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Discrete Dynamics in Nature and Society
VolumeΒ 2012Β (2012), Article IDΒ 760246, 17 pages
doi:10.1155/2012/760246
Research Article

The Number of Chains of Subgroups in the Lattice of Subgroups of the Dicyclic Group

Ju-Mok Oh,1Β Yunjae Kim,2Β and Kyung-Won Hwang2

1Department of Mathematics, Kangnung-Wonju National University, Kangnung 210-702, Republic of Korea
2Department of Mathematics, Dong-A University, Busan 604-714, Republic of Korea

Received 9 May 2012; Accepted 25 July 2012

Academic Editor: Prasanta K.Β Panigrahi

Copyright Β© 2012 Ju-Mok Oh et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We give an explicit formula for the number of chains of subgroups in the lattice of subgroups of the dicyclic group 𝐡 4 𝑛 of order 4 𝑛 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 [1–4]). 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 𝐷 2 𝑛 of order 2 𝑛 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 𝐡 4 𝑛 of order 4 𝑛 by finding its generating function of multivariables where 𝑛 is an arbitrary integer.

2. Preliminaries

Given a group 𝐺 , let π’ž ( 𝐺 ) , 𝒰 ( 𝐺 ) , and β„› ( 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 𝐢 ( 𝐺 ) ∢ = | π’ž ( 𝐺 ) | , π‘ˆ ( 𝐺 ) ∢ = | 𝒰 ( 𝐺 ) | , and 𝑅 ( 𝐺 ) ∢ = | β„› ( 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 ) = π‘ˆ ( 𝐺 ) + 1 and 𝐢 ( 𝐺 ) = 𝑅 ( 𝐺 ) + π‘ˆ ( 𝐺 ) = 2 𝑅 ( 𝐺 ) βˆ’ 1 .

For a fixed positive integer π‘˜ , we define a function πœ† as follows: πœ† ξ€· π‘₯ π‘˜ ξ€Έ ∢ = 1 βˆ’ 2 π‘₯ π‘˜ , πœ† ξ€· π‘₯ π‘˜ , π‘₯ π‘˜ βˆ’ 1 , … , π‘₯ 𝑗 ξ€Έ ξ€· π‘₯ ∢ = πœ† π‘˜ , π‘₯ π‘˜ βˆ’ 1 , … , π‘₯ 𝑗 + 1 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ† π‘˜ , π‘₯ π‘˜ βˆ’ 1 , … , π‘₯ 𝑗 + 1 π‘₯ ξ€Έ ξ€Έ 𝑗 ( 2 . 1 ) for any 𝑗 = π‘˜ βˆ’ 1 , π‘˜ βˆ’ 2 , … , 1 .

Proposition 2.2 (see [5]). Let β„€ 𝑛 be the cyclic group of order 𝑛 = 𝑝 𝛽 1 1 𝑝 𝛽 2 2 β‹― 𝑝 𝛽 π‘˜ π‘˜ , ( 2 . 2 ) where 𝑝 1 , … , 𝑝 π‘˜ are distinct prime numbers and 𝛽 1 , … , 𝛽 π‘˜ are positive integers. Then the number 𝑅 ( β„€ 𝑛 ) of rooted chains of subgroups in the lattice of subgroups of β„€ 𝑛 is the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ . ( 2 . 3 )

Let β„€ be the set of all integer numbers. Given distinct positive integers 𝑖 1 , … , 𝑖 𝑑 , we define a function πœ‹ 𝑖 1 β‹― 𝑖 𝑑 ∢ β„€ π‘˜ ⟼ β„€ π‘˜ , ξ€· π‘₯ 1 , … , π‘₯ π‘˜ ξ€Έ ⟼ ξ€· 𝑦 1 , … , 𝑦 π‘˜ ξ€Έ , ( 2 . 4 ) where 𝑦 β„“ = ξ‚» π‘₯ β„“ , i f β„“ β‰  𝑖 𝑗 π‘₯ βˆ€ 𝑗 = 1 , … , 𝑑 , β„“ βˆ’ 1 , β„“ = 𝑖 𝑗 f o r s o m e 𝑗 s u c h t h a t 1 ≀ 𝑗 ≀ 𝑑 . ( 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 𝐡 4 𝑛

Throughout the section, we assume that 𝑛 ∢ = 𝑝 𝛽 1 1 𝑝 𝛽 2 2 β‹― 𝑝 𝛽 π‘˜ π‘˜ , ( 3 . 1 ) is a positive integer, where 𝑝 1 , … , 𝑝 π‘˜ are distinct prime numbers and 𝛽 1 , … , 𝛽 π‘˜ are nonnegative integers and the dicyclic group 𝐡 4 𝑛 of order 4 𝑛 is defined by the following presentation: 𝐡 4 𝑛  ∢ = π‘Ž , 𝑏 ∣ π‘Ž 2 𝑛 = 𝑒 , 𝑏 2 = π‘Ž 𝑛 , 𝑏 π‘Ž 𝑏 βˆ’ 1 = π‘Ž βˆ’ 1  , ( 3 . 2 ) where 𝑒 is the identity element.

By the elementary group theory, the following is wellknown.

Lemma 3.1. The dicyclic group 𝐡 4 𝑛 has an index 2 subgroup ⟨ π‘Ž ⟩ , which is isomorphic to β„€ 2 𝑛 , and has 𝑝 𝑖 index 𝑝 𝑖 subgroups ⟨ π‘Ž 𝑝 𝑖 , 𝑏 ⟩ , ⟨ π‘Ž 𝑝 𝑖  π‘Ž , π‘Ž 𝑏 ⟩ , … , 𝑝 𝑖 , π‘Ž 𝑝 𝑖 βˆ’ 1 𝑏  , ( 3 . 3 ) which are isomorphic to the dicyclic group 𝐡 4 𝑛 / 𝑝 𝑖 of order 4 𝑛 / 𝑝 𝑖 where 𝑖 = 1 , 2 , … , π‘˜ .

Lemma 3.2. (1) For any 𝑖 = 1 , 2 , … , π‘˜ , ⟨ π‘Ž 𝑝 𝑖 , π‘Ž π‘Ÿ 𝑏 ⟩ ∩ ⟨ π‘Ž 𝑝 𝑖 , π‘Ž 𝑠 𝑏 ⟩ = ⟨ π‘Ž 𝑝 𝑖 ⟩ β‰… β„€ 2 𝑛 / 𝑝 𝑖 , ( 3 . 4 ) where 0 ≀ π‘Ÿ < 𝑠 ≀ 𝑝 𝑖 βˆ’ 1 .
(2) For any distinct prime factors 𝑝 𝑖 1 , 𝑝 𝑖 2 , … , 𝑝 𝑖 𝑑 of 𝑛 , ⟨ π‘Ž 𝑝 𝑖 1 , π‘Ž π‘Ÿ 1 𝑏 ⟩ ∩ ⟨ π‘Ž 𝑝 𝑖 2 , π‘Ž π‘Ÿ 2 𝑏 ⟩ ∩ β‹― ∩ ⟨ π‘Ž 𝑝 𝑖 𝑑 , π‘Ž π‘Ÿ 𝑑 𝑏 ⟩ β‰… 𝐡 4 𝑛 / 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 , ( 3 . 5 ) where π‘Ÿ 1 , … , π‘Ÿ 𝑑 are nonnegative integers.

Proof. (1) To the contrary suppose that ⟨ π‘Ž 𝑝 𝑖 , π‘Ž π‘Ÿ 𝑏 ⟩ ∩ ⟨ π‘Ž 𝑝 𝑖 , π‘Ž 𝑠 𝑏 ⟩ β‰  ⟨ π‘Ž 𝑝 𝑖 ⟩ . ( 3 . 6 ) Then π‘Ž 𝑝 𝑖 𝑒 + π‘Ÿ 𝑏 = π‘Ž 𝑝 𝑖 𝑣 + 𝑠 𝑏 for some integers 𝑒 and 𝑣 . This implies 𝑝 𝑖 ∣ 𝑠 βˆ’ π‘Ÿ . Since 0 ≀ π‘Ÿ < 𝑠 ≀ 𝑝 𝑖 βˆ’ 1 , we have 𝑠 = π‘Ÿ , a contradiction.
(2) We only give its proof when 𝑑 = 2 . The general case can be proved by the inductive process. Let 𝐾 ∢ = ⟨ π‘Ž 𝑝 𝑖 1 , π‘Ž π‘Ÿ 1 𝑏 ⟩ ∩ ⟨ π‘Ž 𝑝 𝑖 2 , π‘Ž π‘Ÿ 2 𝑏 ⟩ . ( 3 . 7 ) Clearly, π‘Ž 𝑝 𝑖 1 𝑝 𝑖 2 ∈ 𝐾 . Since g c d ( 𝑝 𝑖 1 , 𝑝 𝑖 2 ) = 1 , there exist integers 𝑒 and 𝑣 such that 𝑝 𝑖 1 𝑒 + 𝑝 𝑖 2 𝑣 = 1 . Note that π‘Ž 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) + π‘Ÿ 1 𝑏 = π‘Ž 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) π‘Ž π‘Ÿ 1 𝑏 ∈ ⟨ π‘Ž 𝑝 𝑖 1 , π‘Ž π‘Ÿ 1 𝑏 ⟩ . On the other hand, π‘Ž 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) + π‘Ÿ 1 𝑏 = π‘Ž βˆ’ 𝑝 𝑖 1 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) + π‘Ÿ 1 𝑏 = π‘Ž 𝑝 𝑖 2 𝑣 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) βˆ’ ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) + π‘Ÿ 1 𝑏 s i n c e 𝑝 𝑖 1 𝑒 + 𝑝 𝑖 2 𝑣 = 1 = π‘Ž 𝑝 𝑖 2 𝑣 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) + π‘Ÿ 2 𝑏 ∈ ⟨ π‘Ž 𝑝 𝑖 2 , π‘Ž π‘Ÿ 2 𝑏 ⟩ . ( 3 . 8 ) Considering the order of 𝐾 , one can see that 𝐾 = ⟨ π‘Ž 𝑝 𝑖 1 𝑝 𝑖 2 , π‘Ž 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) + π‘Ÿ 1 𝑏 ⟩ . Since ξ€· π‘Ž 𝑝 𝑖 1 𝑝 𝑖 2 ξ€Έ 4 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 ξ€· π‘Ž = 𝑒 , 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) + π‘Ÿ 1 𝑏 ξ€Έ 2 = 𝑏 2 = π‘Ž 𝑛 = ξ€· π‘Ž 𝑝 𝑖 1 𝑝 𝑖 2 ξ€Έ 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 , ξ€· π‘Ž 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) + π‘Ÿ 1 𝑏 π‘Ž ξ€Έ ξ€· 𝑝 𝑖 1 𝑝 𝑖 2 π‘Ž ξ€Έ ξ€· 𝑝 𝑖 1 ( βˆ’ 𝑒 ( π‘Ÿ 1 βˆ’ π‘Ÿ 2 ) ) + π‘Ÿ 1 𝑏 ξ€Έ βˆ’ 1 = ξ€· π‘Ž 𝑝 𝑖 1 𝑝 𝑖 2 ξ€Έ βˆ’ 1 , ( 3 . 9 ) we have 𝐾 β‰… 𝐡 4 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 .

By Lemma 3.1, we have 𝒰 ξ€· 𝐡 4 𝑛 ξ€Έ ξ€· = π’ž ⟨ π‘Ž ⟩ β‰… β„€ 2 𝑛 ξ€Έ π‘˜  𝑝 𝑖 = 0 𝑖 βˆ’ 1  𝑗 = 0 π’ž π‘Ž ξ€·  𝑝 𝑖 , π‘Ž 𝑗 𝑏  β‰… 𝐡 4 𝑛 / 𝑝 𝑖 ξ€Έ . ( 3 . 1 0 ) Using the inclusion-exclusion principle and Lemma 3.2, one can see that the number π‘ˆ ( 𝐡 4 𝑛 ) has the following form: π‘ˆ ξ€· 𝐡 4 𝑛 ξ€Έ ξ€· β„€ = 𝐢 2 𝑛 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ 𝑧 𝑖 1 , … , 𝑖 𝑑 𝐢 ξ‚€ β„€ 2 𝑛 / 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑  +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ 𝑏 𝑖 1 , … , 𝑖 𝑑 𝐢 ξ‚€ 𝐡 4 𝑛 / 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑  ( 3 . 1 1 ) for suitable integers 𝑧 𝑖 1 , … , 𝑖 𝑑 and 𝑏 𝑖 1 , … , 𝑖 𝑑 . In the following, we determine the numbers 𝑧 𝑖 1 , … , 𝑖 𝑑 and 𝑏 𝑖 1 , … , 𝑖 𝑑 explicitly.

Lemma 3.3. (1) 𝑏 𝑖 1 , 𝑖 2 , … , 𝑖 𝑑 = ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 .
(2) 𝑧 𝑖 1 , 𝑖 2 , … , 𝑖 𝑑 = ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 .

Proof. (1) Clearly 𝑏 𝑖 1 = ( βˆ’ 1 ) 1 + 1 𝑝 𝑖 1 = 𝑝 𝑖 1 for any 𝑖 1 = 1 , … , π‘˜ . For any integer 𝑑 β‰₯ 2 , 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 𝐡 4 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 only appears in 𝑑 -intersection of the subgroups  π‘Ž 𝑝 𝑖 1 , π‘Ž 𝑗 1 𝑏  ,  π‘Ž 𝑝 𝑖 2 , π‘Ž 𝑗 2 𝑏   π‘Ž , … , 𝑝 𝑖 𝑑 , π‘Ž 𝑗 𝑑 𝑏  , ( 3 . 1 2 ) where 0 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ βˆ’ 1 and 1 ≀ π‘Ÿ ≀ 𝑑 . Since there are ξ€· 𝑝 𝑖 1 1 ξ€Έ ξ€· 𝑝 𝑖 2 1 ξ€Έ β‹― ξ€· 𝑝 𝑖 𝑑 1 ξ€Έ = 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 such choices, we have 𝑏 𝑖 1 , 𝑖 2 , … , 𝑖 𝑑 = ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 .
(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 β„€ 2 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 only appears one of the following two forms:  π‘Ž ⟨ π‘Ž ⟩ ∩ 𝑝 𝑖 1 , π‘Ž 𝑗 1 𝑏  ∩  π‘Ž 𝑝 𝑖 2 , π‘Ž 𝑗 2 𝑏   π‘Ž ∩ β‹― ∩ 𝑝 𝑖 𝑑 , π‘Ž 𝑗 𝑑 𝑏  ,  π‘Ž 𝑝 𝑖 1 , π‘Ž 𝑗 1 𝑏  ∩  π‘Ž 𝑝 𝑖 2 , π‘Ž 𝑗 2 𝑏   π‘Ž ∩ β‹― ∩ 𝑝 𝑖 𝑑 , π‘Ž 𝑗 𝑑 𝑏  , ( 3 . 1 3 ) where 0 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ βˆ’ 1 and 1 ≀ π‘Ÿ ≀ 𝑑 , 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 β„€ 2 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 obtained from the first form, and let 𝛿 be the number of the groups isomorphic to β„€ 2 𝑛 / 𝑝 𝑖 1 𝑝 𝑖 2 β‹― 𝑝 𝑖 𝑑 obtained from the second form. Then clearly 𝑧 𝑖 1 , 𝑖 2 , β‹― , 𝑖 𝑑 = 𝛾 + 𝛿 . Note that 𝛾 = 𝑝 𝑖 1 + β‹― + 𝑝 𝑖 𝑑 βˆ’ 𝑑  π‘˜ = 0 ( βˆ’ 1 ) 𝑑 + 2 + π‘˜  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 + π‘˜ , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ π‘˜  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ =  π‘˜ β‰₯ 0  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 + π‘˜ , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ , 1 ≀ π‘Ÿ ≀ 𝑑 ( βˆ’ 1 ) 𝑑 𝑑 + π‘˜  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ =  1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ , 1 ≀ π‘Ÿ ≀ 𝑑 ( βˆ’ 1 ) 𝑗 1 + β‹― + 𝑗 𝑑 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ = 𝑑  π‘Ÿ = 1  1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ ( βˆ’ 1 ) 𝑗 1 + β‹― + 𝑗 𝑑 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ = ( βˆ’ 1 ) 𝑑 . ( 3 . 1 4 ) On the other hand, 𝛿 = 𝑝 𝑖 1 + β‹― + 𝑝 𝑖 𝑑 βˆ’ 𝑑 βˆ’ 1  π‘˜ = 0 ( βˆ’ 1 ) 𝑑 + 2 + π‘˜  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 + 1 + π‘˜ , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ = 𝑝 𝑖 1 + β‹― + 𝑝 𝑖 𝑑 βˆ’ 𝑑 βˆ’ 1  π‘˜ = 0 ( βˆ’ 1 ) 𝑑 + 2 + π‘˜  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 + 1 + π‘˜ , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ + ( βˆ’ 1 ) 𝑑 + 1  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βˆ’ ( βˆ’ 1 ) 𝑑 + 1  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ = 𝑝 𝑖 1 + β‹― + 𝑝 𝑖 𝑑 βˆ’ 𝑑  π‘˜ = 0 ( βˆ’ 1 ) 𝑑 + 1 + π‘˜  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 + π‘˜ , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑗 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βˆ’ ( βˆ’ 1 ) 𝑑 + 1  𝑗 1 + β‹― + 𝑗 𝑑 = 𝑑 , 1 ≀ 𝑗 π‘Ÿ ≀ 𝑝 𝑖 π‘Ÿ 𝑑 , 1 ≀ π‘Ÿ ≀ 𝑑  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑝 𝑖 π‘Ÿ 𝑗 π‘Ÿ ⎞ ⎟ ⎟ ⎠ = ( βˆ’ 1 ) 𝑑 βˆ’ ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 . ( 3 . 1 5 ) Therefore, we have 𝑧 𝑖 1 , 𝑖 2 , … , 𝑖 𝑑 = ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 .

By Proposition 2.1 and Lemma 3.3, (3.11) becomes 𝑅 ξ€· 𝐡 4 𝑛 ξ€Έ ξ€· β„€ = 2 𝑅 2 𝑛 ξ€Έ  + 2 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑅 ξ‚€ β„€ 2 𝑛 / 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑   + 2 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑅 ξ‚€ 𝐡 4 𝑛 / 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑  . ( 3 . 1 6 ) Let π‘Ž 𝛽 1 , … , 𝛽 π‘˜ ∢ = 𝑅 ( 𝐡 4 𝑛 ) and let 𝑏 𝛽 1 , … , 𝛽 π‘˜ ∢ = 𝑅 ( β„€ 2 𝑛 ) . Then (3.16) becomes π‘Ž 𝛽 1 , … , 𝛽 π‘˜ = 2 𝑏 𝛽 1 , … , 𝛽 π‘˜  + 2 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑏 πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ )  + 2 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 π‘Ž πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) . ( 3 . 1 7 )

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 𝛽 𝑖 to be zero for computational convenience.

Let πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , π‘₯ π‘˜ βˆ’ 1 , … , π‘₯ 𝑗 ξ€Έ ∢ = ∞  𝛽 𝑗 = 0 β‹― ∞  𝛽 π‘˜ βˆ’ 1 ∞ = 0  𝛽 π‘˜ = 0 π‘Ž 𝛽 1 , … , 𝛽 π‘˜ π‘₯ 𝛽 π‘˜ π‘˜ π‘₯ 𝛽 π‘˜ βˆ’ 1 π‘˜ βˆ’ 1 β‹― π‘₯ 𝛽 𝑗 𝑗 , πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , π‘₯ π‘˜ βˆ’ 1 , … , π‘₯ 𝑗 ξ€Έ ∢ = ∞  𝛽 𝑗 = 0 β‹― ∞  𝛽 π‘˜ βˆ’ 1 ∞ = 0  𝛽 π‘˜ = 0 𝑏 𝛽 1 , … , 𝛽 π‘˜ π‘₯ 𝛽 π‘˜ π‘˜ π‘₯ 𝛽 π‘˜ βˆ’ 1 π‘˜ βˆ’ 1 β‹― π‘₯ 𝛽 𝑗 𝑗 , ( 3 . 1 8 ) where 𝑗 = π‘˜ , π‘˜ βˆ’ 1 , … , 1 .

For a fixed integer 𝑛 = 𝑝 𝛽 1 1 𝑝 𝛽 2 2 β‹― 𝑝 𝛽 π‘˜ π‘˜ such that 𝑝 1 , … , 𝑝 π‘˜ are distinct prime numbers and 𝛽 1 , … , 𝛽 π‘˜ are non-negative integers, we define a function πœ‡ as follows. πœ‡ ξ€· π‘₯ π‘˜ ξ€Έ ∢ = 1 βˆ’ 2 𝑝 π‘˜ π‘₯ π‘˜ , πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ ξ€· π‘₯ ∢ = πœ‡ π‘˜ , … , π‘₯ 𝑗 + 1 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗 + 1 𝑝 ξ€Έ ξ€Έ 𝑗 π‘₯ 𝑗 ( 3 . 1 9 ) for any 𝑗 = π‘˜ βˆ’ 1 , π‘˜ βˆ’ 2 , … , 1 .

Lemma 3.4. Let π‘˜ be a positive integer. If π‘˜ = 1 , then πœ‡ ξ€· π‘₯ 1 ξ€Έ πœ“ 𝛽 1 ξ€· π‘₯ 1 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ 1 πœ™ ξ€Έ ξ€Έ 𝛽 1 ξ€· π‘₯ 1 ξ€Έ . ( 3 . 2 0 ) If π‘˜ β‰₯ 2 , then πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗 Γ— ⎑ ⎒ ⎒ ⎒ ⎣ πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 1 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 1 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ ( 3 . 2 1 ) for any 𝑗 = π‘˜ , π‘˜ βˆ’ 1 , … , 2 .

Proof. Assume first that π‘˜ = 1 . Then (3.17) with π‘˜ = 1 gives us that π‘Ž 𝛽 1 = 2 𝑏 𝛽 1 + 2 𝑝 1 π‘Ž 𝛽 1 βˆ’ 1 βˆ’ 2 𝑝 1 𝑏 𝛽 1 βˆ’ 1 . ( 3 . 2 2 ) Taking βˆ‘ ∞ 𝛽 1 = 1 to both sides of (3.22), we have ξ€· 1 βˆ’ 2 𝑝 1 π‘₯ 1 ξ€Έ πœ“ 𝛽 1 ξ€· π‘₯ 1 ξ€Έ = ξ€· 2 βˆ’ 2 𝑝 1 π‘₯ 1 ξ€Έ πœ™ 𝛽 1 ξ€· π‘₯ 1 ξ€Έ ( 3 . 2 3 ) because π‘Ž 0 = 𝑅 ( 𝐡 4 𝑝 0 1 ) = 𝑅 ( β„€ 2 2 ) = 2 2 and 𝑏 0 = 𝑅 ( β„€ 2 𝑝 0 1 ) = 𝑅 ( β„€ 2 ) = 2 by a direct computation.
From now on, we assume that π‘˜ β‰₯ 2 . We prove (3.21) by double induction on π‘˜ and 𝑗 . Equation (3.17) with π‘˜ = 2 gives us that π‘Ž 𝛽 1 , 𝛽 2 = 2 𝑏 𝛽 1 , 𝛽 2 βˆ’ 2 𝑝 1 𝑏 𝛽 1 βˆ’ 1 , 𝛽 2 βˆ’ 2 𝑝 2 𝑏 𝛽 1 , 𝛽 2 βˆ’ 1 + 2 𝑝 1 𝑝 2 𝑏 𝛽 1 βˆ’ 1 , 𝛽 2 βˆ’ 1 + 2 𝑝 1 π‘Ž 𝛽 1 βˆ’ 1 , 𝛽 2 + 2 𝑝 2 π‘Ž 𝛽 1 , 𝛽 2 βˆ’ 1 βˆ’ 2 𝑝 1 𝑝 2 π‘Ž 𝛽 1 βˆ’ 1 , 𝛽 2 βˆ’ 1 . ( 3 . 2 4 ) Taking βˆ‘ ∞ 𝛽 2 = 1 π‘₯ 𝛽 2 2 of both sides of (3.24), we have ξ€· 1 βˆ’ 2 𝑝 2 π‘₯ 2 ξ€Έ πœ“ 𝛽 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ = ξ€· 2 βˆ’ 2 𝑝 2 π‘₯ 2 𝑝 ξ€Έ ξ€Ί 1 πœ“ 𝛽 1 βˆ’ 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ + πœ™ 𝛽 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ βˆ’ 𝑝 1 πœ™ 𝛽 1 βˆ’ 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ ξ€» ( 3 . 2 5 ) because π‘Ž 𝛽 1 , 0 = π‘Ž 𝛽 1 and 𝑏 𝛽 1 , 0 = 𝑏 𝛽 1 by the definition, and π‘Ž 𝛽 1 , 0 βˆ’ 2 𝑏 𝛽 1 , 0 βˆ’ 2 𝑝 1 π‘Ž 𝛽 1 , 0 + 2 𝑝 1 𝑏 𝛽 1 βˆ’ 1 , 0 = 0 ( 3 . 2 6 ) by (3.17) with π‘˜ = 1 . That is, πœ‡ ξ€· π‘₯ 2 ξ€Έ πœ“ 𝛽 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ 2 πœ™ ξ€Έ ξ€Έ ξ€Ί 𝛽 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ βˆ’ 𝑝 1 πœ™ 𝛽 1 βˆ’ 1 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ + 𝑝 1 πœ“ 𝛽 1 βˆ’ 1 , 𝛽 2 ξ€· π‘₯ 2 . ξ€Έ ξ€» ( 3 . 2 7 ) Thus (3.21) holds for π‘˜ = 2 .
Assume now that (3.21) holds from 2 to π‘˜ βˆ’ 1 and consider the case for π‘˜ . Note that the last two terms of the right-hand side of (3.17) can be divided into three terms, respectively, as follows: 2  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑏 πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) = βˆ’ 2 𝑝 π‘˜ 𝑏 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 𝛽 π‘˜ βˆ’ 1 βˆ’ 2 𝑝 π‘˜  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑏 πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 𝛽 π‘˜ βˆ’ 1 )  + 2 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑏 πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) , 2  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ , 1 ≀ 𝑑 ≀ π‘˜ ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 π‘Ž πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) = 2 𝑝 π‘˜ π‘Ž 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 𝛽 π‘˜ βˆ’ 1 βˆ’ 2 𝑝 π‘˜  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 π‘Ž πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 𝛽 π‘˜ βˆ’ 1 )  + 2 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 π‘Ž πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) . ( 3 . 2 8 ) Taking βˆ‘ ∞ 𝛽 π‘˜ = 1 π‘₯ 𝛽 π‘˜ π‘˜ of both sides of (3.17) and using (3.28), one can see that ξ€· 1 βˆ’ 2 𝑝 π‘˜ π‘₯ π‘˜ ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ ξ€Έ = ξ€· 2 βˆ’ 2 𝑝 π‘˜ π‘₯ π‘˜ ξ€Έ Γ— ⎑ ⎒ ⎒ ⎒ ⎣ πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ ξ€Έ +  1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ ξ€Έ +  1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ ξ€Έ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ + π‘Ž 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0 βˆ’ 2 𝑏 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0  βˆ’ 2 1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑏 πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0 )  βˆ’ 2 1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 π‘Ž πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0 ) . ( 3 . 2 9 ) Further since π‘Ž 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0 βˆ’ 2 𝑏 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0  βˆ’ 2 1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 𝑏 πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0 )  βˆ’ 2 1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 π‘Ž πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 0 ) = 0 ( 3 . 3 0 ) by (3.17), we have ξ€· 1 βˆ’ 2 𝑝 π‘˜ π‘₯ π‘˜ ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ ξ€Έ = ξ€· 2 βˆ’ 2 𝑝 π‘˜ π‘₯ π‘˜ ξ€Έ Γ— ⎑ ⎒ ⎒ ⎒ ⎣ πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ ξ€Έ +  1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ βˆ’ 1 , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ ξ€Έ +  1 ≀ 𝑖 < β‹― < 𝑖 𝑑 ≀ π‘˜ βˆ’ 1 , 1 ≀ 𝑑 ≀ π‘˜ βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ ξ€Έ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ . ( 3 . 3 1 ) Thus (3.21) holds for 𝑗 = π‘˜ . Assume that (3.21) holds from π‘˜ to 𝑗 and consider the case for 𝑗 βˆ’ 1 . Note that the last two terms of the right-hand side of (3.21) can be divided into three terms, respectively, as follows:  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 1 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 1 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ = βˆ’ 𝑝 𝑗 βˆ’ 1 πœ™ 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 𝛽 𝑗 βˆ’ 1 βˆ’ 1 , 𝛽 𝑗 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ βˆ’ 𝑝 𝑗 βˆ’ 1  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 𝛽 𝑗 βˆ’ 1 βˆ’ 1 , 𝛽 𝑗 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ ,  ( 3 . 3 2 ) 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 1 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 1 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ = 𝑝 𝑗 βˆ’ 1 πœ“ 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 𝛽 𝑗 βˆ’ 1 βˆ’ 1 , 𝛽 𝑗 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ βˆ’ 𝑝 𝑗 βˆ’ 1  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 𝛽 𝑗 βˆ’ 1 βˆ’ 1 , 𝛽 𝑗 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ . ( 3 . 3 3 ) Taking βˆ‘ ∞ 𝛽 𝑗 βˆ’ 1 = 1 π‘₯ 𝛽 𝑗 βˆ’ 1 𝑗 βˆ’ 1 of both sides of (3.21), we have πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ 𝑗 βˆ’ 1 , … , π‘₯ π‘˜ ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 Γ— ⎑ ⎒ ⎒ ⎒ ⎣ πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ 𝑗 βˆ’ 1 , … , π‘₯ π‘˜ ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 ξ€Έ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ ξ€· π‘₯ + πœ‡ π‘˜ , … , π‘₯ 𝑗 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 0 , 𝛽 𝑗 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗  ξ€Έ ξ€Έ 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 0 , 𝛽 𝑗 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗  ξ€Έ ξ€Έ 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 0 , 𝛽 𝑗 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ . ( 3 . 3 4 ) Note that πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 0 , 𝛽 𝑗 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗  ξ€Έ ξ€Έ 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 0 , 𝛽 𝑗 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗  ξ€Έ ξ€Έ 1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 𝑗 βˆ’ 2 , 0 , 𝛽 𝑗 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 ξ€Έ = 0 ( 3 . 3 5 ) by induction hypothesis. Thus πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ 𝑗 βˆ’ 1 , … , π‘₯ π‘˜ ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 Γ— ⎑ ⎒ ⎒ ⎒ ⎣ πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ 𝑗 βˆ’ 1 , … , π‘₯ π‘˜ ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ™ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 ξ€Έ +  1 ≀ 𝑖 1 < β‹― < 𝑖 𝑑 ≀ 𝑗 βˆ’ 2 , 1 ≀ 𝑑 ≀ 𝑗 βˆ’ 2 ( βˆ’ 1 ) 𝑑 + 1 𝑝 𝑖 1 β‹― 𝑝 𝑖 𝑑 πœ“ πœ‹ 𝑖 1 𝑑 β‹― 𝑖 ( 𝛽 1 , … , 𝛽 π‘˜ ) ξ€· π‘₯ π‘˜ , … , π‘₯ 𝑗 , π‘₯ 𝑗 βˆ’ 1 ξ€Έ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ . ( 3 . 3 6 ) Therefore, (3.21) holds for 𝑗 βˆ’ 1 .

Equation (3.21) with 𝑗 = 2 gives us that πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 2 Γ— ξ€Ί πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ βˆ’ 𝑝 1 πœ™ 𝛽 1 βˆ’ 1 , 𝛽 2 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ + 𝑝 1 πœ“ 𝛽 1 βˆ’ 1 , 𝛽 2 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 . ξ€Έ ξ€» ( 3 . 3 7 ) Taking βˆ‘ ∞ 𝛽 1 = 1 π‘₯ 𝛽 1 1 of both sides of (3.37), we get that πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 , π‘₯ 1 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 1 πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 , π‘₯ 1 ξ€Έ ξ€· π‘₯ + πœ‡ π‘˜ , … , π‘₯ 2 ξ€Έ πœ“ 0 , 𝛽 2 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 2 πœ™ ξ€Έ ξ€Έ 0 , 𝛽 2 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ . ( 3 . 3 8 )

Lemma 3.5. If π‘˜ β‰₯ 2 , then πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ πœ“ 0 , 𝛽 2 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 2 πœ™ ξ€Έ ξ€Έ 0 , 𝛽 2 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ . ( 3 . 3 9 )

Proof. If π‘˜ = 2 , then since πœ“ 0 , 𝛽 2 ( π‘₯ 2 ) = πœ“ 𝛽 2 ( π‘₯ 2 ) and πœ™ 0 , 𝛽 2 ( π‘₯ 2 ) = πœ™ 𝛽 2 ( π‘₯ 2 ) , the equation πœ‡ ξ€· π‘₯ 2 ξ€Έ πœ“ 0 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ 2 πœ™ ξ€Έ ξ€Έ 0 , 𝛽 2 ξ€· π‘₯ 2 ξ€Έ ( 3 . 4 0 ) holds by (3.20). Assume now that (3.39) holds for π‘˜ . Then by (3.38) we get that πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 , π‘₯ 1 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 1 πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 , π‘₯ 1 ξ€Έ , ( 3 . 4 1 ) which implies that πœ‡ ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ πœ“ 0 , 𝛽 2 , … , 𝛽 π‘˜ + 1 ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ + 1 , … , π‘₯ 2 πœ™ ξ€Έ ξ€Έ 0 , 𝛽 2 , … , 𝛽 π‘˜ + 1 ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ . ( 3 . 4 2 ) Thus (3.39) holds for π‘˜ + 1 .

By Lemmas 3.4 and 3.5 and (3.38), we have πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = ξ€· ξ€· π‘₯ 1 + πœ‡ π‘˜ , … , π‘₯ 1 πœ™ ξ€Έ ξ€Έ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ . ( 3 . 4 3 ) We now need to find the function πœ™ 𝛽 1 , … , 𝛽 π‘˜ ( π‘₯ π‘˜ , … , π‘₯ 1 ) explicitly.

Lemma 3.6. If 𝑝 1 = 2 , then πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = ⎧ βŽͺ βŽͺ ⎨ βŽͺ βŽͺ ⎩ 2 πœ† ξ€· π‘₯ 1 ξ€Έ  1 , 𝑖 𝑓 π‘˜ = 1 , 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ ξƒ­ 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ , 𝑖 𝑓 π‘˜ β‰₯ 2 . ( 3 . 4 4 ) If 𝑝 𝑖 β‰  2 for 𝑖 = 1 , 2 , … , π‘˜ , then πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ =  1 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ ξƒ­ 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ . ( 3 . 4 5 )

Proof. We first assume that 𝑝 1 = 2 . Then by Proposition 2.2, 𝑏 𝛽 1 , 𝛽 2 , … , 𝛽 π‘˜ ξ‚€ β„€ = 𝑅 𝑝 𝛽 1 1 + 1 𝑝 𝛽 2 2 𝑝 𝛽 3 3 β‹― 𝑝 𝛽 π‘˜ π‘˜  ( 3 . 4 6 ) is the coefficient of π‘₯ 𝛽 1 1 + 1 π‘₯ 𝛽 2 2 π‘₯ 𝛽 3 3 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ , ( 3 . 4 7 ) which implies that 𝑏 𝛽 1 , 𝛽 2 , … , 𝛽 π‘˜ is the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 π‘₯ 𝛽 3 3 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 2 πœ† ξ€· π‘₯ 1 ξ€Έ  1 i f π‘˜ = 1 , 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ ξƒ­ 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ i f π‘˜ β‰₯ 2 , ( 3 . 4 8 ) and hence by the definition of πœ™ we get that πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = ⎧ βŽͺ βŽͺ ⎨ βŽͺ βŽͺ ⎩ 2 πœ† ξ€· π‘₯ 1 ξ€Έ  1 , i f π‘˜ = 1 , 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ ξƒ­ 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ , i f π‘˜ β‰₯ 2 . ( 3 . 4 9 )
Assume now that 𝑝 𝑖 β‰  2 for 𝑖 = 1 , 2 , … , π‘˜ . Since 𝑏 𝛽 1 , … , 𝛽 π‘˜ = 𝑅 ( β„€ 2 𝑝 𝛽 1 1 𝑝 𝛽 2 2 β‹― 𝑝 𝛽 π‘˜ π‘˜ ) , by Proposition 2.2 𝑏 𝛽 1 , … , 𝛽 π‘˜ is the coefficient of π‘₯ 1 1 π‘₯ 𝛽 1 2 π‘₯ 𝛽 2 3 β‹― π‘₯ 𝛽 π‘˜ π‘˜ + 1 of 1 πœ† ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 1 ξ€Έ . ( 3 . 5 0 ) Since 1 πœ† ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 1 ξ€Έ = 1 πœ† ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ βˆ’ ξ€· ξ€· π‘₯ 1 + πœ† π‘˜ + 1 , … , π‘₯ 2 π‘₯ ξ€Έ ξ€Έ 1 = 1 πœ† ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ 1 ξ€Ί ξ€· ξ€· π‘₯ 1 βˆ’ 1 + 1 / πœ† π‘˜ + 1 , … , π‘₯ 2 π‘₯ ξ€Έ ξ€Έ ξ€» 1 ( 3 . 5 1 ) by the definition, 𝑏 𝛽 1 , … , 𝛽 π‘˜ is the coefficient of π‘₯ 𝛽 1 2 π‘₯ 𝛽 2 3 β‹― π‘₯ 𝛽 π‘˜ π‘˜ + 1 of 1 πœ† ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ  1 1 + πœ† ξ€· π‘₯ π‘˜ + 1 , … , π‘₯ 2 ξ€Έ ξƒ­ . ( 3 . 5 2 ) By changing the variables π‘₯ 2 , π‘₯ 3 , … , π‘₯ π‘˜ + 1 by π‘₯ 1 , π‘₯ 2 , … , π‘₯ π‘˜ , respectively, we get that 𝑏 𝛽 1 , … , 𝛽 π‘˜ is the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ  1 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ ξƒ­ . ( 3 . 5 3 ) By the definition of πœ™ 𝛽 1 , … , 𝛽 π‘˜ ( π‘₯ π‘˜ , … , π‘₯ 1 ) , we have πœ™ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ  1 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ ξƒ­ . ( 3 . 5 4 )

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

Theorem 3.7. Let 𝑛 ∢ = 𝑝 𝛽 1 1 𝑝 𝛽 2 2 β‹― 𝑝 𝛽 π‘˜ π‘˜ , ( 3 . 5 5 ) be a positive integer such that 𝑝 1 , … , 𝑝 π‘˜ are distinct prime numbers and 𝛽 1 , … , 𝛽 π‘˜ are positive integers. Let 𝐡 4 𝑛  ∢ = π‘Ž , 𝑏 ∣ π‘Ž 2 𝑛 = 𝑒 , 𝑏 2 = π‘Ž 𝑛 , 𝑏 π‘Ž 𝑏 βˆ’ 1 = π‘Ž βˆ’ 1  ( 3 . 5 6 ) be the dicyclic group of order 4 𝑛 . Let 𝑅 ( 𝐡 4 𝑛 ) be the number of rooted chains of subgroups in the lattice of subgroups of 𝐡 4 𝑛 . (1)If 𝑝 1 = 2 , then 𝑅 ( 𝐡 4 𝑛 ) is the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = ⎧ βŽͺ βŽͺ ⎨ βŽͺ βŽͺ ⎩  1 1 + πœ‡ ξ€· π‘₯ 1 ξ€Έ ξƒ­ 2 πœ† ξ€· π‘₯ 1 ξ€Έ  1 , 𝑖 𝑓 π‘˜ = 1 , 1 + πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ 1 ξƒ­  1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ ξƒ­ 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ , 𝑖 𝑓 π‘˜ β‰₯ 2 . ( 3 . 5 7 ) (2)If 𝑝 𝑖 β‰  2 for 𝑖 = 1 , 2 , … , π‘˜ , then 𝑅 ( 𝐡 4 𝑛 ) is the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ =  1 1 + πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ ξƒ­ 1 πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ  1 1 + πœ† ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ ξƒ­ . ( 3 . 5 8 )
Furthermore, the number 𝐢 ( 𝐡 4 𝑛 ) of chains of subgroups in the lattice of subgroups of 𝐡 4 𝑛 is the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 2 πœ“ 𝛽 1 , … , 𝛽 π‘˜ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ βˆ’ π‘˜  𝑖 = 1 1 1 βˆ’ π‘₯ 𝑖 . ( 3 . 5 9 )

We now want to find the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of πœ“ 𝛽 1 , … , 𝛽 π‘˜ ( π‘₯ π‘˜ , … , π‘₯ 1 ) explicitly. Since 1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 1 ξ€Έ = 1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ 1 ξ€Ί ξ€· ξ€· π‘₯ 1 βˆ’ 1 + 1 / πœ‡ π‘˜ , … , π‘₯ 2 𝑝 ξ€Έ ξ€Έ ξ€» 1 π‘₯ 1 , ( 3 . 6 0 ) by the definition, the coefficient of π‘₯ 𝛽 1 1 of 1 / πœ‡ ( π‘₯ π‘˜ , … , π‘₯ 1 ) is 1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ  1 1 + πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ ξƒ­ 𝛽 1 𝑝 𝛽 1 1 = 𝑝 𝛽 1 1 𝛽 1  𝑖 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 1 ⎞ ⎟ ⎟ ⎠  1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 2 ξ€Έ ξƒ­ 𝑖 1 + 1 = 𝑝 𝛽 1 1 𝛽 1  𝑖 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 1 ⎞ ⎟ ⎟ ⎠  1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 3 ξ€Έ ξƒ­ 𝑖 1 + 1  1 ξ€Ί ξ€· ξ€· π‘₯ 1 βˆ’ 1 + 1 / πœ‡ π‘˜ , … , π‘₯ 3 𝑝 ξ€Έ ξ€Έ ξ€» 2 π‘₯ 2 ξƒ­ 𝑖 1 + 1 . ( 3 . 6 1 ) Thus the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 of 1 / πœ‡ ( π‘₯ π‘˜ , … , π‘₯ 1 ) is 𝑝 𝛽 1 1 𝑝 𝛽 2 2 𝛽 1  𝑖 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 1 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝑖 1 + 𝛽 2 𝛽 2 ⎞ ⎟ ⎟ ⎠  1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 3 ξ€Έ ξƒ­ 𝑖 1 + 1  1 1 + πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 3 ξ€Έ ξƒ­ 𝛽 2 = 𝑝 𝛽 1 1 𝑝 𝛽 2 2 𝛽 1  𝑖 1 𝛽 = 0 2  𝑖 2 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 1 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝛽 2 𝑖 2 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝑖 1 + 𝛽 2 𝛽 2 ⎞ ⎟ ⎟ ⎠  1 πœ‡ ξ€· π‘₯ π‘˜ , … , π‘₯ 3 ξ€Έ ξƒ­ 𝑖 1 + 𝑖 2 + 1 . ( 3 . 6 2 ) Continuing this process, one can see that the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 1 / πœ‡ ( π‘₯ π‘˜ , … , π‘₯ 1 ) is 2 𝛽 π‘˜ 𝑝 𝛽 1 1 𝑝 𝛽 2 2 β‹― 𝑝 𝛽 π‘˜ π‘˜ 𝛽 1  𝑖 1 𝛽 = 0 2  𝑖 2 = 0 β‹― 𝛽 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 π‘˜ βˆ’ 1  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝛽 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝛽 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝛽 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ . ( 3 . 6 3 ) Similarly one can see that the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 1 / πœ† ( π‘₯ π‘˜ , … , π‘₯ 1 ) is 2 𝛽 π‘˜ 𝛽 1  𝑖 1 𝛽 = 0 2  𝑖 2 = 0 β‹― 𝛽 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 π‘˜ βˆ’ 1  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝛽 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝛽 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝛽 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ , ( 3 . 6 4 ) the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of [ 1 + ( 1 / πœ† ( π‘₯ π‘˜ , … , π‘₯ 2 ) ) ] ( 1 / πœ† ( π‘₯ π‘˜ , … , π‘₯ 1 ) ) is 2 𝛽 π‘˜ 𝛽 1 + 1  𝑖 1 𝛽 = 0 2  𝑖 2 𝛽 = 0 3  𝑖 3 = 0 β‹― 𝛽 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 + 1 1 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝛽 2 + 𝑖 1 𝛽 2 ⎞ ⎟ ⎟ ⎠ π‘˜ βˆ’ 1  π‘Ÿ = 2 βŽ› ⎜ ⎜ ⎝ 𝛽 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝛽 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝛽 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ ( 3 . 6 5 ) and the coefficient of π‘₯ 𝛽 1 1 π‘₯ 𝛽 2 2 β‹― π‘₯ 𝛽 π‘˜ π‘˜ of 1 / πœ† ( π‘₯ π‘˜ , … , π‘₯ 1 ) 2 is ξ€· 𝛽 1 ξ€Έ 2 + 1 𝛽 π‘˜ 𝛽 1  𝑖 1 𝛽 = 0 2  𝑖 2 = 0 β‹― 𝛽 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 π‘˜ βˆ’ 1  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝛽 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝛽 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝛽 + 1 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ . ( 3 . 6 6 )

Therefore, one can have the following.

Corollary 3.8. Let 𝑛 and 𝐡 4 𝑛 be the positive integer and the dicyclic group, respectively, defined in Theorem 3.7. Let 𝑅 ( 𝐡 4 𝑛 ) be the number of rooted chains of subgroups in the lattice of subgroups of 𝐡 4 𝑛 . (1)If 𝑝 1 = 2 , then 𝑅 ξ€· 𝐡 4 𝑛 ξ€Έ = 2 𝛽 π‘˜ 𝛽 1 + 1  𝑖 1 𝛽 = 0 2  𝑖 2 𝛽 = 0 3  𝑖 3 = 0 β‹― 𝛽 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 + 1 1 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝛽 2 + 𝑖 1 𝛽 2 ⎞ ⎟ ⎟ ⎠ π‘˜ βˆ’ 1  π‘Ÿ = 2 βŽ› ⎜ ⎜ ⎝ 𝛽 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝛽 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝛽 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ + 2 𝛽 π‘˜ 𝛽 1  𝑗 1 𝛽 = 0 2  𝑗 2 = 0 β‹― 𝛽 π‘˜  𝑗 π‘˜ = 0 ⎑ ⎒ ⎒ ⎒ ⎣ ⎑ ⎒ ⎒ ⎒ ⎣ 𝑝 𝑗 1 1 𝑝 𝑗 2 2 β‹― 𝑝 𝑗 π‘˜ π‘˜ 𝑗 1  𝑖 1 𝑗 = 0 2  𝑖 2 = 0 β‹― 𝑗 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 π‘˜ βˆ’ 1  π‘Ÿ = 1 βŽ› ⎜ ⎜ ⎝ 𝑗 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝑗 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝑗 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ Γ— ⎑ ⎒ ⎒ ⎣ 𝛽 1 βˆ’ 𝑗 1 + 1  𝑖 1 𝛽 = 0 2 βˆ’ 𝑗 2  𝑖 2 𝛽 = 0 3 βˆ’ 𝑗 3  𝑖 3 = 0 β‹― 𝛽 π‘˜ βˆ’ 1 βˆ’ 𝑗 π‘˜ βˆ’ 1  𝑖 π‘˜ βˆ’ 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 βˆ’ 𝑗 1 𝑖 + 1 1 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝛽 2 βˆ’ 𝑗 2 + 𝑖 1 𝛽 2 βˆ’ 𝑗 2 ⎞ ⎟ ⎟ ⎠ Γ— π‘˜ βˆ’ 1  π‘Ÿ = 2 βŽ› ⎜ ⎜ ⎝ 𝛽 π‘Ÿ βˆ’ 𝑗 π‘Ÿ 𝑖 π‘Ÿ ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎜ ⎝ 𝛽 π‘Ÿ + 1 βˆ’ 𝑗 π‘Ÿ + 1 + π‘Ÿ βˆ‘ π‘š = 1 𝑖 π‘š 𝛽 π‘Ÿ + 1 βˆ’ 𝑗 π‘Ÿ + 1 ⎞ ⎟ ⎟ ⎟ ⎠ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ ⎀ βŽ₯ βŽ₯ βŽ₯ ⎦ , ( 3 . 6 7 ) where if π‘˜ = 1 , then 𝑅 ( 𝐡 4 β‹… 2 𝛽 1 ) = 2 2 𝛽 1 + 2 and if π‘˜ = 2 , then 𝑅 ξ‚€ 𝐡 4 β‹… 2 𝛽 1 𝑝 𝛽 2 2  = 2 𝛽 2 𝛽 1 + 1  𝑖 1 = 0 βŽ› ⎜ ⎜ ⎝ 𝛽 1 𝑖 + 1 1 ⎞ ⎟ ⎟ ⎠ βŽ› ⎜ ⎜ ⎝ 𝛽 2 + 𝑖 1