Graphs and Combinatorics (2016) 32:1077–1100 DOI 10.1007/s00373-015-1635-1 ORIGINAL PAPER

A Completion of the Spectrum for the Overlarge Sets of Pure Mendelsohn Triple Systems Yuanyuan Liu1

Received: 16 February 2015 / Revised: 17 September 2015 / Published online: 30 September 2015 © Springer Japan 2015

Abstract A pure Mendelsohn triple system of order v, denoted by P M T S(v), is a pair (X, B) where X is a v-set and B is a collection of cyclic triples on X such that every ordered pair of X belongs to exactly one triple of B and if a, b, c ∈ B implies c, b, a ∈ / B. An overlarge set of P M T S(v), denoted by O L P M T S(v), is a collection {(Y \{yi }, Ai )}i , where Y is a (v + 1)-set, yi ∈ Y , each (Y \{yi }, Ai ) is a P M T S(v) and these Ai s form a partition of all cyclic triples on Y . It is shown in [3] that there exists an O L P M T S(v) for v ≡ 1, 3 (mod 6), v > 3, or v ≡ 0, 4 (mod 12). In this paper, we shall discuss the existence problem of O L P M T S(v)s for v ≡ 6, 10 (mod 12) and get the following conclusion: there exists an O L P M T S(v) if and only if v ≡ 0, 1 (mod 3), v > 3 and v = 6. Keywords

Overlarge set · Pure · Mendelsohn triple system

1 Introduction Let X be a finite set. In what follows, an ordered pair of X is always an ordered pair (x, y), where x = y ∈ X . A cyclic triple on X is a set of three ordered pairs (x, y), (y, z) and (z, x) of X , which is denoted by x, y, z (or y, z, x, or z, x, y). A Mendelsohn triple system of order v, denoted by M T S(v), is a pair (X, A) where X is a v-set and B is a collection of cyclic triples on X , called the blocks, such

This research was supported by Natural Science Foundation for the Youth 11101003, NSFC Grant 11171089.

B 1

Yuanyuan Liu [email protected] Department of Fundamental Science, North China Institute of Aerospace Engineering, Langfang 065000, People’s Republic of China

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that every ordered pair of X belongs to exactly one block of A. A holey M T S(v, w) is a trio (X, Y, A) where X is a v-set, Y is its w-subset (called the hole), A is a collection of cyclic triples of X such that every ordered pair of X but those on Y is contained in exactly one block of A and no ordered pair of Y is contained in any block. An M T S(v) (or M T S(v, w)) is called pure and denoted by P M T S(v) (or P M T S(v, w)) if a, b, c ∈ A implies c, b, a ∈ / A. ) is a trio For positive integers n i and gi , 1 ≤ i ≤ r , an M G D D(g1n 1 · · · grnr (X, G, A) satisfying the following conditions: (1) X is a set containing ri=1 n i gi points; (2) G is a partition of X , which consists of n i subsets with size gi (called the groups); (3) A is a family of some cyclic triples of X (called the blocks) such that |A ∩ G| ≤ 1, ∀A ∈ A, G ∈ G and each ordered pair on X from distinct (or the same) groups is contained in exactly one (or no) block. An M G D D(g1n 1 · · · grnr ) (X, G, A) is called pure and denoted by P M G D D(g1n 1 · · · grnr ) if a, b, c ∈ A implies c, b, a ∈ / A. Let Y be a (v + 1)-set and yi ∈ Y . An overlarge set of P M T S(v), denoted by O L P M T S(v), is a collection {(Y \{yi }, Ai )}i such that each (Y \{yi }, Ai ) is a P M T S(v) and these Ai s form a partition of all cyclic triples on Y . For any y ∈ Y , it is not difficult to see that there is one P M T S(v) on the set Y \{y} in an O L P M T S(v). For convenience, we denote the O L P M T S(v) as {(Y \{y}, A y ) : y ∈ Y }. For positive integers w < v, let X be a (v + 1)-set, Y be its (w + 1)-subset. An O L P M T S(v, w) is a partition of all cyclic triples on X but those on Y into disjoint sets Ax (x ∈ X ) such that (i) Each (X \{x}, Ax ) is a P M T S(v) for x ∈ X \Y ; (ii) Each (X \{x}, Y \{x}, Ax ) is a P M T S(v, w) for x ∈ Y . Lemma 1.1 [3] There exists an O L P M T S(v) for v ≡ 1, 3(mod6), v > 3, or v ≡ 0, 4 (mod12). Lemma 1.2 There exists an O L P M T S(v) only if v ≡ 0, 1 (mod3), v ≥ 4 and v = 6. Proof It is well known from [1] that a P M T S(v) exists if and only if v ≡ 0, 1 (mod 3), v ≥ 4 and v = 6. So we obtain the necessary condition.  In this paper, we focus on the existence of O L P M T Ss. Some work on O L P M T Ss has been done (see Lemma 1.1). Thus, in order to get the spectrum of O L P M T Ss, we only need to discuss the existence problem of O L P M T S(v)s for v ≡ 6, 10 (mod 12). In the following sections, we deal with O L P M T S(v)s for v ≡ 6, 10 (mod 12). In Sect. 2, we display some recursive constructions. In Sects. 3–5, we discuss the existence of an O L P M T S(v)s for v = 12k + 10 (k = 11), v = 142, v = 12k + 6 respectively. Finally, the existence spectrum of O L P M T Ss is determined.

2 Recursive Construction In this section, some definitions are described and a series of useful constructions are presented.

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Below, we introduce two definitions about MC S and t-P P MC S, where the definition of MC S is a simple modification to a candelabra t-system in [8]. A Mendelsohn candelabra system (MC S for short) of order v is a quadruple (X, S, G, A) that satisfies the following properties: (1) X is a set of v elements (called the points) and S is an s-subset (called the stem) of X ; (2) G is a collection of subsets (called the groups) of X \S which partition X \S; (3) A is a family of cyclic triples (called the blocks) on X , every cyclic triple a, b, c on X with |{a, b, c} ∩ (S ∪ G)| < 3 for any G ∈ G is contained in exactly one block, and for each G ∈ G no cyclic triple on S ∪ G is contained in any block. By the type of an MC S we mean the list ({|G| : G ∈ G} : s). We also use the exponential notation to denote the type of G and separate the stem size by a colon. Let s and t be two non-negative integers such that t ≤ min{g1 , g2 , . . . , gr } and s + 2t − 2 ≥ 0. An MC S(g1 g2 · · · gr : s) (X, S, G, A) is called t-purely partitionable and denoted by t-P P MC S(g1 g2 · · · gr : s) if the block set A can be partitioned into Ax (x ∈ G, G ∈ G) and Ai (1 ≤ i ≤ s + 2t − 2) with the following properties: G ), (i) For each G ∈ G, there exists a t × |G| Latin rectangle on G, say, L G = (lkx with rows indexed by Z t and columns indexedby G such that each Ax , x ∈ G and G ∈ G, is the block set of a P M G D D(1 1≤m≤r gm −|G| (|G| + s − t)1 ) with G , lG , . . . , lG (G ∪ S)\{l0x 1x t−1,x } as the long group (if t = 0, just take G ∪ S); (ii) For 1 ≤ i ≤ s + 2t − 2, each (X \S, G, Ai ) is a P M G D D(g1 g2 · · · gr ). Construction 2.1 If there exist a 1-PPMCS(g01 g1a1 g2a2 · · · grnr : s), an OLPMTS(g0 + ≤ r , then there exists an s − 1) and anO L P M T S (gi + s − 1, s − 1) for each 1 ≤ i  O L P M T S( 1≤i≤r ai gi + g0 + s − 1) and an O L P M T S( 1≤i≤r ai gi + g0 + s − 1, g0 + s − 1). Proof The proof is similar to [6, Theorem 2.1].



By above construction, 1-P P MC Ss play an important role in constructing O L P M T Ss. In order to display the following three constructions for t-P P MC Ss, we introduce the concepts of s-fan design, P M F, P P M G D D and P P M G D D ∗ . Let X be a v-set. A t-B D of order v is a pair (X, A) such that each t-subset of X is contained in exactly one block of A. An S(t, K , v) denotes a t-B D of order v with block sizes from the set K . Let v be a positive integer, X be a v-set, G be a partition of X , and K 1 , . . . , K s , K T be sets of positive integers. Suppose that B1 , . . . , Bs and T are collections of some subsets of X with size from K 1 , . . . , K s and K T respectively. An s- f an design sF G(3, (K 1 , . . . , K s , K T ), v) is an (s + 3)-tuple (X, G, B1 , . . . , Bs , T ), where (X, G) s B ) ∪ T ) is is a 1-B D, (X, G ∪ Bi ) is a 2-B D for each 1 ≤ i ≤ s, and (X, G ∪ (∪i=1 i a 3-B D. The group type of the s-F G is the multiset {|G| : G ∈ G}. For positive integers n and g, a P M F(g n ) is a trio (X, G, A) where X is a gn-set of points, G is a partition of X into n subsets (called the groups) with size g, A is a collection of all cyclic triples intersecting any given group in at most one point, and A can be partitioned into gn Ax such that each (X \G, G\{G}, Ax ) is a P M G D D(g n−1 ), where x ∈ G ∈ G.

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For positive integers n, g and s ≥ 2, a P P M G D D(g n : s) is a quadruple (X, S, G, A) satisfying the following conditions: (1) X is a set of ng elements, S is a set of s-set and X ∩ S = ∅, the element of X ∩ S is called point; (2) G is a partition of X into n subsets (called the groups) of size g. For x ∈ G ∈ G, denote G = G x . (3) A consists of all cyclic triples (called the blocks) from X ∩ S such that |A ∩ S| ≤ 1, |A ∩ G| ≤ 1, ∀A ∈ A, G ∈ G. And, A can be partitioned into {Ax : x ∈ X } ∪ {Bk : 1 ≤ k ≤ s − 2}, where Ax forms a P M G D D(g n−1 (s + 1)1 ) with the group set (G\G x ) ∪ {S ∪ {x}} and Bk forms a P M G D D(g n ) with the group set G. The definition of P P M G D D ∗ is similar to that of P P M G D D. The difference between them is only the partition of the block set. A P P M G D D ∗ (g n : s) is a quadruple (X, S, G, A) satisfying the following conditions: (1) X is a set of ng elements, S is a set of s-set and X ∩ S = ∅, the element of X ∩ S is called point; (2) G is a partition of X into n subsets (called the groups) of size g. For x ∈ G ∈ G, denote G = G x . (3) A consists of all cyclic triples (called the blocks) from X ∩ S such that |A ∩ S| ≤ 1, |A ∩ G| ≤ 1, ∀A ∈ A, G ∈ G. And, A can be partitioned into {Ax : x ∈ X } ∪ {Bk : 1 ≤ k ≤ s}, where Ax forms a P M G D D(g n−1 s 1 ) with the group set (G\G x ) ∪ S and Bk forms a P M G D D(g n ) with the group set G. Construction 2.2 For g ≥ 3 and g = 4, (1) if there exists a P P M G D D(g n : s) (s ≥ 2), then there exists 0-P P MC S(g n : s); (2) If there exists a P P M G D D ∗ (g n : r ), then there exists 1-P P MC S(g n : r ). Proof (1) Let (X, S, G, A) be the given P P M G D D(g n : s), where X = Z g × Z n , S is an s-set and X ∩ S = ∅, G = {G i : i ∈ Z n }, G i = Z g × {i}. By the definition, A can be partitioned into {A(x, i) : (x, i) ∈ X } ∪ {Bk : 1 ≤ k ≤ s − 2}, where A(x, i) forms a P M G D D(g n−1 (s +1)1 ) with the group set (G\G i )∪{S∪{(x, i)}} and Bk forms a P M G D D(g n ) with the group set G. For g ≥ 3 and g = 4, there exists a 1-P P MC S(g 2 : 0) which is implied in [3, Lemma 2.5]. So for any j = i ∈ Z n , there exists a 1-P P MC S(g 2 : 0) with the points set X j = G i ∪ G j , the group set G j = {G i , G j } and the block set C j which can be partitioned into {C j (x, k) : (x, k) ∈ X j }, where each C j (x, k) forms a P M G D D((g − 1)1 g 1 ) with the group set (G j \G k ) ∪ {G  k \{(x, k)}}, (x, k) ∈ G k . Then it is easy to see that D(x, i) = A(x, i) ∪ ( j=i∈Z n C j (x, i)) forms a P M G D D(1g(n−1) (s + g)1 ) on X ∪ S with the long group G i ∪ S, x ∈ Z g , i ∈ Z n . By the definition, the collection {D(x, i) ∪ Bk : x ∈ Z g , i ∈ Z n , 1 ≤ k ≤ s − 2} forms the block set of a 0-P P MC S(g n : s). (2) The proof is similar to (1). 

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Construction 2.3 [9] If there exist an e-F G(3, (K 1 , K 2 , . . . , K e , K T ), v) (e ≥ 2) of type g1a1 g2a2 · · · grar , a (t + 1)-P P MC S(m k1 : r1 ) (t + 1 ≤ m) for each k1 ∈ K 1 , a P P M G D D(m k2 : r2 ) (r2 ≥ 2) for each k2 ∈ K 2 , a P M F(m ki +1 ) for each ki ∈ K i (3 ≤ i ≤ e) and a P M F(m k ) for each k ∈ K T , then there exists a tP P MC S((mg1 )a1 (mg2 )a2 · · · (mgr )ar : (e − 2)m + r1 + r2 ). By Construction 2.3, It is easy to get the following Construction. Construction 2.4 If there exist an e-F G(3, (K 1 , K 2 , . . . , K e , K T ), v) (e ≥ 1) of type g1a1 g2a2 · · · grar , a t-P P MC S(m k1 : r ) (t + 1 ≤ m) for each k1 ∈ K 1 , a P M F(m ki +1 ) for each ki ∈ K i (2 ≤ i ≤ e) and a P M F(m k ) for each k ∈ K T , then there exists a t-P P MC S((mg1 )a1 (mg2 )a2 · · · (mgr )ar : (e − 1)m + r ). As for the constructions for P P M G D Ds and P M Fs, we employ s-fan G D Ds, below, we explain the definitions and list the constructions. A t-G D D on a v-set X is a triple (X, G, B) satisfying the following properties: (1) (X, G) is a 1-BD; (2) B is a family of subsets of X (called the blocks) and each block intersects any given group in at most one point; (3) each t-subset from t distinct groups is contained in exactly one block. If the block sizes come from a set K , we denote the G D D by G D D(t, K , v). The group type of the G D D means the list {|G| : G ∈ G}. A G D D(3, K , v) (X, G, B) is called an s- f an G D D(3, (K 1 , K 2 , . . . , K s , K T ), v) if K = K 1 ∪ K 2 ∪ · · · ∪ K s ∪ K T and B can be partitioned into disjoint subsets B1 , B2 , . . . , Bs and T such that each (X, G, Bi ) is a G D D(2, K i , v) for 1 ≤ i ≤ s. Construction 2.5 [9] If there exist an e- f an G D D(3, (K 1 , K 2 , . . . , K e , K T ), v) (e ≥ 1) of type g n , a P P M G D D(m k1 : r ) (r ≥ 2) for each k1 ∈ K 1 , a P M F(m ki +1 ) for each ki ∈ K i (2 ≤ i ≤ e) and a P M F(m k ) for each k ∈ K T , then there exists a P P M G D D((mg)n : (e − 1)m + r ). Similar to Construction 2.5, we get easily the following construction. Construction 2.6 If there exist an e- f an G D D(3, (K 1 , K 2 , . . . , K e , K T ), v) (e ≥ 1) of type g n , a P P M G D D ∗ (m k1 : r ) for each k1 ∈ K 1 , a P M F(m ki +1 ) for each ki ∈ K i (2 ≤ i ≤ e) and a P M F(m k ) for each k ∈ K T , then there exists a P P M G D D ∗ ((mg)n : (e − 1)m + r ). Construction 2.7 [9] If there exist a G D D(3, K , v) of type g n and a P M F(m k ) for each k ∈ K , then there exists a P M F((mg)n ). Construction 2.8 If there exists an O L P M T S(v), then there exists a P M F(n v+1 ) for any integer n > 0. Proof Let X be the (v + 1)-set, {(X \{x}, Ax ) : x ∈ X } be the given O L P M T S(v) by the assumption. Now, we construct the desired design P M F(n v+1 ) on X  = X × Z n with the group set G = {{x} × Z n : x ∈ X }. For any x ∈ X, i ∈ Z n , define the block set B(x, i) = {(a, p), (b, q), (c, r ) : a, b, c ∈ Ax , p, q, r ∈ Z n and p +q +r ≡ i (mod n)}. Then ((X \{x}) × Z n , {{y} × Z n : y ∈ X \{x}}, B(x, i)) is a P M G D D(n v ). The union of B(x, i), x ∈ X, i ∈ Z n forms the block set of a P M F(n v+1 ). 

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For later use, we list some results on s-F Gs, s- f an G D Ds and S(3, K , v) as follows. Lemma 2.1 (1) [9] There exists a 2-F G(3, (3, 3, 4), 2k ) for k ≡ 0, 1 (mod 3) and k ≥ 3; There exists a 2-F G(3, (3, 3, {4, 6}), 2k−2 41 ) for k ≡ 2 (mod 3) and k ≥ 5. (2) [10] There exists a 2-F G(3, (4, 3, 4), 34k+1 ) for any integer k ≥ 1. (3) [10] There exists a 2-F G(3, ({3, 5}, {3, 5}, {4, 6}), 2k ) for k ≥ 3. (4) [5] There exists an S(3, {4, 5, 6}, v+2), where v ≡ 0, 2, 3 (mod 4) and v = 7, 11. (5) [4] There exist a 2-F G(3, (4, 4, 4), 65 ) and a g-F G(3, (4, . . . , 4), g 4 ) if g ≡ 0 (mod 4).  (6) [11] Let g = 2i 3 j k pkαk , where pk ≥ 5 is a prime, αk is a non-negative integer and i = 1, j = 1, then there exist a g- f an G D D(3, (4, . . . , 4), g 4 ) and a gf an G D D(3, (5, . . . , 5), g 5 ).

3 Existence of O L P M T S(12k + 10)s In this section we establish the existence of O L P M T S(12k + 10)s for any nonnegative integer k by the constructions in Sect. 2 except the possible O L P M T S(142). Lemma 3.1 There exists a P P M G D D(34 : 2). Proof A P P M G D D(34 : 2) is constructed on Z 12 ∪{12, 13} with group set G = {G i : i ∈ Z 4 } ∪ {{12, 13}}, where G i = {i, i + 4, i + 8}, i ∈ Z 4 . For each x ∈ G i , i ∈ Z 4 , we construct a P M G D D(34 ) with the group set (G\{G i }) ∪ {{12, 13, x}}. Firstly, we list two disjoint block sets Ak (k ∈ Z 2 ) as follows: A0 : 0 1 3 0 2 1 1 2 7 1 6 12 2 9 13 2 11 9 5 7 13 5 12 7 A1 : 0 1 10 0 2 11 123 138 2 8 3 2 12 11 4 13 11 6 8 12

032 176 2 12 5 5 13 6 036 146 2 13 4 6 11 8

0 5 10 1 10 3 2 13 7 6 9 11 0 6 13 167 3 4 10 6 12 7

067 1 11 13 356 6 11 12 0 7 12 174 3 10 12 7 8 13

079 096 1 12 11 1 13 10 3 6 13 3 9 12 7 10 9 7 12 10 0 10 7 0 11 1 1 8 2 1 11 10 3 12 4 3 13 6 7 10 8 8 10 13

0 10 11 2 3 12 3 10 5 9 10 12 0 12 2 247 4 11 6 8 11 12

0 11 5 2 5 11 3 13 9 10 13 11 0 13 3 2 7 13 4 12 10 10 11 13

Then it can be checked that each Ak (k ∈ Z 2 ) is the block set of a P M G D D(34 ) with the group set (G\{G i }) ∪ {{12, 13, k}}, k ∈ G i . For each j ∈ Z 6 , k ∈ Z 2 , define A2 j+k = {B + 2 j : B ∈ Ak }. Clearly, each A2 j+k is the block set of a P M G D D(34 ) with the group set (G\{G i }) ∪ {{12, 13, 2 j + k}}, 2 j + k ∈ G i . Finally, we can check that the collection {Ak : k ∈ Z 12 } contains twelve disjoint block sets and hence forms  the desired P P M G D D(34 : 2). Lemma 3.2 (1) There exist a P M F(126 ) and a P M F(g 4 ) for g ∈ {2, 4, 12}; (2) There exists a P M F(g 5 ) for any positive integer g; (3) There exist a P P M G D D(6k : 5), a P P M G D D(12k : 11) for k ∈ {3, 5} and a P P M G D D(124 : 11);

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(4) There exists a P P M G D D(243 : 23). Proof (1) First we construct a P M F(24 ) on Z 8 with the group set G = {G i : i ∈ Z 4 }, where G i = {i, i +4}, i ∈ Z 4 . Let A0 = {123, 136, 167, 172, 253, 275, 356, 576}. It is easy to see that A0 is the block set of a P M G D D(23 ) with the group set G\{{0, 4}}. For each x ∈ Z 8 , define Ax = {B + x, B ∈ A0 }. Clearly, each Ax is the block set of a P M G D D(23 ) with the group set G\{G i }, x ∈ G i . Finally, we can check that the collection {Ax : x ∈ Z 8 } contains eight disjoint block sets and hence forms the desired P M F(24 ). Since there exist a G D D(3, 4, 24 ) and a G D D(3, {4, 6}, 26 ) by [7,10], a P M F(64 ) and a P M F(66 ) also exist by [9], there exist a P M F(124 ), a P M F(126 ) and a P M F(44 ) by Construction 2.7. (2) Since an O L P M T S(4) exists by Lemma 1.1, there exists a P M F(g n ) for any positive integer g by Construction 2.8. (3) For k ∈ {3, 5}, by deleting a group of a G D D(3, {4, 6}, 2k+1 ), we obtain a 2f an G D D(3, ({3, 5}, {3, 5}, {4, 6}), 2k ). Since there exist a P P M G D D(3k : 2) which is implied in [10, Lemmas 3.3–3.4] and a P M F(34 ) by [9] and a P M F(36 ) which is implied in [10, Lemma 3.8], there exists P P M G D D(6k : 5) by Construction 2.5. There exists a P M F(6k+1 ) by [9], so we also obtain a P P M G D D(12k : 11) by Construction 2.5. For k = 4, there exists a 4- f an G D D(3, (4, 4, 4, 4, 4), 44 ) by Lemma 2.1(6), a P P M G D D(34 : 2) by Lemma 3.1, a P M F(35 ) by above (2) and a P M F(34 ) by [9], so a P P M G D D(124 : 11) exists by Construction 2.5. (4) There exist a 2- f an G D D(3, ({3, 5}, {3, 5}, {4, 6}), 23 ), a P P M G D D(12k : 11) and a P M F(12k+1 ) for k ∈ {3, 5} by the above (3) and (1), so we can get a P P M G D D(243 : 23) by Construction 2.5.  Lemma 3.3 There exists a 2-P P MC S(12k : 0) for k ≡ 0, 1 (mod 3). Proof Let (X, G, B1 , B2 , BT ) be a 2-F G(3, ({3}, {3}, {4, 6}), 2k ) for k ≡ 0, 1 (mod 3), which exists in Lemma 2.1(1). We construct the desired design on X  = Z 6 × X . (1) For each block B = {y0 , y1 , y2 } ∈ B1 , take the point set Z 6 × {y0 , y1 , y2 }. For convenience, the point (a, yi ), a ∈ Z 6 , i ∈ Z 3 is briefly denoted by (a, i). Then we can construct the design on Z 6 × Z 3 . The block set A B (0, 0) is listed as follows, where the point (a, b) is denoted by ab : 11 22 20 21 31 20

22 21 42 51 02 32

20 41 21 02 01 52

11 22 20 21 31 11

20 31 41 01 01 12

31 20 12 12 42 42

11 22 20 21 42

32 42 52 12 41

41 01 01 52 02

11 22 20 32 42

31 41 51 31 51

52 42 42 12 50

11 22 20 32 41

42 52 02 50 52

02 51 32 52 12

11 22 20 32 41

41 50 01 51 01

50 31 41 31 50

11 22 20 32 52

52 51 12 02 50

22 21 51 41 01

11 22 21 32 50

50 02 32 12 51

12 50 01 01 12

11 22 21 31 31

51 01 42 42 51

32 02 50 12 52

11 20 21 31 21

02 21 52 50 50

51 02 41 02 32

It is not difficult to verify that each A B (0, 0) contains the following pairs: (a0 , bi ), (bi , a0 ), (bi , c j ), where a ∈ {2, 5}, b, c ∈ Z 6 , i = j ∈ {1, 2}; (bi , (b + 2)i ), (bi , (b + 3)i ), (bi , (b + 4)i ), where b ∈ Z 6 , i ∈ {1, 2}. Let A B (a, b) = {B + (a, b) : B ∈ A B (0, 0)}, (a, b) ∈ Z 6 × Z 3 , then we can get all A B (a, b).

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Define C B0 = {B + (a, b) : B ∈ H0B , (a, b) ∈ {(0, 0), (1, 0), (2, 0)}}, C B1 = {c, b, a : a, b, c ∈ C B0 }. The blocks in H0B are listed as follows: 40 12 31 40 42 01 10 12 01 10 42 31 30 11 42 30 41 12 00 41 42 00 11 12 30 22 31 30 52 01 00 52 31 00 22 01 40 01 12 40 31 42 10 31 12 10 01 42 30 12 11 30 42 41 00 42 11 00 12 41 30 01 22 30 31 52 00 31 22 00 01 52 j

We can check that each C B ( j ∈ Z 2 ) forms a P M G D D(63 ) with the groups {Z 6 ×{i} : i ∈ Z 3 }. Replacing the point (a, b) by (a, yb ), then we also get the corresponding block j set A B (a, yb ) and C B . (2) For each block B = {y0 , y1 , y2 } ∈ B2 , take the point set Z 6 × {y0 , y1 , y2 }. Similar to (1), we construct the design on Z 6 × Z 3 . The block set D B (0, 0) is listed as follows: 01 12 11 21 30

12 22 30 32 02

11 41 52 42 41

01 12 11 21 31

11 21 42 31 40

22 11 32 02 42

01 12 11 21 31

22 30 40 42 41

40 21 02 52 02

01 12 11 21 31

32 31 52 40 52

30 22 42 32 41

01 12 11 21 42

30 40 02 52 40

42 31 30 31 41

01 12 22 21 42

42 41 21 02 41

51 40 30 40 51

01 12 22 32 40

40 51 32 31 51

52 30 41 30 52

01 12 22 32 52

52 02 30 51 51

02 51 51 41 02

01 11 22 30

51 21 31 31

32 22 32 42

01 11 22 30

02 32 51 41

12 40 40 52

It is not difficult to verify that each D B (0, 0) contains the pairs: (a0 , bi ), (bi , a0 ), (bi , c j ), where a ∈ {3, 4}, b, c ∈ Z 6 , i = j ∈ {1, 2}; (bi , (b + 1)i ), (bi , (b + 5)i ), where b ∈ Z 6 , i ∈ {1, 2}. Let D B (a, b) = {H + (a, b) : H ∈ A B (0, 0)}, (a, b) ∈ Z 6 × Z 3 , then we can get all D B (a, b) and the corresponding D B (a, yb ). (3) For each B ∈ BT , construct a P M F(6|B| ) on Z 6 × B with the group set  B = {G x : x ∈ B}, where G x = Z 6 × {x}. Such a design exists by [10]. Its block set can be partitioned into 6|B| disjoint families E B (i, x), each of which forms a P M G D D(6|B|−1 ) on Z 6 × (B\{x}) with the group set  B \G x , where (i, x) ∈ Z 6 × B. Finally, for each x ∈ X, i ∈ Z 6 , j ∈ Z 2 , define ⎛



F(i, x) = ⎝

x∈B∈B1

F (∞) = j



⎞ A B (i, x)⎠



⎛ ⎝

 x∈B∈B2

⎞ D B (i, x)⎠



⎛ ⎝



⎞ E B (i, x)⎠ ,

x∈B∈B3

j CB .

B∈B1

It is not difficult to verify that all F(i, x) and F j (∞) (x ∈ X, i ∈ Z 6 , j ∈ Z 2 ) are mutually disjoint and each F(i, x) forms a P M G D D(101 112(k−1) ) on X  \{(i, x), (i + 1, x)} with the long group {(Z 6 × G)\{(i, x), (i + 1, x)} : x ∈ G ∈ G} and each F j (∞) forms a P M G D D(12k ) with the groups {Z 6 × G : G ∈ G}. Therefore, by  the definition, these designs form a 2-P P MC S(12k : 0) for k ≡ 0, 1 (mod 3).

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Lemma 3.4 There exists a 2-P P MC S(125 : 0). Proof Deleting two points from an S(3, 6, 22) which exists by [2], we can obtain a 2-F G(3, (5, 5, 6), 45 ), it is also a 1-F G(3, ({5}, {5, 6}), 45 ). There exist a 2P P MC S(35 : 0) and a P M F(36 ) which is implied in [9, Lemmas 4.2 and 3.8], a P M F(35 ) also exists by Lemma 3.2(2), so there exists a 2-P P MC S(125 : 0) by Construction 2.4.  Lemma 3.5 There exists a 1-P P MC S(24a 36b 48c : 11), where v = 2a + 3b + 4c, v ≡ 0, 2, 3 (mod 4) and v = 7, 11. Proof There exists an S(3, {4, 5, 6}, v + 2), where v ≡ 0, 2, 3 (mod 4) and v = 7, 11 by Lemma 2.1(4), deleting any two points, we can obtain a 2F G(3, {3, 4, 5}, {3, 4, 5}, {4, 5, 6}, v) of type 2a 3b 4c . For k ∈ {3, 4, 5}, there exist a 2-P P MC S(12k : 0), a P P M G D D(12k : 11) and a P M F(12k+1 ) by Lemmas 3.2–3.4. So there exists a 1-P P MC S(24a 36b 48c : 11) for v = 2a + 3b + 4c and v ≡ 0, 2, 3 (mod 4), v = 7, 11 by Construction 2.3.  Lemma 3.6 There exist an O L P M T S(10, 4) and an O L P M T S(10). Proof Firstly, we construct an O L P M T S(10, 4) on X = Z 6 ∪ {6, 7, 8, 9, 10} with a hole S = {6, 7, 8, 9, 10}. By the definition, such a design should contain six P M T S(10)s and five P M T S(10, 4)s. The required block sets are described below. Let Bk be the set consisting of the follow blocks, where each Bk contain 30 blocks, k ∈ Z3. B0 : 4 5 1 521 1 9 10 B1 : 3 4 6 405 5 10 6 B2 : 3 4 10 408 5 10 8

435 5 6 10 1 10 8 358 4 2 10 069 350 415 067

413 578 2 3 10 304 478 076 301 4 6 10 0 7 10

429 586 276 320 486 087 314 476 098

467 597 289 3 6 10 497 0 9 10 365 489 0 10 6

4 7 10 5 10 9 2 10 7 372 4 10 9 0 10 8 378 490 168

482 126 639 389 502 268 386 510 179

498 169 683 395 567 279 397 569 1 8 10

4 10 6 173 793 3 10 7 5 7 10 2 8 10 3 10 9 587 196

532 187 8 10 3 452 598 296 457 5 9 10 1 10 7

We can check that each Bk is the block set of a P M T S(10) on X \{k} for k ∈ Z 3 . Define Bk+3 = {B + k : B ∈ Bk } for k ∈ Z 3 . Clearly, each Bk is the block set of a P M T S(10) on X \{k} for k ∈ Z 6 . An O L P M T S(10, 4) also contain five P M T S(10, 4)s. For 6 ≤ k ≤ 10, let Bk = {a, b + i, c + i : i ∈ Z 6 , if a, b, c belongs the not underlined block in Ak ; a, b + 3i, c + 3i : i ∈ Z 2 , if a, b, c belongs the underlined block in Ak }. we list the blocks in Ak below.

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A6 : A7 : A8 : A9 : A10 :

703 605 603 602 702

804 801 705 805 803

901 903 902 10 0 3 905

10 0 5 10 0 2 10 1 2 701 623

024 042 10 0 4 10 2 0 2 3 4 715 720 123 604 615 012

It can be checked that each Bk is the block set of a P M T S(10, 4) on X \{k} with the hole S\{k} for 6 ≤ k ≤ 10. Finally, we can check that the collection {Bk : k ∈ Z 10 } contains 11 disjoint block sets and hence forms the desired O L P M T S(10, 4). By inputting design O L P M T S(4), which exists by Lemma 1.1, into O L P M T S(10, 4), we can obtain an O L P M T S(10).  Lemma 3.7 There exist an O L P M T S(22, 10) and an O L P M T S(22). Proof Firstly, we construct an O L P M T S(22, 10) on X = Z 12 ∪ {12, 13, . . . , 22} with a hole S = {12, 13, . . . , 22}. By the definition, such a design should contain 12 P M T S(22)s and eleven P M T S(22, 10)s. The required block sets are described below. Let B0 be the set consisting of the follow 154 blocks: 124 279 467 264 19 10 5 12 19 7 19 12 8 13 20 7 21 13 10 15 14 6 15 18 8 21 15 6 19 16 10 19 17 2 21 18 9 21 20 3

132 285 4 7 10 3 10 7 20 4 9 12 20 9 20 12 10 13 21 11 22 13 7 16 14 5 15 19 11 22 15 2 20 16 5 20 17 8 22 18 1 22 20 5

145 298 4 11 9 3 11 4 21 9 5 12 21 10 21 12 11 13 22 10 14 15 5 17 14 10 15 20 10 16 17 11 21 16 7 21 17 1 19 20 6 21 22 4

157 2 10 11 5 6 11 12 4 8 22 5 4 12 22 11 22 12 9 14 13 2 14 16 6 18 14 4 15 21 2 16 18 10 22 16 8 22 17 10 19 21 4 22 21 3

169 2 11 6 5 11 7 13 5 9 12 13 1 13 12 2 13 14 1 15 13 3 14 17 4 19 14 3 15 22 9 16 19 1 17 18 2 18 19 5 19 22 1

176 3 4 10 6 8 10 14 9 10 12 14 2 14 12 1 13 15 4 16 13 4 14 18 3 20 14 7 16 15 9 16 20 2 17 19 4 18 20 8 20 19 3

1 8 11 358 6 10 9 15 8 4 12 15 3 15 12 5 13 16 3 17 13 8 14 19 9 21 14 8 17 15 11 16 21 8 17 20 1 18 21 1 21 19 2

193 365 789 16 2 3 12 16 4 16 12 6 13 17 6 18 13 6 14 20 11 22 14 11 18 15 10 16 22 7 17 21 5 18 22 3 22 19 6

1 10 8 386 7 11 8 17 3 7 12 17 5 17 12 3 13 18 5 19 13 9 14 21 7 15 16 1 19 15 7 17 16 9 17 22 6 19 18 11 20 21 6

1 11 10 3 9 11 2 5 10 18 7 2 12 18 6 18 12 7 13 19 8 20 13 11 14 22 8 15 17 7 20 15 1 18 16 11 18 17 9 20 18 4 20 22 2

We can check that B0 is the block set of a P M T S(22) on X \{0}. For each k ∈ Z 12 , define Bk = {B + k : B ∈ B0 }. Clearly, each Bk is the block set of a P M T S(22) on X \{k}. An O L P M T S(22, 10) also contain eleven P M T S(22, 10)s. For 12 ≤ k ≤ 22, let

Bk = {a, b + i, c + i : i ∈ Z 12 , if a, b, c belongs the not underlined block in Ak a, b + 3i, c + 3i : i ∈ Z 4 , if a, b, c belongs the underlined block in Ak }. we list the blocks in Ak below.

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A12 : 13 0 3 15 0 4 A13 : 12 0 6 15 2 6 A14 : 15 0 3 12 2 10 A15 : 12 0 2 20 0 1 A16 : 12 0 9 15 1 5 A17 : 12 0 3 20 1 2 A18 : 14 0 4 12 1 9 A19 : 14 0 7 12 0 8 A20 : 12 0 7 21 0 5 A21 : 12 0 10 20 2 3 A22 : 12 0 1 20 0 2

14 0 2 15 1 2 14 0 9 15 0 1 16 0 4 12 1 0 13 0 6 20 1 5 13 0 10 15 2 3 13 0 2 20 2 6 15 0 10 12 0 11 15 0 9 12 2 1 13 0 1 22 0 8 13 0 9 20 0 4 13 0 7 21 0 11

16 0 5 15 2 9 16 0 11 15 1 8 17 0 7 12 0 5 14 0 11 20 2 9 14 0 3 15 0 7 14 0 5 20 0 7 16 0 7 12 2 7 16 0 2 12 1 6 14 0 10 048 14 0 8 20 1 8 14 0 6 084

17 0 11 21 1 5 17 0 5 21 0 4 18 0 1 13 0 8 16 0 9 22 2 3 17 0 8 21 2 6 15 0 6 22 0 1 17 0 2 13 2 10 17 0 1 13 1 9 15 0 2

1087

18 0 9 21 2 3 18 0 8 21 1 2 19 0 10 13 2 1 17 0 10 22 0 4 18 0 6 21 0 1 16 0 8 22 1 5 19 0 1 13 1 0 18 0 4 13 0 11 16 0 3

19 0 6 21 0 7 19 0 3 21 2 9 20 0 9 13 1 6 18 0 5 22 1 8 19 0 2 21 1 8 18 0 10 22 2 9 20 0 3 13 0 5 20 0 6 13 2 7 17 0 6

15 0 11 16 0 6 17 0 3 22 1 2 22 2 6 22 0 7 15 0 5 16 0 10 17 0 9

20 0 8 267 20 0 10 156 21 0 2 198 19 0 8 126 20 0 11 045 19 0 11 237 21 0 6 087 21 0 10 2 10 9 18 0 11 18 0 2 015 18 0 3

22 0 10 22 0 2 22 0 6 21 0 3 22 0 5 21 0 9 22 0 9 22 0 3 19 0 9 19 0 5 19 0 4

It can be checked that each Bk is the block set of a P M T S(22, 10) on X \{k} with the hole S\{k} for 12 ≤ k ≤ 22. Finally, we can check that the collection {Bk : k ∈ Z 23 } contains 23 disjoint block sets and hence forms the desired O L P M T S(22, 10). By inputting design O L P M T S(10), which exists by Lemma 3.6, into O L P M T S(22, 10), we can obtain an O L P M T S(22).  Lemma 3.8 There exist an O L P M T S(34, 10) and an O L P M T S(34). Proof There exists an S(3, 5, 17) by [2], deleting two points, we can obtain a 2F G(3, (4, 4, 5), 35 ). There exists a P M F(25 ) by Lemma 3.2(2), if there exist a 2P P MC S(24 : 2) and a P P M G D D(24 : 3), then we can get a 1-P P MC S(65 : 5) by Construction 2.3. An O L P M T S(10) and an O L P M T S(10, 4) exist by Lemma 3.6, so there exist an O L P M T S(34) and an O L P M T S(34, 10) by Construction 2.1. The desired 2-P P MC S(24 : 2) and P P M G D D(24 : 3) are listed as follows: 2 − P P MC S(24 : 2) A 2-P P MC S(24 : 2) is constructed on X = Z 8 ∪ {8, 9} with the groups G i = {i, i + 4}, i ∈ Z 4 and the stem S = {8, 9}. For j ∈ Z 8 , the block set of each A j is listed as follows.

123

1088 A0 A1 A2 A3 A4 A5 A6 A7

: : : : : : : :

Graphs and Combinatorics (2016) 32:1077–1100

125 024 014 015 126 026 018 019

137 037 038 029 135 034 039 028

156 046 045 042 157 047 043 041

168 068 059 058 169 069 051 054

179 079 073 064 178 078 074 062

183 083 081 086 182 082 087 085

192 092 097 091 193 093 095 096

236 238 139 128 237 239 138 129

267 263 153 145 256 246 149 148

278 276 175 162 279 273 154 152

285 284 187 184 283 287 173 165

293 297 194 196 295 294 197 186

357 347 349 248 367 368 348 249

386 369 358 259 385 376 359 258

395 394 374 265 396 384 375 264

587 487 478 469 586 486 458 459

596 496 485 495 597 497 479 468

697 678 579 568 687 679 578 569

We can check that each A j is the block set of a P M G D D(16 21 ) on X \{G i } with the long group {8, 9}, where j ∈ G i , i ∈ Z 4 . A 2-P P MC S(24 : 2) also contains four P M G D D(24 )s. Let Bk = {B + 2i : B ∈ Ck , i ∈ Z 4 }, k ∈ Z 4 , where the blocks in Ck are listed below. C0 : 012 163 036 107 C2 : 013 075 124 106

C1 : 123 052 147 076 C3 : 016 072 127 103

It is easy to see that each Bk (k ∈ Z 4 ) is the block set of a P M G D D(24 ) with the groups G i (i ∈ Z 4 ). Finally, we can check that the collection {Ai : i ∈ Z 8 } ∪ {Bk : k ∈ Z 4 } contain 12 disjoint block sets and hence forms the desired 2-P P MC S(24 : 2). P P M G D D(24 : 3) A P P M G D D(24 : 3) is constructed on Z 8 ∪ {8, 9, 10} with the group set G = {G i : i ∈ Z 4 } ∪ {{8, 9, 10}}, where G i = {i, i + 4}. For each x ∈ G i , i ∈ Z 4 , we construct a P M G D D(23 41 ) with the group set (G\{G i }) ∪ {8, 9, 10, x}. Firstly, we list two disjoint block sets Ak (k ∈ Z 2 ) as follows: A0 : 0 1 7 258 A1 : 0 1 3 238

023 2 7 10 021 249

035 295 0 3 10 284

056 2 10 3 069 293

061 369 078 2 10 7

072 386 086 3 4 10

128 3 10 5 097 368

139 578 0 10 2 396

1 6 10 597 127 4 6 10

183 5 10 6 143 479

192 679 164 487

1 10 7 687 176 6 7 10

Then it can be checked that each Ak (k ∈ Z 2 ) is the block set of a P M G D D(23 41 ) with the group set (G\{G i }) ∪ {{8, 9, 10, k}}, k ∈ G i . For each j ∈ Z 4 , k ∈ Z 2 , define A2 j+k = {B + 2 j : B ∈ Ak }. Clearly, each A2 j+k is the block set of a P M G D D(23 41 ) with the group set (G\{G i }) ∪ {{8, 9, 10, 2 j + k}}, 2 j + k ∈ G i . A 1-P P M G D D(24 : 3) also contain one P M G D D(24 ). We list the blocks in B as follows, i ∈ Z 4 . 0, 1, 2 + 2i, 1, 6, 3 + 2i, 0, 3, 6 + 2i, 1, 0, 7 + 2i. Finally, we can check that the collection {Ak : k ∈ Z 8 } ∪ {B} contains nine disjoint  block sets and hence forms the desired P P M G D D(24 : 3). Lemma 3.9 There exist an O L P M T S(v, 10) and an O L P M T S(v) for v ∈ {46, 58}. Proof There exist a 2-F G(3, (4, 4, 5), 35 ) by Lemma 3.8, it is also a 1-F G(3, {4}, {4, 5}, 35 ), and a 4-F G(3, (4, 4, 4, 4, 4), 44 ) by Lemmas 2.1(5). A P M F(34 ) and a P M F(35 ) exist by [10] and Lemma 3.2(2). There exists a 1-P P MC S(34 : 2) which is implied in [10, Lemma 5.1], so a 1-P P MC S(95 : 2) and a 1P P MC S(124 : 11) exist by Constructions 2.4. By Lemmas 3.6 and 3.7, there exist an

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O L P M T S(10), an O L P M T S(22) and an O L P M T S(22, 10), an O L P M T S(10) is also an O L P M T S(10, 1), so there exist an O L P M T S(v) and an O L P M T S(v, 10) for v ∈ {46, 58} by Construction 2.1.  Theorem 3.1 There exists an O L P M T S(v) for v = 48k +t, t ∈ {10, 34, 46}, except possible v = 142. Proof There exists a 1-P P MC S(24a 36b 48c : 11) for n = 2a+3b+4c = 4k+m, m ∈ {0, 2, 3}, n = 7, 11 by Lemma 3.5. Since it is impossible for a, b, c all equal to 0, suppose a = 0 (the case b = 0 or c = 0 is similar), then a 1-P P MC S(24a 36b 48c : 11) is also a 1-P P MC S(241 24a−1 36b 48c : 11). Since an O L P M T S(34) and an O L P M T S(m, 10) (m ∈ {34, 46, 58}) exist by Lemmas 3.8 and 3.9, we can get an O L P M T S(v) for v = 48k + t, t ∈ {10, 34, 46}, except possible v ∈ {94, 142}. For v = 94, there exist a 2-F G(3, (4, 4, 4), 65 ) by Lemma 2.1(5), a 1P P MC S(34 : 2) which is implied in [10, Lemma 5.1], a P M F(35 ) by Lemma 3.2(2) and a P M F(34 ) by [9], so there exist a 1-P P MC S(185 : 5) by Construction 2.4. By Lemmas 3.7 and 3.6, there exist an O L P M T S(22), an O L P M T S(22, 10) and an O L P M T S(10, 4). By inputting design O L P M T S(10, 4) into O L P M T S(22, 10), we can get an O L P M T S(22, 4). Finally we can get an O L P M T S(94) by Construction 2.1.  Lemma 3.10 There exists a 2-P P MC S(243 : 0). Proof Let (X, G, B1 , B2 , BT ) be a 2-F G(3, ({3}, {3}, {4}), 63 ) which exists by [11, Lemma 1.3], where G = {G i : i ∈ Z 3 }. We construct the desired design on X  = X × Z4. Below, we give three corresponding designs for the blocks in B1 , B2 and BT , respectively. In the constructions (1) and (2), for given three elements ri ∈ G i , i ∈ Z 3 , the desired design on the set {r0 , r1 , r2 } × Z 4 is transformed to one on the set Z 12 . The elements 0, 3, 6, 9, 1, 4, 7, 10, 11, 8, 5, 2 in Z 12 represent (r0 , 0), (r0 , 1), (r0 , 2), (r0 , 3), (r1 , 0), (r1 , 1), (r1 , 2), (r1 , 3), (r2 , 0), (r2 , 1), (r2 , 2), (r2 , 3) respectively. (1) For each block B = {r0 , r1 , r2 } ∈ B1 , take the point set {r0 , r1 , r2 }× Z 4 . Construct the following families on Z 12 . A B (0) : 1 2 8 2 11 4 A B (1) : 0 2 3 356

1 4 11 459 0 5 10 389

1 5 10 4 8 10 0 8 11 3 9 11

185 498 098 3 10 8

192 4 10 5 0 10 2 3 11 10

1 11 9 579 0 11 5 586

247 789 253 5 9 10

275 7 10 11 2 6 11 5 11 9

2 9 10 7 11 8 296 6 8 10

2 10 8 9 11 10 2 10 9 6 10 11

Replace the elements in Z 12 by the elements in {r0 , r1 , r2 }× Z 4 . It is not difficult to verify that each A B (r0 , 0) contains the following pairs: ((r1 , i), (r2 , j)), ((r2 , j), ((r0 , 3), (rk , j)), ((rk , j), (r0 , 3)), ((rk , i), (rk , i + 1)), (r1 , i)), ((rk , m), (rk , m + 2)), where i, j ∈ Z 4 , k ∈ {1, 2}, m ∈ {1, 3}. Note the underlined pairs are complement with those in D B (r0 , 0). Similarly, we can easily get the pairs contained in the A B (r1 , 0). Let A B (k + 2i) = A B (k) + 2i, i ∈ Z 6 , k ∈ Z 2 , then we can get all A B (ri , j), where i ∈ Z 3 , j ∈ Z 4 .

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Let C B0 = {0, 1, 2 + 2i, 1, 6, 11 + 2i, 0, 4, 8 + j : i ∈ Z 6 , j ∈ Z 4 }, C B1 = {c, b, a : a, b, c ∈ C B0 }. It is easy to verify that each C iB (i ∈ Z 2 ) forms a G D D(43 ) with the groups {{ri } × Z 4 : i ∈ Z 3 }. (2) For each block B = {r0 , r1 , r2 } ∈ B2 , take the point set {r0 , r1 , r2 }× Z 4 . Construct the following families on Z 12 . D B (0) : 1 2 4

1 5 7 1 6 5 1 7 8 1 8 6 1 10 11 1 11 2 2 5 10 2 6 4 2 7 6 2 10 7 4 5 8 4 6 11 4 8 7 4 11 5 5 6 10 5 11 7 6 7 11 6 8 10 8 11 10

D B (1) : 0 2 9 0 3 5

0 5 2 0 6 8 0 7 11 0 8 7 0 11 6 2 3 8 2 6 7 2 7 3 2 8 6 2 11 9 3 6 11 3 7 5 3 11 8 5 6 9 5 7 6 5 9 8 7 8 9 7 9 11

Replace the elements in Z 12 by the elements in {r0 , r1 , r2 }× Z 4 . It is not difficult to verify that each D B (r0 , 0) contains the following pairs: ((r1 , i), (r2 , j)), ((r2 , j), ((r0 , 2), (rk , j)), ((rk , j), (r0 , 2)), ((rk , i + 1), (rk , i)), (r1 , i)), ((rk , m), (rk , m + 2)), where i, j ∈ Z 4 , k ∈ {1, 2}, m ∈ {0, 2}. Note the underlined pairs are complement with those in A B (r0 , 0). Similarly, we can easily get the pairs contained in the D B (r1 , 0). Let D B (k + 2i) = D B (k) + 2i, i ∈ Z 6 , k ∈ Z 2 , then we can get all D B (ri , j), where i ∈ Z 3 , j ∈ Z 4 . Let H0B = {0, 1, 2 + 2i, 1, 6, 11 + 2i, 0, 4, 8 + j : i ∈ Z 6 , j ∈ Z 4 }, H1B = {c, b, a : a, b, c ∈ H0B }. It is easy to verify that each HiB (i ∈ Z 2 ) forms a G D D(43 ) with the groups {{ri } × Z 4 : i ∈ Z 3 }. (3) For each B ∈ BT , construct a P M F(44 ) on B × Z 4 with the groups  B = {G x : x ∈ B} where G x = {x} × Z 4 . Such a design exists by Lemma 3.2(1). Its block set E B can be partitioned into 16 disjoint families E B (x, i), each of which forms a P M G D D(43 ) on (B\{x}) × Z 4 with the groups  B \G x , where (x, i) ∈ B × Z 4 . Final, for each x ∈ X, i ∈ Z 4 , define ⎛ ⎛ ⎛ ⎞ ⎞      ⎝ ⎝ F(x, i) = ⎝ A B (x, i)⎠ D B (x, i)⎠ ⎛ F0 = ⎝

x∈B∈B1

 B∈B1

⎞

C B0 ⎠



⎛ ⎝

x∈B∈B2

 B∈B2

⎞

H1B ⎠ ,

⎛ F1 = ⎝

x∈B∈BT

 B∈B1

⎞

C B1 ⎠



⎞ E B (x, i)⎠ , ⎛ ⎝



⎞ H0B ⎠ .

B∈B2

For any x ∈ G k ∈ G, i, k ∈ Z 3 , j ∈ Z 2 , it is not difficult to verify that all F(x, i) and F j are mutually disjoint, each F(x, i) forms a P M G D D(221 148 ) on X  \{(x, i), (x, i + 1)} with the long group {(G k × Z 4 )\{(x, i), (x, i + 1)} and each F j forms a P M G D D(243 ) with the groups {G k × Z 4 }. Therefore, these designs form  the desired 2-P P MC S(243 : 0). Lemma 3.11 There exist an O L P M T S(70, 22), an O L P M T S(70) and an O L P M T S(118). Proof There exists a 5- f an G D D(3, (5, . . . , 5), 45 ) by Lemma 2.1, so there exists a 5-F G(3, (5, 5, 5, 5, 5, {4, 5}), 45 ) by [4, Lemma 3.7], it is also a 4F G(3, (5, 5, 5, 5, {4, 5}), 45 ). Since an O L P M T S(4) exists by Lemma 1.1, we can

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get a 1-P P MC S(35 : 2) in which the construction is similar to the construction in [6, Lemma 3.6]. There exists a P M F(35 ) by Lemma 3.2(2) and a P M F(36 ) which is implied in [10, Lemma 3.8], so a 1-P P MC S(125 : 11) exists by Construction 2.4. Inputting designs O L P M T S(22) and an O L P M T S(22, 10) which exist by Lemma 3.7, we can get an O L P M T S(70, 22) and an O L P M T S(70) by construction 2.1. There exists a 1-F G(3, (4, {4, 5, 13}), 313 ) by [4, Lemma 4.14]. For k ∈ {4, 12}, an O L P M T S(k) exists by Lemma 1.1, so a P M F(3k+1 ) also exist by Construction 2.8. There exist a 1-P P MC S(34 : 2) which is implied in [10, Lemma 5.1] and a P M F(34 ) by [9], so we can get a 1-P P MC S(913 : 2) by Construction 2.4. There exists an O L P M T S(10) by Lemma 3.6, it is also an O L P M T S(10, 1). Thus, there exists an O L P M T S(118) by Construction 2.1.  Theorem 3.2 There exists an O L P M T S(48k + 22) for any non-negative integer k. Proof For k ∈ {0, 1, 2}, there exists an O L P M T S(k) by Lemmas 3.7 and 3.11; For k ≥ 3, there exist a 2-F G(3, (3, 3, 4), 2k ) for k ≡ 0, 1 (mod 3) (k ≥ 3) and a 2-F G(3, (3, 3, {4, 6}), 2k−2 41 ) for k ≡ 2 (mod 3) (k ≥ 5) by Lemma 2.1(1). A 2-P P MC S(243 : 0) and a P P M G D D(243 : 23) exist by Lemmas 3.10 and 3.2(4), so there exists a 1-P P MC S(48k : 23) or a 1-P P MC S(48k−2 961 : 23) by Construction 2.3. Since an O L P M T S(70), an O L P M T S(70, 22) and an O L P M T S(118) exist by Lemma 3.11, there exists an O L P M T S(48k + 22) for k ≥ 3 by Construction 2.1. 

4 Existence of O L P M T S(12k + 5)s and O L P M T S(142) Combining Theorems 3.1 and 3.2, we can get that there exists an O L P M T S(12k +11) for any non-negative integer k except possible O L P M T S(142). For the existence of an O L P M T S(142), we doesn’t find out more simple method to construct it. In this section, we give a construction of O L P M T S(142) and also get the existence of an O L P M T S(12k + 5) for any integer k ≥ 3. Lemma 4.1 There exists a 2-P P MC S(6k : 0) for k ∈ {3, 5}. Proof First we construct a 2-P P MC S(23 : 0) on Z 6 with the groups G i = {i, i + 3}, i ∈ Z 3 . By the definition, such a design should contain six P M T S(4)s and two P M G D D(23 )s. Let A0 = {142, 415, 251, 524}; A1 = {032, 305, 250, 523}, A2 j+k = {B + 2 j : B ∈ Ak }, k ∈ Z 2 , j ∈ Z 3 . B0 = {012, 234, 450, 153, 321, 543, 105, 240}; B1 = {cba : abc ∈ B0 }. Then it is easy to see each Ax (x ∈ Z 6 ) is a P M T S(4) on Z 6 \G i , x ∈ G i and each Bk (k ∈ Z 2 ) is a P M G D D(23 ) with the groups G i , i ∈ Z 3 . we can check that the collection {Ak : k ∈ Z 6 } ∪ {Bk : k ∈ Z 2 } contains eight disjoint block sets and hence forms the desired 2-P P MC S(23 : 0). There exists a 2-F G(3, (4, 3, 4), 35 ) by Lemma 2.1(2), it is also a 1-F G(3, (3, 4), 5 3 ). A 1-F G(3, (3, 4), 33 ) also exists by [2]. Since there exist a 2-P P MC S(23 : 0)

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and a P M F(24 ) by Lemma 3.2(1), so there exists a 2-P P MC S(6k : 0) for k ∈ {3, 5} by Construction 2.4.  Lemma 4.2 There exist an O L P M T S(16, 4) and an O L P M T S(16). Proof An O L P M T S(16, 4) is constructed on X = Z 12 ∪ {12, 13, 14, 15, 16} with a hole S = {12, 13, 14, 15, 16}. By the definition, such a design should contain twelve P M T S(16)s and five P M T S(16, 4)s. The required block sets are described below. Let B0 be the set consisting of the follow 80 blocks: 124 1 12 11 297 3 6 16 4 5 14 4 16 8 6 11 12 8 11 9

132 1 13 12 2 10 15 379 4 6 10 578 6 12 15 8 12 13

143 1 14 13 2 11 8 386 4 7 13 5 8 15 6 13 7 8 13 15

156 1 15 14 2 12 5 395 487 596 6 14 16 8 14 12

167 1 16 15 2 13 6 3 10 13 4 9 16 5 12 14 6 15 11 9 11 13

175 2 3 13 2 14 10 3 11 14 4 10 12 5 13 16 7 10 11 9 12 10

189 2 5 11 2 15 9 3 12 7 4 11 5 5 15 13 7 11 15 9 13 14

1 9 10 264 2 16 12 3 14 8 4 12 9 5 16 10 7 12 16 9 15 16

1 10 8 2 7 14 3 4 15 3 15 12 4 13 11 6 8 10 7 15 10 10 14 11

1 11 16 2 8 16 3 5 10 3 16 11 4 14 15 6 9 14 7 16 14 10 16 13

Then we can check that B0 forms the block set of a P M T S(16) on X \{0}. For k ∈ Z 12 , define Bk = {B + k : B ∈ B0 }. Clearly, each Bk is the block set of a P M T S(16) on X \{k}. An O L P M T S(16, 4) also contains five P M T S(16, 4)s. For 12 ≤ k ≤ 16, let Bk = {a, b+i, c+i : i ∈ Z 12 , if a, b, c belongs the not underlined block in Ak ; a, b+3i, c+3i : i ∈ Z 4 , if a, b, c belongs the underlined block in Ak }. we list the blocks in Ak below. A12 A13 A14 A15 A16

: : : : :

13 0 9 12 0 3 12 2 3 12 1 2 12 0 1

14 0 10 14 0 2 12 1 11 12 2 0 12 1 5

15 0 2 15 0 10 12 0 10 12 0 4 12 2 6

16 0 1 16 0 11 012 13 0 10 13 1 11

076 016 13 0 6 13 1 5 13 2 0

038 074 15 0 11 13 2 6 13 0 4

084 048 16 0 2 0 3 7 0 7 3 2 3 4 14 0 6 16 0 9 0 3 5 0 8 1 1 2 3 14 0 9 15 0 6 0 2 5 0 5 1

It is easy to see that each Bk is the block set of an O L P M T S(16, 4) on X \{k} with the hole S\{k} for 12 ≤ k ≤ 16. After checking that all block sets in the collection {Bk : k ∈ Z 17 } are pairwise disjoint, we obtain an O L P M T S(16, 4). By inputting design O L P M T S(4), which exists by Lemma 1.1, into O L P M T S(16, 4), we can obtain an O L P M T S(16).  Theorem 4.1 There exists an O L P M T S(12k + 4) for any integer k ≥ 3. Proof There exists a 2-F G(3, ({3, 5}, {3, 5}, {4, 6}), 2k ) for k ≥ 3 by Lemma 2.1(3). For i ∈ {3, 5}, there exist a 2-P P MC S(6i : 0) by Lemma 4.1, a 1-P P M G D D(6i : 5) by Lemma 3.2(3) and a P M F(6i+1 ) by [9]. Then we can get a 1-P P MC S(12k : 5) for k ≥ 3 by Construction 2.3. By Lemma 4.2, there exist an O L P M T S(16, 4) and an O L P M T S(16), then the conclusion follows by Construction 2.1. 

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Lemma 4.3 There exists an O L P M T S(40, 6). Proof By the construction of an O L P M T S(12k + 5) in Theorem 4.1, there exists an O L P M T S(40, 16). If inputting design O L P M T S(16, 6) into O L P M T S(40, 16), then we can get an O L P M T S(40, 6). The desired O L P M T S(16, 6) is constructed as follows: The point set is X = Z 10 ∪{10, 11, . . . , 16}, the hole is S = {10, 11, . . . , 16}. By the definition, such a design should contain ten P M T S(16)s and seven P M T S(16, 6)s. The required block sets are described below. Let B0 be the set consisting of the follow 80 blocks: 123 1 12 10 295 379 4 5 14 576 6 9 14 7 16 8

135 1 13 14 2 10 12 385 4 6 12 5 9 15 6 10 15 8 12 14

142 1 14 15 2 11 10 394 4 9 10 5 10 14 6 11 9 8 13 9

156 1 15 16 2 12 11 3 10 13 4 10 5 5 11 13 6 14 12 8 14 13

164 1 16 13 2 13 15 3 11 14 4 11 16 5 12 7 6 16 10 8 15 10

1 7 12 247 2 14 16 3 12 16 4 12 13 5 13 16 7 8 10 8 16 12

189 258 2 15 14 3 13 12 4 13 11 5 15 12 7 10 16 9 11 12

197 2 6 13 2 16 9 3 14 10 4 14 7 5 16 11 7 11 15 9 12 15

1 10 11 273 348 3 15 11 4 15 8 6 7 13 7 14 11 9 13 10

1 11 8 286 3 6 15 3 16 6 4 16 15 6 8 11 7 15 13 9 16 14

Then we can check that B0 forms the block set of a P M T S(16) on X \{0}. For k ∈ Z 10 , define Bk = {B + k : B ∈ B0 }. Clearly, each Bk is the block set of a P M T S(16) on X \{k}. An O L P M T S(16, 6) also contains seven P M T S(16, 6)s. For 10 ≤ k ≤ 16, let Bk = {B + i : i ∈ Z 10 , B ∈ Ak }, we list the blocks in Ak below. A10 A11 A12 A13 A14 A15 A16

: : : : : : :

11 0 8 10 0 7 10 0 3 10 0 6 10 0 4 10 0 8 10 0 2

12 0 4 12 0 5 11 0 5 11 0 1 11 0 9 11 0 6 11 0 4

13 0 6 13 0 2 13 0 8 12 0 3 12 0 7 12 0 1 12 0 9

14 0 9 14 0 4 14 0 6 14 0 5 13 0 5 13 0 3 13 0 7

15 0 7 15 0 1 15 0 9 15 0 2 15 0 8 14 0 2 14 0 8

16 0 1 16 0 6 16 0 4 16 0 8 16 0 2 16 0 5 15 0 5

027 032 018 076 034 096 014

It is easy to see that each Bk is the block set of an O L P M T S(16, 6) on X \{k} with the hole S\{k} for 10 ≤ k ≤ 16. After checking that all block sets in the collection  {Bk : k ∈ Z 17 } are pairwise disjoint, we obtain an O L P M T S(16, 6). Theorem 4.2 There exists an O L P M T S(142). Proof Firstly, we construct a 1-P P MC S(344 : 7). There exists a 4- f an G D D(3, (4, 4, 4, 4, 4), 174 ) by Lemma 2.1(6) and a P M F(24 ) by Lemma 3.2(1). If there exists a P P M G D D ∗ (24 : 1), then we can get a P P M G D D ∗ (344 : 7) by Construction 2.6. A 1-P P MC S(344 : 7) is also obtained by Construction 2.2(2). The desired P P M G D D ∗ (24 : 1) is constructed as follows:

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The point set is Z 8 ∪ {∞}, group set is G = {G i : i ∈ Z 4 } ∪ {{∞}}, where G i = {i, i + 4}. For each G i ∈ G, we construct two P M G D D(33 21 )s with the group set (G\{G i }) ∪ {{∞}}. A0 : 1 2 3 1 7 2 1 6 7 2 5 3 3 5 6 5 7 6 3 ∞ 1 1 ∞ 6 7 ∞ 2 2 ∞ 5 6 ∞ 3 5 ∞ 7 A1 : 0 2 3 0 3 6 0 7 2 2 7 4 3 4 6 4 7 6 6 ∞ 0 0 ∞ 7 4 ∞ 2 2 ∞ 3 3 ∞ 4 7 ∞ 6

Then it can be checked that each Ak (k ∈ Z 2 ) is the block set of a P M G D D(23 11 ) with the group set (G\{G i }) ∪ {{∞}}, k ∈ G i . For each j ∈ Z 4 , k ∈ Z 2 , define A2 j+k = {B+2 j : B ∈ Ak }. Clearly, each A2 j+k is the block set of a P M G D D(23 11 ) with the group set (G\{G i }) ∪ {{∞}}, 2 j + k ∈ G i . Let B = {0, 1, 2 + 2i, 0, 3, 6 + 2i, 1, 6, 3 + 2i, 1, 0, 7 + 2i, i ∈ Z 4 } It is easy to see that B is the block set of a P M G D D(24 ) with the group set G. Finally, we can check that the collection {Ak : k ∈ Z 8 } ∪ {B} contains nine disjoint block sets and hence forms the desired P P M G D D ∗ (24 : 1). By the construction above, there exists a 1-P P MC S(344 : 7). An O L P M T S(40, 6) and an O L P M T S(40) also exist by Lemma 4.3 and Theorem 4.1, so we can get an O L P M T S(142) by Construction 2.1. 

5 Existence of O L P M T S(12k + 6)s In this section, we establish the existence of O L P M T S(12k + 6)s for any positive integer k by the constructions in Sect. 2. Lemma 5.1 There exists a 2-P P MC S(63 : 2). Proof Let (X, G, B1 , B2 , BT ) be a 2-F G(3, ({3}, {3}, {4}), 23 ), which exists by Lemma 2.1(1), where G = {G i : i ∈ Z 3 }. We construct the desired design on X  = (X × Z 3 ) ∪ {∞1 , ∞2 }. Below, we give three corresponding designs for the blocks in B1 , B2 and BT , respectively. In the constructions (1) and (2), for given three elements ri ∈ G i , i ∈ Z 3 , the desired design on the set ({r0 , r1 , r2 } × Z 3 ) ∪ {∞1 , ∞2 } is transformed to one on the set F9 ∪ {∞1 , ∞2 }. Here, F9 is a finite field, g is its primitive element and g 2 = 1 + 2g. The elements 0, g 5 , g 1 , g 0 , g 2 , g 7 , g 3 , g 6 , g 4 in F9 represent (r0 , 0), (r0 , 1), (r0 , 2), (r1 , 0), (r1 , 1), (r1 , 2), (r2 , 0), (r2 , 1), (r2 , 2) respectively. (1) For each block B ∈ B1 , take the point set ({r0 , r1 , r2 } × Z 3 ) ∪ {∞1 , ∞2 }. Construct the following families on F9 ∪ {∞1 , ∞2 }. A B (0) : g 0 g 3 g 2 g 0 g 6 g 3 g 0 g 7 g 6 g 2 g 3 g 4 g 2 g 4 g 7 g 4 g 6 g 7 g 3 g 1 g 7 g 0 g 1 g 4 g 4 ∞2 g 0 g 6 ∞2 g 2 g 2 ∞1 g 6 g 7 ∞1 g 3

Replace the elements in F9 by the elements in {r0 , r1 , r2 } × Z 3 . It is not difficult to verify that each A B (r0 , 0) contains the following pairs: ((r1 , i), (r2 , j)), ((r2 , j), (r1 , i)), where i, j ∈ Z 3 , k ∈ {1, 2}.

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((r0 , 2), (rk , 2)), ((rk , 0), (r0 , 2)); (∞1 , (r2 , k −1)), (∞2 , (r1 , k −1)); ((r2 , k), ∞2 ); ((r1 , k), ∞1 ), ((rk , j + 1), (rk , j)).

⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭

1095

These pairs are complemental with those in D B (0).

Then, we construct two P M G D D(33 )s. Let H B = {(0, g 0 , g 4 ), (0, g 2 , g 6 ), (0, g 7 , g 3 ), (0, g 3 , g 0 )}. C B0 = {B + x : B ∈ H B , x ∈ F9 }, C B1 = {c, b, a : a, b, c ∈ C B0 }. Note the underlined blocks in C B generate only three different blocks. It is also easy to verify that each C iB (i ∈ Z 2 ) forms a P M G D D(33 ) with the groups {{ri } × Z 3 : i ∈ Z 3 }. Let A B (x) = A B (0) + x, x ∈ F9 , then we can get all A B (ri , j), where i, j ∈ Z 3 . (2) For each block B = {r0 , r1 , r2 } ∈ B2 , take the point set {r0 , r1 , r2 } × Z 3 . Construct the following families on F9 ∪ {∞1 , ∞2 }. D B (0) : g 0 g 2 g 6 g 0 g 4 g 7 g 1 g 2 g 7 g 1 g 6 g 4 g 2 g 3 g 6 g 3 g 7 g 4 g 0 g 6 g 1 g 1 g 3 g 2 g 4 g 2 ∞2 g 3 g 0 ∞2 g 6 g 7 ∞2 g 7 g 3 ∞2 g 2 g 4 ∞1 g 0 g 3 ∞1 g 7 g 6 ∞1 g 4 g 0 ∞1

Replace the elements in F9 by the elements in {r0 , r1 , r2 } × Z 3 . It is not difficult to verify that each D B (r0 , 0) contains the following pairs: ((r1 , i), (r2 , j)), ((r2 , j), (r1 , i)), where i, j ∈ Z 3 , k ∈ {1, 2}. ⎫ ((r0 , 2), (rk , 0)), ((r0 , 2), (rk , 1)), ((rk , 1), (r0 , 2)) ((rk , 2), (r0 , 2)); ⎪ ⎪ ⎬ (∞1 , (r2 , 2)), (∞2 , (r2 , i)), (∞2 , (r1 , 2)); (∞1 , (r1 , i)), ⇒ ((r1 , 0), ∞1 ), ((r1 , i), ∞2 ), ((r2 , 0), ∞2 ); ⎪ ((r2 , i), ∞1 ), ⎪ ⎭ ((rk , j), (rk , j + 1)). These pairs are complemental with those in A B (0). (3) For each B ∈ BT , construct a P M F(34 ) on B × Z 3 with the groups  B = {G x : x ∈ B} where G x = {x} × Z 3 . Such a design exists by [9]. Its block set E B can be partitioned into 12 disjoint families E B (x, i), each of which forms a P M G D D(33 ) on (B\{x}) × Z 3 with the groups  B \G x , where (x, i) ∈ B × Z 3 . Final, for each x ∈ X, i ∈ Z 3 , j ∈ Z 2 , define ⎛



F(x, i) = ⎝

x∈B∈B1

F (∞) = j



⎞ A B (x, i)⎠



⎛ ⎝

 x∈B∈B2

⎞ D B (x, i)⎠



⎛ ⎝



⎞ E B (x, i)⎠ ,

x∈B∈BT

j CB .

B∈B1

For any x ∈ G k ∈ G, i, k ∈ Z 3 , j ∈ Z 2 , it is not difficult to verify that all F(x, i) and F j (∞) are mutually disjoint, each F(x, i) forms a P M G D D(61 112 ) on X  \{(x, i), (x, i +1)} with the long group {((G k × Z 3 )\{(x, i), (x, i +1)})∪{∞1 , ∞2 } and each F j (∞) forms a P M G D D(63 ) with the groups {G k × Z 3 }. Therefore, these  designs form the desired 2-P P MC S(63 : 2).

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Lemma 5.2 There exists an O L P M T S(18, 6). Proof Firstly, we construct an O L P M T S(18, 6) on X = Z 12 ∪ {12, 13, . . . , 18} with a hole S = {12, 13, . . . , 18}. By the definition, such a design should contain twelve P M T S(18)s and seven P M T S(18, 6)s. The required block sets are described below. Let B0 be the set consisting of the follow 102 blocks: 124 1 12 13 2 7 10 2 17 6 3 14 10 4 10 17 5 8 10 6 10 16 7 13 16 9 11 14 11 15 12

136 1 13 11 2 8 11 2 18 17 3 15 17 4 12 16 598 6 11 13 7 14 18 9 13 10 11 18 16

143 1 14 15 2 9 12 3 4 11 3 16 12 4 13 9 5 10 12 6 12 7 7 15 14 9 15 11

152 1 15 16 2 10 15 3 5 18 3 17 13 4 14 6 5 11 7 6 13 18 7 17 9 9 16 15

165 1 16 17 2 11 4 3 8 16 3 18 9 4 15 10 5 12 17 6 14 12 7 18 13 9 17 16

179 1 17 18 2 12 8 396 4 5 13 4 16 14 5 14 11 6 16 8 8 9 14 9 18 12

187 1 18 14 2 13 14 3 10 7 467 4 17 12 5 15 18 6 17 11 8 12 18 10 11 16

1 9 10 237 2 14 16 3 11 8 478 4 18 15 5 16 13 6 18 10 8 13 17 10 13 12

1 10 8 253 2 15 13 3 12 14 4 8 18 5 6 15 5 17 14 7 11 17 8 14 13 10 14 17

1 11 12 269 2 16 18 3 13 15 495 5 7 16 6 8 15 7 12 15 8 17 15 10 18 11

We can check that B0 is the block set of a P M T S(18) on X \{0}. For each k ∈ Z 12 , define Bk = {B + k : B ∈ B0 }. Clearly, each Bk is the block set of a P M T S(18) on X \{k}. An O L P M T S(18, 6) also contain seven P M T S(18, 6)s. For 12 ≤ k ≤ 18, let Bk = {a, b+i, c+i : i ∈ Z 12 , if  a, b, c belongs the not underlined block in Ak ; a, b + 3i, c + 3i : i ∈ Z 4 , if a, b, c belongs the underlined block in Ak }. we list the blocks in Ak below. A12 : 13 0 9 A13 : 12 0 1 A14 : 12 0 9 1 6 11 A15 : 12 0 8 0 5 10 A16 : 15 0 9 14 1 9 A17 : 15 0 11 14 0 8 A18 : 15 0 7 14 2 10

14 0 11 14 0 7 13 0 4 210 13 0 6 1 0 11 17 0 1 14 0 4 16 0 7 14 2 6 16 0 11 14 1 5

15 0 5 15 0 4 15 0 3 183 14 0 9 072 18 0 5 14 2 5 18 0 9 14 1 4 17 0 9 14 0 3

16 0 8 16 0 9 16 0 6 234 16 0 3 123 0 11 6 246 017 135 016 024

17 0 3 18 0 3 17 2 7 084 17 1 6 048 12 0 2 2 0 10 12 2 4 1 11 9 12 1 3 0 10 8

046 18 1 8 18 0 10 18 2 3 0 10 6 17 0 5 17 1 0 17 2 4 17 0 11 17 1 3 18 0 7 18 2 0

294 345 270 321 18 1 2

17 2 1

17 0 2 18 2 9

18 1 11 18 0 1

12 2 6

12 1 4 13 1 11 13 2 10 13 0 3

12 1 5

12 0 3 13 0 10 13 1 9

13 2 5

12 0 4

12 2 5 13 2 0

13 1 4

13 0 8

It can be checked that each Bk is the block set of a P M T S(18, 6) on X \{k} with the hole S\{k} for 12 ≤ k ≤ 18. Finally, we can check that the collection {Bk : k ∈ Z 19 } contains 19 disjoint block sets and hence forms the desired O L P M T S(18, 6).  Lemma 5.3 There exists an O L P M T S(18). Proof First we construct an O L P M T S(18, 7) on X = Z 11 ∪ {11, 12, . . . , 18} with a hole S = {10, 11, . . . , 18}. By the definition, such a design should contain eleven P M T S(18)s and eight P M T S(18, 7)s. The required block sets are described below.

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Let B0 be the set consisting of the follow 102 blocks: 123 2 10 7 5 9 10 13 10 1 17 4 7 11 16 8 12 13 3 17 12 8 17 13 9 15 16 7 17 18 5

169 357 6 10 8 14 1 10 17 7 6 11 17 9 12 14 9 18 12 3 18 13 4 15 17 8 18 17 1

178 368 11 1 3 14 10 6 17 6 4 11 18 10 12 15 7 13 14 4 14 15 5 15 18 1

185 3 7 10 11 3 4 14 6 1 18 6 7 12 11 6 12 16 10 13 15 8 14 16 4 16 15 1

197 389 11 4 1 15 2 4 18 7 9 13 11 7 12 17 5 13 16 6 14 17 3 17 15 3

256 3 10 4 12 1 4 15 4 9 18 9 6 14 11 9 12 18 8 13 17 2 14 18 2 18 15 10

263 458 12 4 2 15 9 2 11 12 5 15 11 5 13 12 9 13 18 3 15 14 3 16 17 10

275 465 12 2 1 16 3 9 11 13 2 16 11 2 14 12 7 14 13 8 16 14 8 16 18 4

2 8 10 487 13 1 5 16 9 5 11 14 7 17 11 10 15 12 10 15 13 6 17 14 2 17 16 1

298 4 10 9 13 5 10 16 5 3 11 15 6 18 11 8 16 12 6 16 13 7 18 14 5 18 16 2

Then we can check that B0 forms the block set of a P M T S(18) on X \{0}. For k ∈ Z 11 , define Bk = {B + k : B ∈ B0 }. Clearly, each Bk is the block set of a P M T S(18) on X \{k}. An O L P M T S(18, 7) also contains eight P M T S(18, 7)s. For 11 ≤ k ≤ 18, let Bk = {B + i : i ∈ Z 11 , B ∈ Ak }, we list the blocks in Ak below. A11 A12 A13 A14 A15 A16 A17 A18

: : : : : : : :

1 10 6 395 249 1 5 10 476 679 134 142

12 0 2 11 0 10 11 0 7 11 0 6 11 0 5 11 0 9 11 0 3 11 0 4

13 0 10 13 0 8 12 0 6 12 0 8 12 0 1 12 0 5 12 0 4 12 0 7

14 0 3 14 0 1 14 0 10 13 0 9 13 0 7 13 0 3 13 0 6 13 0 1

15 0 1 15 0 3 15 0 9 15 0 10 14 0 8 14 0 4 14 0 5 14 0 2

16 0 8 16 0 4 16 0 1 16 0 3 16 0 2 15 0 6 15 0 7 15 0 8

17 0 5 17 0 2 17 0 8 17 0 1 17 0 4 17 0 7 16 0 10 16 0 5

18 0 4 18 0 5 18 0 3 18 0 7 18 0 6 18 0 10 18 0 9 17 0 6

It is easy to see that each Bk is the block set of an O L P M T S(18, 7) on X \{k} with the hole S\{k}, 11 ≤ k ≤ 18. After checking that the 19 block sets in the collection {Bk : k ∈ Z 19 } are pairwise disjoint, we obtain an O L P M T S(18, 7). By inputting design O L P M T S(7), which exists by Lemma 1.1, into O L P M T S(18, 7), we can obtain an O L P M T S(18).  Lemma 5.4 There exists an O L P M T S(30). Proof First we construct an O L P M T S(30, 13) on X = F17 ∪{17, 18, . . . , 30} with a hole S = {17, 18, . . . , 30}, where g = 3 is a primitive element of the finite field F17 . By the definition, such a design should contain 17 P M T S(30)s and 14 P M T S(30, 13)s. Let B0 be the set consisting of the follow three parts of blocks, where the element g a of F17 is briefly denoted by its index a: Part 1: 0, 2, 10 + 2 j, 14, 3, 11 + 2 j, 10, 8, 11 + 2 j, 12, 13, 8 + 2 j, 6, 13, 15 + 2 j, 9, 7, 3 + 2 j, 8, 4, 3 + 2 j, 11, 12, 7 + 2 j, 1, 14, 8, 2, 15, 9, where a, b, c + 2 j = a + 2 j, b + 2 j, c + 2 j (mod 16), j ∈ Z 8 . Part 2: 17+i, ai , i, 17+i, i, bi , 17+i, bi , ai , where ai = 3+i (mod 16), bi = 10 + i (mod 16), 0 ≤ i ≤ 13.

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Part 3: 17 18 2 17 28 13 25 17 12 18 23 8 21 18 5 19 20 0 19 30 14 29 19 7 20 29 8 29 20 5 21 30 10 22 23 14 25 22 1 23 28 3 24 25 12 29 24 2 29 25 4 27 28 0 29 30 9

17 19 1 17 29 14 26 17 13 18 24 6 22 18 6 19 21 1 20 19 4 30 19 8 20 30 11 30 20 12 22 21 11 22 24 12 26 22 2 23 29 4 24 26 5 30 24 3 30 25 5 27 29 2 30 29 1

17 20 4 17 30 15 27 17 9 18 25 10 23 18 7 19 22 9 21 19 3 20 21 2 21 20 1 21 22 0 23 21 13 22 25 13 27 22 3 23 30 5 24 27 3 25 26 4 26 27 11 27 30 8

17 21 5 18 17 5 28 17 15 18 26 13 24 18 8 19 23 10 22 19 10 20 22 10 22 20 9 21 23 12 24 21 15 22 26 0 28 22 4 24 23 13 24 28 4 25 27 5 26 28 7 28 27 9

17 22 6 19 17 4 29 17 11 18 27 14 25 18 9 19 24 11 23 19 11 20 23 5 23 20 10 21 24 13 25 21 0 22 27 1 29 22 13 25 23 15 24 29 0 25 28 6 26 29 5 29 27 0

17 23 7 20 17 1 30 17 14 18 28 15 26 18 10 19 25 6 24 19 9 20 24 9 24 20 11 21 25 15 26 21 8 22 28 2 30 22 11 26 23 1 24 30 2 25 29 10 26 30 6 30 27 6

17 24 8 21 17 2 18 19 0 18 29 9 27 18 12 19 26 7 25 19 13 20 25 14 25 20 7 21 26 6 27 21 6 22 29 3 23 24 15 27 23 11 25 24 14 25 30 3 27 26 1 28 29 7

17 25 9 22 17 7 18 20 2 18 30 12 28 18 13 19 27 15 26 19 15 20 26 15 26 20 14 21 27 8 28 21 12 22 30 4 23 25 1 28 23 2 26 24 4 26 25 0 28 26 10 28 30 1

17 26 11 23 17 8 18 21 3 19 18 3 29 18 14 19 28 8 27 19 14 20 27 7 27 20 15 21 28 9 29 21 10 23 22 12 23 26 14 29 23 3 27 24 5 27 25 7 29 26 8 29 28 1

17 27 12 24 17 6 18 22 7 20 18 0 30 18 15 19 29 13 28 19 6 20 28 12 28 20 8 21 29 11 30 21 9 24 22 14 23 27 2 30 23 4 28 24 0 28 25 3 30 26 2 30 28 10

Then we can check that B0 forms the block set of a P M T S(30) on X \{0}. For k ∈ F17 , define Bk = {B + k : B ∈ B0 }. Clearly, each Bk is the block set of a P M T S(30) on X \{k}. An O L P M T S(30, 13) also contains fourteen P M T S(30, 13)s. For 17 ≤ k ≤ 30, let Bk = {B + i : i ∈ F17 , B ∈ Ak }, we list the blocks in Ak below. A17 : g 0 g 13 g 7 24 0 g 6 A18 : g 15 g 12 g 6 24 0 g 0 A19 : g 14 g 11 g 5 24 0 g 4 A20 : g 13 g 10 g 4 24 0 g 13 A21 : g 12 g 9 g 3 24 0 g 15 A22 : g 11 g 8 g 2 24 0 g 14 A23 : g 10 g 7 g 1 24 0 g 3 A24 : g 9 g 6 g 0 23 0 g 12 A25 : g 8 g 5 g 15 23 0 g 11 A26 : g 7 g 4 g 14 23 0 g 10

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18 0 g 14 25 0 g 4 17 0 g 13 25 0 g 5 17 0 g 12 25 0 g 1 17 0 g 15 25 0 g 7 17 0 g 9 25 0 g 0 17 0 g 8 25 0 g 15 17 0 g 7 25 0 g 14 17 0 g 6 25 0 g 13 17 0 g 5 24 0 g 12 17 0 g 4 24 0 g 11

19 0 g 15 26 0 g 2 19 0 g 14 26 0 g 1 18 0 g 13 26 0 g 8 18 0 g 0 26 0 g 6 18 0 g 10 26 0 g 5 18 0 g 9 26 0 g 0 18 0 g 8 26 0 g 15 18 0 g 7 26 0 g 14 18 0 g 6 26 0 g 13 18 0 g 5 25 0 g 12

20 0 g 0 27 0 g 3 20 0 g 2 27 0 g 11 20 0 g 15 27 0 g 10 19 0 g 11 27 0 g 9 19 0 g 1 27 0 g 2 19 0 g 10 27 0 g 7 19 0 g 9 27 0 g 6 19 0 g 8 27 0 g 15 19 0 g 7 27 0 g 1 19 0 g 6 27 0 g 0

21 0 g 13 28 0 g 10 21 0 g 3 28 0 g 8 21 0 g 0 28 0 g 7 21 0 g 1 28 0 g 12 20 0 g 12 28 0 g 11 20 0 g 11 28 0 g 4 20 0 g 10 28 0 g 0 20 0 g 9 28 0 g 3 20 0 g 8 28 0 g 2 20 0 g 7 28 0 g 1

22 0 g 1 29 0 g 12 22 0 g 4 29 0 g 9 22 0 g 14 29 0 g 11 22 0 g 2 29 0 g 14 22 0 g 13 29 0 g 8 21 0 g 12 29 0 g 5 21 0 g 11 29 0 g 13 21 0 g 10 29 0 g 2 21 0 g 9 29 0 g 4 21 0 g 8 29 0 g 3

23 0 g 5 30 0 g 9 23 0 g 15 30 0 g 12 23 0 g 2 30 0 g 3 23 0 g 3 30 0 g 10 23 0 g 14 30 0 g 6 23 0 g 13 30 0 g 1 22 0 g 12 30 0 g 4 22 0 g 11 30 0 g 5 22 0 g 10 30 0 g 14 22 0 g 9 30 0 g 13

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A27 : g 6 g 3 g 13 23 0 g 9 A28 : g 5 g 2 g 12 23 0 g 8 A29 : g 4 g 1 g 11 23 0 g 7 A30 : g 3 g 0 g 10 23 0 g 6

17 0 g 3 24 0 g 10 17 0 g 2 24 0 g 9 17 0 g 1 24 0 g 8 17 0 g 0 24 0 g 7

18 0 g 4 25 0 g 11 18 0 g 3 25 0 g 10 18 0 g 2 25 0 g 9 18 0 g 1 25 0 g 8

1099

19 0 g 5 26 0 g 12 19 0 g 4 26 0 g 11 19 0 g 3 26 0 g 10 19 0 g 2 26 0 g 9

20 0 g 6 28 0 g 15 20 0 g 5 27 0 g 14 20 0 g 4 27 0 g 13 20 0 g 3 27 0 g 12

21 0 g 7 29 0 g 0 21 0 g 6 29 0 g 1 21 0 g 5 28 0 g 14 21 0 g 4 28 0 g 13

22 0 g 8 30 0 g 2 22 0 g 7 30 0 g 15 22 0 g 6 30 0 g 0 22 0 g 5 29 0 g 15

It is easy to see that each Bk is the block set of an O L P M T S(30, 13) on X \{k} with the hole S\{k}, 17 ≤ k ≤ 30. After checking that the 31 block sets in the collection {Bk : k ∈ Z 31 } are pairwise disjoint, we obtain an O L P M T S(30, 13). By inputting design O L P M T S(13), which exists by Lemma 1.1, into O L P M T S(30, 13), we can obtain an O L P M T S(30).  Theorem 5.1 There exists an O L P M T S(12k + 6) for any positive integer k. Proof For k ∈ {1, 2}, there exists an O L P M T S(12k + 6) by Lemmas 5.3 and 5.4; For k ≥ 3, there exist a 2-F G(3, (3, 3, 4), 2k ) for k ≡ 0, 1 (mod 3) (k ≥ 3) and a 2-F G(3, (3, 3, {4, 6}), 2k−2 41 ) for k ≡ 2 (mod 3) (k ≥ 5) by Lemma 2.1(1). A 2-P P MC S(63 : 2) and a P P M G D D(63 : 5) exist by Lemmas 5.1 and [9], so there exists a 1-P P MC S(12k : 7) or a 1-P P MC S(12k−2 241 : 7) by Construction 2.3. Since an O L P M T S(18), an O L P M T S(18, 6) and an O L P M T S(30) exist by Lemmas 5.2–5.4, there exists an O L P M T S(12k + 6) for k ≥ 3 by Construction 2.1. 

6 Conclusion Combining Theorems 3.1, 3.2, 4.2 and 5.1, we obtain the existence of O L P M T S(v)s for v = 6, 10 (mod 12). Theorem 6.1 There exists an O L P M T S(v) for v = 6, 10 (mod 12) and v = 6. Finally, by Lemmas 1.1 and 1.2, we get the existence spectrum for overlarge sets of pure Mendelsohn triple systems. Theorem 6.2 There exists an O L P M T S(v) if and only if v = 0, 1 (mod 3), v > 3 and v = 6.

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