The original classiffcation of PBIBDs defined group divisible designs GDD(v = v₁ + v₂ + ⋯ + v_g, g, k, λ₁, λ₂) with λ₁ ≠ 0. In this paper, we prove that the necessary conditions are suffcient for the existence of the group divisible designs with two groups of unequal sizes and block size three with λ₂ = 1.

1. Introduction

A pairwise balanced design is an ordered pair (S, ℬ), denoted PBD(S, ℬ), where S is a finite set of symbols, and ℬ is a collection of subsets of S called blocks, such that each pair of distinct elements of S occurs together in exactly one block of ℬ. Here |S | = v is called the order of the PBD. Note that there is no condition on the size of the blocks in ℬ. If all blocks are of the same size k, then we have a Steiner system S(v, k). A PBD with index λ can be defined similarly; each pair of distinct elements occurs in λ blocks. If all blocks are same size, say k, then we get a balanced incomplete block design BIBD(v, b, r, k, λ). In other words, a BIBD(v, b, r, k, λ) is a set S of v elements together with a collection of b k-subsets of S, called blocks, where each point occurs in r blocks, and each pair of distinct elements occurs in exactly λ blocks (see [1–3]).

Note that in a BIBD(v, b, r, k, λ), the parameters must satisfy the necessary conditions

(1)
vr = bk and
(2)
λ(v − 1) = r(k − 1).

With these conditions, a BIBD(v, b, r, k, λ) is usually written as BIBD(v, k, λ).

A group divisible design GDD(v = v₁ + v₂ + ⋯+v_g, g, k, λ₁, λ₂) is an ordered triple (V, G, ℬ), where V is a v-set of symbols, G is a partition of V into g sets of size v₁, v₂, …, v_g, each set being called group, and ℬ is a collection of k-subsets (called blocks) of V, such that each pair of symbols from the same group occurs in exactly λ₁ blocks, and each pair of symbols from different groups occurs in exactly λ₂ blocks (see [1, 2, 4]). Elements occurring together in the same group are called first associates, and elements occurring in different groups we called second associates. We say that the GDD is defined on the set V. The existence of such GDDs has been of interest over the years, going back to at least the work of Bose and Shimamoto in 1952 who began classifying such designs [5]. More recently, much work has been done on the existence of such designs when λ₁ = 0 (see [6] for a summary), and the designs here are called partially balanced incomplete block designs (PBIBDs) of group divisible type in [6]. The existence question for k = 3 has been solved by Fu and Rodger [1, 2] when all groups are the same size.

In this paper, we continue to focus on blocks of size 3, solving the problem when the required designs having two groups of unequal size, namely, we consider the problem of determining necessary conditions for an existence of GDD(v = m + n, 2,3, λ₁, λ₂) and prove that the conditions are sufficient for some infinite families. Since we are dealing on GDDs with two groups and block size 3, we will use GDD(m, n; λ₁, λ₂) for GDD(v = m + n, 2,3, λ₁, λ₂) from now on, and we refer to the blocks as triples. We denote (X, Y; ℬ) for a GDD(m, n; λ₁, λ₂) if X and Y are m-set and n-set, respectively. Chaiyasena, et al. [7] have written a paper in this direction. In particular, they have completely solved the problem of determining all pairs of integers (n, λ) in which a GDD(1, n; 1, λ) exists. We continue to investigate in this paper all triples of integers (m, n, λ) in which a GDD(m, n; λ, 1) exists. We will see that necessary conditions on the existence of a GDD(m, n; λ₁, λ₂) can be easily obtained by describing it graphically as follows.

Let λK_v denote the graph on v vertices in which each pair of vertices is joined by λ edges. Let G₁ and G₂ be graphs. The graph G₁∨_λG₂ is formed from the union of G₁ and G₂ by joining each vertex in G₁ to each vertex in G₂ with λ edges. A G-decomposition of a graph H is a partition of the edges of H such that each element of the partition induces a copy of G. Thus the existence of a GDD(m, n; λ₁, λ₂) is easily seen to be equivalent to the existence of a K₃-decomposition of

. The graph

is of order m + n and size

. It contains m vertices of degree λ₁(m − 1) + λ₂n and n vertices of degree λ₁(n − 1) + λ₂m. Thus the existence of a K₃-decomposition of

implies

(1)
, and
(2)
2∣λ₁(m − 1) + λ₂n and 2∣λ₁(n − 1) + λ₂m.

2. Preliminary Results

We will review some known results concerning triple designs that will be used in the sequel, most of which are taken from [3].

Theorem 2.1. Let v be a positive integer. Then there exists a BIBD(v, 3,1) if and only if v ≡ 1 or 3 (mod 6).

A BIBD(v, 3,1) is usually called Steiner triple system and is denoted by STS(v). Let (V, ℬ) be an STS(v). Then the number of triples b = |ℬ | = v(v − 1)/6. A parallel class in an STS(v) is a set of disjoint triples whose union is the set V. A parallel class contains v/3 triples, and, hence, an STS(v) having a parallel class can exist only when v ≡ 3 (mod 6). When the set ℬ can be partitioned into parallel classes, such a partition ℛ is called a resolution of the STS(v), and the STS(v) is called resolvable. If (V, ℬ) is an STS(v), and ℛ is a resolution of it, then (V, ℬ, ℛ) is called a Kirkman triple system, denoted by KTS(v), with (V, ℬ) as its underlying STS. It is well known that a KTS(v) exists if and only if v ≡ 3 (mod 6). Thus if (V, ℬ, ℛ) is a KTS(v), then ℛ contains (v − 1)/2 parallel classes.

Theorem 2.2. There exists a PBD(6k + 5) with one block of size 5 and 6k² + 9k blocks of size 3.

Example 2.3. Let S = {1,2, 3, …, 11}. Then PBD(11) is an ordered pair (S, ℬ), where ℬ contains the following blocks:

(2.1)

A factor of a graph G is a spanning subgraph. An r-factor of a graph is a spanning r-regular subgraph, and an r-factorization is a partition of the edges of the graph into disjoint r-factors. A graph G is said to be r-factorable if it admits an r-factorization. In particular, a 1-factor is a perfect matching, and a 1-factorization of an r-regular graph G is a set of 1-factors which partition the egde set of G. The following results are well known.

Theorem 2.4. The complete graph K_2n is 1-factorable, K_2n+1 is 2-factorable, and K_3n+1 is 3-factorable.

The following results on existence of λ-fold triple systems are well known (see, e.g., [3]).

Theorem 2.5. Let n be a positive integer. Then a BIBD(n, 3, λ) exists if and only if λ and n are in one of the following cases:

(a)
λ ≡ 0 (mod 6) and n ≠ 2,
(b)
λ ≡ 1 or 5 ( mod 6) and n ≡ 1 or 3 (mod 6),
(c)
λ ≡ 2 or 4 ( mod 6) and n ≡ 0 or 1 (mod 3), and
(d)
λ ≡ 3 ( mod 6) and n is odd.

The results of Chaiyasena, et al. [7] will be useful, and we will state their results as follows.

Theorem 2.6. Let v be a positive integer with v ≥ 3. The spectrum of λ, denoted S_1,v is defined as

(2.2)

Then

(a)
S_1,v = {1,3, 5, …, v − 1} if v ≡ 0 (mod 6),
(b)
S_1,v = {6,12,18, …, v − 1} if v ≡ 1 (mod 6),
(c)
S_1,v = {1,7, 13, …, v − 1} if v ≡ 2 (mod 6),
(d)
S_1,v = {2,4, 6, …, v − 1} if v ≡ 3 (mod 6),
(e)
S_1,v = {3,9, 15, …, v − 1} if v ≡ 4 (mod 6), and
(f)
S_1,v = {4,10,16, …, v − 1} if v ≡ 5 (mod 6).

The following notations will be used throughout the paper for our constructions.

(1)
Let T = {x, y, z} be a triple and a ∉ T. We use a*T for three triples of the form {a, x, y}, {a, x, z}, {a, y, z}. If 𝒯 is a set of triples, then a*𝒯 is defined as {a*T : T ∈ 𝒯}.
(2)
Let G = 〈V(G), E(G)〉 be a graph. If u, v ∈ V(G), e = uv ∈ E(G), and a ∉ V(G), then we use a + e for the triple {a, u, v}. We further use a + E(G) for the collection of triples a + e for all e ∈ E(G). In other words,
(2.3)
In particular, if ℱ = {x₁y₁, x₂y₂, …, x_ny_n} is a 1-factor of K_2n and a is not in the vertex set of K_2n, then
(2.4)
If C_m : x₁, x₂, …, x_m+1 = x₁ is a cycle in K_n, then
(2.5)
Also if G is a 2-regular graph and a ∉ V(G), then a + E(G) forms a collection of triples such that for each u ∈ V(G), there are exactly two triples in a + E(G) containing a and u. In general if G is an r-regular graph and a ∉ V(G), then a + E(G) forms a collection of triples such that for each u ∈ V(G), there are exactly r triples in a + E(G) containing a and u.
(3)
Let V be a v-set. We use K(V) for the complete graph K_v on the vertex set V.
(4)
Let V be a v-set. Let STS(V) be defined as
(2.6)
KTS(V) and BIBD(V, 3, λ) can be defined similarly, that is,
(2.7)
Let X and Y be disjoint sets of cardinality m and n, respectively. We define GDD(X, Y; λ₁, λ₂) as
(2.8)
(5)
When we say that ℬ is a collection of subsets (blocks) of a v-set V, ℬ may contain repeated blocks. Thus, “∪” in our construction will be used for the union of multisets.

3. GDD(m, n; λ, 1)

Let λ be a positive integer. We consider in this section the problem of determining all pairs of integers (m, n) in which a GDD(m, n; λ, 1) exists. Recall that the existence of GDD(m, n; λ, 1) implies 3∣λ[m(m − 1) + n(n − 1)] + 2mn, 2∣λ(m − 1) + n and 2∣λ(n − 1) + m. Let

(3.1)

By solving systems of linear congruences, we obtain the following necessary conditions.

Lemma 3.1. Let t be a nonnegative integer.

(a)
If (m, n) ∈ S(6t + 1), then there exist nonnegative integers h and k such that {m, n}∈{{6k + 1,6h + 2}, {6k + 1,6h + 6}, {6k + 3,6h + 4}, {6k + 3,6h + 6}, {6k + 5,6h + 2}, {6k + 5,6h + 4}}.
(b)
If (m, n) ∈ S(6t + 2), then there exist nonnegative integers h and k such that {m, n}∈{{6k + 6,6h + 4}, {6k + 6,6h + 6}}.
(c)
If (m, n) ∈ S(6t + 3), then there exist nonnegative integers h and k such that {m, n}∈{{6k + 1,6h + 6}, {6k + 3,6h + 2}, {6k + 3,6h + 4}, {6k + 3,6h + 6}, {6k + 5,6h + 6}}.
(d)
If (m, n) ∈ S(6t + 4), then there exist nonnegative integers h and k such that {m, n}∈{{6k + 2,6h + 2}, {6k + 2,6h + 4}, {6k + 6,6h + 4}, {6k + 6,6h + 6}}.
(e)
If (m, n) ∈ S(6t + 5), then there exist nonnegative integers h and k such that {m, n}∈{{6k + 1,6h + 6}, {6k + 3,6h + 4}, {6k + 3,6h + 6}}.
(f)
If (m, n) ∈ S(6t + 6), then there exist nonnegative integers h and k such that {m, n}∈{{6k + 6,6h + 2}, {6k + 6,6h + 4}, {6k + 6,6h + 6}}.

In order to obtain sufficient conditions on an existence of GDD(m, n; λ, 1), we first observe the following facts.

(1)
Let X and Y be two disjoint sets of size m and n, respectively. Then STS(X ∪ Y) ≠ ∅ if and only if GDD(X, Y; 1,1) ≠ ∅.
(2)
Let X and Y be two disjoint sets of size m and n; respectively, and let λ ∈ {2,3, 4,5, 6}. Then GDD(X, Y; λ, 1) ≠ ∅ if STS(X ∪ Y) ≠ ∅, BIBD(X, 3, λ − 1) ≠ ∅, and BIBD(Y, 3, λ − 1) ≠ ∅.

Thus, we have the following results.

Lemma 3.2. Let h and k be nonnegative integers. Then

(a)
(6k + 1,6h + 6), (6k + 6,6h + 1), (6k + 3,6h + 6), (6k + 6,6h + 3), (6k + 3,6h + 4), (6k + 4,6h + 3), (6k + 1,6h + 2), (6k + 2,6h + 1), (6k + 5,6h + 2), (6k + 2,6h + 5), (6k + 5,6h + 4), (6k + 4,6h + 5) ∈ S(1),
(b)
(6k + 1,6h + 6), (6k + 6,6h + 1), (6k + 3,6h + 6), (6k + 6,6h + 3), (6k + 3,6h + 4),(6k + 4,6h + 3) ∈ S(3), and
(c)
(6k + 1,6h + 6), (6k + 6,6h + 1), (6k + 3,6h + 6), (6k + 6,6h + 3), (6k + 3,6h + 4), (6k + 4,6h + 3) ∈ S(5).

Lemma 3.3. Let h and k be nonnegative integers. Then,

(a)
(6k + 6,6h + 6) ∈ S(2) and
(b)
(6k + 6,6h + 4), (6k + 4,6h + 6) ∈ S(2).

Proof. (a) We first consider an existence of GDD(6,6; 2,1), where the groups are X = {1,2, 3,4, 5,6} and Y = {a₁, a₂, …, a₆}. Let ℱ = {ℱ₁, ℱ₂, …, ℱ₅} be a 1-factorization of K(X). Let ), ℬ₂ ∈ STS(X ∪ {a₆}), and ℬ₃ ∈ BIBD(Y, 3,2). Then (X, Y; ℬ) forms a GDD(6,6; 2,1), where ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃. Thus (6,6) ∈ S(2).

Let X and Y be two sets of size 6k + 6 and 6h + 6, respectively. Suppose that k ≤ h and h ≥ 1. Let a₁, a₂, a₃ ∈ Y and let Y′ = Y − {a₁, a₂, a₃}. Thus, KTS(Y′) ≠ ∅. Let ℬ₁ ∈ KTS(Y′) with 𝒫₁, 𝒫₂, …, 𝒫_3h+1 as its parallel classes. Since STS(X ∪ Y′) and STS(X ∪ {a₁, a₂, a₃}) are not empty, there exist ℬ₂ ∈ STS(X ∪ Y′) and ℬ₃ ∈ STS(X ∪ {a₁, a₂, a₃}). We now let ℬ as

(3.2)

Thus, (X, Y; ℬ) forms a GDD(6k + 6,6h + 6; 2,1) and (6k + 6,6h + 6) ∈ S(2).

(b) Let X and Y be two sets of size 6k + 6 and 6h + 4, respectively, a ∈ Y and let Y′ = Y − {a}. Choose ℬ₁ ∈ KTS(Y′) with 𝒫₁, 𝒫₂, …, 𝒫_3h+1 as its parallel classes. Since STS(X ∪ Y′) and STS(X ∪ {a}) are not empty, there exist ℬ₂ ∈ STS(X ∪ Y′) and ℬ₃ ∈ STS(X ∪ {a}). We now let ℬ as

(3.3)

Thus, (X, Y; ℬ) forms a GDD(6k + 6,6h + 4; 2,1) and (6k + 6,6h + 4) ∈ S(2).

Therefore, the proof is complete.

Part of the proof of the following lemma is based on an existence of GDD(4,4; 2,3) which we now construct. Let A = {a, b, c, d} and B = {1,2, 3,4}. Then it is easy to check that F ∈ GDD(A, B; 2,3), where F = {{1, a, b}, {1, a, c}, {1, a, d}, {2, b, c}, {2, b, d}, {2, b, a}, {3, c, d}, {3, c, a}, {3, c, b}, {4, d, a}, {4, d, b}, {4, d, c}, {a, 2,3}, {a, 2,4}, {a, 3,4}, {b, 1,3}, {b, 1,4}, {b, 3,4}, {c, 1,2}, {c, 1,4}, {c, 2,4}, {d, 1,2}, {d, 1,3}, {d, 2,3}}.

Lemma 3.4. Let h and k be nonnegative integers. Then

(3.4)

Proof.

Case 1. Let X_k be a (6k + 2)-set containing a₁, a₂, and Y_h be a (6h + 3)-set containing 1,2, 3.

Subcase 1 (k = 0). Let ℬ₀ = {{1,2, 3}, {1,2, 3}, {1,2, 3}, {1, a₁, a₂}, {2, a₁, a₂}, {3, a₁, a₂}}, and we can see that ℬ₀ ∈ GDD(X₀, Y₀; 3,1). Suppose that h ≥ 1. Since X₀ ∪ Y_h is a set of size 6h + 5, it follows by Theorem 2.2, that there exists a PBD(6h + 5), (X₀ ∪ Y_h, ℬ₁), in which {1,2, 3, a₁, a₂} ∈ ℬ₁ and 6h² + 9h triples in ℬ₁. Let . Since Y_h is a (6h + 3)-set, it follows, by Theorem 2.5(c), that BIBD(Y_h, 3,2) ≠ ∅. Let ℬ₂ ∈ BIBD(Y_h, 3,2). It is easy to see that (X₀, Y_h; ℬ) forms a GDD(2,6h + 3; 3,1), where

(3.5)

Subcase 2 (k = 1). A GDD(8,6h + 3; 3,1) can be constructed as follows. Let X = A ∪ B, where A and B are sets of size four. It is clear that STS(Y_h), STS(A ∪ Y_h), and STS(B ∪ Y_h) are not empty. It has been shown above that GDD(A, B; 2,3) is not empty. We now choose ℬ₁ ∈ STS(Y_h), ℬ₂ ∈ STS(A ∪ Y_h), ℬ₃ ∈ STS(B ∪ Y_h), and ℬ₄ ∈ GDD(A, B; 2,3), and let ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃ ∪ ℬ₄. Then, (X, Y_h; ℬ) form a GDD(8,6h + 3; 3,1).

Subcase 3 (k = 2). We first consider the existence of GDD(4,10; 2,3) with A = {0,1, 2, …, 9} and B = {a₀, a₁, a₂, a₃}. Let K(A) be the complete graph of order 10 with A as its vertex set. It is well known that K₁₀ is 1-factorable. In other words, K₁₀ can be decomposed as a union of nine edge-disjoint 1-factors. Consequently, K₁₀ can be decomposed as a union of three edge-disjoint 3-factors. Also, K₁₀ can be decomposed as a union of 10C₃ and a 3-factor: ten triples {{x, x + 1, x + 3} : x = 0,1, …, 9} and a 3-factor ℱ₀ of K₁₀, where

(3.6)

reducing arithmetic operations (mod 10). Therefore, 2K₁₀ can be decomposed as a union of 10C₃ and four 3-factors.

Let ℱ₁, ℱ₂, ℱ₃ be a 3-factorization of K₁₀ and 10C₃ and ℱ₀ as described above. Then (A, B; ℬ) forms a GDD(10,4; 2,3), where the collection

(3.7)

with ℬ₁ = {{a₀, a₁, a₂}, {a₁, a₂, a₃}, {a₂, a₃, a₀}, {a₃, a₀, a₁}}.

A GDD(14,6h + 3; 3,1) can be constructed as follows. Let X = A ∪ B, where A and B are sets of size ten and four, respectively. It is clear that STS(Y_h), STS(A ∪ Y_h), and STS(B ∪ Y_h) are not empty. It has been shown above that GDD(A, B; 2,3) is not empty. We now choose ℬ₁ ∈ STS(Y_h), ℬ₂ ∈ STS(A ∪ Y_h), ℬ₃ ∈ STS(B ∪ Y_h), and ℬ₄ ∈ GDD(A, B; 2,3) and let ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃ ∪ ℬ₄. Then (X, Y_h; ℬ) form a GDD(14,6h + 3; 3,1).

Subcase 4 (k ≥ 3). Let A = {a₁, a₂, a₃, a₄, a₅}. Suppose that A⊆X_k and X′ = X_k − A. Since X′ is a (6k − 3)-set, it follows that STS(X′) ≠ ∅ and KTS(X′) ≠ ∅. Choose ℬ₁∈ STS(X′) and let 𝒦 ∈ KTS(X′) with 𝒫₁, 𝒫₂, …, 𝒫_3k−2 as its parallel classes. Let . Since X_k ∪ Y_h is a set of size 6(k + h) + 5, we choose a PBD(6(k + h) + 5), (X_k ∪ Y_h, ℬ₃), as in Theorem 2.2 in which A ∈ ℬ₃. Let . Since A is a 5-set and Y_h is a (6h + 3)-set, it follows, by Theorem 2.5(c) and (d), that there exist ℬ₄ ∈ BIBD(5,3, 3) and ℬ₅ ∈ BIBD(Y_h, 3,2). Thus, we can see that (X_k, Y_h; ℬ) forms a GDD(6k + 2,6h + 3; 3,1), where

(3.8)

Case 2. We now suppose that X and Y be sets of size 6k + 5 and 6h + 6, respectively. We suppose further that a ∈ X and X′ = X − {a}. By Lemma 3.3(b), we have GDD(X′, Y; 2,1) ≠ ∅. Choose ℬ₁ ∈ GDD(X′, Y; 2,1) and ℬ₂ ∈ STS(Y ∪ {a}). By Theorem 2.6(e) that GDD({a}, X′; 1,3) ≠ ∅. Choose ℬ₃ ∈ GDD({a}, X′; 1,3). Let ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃, and it is easy to see that ℬ ∈ GDD(X, Y; 3,1).

Thus, (6k + 5,6h + 6) ∈ S(3).

Lemma 3.5. Let h and k be nonnegative integers. Then

(a)
(6k + 6,6h + 6) ∈ S(4) and (6k + 6,6h + 6) ∈ S(6),
(b)
(6k + 6,6h + 4), (6k + 4,6h + 6) ∈ S(4) and (6k + 6,6h + 4), (6k + 4,6h + 6) ∈ S(6),
(c)
(6k + 2,6h + 2) with h, k not both zero, (6k + 2,6h + 4), (6k + 4,6h + 2) ∈ S(4), and
(d)
(6k + 6,6h + 2), (6k + 2,6h + 6) ∈ S(6).

Proof. The proofs of (a) and (b) follow from the results of Lemma 3.3(a), and (b), respectively, and Theorem 2.5(c).

Case 1 (6k + 2, 6h + 2). Let X_k be a (6k + 2)-set and Y_h be a (6h + 2)-set. It is clear that GDD(X₀, Y₀; 4,1) = ∅. We now construct a GDD(2,8; 4,1), (X₀, Y₁; ℬ), with X₀ = {x, y}, Y₁ = {a₁, a₂, …, a₈}, A = {a₁, a₂, a₃}, , and ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃ ∪ ℬ₄, where ℬ₁ ∈ BIBD(A, 3,4), , , and ℬ₄ = {{a_i, x, y} : i = 1,2, 3}. We now construct a GDD(6k + 2,6h + 2; 4,1), (X_k, Y_h; ℬ), in general case, where k ≥ 0 and h ≥ 1. We first let A = {a₁, a₂, a₃}⊆Y_h, , and we will use a result on the existence of GDD(1,6h − 1; 1,4) which has been shown in Theorem 2.6(f), namely, GDD. Therefore, we can choose , and , where F_i ∈ STS(X_k ∪ {a_i}). We can see that ℬ ∈ GDD(X_k, Y_h; 4,1), where ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃ ∪ ℬ₄ ∪ ℬ₅ ∪ ℬ₆.

Case 2 (6k + 2, 6h + 4). Let X_k be a (6k + 2)-set and Y_h be a (6h + 4)-set. It is easy to see that GDD(X₀, Y₀; 4,1) ≠ ∅ by constructing (X₀, Y₀; ℬ) as follows. Let X₀ = {a, b}, Y₀ = {1,2, 3,4}, and , where 𝒫₂ = {{1,2, 3}, {1,2, 4}, {1,3, 4}, {2,3, 4}}. We now turn to more general cases. Suppose that a ∈ Y_h and . Since is a (6h + 3)-set, it follows that KTS. Choose with parallel classes 𝒫₁, 𝒫₂, …, 𝒫_3h+1. Let . We have shown in Lemma 3.4(d) that GDD. Choose and ℬ₄ ∈ STS(X_k ∪ {a}). We can see that ℬ ∈ GDD(X_k, Y_h; 4,1), where ℬ = ℬ₂ ∪ ℬ₃ ∪ ℬ₄.

(d) Let X_k be a (6k + 6)-set and Y_h be a (6h + 2)-set. Let X₀ = {a₁, a₂, …, a₆} and Y₀ = {a, b}. Let ℬ₁ = {{a_i, a, b} : i = 1,2, …, 6}, ℬ₂ ∈ BIBD(X₀, 3,6). Then ℬ₁ ∪ ℬ₂ ∈ GDD(X₀, Y₀; 6,1).

Next we will show that GDD(X_k, Y₀; 6,1) ≠ ∅ by letting X_k = {a₁, a₂, …, a_6k+6}, Y₀ = {a, b}, A = {a₁, a₂, …, a₅} and . Let ℬ₁ = {{a_i, a, b} : i = 1,2, …, 5}, , and ℬ₃ ∈ BIBD(A, 3,6). Theorem 2.6(b) shows an existence of a GDD(1,6h + 1; 1,6). Let , where . It is easy to check that ℬ ∈ GDD(X_k, Y₀; 6,1), where ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃ ∪ ℬ₄.

Finally, let h ≥ 1, a ∈ Y and . We now choose ℬ₁ ∈ BIBD(X_k, 3,4), , , and ℬ₄ ∈ STS(X_k ∪ {a}). By Theorem 2.6(b) that GDD. Choose . Thus, we can check that ℬ ∈ GDD(X_k, Y_h; 6,1), where ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃ ∪ ℬ₄ ∪ ℬ₅. Thus (6k + 6,6h + 2) ∈ S(6) for all positive integers h, k.

Now we have an existence of a GDD(m, n; r, 1) for r = 1,2, …, 6 whenever m and n are not equal to 2, so we can readily extend to any 6t + r by the following lemma.

Lemma 3.6. Let m and n be positive integers with m ≠ 2 and n ≠ 2. If there exists a GDD(m, n; r, 1) with r ≥ 1, then a GDD(m, n; 6t + r, 1), t ≥ 0, exists

Proof. Let X be an m-set and Y be an n-set. By assumption we have GDD(X, Y; r, 1) ≠ ∅. Choose ℬ₁ ∈ GDD(X, Y; r, 1). Since m and n are not equal to 2, by Theorem 2.5(a) there exist ℬ₂ ∈ BIBD(X, 3,6t) and ℬ₃ ∈ BIBD(Y, 3,6t). It is easy to see that (X, Y; ℬ) forms a GDD(m, n; 6t + r, 1), where ℬ = ℬ₁ ∪ ℬ₂ ∪ ℬ₃. Thus (m, n) ∈ S(6t + r) with r ≥ 1.

Finally, we have the main result as in the following.

Theorem 3.7. Let m and n be positive integers with m ≠ 2 and n ≠ 2. There exists a GDD(m, n; λ, 1), λ ≥ 1 if and only if

(1)
3∣λ[m(m − 1) + n(n − 1)] + 2mn and
(2)
2∣λ(m − 1) + n and 2∣λ(n − 1) + m.

Proof. The proof follows from Lemmas 3.1–3.6.

Achnowledgment

N. Pabhapote is supported by the University of the Thai Chamber of Commerce, and N. Punnim is supported by the Thailand Research Fund.

References

1 Fu H. L. and Rodger C. A., Group divisible designs with two associate classes: n = 2 or m = 2, Journal of Combinatorial Theory A. (1998) 83, no. 1, 94–117, 1629448, https://doi.org/10.1006/jcta.1998.2868, ZBL0911.05024.
10.1006/jcta.1998.2868
Web of Science® Google Scholar
2 Fu H. L., Rodger C. A., and Sarvate D. G., The existence of group divisible designs with first and second associates, having block size 3, Ars Combinatoria. (2000) 54, 33–50, 1742405, ZBL0993.05018.
Web of Science® Google Scholar
3 Lindner C. C. and Rodger C. A., Design Theory, 1997, CRC Press, Boca Raton, Fla, USA, 2469212.
Google Scholar
4 El-Zanati S. I., Punnim N., and Rodger C. A., Gregarious GDDs with two associate classes, Graphs and Combinatorics. (2010) 26, 775–780, 2729021, https://doi.org/10.1007/s00373-010-0951-8.
10.1007/s00373-010-0951-8
Web of Science® Google Scholar
5 Bose R. C. and Shimamoto T., Classification and analysis of partially balanced incomplete block designs with two associate classes, Journal of the American Statistical Association. (1952) 47, 151–184, 0048772, https://doi.org/10.2307/2280741, ZBL0048.11603.
10.1080/01621459.1952.10501161
Web of Science® Google Scholar
6 C. J. Colbourn and D. H. Dinitz, Handbook of Combinatorial Designs, 2007, 2nd edition, Chapman & Hall, CRC Press, Boca Raton, Fla, USA, 2246267.
Google Scholar
7 Chaiyasena A., Hurd S. P., Punnim N., and Sarvate D. G., Group divisible designs with two associattion classes, Journal of Combinatorial Mathematics and Combinatorial Computing. In press.
Google Scholar

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Group Divisible Designs with Two Associate Classes and λ₂ = 1

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1. Introduction

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Group Divisible Designs with Two Associate Classes and λ2 = 1

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Group Divisible Designs with Two Associate Classes and λ₂ = 1