝑟𝑗𝑟+1+𝑟𝑚=1𝑖𝑚𝑗𝑟+1×𝛽1𝑗1+1𝑖1𝛽=02𝑗2𝑖2𝛽=03𝑗3𝑖3=0𝛽𝑘1𝑗𝑘1𝑖𝑘1=0𝛽1𝑗1𝑖+11𝛽2𝑗2+𝑖1𝛽2𝑗2×𝑘1𝑟=2𝛽𝑟𝑗𝑟𝑖𝑟𝛽𝑟+1𝑗𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1𝑗𝑟+1,(3.67)where if 𝑘=1, then 𝑅(𝐵42𝛽1)=22𝛽1+2 and if 𝑘=2, then 𝑅𝐵42𝛽1𝑝𝛽22=2𝛽2𝛽1+1𝑖1=0𝛽1𝑖+11𝛽2+𝑖1𝛽2+2𝛽2𝛽1𝑗1𝛽=02𝑗2=02𝑗1𝑝𝑗22𝑗1𝑖1=0𝑗1𝑖1𝑗2+𝑖1𝑗2×𝛽1𝑗1+1𝑖1=0𝛽1𝑗1𝑖+11𝛽2𝑗2+𝑖1𝛽2𝑗2.(3.68)(2)If 𝑝𝑖2 for 𝑖=1,2,,𝑘, then 𝑅𝐵4𝑛=2𝛽𝑘𝛽1𝑖1𝛽=02𝑖2=0𝛽𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑖𝑟𝛽𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1+𝛽12+1𝛽𝑘𝛽1𝑖1𝛽=02𝑖2=0𝛽𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑖𝑟𝛽𝑟+1+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1+2𝛽𝑘𝛽1𝑗1𝛽=02𝑗2=0𝛽𝑘𝑗𝑘=0𝑝𝑗11𝑝𝑗22𝑝𝑗𝑘𝑘𝑗1𝑖1𝑗=02𝑖2=0𝑗𝑘1𝑖𝑘1=0𝑘1𝑟=1𝑗𝑟𝑖𝑟𝑗𝑟+1+𝑟𝑚=1𝑖𝑚𝑗𝑟+1×𝛽1𝑗1𝑖1𝛽=02𝑗2𝑖2=0𝛽𝑘1𝑗𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑗𝑟𝑖𝑟𝛽𝑟+1𝑗𝑟+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1𝑗𝑟+1+2𝛽𝑘𝛽1𝑗1𝛽=02𝑗2=0𝛽𝑘𝑗𝑘=0𝑝𝑗11𝑝𝑗22𝑝𝑗𝑘𝑘𝑗1𝑖1𝑗=02𝑖2=0𝑗𝑘1𝑖𝑘1=0𝑘1𝑟=1𝑗𝑟𝑖𝑟𝑗𝑟+1+𝑟𝑚=1𝑖𝑚𝑗𝑟+1×𝛽1𝑗1+1𝛽1𝑗1𝑖1𝛽=02𝑗2𝑖2=0𝛽𝑘1𝑗𝑘1𝑖𝑘1=0𝑘1𝑟=1𝛽𝑟𝑗𝑟𝑖𝑟×𝛽𝑟+1𝑗𝑟+1+1+𝑟𝑚=1𝑖𝑚𝛽𝑟+1𝑗𝑟+1,(3.69)where if 𝑘=1, then 𝑅𝐵4𝑝𝛽11=2𝛽1+𝛽12+1𝛽1+2𝛽1𝛽1𝑗1=0𝑝𝑗11+2𝛽1𝛽1𝑗1=0𝑝𝑗11𝛽1𝑗1+1=2𝛽1𝛽1𝑝+2+𝛽11+11𝑝1+𝑝1𝛽11+2𝛽1𝑝+21+𝛽1+1𝑝112.(3.70)

Acknowledgments

The first author was funded by the Korean Government (KRF-2009-353-C00040). In the case of the third author, this research was supported by Basic Science Research Program Through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2011-0025252).