# Constructions of Optimal Optical Orthogonal Codes with Weights Set5,7 via Cyclic Packing

Authors: Bichang Huang; Dan Luo; Wenxing Zhu
Journal: Mathematical Problems in Engineering
 (2022)

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## Abstract

Double-weight optical orthogonal codes are variable-weight optical orthogonal codes (OOCs), which have been widely applied in optical networks and systems. Some works have been devoted to optimaln,W,1,Q-OOCs with maxw:w∈W≤6. So far, there is no explicit construction of optimal n,W,1,Q-OOCs with W=5,7. It is known that heavier-weight codewords have better code performance than lighter-weight codewords. So, in this paper, we use cyclic packing to construct two infinite classes of optimal OOCs with weights set 5,7 explicitly, for any prime p≡3mod4 and p≥7. In addition, for 1≤t<p−1/2, by breaking t blocks of size 7 into 3 of 31p,5,7,1,1/2,1/2-OOCs and 41p,5,7,1,2/3,1/3-OOCs, we obtain new infinite classes of optimal 31p,3,5,7,1,7t/p−1+6t,p−1/2p−1+6t,p−1−2t/2p−1+6t-OOCs and 41p,3,5,7,1,14t/3p−1+4t,2p−1/3p−1+4t,p−1−2t/3p−1+4t-OOCs, respectively.


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## Body

## 1. Introduction

In 1989, Salehi [1] defined the constant-weight optical orthogonal codes (OOCs), and applied them in the optical code division multiple access (OCDMA) network. Later on, Yang et al. [2] introduced the variable-weight optical orthogonal codes (variable-weight OOCs) in order to meet the requirements of differential quality-of-service (QoS). That is, the codewords of low code weight are assigned to the low-QoS (quality-of-services) requirement applications and high code weight codewords are assigned to high-QoS requirement applications. Because of its desirable features, OOCs have found wide range of applications in many communications fields, such as mobile radio, frequency-hopping spread-spectrum communications, radar, sonar signal design, and so on [3–7].In [8], suppose W=w1,w2,…,wr is an ordering of a set of r integers greater than 1 and Λa=λa1,λa2,…λar is an r-tuple (autocorrelation sequence) of positive integers. Moreover, suppose that λc is a positive integer (cross-correlation parameter) and Q=q1,q2⋯,qr is an r-tuple (weight distribution sequence) of positive rational numbers whose sum is 1. Then, an n,W,Λa,λc,Q optical orthogonal code (OOC) (briefly, n,W,Λa,λc,Q-OOC) is a set C of subsets (called codeword-sets) of Zn with sizes (weights) from W satisfying the following three properties:(1)
Weight distribution property: the number of codewords with weightwi is exactly qiC, 1≤i≤r, and ∑i=1rqiC=C, where ∑i=1rqi=1(2)
The autocorrelation property:C∩C+t≤λai for any C=x1,x2,…,xwi∈C with weight wi and t∈Zn/0(3)
The cross-correlation property:C′∩C+t≤λc for any C′=x1,x2,…,xwi∈C, C=y1,y2,…ywj∈C with C′≠C and t∈ZnIfλa1=λa2=⋯=λar=λc=λ, then the notation n,W,λ,Q-OOC is used to denote n,W,Λa,λc,Q-OOC. Q is normalized if it is written in the form Q=a1/b,a2/b,…,ar/b with gcda1,a2,…,ar=1. And, an n,W,Λa,λc,Q-OOC is said to be balanced if Q=1/r,1/r,…,1/r.IfW=1 and λc=1, an n,W,Λa,λc,Q-OOC is a conflict-avoiding code [9].It is well known that constructing the optimal OOCs withλai and λc minimal is preferred due to its correlation property. Namely, the smaller the value of the correlation is, the smaller the interference is.The following lemma provides an upper bound on the size of ann,W,1,Q-OOC.Lemma 1.
([8]) If C is an n,W,1,Q-OOC with W=w1,w2,…,wr and normalized weight distribution sequence Q=a1/b,a2/b,…,ar/b, then we have(1)C≤bn−1∑i=1raiwiwi−1.Ann,W,1,Q-OOC is optimal if C reaches the upper bound.2-CPW,1,Q;n is a generalization of 2-n,w,1 packing. It is a useful tool for constructing n,W,1,Q-OOC. The details of the concept of 2-CP W,1,Q;n were introduced in [10].LetG be an Abelian group, and assume that Bj⊆G, 1≤j≤t, and ℬ=Bj:1≤j≤t. Define ΔBj=x−y:x,y∈Bj,x≠y, 1≤j≤t, and Δℬ=∪i=1tΔBj, where ΔBj△Bj, 1≤j≤t, and Δℬ are multisets.Letℬ=B1,B2,…,Bt be a family of subsets of Zn, Bj∈W=w1,w2,…,wr, 1≤j≤t. A 2-CP (W,1,n) is ℬ (base blocks set), such that the differences in ℬ, i.e., Δℬ, cover each element of Zn/0 at most once. A 2-CP W,1,Q;n is a 2-CP W,1;n with the property that the number of blocks of size wi is qiℬ, where qi∈Q, 1≤i≤r.A 2-CPW,1;gv is g-regular if Δℬ=x−y:x,y∈Bj,x≠y,1≤j≤t covers each element of Zgv/vZgv exactly once, and each element of vZgv is not covered.The following results are the relationships between a 2-CPW,1,Q;n and an n,W,1,Q-OOC.Lemma 2.
An optimal 2-CPW,1,Q;n is equivalent to an optimal n,W,1,Q-OOC.Lemma 3.
Letw=∑i=1raiwiwi−1, where wi∈W. If 1≤g≤w, then a g-regular 2-CP W,1,Q;gv is optimal.Recently, many researchers have tried to construct OOCs via various methods, such as the search methods using the greedy strategy, the outer-product matrix algorithm [11], projective geometry [12], the finite field theory [13, 14], the combinational design theory [15–21], and so on. In particular, researchers have obtained many optimal OOCs with λ=1 and maxw:w∈W≤6 by constructing 2-CP W,1,Q;ns. For the case of constant-weight OOCs, see [22–24] for some of the examples. For the case of double-weight OOCs, some works have been done on W∈3,4,3,5,3,6,4,5,4,6, see [10, 25–27] for some of the examples.For the case of double-weight optimal OOCs withλ=1 and maxw:w∈W≥7, little work has been done. The representative articles are [28–31]. So far, as the authors are aware, no explicit result of OOCs with W=5,7 has been obtained. Based on the results of performance analysis in and [32], it is easy to see that heavier-weight codewords have better code performance than lighter-weight codewords due to higher autocorrelation peaks. So, in this paper, our aim is to present explicit results of optimal OOCs with W=5,7.In [8], Buratti et al. presented many explicit results of optimal OOCs with λ=1 and maxw:w∈W≤6 by using elementary facts of the quadratic residues and cyclic packing. In this paper, continue using the construction method in [8] and further elementary conclusions of the quadratic residues, two new infinite classes of OOCs with W=5,7 are presented. The main results are as follows.Theorem 4.
For any primep≡3mod4 and p≥7, there exist an optimal 2-CP 5,7,1,1/2,1/2;31p and an optimal 31p,5,7,1,1/2,1/2-OOC.Theorem 5.
For any primep≡3mod4 and p≥7, there exist an optimal 2-CP 5,7,1,2/3,1/3;41p and an optimal 41p,5,7,1,2/3,1/3-OOC.An outline of this paper is as follows. In Section2, the basic knowledge of the quadratic residues is introduced. Theorem 4 and Theorem 5 are proved in Section 3 and Section 4, respectively, Conclusions are presented in Section 5.


## 2. Preliminaries

For a fixed primep, let θ be a primitive element of Zp, C02=θ2i:0≤i≤p−1/2 and C12=θC02 denote the quadratic residues and the quadratic nonresidues of Zp, respectively.For anyA=a1,j1,a2,j2,…,ak,jk∈Zp×Zmk, x∈Zp, B⊆Zp, define(1)
x.A=xa1,j1,xa2,j2,…xak,jk(2)
B⋅A=b⋅A:b∈BSuppose that∅≠K⊂Z+ and each element of K is greater than 1. Assume that A=A:A=a1,j1,a2,j2,…ak,jk∈Zp×Zmk,k∈K. The difference lists are defined as follows:Li=al−as:al,jl,as,js⊂A∈A,i≡jl−jsmodm,1≤l,s≤k,k∈K, i=0,…,m−1.In order to prove the main results, we need to present the elementary facts of numbers of−1,2,3,5,7 in Zp. The interested readers may refer to [33] for the details.Lemma 6.
Suppose thatp≡3mod4 is a prime, we have(1)
2∈C02,3∈C02,5∈C02,7∈C02, if and only if p≡71,191,239,359,431,599mod840(2)
2∈C02,3∈C02,5∈C02,7∈C12, if and only if p≡311,479,551,671,719,839mod840(3)
2∈C12,3∈C12,5∈C12,7∈C02, if and only if p≡43,67,163,403,547,667mod840(4)
2∈C12,3∈C12,5∈C12,7∈C12, if and only if p≡187,283,307,523,643,787mod840(5)
2∈C02,3∈C02,5∈C12,7∈C02, if and only if p≡23,263,407,527,743,767mod840(6)
2∈C02,3∈C02,5∈C12,7∈C12, if and only if p≡47,143,167,383,503,647mod840(7)
2∈C12,3∈C12,5∈C02,7∈C02, if and only if p≡211,331,379,499,571,739mod840(8)
2∈C12,3∈C12,5∈C02,7∈C12, if and only if p≡19,139,451,619,691,811mod840(9)
2∈C02,3∈C12,5∈C02,7∈C02, if and only if p≡79,151,319,631,751,799mod840(10)
2∈C02,3∈C12,5∈C02,7∈C12, if and only if p≡31,199,271,391,439,559mod840(11)
2∈C12,3∈C02,5∈C12,7∈C02, if and only if p≡107,323,347,443,683,827mod840(12)
2∈C12,3∈C02,5∈C12,7∈C12, if and only if p≡83,227,467,563,587,803mod840(13)
2∈C12,3∈C02,5∈C02,7∈C02, if and only if p≡11,179,491,611,659,779mod840(14)
2∈C12,3∈C02,5∈C02,7∈C12, if and only if p≡59,131,251,299,419,731mod840(15)
2∈C02,3∈C12,5∈C12,7∈C02, if and only if p≡127,247,463,487,583,823mod840(16)
2∈C02,3∈C12,5∈C12,7∈C12, if and only if p≡103,223,367,607,703,727mod840


## 3. Proof of Theorem 4

In this section, Theorem4 will be proved by considering the prime p≡3mod4 in fourteen cases, where explicit constructions are presented, respectively. According to the Chinese Remainder Theorem, for any p≠31 and p≡3mod4, since gcdp,31=1, there exists a one-to-one and onto mapping from Zp×Z31 to Z31p.


### 3.1. The Cases ofp ≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840

Lemma 7.
Suppose thatp ≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840 is a prime, and ξ=minx:x∈C12. Then, A=C02⋅A1,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(2)A1,1=0,0,5,0,1,1,ξ,3,3,8,2,12,4,18,A1,2=4,0,0,2,2,7,3,11,1,17.Proof.
We compute the difference listsLi, 0≤i≤30, from A1,1 and A1,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the difference lists Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=ξ−1,−4, L3=ξ−5,ξ, L4=L10=L11=1,−1, L5=3−ξ,2, L6=L7=−2,2, L8=−L12=3,−2, L9=3,2−ξ, L14=−3,3, and L15=1,4−ξ.
According to (1) and (2) of Lemma 6, we know that 2∈C02, 3∈C02, 5∈C02, and ξ−l∈C02,1≤l<ξ. It is easy to check that Li∩Ck2=1, 0≤i≤30, k=0,1. Hence, ΔA covers every element in Zp×Z31/0×Z31 exactly once, and each element of 0×Z31 is not covered. So, A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.2. The Cases ofp≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840

Lemma 8.
Suppose thatp≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840 is a prime. Then, A=C02⋅A2,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(3)A2,1=1,0,2,0,3,1,−3,3,0,8,6,12,5,18,A2,2=0,0,−3,2,3,7,−12,11,−2,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A2,1 and A2,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=1,−1, L1=−L8=1,2, L2=−3,−6, L3=−L12=−4,−5, L4=6,−15, L5=3,6, L6=−1,10, L7=−3,3, L9=9,−9, L10=5,−5, L11=3,−12, L13=−3,−4, L14=2,−2, and L15=1,8.
According to (3) and (4) of Lemma 6, we know that 2∈C12, 3∈C12, and 5∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.3. The Cases ofp ≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840

Lemma 9.
Suppose thatp≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840 is a prime. Then, A=C02⋅A3,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(4)A3,1=9,0,5,0,6,1,4,3,8,8,7,12,10,18,A3,2=3,0,4,2,1,7,2,11,0,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A3,1 and A3,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=4,−4, L1=−L8=1,−3, L2=−L10=−2,1, L3=L13=−1,−5, L4=L11=1,−1, L5=−L14=4,−3, L6=L9=3,−2, L7=L12=−2,2, and L15=6,−4.
According to (5) and (6) of Lemma 6, we know that 2∈C02, 3∈C02, and 5∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.4. The Cases ofp ≡ 211, 331, 379, 499, 571, 739 mod840

Lemma 10.
Suppose thatp ≡ 211, 331, 379, 499, 571, 739 mod840 is a prime. Then, A=C02⋅A4,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(5)A4,1=0,0,2,0,1,1,4,3,5,8,3,12,−3,18,A4,2=−1,0,0,2,2,7,4,11,5,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A4,1 and A4,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=L4=2,−2, L1=1,−1, L2=L12=3,1, L3=2,4, L5=1,2, L6=1,−6, L7=3,4, L8=3,5, L9=4,−1, L10=3,−8, L11=2,5, L13=3,5, L14=−6,4, and L15=−7,5.
According to (7) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.5. The Cases ofp ≡ 19, 139, 451, 619, 691, 811 mod840

Lemma 11.
Suppose thatp ≡ 19, 139, 451, 619, 691, 811 mod840 is a prime. Then, A=C02⋅A5,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(6)A5,1=0,0,2,0,1,1,4,3,5,8,3,12,−3,18,A5,2=3,0,0,2,2,7,4,11,−4,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A5,1 and A5,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=L4=2,−2, L1=1,−1, L2=3,−3, L3=2,4, L5=L11=1,2, L6=L10=−8,−6, L7=L9=−1,4, L8=3,5, L12=3,1, L13=3,5, and L14=−L15=7,4.
According to (8) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.6. The Cases ofp ≡ 11, 179, 491, 611, 659, 779 mod840

Lemma 12.
Suppose thatp ≡ 11, 179, 491, 611, 659, 779 mod840 is a prime. Then, A=C02⋅A6,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(7)A6,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A6,2=−3,0,0,2,−1,7,3,11,−4,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A6,1 and A6,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=3,−4, L3=−3,−8, L4=−7,4, L5=−1,9, L6=5,−7, L7=2,5, L8=−L12=1,6, L9=−L10=2,3, L11=−2,6, L14=−3,1, and L15=7,−4.
According to (13) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.7. The Cases ofp ≡ 59, 131, 251, 299, 419, 731 mod840

Lemma 13.
Suppose thatp ≡ 59, 131, 251, 299, 419, 731 mod840 is a prime. Then, A=C02⋅A7,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(8)A7,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A7,2=0,0,3,2,−1,7,−4,11,−5,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A7,1 and A7,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 from as follows:
L0=5,−5, L1=L13=1,−4, L2=3,−4, L3=−3,−8, L4=−7,−3, L5=−4,9, L6=L7=5,−1, L8=−L12=1,6, L9=2,−7, L10=L11=−2,−4, L14=−3,5, and L15=7,−8.
According to (14) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.8. The Cases ofp ≡ 127, 247, 463, 487, 583, 823, 103, 223, 367, 607, 703, 727 mod840

Lemma 14.
Suppose thatp ≡ 127, 247, 463, 487, 583, 823, 103, 223, 367, 607, 703, 727 mod840 is a prime. Then, A=C02⋅A8,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(9)A8,1=7,0,4,0,2,1,9,3,5,8,6,12,−1,18,A8,2=7,0,0,2,1,7,−1,11,6,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A8,1 and A8,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=3,−3, L1=−2,−5, L2=L6=−7,7, L3=2,5, L4=L8=−L12=1,−2, L5=−4,1, L7=3,−6, L9=−L14=−3,−1, L10=−6,5, L11=−8,4, L13=5,8, and L15=6,−10.
According to (15) and (16) of Lemma 6, we know that 2∈C02, 3∈C12, and 5∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.9. The Cases ofp ≡ 79, 151, 319, 631, 751, 799 mod840

Lemma 15.
Suppose thatp ≡ 79, 151, 319, 631, 751, 799 mod840 is a prime. Then, A=C02⋅A9,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(10)A9,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A9,2=−4,0,0,2,−1,7,3,11,6,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A9,1 and A9,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=4,−4, L3=−3,−8, L4=−7,4, L5=−1,9, L6=L7=5,3, L8=−L12=1,6, L9=2,3, L10=L11=−2,7, L14=−3,−10, and L15=7,6.
According to (9) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.10. The Cases ofp ≡ 31, 199, 271, 391, 439, 559 mod840

Lemma 16.
Suppose thatp ≡ 31, 199, 271, 391, 439, 559 mod840 is a prime. Then, A=C02⋅A10,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(11)A10,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A10,2=−4,0,0,2,−1,7,−2,11,1,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A10,1 and A10,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=4,−4, L3=−3,−8, L4=−7,−1, L5=−1,9, L6=L7=5,3, L8=−L12=1,6, L9=L10=L11=−2,2, L14=−3,−5, and L15=7,1.
According to (10) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.11. The Cases ofp ≡ 107, 323, 347, 443, 683, 827 mod840

Lemma 17.
Suppose thatp ≡ 107, 323, 347, 443, 683, 827 mod840 is a prime. Then, A=C02⋅A11,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(12)A11,1=9,0,5,0,1,1,0,3,4,8,6,12,7,18,A11,2=2,0,0,2,−1,7,3,11,−4,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A11,1 and A11,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=4,−4, L1=−8,−4, L2=−1,−2, L3=−9,−5, L4=2,4, L5=−1,4, L6=1,−7, L7=L10=3,−3, L8=−L11=−1,−5, L9=6,3, L12=−3,1, L13=−2,2, L14=−6,6, and L15=7,−4.
According to (11) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


### 3.12. The Cases ofp ≡ 83, 227, 467, 563, 587, 803 mod840

Lemma 18.
Suppose thatp ≡ 83, 227, 467, 563, 587, 803 mod840 is a prime. Then, A=C02⋅A12,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(13)A12,1=9,0,5,0,1,1,0,3,4,8,6,12,7,18,A12,2=9,0,3,2,0,7,1,11,7,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A12,1 and A12,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=4,−4, L1=−8,−4, L2=−1,−6, L3=−9,−5, L4=2,1, L5=−3,4, L6=1,6, L7=3,−9, L8=−1,−5, L9=6,−2, L10=3,7, L11=−8,5, L12=−3,1, L13=−2,2, L14=−6,2, and L15=7,4.
According to (12) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.
We are now in a position to prove Theorem4.Proof of Theorem 4.
For each primep≡3mod4, p>7, and p≠31, we obtain a 2-CP 5,7,1,1/2,1/2;31p from Lemmas 7–18. Let A7=0,7,77,90,107,0,37,80,128,173,0,40,95,152,193,0,1,4,10,12,171,198,0,21,63,115,129,164,202,0,25,51,79,111,150,183,(14)A31=Aj,i:1≤j≤2,0≤j≤14,where A1,i=0,1,3,7,12,37,746×235i and A2,i=0,17,129,306,586235i,0≤i≤14. Then, Ap forms a 2-CP 5,7,1,1/2,1/2;31p for p=7,31.
By Lemmas2 and 3, we obtain an optimal 2-CP 5,7,1,1/2,1/2;31p and an optimal 31p,5,7,1,1/2,1/2-OOC for any prime p≡3mod4 and p≥7. This completes the proof.


## 3.1. The Cases ofp ≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840

Lemma 7.
Suppose thatp ≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840 is a prime, and ξ=minx:x∈C12. Then, A=C02⋅A1,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(2)A1,1=0,0,5,0,1,1,ξ,3,3,8,2,12,4,18,A1,2=4,0,0,2,2,7,3,11,1,17.Proof.
We compute the difference listsLi, 0≤i≤30, from A1,1 and A1,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the difference lists Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=ξ−1,−4, L3=ξ−5,ξ, L4=L10=L11=1,−1, L5=3−ξ,2, L6=L7=−2,2, L8=−L12=3,−2, L9=3,2−ξ, L14=−3,3, and L15=1,4−ξ.
According to (1) and (2) of Lemma 6, we know that 2∈C02, 3∈C02, 5∈C02, and ξ−l∈C02,1≤l<ξ. It is easy to check that Li∩Ck2=1, 0≤i≤30, k=0,1. Hence, ΔA covers every element in Zp×Z31/0×Z31 exactly once, and each element of 0×Z31 is not covered. So, A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.2. The Cases ofp≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840

Lemma 8.
Suppose thatp≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840 is a prime. Then, A=C02⋅A2,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(3)A2,1=1,0,2,0,3,1,−3,3,0,8,6,12,5,18,A2,2=0,0,−3,2,3,7,−12,11,−2,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A2,1 and A2,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=1,−1, L1=−L8=1,2, L2=−3,−6, L3=−L12=−4,−5, L4=6,−15, L5=3,6, L6=−1,10, L7=−3,3, L9=9,−9, L10=5,−5, L11=3,−12, L13=−3,−4, L14=2,−2, and L15=1,8.
According to (3) and (4) of Lemma 6, we know that 2∈C12, 3∈C12, and 5∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.3. The Cases ofp ≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840

Lemma 9.
Suppose thatp≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840 is a prime. Then, A=C02⋅A3,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(4)A3,1=9,0,5,0,6,1,4,3,8,8,7,12,10,18,A3,2=3,0,4,2,1,7,2,11,0,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A3,1 and A3,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=4,−4, L1=−L8=1,−3, L2=−L10=−2,1, L3=L13=−1,−5, L4=L11=1,−1, L5=−L14=4,−3, L6=L9=3,−2, L7=L12=−2,2, and L15=6,−4.
According to (5) and (6) of Lemma 6, we know that 2∈C02, 3∈C02, and 5∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.4. The Cases ofp ≡ 211, 331, 379, 499, 571, 739 mod840

Lemma 10.
Suppose thatp ≡ 211, 331, 379, 499, 571, 739 mod840 is a prime. Then, A=C02⋅A4,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(5)A4,1=0,0,2,0,1,1,4,3,5,8,3,12,−3,18,A4,2=−1,0,0,2,2,7,4,11,5,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A4,1 and A4,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=L4=2,−2, L1=1,−1, L2=L12=3,1, L3=2,4, L5=1,2, L6=1,−6, L7=3,4, L8=3,5, L9=4,−1, L10=3,−8, L11=2,5, L13=3,5, L14=−6,4, and L15=−7,5.
According to (7) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.5. The Cases ofp ≡ 19, 139, 451, 619, 691, 811 mod840

Lemma 11.
Suppose thatp ≡ 19, 139, 451, 619, 691, 811 mod840 is a prime. Then, A=C02⋅A5,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(6)A5,1=0,0,2,0,1,1,4,3,5,8,3,12,−3,18,A5,2=3,0,0,2,2,7,4,11,−4,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A5,1 and A5,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=L4=2,−2, L1=1,−1, L2=3,−3, L3=2,4, L5=L11=1,2, L6=L10=−8,−6, L7=L9=−1,4, L8=3,5, L12=3,1, L13=3,5, and L14=−L15=7,4.
According to (8) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.6. The Cases ofp ≡ 11, 179, 491, 611, 659, 779 mod840

Lemma 12.
Suppose thatp ≡ 11, 179, 491, 611, 659, 779 mod840 is a prime. Then, A=C02⋅A6,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(7)A6,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A6,2=−3,0,0,2,−1,7,3,11,−4,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A6,1 and A6,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=3,−4, L3=−3,−8, L4=−7,4, L5=−1,9, L6=5,−7, L7=2,5, L8=−L12=1,6, L9=−L10=2,3, L11=−2,6, L14=−3,1, and L15=7,−4.
According to (13) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.7. The Cases ofp ≡ 59, 131, 251, 299, 419, 731 mod840

Lemma 13.
Suppose thatp ≡ 59, 131, 251, 299, 419, 731 mod840 is a prime. Then, A=C02⋅A7,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(8)A7,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A7,2=0,0,3,2,−1,7,−4,11,−5,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A7,1 and A7,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 from as follows:
L0=5,−5, L1=L13=1,−4, L2=3,−4, L3=−3,−8, L4=−7,−3, L5=−4,9, L6=L7=5,−1, L8=−L12=1,6, L9=2,−7, L10=L11=−2,−4, L14=−3,5, and L15=7,−8.
According to (14) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.8. The Cases ofp ≡ 127, 247, 463, 487, 583, 823, 103, 223, 367, 607, 703, 727 mod840

Lemma 14.
Suppose thatp ≡ 127, 247, 463, 487, 583, 823, 103, 223, 367, 607, 703, 727 mod840 is a prime. Then, A=C02⋅A8,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(9)A8,1=7,0,4,0,2,1,9,3,5,8,6,12,−1,18,A8,2=7,0,0,2,1,7,−1,11,6,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A8,1 and A8,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=3,−3, L1=−2,−5, L2=L6=−7,7, L3=2,5, L4=L8=−L12=1,−2, L5=−4,1, L7=3,−6, L9=−L14=−3,−1, L10=−6,5, L11=−8,4, L13=5,8, and L15=6,−10.
According to (15) and (16) of Lemma 6, we know that 2∈C02, 3∈C12, and 5∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.9. The Cases ofp ≡ 79, 151, 319, 631, 751, 799 mod840

Lemma 15.
Suppose thatp ≡ 79, 151, 319, 631, 751, 799 mod840 is a prime. Then, A=C02⋅A9,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(10)A9,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A9,2=−4,0,0,2,−1,7,3,11,6,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A9,1 and A9,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=4,−4, L3=−3,−8, L4=−7,4, L5=−1,9, L6=L7=5,3, L8=−L12=1,6, L9=2,3, L10=L11=−2,7, L14=−3,−10, and L15=7,6.
According to (9) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.10. The Cases ofp ≡ 31, 199, 271, 391, 439, 559 mod840

Lemma 16.
Suppose thatp ≡ 31, 199, 271, 391, 439, 559 mod840 is a prime. Then, A=C02⋅A10,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(11)A10,1=4,0,9,0,5,1,1,3,10,8,3,12,8,18,A10,2=−4,0,0,2,−1,7,−2,11,1,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A10,1 and A10,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=5,−5, L1=L13=1,−4, L2=4,−4, L3=−3,−8, L4=−7,−1, L5=−1,9, L6=L7=5,3, L8=−L12=1,6, L9=L10=L11=−2,2, L14=−3,−5, and L15=7,1.
According to (10) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.11. The Cases ofp ≡ 107, 323, 347, 443, 683, 827 mod840

Lemma 17.
Suppose thatp ≡ 107, 323, 347, 443, 683, 827 mod840 is a prime. Then, A=C02⋅A11,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(12)A11,1=9,0,5,0,1,1,0,3,4,8,6,12,7,18,A11,2=2,0,0,2,−1,7,3,11,−4,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A11,1 and A11,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=4,−4, L1=−8,−4, L2=−1,−2, L3=−9,−5, L4=2,4, L5=−1,4, L6=1,−7, L7=L10=3,−3, L8=−L11=−1,−5, L9=6,3, L12=−3,1, L13=−2,2, L14=−6,6, and L15=7,−4.
According to (11) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C02. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.


## 3.12. The Cases ofp ≡ 83, 227, 467, 563, 587, 803 mod840

Lemma 18.
Suppose thatp ≡ 83, 227, 467, 563, 587, 803 mod840 is a prime. Then, A=C02⋅A12,j:j=1,2 forms a 2-CP 5,7,1,1/2,1/2;31p, where(13)A12,1=9,0,5,0,1,1,0,3,4,8,6,12,7,18,A12,2=9,0,3,2,0,7,1,11,7,17.Proof.
We compute the differences ofLi, 0≤i≤30, from A12,1 and A12,2. It is not difficult to see that Ls=−L31−s, 16≤s≤30. So, we only need to compute the differences of Li for 0≤i≤15 as follows:
L0=4,−4, L1=−8,−4, L2=−1,−6, L3=−9,−5, L4=2,1, L5=−3,4, L6=1,6, L7=3,−9, L8=−1,−5, L9=6,−2, L10=3,7, L11=−8,5, L12=−3,1, L13=−2,2, L14=−6,2, and L15=7,4.
According to (12) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C12. It is easy to check that A forms a 2-CP 5,7,1,1/2,1/2;31p.
We are now in a position to prove Theorem4.Proof of Theorem 4.
For each primep≡3mod4, p>7, and p≠31, we obtain a 2-CP 5,7,1,1/2,1/2;31p from Lemmas 7–18. Let A7=0,7,77,90,107,0,37,80,128,173,0,40,95,152,193,0,1,4,10,12,171,198,0,21,63,115,129,164,202,0,25,51,79,111,150,183,(14)A31=Aj,i:1≤j≤2,0≤j≤14,where A1,i=0,1,3,7,12,37,746×235i and A2,i=0,17,129,306,586235i,0≤i≤14. Then, Ap forms a 2-CP 5,7,1,1/2,1/2;31p for p=7,31.
By Lemmas2 and 3, we obtain an optimal 2-CP 5,7,1,1/2,1/2;31p and an optimal 31p,5,7,1,1/2,1/2-OOC for any prime p≡3mod4 and p≥7. This completes the proof.


## 4. Proof of Theorem5

In this section, we prove Theorem5 by considering fourteen cases, and explicit constructions are presented, respectively.For anyp≡3mod4, since gcdp,41=1, according to the Chinese Remainder Theorem, there exists a one-to-one and onto mapping from Zp×Z41 to Z41p.


### 4.1. The Cases ofp ≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840

Lemma 19.
Suppose thatp≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840 is a prime, ξ=minx|x∈C12. Then, ℬ=C02⋅B1,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p where(15)B1,1=0,0,5,0,1,1,ξ,3,3,5,2,12,4,26,B1,2=4,0,0,4,2,10,5,23,1,31,B1,3=2,0,0,6,1,13,3,22,4,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B1,1, B1,2, and B1,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=1,−4, L2=ξ−1,3−ξ, L3=ξ,ξ−5, L4=2,−4, L5=L10=−L12=3,−2, L6=−2,2, L7=L20=−1,1, L8=−L15=4,−1, L9=2,2−ξ, L11=−L14=−2,1, L13=−1,3, L16=3,−3, L17=3,−4, L18=ξ−4,−1, and L19=5,−1.
According to (1) and (2) of Lemma 6, 2∈C02, 3∈C02, 5∈C02, and ξ−l∈C02,1≤l<ξ. It is easy to check that Li∩Ck2=1, 0≤i≤40, k=0,1. Hence, △ℬ covers every element in Zp×Z41/0×Z41 exactly once, and each element of 0×Z41 is not covered. So, ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.2. The Cases ofp ≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840

Lemma 20.
Suppose thatp ≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840 is a prime. Then, ℬ=C02⋅B2,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(16)B2,1=1,0,2,0,3,1,−3,3,6,5,0,12,5,26,B2,2=0,0,4,4,2,10,1,23,3,31,B2,3=6,0,0,6,15,13,12,22,10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B2,1, B2,2, and B2,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=L20=1,−1, L1=−L12=1,2, L2=−6,9, L3=−L5=−4,−5, L4=−L15=3,4, L6=−2,−6, L7=−6,15, L8=2,−2, L9=3,−3, L10=2,−3, L11=−3,−4, L13=−1,9, L14=5,1, L16=12,−2, L17=−10,−5, L18=−8,−1, and L19=−3,−6.
According to (3) and (4) of Lemma 6, we know that 2∈C12, 3∈C12, and 5∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.3. The Cases ofp ≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840

Lemma 21.
Suppose thatp ≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840 is a prime. Then, ℬ=C02⋅B3,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(17)B3,1=9,0,5,0,6,1,4,3,8,5,7,12,10,26,B3,2=4,0,3,4,1,10,2,23,−2,31,B3,3=5,0,0,6,3,13,1,22,9,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B3,1, B3,2, and B3,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=4,−4, L1=−L5=−L7=1,−3, L2=−2,4, L3=L15=−1,−5, L4=−L13=2,−1, L6=−2,−5, L9=L20=−2,3, L8=8,−4, L10=6,−3, L11=L16=−L19=1,−4, L12=−2,2, L14=5,3, L17=−9,6, and L18=−6,2.
According to (5) and (6) of Lemma 6, we know that 2∈C02, 3∈C02, and 5∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.4. The Cases ofp ≡ 211, 331, 379, 499, 571, 739 mod840

Lemma 22.
Suppose thatp ≡ 211, 331, 379, 499, 571, 739 mod840 is a prime. Then, ℬ=C02⋅B4,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(18)B4,1=0,0,2,0,1,1,4,3,5,5,3,12,−3,26,B4,2=1,0,0,4,2,10,6,23,10,31,B4,3=−5,0,0,6,10,13,2,22,16,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B4,1, B4,2, and B4,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=2,−2, L1=1,−1, L2=L12=1,3, L3=L16=2,4, L4=4,−1, L5=L15=5,3, L6=2,5, L7=−2,10, L8=4,14, L9=−1,−8, L10=1,−9, L11=2,−21, L13=4,15, L14=−6,−10, L17=6,−16, L18=−5,7, L19=6,−7, and L20=8,−8.
According to (7) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.5. The Cases ofp ≡ 19, 139, 451, 619, 691, 811 mod840

Lemma 23.
Suppose thatp ≡ 19, 139, 451, 619, 691, 811 mod840 is a prime. Then, ℬ=C02⋅B5,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(19)B5,1=0,0,2,0,1,1,4,3,5,5,3,12,−3,26,B5,2=5,0,0,4,8,10,1,23,7,31,B5,3=2,0,0,6,10,13,7,22,14,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B5,1, B5,2, and B5,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=2,−2, L1=1,−1, L2=−L9=L12=1,3, L3=2,4, L4=4,−5, L5=L15=5,3, L6=8,−2, L7=−2,10, L8=6,7, L10=−2,3, L11=2,−12, L13=−7,8, L14=−6,−7, L16=4,7, L17=−14,4, L18=7,4, L19=1,−5, and L20=8,1.
According to (8) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.6. The Cases ofp ≡ 11, 179, 491, 611, 659, 779 mod840

Lemma 24.
Suppose thatp ≡ 11, 179, 491, 611, 659, 779 mod840 is a prime. Then, ℬ=C02⋅B6,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(20)B6,1=2,0,7,0,1,1,3,3,6,5,8,12,10,26,B6,2=7,0,0,4,6,10,4,23,2,31,B6,3=2,0,0,6,1,13,3,22,5,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B51, B52, and B53. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=−L12=−6,−1, L2=2,3, L3=−L5=1,−4, L4=5,−7, L6=−2,6, L7=−L13=2,1, L8=L14=−2,2, L9=5,2, L10=5,−1, L11=7,−3, L15=−3,−8, L16=−9,3, L17=4,−5, L18=−7,3, L19=4,−1, and L20=4,−4.
According to (13) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.7. The Cases ofp ≡ 59, 131, 251, 299, 419, 731 mod840

Lemma 25.
Suppose thatp ≡ 59, 131, 251, 299, 419, 731 mod840 is a prime. Then, ℬ=C02⋅B7,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(21)B7,1=2,0,7,0,1,1,3,3,6,5,8,12,10,26,B7,2=3,0,0,4,1,10,−4,23,8,31,B7,3=1,0,0,6,4,13,3,22,9,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B7,1, B7,2, and B7,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=−L12=−6,−1, L2=2,3, L3=−L5=1,−4, L4=−L13=5,−3, L6=1,−1, L7=−L19=2,4, L8=6,12, L9=5,−1, L10=−2,−5, L11=7,−8, L14=2,−8, L15=−3,−8, L16=−9,3, L17=5,−9, L18=−7,7, and L20=−7,−4.
According to (14) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.8. The Cases ofp ≡ 127, 247, 463, 487, 583, 823 mod840

Lemma 26.
Suppose thatp ≡ 127, 247, 463, 487, 583, 823 mod840 is a prime. Then, ℬ=C02⋅B8,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(22)B8,1=7,0,4,0,2,1,5,3,9,5,6,12,−1,26,B8,2=9,0,0,4,21,10,1,23,5,31,B8,3=0,0,4,6,3,13,8,22,18,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B8,1, B8,2, and B8,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=3,−3, L1=−5,−2, L2=4,3, L3=−L12=1,−2, L4=7,−9, L5=2,5, L6=21,4, L7=−1,−3, L8=10,4, L9=5,1, L10=12,4, L11=4,−18, L13=−20,3, L14=−7,−5, L15=5,8, L16=3,4, L17=−14,15, L18=6,8, L19=1,−8, and L20=16,10.
According to (15) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C12, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.9. The Cases ofp ≡ 103, 223, 367, 607, 703, 727 mod840

Lemma 27.
Suppose thatp ≡ 103, 223, 367, 607, 703, 727 mod840 is a prime. Then, ℬ=C02⋅B9,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p where(23)B9,1=7,0,4,0,2,1,5,3,9,5,6,12,−1,26,B9,2=7,0,0,4,1,10,−1,23,4,31,B9,3=1,0,0,6,3,13,8,22,10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B9,1, B9,2, and B9,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=L7=3,−3, L1=−L8=−5,−2, L2=4,3, L3=−L12=1,−2, L4=7,−7, L5=2,5, L6=1,−1, L9=5,1, L10=3,−6, L11=4,−9, L13=−2,2, L14=−7,−4, L15=5,8, L16=3,8, L17=−10,7, L18=6,8, L19=−1,−7, and L20=−3,10.
According to (16) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C12, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.10. The Cases ofp ≡ 79, 151, 319, 631, 751, 799 mod840

Lemma 28.
Suppose thatp ≡ 79, 151, 319, 631, 751, 799 mod840 is a prime. Then, ℬ=C02⋅B10,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p where(24)B10,1=4,0,9,0,5,1,1,3,10,5,3,12,8,26,B10,2=1,0,0,4,3,10,4,23,5,31,B10,3=3,0,0,6,9,13,12,22,15,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B10,1, B10,2, and B10,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=L14=5,−5, L1=L15=1,−4, L2=−4,9, L3=−3,−8, L4=5,−1, L5=−L12=L13=1,6, L6=3,−3, L7=−7,9, L8=1,3, L9=2,3, L10=2,−4, L11=−2,−12, L16=−3,12, L17=6,−15, L18=−7,−3, L19=4,−9, and L20=−2,2.
According to (9) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.11. The Cases ofp ≡ 31, 199, 271, 391, 439, 559 mod840

Lemma 29.
Suppose thatp ≡ 31, 199, 271, 391, 439, 559 mod840 is a prime. Then, ℬ=C02⋅B11,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(25)B11,1=4,0,9,0,5,1,1,3,10,5,3,12,8,26,B11,2=1,0,0,4,3,10,5,23,21,31,B11,3=3,0,0,6,27,13,7,22,6,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B11,1, B11,2, and B11,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=L15=1,−4, L2=−4,9, L3=−3,−8, L4=5,−1, L5=−L12=1,6, L6=3,−3, L7=−7,27, L8=16,−1, L9=L10=2,−20, L11=−2,−3, L13=24,2, L14=5,−21, L16=−3,7, L17=−6,−21, L18=−7,−4, L19=−4,5, and L20=−18,2.
According to (10) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.12. The Cases ofp ≡ 107, 323, 347, 443, 683, 827 mod840

Lemma 30.
Suppose thatp ≡ 107, 323, 347, 443, 683, 827 mod840 is a prime. Then, ℬ=C02⋅B12,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(26)B12,1=9,0,5,0,1,1,0,3,4,5,6,12,7,26,B12,2=4,0,0,4,2,10,3,23,7,31,B12,3=−3,0,0,6,−6,13,6,22,−10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B12,1, B12,2, and B12,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=4,−4, L1=−8,−4, L2=−1,4, L3=−5,−9, L4=3,−4, L5=−1,−5, L6=−L10=3,2, L7=−6,2, L8=4,−16, L9=6,12, L11=5,7, L12=L13=1,−3, L14=L18=1,−7, L15=2,−2, L16=−6,6, L17=10,−4, L19=3,−9, and L20=−3,−5.
According to (11) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


### 4.13. The Cases ofp ≡ 83, 227, 467, 563, 587, 803 mod840

Lemma 31.
Suppose thatp ≡ 83, 227, 467, 563, 587, 803 mod840 is a prime. Then, ℬ=C02⋅B13,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(27)B13,1=9,0,5,0,1,1,0,3,4,5,6,12,7,26,B13,2=1,0,3,4,0,10,4,23,6,31,B13,3=2,0,0,6,9,13,18,22,10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B13,1, B13,2, and B13,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:L0=4,−4, L1=−8,−4, L2=−1,4, L3=−5,−9, L4=−L6=3,2, L5=L10=−1,−5, L7=9,2, L8=2,−8, L9=6,9, L11=5,−8, L12=L14=1,−3, L13=4,7, L15=2,−2, L16=−6,18, L17=−10,1, L18=−3,−7, L19=1,−16, and L20=−3,−6.According to (12) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.We are now in a position to prove Theorem5.Proof of Theorem 5.
For each primep≡3mod4 and p>7, we obtain a 2-CP 5,7,1,2/3,1/3;41p from Lemmas 19–31. Let ℬ=0,24,69,142,217,0,40,97,138,210,0,42,85,129,197,0,47,106,154,232,0,37,76,140,241,0,49,105,163,226,0,1,3,7,12,20,30,0,14,35,50,66,88,150,0,25,51,79,111,144,178.
Then,ℬ forms a 2-CP 5,7,1,2/3,1/3;41p for p=7.
By Lemmas2 and 3, we obtain an optimal 2-CP 5,7,1,2/3,1/3;41p and an optimal 41p,5,7,1,2/3,1/3-OOC for any prime p≡3mod4 and p≥7. This completes the proof.


## 4.1. The Cases ofp ≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840

Lemma 19.
Suppose thatp≡ 71, 191, 239, 359, 431, 599, 311, 479, 551, 671, 719, 839 mod840 is a prime, ξ=minx|x∈C12. Then, ℬ=C02⋅B1,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p where(15)B1,1=0,0,5,0,1,1,ξ,3,3,5,2,12,4,26,B1,2=4,0,0,4,2,10,5,23,1,31,B1,3=2,0,0,6,1,13,3,22,4,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B1,1, B1,2, and B1,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=1,−4, L2=ξ−1,3−ξ, L3=ξ,ξ−5, L4=2,−4, L5=L10=−L12=3,−2, L6=−2,2, L7=L20=−1,1, L8=−L15=4,−1, L9=2,2−ξ, L11=−L14=−2,1, L13=−1,3, L16=3,−3, L17=3,−4, L18=ξ−4,−1, and L19=5,−1.
According to (1) and (2) of Lemma 6, 2∈C02, 3∈C02, 5∈C02, and ξ−l∈C02,1≤l<ξ. It is easy to check that Li∩Ck2=1, 0≤i≤40, k=0,1. Hence, △ℬ covers every element in Zp×Z41/0×Z41 exactly once, and each element of 0×Z41 is not covered. So, ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.2. The Cases ofp ≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840

Lemma 20.
Suppose thatp ≡ 43, 67, 163, 403, 547, 667, 187, 283, 307, 523, 643, 787 mod840 is a prime. Then, ℬ=C02⋅B2,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(16)B2,1=1,0,2,0,3,1,−3,3,6,5,0,12,5,26,B2,2=0,0,4,4,2,10,1,23,3,31,B2,3=6,0,0,6,15,13,12,22,10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B2,1, B2,2, and B2,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=L20=1,−1, L1=−L12=1,2, L2=−6,9, L3=−L5=−4,−5, L4=−L15=3,4, L6=−2,−6, L7=−6,15, L8=2,−2, L9=3,−3, L10=2,−3, L11=−3,−4, L13=−1,9, L14=5,1, L16=12,−2, L17=−10,−5, L18=−8,−1, and L19=−3,−6.
According to (3) and (4) of Lemma 6, we know that 2∈C12, 3∈C12, and 5∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.3. The Cases ofp ≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840

Lemma 21.
Suppose thatp ≡ 23, 263, 407, 527, 743, 767, 47, 143, 167, 383, 503, 647 mod840 is a prime. Then, ℬ=C02⋅B3,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(17)B3,1=9,0,5,0,6,1,4,3,8,5,7,12,10,26,B3,2=4,0,3,4,1,10,2,23,−2,31,B3,3=5,0,0,6,3,13,1,22,9,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B3,1, B3,2, and B3,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=4,−4, L1=−L5=−L7=1,−3, L2=−2,4, L3=L15=−1,−5, L4=−L13=2,−1, L6=−2,−5, L9=L20=−2,3, L8=8,−4, L10=6,−3, L11=L16=−L19=1,−4, L12=−2,2, L14=5,3, L17=−9,6, and L18=−6,2.
According to (5) and (6) of Lemma 6, we know that 2∈C02, 3∈C02, and 5∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.4. The Cases ofp ≡ 211, 331, 379, 499, 571, 739 mod840

Lemma 22.
Suppose thatp ≡ 211, 331, 379, 499, 571, 739 mod840 is a prime. Then, ℬ=C02⋅B4,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(18)B4,1=0,0,2,0,1,1,4,3,5,5,3,12,−3,26,B4,2=1,0,0,4,2,10,6,23,10,31,B4,3=−5,0,0,6,10,13,2,22,16,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B4,1, B4,2, and B4,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=2,−2, L1=1,−1, L2=L12=1,3, L3=L16=2,4, L4=4,−1, L5=L15=5,3, L6=2,5, L7=−2,10, L8=4,14, L9=−1,−8, L10=1,−9, L11=2,−21, L13=4,15, L14=−6,−10, L17=6,−16, L18=−5,7, L19=6,−7, and L20=8,−8.
According to (7) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.5. The Cases ofp ≡ 19, 139, 451, 619, 691, 811 mod840

Lemma 23.
Suppose thatp ≡ 19, 139, 451, 619, 691, 811 mod840 is a prime. Then, ℬ=C02⋅B5,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(19)B5,1=0,0,2,0,1,1,4,3,5,5,3,12,−3,26,B5,2=5,0,0,4,8,10,1,23,7,31,B5,3=2,0,0,6,10,13,7,22,14,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B5,1, B5,2, and B5,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=2,−2, L1=1,−1, L2=−L9=L12=1,3, L3=2,4, L4=4,−5, L5=L15=5,3, L6=8,−2, L7=−2,10, L8=6,7, L10=−2,3, L11=2,−12, L13=−7,8, L14=−6,−7, L16=4,7, L17=−14,4, L18=7,4, L19=1,−5, and L20=8,1.
According to (8) of Lemma 6, we know that 2∈C12, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.6. The Cases ofp ≡ 11, 179, 491, 611, 659, 779 mod840

Lemma 24.
Suppose thatp ≡ 11, 179, 491, 611, 659, 779 mod840 is a prime. Then, ℬ=C02⋅B6,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(20)B6,1=2,0,7,0,1,1,3,3,6,5,8,12,10,26,B6,2=7,0,0,4,6,10,4,23,2,31,B6,3=2,0,0,6,1,13,3,22,5,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B51, B52, and B53. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=−L12=−6,−1, L2=2,3, L3=−L5=1,−4, L4=5,−7, L6=−2,6, L7=−L13=2,1, L8=L14=−2,2, L9=5,2, L10=5,−1, L11=7,−3, L15=−3,−8, L16=−9,3, L17=4,−5, L18=−7,3, L19=4,−1, and L20=4,−4.
According to (13) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.7. The Cases ofp ≡ 59, 131, 251, 299, 419, 731 mod840

Lemma 25.
Suppose thatp ≡ 59, 131, 251, 299, 419, 731 mod840 is a prime. Then, ℬ=C02⋅B7,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(21)B7,1=2,0,7,0,1,1,3,3,6,5,8,12,10,26,B7,2=3,0,0,4,1,10,−4,23,8,31,B7,3=1,0,0,6,4,13,3,22,9,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B7,1, B7,2, and B7,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=−L12=−6,−1, L2=2,3, L3=−L5=1,−4, L4=−L13=5,−3, L6=1,−1, L7=−L19=2,4, L8=6,12, L9=5,−1, L10=−2,−5, L11=7,−8, L14=2,−8, L15=−3,−8, L16=−9,3, L17=5,−9, L18=−7,7, and L20=−7,−4.
According to (14) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C02, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.8. The Cases ofp ≡ 127, 247, 463, 487, 583, 823 mod840

Lemma 26.
Suppose thatp ≡ 127, 247, 463, 487, 583, 823 mod840 is a prime. Then, ℬ=C02⋅B8,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(22)B8,1=7,0,4,0,2,1,5,3,9,5,6,12,−1,26,B8,2=9,0,0,4,21,10,1,23,5,31,B8,3=0,0,4,6,3,13,8,22,18,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B8,1, B8,2, and B8,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=3,−3, L1=−5,−2, L2=4,3, L3=−L12=1,−2, L4=7,−9, L5=2,5, L6=21,4, L7=−1,−3, L8=10,4, L9=5,1, L10=12,4, L11=4,−18, L13=−20,3, L14=−7,−5, L15=5,8, L16=3,4, L17=−14,15, L18=6,8, L19=1,−8, and L20=16,10.
According to (15) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C12, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.9. The Cases ofp ≡ 103, 223, 367, 607, 703, 727 mod840

Lemma 27.
Suppose thatp ≡ 103, 223, 367, 607, 703, 727 mod840 is a prime. Then, ℬ=C02⋅B9,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p where(23)B9,1=7,0,4,0,2,1,5,3,9,5,6,12,−1,26,B9,2=7,0,0,4,1,10,−1,23,4,31,B9,3=1,0,0,6,3,13,8,22,10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B9,1, B9,2, and B9,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=L7=3,−3, L1=−L8=−5,−2, L2=4,3, L3=−L12=1,−2, L4=7,−7, L5=2,5, L6=1,−1, L9=5,1, L10=3,−6, L11=4,−9, L13=−2,2, L14=−7,−4, L15=5,8, L16=3,8, L17=−10,7, L18=6,8, L19=−1,−7, and L20=−3,10.
According to (16) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C12, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.10. The Cases ofp ≡ 79, 151, 319, 631, 751, 799 mod840

Lemma 28.
Suppose thatp ≡ 79, 151, 319, 631, 751, 799 mod840 is a prime. Then, ℬ=C02⋅B10,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p where(24)B10,1=4,0,9,0,5,1,1,3,10,5,3,12,8,26,B10,2=1,0,0,4,3,10,4,23,5,31,B10,3=3,0,0,6,9,13,12,22,15,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B10,1, B10,2, and B10,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=L14=5,−5, L1=L15=1,−4, L2=−4,9, L3=−3,−8, L4=5,−1, L5=−L12=L13=1,6, L6=3,−3, L7=−7,9, L8=1,3, L9=2,3, L10=2,−4, L11=−2,−12, L16=−3,12, L17=6,−15, L18=−7,−3, L19=4,−9, and L20=−2,2.
According to (9) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.11. The Cases ofp ≡ 31, 199, 271, 391, 439, 559 mod840

Lemma 29.
Suppose thatp ≡ 31, 199, 271, 391, 439, 559 mod840 is a prime. Then, ℬ=C02⋅B11,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(25)B11,1=4,0,9,0,5,1,1,3,10,5,3,12,8,26,B11,2=1,0,0,4,3,10,5,23,21,31,B11,3=3,0,0,6,27,13,7,22,6,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B11,1, B11,2, and B11,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=5,−5, L1=L15=1,−4, L2=−4,9, L3=−3,−8, L4=5,−1, L5=−L12=1,6, L6=3,−3, L7=−7,27, L8=16,−1, L9=L10=2,−20, L11=−2,−3, L13=24,2, L14=5,−21, L16=−3,7, L17=−6,−21, L18=−7,−4, L19=−4,5, and L20=−18,2.
According to (10) of Lemma 6, we know that 2∈C02, 3∈C12, 5∈C02, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.12. The Cases ofp ≡ 107, 323, 347, 443, 683, 827 mod840

Lemma 30.
Suppose thatp ≡ 107, 323, 347, 443, 683, 827 mod840 is a prime. Then, ℬ=C02⋅B12,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(26)B12,1=9,0,5,0,1,1,0,3,4,5,6,12,7,26,B12,2=4,0,0,4,2,10,3,23,7,31,B12,3=−3,0,0,6,−6,13,6,22,−10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B12,1, B12,2, and B12,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:
L0=4,−4, L1=−8,−4, L2=−1,4, L3=−5,−9, L4=3,−4, L5=−1,−5, L6=−L10=3,2, L7=−6,2, L8=4,−16, L9=6,12, L11=5,7, L12=L13=1,−3, L14=L18=1,−7, L15=2,−2, L16=−6,6, L17=10,−4, L19=3,−9, and L20=−3,−5.
According to (11) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C02. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.


## 4.13. The Cases ofp ≡ 83, 227, 467, 563, 587, 803 mod840

Lemma 31.
Suppose thatp ≡ 83, 227, 467, 563, 587, 803 mod840 is a prime. Then, ℬ=C02⋅B13,j:j=1,2,3 forms a 2-CP 5,7,1,2/3,1/3;41p, where(27)B13,1=9,0,5,0,1,1,0,3,4,5,6,12,7,26,B13,2=1,0,3,4,0,10,4,23,6,31,B13,3=2,0,0,6,9,13,18,22,10,30.Proof.
We compute the differences ofLi, 0≤i≤40, from B13,1, B13,2, and B13,3. It is not difficult to see that Ls=−L41−s, 21≤s≤40. So, we only need to compute the differences of Li for 0≤i≤20 as follows:L0=4,−4, L1=−8,−4, L2=−1,4, L3=−5,−9, L4=−L6=3,2, L5=L10=−1,−5, L7=9,2, L8=2,−8, L9=6,9, L11=5,−8, L12=L14=1,−3, L13=4,7, L15=2,−2, L16=−6,18, L17=−10,1, L18=−3,−7, L19=1,−16, and L20=−3,−6.According to (12) of Lemma 6, we know that 2∈C12, 3∈C02, 5∈C12, and 7∈C12. It is easy to check that ℬ forms a 2-CP 5,7,1,2/3,1/3;41p.We are now in a position to prove Theorem5.Proof of Theorem 5.
For each primep≡3mod4 and p>7, we obtain a 2-CP 5,7,1,2/3,1/3;41p from Lemmas 19–31. Let ℬ=0,24,69,142,217,0,40,97,138,210,0,42,85,129,197,0,47,106,154,232,0,37,76,140,241,0,49,105,163,226,0,1,3,7,12,20,30,0,14,35,50,66,88,150,0,25,51,79,111,144,178.
Then,ℬ forms a 2-CP 5,7,1,2/3,1/3;41p for p=7.
By Lemmas2 and 3, we obtain an optimal 2-CP 5,7,1,2/3,1/3;41p and an optimal 41p,5,7,1,2/3,1/3-OOC for any prime p≡3mod4 and p≥7. This completes the proof.


## 5. Conclusions

In this paper, by constructing explicit 2-CPW,1,Q;n, we have obtained the explicit constructions of optimal 31p,5,7,1,1/2,1/2-OOCs and 41p,5,7,1,2/3,1/3-OOCs, for any prime p≡3mod4 and p≥7. In addition, referring to [34, 35], for 1≤t<p−1/2, if we break t blocks of size 7 into 3 of 31p,5,7,1,1/2,1/2-OOCs and 41p,5,7,1,2/3,1/3-OOCs, then we will get new infinite classes of optimal 31p,3,5,7,1,7t/p−1+6t,p−1/2p−1+6t,p−1−2t/2p−1+6t-OOCs and 41p,3,5,7,1,14t/3p−1+4t,2p−1/3p−1+4t,p−1−2t/3p−1+4t-OOCs, respectively.



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*Source: 3368661-2022-01-03.xml*