the existence of mixed orthogonal arrays with four and five factors of strength two

The Existence of Mixed OrthogonalArrays with Four and Five Factorsof Strength Two

Guangzhou Chen,1 Lijun Ji,2 and Jianguo Lei11College of Mathematics and Information Science, Hebei Normal University,Shijiazhuang 050024, P.R. China, E-mail: [email protected];[email protected]

2Department of Mathematics, Soochow University, Suzhou 215006, P.R. China,E-mail: [email protected]

Received December 13, 2012; revised April 14, 2013

Published online 8 May 2013 in Wiley Online Library (wileyonlinelibrary.com).DOI 10.1002/jcd.21350

Abstract: Symmetric orthogonal arrays and mixed orthogonal arrays are useful in the de-sign of various experiments. They are also a fundamental tool in the construction of variouscombinatorial configurations. In this paper, we investigated the mixed orthogonal arrays withfour and five factors of strength two, and proved that the necessary conditions of such mixedorthogonal arrays are also sufficient with several exceptions and one possible exception. C© 2013Wiley Periodicals, Inc. J. Combin. Designs 22: 323–342, 2014

Keywords: orthogonal arrays; mixed; asymmetric; α-resolvable; transversal designs; Latin squares

1. INTRODUCTION

Orthogonal arrays have been studied extensively, a great deal of methods and results canbe found in the two monographs [7, 11] and the Handbook [6]. Mixed (or asymmetric)orthogonal arrays, introduced by Rao [13], have received much attention in recent years.These arrays play an important role in experimental designs as universally optimalfractions of asymmetric factorials (see Cheng [5] and Mukerjee [12]). Mixed orthogonal

Corresponding author: [email protected] grant sponsor: Natural Science Foundation of China; Contract grant numbers: 61071221; 11271004(J. L.); 11222113 (L. J.); Contract grant sponsor: Youth Science Foundation of China; Contract grant number:11201114 (G. C.).

Journal of Combinatorial DesignsC© 2013 Wiley Periodicals, Inc. 323

324 CHEN, JI, AND LEI

arrays have been used extensively in industrial experiments for quality improvement, andtheir use in other experimental situations has also been widespread.

A mixed (or asymmetric) orthogonal array MOA(N ; s1s2 · · · sk, t) is an array of sizek × N , in which the i-th row have symbols from Si = {0, 1, . . . , si − 1}, 1 ≤ i ≤ k, withthe property that in any t × N subarray, every possible t-tuple occurs an equal number oftimes as a column. In an MOA(N ; s1s2 · · · sk, t), the columns are referred to as runs, soN is the number of runs of the MOA, also called its size. The set Si, 1 ≤ i ≤ k, is calleda factor, so k is the number of factors. Moreover, si is called the level of the i-th factorand s1s2 · · · sk is called type. The parameter t is called the strength.

Note that it is not required that s1, s2, . . . , sk be distinct. Normally we combine termswith equal sl’s, so that 222332 would be replaced by 2532. Note that permuting the rowsdoes not affect the orthogonality of an MOA. Consequently, throughout this paper, werestrict our attention to those types s1s2 · · · sk , where s1 ≤ s2 ≤ · · · ≤ sk . It is easy tosee that the existence of an MOA(N ; 1s2 · · · sk, t) is equivalent to the existence of anMOA(N ; s2 · · · sk, t). For convenience, the MOAs considered here are assume to have nofactor at one level.

Clearly, the run size N of an MOA(N ; s1s2 · · · sk, t) is divisible by the least commonmultiple of all integers

∏i∈I si , I ⊂ {1, 2, . . . , k}, and |I | = t . An MOA(N ; s1s2 · · · sk, t)

is called trivial if every possible k-tuple occurs an equal number of times as a column.A greatest common divisor of integers v1, v2, . . . , vn is denoted by g.c.d. {v1,

v2, . . . , vn}, a least common multiple of integers v1, v2, . . . , vn is denoted by l.c.m.{v1, v2, . . . , vn}. If an integer a is divisible by an integer b, then it is denoted by b|a.

The existence of MOA with two and three factors of strength two was given as anexercise in [11].

Theorem 1.1. Let s1, s2, s3 be positive integers. Then

(1) An MOA(N ; s1s2, 2) exists if and only if s1s2|N .(2) An MOA(N ; s1s2s3, 2) exists if and only if M|N , where M = l.c.m. {s1s2,

s1s3, s2s3}.The problem of the existence of MOA with 4, 5, 6, . . . factors of strength two was posed

by Hedayat, Sloane, and Stufken [11]. In this paper, we will investigate the existence ofMOAs with four and five factors of strength two. We obtain the following results.

Theorem 1.2. Let a, b, c, d be integers and greater than 1. There is anMOA(N ; abcd, 2) of type abcd �∈ {24, 64} ∪ {23d1 : d ≡ 2 (mod 4)} if and only if M|N ,while there is an MOA(N ; abcd, 2) of type abcd ∈ {24, 64} ∪ {23d1 : d ≡ 2 (mod 4)}if and only if M|N and N > M , where M = l.c.m. {αβ : {α, β} ⊂ {a, b, c, d}}.

Theorem 1.3. Let a, b, c, d, e be integers and greater than 1. There is anMOA(N ; abcde, 2) of type abcde �∈ L5 = {25, 35, 6421, 6431, 65, 105} ∪ {24g1 : g ≡ 2(mod 4)} if and only if M|N , while there is an MOA(N ; abcde, 2) of type abcde ∈L5 \ {105} if and only if M|N and N > M , and there is an MOA(100u; 105, 2) for u > 1,where M = l.c.m. {αβ : {α, β} ⊂ {a, b, c, d, e}}.

The paper is organized as follows. In Section 2, we will give some recursive con-structions and some known results. The proofs of Theorems 1.2 and 1.3 will be given inSections 3 and 4, respectively.

Journal of Combinatorial Designs DOI 10.1002/jcd

THE EXISTENCE OF MIXED ORTHOGONAL ARRAYS 325

2. PRELIMINARIES

In this section, we will give some recursive constructions and some known results. Thefollowing three constructions are simple but very useful, the readers are referred to [4,11]and the references therein.

Construction 2.1 ([4]). If there is an MOA(N ; a1a2 · · · ak, t), then there is anMOA(nN; a1a2 · · · (nak), t), where n is a positive integer.

Construction 2.2 ([4, 11]). If there is an MOA (N ′; a1a2 · · · ak, t), an MOA(N ′′; b1a2 · · · ak, t) and N ′/a1 = N ′′/b1, then there is an MOA(N ′ + N ′′; (a1 +b1)a2 · · · ak, t).

Construction 2.3 ([4]). If there is an MOA(N ; ba3 · · · ak, t) and b = a1a2, then thereis an MOA(N ; a1a2 · · · ak, t).

Lemma 2.4. Suppose that k − 1 ≥ t , b = a1a2, and g.c.d. {a1, a2} = 1. If there is anMOA(N; ba3 · · · ak , t), where N = l.c.m. {∏t

i=1 xi : {x1, x2, . . . , xt } ⊂ {b, a3, . . . , ak}},then there is an MOA(N; a1a2 · · · ak , t), where N = l.c.m. {∏t

i=1 yi : {y1, y2, . . . , yt } ⊂{a1, a2, a3, . . . , ak}}.Proof. By applying Construction 2.3, we need only to show that

l.c.m.

{t∏

i=1

yi : {y1, y2, . . . , yt } ⊂ {a1, a2, a3, . . . , ak}}

= l.c.m.

{t∏

i=1

xi : {x1, x2, . . . , xt } ⊂ {b, a3, . . . , ak}}

.

Since

l.c.m.

{t∏

i=1

yi : {y1, y2, . . . , yt } ⊂ {a1, a2, a3, . . . , ak}}

= l.c.m.

{a1a2

t−2∏i=1

yi, a1

t−1∏i=1

yi, a2

t−1∏i=1

yi,

t∏i=1

yi : {y1, y2, . . . , yt } ⊂ {a3, . . . , ak}}

= l.c.m.

{a1a2

t−2∏i=1

yi, a1a2

t−1∏i=1

yi,

t∏i=1

yi : {y1, y2, . . . , yt } ⊂ {a3, . . . , ak}}

= l.c.m.

{a1a2

t−1∏i=1

yi,

t∏i=1

yi : {y1, y2, . . . , yt } ⊂ {a3, . . . , ak}}

= l.c.m.

{b

t−1∏i=1

yi,

t∏i=1

yi : {y1, y2, . . . , yt } ⊂ {a3, . . . , ak}}

= l.c.m.

{t∏

i=1

xi : {x1, x2, . . . , xt } ⊂ {b, a3, . . . , ak}}

,

we get the conclusion. �



Construction 2.5. If there is an MOA(M1; a1a2 · · · ak, t) and an MOA (M2; a1

a2 · · · ak, t), then there is an MOA(xM1 + yM2; a1a2 · · · ak, t), where x, y are nonnega-tive integers.

Proof. Let A be an MOA(M1; a1a2 · · · ak, t) and B be an MOA(M2; a1a2 · · · ak, t)over the same point set. It is readily checked that C = (A A · · · A︸︷︷︸

x

B B · · · B︸︷︷︸y

) is an

MOA(xM1 + yM2; a1a2 · · · ak, t). �

Construction 2.6. If there is an MOA(N ; a1a2 · · · ak, t) and di |ai for 1 ≤ i ≤ k, thenthere is an MOA(N ; d1d2 · · · dk, t).

Proof. Let A = (ai,j ), 1 ≤ i ≤ k, 1 ≤ j ≤ N , be an MOA(N ; a1a2 · · · ak, t) over Za1,

Za2, . . . , Zak. For 1 ≤ i ≤ k, write ai = xidi since di |ai , and define a function fi :

Zai→ Zdi

, s �→ s/xi�, where ·� denotes the floor function. Let B = (bi,j ), wherebi,j = fi(ai,j ). It is readily checked that B is an MOA(N ; d1d2 · · · dk, t). �

The following is a product construction.

Construction 2.7. If there is an MOA(N1; a1a2 · · · ak, t) and an MOA (N2; b1

b2 · · · bk, t), then there is an MOA(N1N2; (a1b1)(a2b2) · · · (akbk), t).

Proof. Let A = (ai,j ), 1 ≤ i ≤ k, 1 ≤ j ≤ N1, be an MOA(N1; a1a2 · · · ak, t) overZa1, Za2, . . . , Zak

, and let B = (bs,t ), 1 ≤ s ≤ k, 1 ≤ t ≤ N2, be an MOA(N2; b1b2 · · ·bk, t) over Zb1, Zb2, . . . , Zbk

. Define an array C = (C1C2 · · ·CN1 ) over Za1 × Zb1, Za2 ×Zb2, . . . , Zak

× Zbkas follows:

Cm =

⎛⎜⎜⎜⎝

(a1,m, b1,1) (a1,m, b1,2) (a1,m, b1,3) · · · (a1,m, b1,N2 )(a2,m, b2,1) (a2,m, b2,2) (a2,m, b2,3) · · · (a2,m, b2,N2 )

......

......

(ak,m, bk,1) (ak,m, bk,2) (ak,m, bk,3) · · · (ak,m, bk,N2 )

⎞⎟⎟⎟⎠ ,

where 1 ≤ m ≤ N1.Take any t-tuple column vector c. Without loss of generality, assume that the t-

tuple column vector c = ((a1, b1), (a2, b2), (a3, b3), . . . , (at , bt ))T ∈ (Za1 × Zb1, Za2 ×Zb2, . . . , Zat

× Zbt)T . The number of occurrence times of column vector (b1, b2, b3, . . . ,

bt )T in B is N2/(b1b2 · · · bt ) since B is an MOA(N2; b1b2 · · · bk, t). The number ofoccurrence times of column vector (a1, a2, a3, . . . , at )T in A is N1/(a1a2 · · · at ) since A

is an MOA(N1; a1a2 · · · ak, t). Then the number of occurrence times of column vector c inC is N1/(a1a2 · · · at ) × N2/(b1b2 · · · bt ) = N1N2/(a1b1a2b2 · · · atbt ) by the constructionabove. It follows that C is an MOA(N1N2; (a1b1)(a2b2) · · · (akbk), t). �

For the results on MOAs with N ≤ 100, the readers are referred to Table 12.7 of themonograph [11] or the webpage http://www.research.att.com/ njas/oadir/ or other papers,such as [15, 16, 18]. Here, we only list a few MOAs in the following which are used inthis paper later.

Lemma 2.8 ([11, 15, 16, 18]). There is an MOA(8; 2441, 2), an MOA(12; 2431, 2),an MOA(18; 3461, 2), an MOA(20; 2451, 2), an MOA(24; 2461, 2), an MOA(27; 3491, 2),an MOA(28; 2471, 2), an MOA(36; 6223, 2), an MOA(36; 622231, 2), an MOA(36; 6233, 2),



an MOA(36; 622132, 2), an MOA(36; 632131, 2), an MOA(36; 6332, 2), anMOA(36; 6322, 2), an MOA(72; 64(12)1, 2).

Lemma 2.9 ([6, 10, 11]). There exists an MOA(gt ; gt+1, t) for any integer g ≥ 2.

It is well known that there is an approach to obtain the least common multiple of someintegers. We list it as follows.

Lemma 2.10. Suppose that m is a common multiple of integers a1, a2, . . . , ak−1 andak . Then l.c.m. {a1, a2, . . . , ak} = m

g.c.d. { ma1

, ma2

,..., mak

} .

Lemma 2.11. There is an MOA(N ; a1a2 · · · at+1, t), where N = l.c.m. {a1a2 · · ·at+1/ai : 1 ≤ i ≤ t + 1}.

Proof. Since a1a2 · · · at+1 is a common multiple of integers a1a2 · · · at+1/ai ,1 ≤ i ≤ t + 1, by Lemma 2.10, we have l.c.m. {a1a2 · · · at+1/ai : 1 ≤ i ≤ t + 1} =

a1a2···at+1

g.c.d. {a1,a2,...,at+1} .

Let g.c.d. {a1, a2, . . . , at+1} = d. If d = 1, an MOA(N ; a1a2 · · · at+1, t) always existssince N = a1a2 · · · at+1, i.e., it is a trivial design. If d ≥ 2, there is an MOA(dt ; dt+1, t)obtained from Lemma 2.9 and there is a trivial MOA(M; a1

da2d

· · · at+1

d, t), where M =

a1d

a2d

· · · at+1

d. Then we obtain an MOA(N ; a1a2 · · · at+1, t) by Construction 2.7, where

N = dtM = l.c.m. {a1a2 · · · at+1/ai : 1 ≤ i ≤ t + 1}. �

In order to state more constructions for MOAs, we need the concept of a resolvableTD. A transversal design TDλ(k, n) is a triple (X ,G,A), where X is a kn-set of points, Gis a collection of n-subsets of X (called groups) which partition X , and A is a collectionof subsets of X (called blocks), each block meeting every group in exactly one point, andsuch that any pair of points from distinct groups occurs in exactly λ blocks of A. In aTDλ(k, n), if λ is omitted it is understood to be one. A TD is said to be α-resolvable if itsblocks can be partitioned into classes (called α-resolution classes) such that each pointof design occurs in precisely α blocks in each class. We simply refer to a 1-resolvableTDλ(k, n) as a resolvable TDλ(k, n), denoted by RTDλ(k, n).

The following theorem is obvious but very useful. Its proof is omitted here.

Theorem 2.12. The existence of an α-resolvable TDλ(k, n) is equivalent to the exis-tence of an MOA(λn2; nk(λn/α)1, 2).

Lemma 2.13 ([9, 17]). There is an RTD2(4, 10) and an RTD3(4, n) for n ∈ {6, 10}.

Lemma 2.14. There is an MOA(108; 64(18)1, 2), an MOA(200; (10)4(20)1, 2), an MOA(300; (10)4(30)1, 2), an MOA(108; 64u1, 2) for any u|18 and an MOA(300; (10)4v1, 2)for any v|30.

Proof. There is an RTD2(4, 10) and an RTD3(4, n) for n ∈ {6, 10} by Lemma 2.13.We get an MOA(200; (10)4(20)1, 2), an MOA(108; 64(18)1, 2), and an MOA(300;(10)4(30)1, 2) by Theorem 2.12. Since u|18 and v|30, we get an MOA(108; 64u1, 2)and an MOA(300; (10)4v1, 2) by Construction 2.6. �

A Latin square of order n is an n × n array in which each cell contains a single symbolfrom an n-set S, such that each symbol occurs exactly once in each row and exactly



once in each column. Suppose that L1 is a Latin square of order n with entries fromX and L2 is a Latin square of order n with entries from Y . We say that L1 and L2

are orthogonal Latin squares provided that, for every x ∈ X and for every y ∈ Y , thereis a unique cell (i, j ) such that L1(i, j ) = x and L2(i, j ) = y. A set of Latin squaresL1, . . . , Lm is mutually orthogonal if for any 1 ≤ i < j ≤ m, Li and Lj are orthogonal.It is well known that a TD(k, n) is equivalent to k − 2 mutually orthogonal Latin squaresof order n. An α-transversal in a Latin square of order n is a set of nα cells, α cells fromeach column, α cells from each row, in which every symbol occur exactly α times. It iswell known that an α-parallel class of TD(k, n) is equivalent to a common α-transversalin the k − 2 mutually orthogonal Latin squares of order n. It follows that there is anα-resolvable TD(k, n) if and only if k − 2 mutually orthogonal Latin squares of size n

have n/α disjoint α-transversals.

Lemma 2.15. There is an MOA(100; (10)451, 2), an MOA(100; (10)421, 2), anMOA(100; (10)251 a1b1, 2), and an MOA(100; (10)221a1b1, 2) for any a|10, b|10.

Proof. To prove that an MOA(100; (10)451, 2) and an MOA(100; (10)421, 2) exist, byapplying Theorem 2.12, it suffices to construct an α-resolvable TD(4, 10) for α ∈ {2, 5}.

Let

A =

0 4 7 6 2 1 9 8 3 52 1 5 0 8 9 4 3 6 73 5 2 8 7 6 0 1 9 41 6 9 3 5 7 8 2 4 07 8 1 2 4 3 5 9 0 68 7 4 9 6 5 1 0 2 35 9 0 1 3 8 6 4 7 29 2 6 5 0 4 3 7 1 86 0 3 4 9 2 7 5 8 14 3 8 7 1 0 2 6 5 9

B =

0 8 6 1 7 9 2 5 4 33 1 4 6 9 0 5 2 7 81 7 2 4 0 8 9 3 5 62 9 8 3 6 4 7 1 0 59 0 5 8 4 7 1 6 3 26 2 1 7 3 5 4 8 9 08 3 7 0 5 2 6 9 1 44 5 0 9 2 3 8 7 6 15 4 9 2 1 6 3 0 8 77 6 3 5 8 1 0 4 2 9

As noted in [8] these two orthogonal Latin squares A and B have four disjoint commontransversals Q1,Q2, Q3, Q4, where

Q1 = {(0, 0), (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9)}.Q2 = {(0, 1), (1, 0), (2, 4), (3, 9), (4, 6), (5, 3), (6, 5), (7, 8), (8, 2), (9, 7)}.Q3 = {(0, 4), (1, 6), (2, 5), (3, 0), (4, 7), (5, 9), (6, 8), (7, 3), (8, 1), (9, 2)}.Q4 = {(0, 9), (1, 8), (2, 0), (3, 2), (4, 1), (5, 6), (6, 7), (7, 4), (8, 5), (9, 3)}.It has been checked that these two orthogonal Latin squares A and B have another

three disjoint common 2-transversals:Q5 = {(0, 2), (0, 3), (1, 2), (1, 9), (2, 6), (2, 7), (3, 6), (3, 8), (4, 5), (4, 9), (5, 0),

(5, 8), (6, 1), (6, 4), (7, 0), (7, 1), (8, 3), (8, 7), (9, 4), (9, 5)},Q6 = {(0, 5), (0, 6), (1, 4), (1, 7), (2, 1), (2, 8), (3, 4), (3, 5), (4, 3), (4, 8), (5, 2),

(5, 7), (6, 3), (6, 9), (7, 2), (7, 9), (8, 0), (8, 6), (9, 0), (9, 1)},Q7 = {(0, 7), (0, 8), (1, 3), (1, 5), (2, 3), (2, 9), (3, 1), (3, 7), (4, 0), (4, 2), (5, 1),

(5, 4), (6, 0), (6, 2), (7, 5), (7, 6), (8, 4), (8, 9), (9, 6), (9, 8)}.Clearly, these two orthogonal Latin squares A and B have five disjoint 2-transversals

Q1 ∪ Q2, Q3 ∪ Q4, Q5, Q6, and Q7. So there is a 2-resolvable TD(4, 10). Simi-larly, Q1 ∪ Q6 ∪ Q7 and Q2 ∪ Q3 ∪ Q4 ∪ Q5 are two disjoint 5-transversals, then a



5-resolvable TD(4, 10) also exists. It follows that there is an MOA(100; (10)451, 2) andan MOA(100; (10)421, 2). Since a|10, b|10, we get an MOA(100; (10)251a1b1, 2) and anMOA(100; (10)221a1b1, 2) by Construction 2.6. The proof is completed. �

3. THE PROOF OF THEOREM 1.2

In this section, we will prove Theorem 1.2.

Lemma 3.1. Let a, b, c, d be integers and greater than 1. Suppose that thereexists a subset {x, y} ⊂ {a, b, c, d} such that g.c.d. {x, y} = 1. Then there is anMOA(N ; abcd, 2), where N = l.c.m. {uv : {u, v} ⊂ {a, b, c, d}}.

Proof. Without loss of generality, let x = a, y = b. By Theorem 1.1 there isan MOA(N ; (ab)cd, 2), where N = l.c.m. {abc, abd, cd}. Clearly, N = l.c.m. {uv :{u, v} ⊂ {a, b, c, d}}. Then by Lemma 2.4 we get an MOA(N ; abcd, 2). �

In the following, we only consider the case that for any positive integer a, b, c, d andfor any subset {x, y} ⊂ {a, b, c, d}, g.c.d. {x, y} > 1. We will divide it into four cases:(1) a = b = c = d, (2) a = c = d �= b, (3) a = d, a �= b and a �= c, and (4) a, b, c, d

are pairwise distinct.

Case 1: a = b = c = d.

For a = b = c = d, we have the following theorem.

Theorem 3.2 ([3, 6, 10]). (1) There is an MOA(a2; a4, 2) for any positive integer a

with two exceptions a = 2, 6.(2) There is an MOA(ka2; a4, 2) for a ∈ {2, 6}, where k ∈ {2, 3}.

Lemma 3.3. There is an MOA(4u; 24, 2) and an MOA(36u; 64, 2) for any integeru > 1.

Proof. For each u > 1, it can be written as u = 2x + 3y, where x, y are non-negative integers and x + y ≥ 1. By Theorem 3.2, there is an MOA(ka2; a4, 2) fora ∈ {2, 6}, where k ∈ {2, 3}. Then we obtain an MOA(4u; 24, 2) and an MOA(36u; 64, 2)by Construction 2.5. �

Case 2: a = c = d �= b.

In this case, we consider the existence of MOA(N ; a3b, 2), where a �= b and N =l.c.m.{a2, ab}. The following result is obvious.

Lemma 3.4. If there is an MOA(N ; a1a2 · · · ak, t), then there is an MOA(N ; ai1

ai2 · · · ais , t), where s ≤ k and {i1, i2, . . . , is} ⊆ {1, 2, . . . , k}.

Lemma 3.5. Let b be a positive integer. If b ≡ 0 (mod 4), then there is anMOA(2b; 23b, 2). If b ≡ 2 (mod 4), then there does not exist an MOA(2b; 23b, 2), whilethere is an MOA(2bu; 23b, 2) for any integer u > 1.

Proof. When b ≡ 0 (mod 4), we start with an MOA(8; 234, 2) from Lemma 2.8 andLemma 3.4. Applying Construction 2.1 with n = b/4, we get an MOA(2b; 23b, 2).



When b ≡ 2 (mod 4), for each u > 1, it can be written as u = 2x + 3y, wherex, y are nonnegative integers and x + y > 0. We start with an MOA(8; 24, 2) and anMOA(12; 24, 2) from Theorem 3.2. By applying Construction 2.1 with n = b/2, weget an MOA(4b; 23b, 2) and an MOA(6b; 23b, 2). Then by Construction 2.5, we get anMOA(2bu; 23b, 2) for any integer u > 1.

It is left to show that there does not exist an MOA(2b; 23b, 2) for b ≡ 2 (mod 4).Assume that there is an MOA(2b; 23b, 2) A = (ai,j ) of the following form over

Z2, Z2, Z2, Zb.

A =

⎛⎜⎜⎝

a1,1 a1,2 · · · a1,b a1,b+1 a1,b+2 · · · a1,2b

a2,1 a2,2 · · · a2,b a2,b+1 a2,b+2 · · · a2,2b

0 0 · · · 0 1 1 · · · 10 1 · · · b − 1 0 1 · · · b − 1

⎞⎟⎟⎠ =

⎛⎝A1 A2

0 1α α

⎞⎠ .

Consider the subarray consisting of the first two rows of A, i.e., (A1, A2). Suppose thatthe numbers of occurrence times of column vectors (0, 0)T and (1, 0)T in A1 are x and y,respectively. Since the first three rows form an MOA(2b; 23, 2), we have that x + y = b

2 ,the numbers of occurrence times of column vectors (0, 1)T and (1, 1)T in A1 are b

2 − x

and b2 − y, respectively, and the numbers of occurrence times of column vectors (0, 0)T ,

(0, 1)T , (1, 0)T and (1, 1)T in A2 are b2 − x, x, b

2 − y, and y, respectively. Since A isan MOA(2b; 23b, 2), the column vectors (c1, c2, c3, i)T and (1 − c1, 1 − c2, 1 − c3, i)T

occur simultaneously. It follows that the number of occurrence times of column vectors(0, 0)T in A1 is equal to the number of occurrence times of column vectors (1, 1)T in A2,i.e., x = y. So, 2x = b

2 and b is divisible by 4, which is a contradiction. �

Lemma 3.6. For any positive integer b �= 6 with g.c.d. {6, b} �= 1, there is anMOA(N ; 63b, 2), where N = l.c.m. {36, 6b}.

Proof. If b ≡ 3 (mod 6), by simple calculation, we have N = l.c.m. {36, 6b} =36b

3 = 12b. Start with an MOA(36; 633, 2) by Lemmas 2.8 and 3.4. By applying Con-struction 2.1 with n = b

3 , we obtain an MOA(12b; 63b, 2).If b ≡ 2, 4 (mod 6), we have N = l.c.m. {36, 6b} = 36b

2 = 18b. Start with anMOA(36; 632, 2) by Lemma 2.8 and Lemma 3.4. By applying Construction 2.1 withn = b

2 , we obtain an MOA(18b; 63b, 2).If b ≡ 0 (mod 6) and b �= 6, then we write b = 12x + 18y, where x, y are nonnega-

tive integers. There is an MOA(72; 63(12), 2) and an MOA(108; 63(18), 2) obtained fromLemma 2.8 and Lemma 2.14, respectively. By applying Construction 2.1 with n = x andn = y, respectively, we obtain an MOA(72x; 63(12x), 2) and an MOA(108y; 63(18y), 2).Then we get an MOA(6b; 63b, 2) by Construction 2.2. The proof is completed. �

Lemma 3.7. Let a, b be positive integers, g.c.d. {a, b} = d > 1 and a �= 2, 6. Thenthere is an MOA(N ; a3b, 2), where N = l.c.m. {a2, ab} = a2b/d.

Proof. For a �= 2, 6, there exists an MOA(a2; a4, 2) obtained from Theorem 3.2. Sinced|a, by Construction 2.6, there is an MOA(a2; a3d, 2). By applying Construction 2.1with n = b

d, we get an MOA(a2b/d; a3b, 2). �

Case 3: a = d, a �= b, and a �= c.



In this case, we consider the existence of MOA(N ; a2bc, 2), where a �=b, a �= c, and N =l.c.m. {a2, ab, ac, bc} = l.c.m. {l.c.m. {a2, ab}, l.c.m. {ac, bc}} =l.c.m. {a, b} · l.c.m. {a, c}.Lemma 3.8. Let a, b, c be positive integers with a �= b, a �= c. Suppose that for anysubset {x, y} ⊂ {a, b, c}, g.c.d. {x, y} > 1. Then there is an MOA(N ; a2bc, 2), whereN = l.c.m.{a2, ab, ac, bc}.

Proof. For a = 2, by assumption, both b and c are even, if b ≡ 0 (mod 4)(or c ≡ 0 (mod 4)), then there is an MOA(2b; 23b, 2) (or MOA(2c; 23c, 2) ) byLemma 3.5. By applying Construction 2.1 with n = c/2 (or n = b/2), we obtainan MOA(bc; 22bc, 2). If b ≡ c ≡ 2 (mod 4) and b �= 2, c �= 2, then we write b =6 + 4x, c = 6 + 4y, where x and y are nonnegative integers. When x = y = 0, thereis an MOA(36; 2262, 2) obtained from Lemma 2.8 and Lemma 3.4. Since there is anMOA(8; 234, 2) obtained from Lemma 2.8 and Lemma 3.4. By applying Construc-tion 2.1 with n = 3 and n = b/2, respectively, we get an MOA(24; 226141, 2) and anMOA(4b; 22b141, 2). Further, by applying Construction 2.1 with n = x and n = y,respectively, we obtain an MOA(24x; 2261(4x)1, 2) and an MOA(4yb; 22b1(4y)1, 2).For an MOA(24x; 2261(4x)1, 2), by Construction 2.2 with an MOA(36; 2262, 2), thereis an MOA(24x + 36; 2261(4x + 6)1, 2), i.e., an MOA(6b; 2261b1, 2). Further, by ap-plying Construction 2.2 with an MOA(4yb; 22b1(4y)1, 2), we obtain an MOA(6b +4yb; 22b1(4y + 6)1, 2), i.e., an MOA(bc; 22bc, 2).

For a = 6, if 6|b and 6|c, we start with an MOA(6b; 63b, 2) by Lemma 3.6. ApplyingConstruction 2.1 with n = c/6 yields an MOA(bc; 62cb, 2). Otherwise, start with anMOA(36, 6222, 2), an MOA(36, 622131, 2), an MOA(36, 6232, 2), an MOA(36, 6321, 2),and an MOA(36, 6331, 2) by Lemma 2.8 and Lemma 3.4. By Construction 2.1, we getan MOA(N ; 62bc, 2).

For a �= 2, 6, let g.c.d. {a, b} = d1 > 1, g.c.d. {a, c} = d2 > 1, we get N =l.c.m. {a2, ab, ac, bc} = a2bc

d1d2. There exists an MOA(a2; a4, 2) by Theorem 3.2. Since

d1|a and d2|a, by Construction 2.6 there is an MOA(a2; a2d1d2, 2). By apply-ing Construction 2.7 with a trivial MOA( bc

d1d2; 1111( b

d1)1( c

d2)1, 2), then we obtain an

MOA(a2bc/d1d2; a2bc, 2). �

Case 4: a, b, c, d are pairwise distinct.

In this case, we consider the existence of MOA(N ; abcd, 2), where a, b, c, d arepairwise distinct and N = l.c.m. {ab, ac, ad, bc, bd, cd}.Lemma 3.9. Let a, b, c, d be positive integers. Suppose that g.c.d. {a, b} = r1, g.c.d.{r1, c} = r2, and g.c.d. {r1, d} = r3. Let N∗ = l.c.m. {αβ : {α, β} ⊂ { a

r1, b

r1, c

r2, d

r3}}. Then

N = r21 N∗.

Proof. By Lemma 2.10, we have thatN = l.c.m. {xy : {x, y} ⊂ {a, b, c, d}} = abcd

P, where P = g.c.d. { ab, ac, ad, bc, bd,

cd}.Clearly, N∗ = l.c.m. {αβ : {α, β} ⊂ { a

r1, b

r1, c

r2, d

r3} =

ar1

br1

cr2

dr3

Q, where Q = g.c.d.

{αβ : {α, β} ⊂ { ar1

, br1

, cr2

, dr3

}}. So, to prove N = r21 N∗, we need only to show that



P = r2r3Q. Since

P = g.c.d. {ab, ac, ad, bc, bd, cd}= g.c.d. {ab, g.c.d. {ac, bc}, g.c.d. {ad, , bd}, cd}= g.c.d. {ab, r1c, r1d, cd}= g.c.d. {ab, g.c.d. {r1c, cd}, g.c.d. {r1d, cd}}= g.c.d. {ab, r3c, r2d}= r2r3 × g.c.d.

{a

r1

b

r1

r1

r2

r1

r3,

c

r2,

d

r3

}

= r2r3 × g.c.d.

{a

r1

b

r1,

c

r2,

d

r3

},

and

Q = g.c.d.

{αβ : {α, β} ⊂

{a

r1,

b

r1,

c

r2,

d

r3

}}

= g.c.d.

{a

r1

b

r1,

a

r1

c

r2,

a

r1

d

r3,

b

r1

c

r2,

b

r1

d

r3,

c

r2

d

r3

}

= g.c.d.

{a

r1

b

r1, g.c.d.

{a

r1

c

r2,

b

r1

c

r2

}, g.c.d.

{a

r1

d

r3,

b

r1

d

r3

},

c

r2

d

r3

}

= g.c.d.

{a

r1

b

r1,

c

r2,

d

r3,

c

r2

d

r3

}

= g.c.d.

{a

r1

b

r1,

c

r2,

d

r3

},

the conclusion holds. �

Lemma 3.10. Let a, b, c, d be four distinct positive integers and for any {x, y} ⊆{a, b, c, d}, g.c.d. {x, y} > 1. Then there is an MOA(N ; abcd, 2).

Proof. Without loss of generality, let g.c.d. {a, b} = r1 =max{g.c.d. {x, y} : {x, y} ⊆{a, b, c, d}} and let g.c.d. {r1, c} = r2 and g.c.d. {r1, d} = r3.

If r1 = r2 = r3, then g.c.d. {x, y} = r1 for any {x, y} ⊆ {a, b, c, d}. Since a, b, c, d

are distinct, there must be at least two of a, b, c, d greater than r1. Let c, d be two suchintegers. By Lemma 3.8, there is an MOA(cd; r2

1 cd, 2). By applying Construction 2.7with a trivial MOA( ab

r21

; ( ar1

)1( br1

)11111, 2), then we obtain an MOA(N ; abcd, 2).

If r2 = 1 or r3 = 1 or r2 = r3 = 1, there is an MOA(r21 ; r2

1 r2r3, 2) by Theorem 1.1. If1 < r2, 1 < r3, and r1 = r2 �= r3 (or r1 = r3 �= r2), then r3 < r1 and r3|r1 (or r2 < r1 andr2|r1). There is an MOA(r2

1 ; r21 r2r3, 2) by Lemma 3.1 and Lemmas 3.5–3.7. If 1 < r2 < r1

and 1 < r3 < r1, then by Lemmas 3.1 and 3.8 there is an MOA(r21 ; r2

1 r2r3, 2). Since g.c.d.{a/r1, b/r1} = 1, by Lemma 3.1 and Theorem 1.1 there is an MOA(N∗; a

r1

br1

cr2

dr3

, 2),

where N∗ = l.c.m. {αβ : {α, β} ⊂ { ar1

, br1

, cr2

, dr3

}}. Then we get an MOA(N ; abcd, 2) byConstruction 2.7 and Lemma 3.9. �

Combining Lemma 3.1, Theorem 3.2, Lemma 3.3, Lemmas 3.5–3.8, and Lemma 3.10,we obtain the following theorem.



Theorem 3.11. (1) For any integers a, b, c, d and greater than 1, there is anMOA(N ; abcd, 2), where N = l.c.m. {αβ : {α, β} ⊂ {a, b, c, d}}, except an MOA(4;24, 2), an MOA(36; 64, 2), and an MOA(2d; 23d1, 2), where d ≡ 2 (mod 4).

(2) There is an MOA(4u; 24, 2), an MOA(36u; 64, 2), and an MOA(2du; 23d1, 2), whered ≡ 2 (mod 4) and u > 1.

The proof of Theorem 1.2. By Theorem 3.11 and Construction 2.5, we complete theproof of Theorem 1.2.

4. THE PROOF OF THEOREM 1.3

In this section, we will prove Theorem 1.3.

Lemma 4.1. Let a, b, c, d, e be integers and greater than 1. Suppose that thereexists a subset {x, y} ⊂ {a, b, c, d, e} such that g.c.d. {x, y} = 1. Then there is anMOA(N ; abcde, 2), where N = l.c.m. {uv : {u, v} ⊂ {a, b, c, d, e}}.

Proof. Without loss of generality, let x = a, y = b. By Theorem 1.2, there isan MOA(N ; cde(ab), 2), where N = l.c.m. {αβ : {α, β} ⊂ {a, b, c, d, e}}, except anMOA(36; 64, 2) and an MOA(2(4k + 2); 23(4k + 2)1, 2), where k ≥ 1. Then by Lemma2.4, we get an MOA(N ; cdeab, 2), where N = l.c.m. {uv : {u, v} ⊂ {a, b, c, d, e}}, ex-cept an MOA(36; 632131, 2) and an MOA(2(4k + 2); 24(2k + 1)1, 2), where k ≥ 1. AnMOA(36; 632131, 2) exists by Lemma 2.8.

It is left to show that there is an MOA(2(4k + 2); 24(2k + 1)1, 2), where k ≥ 1. For k =1, there is an MOA(12; 2431, 2) by Lemma 2.8. For k > 1, there is an MOA(8; 2441, 2)by Lemma 2.8. Applying Construction 2.6 gives an MOA(8; 25, 2). By applying Con-struction 2.1 with n = k − 1, we get an MOA(8(k − 1); 24(2(k − 1))1, 2). Further, byapplying Construction 2.2 with an MOA(12; 2431, 2), we get an MOA(2(4k + 2); 24

(2k + 1)1, 2). �

In the following, we only consider the case that for any positive integer a, b, c, d, e

and for any subset {x, y} ⊂ {a, b, c, d, e}, g.c.d. {x, y} > 1. We will divide it into fivecases: (1) a = b = c = d = e, (2) a = c = d = e �= b, (3) a = d = e, a �= b, and a �= c,(4) a �= b, a �= c, a �= d, and b, c, d are not all equal, and (5) a, b, c, d, e are pairwisedistinct.

Case 1: a = b = c = d = e.

For a = b = c = d = e, we have the following theorem.

Theorem 4.2 ([1, 2, 6, 14]). (1) There is an MOA(a2; a5, 2) for any positive integer a

with three exceptions a = 2, 3, 6 and one possible exception a = 10.(2) There is an MOA(2a2; a5, 2) for a ∈ {2, 3, 6, 10} and an MOA(3a2; a5, 2) for a ∈

{2, 3, 6}.Lemma 4.3. There is an MOA(4u; 25, 2), an MOA(9u; 35, 2), an MOA(36u; 65, 2), andan MOA(100u; (10)5, 2) for any integer u > 1.

Proof. For each u > 1, it can be written as u = 2x + 3y, where x, y are non-negative integers and x + y ≥ 1. By Theorem 4.2, there is an MOA(ka2; a5, 2) fora ∈ {2, 3, 6}, where k ∈ {2, 3}. Then we obtain an MOA(4u; 25, 2), an MOA(9u; 35, 2),



and an MOA(36u; 65, 2) by Construction 2.5. Since there is an MOA(4u; 25, 2) asabove and an MOA(25; 55, 2) by Theorem 4.2, we get an MOA(100u; (10)5, 2) byConstruction 2.7. �

Case 2: a = c = d = e �= b.

In this case, we consider the existence of MOA(N ; a4b, 2), where a �= b and N =l.c.m.{a2, ab}.Lemma 4.4.

(1) If b ≡ 0 (mod 4), then there is an MOA(2b; 24b, 2); if b ≡ 2 (mod 4), thenthere does not exist an MOA(2b; 24b, 2), while there is an MOA(2bu; 24b, 2) forany integer u > 1.

(2) If a ∈ {3, 6, 10}, b �= a and g.c.d. {a, b} = d > 1, then there is anMOA(N ; a4b, 2), where N = l.c.m. {a2, ab}, and except an MOA(36; 6421, 2) andan MOA(36; 6431, 2).

(3) There is an MOA(36u; 6421, 2) and an MOA(36u; 6431, 2) for any integer u > 1.

Proof.

(1) For b ≡ 0 (mod 4), by Lemma 2.8 there is an MOA(8; 2441, 2). By applyingConstruction 2.1 with n = b

4 , we get an MOA(2b; 24b, 2).For b ≡ 2 (mod 4), we first show that there does not exist an MOA(2b; 24b, 2).

Assume that there is an MOA(2b; 24b, 2), then we obtain an MOA(2b; 23b, 2)by Lemma 3.4, which contradicts to Theorem 3.11. So, there does not exist anMOA(2b; 24b, 2). Second, we show that there is an MOA(2bu; 24b, 2) for any integeru > 1. For each integer u > 1, write u = 2x + 3y, where x, y are nonnegative inte-gers and x + y ≥ 1. Since there is an MOA(8; 25, 2) and an MOA(12; 25, 2) by The-orem 4.2, by applying Construction 2.1 with n = b

2 , we get an MOA(4b; 24b, 2) andan MOA(6b; 24b, 2). Then by Construction 2.5, we get an MOA(4bx + 6by; 24b, 2),i.e., an MOA(2bu; 24b, 2).

(2) For a = 3, since b �= a and g.c.d. {a, b} = d > 1, let b = 3k, k ≥ 2. Writek = 2x + 3y, where x, y are nonnegative integers and x + y ≥ 1. There is anMOA(18; 3461, 2) and an MOA(27; 3491, 2) by Lemma 2.8. By applying Construc-tion 2.1 with n = x and n = y, respectively, we obtain an MOA(18x; 34(6x)1, 2)and an MOA(27y; 34(9y)1, 2). Further, by applying Construction 2.2, we obtain anMOA(18x + 27y; 34(6x + 9y)1, 2), i.e., an MOA(3b; 34b1, 2).

For a = 6, since b �= a and g.c.d. {a, b} = d > 1, we need only to con-sider that b ≡ 0, 2, 3, 4, 6 (mod 6). If b ≡ 0 (mod 6), let b = 6k, k ≥ 2. Writek = 2x + 3y, where x, y are nonnegative integers and x + y ≥ 1. There is anMOA(72; 64(12)1, 2) and an MOA(108; 64(18)1, 2) obtained from Lemma 2.8 andLemma 2.14, respectively. By applying Construction 2.1 with n = x and n = y, re-spectively, we get an MOA(72x; 64(12x)1, 2) and an MOA(108y; 64(18y)1, 2). Thenby applying Construction 2.2, we get an MOA(72x + 108y; 64(12x + 18y)1, 2), i.e.,an MOA(6b; 64b1, 2).

If b ≡ 2 (mod 6), let b = 6k + 2, by simple calculation, we have N = l.c.m.

{36, 6b} = 36b2 = 18b. We first show that there does not exist an MOA(36; 6421, 2).

Assume that there is an MOA(36; 6421, 2), then we obtain an MOA(36; 64, 2) byLemma 3.4, which contradicts to Theorem 3.11. Hence, there does not exist an



MOA(36; 6421, 2). Second, for k ≥ 1, by Theorem 4.2 and Lemma 2.8 there is anMOA(108; 6461, 2) and an MOA(72; 6441, 2). By applying Construction 2.1 withn = k − 1 and n = 2, respectively, we get an MOA(108(k − 1); 64(6(k − 1))1, 2)and an MOA(144; 6481, 2). Then by applying Construction 2.2, we can get anMOA(18 · (6k + 2); 64(6k + 2)1, 2), i.e., an MOA(18b; 64b1, 2).

If b ≡ 3 (mod 6), then we have N = l.c.m. {36, 6b} = 36b3 = 12b. Write b =

6k + 3. We first show that there does not exist an MOA(36; 6431, 2). Assume thatthere is an MOA(36; 6431, 2), then we obtain an MOA(36; 64, 2) by Lemma 3.4,which contradicts to Theorem 3.11. Hence, there does not exist an MOA(36; 6431, 2).Second, for k ≥ 1, there is an MOA(72; 6461, 2) by Theorem 4.2. Applying Con-struction 2.1 with n = k − 1 gives an MOA(72(k − 1); 64(6(k − 1))1, 2). There isan MOA(108; 6491, 2) by Lemma 2.14. Then by applying Construction 2.2, we canget an MOA(12 · (6k + 3); 64(6k + 3)1, 2), i.e., an MOA(12b; 64b1, 2).

If b ≡ 4 (mod 6), we have N = l.c.m. {36, 6b} = 36b2 = 18b. Write b = 6k +

4. For k = 0, there is an MOA(72; 6441, 2) by Lemma 2.8. For k ≥ 1, thereis an MOA(108; 6461, 2) by Theorem 4.2. Applying Construction 2.1 withn = k yields an MOA(108k; 64(6k)1, 2). Further, by applying Construction 2.2with an MOA(72; 6441, 2), we get an MOA(18 · (6k + 4); 64(6k + 4)1, 2), i.e., anMOA(18b; 64b1, 2).

For a = 10, since b �= a and g.c.d. {a, b} = d > 1, we need only to con-sider that b ≡ 0, 2, 4, 5, 6, 8 (mod 10). If b ≡ 0 (mod 10), let b = 10k, k ≥ 2.Write k = 2x + 3y, where x, y are nonnegative integers and x + y ≥ 1. Thereis an MOA(200; (10)4(20)1, 2) and an MOA(300; (10)4(30)1, 2) by Lemma 2.14.By applying Construction 2.1 with n = x and n = y, respectively, we get anMOA(200x; (10)4(20x)1, 2) and an MOA(300y; (10)4(30y)1, 2). Further, by apply-ing Construction 2.2, we get an MOA(200x + 300y; (10)4(20x + 30y)1, 2), i.e.,MOA(10b; (10)4b1, 2).

If b ≡ w (mod 10), where w ∈ {2, 4, 6, 8}, by simple calculation, we have N =l.c.m. {100, 10b} = 100b

2 = 50b. Start with an MOA(100; (10)421, 2) obtained fromLemma 2.15 and apply Construction 2.1 with n = b

2 . We then get an MOA(50b;(10)4b1, 2).

If b ≡ 5 (mod 10), we have N = l.c.m. {100, 10b}= 100b5 = 20b. Start with an

MOA(100; (10)451, 2) from Lemma 2.15 and apply Construction 2.1 with n = b5 ,

an MOA(20b; (10)4b1, 2) is then obtained.(3) There is an MOA(36u; 65, 2) by Lemma 4.3. By applying Construction 2.6, we get

an MOA(36u; 6421, 2) and an MOA(36u; 6431, 2). The proof is completed. �

Lemma 4.5. Suppose that a, b be positive integers, a �∈ {2, 3, 6, 10} and g.c.d. {a, b}= d > 1. Then there is an MOA(N ; a4b, 2), where N = l.c.m. {a2, ab} = a2b/d.

Proof. For a �∈ {2, 3, 6, 10}, there exists an MOA(a2; a5, 2) by Theorem 4.2. Sinced|a, by Construction 2.6 there is an MOA(a2; a4d, 2). By applying Construction 2.1 withn = b

d, we get an MOA(a2b/d; a3b, 2). �

Case 3: a = d = e, a �= b, and a �= c.

In this case, we consider the existence of MOA(N ; a3bc, 2), where a �=b, a �= c and N =l.c.m.{a2, ab, ac, bc} = l.c.m. {l.c.m. {a2, ab}, l.c.m. {ac, bc}} =l.c.m. {a, b} · l.c.m. {a, c}.



Lemma 4.6. Let a, b, c be positive integers, a �= b, a �= c. Suppose that for anysubset {x, y} ⊂ {a, b, c}, g.c.d. {x, y} > 1. Then there is an MOA(N ; a3bc, 2), whereN = l.c.m. {a2, ab, ac, bc}.

Proof.

(1) For a = 2, we have that b �= 2, c �= 2, and b ≡ c ≡ 0 (mod 2).If b ≡ 0 (mod 4) (or c ≡ 0 (mod 4)), then there is an MOA(2b; 24b, 2) (or

MOA(2c; 24c, 2)) by Lemma 4.4. By applying Construction 2.1 with n = c2 (or

n = b2 ), we get an MOA(bc; 23bc, 2).

If b ≡ c ≡ 2 (mod 4), then we write b = 6 + 4x, c = 6 + 4y where x and y

are nonnegative integers. When x = y = 0, there is an MOA(36; 236161, 2) byLemma 2.8. When x + y ≥ 1, by Lemma 2.8 there is an MOA(24; 234161, 2).By applying Construction 2.1 with n = x, we get an MOA(24x; 23(4x)161, 2).Apply Construction 2.2 to an MOA(36; 236161, 2). Then we get anMOA(24x + 36; 23(4x + 6)161, 2). For y > 0, since 4y ≡ 0 (mod 4), there isan MOA((4x + 6)(4y); 23(4x + 6)1(4y)1, 2) as above. By applying Construction2.2, we get an MOA(24x + 36 + (4x + 6)(4y); 23(4x + 6)1(6 + 4y)1, 2), i.e., anMOA(bc; 23bc, 2).

(2) For a = 3, we have b �= 3, c �= 3, and 3|b, 3|c. Write b = 3x, c = 3y, where x > 1,y > 1, both x and y are integers. By Lemma 4.4 there is an MOA(9x; 34(3x)1, 2).By applying Construction 2.1 with n = y, we get an MOA(9xy; 33(3x)1(3y)1, 2),i.e., an MOA(bc; 33b1c1, 2).

(3) For a = 6, we have b �= 6, c �= 6, and g.c.d. {6, b} ∈ X, g.c.d. {6, c} ∈ X, whereX = {2, 3, 6}.

If 6|b and 6|c, we start with an MOA(6b; 64b, 2) obtained from Lemma 4.4. Byapplying Construction 2.1 with n = c

6 , we get an MOA(bc; 63bc, 2).If exact one of b, c is not divisible by 6, without loss of generality, let 6|b,

g.c.d. {6, c} = s, s ∈ {2, 3}. By simple calculation, we have N = l.c.m. {6, b} ·l.c.m. {6, c} = 6bc

s. Write b = 6(2x + 3y), where x and y are nonnegative inte-

gers and x + y ≥ 1. There is an MOA(72; 64(12)1, 2) and an MOA(108; 64(18)1, 2)by Lemmas 2.8 and 2.14, respectively. By Construction 2.6, there is anMOA(72; 63(12)1s1, 2) and an MOA(108; 63(18)1s1, 2). By applying Construc-tion 2.1 with n = x and n = y, respectively, we get an MOA(72x; 63(12x)1s1, 2)and an MOA(108y; 63(18y)1s1, 2). By Construction 2.2, we get an MOA(72x +108y; 63(12x + 18y)1s1, 2), i.e., an MOA(6b; 63b1s1, 2). Further, by applying Con-struction 2.1 with n = c

s, we get an MOA(6bc/s; 63b1c1, 2).

If neither of b, c is divisible by 6, there is an MOA(36, 6322, 2), anMOA(36, 632131, 2), and an MOA(36, 6332, 2) by Lemma 2.8. We get anMOA(N ; 63bc, 2) by Construction 2.1, where N = l.c.m. {a2, ab, ac, bc}.

(4) For a = 10, we have b �= 10, c �= 10, and g.c.d. {10, b} ∈ X, g.c.d. {10, c} ∈ X,where X = {2, 5, 10}.

If 10|b and 10|c, we start with an MOA(10b; (10)4b, 2) by Lemma 4.4. ApplyingConstruction 2.1 with n = c

10 yields an MOA(bc; (10)3bc, 2). Otherwise, we startwith an MOA(100; (10)451, 2), an MOA(100; (10)421, 2), an MOA(100; (10)322, 2),an MOA(100; (10)32151, 2), and an MOA(100; (10)352, 2) by Lemma 2.15. Then weget an MOA(N ; (10)3bc, 2) by Construction 2.1, where N = l.c.m. {a2, ab, ac, bc}.



(5) For a �∈ {2, 3, 6, 10}, let g.c.d. {a, b} = d1 and g.c.d. {a, c} = d2. There existsan MOA(a2; a5, 2) by Theorem 4.2. Since d1|a and d2|a, by Construction 2.6there is an MOA(a2; a3d1d2, 2). By applying Construction 2.1, we get anMOA(a2bc/(d1d2); a3bc, 2). The proof is completed. �

Case 4: a �= b, a �= c, a �= d, and b, c, d are not all equal.

In this case, we consider the existence of MOA(N ; a2bcd, 2), where a �= b, a �= c,a �= d, and b, c, d are not all equal and N = l.c.m. {a2, ab, ac, ad, bc, bd, cd} whichruns through this case.

Lemma 4.7. Let a, b, c, d be positive integers. Suppose that g.c.d. {a, b} = r1, g.c.d.{a, c} = r2, and g.c.d. {a, d} = r3. Then N = a2bcd

r1r2r3r4, where r4 = g.c.d{ b

r1, c

r2, d

r3}.

Proof. By assumption, we have that

N = l.c.m. {a2, ab, ac, ad, bc, bd, cd}= l.c.m. {a2, l.c.m. {ab, bd}, l.c.m. {ac, bc}, l.c.m. {ad, cd}}= l.c.m.

{a2,

abd

r3,abc

r1,acd

r2

}

= a · l.c.m.

{a,

bd

r3,bc

r1,cd

r2

}

= a · l.c.m.

{l.c.m.

{a,

bd

r3

}, l.c.m.

{a,

bc

r1

}, l.c.m.

{a,

cd

r2

}}

= a · l.c.m.

{abd

r1r3,abc

r1r2,acd

r2r3

}.

Since abcdr1r2r3

is a common multiple of integers abdr1r3

, acdr2r3

, abcr1r2

, by Lemma 2.10, we have

l.c.m.

{abd

r1r3,abc

r1r2,acd

r2r3

}=

abcd

r1r2r3

g.c.d.

{c

r2,

d

r3,

b

r1

}

= abcd

r1r2r3r4.

So, N = a2bcdr1r2r3r4

. �

Lemma 4.8. Suppose that a, b, c, d be positive integers, a �= b, a �= c, a �= d, b, c, d

are not all equal, and for any {x, y} ⊆ {a, b, c, d}, g.c.d. {x, y} > 1. Then there is anMOA(N ; a2bcd, 2).

Proof. Without loss of generality, let g.c.d. {a, b} = r1, g.c.d. {a, c} = r2, and g.c.d.{a, d} = r3 and let g.c.d. {b, c, d} = s1.

(1) For a ∈ {2, 3}, we have that a|b, a|c, a|d, b �= a, c �= a, d �= a, and b, c, d arenot all equal. Clearly, r1 = r2 = r3 and a|s1. If s1 = a, there is an MOA(cd;a3cd, 2) by Lemma 4.6. By applying Construction 2.1 with n = b

a, we get an



MOA(bcd/a; a2bcd, 2). If s1 > a, there is an MOA(s21 ; a2s3

1 , 2) by Lemma 4.6 andthere is a trivial MOA(bcd/s3

1 ; 1111 bs1

cs1

ds1

, 2). Then we get an MOA(N ; a2bcd, 2)by Construction 2.7, here, N = bcd/s1.

(2) For a = 6, we have b �= 6, c �= 6, d �= 6, ri ∈ {2, 3, 6}, i = 1, 2, 3, and b, c, d arenot all equal.

If r1 = r2 = r3 = 6, i.e., 6|s1, there is an MOA(cd; s31cd, 2) obtained from Lemma

4.6. By Construction 2.1 with n = bs1

, we get an MOA(bcd/s1; s21bcd, 2). Since 6|s1,

we obtain an MOA(bcd/s1; 62bcd, 2) by Construction 2.6.If r1 = r2 = 6 and r3 ∈ {2, 3}, then g.c.d. { 6

r3, d} = 1, 6

r3|b, 6

r3|c, and by Lemma

2.10, we have

l.c.m.

{bc, bd

6

r3, cd

6

r3

}=

6

r3bcd

g.c.d.

{6

r3d, c, b

} =6

r3bcd

6

r3g.c.d. {d, c, b}

= bcd

s1.

Since g.c.d. {6, 6r3

d} = 6, there is an MOA( bcds1

; 62bc( 6r3

d), 2) as above. Then by

Construction 2.6, we obtain an MOA(N ; 62bcd, 2), where N = bcds1

.If at most one of a, b, c equals 6, let A = {{2, 2, 2}, {2, 2, 3}, {2, 2, 6}, {2, 3, 3},

{2, 3, 6}, {3, 3, 3}, {3, 3, 6}}. If {r1, r2, r3} ∈ A, there is an MOA(36; 62r11 r1

2 r13 , 2)

by Lemma 2.8. Since there is an MOA(M; 1111 br1

cr2

dr3

, 2) which is equivalent

to an MOA(M; br1

cr2

dr3

, 2) obtained from Theorem 1.1, where M = l.c.m. { br1

cr2

,br1

dr3

, cr2

dr3

}. Then applying Construction 2.7 and Lemma 4.7, we get an MOA(N ;

62bcd, 2), where N = 62bcdr1r2r3r4

, d4 = g.c.d{ br1

, cr2

, dr3

}.(3) For a = 10, we have b �= 10, c �= 10, d �= 10, ri ∈ {2, 5, 10}, i = 1, 2, 3, and b, c, d

are not all equal.If r1 = r2 = r3 = 10, then 10|s1. When s1 = 10, there is an

MOA(cd; (10)3cd, 2) by Lemma 4.6. By Construction 2.1 with n = b10 , we get

an MOA(bcd/10; (10)2bcd, 2). When s1 > 10, there is an MOA(s21 ; (10)2s3

1 , 2) byLemma 4.6. Then by applying Construction 2.7 with a trivial MOA( bcd

s31

; 1111 bs1

cs1

ds1

),

then we get an MOA( bcds1

; (10)2bcd, 2).If r1, r2, r3 are not all equal to 10, let A = {{2, 2, 2}, {2, 2, 5}, {2, 2, 10},

{2, 5, 5}, {2, 5, 10}, {2, 10, 10}, {5, 5, 5}, {5, 5, 10}, {5, 10, 10}}. If {r1, r2, r3} ∈ A,there is an MOA(100; (10)2r1

1 r12 r1

3 , 2) by Lemma 2.15. By Theorem 1.1 there isan MOA(M; b

r1

cr2

dr3

, 2), which is equivalent to an MOA(M; 1111 br1

cr2

dr3

, 2), where

M = bcd/(r1r2r3r4) and r4 = g.c.d{ br1

, cr2

, dr3

}. By applying Construction 2.7 and

Lemma 4.7, we get an MOA(N ; (10)2bcd, 2).(4) For a �∈ {2, 3, 6, 10}, there exists an MOA(a2; a5, 2) by Theorem 4.2. Since r1|a,

r2|a and r3|a, there is an MOA(a2; a2r1r2r3, 2) by Construction 2.6. By Theorem 1.1there is an MOA(M; b

r1

cr2

dr3

, 2), where M = bcd/(r1r2r3r4), r4 = g.c.d{ br1

, cr2

, dr3

}.So, an MOA(M; 1111 b

r1

cr2

dr3

, 2) also exists. By applying Construction 2.7 and

Lemma 4.7, we get an MOA(N ; a2bcd, 2). The proof is completed. �

Case 5: a, b, c, d, e are pairwise distinct.

In this case, we consider the existence of MOA(N ; abcde, 2), where a, b, c, d, e arepairwise distinct and N = l.c.m. {xy : {x, y} ⊂ {a, b, c, d, e}}.



Lemma 4.9. Let a, b, c, d, e be positive integers, suppose that g.c.d. {d, e} = r1, g.c.d.{r1, a} = r2, g.c.d. {r1, b} = r3, and g.c.d. {r1, c} = r4. Let M = l.c.m. {αβ : {α, β} ⊂{ a

r2, b

r3, c

r4, d

r1, e

r1}}. Then N = r2

1 M .

Proof. The proof is presented in the Appendix. �

Lemma 4.10. Let a, b, c, d, e be five distinct positive integers. Suppose that for any{x, y} ⊆ {a, b, c, d, e}, g.c.d. {x, y} > 1 and there is a subset {x, y} ⊆ {a, b, c, d, e} suchthat g.c.d. {x, y} = r1 �∈ {2, 3, 6, 10}. Then there is an MOA(N ; abcde, 2).

Proof. Without loss of generality, let g.c.d. {d, e} = r1 > 1, r1 �∈ {2, 3, 6, 10},g.c.d. {r1, a} = r2, g.c.d. {r1, b} = r3, g.c.d. {r1, c} = r4. Since there is an MOA(r2

1 ; r51 , 2) by Theorem 4.2, there is an MOA(r2

1 ; r21 r2r3r4, 2) by Construction 2.6. Since

g.c.d. { dr1

, er1

} = 1, by Lemma 4.1 there is an MOA(N∗; dr1

er1

ar2

br3

cr4

, 2) if all of dr1

, er1

, ar2

,br3

and cr4

are greater than 1, where N∗ = l.c.m. {x ′y ′ : {x ′, y ′} ⊂ { dr1

, er1

, ar2

, br3

, cr4

}}. By

Lemma 3.1, there is an MOA(N∗; dr1

er1

ar2

br3

cr4

, 2) if one of ar2

, br3

, and cr4

equals 1. By

Theorem 1.2, there is an MOA(N∗; dr1

er1

ar2

br3

cr4

, 2) if dr1

= 1 and the type er1

ar2

br3

cr4

�∈ E ={64, 23(4k + 2)1 : k ≥ 0}( or e

r1= 1 and the type d

r1

ar2

br3

cr4

�∈ E). It follows that there isan MOA(N ; abcde, 2) by Construction 2.7 and Lemma 4.9.

Next, without loss of generality, suppose that dr1

= 1, i.e., d = r1, we only consider the

type er1

ar2

br3

cr4

∈ {64, 23(4k + 2)1 : k ≥ 0}.For { e

r1, a

r2, b

r3, c

r4} = {2, 2, 2, 2}. We have e

r1= 2, a

r2= 2, b

r3= 2, c

r4= 2. Then

e = 2r1 = 2d, g.c.d. {e, a} = g.c.d. {2r1, 2r2} = 2r2, g.c.d. {e, b} = g.c.d. {2r1, 2r3} =2r3,g.c.d. {e, c} = g.c.d. {2r1, 2r4} = 2r4. Since a, b, c are pairwise distinct, i.e.,2r2, 2r3, 2r4 are pairwise distinct, and r2 = g.c.d. {d, a} > 1, r3 = g.c.d. {d, b} >

1, r4 = g.c.d. {d, c} > 1. Hence, there is at least one i such that 2ri �∈ {2, 3, 6, 10}, i =2, 3, 4. Assume that i = 2, then g.c.d. {e, a} = 2r2, g.c.d. {2r2, d} = g.c.d. {a, d} = r2,and let g.c.d. {2r2, b} = s2, g.c.d. {2r2, c} = s3. We have 2r2 �∈ {2, 3, 6, 10}, d

r2= r1

r2

is not divisible by 2 since g.c.d. {2r2, d} = r2 and e2r2

a2r2

bs2

cs3

dr2

= e2r2

1 bs2

cs3

dr2

, wheree

2r2= 2r1

2r2= r1

r2is odd. There is an MOA ((2r2)2; (2r2)2s2s3r2, 2) by Lemmas 4.1, 4.5, 4.6,

4.8 and Theorems 1.1–1.2 since 2r2 �∈ {2, 3, 6, 10}. By Theorems 1.1 and 1.2, there isan MOA(N∗; e

2r2

a2r2

bs2

cs3

dr2

, 2), where N∗ = l.c.m. {x ′y ′ : {x ′, y ′} ⊂ { e2r2

, a2r2

, bs2

, cs3

, dr2

}.It follows that there is an MOA(N ; abcde, 2) by Construction 2.7 and Lemma 4.9.Similarly, for { e

r1, a

r2, b

r3, c

r4} = {6, 6, 6, 6}, there is an MOA(N ; abcde, 2).

For { er1

, ar2

, br3

, cr4

} = {2, 2, 2, 4k + 2}, k > 0, let er1

= 2, ar2

= 4k + 2, br3

= 2, cr4

=2. Then we get e = 2r1 = 2d, g.c.d. {e, b} = g.c.d. {2r1, 2r3} = 2r3, g.c.d. {e, c} =g.c.d. {2r1, 2r4} = 2r4. Since b �= c, i.e., 2r3 �= 2r4, and r1 = g.c.d. {d, e} > 1, r3 =g.c.d. {d, b} > 1, r4 = g.c.d. {d, c} > 1, there is at least one u ∈ {r3, r4, 2r3, 2r4}, suchthat u �∈ {2, 3, 6, 10}. Assume that u = r3. Then g.c.d. {b, d} = r3, g.c.d. {r3, e} =g.c.d. {r3, 2r1} = r3, and let g.c.d. {r3, a} = s1, g.c.d. {r3, c} = s2. We have r3 �∈{2, 3, 6, 10}, d

r3= r1

r3is not divisible by 2, r1 �= r3 and b

r3

dr3

as1

cs2

er3

= 2r3r3

r1r3

as1

cs2

2r1r3

. There is

an MOA ((r3)2; (r3)2r3s1s2, 2) by Lemmas 4.1, 4.6 and Theorem 1.2. By Lemma 4.1 andTheorem 1.2 there is an MOA(N∗; 2r3

r3

r1r3

as1

cs2

2r1r3

, 2), where N∗ = l.c.m. {x ′y ′ : {x ′, y ′} ⊂{ 2r3

r3, r1

r3, a

s1, c

s2, 2r1

r3}. It follows that there is an MOA(N ; abcde, 2) by Construction 2.7

and Lemma 4.9. The proof is completed. �



Lemma 4.11. Let a, b, c, d, e be five distinct positive integers. Suppose that for any{x, y} ⊆ {a, b, c, d, e}, g.c.d. {x, y} ∈ {2, 3, 6, 10}. Then there is an MOA(N ; abcde, 2).

Proof. Without loss of generality, let g.c.d. {d, e} = r1 = max{g.c.d. {x, y} : {x, y} ⊆{a, b, c, d, e}}, g.c.d. {r1, a} = r2, g.c.d. {r1, b} = r3, and g.c.d. {r1, c} = r4.

For r1 = h, h ∈ {2, 3}, since for any {x, y} ⊆ {a, b, c, d, e}, g.c.d. {x, y} > 1and max{g.c.d. {x, y} : {x, y} ⊆ {a, b, c, d, e}} = h, we have g.c.d. {x, y} = h forany {x, y} ⊆ {a, b, c, d, e}. It follows that N = l.c.m. {ab, ac, ad, ae, bc, bd, be, cd,

ce, de} = abcde/h3. Since c, d, e are pairwise distinct, there is an MOA( cdeh

; h2cde, 2)by Lemma 4.8. By applying Construction 2.7 with a trivial MOA( ab

h2 ; ( ah

)1( bh

)1111111, 2),then we obtain an MOA(N ; abcde, 2), where N = abcde/h3.

For r1 = 6, if r2 = r3 = r4 = 6, we have g.c.d. {x, y} = 6 for any {x, y} ⊆{a, b, c, d, e}. In a similar way to the case r1 = 2, we can obtain an MOA(N ; abcde, 2),where N = abcde/63. If r2 = r3 = 6, r4 = 2 or 3, there is an MOA(N ; deab(6c/r4), 2)as above since g.c.d. {r1, 6c/r4} = 6. By applying Construction 2.6, we get anMOA(N ; deabc, 2), where N = l.c.m. {αβ : {α, β} ⊂ {a, b, c, d, e}}. If {r2, r3, r4} ∈ A,where A = {{2, 2, 2}, {2, 2, 3}, {2, 2, 6}, {2, 3, 3}, {2, 3, 6}, {3, 3, 3}, {3, 3, 6}}, thereis an MOA(r2

1 ; r21 r1

2 r13 r1

4 , 2) by Lemma 2.8. Since g.c.d. { dr1

, er1

} = 1, by Lemma 4.1

there is an MOA(N∗; dr1

er1

ar2

br3

cr4

, 2) if all of dr1

, er1

, ar2

, br3

, and cr4

are greater than

1, where N∗ = l.c.m. {x ′y ′ : {x ′, y ′} ⊂ { dr1

, er1

, ar2

, br3

, cr4

}}. By Lemma 3.1, there is an

MOA(N∗; dr1

er1

ar2

br3

cr4

, 2) if one of ar2

, br3

, and cr4

equals 1. By Theorem 1.2, there is

an MOA(N∗; dr1

er1

ar2

br3

cr4

, 2) if dr1

= 1 and the type er1

ar2

br3

cr4

�∈ E = {64, 23(4k + 2)1 :

k ≥ 0} (or er1

= 1 and the type dr1

ar2

br3

cr4

�∈ E). Note that if dr1

= 1, then the typeer1

ar2

br3

cr4

�∈ E since a, b, c, e are pairwise distinct. Then by Construction 2.7, we get

an MOA(r21 N∗; deabc, 2), which is also an MOA(N ; deabc, 2) by Lemma 4.9.

For r1 = 10, if r2 = r3 = r4 = 10, we have g.c.d. {x, y} = 10 for any {x, y} ⊆{a, b, c, d, e}. In a similar way to the case r1 = 2, we can obtain an MOA(N ; abcde, 2),where N = abcde/(10)3. If {r2, r3, r4} ∈ A, where A = {{2, 2, 2}, {2, 2, 5}, {2, 2, 10},{2, 5, 5}, {2, 5, 10}, {2, 10, 10}, {5, 5, 5}, {5, 5, 10}, {5, 10, 10}}, there is anMOA(r2

1 ; r21 r1

2 r13 r1

4 , 2) by Lemma 2.15. Similarly there is an MOA(N∗; dr1

er1

ar2

br3

cr4

, 2),

where N∗ = l.c.m. {x ′y ′ : {x ′, y ′} ⊂ { dr1

, er1

, ar2

, br3

, cr4

}}. Then by Construction 2.7, we

get an MOA(r21 N∗; deabc, 2), which is also an MOA(N ; deabc, 2) by Lemma 4.9. The

proof is completed. �

Combining Lemma 4.1, Theorem 4.2, Lemmas 4.3–4.6, Lemma 4.8, andLemmas 4.10–4.11, we get the following theorem.

Theorem 4.12.

(1) Let a, b, c, d, e be integers and greater than 1, there is an MOA(N ; abcde, 2),where N = l.c.m. {αβ : {α, β} ⊂ {a, b, c, d, e}}, except an MOA(4; 25, 2), anMOA(9; 35, 2), an MOA(36; 6421, 2), an MOA(36; 6431, 2), an MOA(36; 65, 2),and an MOA(2e; 24e1, 2) where e ≡ 2 (mod 4), and possibly except anMOA(100; (10)5, 2).

(2) There exists an MOA(4u; 25, 2), an MOA(36u; 65, 2), an MOA(36u; 6421, 2), anMOA(36u; 6431, 2), an MOA(100u; (10)5, 2), and an MOA(2du; 24d1, 2), whered ≡ 2 (mod 4) and u > 1.



The proof of Theorem 1.3. By Theorem 4.12 and Construction 2.5, we complete theproof of Theorem 1.3.

ACKNOWLEDGMENT

The authors would like to thank the referees for many helpful comments.

REFERENCES

[1] R. J. R. Abel, F. E. Bennett, and G. Ge, Super-simple holey Steiner pentagon systems andrelated designs, J Combin Designs 16 (2008), 301–328.

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APPENDIX

The proof of the Lemma 4.9. By Lemma 2.10, we haveN = l.c.m. {xy : {x, y} ⊂ {a, b, c, d, e}} = abcde

P, where P = g.c.d. {xyz : {x, y, z}

⊂ {a, b, c, d, e}}.M = l.c.m. {αβ : {α, β} ⊂ { a

r2, b

r3, c

r4, d

r1, e

r1}} =

abcder1r1r2r3r4

Q, where g.c.d. {d, e} = r1,

g.c.d. {r1, a} = r2, g.c.d. {r1, b} = r3, g.c.d. {r1, c} = r4, and Q = g.c.d. {αβγ :{α, β, γ } ⊂ { a

r2, b

r3, c

r4, d

r1, e

r1}}.

So, to prove N = r21 M , we need only to show that P = r2r3r4Q. Since

P = g.c.d. {cde, bde, bce, bcd, ade, ace, acd, abe, abd, abc}= g.c.d. {cde, bde, g.c.d. {bce, bcd}, ade, g.c.d. {ace, acd}, g.c.d. {abe, abd}, abc}= g.c.d. {cde, bde, bcr1, ade, acr1, abr1, abc}= g.c.d. {cde, bde, ade, g.c.d. {bcr1, abc}, g.c.d. {acr1, abc}, g.c.d. {abr1, abc}}= g.c.d. {cde, bde, ade, bcr2, acr3, abr4}}= r2r3r4 × g.c.d.

{c

r4

d

r2

e

r3,

b

r3

d

r2

e

r4,

a

r2

d

r3

e

r4,

b

r3

c

r4,

a

r2

c

r4,

a

r2

b

r3

}

= r2r3r4 × g.c.d.

{c

r4

d

r1

e

r1

r1

r2

r1

r3,

b

r3

d

r1

e

r1

r1

r2

r1

r4,

a

r2

d

r1

e

r1

r1

r3

r1

r4,

b

r3

c

r4,

a

r2

c

r4,

a

r2

b

r3

}

= r2r3r4 × g.c.d.

{g.c.d.

{c

r4

d

r1

e

r1

r1

r2

r1

r3,

b

r3

c

r4

}, g.c.d.

{b

r3

d

r1

e

r1

r1

r2

r1

r4,

a

r2

b

r3

},

g.c.d.

{a

r2

d

r1

e

r1

r1

r3

r1

r4,

a

r2

c

r4

}}

= r2r3r4 × g.c.d.

{c

r4

d

r1

e

r1,

b

r3

d

r1

e

r1,

a

r2

d

r1

e

r1,

b

r3

c

r4,

a

r2

c

r4,

a

r2

b

r3

}

and

Q = g.c.d.

{abc

r2r3r4,

bce

r3r4r1,

ace

r2r4r1,

abe

r2r3r1,

bcd

r3r4r1,

acd

r2r4r1,

abd

r2r3r1,

cde

r4r1r1,

bde

r3r1r1,

ade

r2r1r1

}

= g.c.d.

{abc

r2r3r4, g.c.d.

{bce

r3r4r1,

bcd

r3r4r1

}, g.c.d.

{ace

r2r4r1,

acd

r2r4r1

},

g.c.d.

{abe

r2r3r1,

abd

r2r3r1

},

cde

r4r1r1,

bde

r3r1r1,

ade

r2r1r1

}

= g.c.d.

{abc

r2r3r4,

bc

r3r4,

ac

r2r4,

ab

r2r3,

cde

r4r1r1,

bde

r3r1r1,

ade

r2r1r1

}

= g.c.d.

{bc

r3r4,

ac

r2r4,

ab

r2r3,

cde

r4r1r1,

bde

r3r1r1,

ade

r2r1r1

},

we have that P = r2r3r4Q. The proof is completed.


the existence of mixed orthogonal arrays with four and five factors of strength two

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