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

The Spectrum of BSA(v, 3, λ; α) withα = 2, 3

Jing Zhang, Yanxun ChangInstitute of Mathematics, Beijing Jiaotong University, Beijing 100044, China,E-mail: [email protected] June 27, 2005; revised December 15, 2005

Published online 24 March 2006 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/jcd.20104

Abstract: In this article, a kind of auxiliary design BSA∗ for constructing BSAs is introducedand studied. Two powerful recursive constructions on BSAs from 3-IGDDs and BSA∗s areexploited. Finally, the necessary and sufficient conditions for the existence of a BSA(v, 3, λ; α)with α = 2, 3 are established. © 2006 Wiley Periodicals, Inc. J Combin Designs 15: 61–76, 2007

Keywords: 3-IGDD; balanced sampling plan avoiding adjacent units (or BSA); automorphismgroup; cyclic BSA; BSA∗(v, {2, 3}, λ; α, t)

1. INTRODUCTION

A balanced sampling plan avoiding adjacent units (or BSA) is designed for a survey planwhen several adjacent units provide similar information (see [2,4]). We introduce the defi-nition of BSA as below.

Let X = {x1, . . . , xv} be a set of v-elements (or units) with a cyclic order x1 ≺ x2 ≺· · · ≺ xv ≺ x1 (or denoted by (x1, x2, . . . , xv)). The pair (xi, xi+s) is said to be s-adjacentunits of X in a cyclic v-tuple (x1, x2, . . . , xv) where i + s is taken module v.

Definition 1.1. A k-sized balanced sampling plan avoiding adjacent units is a pair (X,B),where X is a v-set with a cyclic order (x1, x2, . . . , xv) and B is a collection (so repetitionsare allowed) of k-subsets of X called blocks, such that no pair of s-adjacent units (xi, xi+s)appears in any block where s = 1, 2, . . . , α, while any other pair of units appears in exactlyλ blocks.

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

© 2006 Wiley Periodicals, Inc.

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62 ZHANG AND CHANG

We will use the notation BSA(v, k, λ; α) to denote such a design. Let X = Zv denotethe cyclic additive group of order v with the usual cyclic order (0, 1, . . . , v − 1) and (X,B)a BSA(v, k, λ; α). If Zv is an automorphism group of the BSA(v, k, λ; α), then we call(X,B) cyclic and denote it by CBSA(v, k, λ; α). It is easy to see that a BSA(v, k, λ; 1) isactually a BSEC(v, k, λ) (see [5]). Colbourn and Ling [1] established the necessary andsufficient conditions for the existence of a BSEC(v, k, λ). In [7], the authors establishedthe necessary and sufficient conditions for the existence of a cyclic BSA(v, 3, λ; α) withα = 2, 3. We quote the result as follows.

Theorem 1.2 (Zhang and Chang [7]). A CBSA(v, 3, λ; α) with α = 2, 3 exists if and onlyif v ≥ 3(2α + 1) and

(1) v ≡ 3, 2α + 1 (mod 6) when λ ≡ 1, 5 (mod 6) but v �≡ 3 (mod 6) when λ = 1 andα �≡ 0 (mod 3);

(2) v ≡ 0, 2α + 1 (mod 3) and v �≡ 2 (mod 4) when λ ≡ 2, 10 (mod 12);(3) v ≡ 1 (mod 2) when λ ≡ 3 (mod 6);(4) v ≡ 0, 2α + 1 (mod 3) when λ ≡ 4, 8 (mod 12);(5) v �≡ 2 (mod 4) when λ ≡ 6 (mod 12);(6) v is unrestricted when λ ≡ 0 (mod 12).

First we give the necessary conditions for the existence of a BSA(v, 3, λ; α). From [4],we have v ≥ 3(2α + 1). Through simple calculation, we obtain the necessary conditionsfor the existence of a BSA(v, 3, λ; α) as follows.

Lemma 1.3. If a BSA(v, 3, λ; α) exists, then λv(v − 2α − 1) ≡ 0 (mod 6), λ(v − 2α −1) ≡ 0 (mod 2), and v ≥ 3(2α + 1).

In this article we prove the following

Main result. Previous necessary conditions are also sufficient when α = 2, 3.

First, let us note that if (Zv,B1) is a BSA(v, k, λ1; α) and (Zv,B2) is a BSA(v, k, λ2; α),then (Zv,B) is a BSA(v, k, λ1 + λ2; α) where B is the multiset union of B1 and B2. Thenby Theorem 1.2 and Lemma 1.3, to prove the main result it is necessary and sufficient toestablish the existence of a

(i) BSA(v, 3, 1; 2) with v ≡ 3 (mod 6) and v ≥ 15;(ii) BSA(v, 3, 2; 2) with v ≡ 2, 6 (mod 12) and v ≥ 15;

(iii) BSA(v, 3, 6; 2) with v ≡ 10 (mod 12) and v ≥ 15;(iv) BSA(v, 3, 2; 3) with v ≡ 6, 10 (mod 12) and v ≥ 21;(v) BSA(v, 3, 6; 3) with v ≡ 2 (mod 12) and v ≥ 21.

Let [a, b] denote the set of all integers n such that a ≤ n ≤ b. The following two lemmasare from [6] by utilizing Langford sequences. For more information on Langford sequences,the reader is referred to [3].

Lemma 1.4. Let (m, d) ≡ (0, 1), (1, 1), (0, 0), (3, 0) (mod (4, 2)) such that m ≥ 2d − 1.Then [d, d + 3m − 1] can be partitioned into triples {ai, bi, ci}, 1 ≤ i ≤ m, such that ai +bi = ci.

Lemma 1.5. Let (m, d) ≡ (2, 0), (1, 0), (2, 1), (3, 1) (mod (4, 2)) such that m(m − 2d +1) + 2 ≥ 0. Then [d, d + 3m] \ {d + 3m − 1} can be partitioned into triples {ai, bi, ci},1 ≤ i ≤ m, such that ai + bi = ci.

SPECTRUM OF BSA(ν, 3, λ; α) 63

An incomplete group (or groop) divisible design with block-size 3 (or 3-IGDD) of type(g, h)u and index λ is a quadruple (X, H,G,B) where:

� X is a finite set of cardinality gu;� G is a partition of X into g-subsets (called groups or groops);� H is a subset (called hole) of X such that |H ∩ G| = h for each G ∈ G;� B is a collection of 3-subsets (called blocks) of X such that each pair of distinct

elements x and y of X occurs in exactly λ blocks if x and y are not in the same groupand {x, y} �⊆ H , otherwise it occurs in no block.

Miao and Zhu [8] settled the existence of 3-IGDDs of index one, which can be stated asfollows.

Lemma 1.6 (Miao and Zhu [8]). A 3-IGDD of type (g, h)u and index one exists if andonly if the following conditions hold:

(1) g ≥ 2h and u ≥ 3;(2) g(u − 1) ≡ 0 (mod 2);(3) (g − h)(u − 1) ≡ 0 (mod 2);(4) u(u − 1)(g2 − h2) ≡ 0 (mod 6).

Now we give a kind of auxiliary design BSA∗(v, {2, 3}, λ; α, t) for later use.

Definition 1.7. A BSA∗(v, {2, 3}, λ; α, t) is a pair (X,B), where X = {x1, . . . , xv}with a cyclic order (x1, x2, . . . , xv) and B is a collection (so repetitions are allowed)of k-subsets of X called blocks, where k ∈ {2, 3}, which satisfies the followingconditions:

(1) no pair of s-adjacent units (xi, xi+s) appears in any block where s = 1, 2, . . . , α,while any other pair of units appears in exactly λ blocks;

(2) all blocks with block size 2 can be partitioned into exactly t λ-parallel classes,where λ-parallel class means that every unit of X appears exactly λ times in theclass.

In the case of t = 0, a BSA∗(v, {2, 3}, λ; α, t) is actually a BSA(v, 3, λ; α).

2. CONSTRUCTIONS OF BSA(v, 3, λ; 2)

Lemma 2.1. There exists a BSA∗(v, {2, 3}, 1; 2, t) for the following cases:

(1) v = 24s + 2 where s ≥ 2 and t = 15;(2) v = 24s + 8 where s ≥ 1 and t = 15;(3) v = 24s + 14 where s ≥ 3 and t = 33;(4) v = 24s + 20 where s ≥ 5 and t = 69.

Proof. (1) v = 24s + 2 where s ≥ 2 and t = 15. When s ≥ 4, [3, 12s] \ {3, 5, 4s + 1,

6s − 3,6s + 1,8s − 1,12s − 1} can be partitioned into triples {ai, bi, ci} such thatai + bi =ci as follows.

64 ZHANG AND CHANG

ai bi ci j ∈6s − 4 − j 7 + 2j 6s + 3 + j [0, s − 4]5s − 1 − j 2s + 2 + 2j 7s + 1 + j [0, s − 3]10s − 3 − j 4 + 2j 10s + 1 + j [0, s − 3]9s − 1 − j 2s + 1 + 2j 11s + j [0, s − 2]

The other triples are: {6s − 2, 6s + 2, 12s}, {4s − 1, 7s, 11s − 1}, {4s − 2, 6s, 10s − 2},{2s, 8s, 10s}, {4s, 6s − 1, 10s − 1}.

When s = 2, the triples {ai, bi, ci} are: {4, 20, 24}, {3, 11, 14}, {5, 7, 12}, {6, 16, 22}, {8,

10, 18}, which partition the set [3, 24]\ {9, 13, 15, 17, 19, 21, 23}.When s = 3, the triples {ai, bi, ci} are: {4, 12, 16}, {5, 29, 34}, {6, 26, 32}, {7, 23, 30},

{8, 20, 28}, {9, 18, 27}, {10, 15, 25}, {11, 13, 24}, {14, 22, 36}, which partition the set[3, 36] \ {3, 17, 19, 21, 31, 33, 35}.

So, [3, v/2 − 1] \ S, where S is a set of 7 odd integers indicated as above, can be parti-tioned into triples {ai, bi, ci} such that ai + bi = ci, 1 ≤ i ≤ 4s − 3, when v = 24s + 2 ands ≥ 2.

We can easily check that {l, 12s + 1 + l}, {2l, d + 2l}, {1 + 2l, d + 1 + 2l}, 0 ≤ l ≤v/2 − 1, d ∈ S, are all the 15 1-parallel classes, while {j, ai + j, ci + j}, 1 ≤ i ≤ 4s − 3,j ∈ Zv, are all the blocks with block size 3 of a BSA∗(v, {2, 3}, 1; 2, 15), where the additionis taken in Zv. So there exists a BSA∗(v, {2, 3}, 1; 2, 15) for v = 24s + 2 and s ≥ 2.

The other cases can be proved in a similar way. For abbreviation, we only list the oddintegers of S and the triples {ai, bi, ci} such that ai + bi = ci.

(2) v = 24s + 8 where s ≥ 1 and t = 15. When s ≥ 4, [3, 12s + 3] \ {6s − 1, 6s +1, 6s + 5, 8s + 5, 10s + 3, 12s + 1, 12s + 3} can be partitioned into triples {ai, bi, ci} suchthat ai + bi = ci as follows.

ai bi ci j ∈6s − j 4 + 2j 6s + 4 + j [0, s − 3] \ {1}5s + 2 − j 2s + 1 + 2j 7s + 3 + j [0, s − 1]10s + 1 − j 3 + 2j 10s + 4 + j [0, s − 2]9s + 2 − j 2s + 2 + 2j 11s + 4 + j [0, s − 2] \ {s − 3}

The other triples are: {2s, 6s + 3, 8s + 3}, {4s + 1, 7s + 2, 11s + 3}, {6, 4s − 4, 4s + 2},{4s, 6s + 2, 10s + 2}.

When s = 2, the triples {ai, bi, ci} are: {4, 18, 22}, {6, 8, 14}, {10, 16, 26}, {5, 7, 12}, {9,

11, 20}, {3, 21, 24}, which partition the set [3, 27] \ {13, 15, 17, 19, 23, 25, 27}.When s = 3, the triples {ai, bi, ci} are: {6, 30, 36}, {8, 24, 32}, {10, 18, 28}, {12, 14, 26},

{4, 25, 29}, {16, 21, 37}, {7, 31, 38}, {17, 22, 39}, {13, 20, 33}, {11, 23, 34}, which partitionthe set [3, 39] \ {3, 5, 9, 15, 19, 27, 35}.

So, [3, v/2 − 1] \ S, where S is a set of 7 odd integers, can be partitioned into triples{ai, bi, ci} such that ai + bi = ci, 1 ≤ i ≤ 4s − 2, when v = 24s + 8 and s ≥ 2.

When s = 1, the blocks of a BSA∗(32, {2, 3}, 1; 2, 15) are as follows: {l, 6 + l, 18 +l}, {l, 3 + l, 10 + l}, {j, 16 + j}, {2j, d + 2j}, {1 + 2j, d + 1 + 2j}, {x, x + 4}, {x +4, x + 8}, {y, y + 8}, {y + 8, y + 16}, where the addition is taken in Z32, 0 ≤ j ≤ 15,x ∈ {8a + b : 0 ≤ a ≤ 3, 0 ≤ b ≤ 3}, y ∈ {16a + b : a = 0, 1, 0 ≤ b ≤ 7}, l ∈ Z32, d ∈{5, 9, 11, 13, 15}.

Hence, there exists a BSA∗(v, {2, 3}, 1; 2, 15) for v = 24s + 8 and s ≥ 1.


(3) v = 24s + 14 where s ≥ 3 and t = 33. When s ≥ 7, [3, 12s + 6] \ S, where S ={3, 7, 9, 11, 4s − 7, 4s − 5, 4s − 1, 4s + 1, 4s + 7, 6s − 1, 6s + 5, 6s + 7, 6s + 9, 8s + 1,

8s + 3, 12s + 5}, can be partitioned into triples {ai, bi, ci} such that ai + bi = ci asfollows.

ai bi ci j ∈6s + 1 − j 5 + 2j 6s + 6 + j [4, s − 3] ∪ {0}5s + 2 − j 2s + 3 + 2j 7s + 5 + j [0, s − 3] \ {s − 5, s − 4}10s + 3 − j 4 + 2j 10s + 7 + j [0, s − 3]9s + 3 − j 2s + 2 + 2j 11s + 5 + j [0, s − 2]

The other triples are: {2s, 7s + 4, 9s + 4}, {4s + 2, 5s + 3, 9s + 5}, {4s, 6s + 4, 10s +4}, {4s + 6, 6s, 10s + 6}, {4s + 4, 8s, 12s + 4}, {6s − 2, 6s + 8, 12s + 6}, {2s + 1, 6s +3, 8s + 4}, {4s + 3, 6s + 2, 10s + 5}.

When s = 3, the triples {ai, bi, ci} are: {4, 5, 9}, {3, 17, 20}, {14, 24, 38}, {16, 26, 42},{6, 34, 40}, {8, 28, 36}, {10, 22, 32}, {12, 18, 30}, which partition the set [3, 42] \ S whereS = {16 odd integers in [7, 41] \ {9, 17}}.

When s = 4, the triples {ai, bi, ci} are: {4, 50, 54}, {6, 46, 52}, {8, 40, 48}, {10, 34, 44},{12, 30, 42}, {14, 24, 38}, {16, 20, 36}, {3, 29, 32}, {5, 13, 18}, {7, 21, 28}, {9, 17, 26},{22, 31, 53}, which partition the set [3, 54] \ S where S = {16 odd integers in [11, 51] \{13, 17, 21, 29, 31}}.

When s = 5, the triples {ai, bi, ci} are: {4, 62, 66}, {6, 58, 64}, {8, 52, 60}, {10, 46, 56},{12, 42, 54}, {14, 36, 50}, {16, 32, 48}, {18, 26, 44}, {3, 35, 38}, {5, 29, 34}, {7, 23, 30},{9, 19, 28}, {11, 13, 24}, {15, 25, 40}, {17, 20, 37}, {22, 43, 65}, which partition the set[3, 66] \ S where S = {16 odd integers in [21, 63] \ {23, 25, 29, 35, 37, 43}}.

When s = 6, the triples {ai, bi, ci} are: {3, 49, 52}, {5, 50, 55}, {9, 32, 41}, {4, 56, 60},{6, 36, 42}, {8, 68, 76}, {10, 65, 75}, {12, 61, 73}, {14, 58, 72}, {16, 54, 70}, {18, 51, 69},{22, 44, 66}, {24, 40, 64}, {26, 37, 63}, {20, 47, 67}, {28, 34, 62}, {30, 48, 78}, {31, 46, 77},{33, 38, 71}, {35, 39, 74}, which partition the set [3, 78] \ S where S = {16 odd integers in[11, 29] ∪ {7, 43, 45, 53, 57, 59}}.

So, [3, v/2 − 1] deleting 16 odd integers can be partitioned into triples {ai, bi, ci} suchthat ai + bi = ci, 1 ≤ i ≤ 4s − 4, when v = 24s + 14 and s ≥ 3. Hence, there exists aBSA∗(v, {2, 3}, 1; 2, 33) for v = 24s + 14 and s ≥ 3.

(4) v = 24s + 20 where s ≥ 5 and t = 69. When s ≥ 8, [3, 12s + 9] \ S, whereS = {3, 5, 7, 11, 13, 4s − 13, 4s − 9, 4s − 7, 4s − 5, 4s − 3, 4s − 1, 4s + 1, 4s + 3, 4s +5, 4s + 7, 6s − 3, 6s + 1, 6s + 5, 6s + 7, 6s + 9, 8s − 3, 8s − 1, 8s + 1, 8s + 3, 8s +5, 10s − 1, 10s + 1, 10s + 5, 10s + 9, 10s + 11, 12s + 3, 12s + 5, 12s + 7, 12s + 9}, canbe partitioned into triples {ai, bi, ci} such that ai + bi = ci as follows.

ai bi ci j ∈6s + 1 − j 5 + 2j 6s + 6 + j [2, s − 3] \ {3, 4}5s + 2 − j 2s + 3 + 2j 7s + 5 + j [0, s − 7] \ {s − 8}10s + 3 − j 4 + 2j 10s + 7 + j [0, s − 3] \ {2, 4}9s + 3 − j 2s + 2 + 2j 11s + 5 + j [0, s − 3]

The other triples are: {2s, 7s + 4, 9s + 4}, {4s + 2, 5s + 3, 9s + 5}, {4s, 6s + 4, 10s +4}, {4s + 4, 8s, 12s + 4}, {4s + 6, 8s + 2, 12s + 8}, {6s, 6s + 6, 12s + 6}, {8, 6s + 2, 6s +10}, {4s + 8, 6s − 2, 10s + 6}, {12, 4s − 2, 4s + 10}, {2s + 1, 6s + 3, 8s + 4}.

66 ZHANG AND CHANG

When s = 5, the triples {ai, bi, ci} are: {4, 58, 62}, {6, 26, 32}, {8, 52, 60}, {10, 38, 48},{12, 44, 56}, {14, 54, 68}, {16, 34, 50}, {18, 28, 46}, {20, 22, 42}, {24, 40, 64}, {30, 36, 66},which partition the set [3, 69] \ S where S = {34 odd integers in [3, 69]}.

When s = 6, the triples {ai, bi, ci} are: {4, 76, 80}, {6, 72, 78}, {8, 66, 74}, {10, 38, 48},{12, 56, 68}, {14, 50, 64}, {16, 46, 62}, {18, 40, 58}, {20, 34, 54}, {22, 30, 52}, {26, 44, 70},{28, 32, 60}, {3, 21, 24}, {5, 31, 36}, {7, 35, 42}, which partition the set [3, 81] \ S whereS = {34 odd integers in [9, 81] \ {21, 31, 35}}.

When s = 7, the triples {ai, bi, ci} are: {4, 28, 32}, {6, 84, 90}, {8, 78, 86}, {10, 72, 82},{12, 68, 80}, {14, 62, 76}, {16, 58, 74}, {18, 52, 70}, {20, 46, 66}, {22, 42, 64}, {24, 36, 60},{26, 30, 56}, {3, 34, 37}, {5, 45, 50}, {7, 41, 48}, {9, 31, 40}, {11, 33, 44}, {13, 75, 88},{38, 54, 92}, which partition the set [3, 93] \ S where S = {34 odd integers in [15, 93] \{31, 33, 37, 41, 45, 75}}.

So, [3, v/2 − 1] deleting 34 odd integers can be partitioned into triples {ai, bi, ci} suchthat ai + bi = ci, 1 ≤ i ≤ 4s − 9, when v = 24s + 20 and s ≥ 5. Hence, there exists aBSA∗(v, {2, 3}, 1; 2, 69) for v = 24s + 20 and s ≥ 5. �

Lemma 2.2. There exists a BSA∗(v, {2, 3}, 2; 2, t) for the following cases:

(1) v = 12s + 1 and t = 3x + 2, where s ≥ 3 if x = 3, 7, or s ≥ 4 if x = 11;(2) v = 12s + 5 and t = 3x, where s ≥ 2 if x = 1, 5.

Proof. (1) v = 12s + 1 and t = 3x + 2, where s ≥ 3 if x = 3, 7, or s ≥ 4 if x = 11.Using Lemma 1.5 with (m, d) = (4s − 2, 4), [4, 12s − 2] \ {12s − 3} can be partitioned

into triples {ai, bi, ci} such that ai + bi = ci, 1 ≤ i ≤ 4s − 2 where s ≥ 3. For x = 3, 7 or11, letB1 = {{l, 3 + l}, {l, 12s − 3 + l}, {l, ai + l}, {l, bi + l}, {l, ci + l} : l ∈ Zv, 1 ≤ i ≤ x},B2 = {{l, ai + l, ci + l} : l ∈ Zv, x + 1 ≤ i ≤ 4s − 2},

where the addition is taken in Zv. The inequality x + 1 ≤ 4s − 2 implies that s ≥ 4 ifx = 11. It is easily checked that (Zv,B1 ∪ B2) is a BSA∗(v, {2, 3}, 2; 2, t).

(2) v = 12s + 5 and t = 3x, where s ≥ 2 if x = 1, 5.Using Lemma 1.4 with (m, d) = (4s, 3), [3, 12s + 2] can be partitioned into triples

{ai, bi, ci} such that ai + bi = ci, 1 ≤ i ≤ 4s where s ≥ 2. For x = 1, 5, letB1 = {{l, ai + l}, {l, bi + l}, {l, ci + l} : 1 ≤ i ≤ x, l ∈ Zv},B2 = {{l, ai + l, ci + l} : x + 1 ≤ i ≤ 4s, l ∈ Zv}.

It is easily checked that (Zv,B1 ∪ B2) is a BSA∗(v, {2, 3}, 2; 2, t). �

Lemma 2.3. There exists a BSA∗(v, {2, 3}, 6; 2, t) for v = 12s + 3 and t = 3x + 1, wheres ≥ 2 if x = 0, 4, or s ≥ 3 if x = 8.

Proof. For s ≥ 2, using Lemma 1.4 with (m, d) = (4s − 1, 4), [4, 12s] can be partitionedinto triples {ai, bi, ci}, 1 ≤ i ≤ 4s − 1. For x = 0, 4 or 8, letB1 = {{l, 3 + l}, {l, ai + l}, {l, bi + l}, {l, ci + l} : l ∈ Zv, 1 ≤ i ≤ x},B2 = {{l, ai + l, ci + l} : x + 1 ≤ i ≤ 4s − 1, l ∈ Zv}.

The inequality x + 1 ≤ 4s − 1 implies that s ≥ 3 if x = 8. It is easily checked that(Zv, 3(B1 ∪ B2)) is a BSA∗(v, {2, 3}, 6; 2, t). �

Lemma 2.4. Suppose that there exists a 3-IGDD of type (g, 5)u with index one. If thereexist a BSA∗(g, {2, 3}, λ; 2, t), a BSA(g + t, 3, λ; 2) and a BSA(5u, 3, λ; 2), then thereexists a BSA(gu + t, 3, λ; 2).


Proof. Let X = {1, 2, . . . , u} × Zg, H = {1, 2, . . . , u} × {0, 1, g − 3, g − 2, g − 1}, andG = {{i} × Zg : 1 ≤ i ≤ u}. By assumption let (X, H,G,A) be the given 3-IGDD oftype (g, 5)u with index one. We will construct a BSA(gu + t, 3, λ; 2) on the set X ∪{∞1, . . . ,∞t} with a cyclic order (10, 11, . . . , 1g−1, 20, 21, . . . , 2g−1, . . . , u0, u1, ∞1,

∞2, . . . ,∞t , u2, u3, . . . , ug−1). For each 1 ≤ i ≤ u − 1, by assumption there exists aBSA∗(g, {2, 3}, λ; 2, t) ({i} × Zg,Bi) on {i} × Zg with a cyclic order (i0, i1, . . . , ig−1), inwhich there are exactly t λ-parallel classes, say P1, . . . ,Pt , and the blocks of size 3 in Bi isdenoted byBi. LetB(i) = Bi ∪ {{∞j} ∪ P : P ∈ Pj, 1 ≤ j ≤ t}. For theu-th group of the 3-IGDD, we construct a BSA(g + t, 3, λ; 2) (({u} × Zg) ∪ {∞1, . . . ,∞t}, C) on ({u} × Zg) ∪{∞1, . . . ,∞t} with a cyclic order (∞1, . . . , ∞t , u2, u3, . . . , ug−1, u0, u1). For the hole H ,we have a BSA(5u, 3, λ; 2) (H, C′) on H with a cyclic order (10, 11, 1g−3, 1g−2, 1g−1, 20,

21, 2g−3, 2g−2, 2g−1, . . . , u0, u1, ug−3, ug−2, ug−1). Let D = (λA) ∪ (∪1≤i≤u−1B(i)) ∪C ∪ C′. It is easily but tediously checked that (X ∪ {∞1, . . . ,∞t}, D) is the desiredBSA(gu + t, 3, λ; 2).

Figure 1 explains the proof. �Lemma 2.5. There exists a BSA(v, 3, 1; 2) for v ≡ 3 (mod 18) and v ≥ 21.

Proof. Let (g, t) = (24s + 2, 15) where s ≥ 2, or (g, t) = (24s + 8, 15) where s ≥ 1, or(g, t) = (24s + 14, 33) where s ≥ 3, or (g, t) = (24s + 20, 69) where s ≥ 5. Then g + t ≡5 (mod 6) and g + t ≥ 15. In each case, there exist a BSA∗(g, {2, 3}, 1; 2, t) by Lemma 2.1, aBSA(g + t, 3, 1; 2) by Theorem 1.2 and a BSA(15, 3, 1; 2) by Appendix B. By Lemma 1.6,there exists a 3-IGDD of type (g, 5)3 with g = 24s + 2, 24s + 8, 24s + 14, 24s + 20. Ap-plying Lemma 2.4 with u = 3 we get a BSA(3g + t, 3, 1; 2). This produces a BSA(v, 3, 1; 2)for any integer v ≡ 3 (mod 18), v ≥ 21, and v �∈ {21, 39, 57, 75, 93, 129, 147, 201, 219,

273, 345, 417}.For v ∈ {39, 57, 75, 93, 129, 219}, there exists a BSA(v, 3, 1; 2) by Appendix A. For

v = 21, a BSA(21, 3, 1; 2) can be obtained by Appendix B.For v ∈ {147, 201, 273, 345, 417}, v can be written as v = gu + t by taking (g, u, t) =

(21, 7, 0), (28, 7, 5), (39, 7, 0), (23, 15, 0), (58, 7, 11), respectively. By Lemma 2.4 withthe parameters of g, u, and t, there exists a BSA(v, 3, 1; 2). The required 3-IGDDsof types (g, 5)u come from Lemma 1.6. The required BSA∗(g, {2, 3}, 1; 2, t)s withpairs (g, t) as above come from Appendixes A, B, C, and Theorem 1.2 (note that aBSA∗(g, {2, 3}, 1; 2, t) for (g, t) = (21, 0), (23, 0), (39, 0) is actual a BSA(g, 3, 1; 2)). Therequired BSA(5u, 3, 1; 2) with u = 7, 15 come from Theorem 1.2 and Appendix A. Therequired BSA(g + t, 3, 1; 2) with pairs (g, t) as above but (g, t) �= (58, 11) come fromTheorem 1.2, Appendixes A and B. For pair (g, t) = (58, 11), the required BSA(g + t,

3, 1; 2) can be constructed as follows: By Lemma 1.6 there exists a 3-IGDD of type (23, 5)3;there exists a BSA(23, 3, 1; 2) by Theorem 1.2 and a BSA(15, 3, 1; 2) by Appendix B; byLemma 2.4 there exists a BSA(69, 3, 1; 2). �

FIGURE 1.

68 ZHANG AND CHANG

Lemma 2.6. There exists a BSA(v, 3, 1; 2) for v ≡ 3 (mod 6) and v ≥ 15.

Proof. For s ≥ 2, by Lemma 1.6, there exists a 3-IGDD of type (6s + 5, 5)3 with index one.There exists a BSA(6s + 5, 3, 1; 2) by Theorem 1.2 and a BSA(15, 3, 1; 2) by Appendix B.By Lemma 2.4 with (g, u, t) = (6s + 5, 3, 0) there exists a BSA(18s + 15, 3, 1; 2) for s ≥ 2.For s = 0, 1, there exists a BSA(18s + 15, 3, 1; 2) by Appendix A. Hence, a BSA(18s +15, 3, 1; 2) exists for any integer s ≥ 0.

By Lemma 1.6, there exists a 3-IGDD of type (6s + 3, 5)3 where s ≥ 2. There exists aBSA(15, 3, 1; 2) by Appendix B. Suppose that there exists a BSA(6s + 3, 3, 1; 2) wheres ≥ 2, by Lemma 2.4 with (g, u, t) = (6s + 3, 3, 0) there exists a BSA(18s + 9, 3, 1; 2)where s ≥ 2. Combining with Lemma 2.5, we have the iterative result: if there exists aBSA(6s + 3, 3, 1; 2) where s ≥ 2, then there exists a BSA(v, 3, 1; 2) for v = 18s + 9, 18s +15, 18s + 21, and v ≥ 15. While there exists a BSA(v, 3, 1; 2) for v ∈ {15, 21, 27, 33, 39}by Appendixes A and B, so does a BSA(v, 3, 1; 2) for any integer v ≡ 3 (mod 6) andv ≥ 15. �Lemma 2.7. There exists a BSA(v, 3, 2; 2) for v ≡ 2, 6 (mod 12) and v ≥ 15.

Proof. We divide the problem into two cases:

Case 1. v ≡ 2 (mod 12) and v ≥ 15.

By Lemma 1.6, there exists a 3-IGDD of type (12s + 1, 5)3 where s ≥ 1. By Lemma 2.2,there exists a BSA∗(12s + 1, {2, 3}, 2; 2, 3x + 2) where s ≥ 3 if x = 3, 7, or s ≥ 4 ifx = 11. By Theorem 1.2, there exist a BSA(12s + 3x + 3, 3, 2; 2) where x = 3, 7, 11and a BSA(15, 3, 2; 2). By Lemma 2.4 with (g, u, t) = (12s + 1, 3, 3x + 2), there existsa BSA(36s + 3x + 5, 3, 2; 2) where s ≥ 3 if x = 3, 7 or s ≥ 4 if x = 11. This produces aBSA(v, 3, 2; 2) for v ≡ 2 (mod 12), v ≥ 15 and v �∈ {26, 38, 50, 62, 74, 86, 98, 110, 146}.

For v ∈ {26, 38}, there exists a BSA(v, 3, 2; 2) by Appendix A.For v ∈ {50, 62, 74, 86, 98, 110, 146}, v can be written as v = 3g + t by taking (g, t) =

(16, 2), (19, 5), (22, 8), (28, 2), (28, 14), (34, 8), (43, 17), respectively. By Lemma 2.4 withg, t, and u = 3, there exists a BSA(v, 3, 2; 2). The required 3-IGDDs of types (g, 5)3 withg = 16, 19, 22, 28, 34, 43 come from Lemma 1.6. The required BSA∗(g, {2, 3}, 2; 2, t)swith the pairs (g, t) come from Appendix C. The required BSA(g + t, 3, 2; 2)s comefrom Theorem 1.2 and Appendix B. The required BSA(5u, 3, 2; 2) with u = 3 come fromTheorem 1.2.

Hence, there exists a BSA(v, 3, 2; 2) for v ≡ 2 (mod 12) and v ≥ 15.

Case 2. v ≡ 6 (mod 12) and v ≥ 15.

By Lemma 1.6, there exists a 3-IGDD of type (12s + 5, 5)3 where s ≥ 2. By Lemma 2.2,there exists a BSA∗(12s + 5, {2, 3}, 2; 2, 3x) where s ≥ 2 if x = 1, 5. By Theorem 1.2, thereexist a BSA(12s + 5 + 3x, 3, 2; 2) and a BSA(15, 3, 2; 2). By Lemma 2.4 with (g, u, t) =(12s + 5, 3, 3x), there exists a BSA(36s + 3x + 15, 3, 2; 2) where s ≥ 2 and x = 1, 5. Thisproduces a BSA(v, 3, 2; 2) for v = 36s + 18, v = 36s + 30, and s ≥ 2. By Lemma 1.6, thereexists a 3-IGDD of type (12s + 2, 5)3 where s ≥ 2. Since there exist a BSA(12s + 2, 3, 2; 2)and a BSA(15, 3, 2; 2), by Lemma 2.4 with (g, u, t) = (12s + 2, 3, 0), so does a BSA(36s +6, 3, 2; 2) for any integer s ≥ 2. Hence, there exists a BSA(v, 3, 2; 2) for v ≡ 6 (mod 12),v ≥ 15, and v �∈ {18, 30, 42, 54, 66}.

For v = 18, 30, 42, there exists a BSA(v, 3, 2; 2) by Appendix B.For v = 54, 66, v can be written as v = 3g + t by taking (g, t) = (18, 0) and (20, 6),

respectively. By Lemma 2.4 with g, t and u = 3, there exists a BSA(v, 3, 2; 2). The


required 3-IGDDs of type (g, 5)3 with g = 18, 20 come from Lemma 1.6 . The requiredBSA∗(g, {2, 3}, 2; 2, t)s with (g, t) = (18, 0), (20, 6) come from Appendixes B and C. Therequired BSA(g + t, 3, 2; 2)s with (g, t) = (18, 0), (20, 6) come from Appendixes A and B.The required BSA(5u, 3, 2; 2) with u = 3 comes from Theorem 1.2.

Hence, there exists a BSA(v, 3, 2; 2) for v ≡ 6 (mod 12) and v ≥ 15. �Lemma 2.8. There exists a BSA(v, 3, 6; 2) for v ≡ 10 (mod 12) and v ≥ 15.

Proof. By Lemma 1.6, there exists a 3-IGDD of type (12s + 3, 5)3 for s ≥ 2. ByLemma 2.3, there exists a BSA∗(12s + 3, {2, 3}, 6; 2, 3x + 1) where s ≥ 2 if x = 0, 4,or s ≥ 3 if x = 8. By Theorem 1.2, there exist a BSA(12s + 3x + 4, 3, 6; 2) and aBSA(15, 3, 6; 2). By Lemma 2.4 with (g, u, t) = (12s + 3, 3, 3x + 1) there exists aBSA(36s + 3x + 10, 3, 6; 2) where s ≥ 2 if x = 0, 4, or s ≥ 3 if x = 8. This produces aBSA(v, 3, 6; 2) for v ≡ 10 (mod 12), v ≥ 15, and v �∈ {22, 34, 46, 58, 70, 106}.

For v = 22, 34, there exists a BSA(v, 3, 6; 2) by Appendix A.For v = 46, 58, 70, 106, v can be written as v = 3g + t by taking (g, t) = (15, 1),

(19, 1), (23, 1), (35, 1), respectively. By Lemma 2.4 with g, t and u = 3, there ex-ists a BSA(v, 3, 6; 2). The required 3-IGDDs come from Lemma 1.6. The requiredBSA∗(g, {2, 3}, 6; 2, t)s with the pairs (g, t) come from Appendix C. The required BSA(g +t, 3, 6; 2)s with (g, t) = (15, 1), (19, 1), (23, 1) and (35, 1) come from Theorem 1.2. Therequired BSA(5u, 3, 6; 2) with u = 3 comes from Theorem 1.2. �

3. CONSTRUCTIONS OF BSA(v, 3, λ; 3)

Lemma 3.1. There exists a BSA∗(v, 3, 2; 3, 3x) for the following cases:

(1) v = 12s + 9 where s ≥ 1 if x = 1, or s ≥ 2 if x = 5.(2) v = 12s + 1 where s ≥ 2 if x = 5.

Proof. (1) v = 12s + 9 where s ≥ 1 if x = 1, or s ≥ 2 if x = 5.By Theorem 1.2, there exists a cyclic BSA(12s + 9, 3, 2; 3) where s ≥ 1, which con-

tains 4s + 1 base blocks (only one of them possesses short orbit). Choose x base blocks{0, ai, bi}, x ≤ 4s, and denote by A the blocks generated by other base blocks ofthe cyclic BSA(12s + 9, 3, 2; 3). Let B = A ∪ {{l, ai + l}, {l, bi + l}, {l, ai − bi + l} : l ∈Zv, 1 ≤ i ≤ x}. It is easily checked that (Zv,B) is a BSA∗(12s + 9, {2, 3}, 2; 3, 3x) wheres ≥ 1 if x = 1 or s ≥ 2 if x = 5.

(2) v = 12s + 1 where s ≥ 2 if x = 5.By Theorem 1.2, there exists a cyclic BSA(12s + 1, 3, 2; 3) where s ≥ 2, which con-

tains 4s − 2 base blocks possessing full orbits. Choose 5 base blocks {0, ai, bi}, 5 ≤4s − 2, and denote by A the blocks generated by other base blocks of the cyclicBSA(12s + 1, 3, 2; 3). Let B = A ∪ {{l, ai + l}, {l, bi + l}, {l, ai − bi + l} : l ∈ Zv, 1 ≤i ≤ 5}. It is easily checked that (Zv,B) is a BSA∗(12s + 1, 3, 2; 3, 3x) where s ≥ 2 ifx = 5. �Lemma 3.2. There exists a BSA∗(12s + 2, {2, 3}, 2; 3, 3x + 1) for s ≥ 3 if x = 1, 5, 9.

Proof. Using Lemma 1.5 with (m, d) = (4s − 2, 4), [4, 12 − 2] \ {12s − 3} can be par-titioned into triples {ai, bi, ci} such that ai + bi = ci, 1 ≤ i ≤ 4s − 2 where s ≥ 3. Forx = 1, 5, 9, let

70 ZHANG AND CHANG

B1 = {{l, 12s − 3 + l}, {l, ai + l}, {l, bi + l}, {l, ci + l} : 1 ≤ i ≤ x, l ∈ Zv},B2 = {{l, ai + l, ci + l} : x + 1 ≤ i ≤ 4s − 2, l ∈ Zv}.

It is easily checked that (Zv,B1 ∪ B2) is a BSA∗(12s + 2, {2, 3}, 2; 3, 3x + 1) where x =1, 5, 9 if s ≥ 3. �Lemma 3.3. There exists a BSA∗(12s + 1, {2, 3}, 6; 3, 3x + 2) for s ≥ 2 if x = 3, s ≥ 3if x = 7, or s ≥ 4 if x = 11.

Proof. By Theorem 1.2 there exists a cyclic BSA(12s + 1, 3, 6; 3) with s ≥ 2, which con-tains 12s − 6 base blocks, say B1, . . ., B12s−6, possessing full orbits where Bi = {0, ai, bi}for i = 1, 2, . . . , 12s − 6. For x = 3, 7, 11, letB1 = {{l, ai + l}, {l, bi + l}, {l, ai − bi + l} : 1 ≤ i ≤ 3x + 2, l ∈ Zv},B2 = {{l, ai + l, bi + l} : 3x + 3 ≤ i ≤ 12s − 6, l ∈ Zv}.

The inequality 3x + 3 ≤ 12s − 6 implies that s ≥ 2 if x = 3, s ≥ 3 if x = 7 or s ≥ 4 ifx = 11. Then it is easily checked that (Zv,B1 ∪ B2) is a BSA∗(12s + 1, 3, 6; 3, 3x + 2)where s ≥ 2 if x = 3, s ≥ 3 if x = 7, or s ≥ 4 if x = 11. �Lemma 3.4. Suppose that there exists a 3-IGDD of type (g, 7)u with index one. If thereexist a BSA∗(g, {2, 3}, λ; 3, t), a BSA(g + t, 3, λ; 3), and a BSA(7u, 3, λ; 3), then thereexists a BSA(gu + t, 3, λ; 3).

Proof. Let X = {1, 2, . . . , u} × Zg, H = {1, 2, . . . , u} × {0, 1, 2, g − 4, g − 3, g − 2,

g − 1}, and G = {{i} × Zg : 1 ≤ i ≤ u}. By assumption let (X, H,G,A) be thegiven 3-IGDD of type (g, 7)u with index one. We will construct a BSA(gu +t, 3, λ; 3) on the set X ∪ {∞1, . . . , ∞t} with a cyclic order (10, 11, . . . , 1g−1,

20, 21, . . . , 2g−1, . . . , u0, u1, u2, ∞1, ∞2, . . . ,∞t , u3, . . . , ug−1). For each 1 ≤ i ≤ u −1, by assumption there exists a BSA∗(g, {2, 3}, λ; 3, t) ({i} × Zg,Bi) on {i} × Zg

with a cyclic order (i0, i1, . . . , ig−1), in which there are exactly t λ-parallel classes,say P1, . . . ,Pt , and the blocks of size 3 in Bi is denoted by Bi. Let B(i) = Bi ∪{{∞j} ∪ P : P ∈ Pj, 1 ≤ j ≤ t}. For the u-th group of the 3-IGDD, we constructa BSA(g + t, 3, λ; 3) (({u} × Zg) ∪ {∞1, . . . ,∞t}, C) on ({u} × Zg) ∪ {∞1, . . . ,∞t}with a cyclic order (∞1, . . . ,∞t , u3, u4, . . . , ug−1, u0, u1, u2). For the hole H ,we have a BSA(7u, 3, λ; 3) (H, C′) on H with a cyclic order (10, 11, 12, 1g−4,

1g−3, 1g−2, 1g−1, 20, 21, 22, 2g−4, 2g−3, 2g−2, 2g−1, . . . , u0, u1, u2, ug−4, ug−3, ug−2,

ug−1). Let D = (λA) ∪ (∪1≤i≤u−1B(i)) ∪ C ∪ C′. It is easily but tediously checked that(X ∪ {∞1, . . . ,∞t},D) is the desired BSA(gu + t, 3, λ; 3).

Figure 2 explains the proof. �

Lemma 3.5. There exists a BSA(v, 3, 2; 3) for v ≡ 6, 10 (mod 12) and v ≥ 21.

Proof. We divide the problem into two cases:

FIGURE 2.


Case 1. v ≡ 6 (mod 12) and v ≥ 21.

By Lemma 1.6, there exists a 3-IGDD of type (12s + 9, 7)3 where s ≥ 1 with index one.By Theorem 1.2, there exist a BSA(12s + 9 + 3m, 3, 2; 3) and a BSA(21, 3, 2; 3) where s ≥1 if m = 1 or s ≥ 0 if m = 5. By Lemma 3.1, there exists a BSA∗(12s + 9, {2, 3}, 2; 3, 3m)where s ≥ 1 if m = 1 or s ≥ 2 if m = 5. By Lemma 3.4 with (g, u, t) = (12s + 9, 3, 3m)there exists a BSA(36s + 27 + 3m, 3, 2; 3) where s ≥ 1 if m = 1 or s ≥ 2 if m = 5. Thisproduces a BSA(v, 3, 2; 3) for v = 36s + 30 where s ≥ 1 and v = 36s + 42 where s ≥ 2.By Lemma 1.6 there exists a 3-IGDD of type (12s + 1, 7)3 where s ≥ 2 with index one.By Theorem 1.2 there exist a BSA(12s + 16, 3, 2; 3) and a BSA(21, 3, 2; 3). By Lemma3.1, there exists a BSA∗(12s + 1, 3, 2; 3, 15) where s ≥ 2. By Lemma 3.4 with (g, u, t) =(12s + 1, 3, 15) there exists a BSA(36s + 18, 3, 2; 3) where s ≥ 2.

Hence, there exists aBSA(v, 3, 2; 3) forv ≡ 6 (mod 12),v ≥ 21, andv �∈ {30, 42, 54, 78}.For v = 30, 42, a BSA(v, 3, 2; 3) comes from Appendix B. For v = 54, 78, v can be writtenas v = 3g + t by taking (g, t) = (16, 6) and (25, 3), respectively. By Lemma 3.4 with g, t andu = 3 there exists a BSA(v, 3, 2; 3). The required 3-IGDDs of types (g, 7)3 with g = 16, 25come from Lemma 1.6. The required BSA∗(g, {2, 3}, 2; 3, t)s with (g, t) = (16, 6), (25, 3)come from Appendix C. The required BSA(g + t, 3, 2; 3)s with (g, t) = (16, 6), (25, 3)come from Theorem 1.2 and Appendix A. The required BSA(7u, 3, 2; 3) with u = 3 comesfrom Theorem 1.2. So, there exists a BSA(v, 3, 2; 3) for v ≡ 6 (mod 12) and v ≥ 21.

Case 2. v ≡ 10 (mod 12) and v ≥ 21.

By Lemma 1.6, there exists a 3-IGDD of type (12s + 2, 7)3 where s ≥ 3 with index one.By Lemma 3.2 there exists a BSA∗(12s + 2, {2, 3}, 2; 3, 3x + 1) where s ≥ 3 if x = 1, 5, 9.There exists a BSA(12s + 3 + 3x, 3, 2; 3) for x = 1, 5, 9 by Case 1. By Theorem 1.2 thereexists a BSA(21, 3, 2; 3). By Lemma 3.4 with (g, u, t) = (12s + 2, 3, 3x + 1) there existsa BSA(36s + 7 + 3x, 3, 2; 3) where s ≥ 3 if x = 1, 5 or 9. This produces a BSA(v, 3, 2; 3)for v ≡ 10 (mod 12), v ≥ 21, and v �∈ {22, 34, 46, 58, 70, 82, 94, 106}.

For v = 22, 34, 46, a BSA(v, 3, 2; 3) comes from Appendix A.For v ∈ {58, 70, 82, 94, 106}, v can be written as v = 3g + t by taking (g, t) = (17, 7),

(23, 1), (26, 4), (29, 7), (35, 1), respectively. By Lemma 3.4 with g, t and u = 3 thereexists a BSA(v, 3, 2; 3). The required 3-IGDDs of types (g, 7)3 with g = 17, 23, 26, 29, 35come from Lemma 1.6. The required BSA∗(g, {2, 3}, 2; 3, t)s with pairs (g, t) as abovecome from Appendix C. The required BSA(g + t, 3, 2; 3)s with pairs (g, t) as above comefrom Theorem 1.2 and Appendix B. The required BSA(7u, 3, 2; 3) with u = 3 comes fromTheorem 1.2. So, there exists a BSA(v, 3, 2; 3) for v ≡ 10 (mod 12) and v ≥ 21. �Lemma 3.6. There exists a BSA(v, 3, 6; 3) for v ≡ 2 (mod 12) and v ≥ 21.

Proof. By Lemma 1.6, there exists a 3-IGDD of type (12s + 1, 7)3 where s ≥ 2 with indexone. By Lemma 3.3, there exists a BSA∗(12s + 1, {2, 3}, 6; 3, 3x + 2) for s ≥ 2 if x = 3 ors ≥ 3 if x = 7 or s ≥ 4 if x = 11. By Theorem 1.2, there exist a BSA(12s + 3x + 3, 3, 6; 3)and a BSA(21, 3, 6; 3). By Lemma 3.4 with (g, u, t) = (12s + 1, 3, 3x + 2), there exists aBSA(36s + 5 + 3x, 3, 6; 3) for s ≥ 2 if x = 3 or s ≥ 3 if x = 7 or s ≥ 4 if x = 11. Thisproduces a BSA(v, 3, 6; 3) for v ≡ 2 (mod 12), v ≥ 21 and v �∈ {26, 38, 50, 62, 74, 98,

110, 146}.For v = 26, 38, 50, there exists a BSA(v, 3, 6; 3) by Appendix A.For v ∈ {62, 74, 98, 110, 146}, v can be written as v = 3g + t by taking (g, t) = (20, 2),

(24, 2), (32, 2), (36, 2), (48, 2), respectively. Lemma 3.4 with g, t and u = 3 givesa BSA(v, 3, 2; 3). The required 3-IGDDs of types (g, 7)3 with g = 20, 24, 32, 36, 48

72 ZHANG AND CHANG

come from Lemma 1.6. The required BSA∗(g, {2, 3}, 6; 3, t)s with pairs (g, t) come fromAppendix C. The required BSA(g + t, 3, 6; 3) with pairs (g, t) come from Theorem 1.2 andAppendix A. So, there exists a BSA(v, 3, 6; 3) for v ≡ 2 (mod 12) and v ≥ 21. �

4. CONCLUSION

Combining Theorem 1.2 and Lemmas 1.3, 2.6, 2.7, 2.8, 3.5, 3.6, we obtain the main resultof the article.

Theorem 4.1. For α = 2, 3, the necessary and sufficient conditions for the existence ofa BSA(v, 3, λ; α) are λv(v − 2α − 1) ≡ 0 (mod 6), λ(v − 2α − 1) ≡ 0 (mod 2), and v ≥3(2α + 1).

APPENDIX A

Each BSA(v, 3, λ; α) is constructed on Zv with the usual cyclic order (0, 1, . . . , v − 1). Thebase blocks are listed below. All of the base blocks are partitioned into two parts. Each baseblock in the first part is developed in tZv where tZv means the subgroup of order v/t in Zv.Each base block in the second part is developed in Zv.

74 ZHANG AND CHANG

APPENDIX B

Each BSA(mn, 3, λ; α) (X,B) is constructed on X with the cyclic order (00, 01, . . . , 0n−1,

10, 11, . . . , 1n−1, . . . , (m − 1)0, (m − 1)1, . . . , (m − 1)n−1) where X = Zm × In and Bis obtained by developing the base blocks in (Zm, −). We mark the base blocks with shortorbits by “∗”.

APPENDIX C

The base blocks of each BSA∗(g, {2, 3}, λ; α, t) (Zg,B) are listed below. The base blocksof size 2 are developed and partitioned into t λ-parallel classes. The base blocks of size 3are developed in Zg. The base blocks of size 2 marked with “∗” are developed (+l mod g)where 0 ≤ l ≤ g/2 − 1, and the other base blocks of size 2 are developed in mZg.

76 ZHANG AND CHANG

ACKNOWLEDGMENTS

The authors thank the referees for their comments and suggestions that much improved thequality of this article.

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

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

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

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

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