the spectrum of cyclic bsa(ν, 3, λ; α) with α = 2, 3

The Spectrum of Cyclic BSA(vv, 3, ��; ��)with �� ¼ 2, 3

Jing Zhang, Yanxun ChangInstitute of Mathematics, Beijing Jiaotong University, Beijing 100044,People’s Republic of China, E-mail: [email protected]; [email protected]

Received April 18, 2004; revised January 13, 2005

Published online 19 May 2005 in Wiley InterScience (www.interscience.wiley.com).

DOI 10.1002/jcd.20049

Abstract: Balanced sampling plans excluding contiguous units (or BSEC) were first

introduced by Hedayat, Rao, and Stufken in 1988. In this paper, we generalize the concept

of a cyclic BSEC to a cyclic balanced sampling plan to avoid the selection of adjacent units (or

CBSA for short) and use Langford and extended Langford sequences to construct a cyclic

BSA(vv, 3, ��; ��) with �� ¼ 2; 3. We finally establish the necessary and sufficient conditions for

the existence of a cyclic BSA(vv, 3, ��; ��) where �� ¼ 2; 3. # 2005 Wiley Periodicals, Inc. J Combin

Designs 13: 313–335, 2005

Keywords: Langford sequence; extended Langford sequence; balanced sampling plans avoiding

adjacent units (or BSA); automorphism group; BSEC

1. INTRODUCTION

A balanced sampling plan avoiding adjacent units (or BSA) is designed for a surveyplan when several adjacent units provide similar information (see [3], [11]). Weintroduce the definition of BSA as below.

Definition 1.1. A k-sized balanced sampling plan avoiding adjacent units is a pairðX;BÞ, where X ¼ Zv and B is a collection (so repetitions are allowed) of k-subsetsof X called blocks, such that no pair of points ði; jÞ appears in any block ifi� j � �1; . . . ;�� ðmod vÞ, while any other pair of points appears in exactly �blocks.

We will use the notation BSAðv; k; �;�Þ to denote such a design. Let Zv ¼ f0;1; . . . ; v � 1g denote the cyclic additive group of order v and ðX;BÞ a BSAðv; k; �;�Þ.If Zv is an automorphism group of the BSAðv; k; �;�Þ, then we call ðX;BÞ cyclic and

Contract grant sponsor: TRAPOYT; Contract grant sponsor: NSFC; Contract grant number: 10371002.

# 2005 Wiley Periodicals, Inc.

313

denote it by CBSAðv; k; �;�Þ. It is easy to see that a CBSAðv; k; �; 1Þ is equivalent to aCBSECðv; k; �Þ (see [11]). For more backgrounds on BSECs, the reader is referred to[2, 4, 5, 8]. In [12], the authors established the necessary and sufficient conditions forthe existence of a CBSECðv; 3; �Þ. We quote the result as follows.

Theorem 1.2. The necessary and sufficient conditions for the existence of aCBSECðv; 3; �Þ are either v 2 f1; 3g, or v � 9 and �ðv � 3Þ � 0 ðmod 6Þ butv 6� 2 ðmod 4Þ when � � 2 ðmod 4Þ.

Wei [11] has treated CBSAs of block size three with � ¼ 1; 2 and arbitrary �.

Theorem 1.3 (Wei [11]). (1) There exists a CBSAð6t þ 2�þ 1; 3; 1;�Þ for allinteger t � 2�þ 1. (2) There exists a CBSAðv; 3; 2;�Þ for the following values of vand � :

(i) � is even, v ¼ 12t þ 2�þ 4 and t � 2�þ 1;(ii) � is odd, v ¼ 12t þ 2�þ 10 and t � 2�þ 1;

(iii) any � � 1, v ¼ 6t þ 2�þ 1 and t � 2�þ 1:

First we give the necessary conditions for the existence of a CBSAðv; 3; �;�Þ.We note that every element x lies in v � 2�� 1 distinct pairs of the form fx; ygin which x� y 6� �1;�2; . . . ;�� ðmod vÞ, it follows that �ðv � 2�� 1Þ � 0ðmod 2Þ. Considering the collection of triples in a CBSAðv; 3; �;�Þ, we find�vðv � 2�� 1Þ=2 pairs in total, three per triple, so that �vðv � 2�� 1Þ� 0 ðmod 6Þ.From [10], we have v � 3ð2�þ 1Þ. Through simple calculation, we know thatf0; v=3; 2v=3g is the only possible short orbit for a CBSAðv; 3; �;�Þ. Let a; b denotethe number of full orbits and short orbits in a CBSAðv; 3; �;�Þ, respectively, then6aþ 2b ¼ �ðv � 2�� 1Þ and b � �. In [1], it is proved that there does not exist aCBSAðv; 3; �;�Þ when v � 2 ðmod 4Þ and � � 2 ðmod 4Þ. So, we have the followinglemma.

Lemma 1.4. If a CBSAðv; 3; �;�Þ exists, then v � 3ð2�þ 1Þ and(1) v � 3; 2�þ 1 ðmod 6Þ when � � 1; 5 ðmod 6Þ but v 6� 3 ðmod 6Þ when

� ¼ 1 and � 6� 0 ðmod 3Þ;(2) v � 0; 2�þ 1 ðmod 3Þ and v 6� 2 ðmod 4Þ when � � 2; 10 ðmod 12Þ;(3) v � 1 ðmod 2Þ when � � 3 ðmod 6Þ;(4) v � 0; 2�þ 1 ðmod 3Þ when � � 4; 8 ðmod 12Þ;(5) v 6� 2 ðmod 4Þ when � � 6 ðmod 12Þ;(6) v is unrestricted when � � 0 ðmod 12Þ:

If ðZv;B1Þ is a CBSAðv; k; �1;�Þ and ðZv;B2Þ is a CBSAðv; k; �2;�Þ, then ðZv;BÞis a CBSAðv; k; �1 þ �2;�Þ where B is the multiset union of B1 and B2. ByLemma 1.4, it suffices to establish the existence of a CBSAðv; 3; �;�Þ with� ¼ 1; 2; 3; 4; 6; 12. Then we get the necessary and sufficient conditions for theexistence of a CBSAðv; 3; �;�Þ with � ¼ 2; 3.

Suppose that B � Zv . Let �B ¼ fbi � bjðmod vÞ : bi; bj 2 B; bi 6¼ bjg. Let ½a; b�denote the set of all integers n such that a � n � b, and �½a; b� denote the multisetcontaining each element of ½a; b� exactly � times. The following lemma is from [3].

Lemma 1.5. Suppose there exist k-subsets B1, B2; . . . ;Bt of Zv such that the multisetunion [t

i¼1�Bi ¼ �½�þ 1; v � �� 1�, then there exists a CBSAðv; k; �;�Þ.

314 ZHANG AND CHANG

The subsets B1; . . . ;Bt in Lemma 1.5 are called base blocks of theCBSAðv; k; �;�Þ. The above lemma enables us to use Langford sequences andextended Langford sequences to construct CBSAðv; 3; �;�Þ.

The problem of partitioning the set S ¼ f1; 2; . . . ; 2mg into differences d; dþ1; . . . ; d þ m� 1 is known as the problem of constructing a Langford sequence ofdefect d and length m. In what follows we mean that ðx; yÞ � ðu; vÞ ðmod ðm; nÞÞ ifand only if x � u ðmod mÞ and y � v ðmod nÞ. For arbitrary d, the following resultshave been obtained in [9].

Theorem 1.6. If m � 2d � 1 then ½1; 2m� can be partitioned into differences½d; d þ m� 1� whenever ðm; dÞ � ð0; 1Þ; ð1; 1Þ; ð0; 0Þ; ð3; 0Þ ðmod ð4; 2ÞÞ.

If mðm� 2d þ 1Þ þ 2 � 0 then ½1; 2mþ 1�nf2mg can be partitioned intodifferences ½d; d þ m� 1� whenever ðm; dÞ � ð2; 0Þ; ð1; 0Þ; ð2; 1Þ; ð3; 1Þ ðmod ð4; 2ÞÞ.

The following lemmas on Langford sequences are from [12].

Lemma 1.7. Let ðm; dÞ � ð0; 1Þ; ð1; 1Þ; ð0; 0Þ; ð3; 0Þ ðmod ð4; 2ÞÞ such that m �2d � 1. Then ½d; d þ 3m� 1� can be partitioned into triples fai; bi; cig, 1 � i � m,such that ai þ bi ¼ ci.

Lemma 1.8. Let ðm; dÞ � ð2; 0Þ; ð1; 0Þ; ð2; 1Þ; ð3; 1Þ ðmod ð4; 2ÞÞ such thatmðm� 2d þ 1Þ þ 2 � 0. Then ½d; d þ 3m�nfd þ 3m� 1g can be partitioned intotriples fai; bi; cig, 1 � i � m, such that ai þ bi ¼ ci.

A k-extended Langford sequence of defect d and length m is a partition of½1; 2mþ 1�nfkg into differences ½d; d þ m� 1�.Theorem 1.9 ([6], [7]). For 1 � d � 4, a k-extended Langford sequence of defect dand length m exists if and only if ðm; kÞ � ð0; 1Þ; ð1; dÞ; ð2; 0Þ; ð3; d þ 1Þ ðmod ð4; 2ÞÞsuch that m � 2d � 3 and m

2ð2d � 1 � mÞ þ 1 � k � m

2ðm� 2d þ 5Þ þ 1.

Similar to Lemmas 1.7 and 1.8, we have the following lemma on k-extendedLangford sequences.

Lemma 1.10. For 1 � d � 4, if ðm; kÞ� ð0; 1Þ; ð1; dÞ; ð2; 0Þ; ð3; d þ 1Þ ðmod ð4; 2ÞÞsuch that m � 2d � 3 and ðm=2Þð2d � 1 � mÞ þ 1 � k � ðm=2Þðm� 2d þ 5Þ þ 1,then ½d; d þ 3m�nfk þ d þ m� 1g can be partitioned into triples fai; bi; cig; 1 � i� m, such that ai þ bi ¼ ci.

Proof. By Theorem 1.9, there exists a k-extended Langford sequence and hencethe set f1; 2; . . . ; 2mþ 1gnfkg can be partitioned into pairs fb0i; c0ig such that[mi¼1fc0i � b0ig ¼ fd; d þ 1; . . . ; d þ m� 1g. Let ai ¼ c0i � b0i, bi ¼ b0i þ d þ m� 1,

and ci ¼ c0i þ d þ m� 1 for 1 � i � m. It is easy to see that ½d; d þ 3m�nfk þ d þ m� 1g can be partitioned into triples fai; bi; cig, 1 � i � m, such thatai þ bi ¼ ci. &

2. CONSTRUCTIONS OF CBSA(vv, 3, ��; 2)

Lemma 2.1. If v � 0; 3; 9 ðmod 12Þ and v � 15 but v 6¼ 15; 27; 39; then ½3; v � 3�nf2v=3g can be partitioned into triples fai; bi; cig, 1 � i � ðv � 6Þ=3, such thatai þ bi � ci ðmod vÞ.

THE SPECTRUM OF CYCLIC BSAðv; 3; �;�Þ WITH � ¼ 2; 3 315

Proof. When v ¼ 12s and v � 24, by Lemma 1.10 with ðd;m; kÞ ¼ ð3; 4s� 2; 4sÞ,the set ½3; 12s� 3�nf8sg can be partitioned into triples fai; bi; cig, 1 � i � 4s� 2,such that ai þ bi ¼ ci.

When v ¼ 12sþ 9 and v � 21, by Lemma 1.10 with ðd;m; kÞ ¼ ð3; 4sþ 1;4sþ 3Þ, the set ½3; 12sþ 6�nf8sþ 6g can be partitioned into triples fai; bi; cig,1 � i � 4sþ 1, such that ai þ bi ¼ ci.

When v ¼ 12sþ 3 and v � 51, we divide the problem into 2 cases as follows.

Case 1. v ¼ 24aþ 3 and a � 2. By Lemma 1.7 with ðd;mÞ ¼ ð3; 4aÞ, ½3; 12aþ 2�can be partitioned into triples fai; bi; cig, 1 � i � 4a, such that ai þ bi ¼ ci. ByLemma 1.10 with ðd;m; kÞ ¼ ð3; 4a� 1; 4aÞ, ½3; 12a�nf8aþ 1g can be partitionedinto triples fa0i; b0i; c0ig, 4aþ 1 � i � 8a� 1, such that a0i þ b0i ¼ c0i. Let ai ¼ v � a0i,bi¼v � b0i, and ci ¼ v � c0i for 4aþ 1 � i � 8a� 1. Then ½12aþ 3; 24a�nf16aþ 2gcan be partitioned into triples fai; bi; cig such that ai þ bi � ci ðmod vÞ. So, the set½3; 24a�nf16aþ 2g can be partitioned into triples fai; bi; cig, 1 � i � 8a� 1, suchthat ai þ bi � ci ðmod vÞ.Case 2. v ¼ 24aþ 15 and a � 2. By Lemma 1.7 with ðd;mÞ ¼ ð3; 4aþ 4Þ,½3; 12aþ 14� can be partitioned into triples fai; bi; cig, 1 � i � 4aþ 4, such thatai þ bi ¼ ci. By Lemma 1.10 with ðd;m; kÞ ¼ ð3; 4a� 1; 4aþ 4Þ, ½3; 12a�nf8aþ 5gcan be partitioned into triples fa0i; b0i; c0ig, 4aþ 5 � i � 8aþ 3, such that a0i þ b0i ¼ c0i.Let ai ¼ v � a0i, bi ¼ v � b0i, and ci ¼ v � c0i for 4aþ 5 � i � 8aþ 3. Then½12aþ 15; 24aþ 12�nf16aþ 10g can be partitioned into triples fai; bi; cig such thatai þ bi � ci ðmod vÞ. So, the set ½3; 24aþ 12�nf16aþ 10g can be partitioned intotriples fai; bi; cig, 1 � i � 8aþ 3, such that ai þ bi � ci ðmod vÞ. &

Given v we will use 4 to denote a triple fa; b; cg that satisfies aþ b ¼ c oraþ bþ c ¼ v.

Lemma 2.2. Set Dv ¼ ½3; ðv � 1Þ=2� for each v � 1; 3 ðmod 6Þ and v � 19. Thenthe multiset 3Dv can be partitioned into triples 4i ¼ fai; bi; cig, 1 � i � ðv � 5Þ=2,such that for each i either ai þ bi ¼ ci or ai þ bi þ ci ¼ v.

Proof. When v � 1 ðmod 6Þ and v � 19, we divide the problem into 4 cases asfollows:

Case 1. v ¼ 24sþ 7 and s � 1. For s ¼ 1, the conclusion follows by Appendix A.Next we consider s � 2. By Lemma 1.7 with ðd;mÞ ¼ ð4; 4sÞ (twice), the set2ðDvnf3gÞ can be partitioned into triples 4i ¼ fai; bi; cig, 1 � i � 8s, such thatai þ bi ¼ ci. By Lemma 1.8 with ðd;mÞ ¼ ð3; 4s� 1Þ, the set Dvnf12s� 1; 12sþ 1;12sþ 2; 12sþ 3g can be partitioned into triples 4i ¼ fai; bi; cig, 8sþ 1 � i �12s� 1, such that ai þ bi ¼ ci. Set T ¼ f12s� 1; 12sþ 1; 12sþ 2; 12sþ 3; 3; 3g.Hence ð3DvÞnT can be partitioned into triples 4i ¼ fai; bi; cig, 1 � i � 12s� 1,such that ai þ bi ¼ ci. It is easy to see that T can be partitioned into triples412s ¼ f12s� 1; 3; 12sþ 2g, 412sþ1 ¼ f12sþ 1; 3; 12sþ 3g. Therefore, the multi-set 3Dv can be partitioned into 4s.

The other cases can be proved in a similar way. We will only list the elements of T .

Case 2. v ¼ 24sþ 13 and s � 1. For s ¼ 1; 2; 3, the conclusion follows byAppendix A. Next we consider s � 4. By Lemma 1.7 with ðd;mÞ ¼ ð3; 4sþ 1Þ andð7; 4sÞ, Dvnf12sþ 6g and Dvnf3; 4; 5; 6g can be partitioned into 4s. By Lemma 1.8

316 ZHANG AND CHANG

with ðd;mÞ ¼ ð3; 4s� 1Þ, Dvnf12s� 1; 12sþ 1; . . . ; 12sþ 6g can be partitionedinto 4s. In this case, T ¼ f12sþ 6; 3; 4; 5; 6; 12s� 1; 12sþ 1; . . . ; 12sþ 6g can bepartitioned into 4s: f3; 12sþ 4; 12sþ 6g, f4; 12s� 1; 12sþ 3g, f5; 12sþ 1;12sþ 6g, f6; 12sþ 2; 12sþ 5g. So, 3Dv can be partitioned into 4s.

Case 3. v ¼ 24sþ 19 and s � 0. For s ¼ 0; 1; 2, the conclusion follows byAppendix A. Next we consider s � 3. By Lemma 1.8 with ðd;mÞ ¼ ð3; 4sþ 2Þ,Dvnf12sþ 8g can be partitioned into 4s. By Lemma 1.7 with ðd;mÞ ¼ ð5; 4sþ 1Þand ð5; 4sÞ, Dvnf3; 4; 12sþ 8; 12sþ 9g and Dvnf3; 4; 12sþ 5; . . . ; 12sþ 9g can bepartitioned into 4s. In this case, T ¼ f12sþ 8; 3; 4; 12sþ 8; 12sþ 9; 3; 4; 12sþ5; . . . ; 12sþ 9g can be partitioned into 4s: f3; 12sþ 8; 12sþ 8g, f3; 12sþ 6;12sþ 9g, f4; 12sþ 7; 12sþ 8g, f4; 12sþ 5; 12sþ 9g. So, 3Dv can be partitionedinto 4s.

Case 4. v ¼ 24sþ 1 and s � 1. For s ¼ 1, the conclusion follows by Appendix A.Next we consider s � 2. By Lemma 1.8 with ðd;mÞ ¼ ð3; 4s� 1Þ (twice) andLemma 1.7 with ð4; 4s� 1Þ, 2ðDvnf12s� 1gÞ and Dvnf3g can be partitioned into4s. In this case, T ¼ f3; 12s� 1; 12s� 1g is itself a 4. So, 3Dv can be partitionedinto 4s.

When v � 3 ðmod 6Þ and v � 19, we also divide the problem into 4 cases asfollows:

Case 1: v ¼ 24sþ 3, s � 1. For s ¼ 1; 2, the conclusion follows by Appendix A.Next we consider s � 3. By Lemma 1.8 with ðd;mÞ ¼ ð3; 4s� 1Þ and ð5; 4s� 2Þ(twice), Dvnf12s� 1; 12sþ 1g and 2ðDvnf3; 4; 12s� 2; 12s; 12sþ 1gÞ can bepartitioned into 4s. In this case, T ¼ f12s� 1; 12sþ 1g [ 2f3; 4; 12s� 2; 12s;12sþ 1g can be partitioned into 4s: f3; 12s; 12sg, f3; 12s� 1; 12sþ 1g,2f4; 12s� 2; 12sþ 1g. So, 3Dv can be partitioned into 4s.

Case 2. v ¼ 24sþ 9, s � 1. For s ¼ 1; 2, the conclusion follows by Appendix A.Next we consider s � 3. By Lemma 1.7 with ðd;mÞ ¼ ð4; 4sÞ and ð5; 4sÞ,Dvnf3; 12sþ 4g and Dvnf3; 4g can be partitioned into 4s. By Lemma 1.8 withðd;mÞ ¼ ð3; 4s� 1Þ, Dvnf12s� 1; 12sþ 1; . . . ; 12sþ 4g can be partitioned into 4s.In this case, T ¼ f3; 12sþ 4; 3; 4; 12s� 1; 12sþ 1; . . . ; 12sþ 4g can be partitioned4s: f3; 12sþ 2; 12sþ 4g, f3; 12sþ 1; 12sþ 4g, f4; 12s� 1; 12sþ 3g. So, 3Dv canbe partitioned into 4s.

Case 3. v ¼ 24sþ 15, s � 1. For s ¼ 1, the conclusion follows by Appendix A.Next we consider s � 2. By Lemma 1.8 with ðd;mÞ ¼ ð4; 4sþ 1Þ, Dvnf3; 12sþ 6gcan be partitioned into 4s. By Lemma 1.7 with ðd;mÞ ¼ ð5; 4sþ 1Þ and ð3; 4sÞ,Dvnf3; 4g and Dvnf12sþ 3; . . . ; 12sþ 7g can be partitioned into 4s. In this case,T ¼ f3; 12sþ 6; 3; 4; 12sþ 3; . . . ; 12sþ 7g can be partitioned into 4s: f3; 12sþ 3;12sþ 6g, f3; 12sþ 4; 12sþ 7g, f4; 12sþ 5; 12sþ 6g. So, 3Dv can be partitionedinto 4s.

Case 4. v ¼ 24sþ 21, s � 0. For s ¼ 0; 1, the conclusion follows by Appendix A.Next we consider s � 2. By Lemma 1.8 with ðd;mÞ ¼ ð4; 4sþ 2Þ (twice),2ðDvnf3; 12sþ 9gÞ can be partitioned into 4s. By Lemma 1.8 with ðd;mÞ ¼


ð3; 4sþ 2Þ, Dvnf12sþ 8; 12sþ 10g can be partitioned into 4s. In this case,T ¼ 2f3; 12sþ 9g [ f12sþ 8; 12sþ 10g can be partitioned into 4s: f3; 12sþ 9;12sþ 9g, f3; 12sþ 8; 12sþ 10g. So, 3Dv can be partitioned into 4s. &

Lemma 2.3. (1) If v � 2 ðmod 12Þ and v � 15, then the multiset 2½3; v � 3� can bepartitioned into triples ti, 1 � i � 2ðv � 5Þ=3, such that if ti ¼ fai; bi; cig thenai þ bi ¼ ci.

(2) If v � 6 ðmod 12Þ and v � 30, then the multiset 2ð½3; v � 3�nf2v=3gÞ can bepartitioned into triples ti, 1 � i � 2ðv � 6Þ=3, such that if ti ¼ fai; bi; cig thenai þ bi � ci ðmod vÞ.Proof. When v ¼ 12sþ 2, set Dv ¼ ½3; v � 3�. We consider the case s � 4. ByLemma 1.7 with ðd;mÞ ¼ ð7; 4s� 3Þ and ð4; 4s� 4Þ, Dvnf3; 4; 5; 6; 12s� 2;12s� 1g and Dvnf3; 12s� 8; . . . ; 12s� 1g can be partitioned into triples ti,1 � i � 2ðv � 5Þ=3 � 5, such that if ti ¼ fai; bi; cig then ai þ bi ¼ ci. Note thatT ¼ f3; 4; 5; 6; 12s� 2; 12s� 1; 3; 12s� 8; . . . ; 12s� 1g can be partitioned intotriples: f3; 12s� 4; 12s� 1g, f6; 12s� 7; 12s� 1g, f3; 12s� 5; 12s� 2g,f4; 12s� 6; 12s� 2g, f5; 12s� 8; 12s� 3g. The conclusion then follows whens � 4.

For v ¼ 26; 38, the multiset 2½3; v � 3� can be partitioned into triples as follows:

v ¼ 26:

f6; 16; 22g; f7; 14; 21g; f8; 13; 21g; f9; 11; 20g; f3; 16; 19g; f4; 14; 18g;f5; 13; 18g; f7; 8; 15g; f9; 10; 19g; f4; 11; 15g; f3; 20; 23g; f6; 17; 23g;f10; 12; 22g; f5; 12; 17g:

v ¼ 38:

f5; 29; 34g; f6; 28; 34g; f7; 26; 33g; f8; 25; 33g; f9; 23; 32g; f10; 21; 31g;f11; 19; 30g; f12; 18; 30g; f13; 16; 29g; f4; 23; 27g; f6; 20; 26g; f8; 16; 24g;f9; 13; 22g; f14; 18; 32g; f15; 20; 35g; f3; 14; 17g; f4; 11; 15g; f12; 19; 31g;f10; 25; 35g; f5; 17; 22g; f3; 24; 27g; f7; 21; 28g:

When v � 6 ðmod 12Þ and v � 30, we divide the problem into 2 cases as follows:

Case 1. v ¼ 24aþ 6 where a � 1. We consider the case a � 3. By Lemma 1.8 withðd;mÞ ¼ ð4; 4a� 2Þ and ð4; 4a� 3Þ, ½4; 12a� 2�nf12a� 3g and ½4; 12a� 5�nf12a� 6g can be partitioned into triples fai; bi; cig, 1 � i � 8a� 5, such thatai þ bi ¼ ci. By Lemma 1.10 with ðd;m; kÞ ¼ ð3; 4aþ 1; 4a� 1Þ and ð3; 4aþ 2;4a� 2Þ, ½3; 12aþ 6�nf8aþ 2g and ½3; 12aþ 9�nf8aþ 2g can be partitioned intotriples fa0i; b0i; c0ig, 8a� 4 � i � 16a� 2, such that a0i þ b0i ¼ c0i. Let ai ¼ v � a0i,bi ¼ v � b0i, and ci ¼ v � c0i for 8a� 4 � i � 16a� 2. Then ½12a; 24aþ 3�nf16aþ 4g and ½12a� 3; 24aþ 3�nf16aþ 4g can be partitioned into triplesfai; bi; cig, 8a� 4 � i � 16a� 2, such that ai þ bi � ci ðmod vÞ. Considering alsothe triples f3; 12a� 4; 12a� 1g, f3; 12a� 6; 12a� 3g, we know that 2ð½3; 24aþ 3�nf16aþ 4gÞ can be partitioned into triples fai; bi; cig, 1 � i � 16a� 2, such thatai þ bi � ci ðmod vÞ.

318 ZHANG AND CHANG

For v ¼ 30; 54, the multiset 2ð½3; v � 3�nf2v=3gÞ can be partitioned into triples asfollows:

v ¼ 30 :

f6; 13; 19g; f12; 12; 24g; f4; 21; 25g; f6; 17; 23g; f10; 16; 26g; f13; 14; 27g;f5; 17; 22g; f7; 14; 21g; f7; 18; 25g; f3; 19; 22g; f4; 23; 27g; f3; 8; 11g;f8; 16; 24g; f5; 10; 15g; f11; 15; 26g; f9; 9; 18g:

v ¼ 54 :

f5; 44; 49g, f6; 41; 47g, f7; 38; 45g, f8; 35; 43g, f9; 33; 42g, f10; 30; 40g;f11; 28; 39g, f12; 25; 37g, f14; 18; 32g, f15; 16; 31g, f3; 26; 29g, f4; 23; 27g;f7; 43; 50g, f8; 41; 49g, f9; 39; 48g, f11; 35; 46g, f12; 32; 44g, f14; 26; 40g;f15; 19; 34g, f16; 17; 33g, f18; 20; 38g, f5; 17; 22g, f21; 27; 48g, f31; 20; 51g;f6; 24; 30g, f23; 28; 51g, f3; 22; 25g, f21; 24; 45g, f13; 34; 47g, f10; 19; 29g;f13; 37; 50g, f4; 42; 46g:Case 2. v ¼ 24aþ 18 where a � 1. We consider the case a � 2. By Lemma 1.8

with ðd;mÞ ¼ ð4; 4aþ 2Þ and ð4; 4aþ 1Þ, ½4; 12aþ 10�nf12aþ 9g and½4; 12aþ 7�nf12aþ 6g can be partitioned into triples fai; bi; cig, 1 � i � 8aþ 3,such that ai þ bi ¼ ci. By Lemma 1.10 with ðd;m; kÞ ¼ ð3; 4aþ 1; 4aþ 3Þ andð3; 4aþ 2; 4aþ 2Þ, ½3; 12aþ 6�nf8aþ 6g and ½3; 12aþ 9�nf8aþ 6g can be parti-tioned into triples fa0i; b0i; c0ig, 8aþ 4 � i � 16aþ 6, such that a0i þ b0i ¼ c0i.Let ai ¼ v � a0i, bi ¼ v � b0i, and ci ¼ v � c0i for 8aþ 4 � i � 16aþ 6. Then½12aþ 12; 24aþ 15�nf16aþ 12g and ½12aþ 9; 24aþ 15�nf16aþ 12g can bepartitioned into triples fai; bi; cig, 8aþ 4 � i � 16aþ 6, such that ai þ bi �ci ðmod vÞ. Together with the triples f3; 12aþ 8; 12aþ 11g, f3; 12aþ 6;12aþ 9g,the multiset 2ð½3; 24aþ 15�nf16aþ 12gÞ can be partitioned into triples fai; bi; cig,1 � i � 16aþ 6, such that ai þ bi � ci ðmod vÞ.

For v ¼ 42, the multiset 2ð½3; 39� n f28gÞ can be partitioned into triples as follows:

f5; 32; 37g, f6; 29; 35g, f9; 21; 30g, f11; 16; 27g, f12; 13; 25g, f3; 21; 24g;f4; 18; 22g, f5; 34; 39g, f7; 30; 37g, f9; 26; 35g, f10; 23; 33g, f11; 20; 31g;f12; 17; 29g, f14; 22; 36g, f8; 19; 27g, f10; 15; 25g, f19; 20; 39g, f3; 33; 36g;f6; 26; 32g, f8; 15; 23g, f7; 31; 38g, f14; 24; 38g, f4; 13; 17g, f16; 18; 34g:

&

Lemma 2.4. If v � 4 ðmod 12Þ and v � 28, then the multiset 3½3; v � 3� can bepartitioned into triples ti, 1 � i � v � 5, such that if ti ¼ fai; bi; cig then ai þ bi ¼ ci.

Proof. Set v ¼ 12sþ 16 and Dv ¼ ½3; 12sþ 13�. We consider the case s � 3, byLemma 1.7 with ðd;mÞ ¼ ð3; 4sþ 1Þ and ð4; 4sþ 3Þ, Dvnf12sþ 6; . . . ; 12sþ 13gÞand Dvnf3; 12sþ 13g can be partitioned into triples ti, 1 � i � 8sþ 4, such thatif ti ¼ fai; bi; cig then ai þ bi ¼ ci. By Lemma 1.8 with ðd;mÞ ¼ ð7; 4sþ 2Þ,Dvnf3; 4; 5; 6; 12sþ 12g can be partitioned into triples ti, 8sþ 5 � i � 12sþ 6,such that if ti ¼ fai; bi; cig then ai þ bi ¼ ci. Note that T ¼ f3; 12sþ 13; 3; 4; 5; 6;12sþ 12; 12sþ 6; . . . ; 12sþ 13g can be partitioned into triples: f3; 12sþ 9;12sþ 12g, f3; 12sþ 10; 12sþ 13g, f4; 12sþ 8; 12sþ 12g, f5; 12sþ 6; 12sþ 11g,f6; 12sþ 7; 12sþ 13g.


For v ¼ 28; 40, the multiset 3½3; v � 3� can be partitioned into triples as follows:

v ¼ 28 :

f3; 22; 25g; f4; 21; 25g; f6; 18; 24g; f7; 17; 24g; f8; 16; 24g; f9; 14; 23g;f10; 13; 23g; f11; 12; 23g; f3; 19; 22g; f4; 18; 22g; f5; 16; 21g; f6; 15; 21g;f7; 13; 20g; f8; 12; 20g; 2f9; 10; 19g; f4; 14; 18g; f5; 12; 17g; f6; 11; 17g;f7; 8; 15g; f3; 13; 16g; f5; 15; 20g; f11; 14; 25g:

v ¼ 40 :

f3; 34; 37g; f4; 33; 37g; f5; 32; 37g; f6; 30; 36g; f7; 29; 36g; f9; 26; 35g;f10; 25; 35g; f11; 24; 35g; f14; 20; 34g; f15; 18; 33g; f16; 17; 33g; f4; 28; 32g;f6; 25; 31g; f7; 24; 31g; f8; 22; 30g; f9; 21; 30g; f10; 19; 29g; f11; 18; 29g;f12; 16; 28g; f13; 14; 27g; f12; 15; 27g; f3; 24; 27g; f4; 22; 26g; f5; 21; 26g;f7; 18; 25g; f8; 15; 23g; f9; 14; 23g; f10; 13; 23g; f3; 16; 19g; f12; 20; 32g;f5; 17; 22g; f11; 20; 31g; f17; 19; 36g; f6; 28; 34g; f8; 13; 21g:

&

Lemma 2.5. If v � 10 ðmod 12Þ and v � 22, then the multiset 6½3; v � 3� can bepartitioned into triples ti, 1 � i � 2ðv � 5Þ, such that if ti ¼ fai; bi; cig thenai þ bi ¼ ci.

Proof. Set v ¼ 12sþ 10 and Dv ¼ ½3; 12sþ 7�. For s ¼ 1, the conclusion followsby (2) of Appendix A. Next we consider s � 2. By Lemma 1.8 withðd;mÞ ¼ ð4; 4sþ 1Þ and ð3; 4s� 1Þ, 2Dvnf3; 12sþ 6; 12s� 1; 12sþ 1; . . . ;12sþ 7g can be partitioned into triples ti, 1 � i � 8s, such that if ti ¼ fai; bi; cigthen ai þ bi ¼ ci. Similarly, by Lemma 1.7 with ðd;mÞ ¼ ð5; 4sþ 1Þ (three times)and ð4; 4s� 1Þ, 3ðDvnf3; 4gÞ [ Dvnf3; 12sþ 1; . . . ; 12sþ 7g can be partitioned intotriples ti, 8sþ 1 � i � 24sþ 2 such that if ti ¼ fai; bi; cig then ai þ bi ¼ ci. Notethat T ¼ f3; 12sþ 6; 12s� 1; 12sþ 1; . . . ; 12sþ 7g [ 3f3; 4g [ f3; 12sþ 1; . . . ;12sþ 7g can be partitioned into triples: f3; 12s� 1; 12sþ 2g, 2f3; 12sþ 3;12sþ 6g, 2f3; 12sþ 4; 12sþ 7g, 2f4; 12sþ 1; 12sþ 5g, f4; 12sþ 2; 12sþ 6g. So6Dv can be partitioned into 4s. &

Theorem 2.6. Set v � 15.

(1) There exists a CBSAðv; 3; 1; 2Þ for v � 5 ðmod 6Þ;(2) There exists a CBSAðv; 3; 2; 2Þ for v � 0; 3; 8; 9 ðmod 12Þ;(3) There exists a CBSAðv; 3; 3; 2Þ for v � 1; 3 ðmod 6Þ;(4) There exists a CBSAðv; 3; 4; 2Þ for v � 2; 6 ðmod 12Þ;(5) There exists a CBSAðv; 3; 6; 2Þ for v � 4 ðmod 12Þ;(6) There exists a CBSAðv; 3; 12; 2Þ for v � 10 ðmod 12Þ:

Proof. (1) For v � 5 ðmod 6Þ and v � 35, by Theorem 1.3, there exists aCBSAðv; 3; 1; 2Þ. For each v 2 f17, 23, 29g, a CBSAðv; 3; 1; 2Þ is constructed bylisting base blocks as follows:

v ¼ 17: f0; 3; 8g, f0; 4; 10g.v ¼ 23: f0; 3; 9g, f0; 4; 11g, f0; 5; 13g.

320 ZHANG AND CHANG

v ¼ 29: f0; 3; 10g, f0; 4; 12g, f0; 5; 16g, f0; 6; 15g.(2) For v � 8 ðmod 12Þ and v � 68, by Theorem 1.3, there exists a CBSAðv; 3; 2; 2Þ.

For each v 2 f20; 32; 44; 56g, a CBSAðv; 3; 2; 2Þ is constructed by listing base blocksas follows:

v ¼ 20: f0; 3; 14g, f0; 4; 16g, f0; 5; 13g, f0; 6; 15g, f0; 7; 17g.v ¼ 32: f0; 3; 15g, f0; 4; 28g, f0; 5; 22g, f0; 6; 26g, f0; 7; 25g, f0; 8; 27g,

f0; 9; 23g, f0; 10; 21g, f0; 13; 29g.v ¼ 44: f0; 3; 38g, f0; 4; 40g, f0; 5; 34g, f0; 6; 22g, f0; 7; 28g, f0; 8; 33g,

f0; 9; 32g, f0; 10; 30g, f0; 11; 24g, f0; 12; 39g, f0; 15; 41g, f0; 14; 31g, f0; 18; 37g.v ¼ 56: f0; 3; 50g, f0; 4; 52g, f0; 5; 21g, f0; 6; 49g, f0; 7; 33g, f0; 8; 45g,

f0; 9; 44g, f0; 10; 42g, f0; 11; 41g, f0; 12; 39g, f0; 13; 38g, f0; 14; 36g, f0; 15; 34g,f0; 17; 40g, f0; 18; 46g, f0; 20; 51g, f0; 24; 53g.

When v � 0; 3; 9 ðmod 12Þ and v � 15 but v 6¼ 15; 27; 39, by Lemma 2.1,½3; v � 3�nf2v=3g can be partitioned into triples fai; bi; cig, 1 � i � ðv � 6Þ=3, suchthat ai þ bi � ci ðmod vÞ. Set

Bi ¼ f0; ai; cig; 1 � i � ðv � 6Þ=3:

Then, it is easily checked that [ðv�6Þ=3i¼1 �Bi ¼ 2½3; v � 3�nfv=3; 2v=3g mod v. With

the base block f0; v=3; 2v=3g (short orbit), by Lemma 1.5, there exists a BSAðv; 3;2; 2Þ. When v ¼ 15; 27; 39, the base blocks of a CBSAðv; 3; 2; 2Þ are as follows:

v ¼ 15 : f0; 3; 8g, f0; 4; 11g, f0; 3; 9g, f0; 5; 10g (short orbit).v ¼ 27 : f0; 5; 12g, f0; 4; 10g, f0; 3; 11g, f0; 3; 14g, f0; 4; 10g, f0; 5; 18g,

f0; 7; 15g, f0; 9; 18g (short orbit).v ¼ 39 : f0; 4; 35g, f0; 12; 33g, f0; 7; 32g, f0; 6; 17g, f0; 9; 27g, f0; 14; 30g,

f0; 10; 23g, f0; 8; 36g, f0; 3; 22g, f0; 5; 29g, f0; 5; 20g, f0; 13; 26g (short orbit).(3) By Lemma 2.2, when v � 1; 3 ðmod 6Þ and v � 19, the multiset 3½3; v � 1Þ=2�

can be partitioned into triples fai; bi; cig, 1 � i � ðv � 5Þ=2, such that for each i,either ai þ bi ¼ ci or ai þ bi þ ci ¼ v. For 1 � i � ðv � 5Þ=2, set

Bi ¼f0; ai; cig if ai þ bi ¼ ci;

f0; ai; v � cig if ai þ bi þ ci ¼ v:

�

Then, it is easily checked that [ðv�5Þ=2i¼1 �Bi ¼ 3½3; v � 3� mod v. By Lemma 1.5,

there exists a CBSAðv; 3; 3; 2Þ for v � 19. For the remaining case of v ¼ 15, we listthe 5 base blocks as follows: f0; 3; 10g, f0; 3; 9g, f0; 3; 8g, f0; 4; 8g, f0; 4; 9g.

(4) When v � 2 ðmod 12Þ and v � 15, by Lemma 2.3, the multiset 2½3; v � 3� canbe partitioned into triples fai; bi; cig, 1 � i � 2ðv � 5Þ=3, such that ai þ bi ¼ ci.Set Bi ¼ f0; ai; cig for 1 � i � 2ðv � 5Þ=3. Then, it is easily checked that[2ðv�5Þ=3i¼1 �Bi ¼ 4½3; v � 3� mod v. By Lemma 1.5, there exists a CBSAðv; 3; 4; 2Þ.When v � 6 ðmod 12Þ and v � 30, by Lemma 2.3, the multiset 2ð½3; v � 3�n

f2v=3gÞ can be partitioned into triples fai; bi; cig, 1 � i � 2ðv � 6Þ=3, such thatai þ bi � ci ðmod vÞ. Set Bi ¼ f0; ai; cig for 1 � i � 2ðv � 6Þ=3. Then, it is easilychecked that [2ðv�6Þ=3

i¼1 �Bi ¼ 4½3; v � 3� n 2fv=3; 2v=3g mod v. Considering also thebase blocks 2f0; v=3; 2v=3g (short orbits), by Lemma 1.5, we have that there existsa CBSAðv; 3; 4; 2Þ for v � 30. For v ¼ 18, the base blocks of a CBSAðv; 3; 4; 2Þ


are: f0; 5; 13g, f0; 3; 14g, f0; 5; 15g, f0; 7; 15g, f0; 4; 13g, f0; 4; 11g, f0; 4; 10g,f0; 3; 12g, 2f0; 6; 12g (short orbits).

(5) When v � 4 ðmod 12Þ and v � 28, by Lemma 2.4, the multiset 3½3; v � 3� canbe partitioned into triples fai; bi; cig, 1 � i � v � 5, such that ai þ bi ¼ ci.Set Bi ¼ f0; ai; cig for 1 � i � v � 5. Then, it is easily checked that[v�5i¼1 �Bi ¼ 6½3; v � 3� mod v. By Lemma 1.5, there exists a CBSAðv; 3; 6; 2Þ.For v ¼ 16, the base blocks of a CBSAðv; 3; 6; 2Þ are: f0; 3; 9g; f0; 3; 11g;

f0; 3; 12g; f0; 4; 11g; f0; 4; 12g; f0; 5; 9g; f0; 5; 11g; f0; 5; 13g; f0; 6; 12g;2f0; 3; 10g:

(6) When v � 10 ðmod 12Þ and v � 22, by Lemma 2.5, the multiset 6½3; v � 3�can be partitioned into triples fai; bi; cig, 1 � i � 2ðv � 5Þ, such that ai þ bi ¼ ci.Set Bi ¼ f0; ai; cig for 1 � i � 2ðv � 5Þ. Then, it is easily checked that[2ðv�5Þi¼1 �Bi ¼ 12½3; v � 3� mod v. By Lemma 1.5, there exists a CBSAðv; 3; 12; 2Þ.

&


(1) There exists a CBSAðv; 3; 1; 2Þ if and only if v � 5 ðmod 6Þ;(2) There exists a CBSAðv; 3; 2; 2Þ if and only if v � 0; 2 ðmod 3Þ and

v 6� 2 ðmod 4Þ;(3) There exists a CBSAðv; 3; 3; 2Þ if and only if v � 1 ðmod 2Þ;(4) There exists a CBSAðv; 3; 4; 2Þ if and only if v � 0; 2 ðmod 3Þ;(5) There exists a CBSAðv; 3; 6; 2Þ if and only if v 6� 2 ðmod 4Þ;(6) There exists a CBSAðv; 3; 12; 2Þ if and only if v is a positive integer.

Proof. Necessity follows by Lemma 1.4, so we establish sufficiency in each case.(1) Sufficiency follows by Theorem 2.6 (1).(2) If v � 0; 3; 8; 9 ðmod 12Þ then Theorem 2.6 (2) applies. If v � 5 ðmod 6Þ

a CBSAðv; 3; 1; 2Þ exists by Theorem 2.6 (1), so repeating blocks give aCBSAðv; 3; 2; 2Þ.

(3) If v � 1; 3 ðmod 6Þ then Theorem 2.6 (3) applies. If v � 5 ðmod 6Þ aCBSAðv; 3; 1; 2Þ exists by Theorem 2.6 (1), so repeating blocks three times give aCBSAðv; 3; 3; 2Þ.

(4) If v � 2; 6 ðmod 12Þ then Theorem 2.6 (4) applies. If v � 0; 3; 5; 8; 9;11 ðmod 12Þ a CBSAðv; 3; 2; 2Þ exists by (2) above, so repeating blocks give aCBSAðv; 3; 4; 2Þ.

(5) If v � 4 ðmod 12Þ then Theorem 2.6 (5) applies. All other ordersv 6� 2 ðmod 4Þ are gotten by repeating the blocks of the systems in (2) and (3) above.

(6) If v � 10 ðmod 12Þ then Theorem 2.6 (6) applies. All other orders v are gottenby repeating the blocks of the systems in (3), (4), and (5) above. &

3. CONSTRUCTIONS OF CBSA(vv, 3, ��; 3)

Lemma 3.1. If v � 3 ðmod 6Þ and v � 21, then ½4; ðv � 1Þ=2�nfv=3g can bepartitioned into triples fai; bi; cig, 1 � i � ðv � 9Þ=6, such that either ai þ bi ¼ ci orai þ bi þ ci ¼ v.

Proof. When v ¼ 24sþ 3 and s � 2, by Lemma 1.10 with ðd;m; kÞ ¼ ð4; 4s� 1;4s� 1Þ, ½4; 12sþ 1�nf8sþ 1g can be partitioned into triples. For v ¼ 27, ½4; 13�nf9gcan be partitioned into triples: f7; 8; 12g, f4; 10; 13g, f5; 6; 11g.

322 ZHANG AND CHANG

When v ¼ 24sþ 21 and s � 1, by Lemma 1.10 with ðd;m; kÞ ¼ ð4; 4sþ 2;4sþ 2Þ, ½4; 12sþ 10�nf8sþ 7g can be partitioned into triples. For v ¼ 21,½4; 10�nf7g can be partitioned into triples: f4; 8; 9g, f5; 6; 10g.

When v ¼ 24sþ 9 and s ¼ 1; 2; 3; 4; 5; the conclusion follows by Appendix C.Next we deal with s � 6, the set ½4; ðv � 1Þ=2�nfv=3g can be partitioned into triplesfai; bi; cig such that ai þ bi ¼ ci or ai þ bi þ ci ¼ v as follows:

ai bi ci j 26sþ 1 � j 5 þ 2j 6sþ 6 þ j ½0; s� 3�nf2g5sþ 2 � j 2sþ 3 þ 2j 7sþ 5 þ j ½0; s� 4�10sþ 3 � j 4 þ 2j 10sþ 7 þ j ½0; s� 3�9sþ 3 � j 2sþ 2 þ 2j 11sþ 5 þ j ½0; s� 4�.

The other triples are:

f2s; 7sþ 4; 9sþ 4g; f2sþ 1; 6sþ 3; 8sþ 4g; f4sþ 2; 5sþ 3; 9sþ 5g;f4sþ 3; 6sþ 2; 10sþ 5g; f4s; 6sþ 4; 10sþ 4g; f4sþ 1; 6sþ 5; 10sþ 6g;f4s� 1; 8sþ 5; 12sþ 4g; f4s� 2; 4sþ 4; 8sþ 2g; f4s� 3; 8sþ 6; 12sþ 3g;f9; 4s� 4; 4sþ 5g; f6s� 1; 6sþ 8; 12sþ 2g:

When v ¼ 24sþ 15 and s ¼ 1; 2; 3; 4, the conclusion follows by Appendix C.Next we deal with s � 5. The set ½4; ðv � 1Þ=2�nfv=3g can be partitioned into triplesfai; bi; cig such that ai þ bi ¼ ci or ai þ bi þ ci ¼ v as follows:

ai bi ci j 26s� j 6 þ 2j 6sþ 6 þ j ½0; s� 3�5sþ 2 � j 2sþ 3 þ 2j 7sþ 5 þ j ½0; s� 4�10s� j 9 þ 2j 10sþ 9 þ j ½0; s� 5�9sþ 4 � j 2sþ 2 þ 2j 11sþ 6 þ j ½0; s� 2�.

The other triples are:

f4sþ 1; 7sþ 4; 11sþ 5g; f4s� 1; 6sþ 4; 10sþ 3g; f2sþ 1; 8sþ 4; 10sþ 5g;f4sþ 2; 6sþ 2; 10sþ 4g; f4sþ 3; 6sþ 3; 10sþ 6g; f5; 10sþ 2; 10sþ 7g;f4; 12sþ 5; 12sþ 6g; f4s; 6sþ 1; 10sþ 1g; f6sþ 5; 8sþ 2; 10sþ 8g;f7; 4s� 3; 4sþ 4g; f4sþ 5; 8sþ 3; 12sþ 7g:

This completes the proof. &

Lemma 3.2. If v � 0 ðmod 12Þ and v � 36, then ½4; v � 4�nfv=3; 2v=3g can bepartitioned into triples fai; bi; cig, 1 � i � ðv � 9Þ=3, such that ai þ bi � ci ðmod vÞ.

Proof. Let v ¼ 12s where s � 3. We divide the problem into 2 cases as follows.

Case 1. v ¼ 24a where a � 3. By Lemma 1.10 with ðd;m; kÞ ¼ ð4; 4a; 4a� 3Þ,½4; 12aþ 4�nf8ag can be partitioned into triples fai; bi; cig, 1 � i � 4a, such thatai þ bi ¼ ci. By Lemma 1.10 with ðd;m; kÞ ¼ ð4; 4a� 3; 4aÞ, ½4; 12a� 5�nf8ag canbe partitioned into triples fa0i; b0i; c0ig, 4aþ 1 � i � 8a� 3, such that a0i þ b0i ¼ c0i.Let ai ¼ v � a0i, bi ¼ v � b0i, and ci ¼ v � c0i for 4aþ 1 � i � 8a� 3. Then½12aþ 5; 24a� 4�nf16ag can be partitioned into triples fai; bi; cig, 4aþ 1 � i �


8a� 3, such that ai þ bi � ci ðmod vÞ. So, the set ½4; 24a� 4�nf8a; 16ag can bepartitioned into triples fai; bi; cig, 1 � i � 8a� 3, such that ai þ bi � ci ðmod vÞ.

Case 2. v ¼ 24aþ 12 where a � 2. By Lemma 1.10 with ðd;m; kÞ ¼ð4; 4aþ 1; 4aÞ, ½4; 12aþ 7�nf8aþ 4g can be partitioned into triples fai; bi; cig,1 � i � 4aþ 1, such that ai þ bi ¼ ci. By Lemma 1.10 with ðd;m; kÞ ¼ ð4; 4a;4aþ 1Þ, ½4; 12aþ 4�nf8aþ 4g can be partitioned into triples fa0i; b0i; c0ig,4aþ 2 � i � 8aþ 1, such that a0i þ b0i ¼ c0i. Let ai ¼ v � a0i, bi ¼ v � b0i, andci ¼ v � c0i for 4aþ 2 � i � 8aþ 1. Then ½12aþ 8; 24aþ 8�nf16aþ 8g can bepartitioned into triples fai; bi; cig, 4aþ 2 � i � 8aþ 1, such that ai þ bi �ci ðmod vÞ. So, the set ½4; 24aþ 8� n f8aþ 4; 16aþ 8g can be partitioned intotriples fai; bi; cig, 1 � i � 8aþ 1, such that ai þ bi � ci ðmod vÞ.

For v ¼ 36; 48, the set ½4; v � 4�nfv=3; 2v=3g can be partitioned into triples asfollows:

v ¼ 36 :

f4; 14; 18g; f5; 26; 31g; f6; 17; 23g; f7; 22; 29g; f8; 19; 27g; f9; 16; 25gf10; 20; 30g; f11; 21; 32g; f13; 15; 28g:

v ¼ 48 :

f5; 38; 43g; f6; 36; 42g; f7; 34; 41g; f9; 28; 37g; f10; 25; 35g; f11; 22; 33g;f12; 18; 30g; f14; 15; 29g; f19; 21; 40g; f17; 27; 44g; f4; 20; 24g; f8; 23; 31g;f13; 26; 39g: &

Lemma 3.3. If v � 5 ðmod 6Þ and v � 21, then the multiset 3½4; ðv � 1Þ=2� can bepartitioned into triples fai; bi; cig, 1 � i � ðv � 7Þ=2, such that either ai þ bi ¼ ci orai þ bi þ ci ¼ v.

Proof. Set Dv ¼ ½4; ðv � 1Þ=2� for each v � 5 ðmod 6Þ and v � 21. We divide theproblem into four cases.

Case 1. v ¼ 24sþ 5, s � 1. For s ¼ 1; 2; 3, the conclusion follows by AppendixB. For s � 4, by Lemma 1.7 with ðd;mÞ ¼ ð6; 4s� 1Þ, Dv n f4; 5g can be partitionedinto 4s. By Lemma 1.8 with ðd;mÞ ¼ ð4; 4s� 2Þ and ð6; 4s� 2Þ, Dvnf12s� 3;12s� 1; . . . ; 12sþ 2g and Dv n f4; 5; 12s� 1; 12sþ 1; 12sþ 2g can be partitionedinto 4s. Note that T ¼ f4; 5; 12s� 3; 12s� 1; . . . ; 12sþ 2; 4; 5; 12s� 1; 12sþ 1;12sþ 2g can be partitioned into 4s: f4; 12s; 12sþ 1g, f4; 12s� 1; 12sþ 2g,f5; 12s� 3; 12sþ 2g, f5; 12s� 1; 12sþ 1g. So, 3Dv can be partitioned into 4s.

Case 2. v ¼ 24sþ 11, s � 1. For s ¼ 1; 2, the conclusion follows by Ap-pendix B. For s � 3, by Lemma 1.7 with ðd;mÞ ¼ ð5; 4sÞ, ð6; 4sÞ, and ð4; 4s� 1Þ,Dvnf4; 12sþ 5g, Dvnf4; 5g, and Dvnf12sþ 1; . . . ; 12sþ 5g can be partitioned into4s. Note that T ¼ f4; 12sþ 5; 4; 5; 12sþ 1; . . . ; 12sþ 5g can be partitioned into4s: f4; 12sþ 2; 12sþ 5g, f4; 12sþ 3; 12sþ 4g, f5; 12sþ 1; 12sþ 5g. So, 3Dv canbe partitioned into 4s.

Case 3. v ¼ 24sþ 17, s � 1. For s ¼ 1; 2; 3, the conclusion follows by Ap-pendix B. For s � 4, by Lemma 1.7 with ðd;mÞ ¼ ð7; 4sÞ (twice) and ð4; 4s� 1Þ,2ðDvnf4; 5; 6; 12sþ 7; 12sþ 8gÞ [ Dvnf12sþ 1; . . . ; 12sþ 8g can be partitionedinto 4s. Note that T ¼ 2f4; 5; 6; 12sþ 7; 12sþ 8g [ f12sþ 1; . . . ; 12sþ 8g can be

324 ZHANG AND CHANG

partitioned into 4s: f4; 12sþ 4; 12sþ 8g, f4; 12sþ 6; 12sþ 7g, f5; 12sþ 5;12sþ 7g, f5; 12sþ 3; 12sþ 8g, f6; 12sþ 1; 12sþ 7g; f6; 12sþ 2; 12sþ 8g. So,3Dv can be partitioned into 4s.

Case 4. v ¼ 24sþ 23, s � 0. For s ¼ 0; 1; 2; 3, the conclusion follows by Ap-pendix B. For s � 4, by Lemma 1.7 with ðd;mÞ ¼ ð4; 4sÞ, Dvnf12sþ 4; . . . ; 12sþ11g can be partitioned into 4s. By Lemma 1.8 with ðd;mÞ ¼ ð5; 4sþ 2Þ andð8; 4sþ 1Þ, Dv n f4; 12sþ 10g and Dv n f4; 5; 6; 7; 12sþ 10g can be partitioned into4s. Note that T ¼ f4; 12sþ 10; 4; 5; 6; 7; 12sþ 10; 12sþ 4; . . . ; 12sþ 11g can bepartitioned into 4s: f4; 12sþ 5; 12sþ 9g, f4; 12sþ 6; 12sþ 10g, f5; 12sþ 8;12sþ 10g, f6; 12sþ 7; 12sþ 10g, f7; 12sþ 4; 12sþ 11g. So, 3Dv can be partitionedinto 4s. &

Lemma 3.4. (1) If v � 10 ðmod 12Þ and v � 34, then 2½4; v � 4� can be parti-tioned into triples fai; bi; cig, 1 � i � 2ðv � 7Þ=3, such that ai þ bi ¼ ci.

(2) If v � 6 ðmod 12Þ and v � 21, then the multiset 2½4; v � 4�nf2v=3g can bepartitioned into triples fai; bi; cig, 1 � i � ð2v � 15Þ=3, such that ai þ bi �ci ðmod vÞ.Proof. (1) Set v ¼ 12sþ 10 and Dv ¼ ½4; v � 4�. We consider the case s � 5. ByLemma 1.7 with ðd;mÞ ¼ ð5; 4s� 3Þ, Dv n f4; 12s� 4; . . . ; 12sþ 6g can be parti-tioned into 4s. By Lemma 1.8 with ðd;mÞ ¼ ð9; 4s� 1Þ, Dvnf4; 5; 6; 7; 8; 12sþ 5gcan be partitioned into 4s. Note that T ¼ f4; 5; 6; 7; 8; 12sþ 5; 4; 12s� 4; . . . ;12sþ 6g can be partitioned into 4s: f4; 12s� 1; 12sþ 3g, f4; 12sþ 1; 12sþ5g, f5; 12s� 3; 12sþ 2g, f6; 12s; 12sþ 6g, f7; 12s� 2; 12sþ 5g, f8; 12s� 4;12sþ 4g.

For v ¼ 34; 46; 58, the multiset 2Dv can be partitioned into 4s as follows:

v ¼ 34:

f6; 23; 29g; f8; 20; 28g; f9; 19; 28g; f10; 17; 27g; f11; 16; 27g; f12; 14; 26g;f4; 21; 25g; f5; 19; 24g, f6; 18; 24g; f7; 16; 23g; f8; 14; 22g; f9; 12; 21g;f13; 17; 30g; f5; 13; 18g; f11; 15; 26g; f7; 15; 22g; f4; 25; 29g; f10; 20; 30g:

v ¼ 46:

f4; 38; 42g; f5; 37; 42g; f6; 35; 41g; f8; 32; 40g; f9; 31; 40g; f10; 29; 39g;f11; 28; 39g; f12; 26; 38g; f13; 24; 37g; f14; 22; 36g; f15; 21; 36g; f16; 19; 35g;f17; 17; 34g; f5; 28; 33g; f7; 24; 31g; f8; 22; 30g; f9; 21; 30g; f11; 16; 27g;f12; 15; 27g; f7; 18; 25g; f14; 20; 34g; f18; 23; 41g; f6; 20; 26g; f13; 19; 32g;f4; 25; 29g; f10; 23; 33g:

v ¼ 58:

f5; 49; 54g; f6; 47; 53g; f7; 46; 53g; f8; 44; 52g; f9; 43; 52g; f11; 40; 51g;f12; 38; 50g; f13; 36; 49g; f14; 34; 48g; f15; 33; 48g; f17; 29; 46g; f18; 27; 45g;f18; 26; 44g; f19; 24; 43g; f20; 22; 42g; f4; 38; 42g; f5; 36; 41g; f6; 34; 40g;f7; 32; 39g; f8; 31; 39g; f10; 27; 37g; f11; 26; 37g; f12; 23; 35g; f13; 22; 35g;f9; 19; 28g; f20; 25; 45g; f14; 16; 30g; f21; 29; 50g; f24; 30; 54g; f4; 21; 25g;f28; 23; 51g; f10; 31; 41g; f16; 17; 33g; f15; 32; 47g:

So, 2Dv can be partitioned into 4s when v � 10 ðmod 12Þ and v � 34.(2) Let v ¼ 12sþ 6 and v � 30. We divide the problem into 2 cases:


Case 1. v ¼ 24aþ 6 where a � 3. It follows by Lemma 1.7 with ðd;mÞ ¼ð5; 4aÞ and ð4; 4a� 1Þ that ½5; 12aþ 4� [ ½4; 12a� can be partitioned into triplesfai; bi; cig, 1 � i � 8a� 1, such that ai þ bi ¼ ci. By Lemma 1.7 with ðd;mÞ ¼ð4; 4a� 1Þ, ½4; 12a� can be partitioned into triples fa0i; b0i; c0ig, 8a � i � 12a� 2, suchthat a0i þ b0i ¼ c0i. Taking ai ¼ v � a0i, bi ¼ v � b0i, and ci ¼ v � c0i for 8a � i �12a� 2, we know that ½12aþ 6; 24aþ 2� can be partitioned into triples fai; bi; cig,8a � i � 12a� 2, such that ai þ bi � ci ðmod vÞ. By Lemma 1.10 withðd;m; kÞ ¼ ð4; 4a; 4a� 1Þ, ½4; 12aþ 4� n f8aþ 2g can be partitioned into triplesfa0i; b0i; c0ig, 12a� 1 � i � 16a� 2, such that a0i þ b0i ¼ c0i. Similarly taking ai ¼v � a0i, bi ¼ v � b0i, and ci ¼ v � c0i for 12a� 1 � i � 16a� 2, we then know that½12aþ 2; 24aþ 2�nf16aþ 4g can be partitioned into fai; bi; cig, 12a� 1 � i �16a� 2, such that ai þ bi � ci ðmod vÞ. Considering also the triple f4; 12aþ 1;12aþ 5g, 2½4; v � 4�nf2v=3g can be partitioned into triples fai; bi; cig such thatai þ bi � ci ðmod vÞ.

Case 2. v ¼ 24aþ 18 where a � 2. It follows by Lemma 1.7 with ðd;mÞ ¼ð4; 4aþ 3Þ and ð5; 4aþ 4Þ that ½4; 12aþ 12� [ ½5; 12aþ 16� can be partitioned intotriples fai; bi; cig, 1 � i � 8aþ 7, such that ai þ bi ¼ ci. By Lemma 1.7 withðd;mÞ ¼ ð4; 4a� 1Þ, ½4; 12a� can be partitioned into triples fa0i; b0i; c0ig, 8aþ 8 �i � 12aþ 6, such that a0i þ b0i ¼ c0i. Taking ai ¼ v � a0i, bi ¼ v � b0i, and ci ¼ v � c0ifor 8aþ 8 � i � 12aþ 6, we know that ½12aþ 18; 24aþ 14� can be partitioned intotriples fai; bi; cig, 8aþ 8 � i � 12aþ 6, such that ai þ bi � ci ðmod vÞ. By Lemma1.10 with ðd;m; kÞ ¼ ð4; 4a; 4aþ 3Þ, ½4; 12aþ 4�nf8aþ 6g can be partitioned intotriples fa0i; b0i; c0ig, 12aþ 7 � i � 16aþ 6, such that a0i þ b0i ¼ c0i. Similarly takingai ¼ v � a0i, bi ¼ v � b0i, and ci ¼ v � c0i for 12aþ 7 � i � 16aþ 6, we then knowthat ½12aþ 14; 24aþ 14�nf16aþ 12g can be partitioned into triples fai; bi; cig,12aþ 7 � i � 16aþ 6, such that ai þ bi � ci ðmod vÞ. Note that 2½4; v � 4�nf2v=3g ¼ ð½4; 12a þ 12� [ ½5; 12a þ 16�Þ [ ð½12a þ 14; 24a þ 14�nf16aþ 12gÞ [½12aþ 18; 24aþ 14� [ f4; 12aþ 13; 12aþ 17g. So, 2½4; v � 4�nf2v=3g can be parti-tioned into triples fai; bi; cig such that ai þ bi � ci ðmod vÞ.

For v ¼ 30; 42; 54, 2½4; v � 4�nf2v=3g can be partitioned into triples as follows:

v ¼ 30 :

f6; 18; 24g; f10; 16; 26g; f5; 14; 19g; f7; 16; 23g; f7; 17; 24g; f8; 13; 21g;f8; 15; 23g; f9; 10; 19g; f4; 18; 22g; f4; 11; 15g; f11; 14; 25g; f9; 12; 21g;f5; 17; 22g; f12; 13; 25g; f6; 20; 26g:

v ¼ 42 :

f4; 34; 38g; f5; 32; 37g; f6; 30; 36g; f7; 28; 35g; f8; 25; 33g; f9; 22; 31g;f10; 19; 29g; f11; 16; 27g; f4; 20; 24g; f5; 18; 23g; f6; 15; 21g; f7; 18; 25g;f9; 20; 29g; f12; 24; 36g; f12; 26; 38g; f13; 14; 27g; f14; 21; 35g; f17; 17; 34g;f13; 19; 32g; f15; 16; 31g; f10; 23; 33g; f8; 22; 30g; f11; 26; 37g:

v ¼ 54 :

f4; 46; 50g; f5; 44; 49g; f6; 42; 48g; f7; 40; 47g; f9; 34; 43g; f10; 31; 41g;f11; 28; 39g; f12; 26; 38g; f13; 23; 36g; f14; 21; 35g; f15; 18; 33g; f20; 24; 44g;f5; 25; 30g; f7; 20; 27g; f9; 15; 24g; f11; 39; 50g; f18; 28; 46g; f19; 26; 45g;f21; 22; 43g; f17; 32; 49g; f13; 27; 40g; f10; 37; 47g; f19; 22; 41g; f8; 29; 37g;

326 ZHANG AND CHANG

f12; 33; 45g; f6; 29; 35g; f8; 23; 31g; f17; 25; 42g; f4; 34; 38g; f16; 32; 48g;f14; 16; 30g:

This completes the proof. &

Lemma 3.5. If v � 8 ðmod 12Þ and v � 21, then 3½4; v � 4� can be partitioned intotriples fai; bi; cig, 1 � i � v � 7, such that ai þ bi ¼ ci.

Proof. Set v ¼ 12sþ 8 where s � 2 and Dv ¼ ½4; v � 4�. When s � 4, by Lemma 1.7with ðd;mÞ ¼ ð5; 4sÞ and ð8; 4s� 1Þ, Dvnf4g and Dvnf4; 5; 6; 7g can be partitionedinto 4s. By Lemma 1.8 with ðd;mÞ ¼ ð4; 4s� 3Þ, Dvnf12s� 6;12s� 4; . . . ; 12sþ 4g can be partitioned into 4s. Note that T ¼ f4; 4; 5; 6; 7;12s� 6; 12s� 4; . . . ; 12sþ 4g can be partitioned into 4s: f4; 12s� 2; 12sþ 2g,f4; 12s� 3; 12sþ 1g, f5; 12s� 1; 12sþ 4g, f6; 12s� 6; 12sg, f7; 12s� 4; 12sþ 3g.So, 3½4; v � 4� can be partitioned into 4s.

For v ¼ 32; 44, 3½4; v � 4� can be partitioned into 4s as follows:

v ¼ 32:

f4; 12; 16g; f4; 21; 25g; f4; 18; 22g; f5; 18; 23g; f5; 17; 22g; f5; 20; 25g;f6; 22; 28g; f6; 19; 25g; f6; 15; 21g; f7; 20; 27g; f7; 17; 24g; f7; 14; 21g;f8; 19; 27g; f8; 16; 24g; f8; 12; 20g; f9; 18; 27g; f9; 14; 23g; f9; 10; 19g;f10; 16; 26g; f10; 13; 23g; f11; 15; 26g; f11; 13; 24g; f11; 17; 28g; f12; 14; 26g;f13; 15; 28g:

v ¼ 44:

f4; 31; 35g; f4; 26; 30g; f4; 15; 19g; f5; 30; 35g; f5; 35; 40g; f5; 15; 20g;f6; 22; 28g; f6; 23; 29g; f6; 34; 40g; f7; 32; 39g; f7; 27; 34g; f7; 22; 29g;f8; 20; 28g; f8; 21; 29g; f8; 25; 33g; f9; 30; 39g; f9; 24; 33g; f9; 27; 36g;f10; 28; 38g; f10; 23; 33g; f10; 16; 26g; f11; 14; 25g; f11; 27; 38g; f11; 21; 32g;f12; 26; 38g; f12; 13; 25g; f12; 20; 32g; f13; 24; 37g; f13; 18; 31g; f14; 23; 37g;f14; 17; 31g; f15; 22; 37g; f16; 24; 40g; f16; 18; 34g; f18; 21; 39g; 2f17; 19; 36g:

&

Lemma 3.6. If v � 2 ðmod 12Þ and v � 38, then 6½4; v � 4� can be partitioned intotriples fai; bi; cig, 1 � i � 2ðv � 7Þ, such that ai þ bi ¼ ci.

Proof. Set v ¼ 12sþ 14 where s � 2 and Dv ¼ ½4; v � 4�. For s ¼ 2; 3; 4, the con-clusion follows by (2) of Appendix B. Next we deal with s � 5. By Lemma 1.7with ðd;mÞ ¼ ð7; 4sþ 1Þ (twice) and ð10; 4sÞ, 2ðDvnf4; 5; 6; 12sþ 10gÞ [ Dvnf4; 5; 6; 7; 8; 9; 12sþ 10g can be partitioned into 4s. By Lemma 1.8 withðd;mÞ ¼ ð5; 4s� 2Þ (twice) and ð4; 4sþ 2Þ, 2ðDvn f4; 12s� 2; 12s; . . . ; 12sþ10gÞ [ Dvnf12sþ 9g can be partitioned into 4s. Note that T ¼ f12sþ9g [ ð2f4; 5; 6; 12sþ 10gÞ [ f4; 5; 6; 7; 8; 9; 12sþ 10g[ ð2f4; 12s� 2; 12s; . . . ;12sþ 10gÞ can be partitioned into 4s: f4; 12s� 2; 12sþ 2g (twice), f4; 12sþ3; 12sþ 7g, f4; 12sþ 6; 12sþ 10g (twice), f5; 12sþ 5; 12sþ 10g (twice),f5; 12sþ 4; 12sþ 9g, f6; 12sþ 4; 12sþ 10g, f6; 12sþ 3; 12sþ 9g, f6; 12sþ1; 12sþ 7g, f7; 12sþ 1; 12sþ 8g, f8; 12s; 12sþ 8g, f9; 12s; 12sþ 9g. &



(1) There exists a CBSAðv; 3; 1; 3Þ for v � 1; 3 ðmod 6Þ;(2) There exists a CBSAðv; 3; 2; 3Þ for v � 0; 4 ðmod 12Þ;(3) There exists a CBSAðv; 3; 3; 3Þ for v � 5 ðmod 6Þ;(4) There exists a CBSAðv; 3; 4; 3Þ for v � 6; 10 ðmod 12Þ;(5) There exists a CBSAðv; 3; 6; 3Þ for v � 8 ðmod 12Þ;(6) There exists a CBSAðv; 3; 12; 3Þ for v � 2 ðmod 12Þ:

Proof. (1) For v � 1 ðmod 6Þ and v � 49, by Theorem 1.3, there exists aCBSAðv; 3; 1; 3Þ. For each v 2 f25; 31; 37; 43g, a CBSAðv; 3; 1; 3Þ is constructed bylisting base blocks as follows:

v ¼ 25: f0; 4; 11g; f0; 5; 13g; f0; 6; 15g:v ¼ 31: f0; 4; 12g; f0; 5; 14g; f0; 6; 16g; f0; 7; 18g:v ¼ 37: f0; 4; 11g; f0; 5; 19g; f0; 6; 16g; f0; 8; 20g; f0; 9; 22g:v ¼ 43: f0; 4; 21g; f0; 5; 15g; f0; 6; 19g; f0; 7; 16g; f0; 8; 20g; f0; 11; 25g:

For v � 3 ðmod 6Þ and v � 21, by Lemma 3.1 the multiset ½4; ðv � 1Þ=2�nfv=3gcan be partitioned into triples fai; bi; cig, 1 � i � ðv � 9Þ=6, such that either ai þbi ¼ ci or ai þ bi þ ci ¼ v. For 1 � i � ðv � 9Þ=6; set



�

Then, it is easily checked that [ðv�9Þ=6i¼1 �Bi ¼ ½4; v � 4�nfv=3; 2v=3g mod v. Con-

sidering also the base block f0; v=3; 2v=3g (short orbit), by Lemma 1.5, we havethat a CBSAðv; 3; 1; 3Þ exists.

(2) For v � 4 ðmod 12Þ and v � 100, there exists a CBSAðv; 3; 2; 3Þ by Theorem1.3. For each v 2 f28; 40; 52; 64; 76; 88g, a CBSAðv; 3; 2; 3Þ is constructed by listingbase blocks as follows:

v ¼ 28 :f0; 4; 20g; f0; 5; 18g; f0; 6; 23g; f0; 7; 21g; f0; 8; 19g; f0; 9; 24g;f0; 10; 22g:

v ¼ 40 :f0; 4; 20g; f0; 5; 35g; f0; 6; 28g; f0; 7; 33g; f0; 8; 31g; f0; 9; 27g;f0; 10; 34g; f0; 11; 25g; f0; 12; 29g; f0; 13; 32g; f0; 15; 36g:

v ¼ 52 : f0; 4; 25g; f0; 5; 47g; f0; 6; 30g; f0; 7; 45g; f0; 8; 43g; f0; 9; 41g;f0; 10; 39g; f0; 11; 37g; f0; 12; 46g; f0; 13; 36g; f0; 14; 33g; f0; 15; 31g;f0; 17; 44g; f0; 18; 40g; f0; 20; 48g:

v ¼ 64 :

f0; 4; 52g; f0; 5; 27g; f0; 6; 32g; f0; 7; 57g; f0; 8; 55g; f0; 9; 53g;f0; 10; 51g; f0; 11; 49g; f0; 12; 36g; f0; 13; 46g; f0; 14; 45g; f0; 15; 43g;f0; 16; 58g; f0; 17; 40g; f0; 18; 39g; f0; 19; 56g; f0; 20; 54g; f0; 25; 60g;f0; 29; 59g:

v ¼ 76 :

f0; 4; 70g; f0; 5; 39g; f0; 6; 64g; f0; 7; 69g; f0; 8; 67g; f0; 9; 65g;f0; 10; 63g; f0; 11; 61g; f0; 12; 48g; f0; 13; 45g; f0; 14; 57g; f0; 15; 55g;f0; 16; 38g; f0; 17; 52g; f0; 18; 51g; f0; 19; 49g; f0; 20; 47g; f0; 21; 46g;f0; 23; 60g; f0; 24; 68g; f0; 26; 54g; f0; 29; 71g; f0; 31; 72g:

328 ZHANG AND CHANG

f0; 4; 44g; f0; 5; 83g; f0; 6; 82g; f0; 7; 46g; f0; 8; 71g; f0; 9; 68g;f0; 10; 75g; f0; 11; 73g; f0; 12; 72g; f0; 13; 70g; f0; 14; 69g; f0; 15; 67g;

v ¼ 88 : f0; 16; 66g; f0; 17; 37g; f0; 18; 45g; f0; 19; 61g; f0; 21; 79g; f0; 22; 56g;f0; 23; 54g; f0; 24; 53g; f0; 25; 51g; f0; 28; 64g; f0; 30; 77g; f0; 32; 80g;f0; 33; 74g; f0; 35; 84g; f0; 38; 81g:

For v � 0 ðmod 12Þ and v � 36, by Lemma 3.2, the multiset ½4; v � 4�nfv=3;2v=3g can be partitioned into triples fai; bi; cig, 1 � i � ðv � 9Þ=3, such thatai þ bi � ci ðmod vÞ. Set Bi ¼ f0; ai; cig; 1 � i � ðv � 9Þ=3. Then, it is easilychecked that [ðv�9Þ=3

i¼1 �Bi ¼ 2½4; v � 4� � 2fv=3; 2v=3g mod v. Considering also thetwo base blocks 2f0; v=3; 2v=3g (short orbits), by Lemma 1.5, we have that thereexists a CBSAðv; 3; 2; 3Þ for v � 0 ðmod 12Þ and v � 36. Now we deal with the casev ¼ 24. The base blocks of a BSAð24; 3; 2; 3Þ are: f0; 5; 19g, f0; 6; 18g, f0; 7; 17g,f0; 9; 20g, f0; 4; 15g, 2f0; 8; 16g (short orbits).

(3) For v � 5 ðmod 6Þ and v � 21, by Lemma 3.3, the multiset 3½4; ðv � 1Þ=2� canbe partitioned into triples fai; bi; cig, 1 � i � ðv � 7Þ=2, such that either ai þ bi ¼ cior ai þ bi þ ci ¼ v. For 1 � i � ðv � 7Þ=2; set



�

Then, it is easily checked that [ðv�7Þ=2i¼1 �Bi ¼ 3½4; v � 4� mod v. By Lemma 1.5,

there exists a CBSAðv; 3; 3; 3Þ.(4) For v � 10 ðmod 12Þ and v � 34, by Lemma 3.4, the multiset 2½4; v � 4�

can be partitioned into triples fai; bi; cig, 1 � i � 2ðv � 7Þ=3, such that ai þ bi ¼ ci.Set Bi ¼ f0; ai; cig for 1 � i � 2ðv � 7Þ=3. Then, it is easily checked that[2ðv�7Þ=3i¼1 �Bi ¼ 4½4; v � 4� mod v. By Lemma 1.5, there exists a CBSAðv; 3; 4; 3Þ

for each v � 10 ðmod 12Þ and v � 34. For the remaining case of v ¼ 22, we list the10 base blocks as follows: f0; 4; 14g; f0; 4; 15g; f0; 4; 13g; f0; 5; 16g; f0; 5; 17g;f0; 6; 16g; f0; 6; 14g; f0; 7; 15g; f0; 7; 12g; f0; 9; 18g:

For v � 6 ðmod 12Þ and v � 30, by Lemma 3.4, the multiset 2½4; v � 4�nf2v=3gcan be partitioned into triples fai; bi; cig, 1 � i � ð2v � 15Þ=3, such that ai þ bi �ci ðmod vÞ. Set Bi ¼ f0; ai; cig for 1 � i � ð2v � 15Þ=3. Then, it is readily checkedthat [ð2v�15Þ=3

i¼1 �Bi ¼ 4½4; v � 4� � fv=3; 2v=3g mod v. Considering also the baseblock f0; v=3; 2v=3g (short orbit), by Lemma 1.5, we know that a CBSAðv; 3; 4; 3Þexists for v � 6 ðmod 12Þ and v � 21.

By Lemma 3.5 and Lemma 3.6, similar arguments, as in (4), give the results of (5)and (6), respectively. For the remaining case of v ¼ 26, we list the 38 base blocks asfollows: f0; 5; 21g; f0; 6; 19g; 2f0; 7; 17g; 2f0; 9; 14g; f0; 4; 21g; 2f0; 5; 19g;3f0; 6; 18g; f0; 7; 16g; f0; 8; 14g; f0; 9; 15g; f0; 10; 20g; f0; 4; 20g; f0; 8; 15g;f0; 10; 21g; f0; 9; 22g; 2f0; 4; 15g; f0; 5; 16g; f0; 7; 18g; f0; 8; 19g; f0; 9; 21g;f0; 10; 22g; f0; 5; 22g; 2f0; 7; 20g; f0; 8; 20g; f0; 8; 16g; f0; 6; 17g; f0; 4; 17g;f0; 5; 18g; f0; 4; 16g; f0; 4; 8g; f0; 7; 22g: &


(1) There exists a CBSAðv; 3; 1; 3Þ if and only if v � 1; 3 ðmod 6Þ;(2) There exists a CBSAðv; 3; 2; 3Þ if and only if v � 0; 1 ðmod 3Þ and

v 6� 2 ðmod 4Þ;


(3) There exists a CBSAðv; 3; 3; 3Þ if and only if v � 1 ðmod 2Þ;(4) There exists a CBSAðv; 3; 4; 3Þ if and only if v � 0; 1 ðmod 3Þ;(5) There exists a CBSAðv; 3; 6; 3Þ if and only if v 6� 2 ðmod 4Þ;(6) There exists a CBSAðv; 3; 12; 3Þ if and only if v is a positive integer.

Proof. Necessity follows by Lemma 1.4, so we establish sufficiency in each case.

(1) Sufficiency follows by Theorem 3.7 (1).(2) If v � 0; 4 ðmod 12Þ then Theorem 3.7 (2) applies. If v � 1; 3 ðmod 6Þ,

a CBSAðv; 3; 1; 3Þ exists by Theorem 3.7 (1), so repeating blocks give aCBSAðv; 3; 2; 3Þ.

(3) If v � 5 ðmod 6Þ then Theorem 3.7 (3) applies. If v � 1; 3 ðmod 6Þ, aCBSAðv; 3; 1; 3Þ exists by Theorem 3.7 (1), so repeating blocks three times give aCBSAðv; 3; 3; 3Þ.

For each case of (4)–(6) sufficiency follows by similar arguments as in (3). &

4. CONCLUSIONS

From Theorem 2.7 and Theorem 3.8, we get the main results of this paper.

Theorem 4.1. A CBSAðv; 3; �; 2Þ exists if and only if v � 15 and

(1) v � 3; 5 ðmod 6Þ when � � 1; 5 ðmod 6Þ but v 6� 3 ðmod 6Þ when � ¼ 1;(2) v � 0; 2 ðmod 3Þ and v 6� 2 ðmod 4Þ when � � 2; 10 ðmod 12Þ;(3) v � 1 ðmod 2Þ when � � 3 ðmod 6Þ;(4) v � 0; 2 ðmod 3Þ when � � 4; 8 ðmod 12Þ;(5) v 6� 2 ðmod 4Þ when � � 6 ðmod 12Þ;(6) v is unrestricted when � � 0 ðmod 12Þ:

Theorem 4.2. A CBSAðv; 3; �; 3Þ exists if and only if v � 21 and

(1) v � 1; 3 ðmod 6Þ when � � 1; 5 ðmod 6Þ;(2) v � 0; 1 ðmod 3Þ and v 6� 2 ðmod 4Þ when � � 2; 10 ðmod 12Þ;(3) v � 1 ðmod 2Þ when � � 3 ðmod 6Þ;(4) v � 0; 1 ðmod 3Þ when � � 4; 8 ðmod 12Þ;(5) v 6� 2 ðmod 4Þ when � � 6 ðmod 12Þ;(6) v is unrestricted when � � 0 ðmod 12Þ:

ACKNOWLEDGMENTS

The authors thank Dr. Linek for providing them with reference [7], and the anony-mous referees for their careful reading and useful comments.

REFERENCES

[1] M. J. Colbourn and C. J. Colbourn, Cyclic block designs with block size 3, Eur J Combin2 (1981), 21–26.

330 ZHANG AND CHANG

[2] C. J. Colbourn and A. C. H. Ling, A class of partial triple system with applications insurvey sampling, Comm Statist Theory Methods 27 (1998), 1009–1018.

[3] A. S. Hedayat, C. R. Rao, and J. Stufken, Sampling plans excluding contiguous units,J Statist Plann Inference 19 (1988), 159–170.

[4] A. S. Hedayat, C. R. Rao, and J. Stufken, Designs in survey sampling avoiding contiguousunits, In: Handbook of Statistics, Vol. 6, P. R. Krishnaiah and C. R. Rao, (Editors),Elsevier, Amsterdam, 1988, pp. 575–583.

[5] D. G. Horvitz and D. J. Tompson, A generalization of sampling without replacement froma finite universe, J Amer Statist Assoc 47 (1952), 663–685.

[6] V. Linek and Z. Jiang, Extended Langford sequence with small defects, J Combin TheorySer A 84 (1998), 38–54.

[7] V. Linek and S. Mor, On partitions of f1; . . . ; 2mþ 1g � fkg into differencesd; . . . ; d þ m� 1 : Extended Langford sequence of large defect, J Combin Designs 12(2004), 421–442.

[8] K. See, S. Y. Song, and J. Stufken, On a class of partially incomplete block designs withapplications in survey sampling, Comm Statist Theory Methods 26 (1997), 1–13.

[9] J. E. Simpson, Langford sequences: Perfect and hooked, Discrete Math 44 (1983), 97–104.

[10] J. Stufken, Combinatorial and statistical aspects of sampling plans to avoid the selectionof adjacent units, J Comb Inform System Sci 18 (1993), 149–160.

[11] R. Wei, Cyclic BSEC of block size 3, Discrete Math 250 (2002), 291–298.

[12] J. Zhang and Y. Chang, The spectrum of cyclic BSEC with block size three, preprint.

APPENDIX A

(1) For each integer v 2 f19; 21; 25; 27; 31; 33; 37; 39; 43; 45; 51; 57; 61; 67; 85g, themultiset 3½3; ðv � 1Þ=2� can be partitioned into triples as follows:

v ¼ 19 : f4; 7; 8g; f7; 7; 5g; f3; 6; 9g; 2f3; 5; 8g; 2f4; 6; 9g:

v ¼ 21 : f4; 6; 10g; f3; 7; 10g; f5; 7; 9g; f4; 8; 9g; f6; 6; 9g; f4; 7; 10g;2f3; 5; 8g:

v ¼ 25 : f3; 9; 12g; f3; 7; 10g; f5; 7; 12g; f3; 8; 11g; f4; 7; 11g; f5; 6; 11g;f4; 4; 8g; f5; 8; 12g; 2f6; 9; 10g:

v ¼ 27 : f3; 10; 13g; f4; 9; 13g; f5; 8; 13g; f6; 6; 12g; f3; 9; 12g; f4; 8; 12g;f5; 6; 11g; f4; 7; 11g; f7; 10; 10g; f3; 5; 8g; f7; 9; 11g:

v ¼ 31 : f4; 11; 15g; f5; 10; 15g; f6; 8; 14g; f7; 7; 14g; f3; 11; 14g; f6; 7; 13g;f4; 5; 9g; f8; 10; 13g; f9; 12; 10g; f3; 13; 15g; f3; 9; 12g; f4; 8; 12g;f5; 6; 11g:

v ¼ 33 : f5; 12; 16g; f5; 11; 16g; f6; 9; 15g; f7; 8; 15g; f3; 12; 15g; f4; 10; 14g;f5; 9; 14g; f6; 8; 14g; f4; 9; 13g; f3; 7; 10g; f10; 11; 12g; f3; 8; 11g;f4; 13; 16g; f6; 7; 13g:

v ¼ 37 : f4; 13; 17g; f5; 11; 16g; f3; 13; 16g; f3; 12; 15g; f6; 9; 15g; f4; 6; 10g;f8; 12; 17g; f4; 7; 11g; f8; 10; 18g; f8; 14; 15g; f3; 7; 10g; f9; 12; 16g;f6; 7; 13g; 2f5; 14; 18g; f9; 11; 17g:


v ¼ 39 : f3; 16; 19g; f6; 12; 18g; f7; 11; 18g; f8; 10; 18g; f3; 14; 17g; f4; 13; 17g;f5; 12; 17g; f6; 10; 16g; f7; 9; 16g; f7; 8; 15g; f3; 12; 15g; f4; 10; 14g;f5; 8; 13g; f6; 13; 19g; f4; 11; 15g; f5; 9; 14g; f9; 11; 19g:

v ¼ 43 : f4; 17; 21g; f5; 16; 21g; f6; 14; 20g; f7; 13; 20g; f8; 12; 20g; f9; 10; 19g;f3; 16; 19g; f4; 15; 19g; f5; 13; 18g; f7; 10; 17g; f8; 9; 17g; f3; 13; 16g;f4; 11; 15g; f5; 10; 15g; f3; 9; 12g; f7; 11; 18g; f11; 14; 18g; f6; 6; 12g;f8; 14; 21g:

v ¼ 45 : f5; 17; 22g; f6; 15; 21g; f7; 14; 21g; f8; 13; 21g; f9; 11; 20g; f3; 17; 20g;f4; 16; 20g; f7; 11; 18g; f8; 10; 18g; f3; 14; 17g; f4; 12; 16g; f5; 11; 16g;f7; 8; 15g; f9; 10; 19g; f10; 12; 22g; f13; 13; 19g; f6; 6; 12g; f3; 15; 18g;f5; 9; 14g; f4; 19; 22g:

v ¼ 51 : f3; 22; 25g; f7; 17; 24g; f8; 16; 24g; f7; 16; 23g; f8; 15; 23g; f9; 14; 23g;f10; 12; 22g; f3; 19; 22g; f6; 14; 20g; f8; 11; 19g; f9; 10; 19g; f3; 15; 18g;f4; 14; 18g; f9; 11; 20g; f12; 13; 25g; f13; 17; 21g; f6; 7; 13g; f6; 12; 18g;f4; 17; 21g; f5; 10; 15g; f5; 11; 16g; f4; 20; 24g; f5; 21; 25g:

v ¼ 57 : f3; 25; 28g; f4; 24; 28g; f5; 23; 28g; f6; 21; 27g; f7; 20; 27g; f8; 19; 27g;f9; 17; 26g; f11; 15; 26g; f12; 13; 25g; f3; 22; 25g; f5; 19; 24g; f6; 17; 23g;f7; 16; 23g; f8; 14; 22g; f9; 13; 22g; f10; 11; 21g; f5; 15; 20g; f6; 13; 19g;f7; 11; 18g; f8; 10; 18g; f4; 16; 20g; f10; 14; 24g; f3; 15; 18g; f14; 17; 26g;f4; 12; 16g; f9; 12; 21g:

v ¼ 61 : f4; 26; 30g; f6; 23; 29g; f7; 22; 29g; f9; 19; 28g; f10; 18; 28g; f11; 17; 28g;f13; 14; 27g; f12; 14; 26g; f3; 23; 26g; f4; 21; 25g; f5; 20; 25g; f6; 18; 24g;f7; 17; 24g; f8; 15; 23g; f9; 13; 22g; f10; 12; 22g; f4; 16; 20g; f5; 15; 20g;f7; 10; 17g; f3; 8; 11g; f11; 19; 30g; f9; 16; 25g; f16; 24; 21g; f6; 8; 14g;f5; 27; 29g; f3; 15; 18g; f12; 19; 30g; f13; 21; 27g:

v ¼ 67 : f3; 30; 33g; f4; 29; 33g; f5; 28; 33g; f7; 25; 32g; f8; 24; 32g; f3; 13; 16g;f12; 26; 29g; f9; 22; 31g; f10; 21; 31g; f11; 20; 31g; f12; 18; 30g; f13; 17; 30g;f14; 15; 29g; f5; 23; 28g; f6; 21; 27g; f7; 20; 27g; f8; 19; 27g; f9; 17; 26g;f10; 15; 25g; f11; 14; 25g; f4; 19; 23g; f5; 18; 23g; f6; 16; 22g; f7; 15; 22g;f8; 13; 21g; f9; 11; 20g; f10; 14; 24g; f3; 16; 19g; f6; 12; 18g; f4; 28; 32g;f17; 24; 26g:

v ¼ 85 : f5; 37; 42g; f6; 35; 41g; f7; 34; 41g; f8; 33; 41g; f9; 31; 40g; f11; 29; 40g;f13; 26; 39g; f14; 24; 38g; f15; 23; 38g; f3; 34; 37g; f4; 33; 37g; f5; 31; 36g;f6; 30; 36g; f7; 29; 36g; f9; 26; 35g; f10; 24; 34g; f12; 20; 32g; f13; 19; 32g;f14; 18; 32g; f15; 16; 31g; f14; 16; 30g; f3; 26; 29g; f4; 24; 28g; f5; 23; 28g;f6; 22; 28g; f7; 20; 27g; f8; 17; 25g; f9; 16; 25g; f10; 15; 25g; f11; 12; 23g;f18; 22; 40g; f3; 18; 21g; f4; 35; 39g; f17; 21; 38g; f12; 21; 33g; f19; 27; 39g;f20; 22; 42g; f10; 17; 27g; f13; 30; 42g; f8; 11; 19g:

(2) For v ¼ 22; the multiset 6½3; v � 3� can be partitioned into triples as follows:

f3; 14; 17g; f3; 15; 18g; f3; 14; 17g; f3; 12; 15g; 2f3; 10; 13g; f4; 15; 19g;f4; 14; 18g; f4; 13; 17g; f4; 12; 16g; f4; 11; 15g; f4; 12; 16g; 2f5; 11; 16g;f5; 13; 18g; f5; 10; 15g; 2f5; 9; 14g; f6; 8; 14g; f6; 13; 19g; f6; 12; 18g;f6; 11; 17g; f6; 10; 16g; f6; 7; 13g; f9; 10; 19g; 2f7; 12; 19g; f7; 11; 18g;f7; 10; 17g; f7; 8; 15g; f8; 11; 19g; f8; 9; 17g; f8; 8; 16g; f9; 9; 18g:

332 ZHANG AND CHANG

APPENDIX B

(1) For each v 2 f23; 29; 35; 41; 47; 53; 59; 65; 71; 77; 89; 95g, the multiset 3½4;ðv � 1Þ=2� can be partitioned into triples as follows:

v ¼ 23 : f4; 7; 11g; f4; 6; 10g; f4; 8; 11g; f5; 9; 9g; f7; 7; 9g; f6; 6; 11g;2f5; 8; 10g:

v ¼ 29 : f4; 10; 14g; f4; 8; 12g; f4; 9; 13g; f5; 9; 14g; f5; 8; 13g; f5; 7; 12g;f6; 8; 14g; f6; 7; 13g; f6; 11; 12g; f7; 11; 11g; f9; 10; 10g:

v ¼ 35 : f4; 9; 13g; f4; 7; 11g; f4; 12; 16g; f5; 11; 16g; f5; 12; 17g; f5; 10; 15g;f6; 10; 16g; f6; 11; 17g; f6; 9; 15g; f7; 8; 15g; f7; 10; 17g; f9; 12; 14g;2f8; 13; 14g:



v ¼ 53 : f4; 22; 26g; f5; 21; 26g; f6; 20; 26g; f7; 18; 25g; f8; 17; 25g; f9; 16; 25g;f10; 14; 24g; f11; 13; 24g; f12; 12; 24g; f5; 18; 23g; f6; 17; 23g; f7; 15; 22g;f8; 14; 22g; f9; 12; 21g; f10; 11; 21g; f5; 15; 20g; f6; 13; 19g; f8; 11; 19g;f7; 9; 16g; f10; 13; 23g; f4; 14; 18g; f4; 15; 19g; f16; 17; 20g:

v ¼ 59 : f4; 25; 29g; f6; 23; 29g; f7; 21; 28g; f8; 20; 28g; f9; 19; 28g; f10; 17; 27g;f11; 16; 27g; f12; 15; 27g; f4; 22; 26g; f5; 21; 26g; f7; 18; 25g; f8; 17; 25g;f9; 15; 24g; f10; 14; 24g; f11; 12; 23g; f4; 19; 23g; f5; 17; 22g; f6; 16; 22g;f7; 14; 21g; f8; 12; 20g; f9; 10; 19g; f5; 13; 18g; f11; 13; 24g; f13; 16; 29g;f6; 14; 20g; f15; 18; 26g:

v ¼ 65 : f5; 27; 32g; f6; 26; 32g; f7; 24; 31g; f8; 23; 31g; f10; 20; 30g; f11; 19; 30g;f12; 18; 30g; f13; 16; 29g; f14; 15; 29g; f4; 25; 29g; f5; 23; 28g; f6; 22; 28g;f7; 20; 27g; f8; 19; 27g; f9; 17; 26g; f10; 16; 26g; f11; 14; 25g; f4; 20; 24g;f5; 19; 24g; f6; 17; 23g; f7; 15; 22g; f8; 13; 21g; f9; 12; 21g; f10; 11; 21g;f4; 14; 18g; f15; 17; 32g; f9; 13; 22g; f12; 25; 28g; f16; 18; 31g:

v ¼ 71 : f5; 30; 35g; f7; 27; 34g; f9; 25; 34g; f10; 23; 33g; f11; 22; 33g; f12; 21; 33g;f13; 19; 32g; f14; 18; 32g; f15; 17; 32g; f4; 27; 31g; f5; 26; 31g; f6; 24; 30g;f7; 23; 30g; f8; 21; 29g; f9; 20; 29g; f10; 18; 28g; f11; 17; 28g; f12; 16; 28g;f13; 14; 27g; f4; 22; 26g; f5; 20; 25g; f6; 19; 25g; f7; 17; 24g; f8; 16; 24g;f9; 14; 23g; f10; 12; 22g; f4; 11; 15g; f13; 18; 31g; f16; 19; 35g; f8; 34; 29g;f15; 21; 35g; f6; 20; 26g:

v ¼ 77 : f4; 34; 38g; f5; 33; 38g; f6; 32; 38g; f7; 30; 37g; f9; 28; 37g; f10; 26; 36g;f12; 24; 36g; f13; 22; 35g; f14; 21; 35g; f15; 20; 35g; f16; 18; 34g; f4; 30; 34g;f6; 27; 33g; f7; 25; 32g; f8; 24; 32g; f9; 22; 31g; f10; 21; 31g; f11; 20; 31g;f12; 18; 30g; f13; 16; 29g; f14; 15; 29g; f4; 24; 28g; f5; 22; 27g; f6; 21; 27g;f7; 19; 26g; f9; 16; 25g; f10; 13; 23g; f11; 12; 23g; f8; 17; 25g; f17; 20; 37g;f14; 15; 29g; f17; 19; 36g; f5; 23; 28g; f18; 26; 33g; f8; 11; 19g:

v ¼ 89 : f5; 39; 44g; f6; 38; 44g; f7; 36; 43g; f8; 35; 43g; f9; 34; 43g; f10; 32; 42g;f11; 31; 42g; f12; 30; 42g; f13; 28; 41g; f14; 27; 41g; f15; 26; 41g; f16; 24; 40g;


f17; 23; 40g; f19; 20; 39g; f4; 34; 38g; f6; 31; 37g; f8; 29; 37g; f9; 27; 36g;f10; 26; 36g; f11; 24; 35g; f12; 23; 35g; f13; 21; 34g; f14; 19; 33g; f15; 18; 33g;f4; 28; 32g; f5; 27; 32g; f6; 25; 31g; f7; 23; 30g; f8; 22; 30g; f9; 20; 29g;f10; 19; 29g; f11; 17; 28g; f12; 14; 26g; f7; 15; 22g; f16; 21; 37g; f4; 13; 17g;f20; 24; 44g; f18; 22; 40g; f25; 25; 39g; f5; 16; 21g; f18; 33; 38g:

v ¼ 95 : f5; 42; 47g; f6; 41; 47g; f7; 39; 46g; f8; 38; 46g; f9; 37; 46g; f10; 35; 45g;f11; 34; 45g; f12; 33; 45g; f13; 31; 44g; f14; 30; 44g; f16; 27; 43g; f17; 26; 43g;f18; 24; 42g; f19; 23; 42g; f20; 21; 41g; f4; 37; 41g; f6; 34; 40g; f7; 33; 40g;f8; 31; 39g; f10; 28; 38g; f11; 27; 38g; f12; 25; 37g; f13; 23; 36g; f15; 21; 36g;f16; 19; 35g; f5; 28; 33g; f6; 26; 32g; f7; 25; 32g; f8; 24; 32g; f9; 21; 30g;f10; 19; 29g; f11; 18; 29g; f12; 16; 28g; f13; 14; 27g; f17; 26; 43g; f22; 25; 47g;f4; 20; 24g; f5; 18; 23g; f20; 35; 40g; f9; 22; 31g; f4; 30; 34g; f17; 22; 39g;f15; 44; 36g; f14; 15; 29g:

(2) For each v 2 f38; 50; 62g, the multiset 6½4; v � 4� can be partitioned into triples as follows:

v ¼ 38 : f4; 24; 28g; f5; 23; 28g; f6; 22; 28g; f7; 21; 28g; f8; 19; 27g; f9; 18; 27g;f10; 17; 27g; f11; 16; 27g; f12; 14; 26g; f13; 15; 28g; f14; 15; 29g; f15; 17; 32g;f4; 22; 26g; f5; 21; 26g; f6; 19; 25g; f8; 17; 25g; f9; 15; 24g; f10; 14; 24g;f11; 13; 24g; f7; 11; 18g; f4; 16; 20g; f5; 18; 23g; f6; 17; 23g; f7; 16; 23g;f8; 12; 20g; f9; 13; 22g; f10; 19; 29g; f4; 30; 34g; f5; 15; 20g; f6; 28; 34g;f7; 27; 34g; f8; 26; 34g; f9; 25; 34g; f10; 23; 33g; f11; 22; 33g; f12; 21; 33g;f13; 20; 33g; f14; 19; 33g; f15; 18; 33g; f16; 16; 32g; f4; 21; 25g; f5; 27; 32g;f6; 26; 32g; f7; 25; 32g; f8; 24; 32g; f9; 22; 31g; f10; 21; 31g; f11; 20; 31g;f12; 19; 31g; f13; 18; 31g; f14; 17; 31g; f4; 26; 30g; f5; 25; 30g; f6; 24; 30g;f7; 23; 30g; f8; 22; 30g; f9; 12; 21g; f10; 19; 29g; f11; 18; 29g; f12; 17; 29g;f13; 16; 29g; f14; 20; 34g:

v ¼ 50 : f5; 41; 46g; f6; 40; 46g; f7; 39; 46g; f8; 38; 46g; f9; 37; 46g; f10; 35; 45g;f11; 34; 45g; f12; 33; 45g; f13; 32; 45g; f14; 31; 45g; f15; 30; 45g; f16; 28; 44g;f17; 27; 44g; f18; 26; 44g; f19; 25; 44g; f20; 24; 44g; f21; 23; 44g; f4; 39; 43g;f5; 38; 43g; f7; 36; 43g; f8; 35; 43g; f9; 34; 43g; f10; 32; 42g; f11; 31; 42g;f12; 30; 42g; f13; 29; 42g; f14; 28; 42g; f15; 26; 41g; f16; 25; 41g; f17; 24; 41g;f18; 23; 41g; f19; 22; 41g; f4; 36; 40g; f5; 35; 40g; f6; 34; 40g; f7; 33; 40g;f8; 32; 40g; f9; 30; 39g; f10; 29; 39g; f12; 27; 39g; f13; 25; 38g; f14; 24; 38g;f15; 23; 38g; f16; 22; 38g; f17; 20; 37g; f18; 19; 37g; f4; 33; 37g; f5; 32; 37g;f6; 30; 36g; f7; 29; 36g; f8; 28; 36g; f9; 27; 36g; f10; 25; 35g; f11; 24; 35g;f12; 23; 35g; f13; 21; 34g; f14; 20; 34g; f15; 19; 34g; f16; 17; 33g; f4; 29; 33g;f5; 28; 33g; f6; 26; 32g; f7; 25; 32g; f8; 23; 31g; f9; 22; 31g; f10; 21; 31g;f11; 20; 31g; f12; 18; 30g; f13; 17; 30g; f14; 15; 29g; f4; 25; 29g; f5; 23; 28g;f6; 22; 28g; f7; 20; 27g; f9; 18; 27g; f10; 16; 26g; f11; 15; 26g; f12; 14; 26g;f19; 20; 39g; f22; 24; 46g; f4; 13; 17g; f18; 24; 42g; f21; 22; 43g; f16; 21; 37g;f8; 11; 19g; f6; 21; 27g:

v ¼ 62 : f4; 54; 58g; f5; 53; 58g; f6; 52; 58g; f7; 51; 58g; f8; 50; 58g; f9; 49; 58g;f10; 47; 57g; f11; 46; 57g; f12; 45; 57g; f13; 44; 57g; f14; 43; 57g; f15; 42; 57g;f16; 40; 56g; f17; 39; 56g; f18; 38; 56g; f19; 37; 56g; f20; 36; 56g; f21; 35; 56g;f23; 32; 55g; f24; 31; 55g; f25; 30; 55g; f26; 29; 55g; f27; 28; 55g; f4; 50; 54g;f5; 49; 54g; f6; 48; 54g; f7; 47; 54g; f8; 46; 54g; f9; 44; 53g; f10; 43; 53g;f11; 42; 53g; f12; 41; 53g; f13; 40; 53g; f14; 38; 52g; f16; 36; 52g; f17; 35; 52g;f18; 34; 52g; f19; 32; 51g; f20; 31; 51g; f21; 30; 51g; f22; 29; 51g; f23; 28; 51g;f24; 26; 50g; f4; 46; 50g; f5; 45; 50g; f6; 44; 50g; f7; 42; 49g; f8; 41; 49g;f9; 40; 49g; f11; 37; 48g; f12; 36; 48g; f13; 35; 48g; f14; 34; 48g; f16; 31; 47g;

334 ZHANG AND CHANG

f17; 30; 47g; f18; 29; 47g; f19; 28; 47g; f20; 26; 46g; f21; 25; 46g; f22; 24; 46g;f4; 41; 45g; f5; 40; 45g; f6; 39; 45g; f7; 38; 45g; f9; 35; 44g; f10; 34; 44g;f11; 32; 43g; f12; 31; 43g; f13; 30; 43g; f14; 29; 43g; f15; 27; 42g; f16; 26; 42g;f17; 25; 42g; f18; 23; 41g; f19; 22; 41g; f20; 21; 41g; f4; 36; 40g; f5; 35; 40g;f6; 33; 39g; f7; 32; 39g; f8; 31; 39g; f9; 30; 39g; f10; 28; 38g; f11; 27; 38g;f12; 26; 38g; f13; 24; 37g; f14; 23; 37g; f15; 22; 37g; f16; 20; 36g; f17; 18; 35g;f4; 30; 34g; f5; 29; 34g; f6; 28; 34g; f7; 26; 33g; f8; 25; 33g; f9; 24; 33g;f10; 23; 33g; f11; 21; 32g; f12; 20; 32g; f13; 18; 31g; f14; 15; 29g; f27; 28; 55g;f22; 27; 49g; f10; 17; 27g; f8; 16; 24g; f19; 25; 44g; f23; 25; 48g; f15; 21; 36g;f15; 22; 37g; f19; 33; 52g:

APPENDIX C

(1) For v 2 f33; 39; 57; 63; 81; 87; 105; 111; 129g, the set ½4; ðv � 1Þ=2� n fv=3g canbe partitioned into triples as follows:

v ¼ 33 : f7; 9; 16g; f4; 8; 12g; f5; 10; 15g; f6; 13; 14g:

v ¼ 39 : f4; 15; 19g; f5; 12; 17g; f6; 8; 14g; f7; 9; 16g; f10; 11; 18g:

v ¼ 57 : f4; 24; 28g; f5; 17; 22g; f6; 10; 16g; f7; 18; 25g; f8; 15; 23g; f9; 12; 21g;f13; 14; 27g; f11; 20; 26g:

v ¼ 63 : f4; 27; 31g; f8; 18; 26g; f9; 15; 24g; f7; 13; 20g; f14; 19; 30g; f5; 11; 16g;f6; 17; 23g; f10; 25; 28g; f12; 22; 29g:

v ¼ 81 : f5; 34; 39g; f6; 32; 38g; f9; 26; 35g; f10; 21; 31g; f11; 18; 29g; f17; 23; 40g;f14; 22; 36g; f4; 15; 19g; f7; 30; 37g; f8; 16; 24g; f12; 13; 25g; f20; 28; 33g:

v ¼ 87 : f6; 35; 41g; f7; 33; 40g; f8; 30; 38g; f9; 27; 36g; f10; 24; 34g; f12; 20; 32g;f4; 22; 26g; f15; 28; 43g; f19; 23; 42g; f16; 21; 37g; f5; 13; 18g; f11; 14; 25g;f17; 31; 39g:

v ¼ 105 : f5; 46; 51g; f7; 42; 49g; f8; 39; 47g; f10; 33; 43g; f11; 30; 41g; f12; 28; 40g;f13; 25; 38g; f14; 23; 37g; f15; 19; 34g; f4; 22; 26g; f20; 32; 52g; f17; 31; 48g;f6; 18; 24g; f21; 29; 50g; f9; 27; 36g; f16; 44; 45g:




the spectrum of cyclic bsa(ν, 3, λ; α) with α = 2, 3

Documents