Keywords:

been proved in [15] that f (S) is not algebraic in the set of generators of S.

 1978 H. S. Wilf proposed a conjecture suggesting a regularity in the set N \ S. It says the following :

n(S).

 ), only special cases have been solved and it remains wide open. In [9], D. Dobbs and G. Matthews proved Wilf's Definition 1.0.9. Let n ∈ S * . We define the Apéry set of S with respect to n, denoted by Ap(S, n), to be the set Ap(S, n) = {s ∈ S : sn / ∈ S}.

Remark 1.0.10. Given a non zero integer n and two integers a and b, we write a ≡ b mod (n) to denote that n divides a -b. We denote by b mod n the remainder of the division of b by n.

From Definition 1.0.9, we can easily see the following.

Lemma 1.0.11. Let S be a numerical semigroup and let n ∈ S * . For all 1 ≤ i ≤ n, let w(i) be the smallest element of S such that w(i) ≡ i mod (n). We have the following :

Ap(S, n) = {0, w(1), . . . , w(n -1)}.

Proposition 1.0.12. Let S be a numerical semigroup. Let n ∈ S * and let Ap(S, n) = {w 0 < w 1 . . . < w n-1 } be the Apéry set of S with respect to n. We have the following :

Proposition 1.0.13. (See Lemma 2.6 in [18]) Let S be a numerical semigroup and let n ∈ S * . For all s ∈ S, there exists a unique (k, w) ∈ N × Ap(S, n) such that s = kn + w.

As a consequence of Proposition 1.0.13, we obtain the following property.

Corollary 1.0.14. (Theorem 2.7 in [18]) Let S be a numerical semigroup. Then, S is finitely generated.

Definition 1.0.15. Let S be a numerical semigroup and let A ⊆ S * . We say that A is a minimal set of generators of S if S =< A > and for all x ∈ A, x cannot be written as a linear combination with nonnegative integer coefficients of other elements in A.

Corollary 1.0.16. (See Corollary 2.8 in [18]) Let S be a numerical semigroup. Then, S has a minimal set of generators. This set is finite and unique.

Definition 1.0.17. Let S be a numerical semigroup. We define the following invariants :

• The embedding dimension of S denoted by ν(S), or ν for simplicity, is the cardinality of the minimal set of generators of S.

• The multiplicity of S denoted by m(S), or m for simplicity, is the smallest non zero element of S.

Lemma 1.0.18. (See Proposition 2.10 in [18]) Let S be a numerical semigroup with multiplicity m and embedding dimension ν. We have ν ≤ m.

Let us recall some basic and important invariants of numerical semigroups.

Definition 1.0.19. Let S be a numerical semigroup. We introduce some invariants associated to a numerical semigroup S :

• We define the Frobenius number of S, denoted by f or f (S) to be max (Z \ S).

• We define the conductor of S, denoted by c or c(S) to be f (S) + 1.

Definition 1.0.22. Let S be a numerical semigroup. We say that x ∈ N is a pseudo-Frobenius number if x / ∈ S and x + s ∈ S for all s ∈ S * . We denote by P F (S) the set of all pseudo-Frobenius numbers of S. We denote the cardinality of P F (S) by t(S) and we call it the type of S. It results from the definition of f (S) that f (S) ∈ P F (S), and also f (S) = max (P F (S)).

Corollary 1.0.23. (See Theorem 20 in [11]) Let S be a numerical semigroup with Frobenius number f (S), type t(S) and n(S) = |{s ∈ S : s < f (S)}|. Then, we have

. For some families of numerical semigroups this conjecture is known to be true, but the general case remains unsolved. Remark 1.0.24. By Corollary 1.0.23, if t(S) ≤ ν(S) -1, then S satisfies Wilf's conjecture. Definition 1.0.25. Let a, b ∈ N. We define ≤ S as follows : a ≤ S b if and only if b -a ∈ S.

Remark 1.0.26. As S is a numerical semigroup, it easily follows that ≤ S is an order relation over S (reflexive, transitive and anti symmetric).

Definition 1.0.27. Let S be a numerical semigroup and n ∈ S * . Let Ap(S, n) = {w 0 = 0 < w 1 < w 2 < . . . < w n-1 } be the Apéry set of S with respect to n. Then, define the following sets :

Lemma 1.0.28. (See Lemma 6 in [11]) Let S be a numerical semigroup, n ∈ S * and Ap(S, n) be the Apéry set of S with respect to n. Let w ∈ Ap(S, n) and u ∈ S. If there exist v ∈ S such that w = u + v, then u ∈ Ap(S, n).

)) if and only if x = w i + w j for all w i , w j ∈Ap(S, n) * .

• x ∈ max ≤ S (Ap(S, n)) if and only if w i = x + w j for all w i , w j ∈Ap(S, n) * .

Proof. Let x ∈ Ap(S, n) * .

• Let x ∈ min ≤ S (Ap(S, n)). Suppose by the way of contradiction that x = w i + w j for some w i , w j ∈ Ap(S, n) * . Then, x = w i + w j with w i ∈ Ap(S, n) and w j ∈ S which implies that x / ∈ min ≤ S (Ap(S, n)) and we get a contradiction. Conversely, suppose that x = w i + w j for all w i , w j ∈Ap(S, n) * . Suppose by the way of contradiction that x / ∈ min ≤ S (Ap(S, n)), then there exist w i ∈ Ap(S, n) and s ∈ S such that x = w i + s. By Lemma 1.0.28, it follows that s ∈ Ap(S, n). Thus, x = w i + s such that w i , s ∈ Ap(S, n) which gives a contradiction.

• Let x ∈ max ≤ S (Ap(S, n)). Suppose by the way of contradiction that w i = x + w j for some w i , w j ∈ Ap(S, n) * . Then, w i = x + w j with w i ∈ Ap(S, n) and w j ∈ S which implies that x / ∈ max ≤ S (Ap(S, n)) and we get a contradiction. Conversely, suppose that w i = x + w j for all w i , w j ∈ Ap(S, n) * . Suppose by the way of contradiction that x / ∈ max ≤ S (Ap(S, n)), then there exist w i ∈ Ap(S, m) and s ∈ S such that w i = x + s. By Lemma 1.0.28, it follows that s ∈ Ap(S, n). Thus, w i = x + s such that w i , s ∈ Ap(S, n) which gives a contradiction.

Thus, the proof is complete.

Proposition 1.0.30. (See Lemma 3.2 in [7] ) Let S be a numerical semigroup with multiplicity m and embedding dimension ν and let n ∈ S * . Let Ap(S, n) be the Apéry set of S with respect to n and let {g 1 < g 2 < . . . < g ν } be the minimal set of generators of S. We have the following :

From Proposition 1.0.30, it follows Corollary 1.0.31.

Corollary 1.0.31. Let S be a numerical semigroup with multiplicity m, embedding dimension ν and {g 1 = m, g 2 , . . . , g ν } the minimal system of generators of S. Let n ∈ S * and Ap(S, n) be the Apéry set of S with respect to n. We have the following :

We introduce in Definitions 1.0.32 and 1.0.33 special kind of numerical semigroups and give some properties of this kind in Lemma 1.0.34. Definition 1.0.32. A numerical semigroup is said to irreducible if and only if S cannot be expressed as the intersection of two numerical semigroups S 1 , S 2 such that S ⊂ S 1 , S ⊂ S 2 . Definition 1.0.33. Let S be a numerical semigroup. We have the following :

• S is said to be symmetric if and only if S is irreducible and f (S) is odd.

• S is said to be pseudo-symmetric if and only if S is irreducible and f (S) is even.

Lemma 1.0.34. (See Corollary 4.5 in [18]) Let S be a numerical semigroup with Frobenius number f (S) and genus g(S). We have the following :

ν ≥ m and w m-1 -m ≥ w x + w y 2. [START_REF][END_REF] Numerical semigroups with m -ν > (n-2)(n-3)

Introduction

Let N denote the set of natural numbers, including 0. A semigroup S is an additive submonoid of (N, +), that is 0 ∈ S and if a, b ∈ S, then a + b ∈ S. A numerical semigroup S is a submonoid of N of finite complement, i.e., N \ S is a finite set. It can be shown that a submonoid of N is a numerical semigroups if and only if the group generated by S in Z (namely the set of elements s i=1 λ i a i , λ i ∈ Z, a i ∈ S) is Z. There are many invariants associated to a numerical semigroup S. The Apéry set of S with respect to an element a ∈ S is defined as Ap(S, a) = {s ∈ S; s -a / ∈ S}.

The elements of N \ S are called the gaps of S. The largest gap is denoted by

f = f (S) = max(N \ S)
and is called the Frobenius number of S. The number f (S)+1 is known as the conductor of S and denoted by c or c(S).The number of gaps g = g(S) = |N \ S| is known as genus of S. The smallest non zero element m = m(S) of S is called the multiplicity of S and the set {s ∈ S; s < f (S)} is denoted by n(S). Every numerical semigroup S is finitely generated, i.e., S is of the form S =< g 1 , . . . , g ν >= Ng 1 + . . . + Ng ν for suitable unique coprime integers g 1 , . . . , g ν . The number of minimal set of generators of S is denoted by

ν = ν(S)
and is called the embedding dimension of S. An integer x ∈ N\S is called a pseudo-Frobenius number if x + S \ 0 ⊆ S. The type of the semigroup, denoted by t(S) is the cardinality of set of pseudo-frobenius numbers. We have formulas linking these invariants.

Frobenius in his lectures proposed the problem of giving a formula for the largest integer that is not representable as a linear combination with nonnegative integer coefficients of a given set of positive integers whose greater common divisor is one. He also threw the question of determining how many positive integers do not have such a representation. This problem is known as Diophantine Frobenius Problem. Using the terminology of numerical semigroups, the problem is to give a formula, in terms of the elements in a minimal system of generators of a numerical semigroup S, for the greatest integer not in S. This problem, introduced and solved by Sylvester for the case ν = 2 (see [START_REF] Sylvester | Mathematical questions with their solutions[END_REF]), has been widely studied. For ν = 3, in 1962 Brauer and Shockly (see [START_REF][END_REF]) found a formula for the Frobenius number but their solution was not a polynomial in the generators and it involved magnitudes which could not be expressed by the generators. Later on, more solutions to this case were found by using different methods (for example see [START_REF] Selmer | On the linear diophantine problem of frobenius in three variables[END_REF]). However, all of these methods do not give explicit formula of the Frobenius number in terms of the generators. Generally, it has conjecture for ν ≤ 3. In [14], N. Kaplan proved it for c ≤ 2m and in [START_REF] Eliahou | Wilf's conjecture and macaulay's theorem[END_REF] S. Eliahou extended Kaplan's work for c ≤ 3m.

In Chapter 1, we recall some basics about numerical semigroups that will be used through the thesis.

In Chapter 2, we generalize the case covered by A. Sammartano in [START_REF] Sammartano | Numerical semigroups with large embedding dimension satisfy wilf ?s conjecture[END_REF], who showed that Wilf's conjecture holds for 2ν ≥ m, and m ≤ 8, based on the idea of counting the elements of S in some intervals of length m. We use different intervals in order to get an equivalent form of Wilf's conjecture and then we prove it in some relevant cases. In particular our calculations cover the case where 2ν ≥ m, proved by Sammartano in [START_REF] Sammartano | Numerical semigroups with large embedding dimension satisfy wilf ?s conjecture[END_REF].

Here are few more details on the contents of this Chapter. Section 2.1 is devoted to give some notations that will enable us in the same Section to give an equivalent form of Wilf's conjecture. In Section 2.2, we give some technical results needed in the Chapter. Let Ap(S, m) = {0 = w 0 < w 1 < • • • < w m-1 }. In Section 2.3, first, we show that Wilf's conjecture holds for numerical semigroups that satisfy w m-1 ≥ w 1 + w α and

(2 + α-3 q )ν ≥ m for some 1 < α < m -1 where c = qm -ρ for some q ∈ N, 0 ≤ ρ ≤ m -1. Then, we prove Wilf's conjecture for numerical semigroups with m -ν ≤ 4 in order to cover the case where 2ν ≥ m. We also show that a numerical semigroup with m -ν = 5 verify Wilf's conjecture in order to prove the conjecture for m = 9. Finally, we show in this Section, using the previous cases, that Wilf's conjecture holds for numerical semigroups with (2 + 1 q )ν ≥ m. In Section 2.4, we prove Wilf's conjecture for numerical semigroups with w m-1 ≥ w α-1 + w α and ( α+33 )ν ≥ m for some 1 < α < m -1. In Section 2.5, we show Wilf's conjecture holds for numerical semigroups with w m-1 -m ≥ w x + w y and 2 + ν ≥ m. The last Section 2.6 aims to verify the conjecture in the case m -ν > (n-2)(n-3) 2 and also in the case n ≤ 5.

Exact determination of Ap(S, m), f (S), g(S) and P F (S) is a difficult problem. When S is generated by an arithmetic sequence < m, m + 1, . . . , m + l >, Brauer [START_REF][END_REF] gave a formula for f (S). Roberts [17] extended this result to generators in arithmetic progression (see also [3], [START_REF] Bras-Amorós | Fibonacci-like behavior of the number of numerical semigroups of a given genus[END_REF]). Selmer [START_REF] Selmer | On the linear diophantine problem of frobenius in three variables[END_REF] and Grant [START_REF] Bateman | Remark on a recent note on linear forms[END_REF] generalized this to the case S =< m, hm + d, hm + 2d, . . . , hm + ld >. In [16], it has been considered the case of semigroups generated by {m, m + d, . . . , m + ld, c} (called almost arithmetic semigroups) where it has been given a method to determine Ap(S, m) and also symmetric almost arithmetic semigroups. In [START_REF] García-Marco | Numerical semigroups ii : pseudo-symmetric aa-semigroups[END_REF], pseudo symmetric almost arithmetic semigroups have been characterized. In Chapter 3, we focus our attention on numerical semigroup consisting of all non-negative integer linear combinations of relatively prime positive integers m, m + 1, . . . , m + l, k(m + l) + r where k, m, l, r are positive integers and r ≤ (k + 1)l + 1. We give formulas for Ap(S, m), f (S), g(S) and P F (S). We also determine the symmetric and the pseudo symmetric numerical semigroups of this form. Note that our semigroups < m, m + 1, . . . , m + l, k(m + l) + r > are almost arithmetic semigroups. The advantage is that we are able for this class of semigroups to determine all the invariants with simple formulas. Good references on numerical semigroups are [18] and [1].

Basics and notations

Definition 1.0.1. Let S be a subset of N. We say that S is a submonoid of (N, +) if the following holds : • {0} and N are trivially submonoids of N.

• 0 ∈ S. • If a, b ∈ S, then a + b ∈ S.
• Let d be an element of N, the set dN = {da : a ∈ N} is a submonoid of N. Definition 1.0.4. Let S be a submonoid of N. If N \ S is a finite set, then S is said to be a numerical semigroup.

We have the following characterization of numerical semigroups : Proposition 1.0.5. (See Lemma 2.1 in [18]) Let S = {0}, and S = N be a semigroup of N and let G be the subgroup of Z generated by S, i.e., (G = { s i=1 λ i a i , s ∈ N, λ i ∈ Z, a i ∈ S}). Then, S is a numerical semigroup if and only if G = Z, i.e., (gcd(S)=1). Proposition 1.0.6. (See Proposition 2.2 in [18]) Let S be a semigroup of N. Then, S is isomorphic to a numerical semigroup. Definition 1.0.7. Let S be a numerical semigroup and let A ⊆ S. We say that S is generated by A and we write S =< A > if for all s ∈ S, there exist a 1 , . . . , a r ∈ A and λ 1 , . . . , λ r ∈ N such that a = r i=1 λ i a i . We say that S is finitely generated if S =< A > with A ⊆ S and A is a finite set. Remark 1.0.8. Through this thesis X * will stand for X \ {0}.

Next, we introduce an important tool associated to a numerical semigroup.

• We define the set of gaps of S, denoted by G(S) to be N \ S.

• We define the genus of S, denoted by g(S) to be the cardinality of G(S).

• We denote by n(S), the cardinality of {s ∈ S : s ≤ f (S)}.

Remark 1.0.20. Note that f (S) ≥ 1 for all non trivial numerical semigroups. Lemma 1.0.21. (See in [START_REF][END_REF], [START_REF] Selmer | On the linear diophantine problem of frobenius in three variables[END_REF]) Let S be a numerical semigroup and let n ∈ S. Then,

• f (S) = max(Ap(S, n)) -n.

• g(S) = 1 n w∈Ap(S,n) w -1 2 (n -1).

• S is symmetric if and only if g(S) = f (S) + 1 2 .

• S is pseudo-symmetric if and only if g(S) = f (S) + 2 2 .

Remark 1.0.35. Consider the following notation that will be used throug this thesis :

• We denote by floor (x) = x the largest integer less than or equal to x.

• We denote by ceil (x)= x the smallest integer greater than or equal to x.

Wilf's conjecture

In this chapter, we give an equivalent form of Wilf's conjecture in terms of the elements of the Apéry set of S, embedding dimension and the multiplicity. We also give an affirmative answer to Wilf's conjecture in some cases.

Equivalent form of Wilf's conjecture

Let the notations be as in the introduction. For the sake of clarity we shall use the notations ν, f, n, c... for ν(S), f (S), n(S), c(S).... In this Section, we will introduce some notations and family of numbers that will enable us to give an equivalent form of Wilf's conjecture at the end of this Section.

Notation. Let S be a numerical semigroup with multiplicity m and conductor c = f + 1. Denote by q = c m .

Thus, qm ≥ c and c = qm -ρ with 0 ≤ ρ < m.

Given a non negative integer k, we define the kth interval of length m,

I k = [km -ρ, (k + 1)m -ρ[= {km -ρ, km -ρ + 1, . . . , (k + 1)m -ρ -1}.
We denote by

n k = |S ∩ I k |.
For j ∈ {1, . . . , m -1}, we define η j to be the number of intervals I k with n k = j. Proposition 2.1.1. Under the previous notations, we have :

i) 1≤ n k ≤ m -1 for all 0 ≤ k ≤ q -1.
ii) n k = m for all k ≥ q.

iii)

q-1 k=0 n k = n(S) = n. iv) m-1 j=1 η j = q. v) m-1 j=1 jη j = q-1 k=0 n k = n.
Proof.

i)

We can easily verify that if S contains m consecutive elements a, a + 1, . . . a + m -1, then for all

n ≥ a + m, n ∈ S. Since (q -1)m -ρ < f < qm -ρ, then it follows that n k ≤ m -1 for all 0 ≤ k ≤ q -1. Moreover, km ∈ S ∩ I k for all 0 ≤ k ≤ q -1, thus n k ≥ 1. ii) We have f = qm -ρ -1 ∈ I q-1 .
From the definition of the Frobenius number, it follows that n k = m for all k ≥ q.

iii)

q-1 k=0 n k is nothing but the cardinality of {s ∈ S; s < f } which is n(S) by definition. iv) We have 1 ≤ S ∩ I k ≤ m -1 if and only if 0 ≤ k ≤ q -1
. This implies our assertion.

v) The sum m-1 j=1 jη j is nothing but the cardinality of | ∪ q-1 k=0 S ∩ I k | = n. This proves our assertion. Thus, the proof is complete.

Next, we will express η j in terms of th Apéry set.

Proposition 2.1.2. Let Ap(S, m) = {w 0 = 0 < w 1 < w 2 < . . . < w m-1 }. Under the previous notations, for all 1 ≤ j ≤ m -1, we have

η j = w j + ρ m - w j-1 + ρ m .
Proof. Fix 0 ≤ k ≤ q -1 and let 1 ≤ j ≤ m -1. We will show that the interval I k contains exactly j elements of S if and only if w j-1 < (k + 1)m -ρ ≤ w j .

Suppose that I k contains j elements. Suppose, by contradiction, that w j-1 ≥ (k + 1)m -ρ. We have

w m-1 > . . . > w j-1 ≥ (k + 1)m -ρ, thus w m-1 , . . . , w j-1 ∈ ∪ q t=k+1 I t .
Hence, I k contains at most j -1 elements of S (namely w 0 +km = km, w 1 +k 1 m, w 2 +k 2 m, . . . , w j-2 +k j-2 m for some k 1 , . . . , k j-2 ∈ {0, . . . , k -1}). This contradicts the fact that I k contains exactly j elements of S. Hence,

w j-1 < (k + 1)m -ρ. If w j < (k + 1)m -ρ, then w 0 < . . . < w j < (k + 1)m -ρ, thus w 0 , . . . , w j ∈ ∪ k t=0 I t .
Then, I k contains at least j + 1 elements of S which are :

w 0 + km = km, w 1 + k 1 m, w 2 + k 2 m, . . . , w j + k j m
for some k 1 , . . . , k j ∈ {0, . . . , k -1}, which contradicts the fact that I k contains exactly j elements of S. Hence, w j ≥ (k+1)m-ρ. Consequently, if I k contains exactly j elements of S, then w j-1 < (k+1)m-ρ ≤ w j .

Conversely,

w j-1 < (k + 1)m -ρ implies that w 0 < . . . < w j-1 < (k + 1)m -ρ, then w 0 , . . . , w j-1 ∈ ∪ k t=0 I t .
Hence, I k contains at least j elements of S which are

w 0 + km = km, w 1 + k 1 m, w 2 + k 2 m, . . . , w j-1 + k j-1 m for some k 1 , . . . , k j-1 ∈ {0, . . . , k -1}. On the other hand, w j ≥ (k + 1)m -ρ implies that w m-1 > . . . > w j ≥ (k + 1)m -ρ, then w m-1 , . . . , w j ∈ ∪ q t=k+1 I t .
Thus, I k contains at most j elements of S which are :

w 0 + km = km, w 1 + k 1 m, w 2 + k 2 m, . . . , w j-1 + k j-1 m
for some k 1 , . . . , k j-1 ∈ {0, . . . , k -1}. Hence, if w j-1 < (k + 1)m -ρ ≤ w j , then I k contains exactly j elements of S and this proves our assertion. Consequently,

η j = |{k ∈ N such that |I k ∩ S| = j}| = |{k ∈ N such that w j-1 < (k + 1)m -ρ ≤ w j }| = |{k ∈ N such that w j-1 +ρ m < (k + 1) ≤ w j +ρ m }| = |{k ∈ N such that w j-1 +ρ m -1 < k ≤ w j +ρ m -1}| = |{k ∈ N such that w j-1 +ρ m ≤ k ≤ w j +ρ m -1}| = w j +ρ m - w j-1 +ρ m .
Thus, the proof is complete.

Proposition 2.1.3 gives an equivalent form of Wilf's conjecture using Propositions 2.1.1 and 2.1.2.

Proposition 2.1.3. Let S be a numerical semigroup with multiplicity m, embedding dimension ν and conductor f + 1 = qm -ρ for some q ∈ N and 0 ≤ ρ ≤ m -1. Let w 0 = 0 < w 1 < w 2 < . . . < w m-1 be the elements of Ap(S, m). Then, S satisfies Wilf's conjecture if and only if

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ 0.
Proof. By Proposition 2.1.1, we have

f + 1 ≤ nν ⇔ qm -ρ ≤ ν q-1 k=0 n k ⇔ q-1 k=0 m -ρ ≤ q-1 k=0 n k ν ⇔ q-1 k=0 (n k ν -m) + ρ ≥ 0.
Equivalently, we obtain

m-1 j=1 η j (jν -m) + ρ ≥ 0.
By applying Proposition 2.1.2, we get

m-1 j=1 η j (jν -m) + ρ ≥ 0 ⇔ m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ 0.
Thus, the proof is complete.

Technical results

Let S be a numerical semigroup and let the notations be as in Section 2.1. In this Section, we give some technical results will be used through the Chapter.

Remark 2.2.1. Let Ap(S, m) = {w 0 = 0 < w 1 < . . . < w m-1 }. The following technical remarks will be used through the Chapter :

i) w 0 + ρ m = 0.
ii) For all 1 ≤ i ≤ m -1, we have

w i + ρ m ≥ 1.
iii) For all 1 ≤ i ≤ m -1, we have either

w i + ρ m = w i m or w i + ρ m = w i m + 1. iv) If w i + ρ m = w i m + 1, then w i + ρ m ≥ 2 and ρ ≥ 1. v) For all 0 ≤ i < j ≤ m -1, we have w i + ρ m ≤ w j + ρ m . vi) w m-1 + ρ m = q.
Proof.

i) This is because w 0 = 0 and 0 ≤ ρ < m.

ii) We have m < w i for all 1 ≤ i ≤ m -1. This implies the result since ρ ≥ 0.

iii) For all 1 ≤ i ≤ m -1, let w i = q i m + r i such that q i , r i ∈ N and r i < m. We have w i m = q i . Therefore,

w i + ρ m = q i m + r i + ρ m = q i + r i + ρ m = q i + r i + ρ m = w i m + r i + ρ m . Since 0 ≤ ρ, r i < m, it follows that 0 ≤ r i +ρ m < 2. Consequently, 0 ≤ r i +ρ m ≤ 1. Hence, w i m ≤ w i + ρ m ≤ w i m + 1.
Equivalently, Thus, the proof is complete.

w i + ρ m = w i m or w i + ρ m = w i m + 1. iv) Suppose that w i +ρ m = w i m + 1. By using part ii), we get w i +ρ m ≥ 2. In this case ρ ≥ 1 (as ρ ≥ 0).
Let 1 < α < m -1. Using Remark 2.2.1, we get the following inequalities which will be used later in the Chapter :

α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) = α j=1 w j + ρ m (jν -m) - α j=1 w j-1 + ρ m (jν -m) = α j=1 w j + ρ m (jν -m) - α-1 j=0 w j + ρ m (j + 1)ν -m = α-1 j=1 w j + ρ m (jν -m) + w α + ρ m (αν -m)- w 0 + ρ m (ν -m) - α-1 j=1 w j + ρ m (j + 1)ν -m = w α + ρ m (αν -m) - w 0 + ρ m (ν -m) - α-1 j=1 w j + ρ m ν = w α + ρ m (αν -m) - w 0 + ρ m (ν -m) - w 1 + ρ m ν- α-1 j=2 w j + ρ m ν = w α + ρ m (αν -m) - w 1 + ρ m ν - α-1 j=2 w j + ρ m ν (as w 0 + ρ m = 0).
From Remark 2.2.1 (v), we have

w j +ρ m ≤ wα+ρ m ∀ 2 ≤ j ≤ α -1. Hence, α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ w α + ρ m (αν -m) - w 1 + ρ m ν - α-1 j=2 w α + ρ m ν = w α + ρ m (αν -m) - w 1 + ρ m ν - w α + ρ m (α -2)ν = - w 1 + ρ m ν + w α + ρ m (2ν -m).
Consequently, we have

α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ - w 1 + ρ m ν + w α + ρ m (2ν -m). (2.2.1) Therefore, m-1 j=α+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ m-1 j=α+1 w j + ρ m - w j-1 + ρ m (α + 1)ν -m (as j ≥ α + 1 and w j + ρ m ≥ w j-1 + ρ m ) = (α + 1)ν -m m-1 j=α+1 w j + ρ m - w j-1 + ρ m = (α + 1)ν -m m-1 j=α+1 w j + ρ m - m-1 j=α+1 w j-1 + ρ m = (α + 1)ν -m m-1 j=α+1 w j + ρ m - m-2 j=α w j + ρ m = (α + 1)ν -m m-2 j=α+1 w j + ρ m + w m-1 + ρ m - w α + ρ m - m-2 j=α+1 w j + ρ m = w m-1 + ρ m - w α + ρ m (α + 1)ν -m .
Hence, we obtain

m-1 j=α+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ w m-1 + ρ m - w α + ρ m (α + 1)ν -m . (2.2.2)
The following technical Lemma will be used through the Chapter :

Lemma 2.2.2. Let Ap(S, m) = {w 0 = 0 < w 1 < . . . < w m-1 } and suppose that w i ≥ w j + w k . We have the following :

i) w i + ρ m ≥ w j + ρ m + w k + ρ m -1. ii) If w i + ρ m - w j + ρ m - w k + ρ m = -1, then w j + ρ m = w j m + 1, w k + ρ m = w k m + 1 and ρ ≥ 1.
In particular,

w j + ρ m ≥ 2, w k + ρ m ≥ 2 and ρ ≥ 1.

Proof.

i) Assume that w i ≥ w j + w k . Then, w i + ρ ≥ w j + w k + ρ. Consequently,

w i + ρ m ≥ w j + w k + ρ m ⇒ w i + ρ m ≥ w j + w k + ρ m .
Therefore, we have

w i + ρ m ≥ w j + ρ m + w k m . By Remark 2.2.1 (iii), w k m ≥ w k +ρ m -1. Hence, w i + ρ m ≥ w j + ρ m + w k + ρ m -1.
ii) Suppose that w i ≥ w j + w k and that w i +ρ m -

w j +ρ m -w k +ρ m = -1.
Suppose by the way of contradiction that

w j +ρ m = w j m + 1 or w k +ρ m = w k m + 1 or ρ < 1. By Remark 2.2.1 (iii) and that ρ ≥ 0, it follows that w j + ρ m = w j m or w k + ρ m = w k m or ρ = 0.
Since w i ≥ w j + w k , we have

w i + ρ m ≥ w j + w k + ρ m .
Since

w j +ρ m = w j m or w k +ρ m = w k m or ρ = 0, it follows that w i + ρ m ≥ w j + ρ m + w k + ρ m ,
which contradicts the hypothesis. Hence,

w j + ρ m = w j m + 1, w k + ρ m = w k m + 1 and ρ ≥ 1.
Using Remark 2.2.1 (ii), we get that

w j +ρ m = w j m + 1 ≥ 2, w k +ρ m = w k m + 1 ≥ 2 and ρ ≥ 1.
Thus, the proof is complete.

Numerical semigroups with w

m-1 ≥ w 1 + w α and (2 + α-3 q )ν ≥ m
In this Section, we show that Wilf's conjecture holds for numerical semigroups in the following cases :

1. w m-1 ≥ w 1 + w α and (2 + α-3 q )ν ≥ m for some 1 < α < m -1.
2. m -ν ≤ 5. (Note that the case m -ν ≤ 3 results from the fact that Wilf's conjecture holds for 2ν ≥ m. This case has been proved in [START_REF] Sammartano | Numerical semigroups with large embedding dimension satisfy wilf ?s conjecture[END_REF]), however we shall give a proof in order to cover it through our techniques).

Then, we deduce the conjecture for m = 9 and for (2 + 1 q )ν ≥ m. Next, we will show that Wilf's conjecture holds for numerical semigroups with

w m-1 ≥ w 1 + w α and (2 + α -3 q )ν ≥ m.
Theorem 2.3.1. Let S be a numerical semigroup with multiplicity m, embedding dimension ν and conductor f + 1 = qm -ρ for some q, ρ ∈ N ; 0 ≤ ρ ≤ m -1. Let w 0 = 0 < w 1 < w 2 < . . . < w m-1 be the elements of Ap(S, m). Suppose that w m-1 ≥ w 1 + w α for some 1 < α < m -1. If (2 + α-3 q )ν ≥ m, then S satisfies Wilf's conjecture.

Proof. We are going to use the equivalent form of Wilf's conjecture given in Proposition 2.1.3. Since w m-1 ≥ w 1 + w α , by Lemma 2.2.2, it follows that

w m-1 + ρ m ≥ w 1 + ρ m + w α + ρ m -1. Let x = w m-1 +ρ m -w 1 +ρ m -wα+ρ m . Then, x ≥ -1 and w 1 +ρ m + wα+ρ m = w m-1 +ρ m -x = q -x (
( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + m-1 j=α+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ - w 1 + ρ m ν + w α + ρ m (2ν -m)+ w m-1 + ρ m - w α + ρ m (α + 1)ν -m + ρ (by (2.2.1) and (2.2.2)) = w 1 + ρ m -ν + (α + 1)ν -m -(α + 1)ν -m + w α + ρ m (2ν -m) + w m-1 + ρ m - w α + ρ m (α + 1)ν -m +ρ = w 1 + ρ m (αν -m) + w α + ρ m (2ν -m)+ w m-1 + ρ m - w α + ρ m - w 1 + ρ m (α + 1)ν -m +ρ = ( w 1 + ρ m + w α + ρ m )(2ν -m) + w 1 + ρ m (α -2)ν + w m-1 + ρ m - w α + ρ m - w 1 + ρ m (α + 1)ν -m +ρ = (q -x)(2ν -m) + w 1 + ρ m (α -2)ν+x (α + 1)ν -m + ρ. Consequently, m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ (q -x)(2ν -m) + w 1 + ρ m (α -2)ν+x (α + 1)ν -m + ρ. (2.3.1) Since x = w m-1 + ρ m - w 1 + ρ m - w α + ρ m ≥ -1,
then we have two cases : Note that 3ν < m. We have w 1 = 21, w 14 = 56 and w m-1 = 83 that is w m-1 ≥ w 1 + w 14 . In addition, (2 + α-3 q )ν = (2 + 14-3 4 )6 ≥ 19 = m. Thus, the conditions of Theorem 2.3.1 are valid.

• If x = -1,
( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ (q + 1)(2ν -m) + 2(α -2)ν -(α + 1)ν -m + ρ = ν(2q + 2 + 2α -4 -α -1) -qm + ρ = ν(2q + α -3) -qm + ρ = q ν(2 + α -3 q ) -m + ρ ≥ 0 (
( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ (q -x)(2ν -m) + (α -2)ν + x (α + 1)ν -m + ρ = ν 2q + (α -2)(x + 1) + x -qm + ρ > ν(2q + α -3) -qm + ρ (as x ≥ 0) = q ν(2 + α -3 q ) -m + ρ ≥ 0 (
In the following we shall deduce some cases where Wilf's conjecture holds. We start with the following technical Lemma.

Lemma 2.3.3. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let w 0 = 0 < w 1 < w 2 < . . . < w m-1 be the elements of Ap(S, m).

If m -ν > ( α 2 ) = α(α-1)
2 for some α ∈ N * , then w m-1 ≥ w 1 + w α .

Proof. Suppose by contradiction that w

m-1 < w 1 + w α . Let w ∈ Ap(S, m) * \ min ≤ S (Ap(S, m)).
Thus, w ≤ w m-1 and w = w i + w j for some w i , w j ∈ Ap(S, m) * this follows from Corollary 1.0.29. Hence, w ≤ w m-1 < w 1 + w α . Thus, the only possible values for w are included in

{w i + w j ; 1 ≤ i ≤ j ≤ α -1}. By Corollary 1.0.31, we have m -ν = |Ap(S, m) * \min ≤ S (Ap(S, m))|. Therefore, m -ν ≤ ( α 2 ) = α(α-1) 2 , which is impossible. Hence, w m-1 ≥ w 1 + w α .
Thus, the proof is complete.

Next, we will deduce Wilf's conjecture for numerical Semigroups with

m -ν > α(α -1) 2 and (2 + α -3 q )ν ≥ m.
It will be used later to show that the conjecture holds for those with (2 + 1 q )ν ≥ m, and in order also to cover the result in [START_REF] Sammartano | Numerical semigroups with large embedding dimension satisfy wilf ?s conjecture[END_REF] saying that the conjecture is true for 2ν ≥ m. Corollary 2.3.4. Let S be a numerical semigroup with multiplicity m, embedding dimension ν and conductor f +1 = qm-ρ for some q ∈ N,

0 ≤ ρ ≤ m-1. Suppose that m-ν > ( α 2 ) = α(α-1)
Proof. It follows from Lemma 2.3.

3 that if m -ν > α(α-1)

2

, then w m-1 ≥ w 1 + w α . Now, use Theorem 2.3.1. Thus, the proof is complete.

As a direct consequence of Theorem 2.3.1, we get the following Corollary.

Corollary 2.3.5. Let S be a numerical semigroup with a given multiplicity m and conductor f +1 = qm-ρ for some q ∈ N, 0 ≤ ρ ≤ m -1. Let w 0 = 0 < w 1 < . . . < w m-1 be the elements of Ap(S, m). If w m-1 ≥ w 1 + w α for some 1 < α < m -1 and m ≤ 8 + 4( α-3 q ), then S satisfies Wilf's conjecture.

Proof. By Theorem 2.3.1, we may assume that (2 + α-3 q )ν < m. Therefore,

ν < qm 2q + α -3 ≤ 8q + α -12 2q + α -3 .
Hence, ν < 4. Consequently, S satisfies Wilf's conjecture (see [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]). Thus, the proof is complete.

In the following Lemma, we will show that Wilf's conjecture holds for numerical semigroups with m -ν ≤ 3. This will enable us later to prove the conjecture for numerical semigroups with (2 + 1 q )ν ≥ m and cover the result in [START_REF] Sammartano | Numerical semigroups with large embedding dimension satisfy wilf ?s conjecture[END_REF] saying that the conjecture is true for 2ν ≥ m. Lemma 2.3.6. Let S be a numerical Semigroup with multiplicity m and embedding dimension ν. If m-ν ≤ 3, then S satisfies Wilf's conjecture.

Proof. We may assume that ν ≥ 4 (ν ≤ 3 is solved [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]). We are going to show that S satisfies Wilf's conjecture by means of Proposition 2.1.3.

Case 1.

If m -ν = 0 (S is said to be a numerical semigroup with maximal embedding dimension). Then,

t(S) = m -1 = ν -1 (Corollary 3.2 [18]). Consequently, S satisfies Wilf's conjecture ( [9] Proposition 2.3).
Case 2. If m -ν = 1, then we may assume that m = ν + 1 ≥ 5 (ν ≥ 4). By taking α = 1 in (2.2.2), we get

m-1 j=2 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ ( w m-1 + ρ m - w 1 + ρ m )(2ν -m). (2.3.2)
Hence, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = 1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m)+ m-1 j=2 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = ( w 1 + ρ m - w 0 + ρ m )(ν -m) + m-1 j=2 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = w 1 + ρ m (ν -m) + m-1 j=2 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ (as w 0 +ρ m = 0) ≥ w 1 + ρ m (ν -m) + ( w m-1 + ρ m - w 1 + ρ m )(2ν -m) + ρ (by (2.3.2)) = w 1 + ρ m ν -m + (2ν -m) -(2ν -m) + ( w m-1 + ρ m - w 1 + ρ m )(2ν -m) + ρ = w 1 + ρ m (3ν -2m) + ( w m-1 + ρ m - w 1 + ρ m - w 1 + ρ m )(2ν -m) + ρ = w 1 + ρ m (m -3) + ( w m-1 + ρ m - w 1 + ρ m - w 1 + ρ m )(m -2) + ρ (as m -ν = 1).
Therefore, we get

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 1 + ρ m (m -3) + ( w m-1 + ρ m - w 1 + ρ m - w 1 + ρ m )(m -2) + ρ. (2.3.3) Since m -ν = 1 > 0 = 1(0)
2 , then by Lemma 2.3.3, it follows that w m-1 ≥ w 1 + w 1 . Consequently, by Lemma 2.2.2 (i), we have

w m-1 + ρ m ≥ w 1 + ρ m + w 1 + ρ m -1. • If w m-1 +ρ m -w 1 +ρ m -w 1 +ρ m = -1.
( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 1 + ρ m (m -3) + ( w m-1 + ρ m - w 1 + ρ m - w 1 + ρ m )(m -2) + ρ ≥ 2(m -3) -(m -2) + ρ (as w 1 + ρ m ≥ 2) ≥ 0 (as m ≥ 5). • If w m-1 +ρ m -w 1 +ρ m -w 1 +ρ m ≥ 0. From (2.3.3), we get m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 1 + ρ m (m -3) + ( w m-1 + ρ m - w 1 + ρ m - w 1 + ρ m )(m -2) + ρ ≥ (m -3) + ρ ≥ 0 (as m ≥ 5).
Using Proposition 2.1.3, we get that S satisfies Wilf's conjecture if m -ν = 1.

Case 3. If m -ν ∈ {2, 3}. We have m -ν > 1 = 2(1)
2 . If (2 -1 q )ν ≥ m, then by Corollary 2.3.4 S satisfies Wilf's conjecture. Now, suppose that (2 -1 q )ν < m. Since Wilf's conjecture holds for q ≤ 3 (see [14], [START_REF] Eliahou | Wilf's conjecture and macaulay's theorem[END_REF]), we may assume that q ≥ 4.

• If m -ν = 2. Then, (2 -1 q )ν < ν + 2. Hence, ν < 2( q q-1 ) ≤ 8 3
. By [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF], S satisfies Wilf's conjecture.

• If m -ν = 3. Then, (2 -1 q )ν < ν + 3. Hence, ν < 3( q q-1 ) ≤ 4. By [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF], S satisfies Wilf's conjecture. Thus, Wilf's conjecture holds if m -ν ≤ 3. Thus, the proof is complete.

Proof. If m -ν > 3 = 3(2)
2 and 2ν ≥ m, then by Corollary 2.3.4 Wilf's conjecture holds. If m -ν ≤ 3, by Lemma 2.3.6, S satisfies Wilf's conjecture. Thus, the proof is complete.

In the following Corollary we will deduce Wilf's conjecture for numerical semigroups with m -ν = 4. This will enable us later to prove the conjecture for those with (2 + 1 q )ν ≥ m.

Corollary 2.3.8. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. If m -ν = 4, then S satisfies Wilf's conjecture.

Proof. Since Wilf's conjecture holds for ν ≤ 3 (see [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]), then we may assume that ν ≥ 4. Therefore, ν ≥ m -ν. Consequently, 2ν ≥ m. Hence, S satisfies Wilf's conjecture. Thus, the proof is complete.

The following technical Lemma will be used through the paper.

Lemma 2.3.9. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let w 0 = 0 < w 1 < . . . < w m-1 be the elements of Ap(S, m).

If m -ν ≥ ( α 2 ) -1 = α(α-1) 2 -1 for some 3 ≤ α ≤ m -2, then w m-1 ≥ w 1 + w α or w m-1 ≥ w α-2 + w α-1 .
Proof. Suppose by the way of contradiction that w m-1 < w 1 + w α and w m-1 < w α-2 + w α-1 . Let

w ∈ A(S, m) * \ min ≤ S (Ap(S, m)).
Then, w ≤ w m-1 and w = w i + w j for some w i , w j ∈Ap(S, m) * (Corollary 1.0.29). In this case, the only possible values of w are included in {w i +w j ; 1

≤ i ≤ j ≤ α-1}\{w α-2 +w α-1 , w α-1 +w α-1 }. Consequently, m -ν = |Ap(S, m) * \ min ≤ S (Ap(S, m))| ≤ α(α-1) 2 -2. But α(α-1) 2 -2 < α(α-1) 2 -1, which contradicts the hypothesis. Hence, w m-1 ≥ w 1 + w α or w m-1 ≥ w α-2 + w α-1 .
In the next theorem, we will show that Wilf's conjecture holds for numerical semigroups with m -ν = 5.

Theorem 2.3.10. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. If m -ν = 5, then S satisfies Wilf's conjecture.

Proof. Let m -ν = 5. Since Wilf's conjecture holds for 2ν ≥ m, then we may assume that 2ν < m. This implies that ν < m 2 = ν+5 2 i.e., ν < 5. Since the case ν ≤ 3 is known [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF], then we shall assume that ν = 4. This also implies that m = ν + 5 = 9.

Since m -ν = 5 = 4(3) 2 -1, by Lemma 2.3.9, it follows that

w 8 ≥ w 2 + w 3 or w 8 ≥ w 1 + w 4 . Case 1. If w 8 ≥ w 2 + w 3 . By taking α = 3 in (2.2.2) (m = 9, ν = 4), we get 8 j=4 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) ≥ ( w 8 + ρ 9 - w 3 + ρ 9 )(7). (2.3.4) Hence, m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = 8 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ (m = 9) = 3 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + 8 j=4 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ = ( w 1 + ρ 9 - w 0 + ρ 9 )(-5)+( w 2 + ρ 9 - w 1 + ρ 9 )(-1)+( w 3 + ρ 9 - w 2 + ρ 9 ) (3)
+ 8 j=4 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ = w 1 + ρ 9 (-4) + w 2 + ρ 9 (-4)+ w 3 + ρ 9 (3) + 8 j=4 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ ( w 0 +ρ m = 0) ≥ w 1 + ρ 9 (-4) + w 2 + ρ 9 (-4)+ w 3 + ρ 9 (3)+( w 8 + ρ 9 - w 3 + ρ 9 )(7) + ρ (Using (2.3.4)).
On the other hand, as w 1 +ρ 9 ≤ w 2 +ρ 9 and w 1 +ρ 9

≤ w 3 +ρ 9 , then 4 w 1 + ρ 9 ≤ 3 w 2 + ρ 9 + w 3 + ρ 9 .
Consequently, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 2 + ρ 9 ( - 3
4 )4 + w 3 + ρ 9 ( - 1
4 )4 + w 2 + ρ 9 (-4)+ w 3 + ρ 9 (3) +( w 8 + ρ 9 - w 3 + ρ 9 )(7) + ρ = w 2 + ρ 9 (-7) + w 3 + ρ 9 (2) + ( w 8 + ρ 9 - w 3 + ρ 9 )(7) + ρ = w 3 + ρ 9 (2) + ( w 8 + ρ 9 - w 2 + ρ 9 - w 3 + ρ 9 )(7) + ρ.
Then,

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 3 + ρ 9 (2) + ( w 8 + ρ 9 - w 2 + ρ 9 - w 3 + ρ 9 )(7) + ρ.
(2.3.5) Since w 8 ≥ w 2 + w 3 , by Lemma 2.2.2, it follows that w 8 +ρ 9

≥ w 2 +ρ 9 + w 3 +ρ 9 -1. • If w 8 + ρ 9 - w 2 + ρ 9 - w 3 + ρ 9 ≥ 0, then (2.3.5) gives 8 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ ≥ 0. • If w 8 + ρ 9 - w 2 + ρ 9 - w 3 + ρ 9 = -1.
By Lemma 2.2.2, we have ρ ≥ 1. Since for q ≤ 3 Wilf's conjecture is solved (see [START_REF] Eliahou | Wilf's conjecture and macaulay's theorem[END_REF], [14]), then may assume that q ≥ 4. Since

w 2 + ρ 9 ≤ w 3 + ρ 9 and w 2 + ρ 9 + w 3 + ρ 9 = w 8 + ρ 9 + 1 = q + 1,
in this case, it follows that

w 3 + ρ 9 + w 3 + ρ 9 ≥ w 2 + ρ 9 + w 3 + ρ 9 = q + 1 ≥ 5.
Hence,

w 3 + ρ 9 ≥ 3. Now, (2.3.5) gives 8 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -m) + ρ ≥ 3(2) -7 + 1 ≥ 0.
Using Proposition 2.1.3, we get that S satisfies Wilf's conjecture in this case.

Case 2. If w 8 ≥ w 1 + w 4 . We may assume that w 8 < w 2 + w 3 , since otherwise we are back to case 1. Hence, the possible values of w ∈ Ap(S, 9) * \min ≤ S (Ap(S, 9)) are

{w 1 + w j ; 1 ≤ j ≤ 7} ∪ {w 2 + w 2 }. • If Ap(S, 9) * \min ≤ S (Ap(S, 9)) ⊆ {w 1 + w j ; 1 ≤ j ≤ 7}. We have 5 = m -ν = |Ap(S, 9) * \ min ≤ S (Ap(S, 9))|.
Then, there exist five elements in Ap(S, 9) * included in {w 1 + w j ; 1 ≤ j ≤ 7}. By Corollary 1.0.29, an element x of the Apéry set of S belongs to max ≤ S (Ap(S, m)) if and only if w i = x + w j for all w i , w j ∈ Ap(S, m) * , then there exists at least five elements in Ap(S, 9) * that are not maximal (five elements from {w 1 . . . , w 7 }), hence,

t(S) = |{max ≤ S (Ap(S, 9)) -9}| ≤ 3 = ν -1.
Consequently, S satisfies Wilf's conjecture (Proposition 2.3 [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]).

• If w 2 + w 2 ∈ Ap(S, 9) * \ min ≤ S (Ap(S, 9)), then w 2 + w 2 ∈ Ap(S, 9)
namely w 8 ≥ w 2 + w 2 . By Lemma 2.2.2, we have

w 8 + ρ 9 ≥ 2 w 2 + ρ 9 -1.
In particular,

w 2 + ρ 9 ≤ q + 1 2 . (2.3.6) By taking α = 4 in (2.2.2) (m = 9, ν = 4), we get 8 j=5 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) ≥ ( w 8 + ρ 9 - w 4 + ρ 9 )(11). (2.3.7)
We have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = 8 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ (m = 9) = 4 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9)+ 8 j=5 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ = ( w 1 + ρ 9 - w 0 + ρ 9 )(-5)+( w 2 + ρ 9 - w 1 + ρ 9 )(-1)+( w 3 + ρ 9 - w 2 + ρ m ) (3)
+(

w 4 + ρ 9 - w 3 + ρ 9 )(7) + 8 j=5 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ = w 1 + ρ 9 (-4) + w 2 + ρ 9 (-4)+ w 3 + ρ 9 (-4) + w 4 + ρ 9 (7) + 8 j=5 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ (as w 0 +ρ m = 0) ≥ w 1 + ρ 9 (-4) + w 2 + ρ 9 (-4)+ w 3 + ρ 9 (-4) + w 4 + ρ 9 (7)
+(

w 8 + ρ 9 - w 4 + ρ 9 )(11) + ρ (by (2.3.7)) ≥ w 1 + ρ 9 (-4) + ( q + 1 2 )(-4)+ w 4 + ρ 9 (-4) + w 4 + ρ 9 (7)+( w 8 + ρ 9 - w 4 + ρ 9 ) (11)
+ ρ (by using (2.3.6) and w 3 +ρ 9

≤ w 4 +ρ 9 ) = w 1 + ρ 9 (-4) -2(q + 1) + w 4 + ρ 9 (3)+( w 8 + ρ 9 - w 4 + ρ 9 )(11) + ρ = w 1 + ρ 9 (-4 + 11 -11) -2(q + 1)+ w 4 + ρ 9 (3) +( w 8 + ρ 9 - w 4 + ρ 9 )(11) + ρ = w 1 + ρ 9 (7) -2(q + 1) + w 4 + ρ 9 (3)+( w 8 + ρ 9 - w 4 + ρ 9 - w 1 + ρ 9 )(11) + ρ = ( w 1 + ρ 9 + w 4 + ρ 9 )(3) + w 1 + ρ 9 (4) -2(q + 1) + ( w 8 + ρ 9 - w 1 + ρ 9 - w 4 + ρ 9 )(11) + ρ. Therefore, m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ ( w 1 + ρ 9 + w 4 + ρ 9 )(3) + w 1 + ρ 9 (4) -2(q + 1) + ( w 8 + ρ 9 - w 1 + ρ 9 - w 4 + ρ 9 )(11) + ρ.
(2.3.8) We have w 8 ≥ w 1 + w 4 , then by Lemma 2.2.2 (i)

w 8 + ρ 9 ≥ w 1 + ρ 9 + w 4 + ρ 9 -1. • If w 8 +ρ 9 -w 1 +ρ 9 -w 4 +ρ 9 ≥ 0. Let x = w 8 +ρ 9 -w 1 +ρ 9 -w 4 +ρ 9
. Hence, x ≥ 0 and

w 1 +ρ 9 + w 4 +ρ 9 = w 8 +ρ 9 -x = q -x (Remark 2.2.1 (vi)). Then, (2.3.8) gives 8 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ ≥ (q -x)(3) + 4 -2(q + 1) + 11x + ρ = q + 8x + 2 + ρ ≥ 0. • If w 8 +ρ 9 -w 1 +ρ 9 -w 4 +ρ 9 = -1. Then, w 1 +ρ m + w 4 +ρ 9 = w 8 +ρ 9 + 1 = q + 1 (Remark 2.2.1 (vi)). By Lemma 2.2.2, we have w 1 +ρ 9 ≥ 2 and ρ ≥ 1. Since q ≥ 1 (S = N i.e., f ≥ 1), then (2.3.8) gives 8 j=1 ( w j + ρ 9 - w j-1 + ρ 9 )(4j -9) + ρ ≥ (q + 1)(3) + 8 -2(q + 1) -11 + 1 = q -1 ≥ 0.
Using Proposition 2.1.3, we get that S satisfies Wilf's conjecture in this case.

Thus, Wilf's conjecture holds if m -ν = 5. Thus, the proof is complete.

In the next corollary, we will deduce the conjecture for m = 9.

Corollary 2.3.11. If S is a numerical Semigroup with multiplicity m = 9, then S satisfies Wilf's conjecture.

Proof. By Lemma 2.3.6, Corollary 2.3.8 and Theorem 2.3.10, we may assume that m -ν > 5, hence, ν < m -5 = 4. By [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF], S satisfies Wilf's conjecture. Thus, the proof is complete.

The following Lemma will enable us later to show that Wilf's conjecture holds for numerical semigroups with (2 + 1 q )ν ≥ m. Lemma 2.3.12. Let S be a numerical Semigroup with multiplicity m, embedding dimension ν and conductor

f + 1 = qm -ρ for some q ∈ N, 0 ≤ ρ ≤ m -1. If m -ν = 6 and (2 + 1 q )ν ≥ m, then S satisfies Wilf's conjecture. Proof. Since m -ν = 6 ≥ 4(3)
2 -1, by Lemma 2.3.9, it follows that

w m-1 ≥ w 1 + w 4 or w m-1 ≥ w 2 + w 3 . Case 1. If w m-1 ≥ w 1 + w 4 . By hypothesis (2 + 1 q )ν ≥ m and Theorem 2.3.1 Wilf's conjecture holds in this case. Case 2. If w m-1 ≥ w 2 + w 3 . We may assume that w m-1 < w 1 + w 4 , since otherwise we are back to case i. Hence, Ap(S, m) * \min ≤ S (Ap(S, m)) = {w 1 + w 1 , w 1 + w 2 , w 1 + w 3 , w 2 + w 2 , w 2 + w 3 , w 3 + w 3 } (as 6 = m -ν = |Ap(S, m) * \min ≤ S (Ap(S, m))|). Consequently, m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = 3 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + m-1 j=4 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = ( w 1 + ρ m - w 0 + ρ m )(ν -m)+( w 2 + ρ m - w 1 + ρ m )(2ν -m)+( w 3 + ρ m - w 2 + ρ m )(3ν -m) + m-1 j=4 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = w 1 + ρ m (-ν)+ w 2 + ρ m (-ν) + w 3 + ρ m (3ν -m)+ m-1 j=4 ( w j + ρ m - w j-1 + ρ m )(jν -m) +ρ (as w 0 +ρ m = 0) ≥ w 1 + ρ m (-ν)+ w 2 + ρ m (-ν)+ w 3 + ρ m (3ν -m)+( w m-1 + ρ m - w 3 + ρ m )(4ν -m)+ρ (by (2.3.4)).
On the other hand, as

w 1 +ρ m ≤ w 2 +ρ m and w 1 +ρ m ≤ w 3 +ρ m , then 2 w 1 + ρ m ≤ w 2 + ρ m + w 3 + ρ m .
Consequently, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 2 + ρ m ( -ν 2 ) + w 3 + ρ m ( -ν 2 ) + w 2 + ρ m (-ν)+ w 3 + ρ m (3ν -m) +( w m-1 + ρ m - w 3 + ρ m )(4ν -m) + ρ = w 2 + ρ m ( -3ν 2 ) + w 3 + ρ m ( 5ν 2 -m)+( w m-1 + ρ m - w 3 + ρ m )(4ν -m) + ρ = w 2 + ρ m -3ν 2 + (4ν -m) -(4ν -m) + w 3 + ρ m ( 5ν 2 -m)+( w m-1 + ρ m - w 3 + ρ m )(4ν -m) + ρ = w 2 + ρ m ( 5ν 2 -m) + w 3 + ρ m ( 5ν 2 -m) + ( w m-1 + ρ m - w 2 + ρ m - w 3 + ρ m )(4ν -m) + ρ = w 2 + ρ m ( 3ν 2 -6)+ w 3 + ρ m ( 3ν 2 -6)+( w m-1 + ρ m - w 2 + ρ m - w 3 + ρ m )(3ν -6)+ρ (as m-ν = 6).
Hence,

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w 2 + ρ m ( 3ν 2 -6) + w 3 + ρ m ( 3ν 2 -6) + ( w m-1 + ρ m + ρ- w 2 + ρ m - w 3 + ρ m )(3ν -6).
(2.3.9)

We have w m-1 ≥ w 2 + w 3 , by Lemma 2.2.2, it follows that

w m-1 + ρ m ≥ w 2 + ρ m + w 3 + ρ m -1. • If w m-1 + ρ m - w 2 + ρ m - w 3 + ρ m ≥ 0, using ν ≥ 4 in (2.3.9) (ν ≤ 3 is solved [9]), we get m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ 0. • If w m-1 + ρ m - w 2 + ρ m - w 3 + ρ m = -1, then w 2 + ρ m + w 3 + ρ m = w m-1 + ρ m + 1, that is w 2 + ρ m + w 3 + ρ m = q + 1.
(2.3.10)

We have

w 3 + w 3 ∈ Ap(S, m) * \min ≤ S (Ap(S, m)) namely w 3 + w 3 ∈Ap(S, m), then w m-1 ≥ w 3 + w 3 . By Lemma 2.2.2, we have w m-1 +ρ m ≥ 2 w 3 +ρ m -1.
In particular,

w 3 + ρ m ≤ q + 1 2 . ( 2
= w 3 +ρ m = q+1
2 , in particular q is odd, then we have to assume that q ≥ 5. Now, using (2.3.10), q ≥ 5 and the hypothesis (2

+ 1 q )ν ≥ m = ν + 6 in (2.3.9), we get m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ ( w 2 + ρ m + w 3 + ρ m )( 3ν 2 -6)+ w m-1 + ρ m - w 2 + ρ m - w 3 + ρ m 3ν -6) + ρ = (q + 1)( 3ν 2 -6) -(3ν -6) + ρ = ν( 3q 2 + 3 2 -3) -6q + ρ ≥ ν( 3q 2 - 3 2 ) -qν -ν + ρ (as 6q ≤ qν + ν) = ν( q 2 - 5 2 ) + ρ ≥ 0 (as q ≥ 5).
Using Proposition 2.1.3, we get that S satisfies Wilf's conjecture in this case.

Therefore, Wilf's conjecture holds if m -ν = 6 and (2 + 1 q )ν ≥ m. Thus, the proof is complete. Next, we will generalize a result for Sammartano ( [19]) and show that Wilf's conjecture holds for numerical semigroups satisfying (2 + 1 q )ν ≥ m, using Lemma 2.

f + 1 = qm -ρ for some q ∈ N, 0 ≤ ρ ≤ m -1. If (2 + 1 q )ν ≥ m, then S satisfies Wilf's conjecture.
Proof.

• If m -ν ≤ 3, then by Lemma 2.3.6 Wilf's conjecture holds.

• If m -ν = 4, then by Corollary 2.3.8 Wilf's conjecture holds.

• If m -ν = 5, then by Theorem 2.3.10 Wilf's conjecture holds.

• If m -ν = 6 and (2 + 1 q )ν ≥ m, then by Lemma 2.3.12 Wilf's conjecture holds. Note that 2ν < m. We have (2 + 1 q )ν = (2 + 1 4 )6 ≥ 13 = m. Thus, the conditions of Theorem 2.3.13 are valid.

• If m -ν > 6 and (2 + 1 q )ν ≥ m,
Corollary 2.3.15. Let S be a numerical semigroup with multiplicity m and conductor

f + 1 = qm -ρ for some q ∈ N, 0 ≤ ρ ≤ m -1. If m ≤ 8 + 4
q , then S satisfies Wilf's conjecture.

Proof. If ν < 4, then S satisfies Wilf's conjecture (see [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]). Hence, we can suppose that ν ≥ 4. Thus,

(2 + 1 q )ν ≥ (2 + 1 q )4 ≥ m.
By using Theorem 2.3.13 S satisfies Wilf's conjecture. Thus, the proof is complete.

Numerical semigroups with w

m-1 ≥ w α-1 + w α and ( α+3 3 )ν ≥ m
In this Section, we will show that if S is a numerical semigroup such that

w m-1 ≥ w α-1 + w α and ( α + 3 3 )ν ≥ m,
then S satisfies Wilf's conjecture.

Theorem 2.4.1. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let w 0 = 0 < w 1 < w 2 < . . . < w m-1 be the elements of Ap(S, m). Suppose that w m-1 ≥ w α-1 + w α for some 1 < α < m -1. If ( α+33 )ν ≥ m, then S satisfies Wilf's conjecture.

Proof. We may assume that ρ ≥ (3-q)αm 2α+6 . Indeed, if 0 ≤ ρ < (3-q)αm 2α+6 , then q < 3 and Wilf's conjecture holds for this case (see [14]). We are going to show that S satisfies Wilf's conjecture by means of Proposition 2.1.3. We have

α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) = α j=1 w j + ρ m (jν -m) - α j=1 w j-1 + ρ m (jν -m) = α j=1 w j + ρ m (jν -m) - α-1 j=0 w j + ρ m (j + 1)ν -m = w α + ρ m (αν -m) - w 0 + ρ m (ν -m) - α-1 j=1 w j + ρ m ν = w α + ρ m (αν -m) - w α-1 + ρ m ν - α-2 j=1 w j + ρ m ν (as w 0 +ρ m = 0) ≥ w α + ρ m (αν -m) - w α-1 + ρ m ν - α-2 j=1 1 2
w α + ρ m + w α-1 + ρ m ν (by Remark 2.2.1 (v)) = w α + ρ m (αν -m) - w α-1 + ρ m ν- w α + ρ m + w α-1 + ρ m (α -2)ν 2 .
Hence,

α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ w α + ρ m ( α + 2 2 )ν -m - w α-1 + ρ m ( αν 2 ).
(2.4.1)

By (2.2.2), we have m-1 j=α+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ ( w m-1 + ρ m - w α + ρ m ) (α + 1)ν -m .
Since w m-1 ≥ w α-1 + w α , by Lemma 2.2.2, it follows that

w m-1 + ρ m ≥ w α-1 + ρ m + w α + ρ m -1. Let x = w m-1 +ρ m w α-1 +ρ m -wα+ρ m . Then, x ≥ -1 and w α-1 +ρ m + wα+ρ m = w m-1 +ρ m -x = q -x (by Remark 2.2.1 vi). Now, using ρ ≥ (3-q)αm 2α+6 and ( α+3 3 )ν ≥ m, we get m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = α j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m)+ m-1 j=α+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w α + ρ m ( α + 2 2 )ν -m - w α-1 + ρ m ( αν 2 ) + w m-1 + ρ m - w α + ρ m (α + 1)ν -m + ρ (by (2.4.1) and (2.2.2)) = w α-1 + ρ m -αν 2 + (α + 1)ν -m -(α + 1)ν -m + w α + ρ m ( α + 2 2 )ν -m + w m-1 + ρ m - w α + ρ m (α + 1)ν -m + ρ = w α-1 + ρ m + w α + ρ m ( α + 2 2 )ν -m + w m-1 + ρ m - w α-1 + ρ m - w α + ρ m (α + 1)ν -m +ρ ≥ q -x ( α + 2 2 )ν -m + x (α + 1)ν -m + ρ = ν(q + qα 2 + xα 2 ) -qm + ρ ≥ ν(q + qα 2 - α 2 ) -qm + (3 -q)αm 2α + 6 (as ρ ≥ (3-q)αm 2α+6 ) = ν(q + qα 2 - α 2 ) -m q(2α + 6) + (q -3)α 2α + 6 = ν(q + qα 2 - α 2 ) -m 3q α + 3 + 3qα 2(α + 3) - 3α 2(α + 3) = q + qα 2 - α 2 3 α + 3 ( α + 3 3 )ν -m ≥ 0 (
m -ν ≥ α(α-1) 2 -1 for some 7 ≤ α ≤ m -2. If (2 + α-3 q )ν ≥ m, then S satisfies Wilf's conjecture. Proof. Since m -ν ≥ α(α-1)
2 -1, then by Lemma 2.3.9, we have

w m-1 ≥ w 1 + w α or w m-1 ≥ w α-2 + w α-1 . Suppose that w m-1 ≥ w 1 + w α . Since (2 + α-3
q )ν ≥ m, by applying Theorem 2.3.1, S satisfies Wilf's conjecture. Now, suppose that w m-1 ≥ w α-2 + w α-1 . We may assume that q ≥ 4 (q ≤ 3 is solved [14], [START_REF] Eliahou | Wilf's conjecture and macaulay's theorem[END_REF]). Then, for α ≥ 7, we have ( α-1+3

3

)ν ≥ (2 + α-3 q )ν. Consequently, ( α-1+3
3

)ν ≥ m. Next, by applying Theorem 2.4.1, S satisfies Wilf's conjecture. Thus, the proof is complete. As a direct consequence of Theorem 2.4.1, we get the following Corollary.

Corollary 2.4.4. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let w 0 = 0 < w 1 < w 2 < . . . < w m-1 be the elements of Ap(S, m). Suppose that w m-1 ≥ w α-1 + w α for some

1 < α < m -1. If m ≤ 4(α+3)
Proof. If ν < 4, then S satisfies Wilf's conjecture (see [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]). Hence, we can suppose that ν ≥ 4. Thus, ( α+33 )(ν) ≥ 4(α+3) 3 ≥ m. By applying Theorem 2.4.1 S satisfies Wilf's conjecture. Thus, the proof is complete. Proof. We are going to show that S satisfies Wilf's conjecture by means of Proposition 2.1.3. We have

Numerical semigroups with 2 +

x j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) = x j=1 w j + ρ m (jν -m) - x j=1 w j-1 + ρ m (jν -m) = x j=1 w j + ρ m (jν -m) - x-1 j=0 w j + ρ m (j + 1)ν -m = x-1 j=1 w j + ρ m (jν -m) + w x + ρ m (xν -m) - w 0 + ρ m (ν -m)- x-1 j=1 w j + ρ m (j + 1)ν -m = w x + ρ m (xν -m) - w 0 + ρ m (ν -m) - x-1 j=1 w j + ρ m ν = w x + ρ m (xν -m) - x-1 j=1 w j + ρ m ν (as w 0 +ρ m = 0) ≥ w x + ρ m (xν -m) - x-1 j=1 w x + ρ m ν (by Remark 2.2.1 (v)) = w x + ρ m (xν -m) - w x + ρ m (x -1)ν = w x + ρ m (ν -m).
Therefore,

x j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ w x + ρ m (ν -m).
(2.5.1)

In addition,

y j=x+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) = y j=x+1 w j + ρ m (jν -m) - y j=x+1 w j-1 + ρ m (jν -m) = y j=x+1 w j + ρ m (jν -m) - y-1 j=x w j + ρ m (j + 1)ν -m = y-1 j=x+1 w j + ρ m (jν -m) + w y + ρ m (yν -m)- w x + ρ m (x + 1)ν -m - y-1 j=x+1 w j + ρ m (j + 1)ν -m = w y + ρ m (yν -m) - w x + ρ m (x + 1)ν -m - y-1 j=x+1 w j + ρ m ν ≥ w y + ρ m (yν -m) - w x + ρ m (x + 1)ν -m - y-1 j=x+1 w y + ρ m ν (using Remark 2.2.1 (v)) = w y + ρ m (yν -m) - w x + ρ m (x + 1)ν -m - w y + ρ m (y -x -1)ν = w y + ρ m (x + 1)ν -m - w x + ρ m (x + 1)ν -m .
Hence,

y j=x+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ w y + ρ m (x + 1)ν -m - w x + ρ m (x + 1)ν -m . (2.5.2)
Moreover, we have

m-1 j=y+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ m-1 j=y+1 ( w j + ρ m - w j-1 + ρ m ) (y + 1)ν -m (using Remark 2.2.1 (v)) = (y + 1)ν -m m-1 j=y+1 w j + ρ m - m-1 j=y+1 w j-1 + ρ m = (y + 1)ν -m m-1 j=y+1 w j + ρ m - m-2 j=y w j + ρ m = w m-1 + ρ m - w y + ρ m (y + 1)ν -m . Therefore, m-1 j=y+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) ≥ w m-1 + ρ m - w y + ρ m (y + 1)ν -m . (2.5.3) Consequently, m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ = x j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + y j=x+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + m-1 j=y+1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x + ρ m (ν -m)+ w y + ρ m (x + 1)ν -m - w x + ρ m (x + 1)ν -m +( w m-1 + ρ m - w y + ρ m ) (y + 1)ν -m + ρ (using (2.5.1), (2.5.2) and (2.5.3)) = w x + ρ m (-xν)+ w y + ρ m (x + 1)ν -m + w m-1 + ρ m - w y + ρ m (y + 1)ν -m + ρ = w x + ρ m -xν + (y + 1)ν -m -(y + 1)ν -m + w y + ρ m (x + 1)ν -m + w m-1 + ρ m - w y + ρ m (y + 1)ν -m + ρ = w x + ρ m (y -x + 1)ν -m + w y + ρ m (x + 1)ν -m + w m-1 + ρ m - w y + ρ m - w x + ρ m (y + 1)ν -m +ρ. Consequently, m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x + ρ m (y -x + 1)ν -m + w y + ρ m (x + 1)ν -m + w m-1 + ρ m - w y + ρ m - w x + ρ m (y + 1)ν -m +ρ.
(2.5.4)

Since w m-1 -m ≥ w x + w y , it follows w m-1 + ρ m > w x + w y + ρ m .
(2.5.5)

Consider the following cases :

Case 1. If wx+ρ m = wx m + 1 and wy+ρ m = wy m + 1, then (2.5.5) gives w m-1 + ρ m ≥ w x + ρ m + w y + ρ m .
Then, from (2.5.4) and the hypothesis, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x + ρ m (y -x + 1)ν -m + w y + ρ m (x + 1)ν -m + ρ = ( w x m + 1) (y -x + 1)ν -m) + ( w y m + 1) (x + 1)ν -m + ρ = w x m + w y m + 2 (2 + wx m (y -x -1) + (y -2) + wy m (x -1) wx m + wy m + 2 )ν -m + ρ ≥ 0.
By Proposition 2.1.3, we get that Wilf's conjecture holds in this case.

m-1 + ρ m > w x + ρ m + w y + ρ m .
Then, from (2.5.4) and the hypothesis, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x + ρ m (y -x + 1)ν -m + w y + ρ m (x + 1)ν -m + (y + 1)ν -m + ρ = w x m (y -x + 1)ν -m + w y m + 1 (x + 1)ν -m + (y + 1)ν -m + ρ = w x m + w y m + 2 (2 + wx m (y -x -1) + (y -2) + wy m (x -1) + x wx m + wy m + 2 )ν -m + ρ > w x m + w y m + 2 (2 + wx m (y -x -1) + (y -2) + wy m (x -1) wx m + wy m + 2 )ν -m + ρ ≥ 0.
By Proposition 2.1.3, we get that Wilf's conjecture holds in this case.

w m-1 + ρ m > w x + ρ m + w y + ρ m .
Then, from (2.5.4) and the hypothesis, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x + ρ m (y -x + 1)ν -m + w y + ρ m (x + 1)ν -m + (y + 1)ν -m + ρ = w x m + 1 (y -x + 1)ν -m + w y m (x + 1)ν -m + (y + 1)ν -m + ρ = P xy (2 + wx m (y -x -1) + (y -2) + wy m (x -1) + (y -x) wx m + wy m + 2 )ν -m + ρ ≥ P xy (2 + wx m (y -x -1) + (y -2) + wy m (x -1) wx m + wy m + 2 )ν -m + ρ ≥ 0,
where

P xy = w x m + w y m + 2 .
By Proposition 2.1.3, we get that Wilf's conjecture holds in this case.

m-1 + ρ m > w x + ρ m + w y + ρ m .
Then, from (2.5.4) and the hypothesis, we have

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x + ρ m (y -x + 1)ν -m + w y + ρ m (x + 1)ν -m + (y + 1)ν -m + ρ = w x m (y -x + 1)ν -m + w y m (x + 1)ν -m + (y + 1)ν -m + ρ.
Hence,

m-1 j=1 ( w j + ρ m - w j-1 + ρ m )(jν -m) + ρ ≥ w x m + w y m + 1 (2 + wx m (y -x -1) + (y -1) + wy m (x -1) wx m + wy m + 1 )ν -m + ρ = w x m + w y m + 1 (2 + wx m (y -x -1) + (y -2) + wy m (x -1) wx m + wy m + 2 )ν -m + ρ ≥ 0.
By Proposition 2.1.3, we get that Wilf's conjecture holds in this case. Thus, Wilf's conjecture holds in all cases. Thus, the proof is complete.

As a direct consequence of Theorem 2.5.1, we get the following Corollaries.

Corollary 2.5.2. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let By Theorem 2.5.1, S satisfies Wilf's conjecture. Thus, the proof is complete.

w 0 = 0 < w 1 < . . . < w m-1 be the elements of Ap(S, m). Suppose that w m-1 -m ≥ w x + w y , x ≥ α + 1, y -x ≥ α + 1 for some α ∈ N. If (2 + α)ν ≥ m,
Corollary 2.5.3. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let w 0 = 0 < w 1 < . . . < w m-1 be the elements of Ap(S, m). Suppose that w m-1 -m ≥ w x + w y , for some

x ≥ α + 1, y -x ≥ α + 1 and α ∈ N. If m ≤ 4(2 + α), then S satisfies Wilf's conjecture.
Proof. We may assume that ν ≥ 4 (ν ≤ 3 is solved [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF]), then (2 + α)ν ≥ (2 + α)4 ≥ m. By applying Corollary 2.5.2, S satisfies Wilf's conjecture. Thus, the proof is complete. Thus, the conditions of Theorem 2.5.1 are valid.

Numerical semigroups with

m -ν > (n-2)(n-3) 2
In this Section, we show that Wilf's conjecture holds for numerical semigroups with m -ν > (n-2)(n-3) 2 and also the conjecture holds for those with n ≤ 5.

Lemma 2.6.1. Let S be a numerical semigroup with multiplicity m, embedding dimension ν, Frobenius number f and n = |{s ∈ S; s < f }|. If m -ν > α(α-1) 2 for some 0 ≤ α ≤ m -2, then w α < f . In particular, n ≥ α + 2.

Proof. We claim that w α < f . Suppose by the way of contradiction that w α > f (w α = f ), and let w ∈ Ap(S, m) * \min ≤ S (Ap(S, m)). Then, there exists w i , w j ∈ Ap(S, m) * such that w = w i + w j (Corollary 1.0.29). Suppose that at least one of the two indicies, let's say i, is greater than or equal to α. Then, w = w i +w j ≥ w α +m ≥ f +1+m. Hence, w-m ∈ S which contradicts the fact that w ∈ Ap(S, m). Consequently, the two indicies are necessarly less than or equal to α -

1. Since |Ap(S, m) * \min ≤ S (Ap(S, m))| = m -ν (Corollary 1.0.31), we deduce that m -ν ≤ α(α-1)
2 which is impossible. Consequently, w α < f . Therefore, we get that {0, m, w 1 , w 2 , . . . , w α } ⊆ {s ∈ S; s < f }. Hence, n ≥ α + 2.

Theorem 2.6.2. Let S be a numerical semigroup with multiplicity m, embedding dimension ν and n

= |{s ∈ S; s < f }|. If m -ν > (n-2)(n-3) 2 with 2 ≤ n ≤ m, then S satisfies Wilf's conjecture. Proof. By Lemma 2.6.1, the condition m -ν > (n-2)(n-3) 2 , gives that {0, m, w 1 , w 2 , . . . , w n-2 } ⊆ {s ∈ S; s < f }.
Therefore, {0, m, w 1 , w 2 , . . . , w n-2 } = {s ∈ S; s < f }. Hence, 2m > f . By [14], it follows that S satisfies Wilf's conjecture.

In [START_REF] Dobbs | On a question of wilf concerning numerical semigroups[END_REF], D. Dobbs and G. Matthews proved Wilf's conjecture for n ≤ 4 in a long technical proof. In [START_REF] Eliahou | Wilf's conjecture and macaulay's theorem[END_REF], S. Eliahou showed Wilf's conjecture has a positive answer for n ≤ 6. We are going to introduce a simpler proof for n ≤ 5 using the previous theorem (note that If n ≤ 5, then we can assume 2 ≤ n ≤ 5).

Corollary 2.6.3. Let S be a numerical semigroup with multiplicity m, embedding dimension ν and n = |{s ∈ S; s < f }|. If n ≤ 5, then S satisfies Wilf's conjecture.

Proof. By Theorem 2.3.6 and Theorem 2.3.10, we will assume that m -ν > 5 which is strictly greater than α(α-1) 2

for α ∈ {0, 1, 2, 3}. Hence, by applying Theorem 2.6.2 for n = α + 2 ∈ {2, 3, 4, 5} Wilf's conjecture holds.

3

Numerical semigroup of the form < m, m + 1, . . . , m + l, k(m + l) + r > Throughout this chapter we suppose that S is a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with k, l, m, r ∈ N * and r ≤ (k + 1)l + 1. The aim of this chapter is to determine the Frobenius number f (S) and the genus g(S). Also, it aims to characterize those numerical semigroups which are symmetric (resp. pseudo-symmetric) and to determine the set of pseudo-Frobenius numbers P F (S). Definition 3.0.1. Let k, l, m, r ∈ N * . For every 1 ≤ i ≤ m -1, write, by the euclidean division, i = α i (kl + r) + β i l + t i with 0 ≤ β i l + t i < kl + r and 0 ≤ t i < l. In particular

α i = i kl + r , β i = i -α i (kl + r) l and t i = i -α i (kl + r) -β i l.
For the convenience of the statement we will use the following notation :

i = α i (kl + r) + β i l + i t i where i = 1 if t i = 0, 0 if t i = 0.
Clearly i t i = t i but we shall use i later in the notations.

Proposition 3.0.2 and 3.0.3 give some properties that will be used in this Chapter using the notations used in Definition 3.0.1.

Proposition 3.0.2. Let the notations be as in Definition 3.0.1 and suppose that r ≤ (k + 1)l + 1, we have

β i + i ≤ 2k + 1.
Proof. By using Definition 3.0.1 and r ≤ (k + 1)l + 1, it follows that

β i l + i t i ≤ kl + r -1 ≤ kl + (k + 1)l = (2k + 1)l.
Case 1. If i = 0. We have

β i l = β i l + i t i ≤ (2k + 1)l. Consequently, β i ≤ 2k + 1. Hence, β i + i ≤ 2k + 1.
Case 2. If i = 1. We have β i l + i t i ≤ (2k + 1)l and t i ≥ 1 (as i = 1). If β i ≥ 2k + 1, then β i l + i t i ≥ (2k + 1)l + 1, which gives a contradiction. Consequently, β i ≤ 2k. Then,

β i + i ≤ 2k + 1.
Proposition 3.0.3. Let the notations be as in Definition 3.0.1 and suppose that r ≤ (k + 1)l + 1, we may assume that r ≤ min((k + 1)l + 1, m + l -1).

Proof. We claim that we may assume that r < m + l. Indeed, if r ≥ m + l, then there exist q , r ∈ N such that r = q (m + l) + r with r < m + l. Let k = k + q , then k(m + l) + r = (k + q )(m + l) + r = k (m + l) + r with r < m + l and r = r -q (m + l) ≤ (k + 1)l + 1 ≤ (k + q + 1)l + 1 = (k + 1)l + 1. Hence, S is a numerical semigroup generated by m, m + 1, . . . , m + l, k (m + l) + r with r ≤ (k + 1)l + 1.

Consequently, we may assume that r ≤ m + l -1. By hypothesis, we have r ≤ (k + 1)l + 1. Hence, we get our assumption. Thus, the proof is complete.

Apéry set of S

Let Ap(S, m) = {0, w(1), . . . , w(m -1)} be the Apéry set of S with respect to m, where w(i) is the smallest element of S which is congruent to i mod m. The following theorem gives a formula for the Apéry set of S. For all 1 ≤ i ≤ m -1 where i is written as in Definition 3.0.1, we have :

w(i) = m(kα i + β i + i ) + i.
Proof. Let λ i = m(kα i + β i + i ) + i where i is written as in Definition 3.0.1. We are going to show that λ i = w(i) for all 1 ≤ i ≤ m -1. To this end, we will show that λ i ∈ S, λ i is congruent to i mod m and λ i -m / ∈ S. • We have λ i ∈ S and λ i is congruent to i mod m. It follows from

λ i = m(kα i + β i + i ) + i = m(kα i + β i + i ) + α i (kl + r) + β i l + i t i = α i (k(m + l) + r) + β i (m + l) + i (m + t i ). (3.1.1)
• We will prove that λ i -m / ∈ S by the way of contradiction. From (3.1.1), we have

λ i -m = α i (k(m + l) + r) + β i (m + l) + ( i -1)m + i t i .
Suppose by the way of contradiction that λ i -m ∈ S. By Definition 1.0.7, there exist x, x l , . . . , x 0 ∈ N such that

λ i -m = x(k(m + l) + r) + x l (m + l) + . . . + x 1 (m + 1) + x 0 m. Thus, α i (k(m + l) + r) + β i (m + l) + ( i -1)m + i t i = x(k(m + l) + r) + x l (m + l) + . . . + x 1 (m + 1) + x 0 m.

In particular, m(kα

i + β i + i -1) + α i (kl + r) + β i l + i t i = m(kx + x l + . . . + x 1 + x 0 ) + x(kl + r) + x l l + x l-1 (l -1) + . . . + x 1 . (3.1.2)
To show that λ i -m / ∈ S, we are going to show first that x -α i > 0, then we will show that x -α i ≥ 2 and conclude our assertion from (3.1.2). Since 1 ≤ i ≤ m -1 and i = α i (kl + r) + β i l + i t i , it follows that

α i (kl + r) + β i l + i t i ≤ m -1. (3.1.3)
We claim that

kα i + β i + i -1 ≥ kx + x l + x l-1 + . . . + x 1 + x 0 .
Suppose by the way of contradiction that

kα i + β i + i -1 < kx + x l + x l-1 + . . . + x 1 + x 0 .
Then, from (3.1.3), we get

m(kα i + β i + i -1) + α i (kl + r) + β i l + i t i ≤ m(kx + x l + x l-1 + . . . + x 1 + x 0 -1) + m -1 = m(kx + x l + x l-1 + . . . + x 1 + x 0 ) -1 ≤ m(kx + x l + . . . + x 1 + x 0 ) + x(kl + r) + x l l + x l-1 (l -1) + . . . + x 1 -1.
This contradicts (3.1.2). Consequently,

kα i + β i + i -1 ≥ kx + x l + x l-1 + . . . + x 1 + x 0 . ( 3
α i (kl + r) + β i l + i t i ≤ x(kl + r) + x l l + x l-1 (l -1) + . . . + x 1 . (3.1.5)
If we multiply (3.1.4) by l, we get

α i kl + β i l ≥ (kx + x l + x l-1 + . . . + x 1 + x 0 )l + (1 -i )l. ( 3
xkl + (x -α i )r + x l l + x l-1 (l -1) + . . . + x 1 -i t i ≥ α i kl + β i l ≥ xkl + (x l + . . . + x 0 )l + (1 -i )l.
Consequently, (x -α i )r ≥ x 0 l + x 1 (l -1) + . . .

+ x l-1 + (1 -i )l + i t i .
Since x 0 , . . . , x l-1 , r ∈ N, l ∈ N * and i ∈ {0, 1}, we get that

x -α i > 0. (3.1.7)
Next, we aim to show that x -α i ≥ 2. From (3.1.2), we have

m(β i ) = m(k(x -α i ) + 1 -i + x l + . . . + x 1 + x 0 ) + (x -α i )(kl + r) -β i l -i t i + x l l + x l-1 (l -1) + . . . + x 1 . (3.1.8) Then, (x -α i )(kl + r) -β i l -i t i + x l l + x l-1 (l -1) + . . . + x 1 is divisible by m. Since x -α i > 0 (by (3.1. 7
)), β i l + i t i < kl + r (by definition) and x 1 , . . . , x l ∈ N, then there exists p ∈ N * such that

(x -α i )(kl + r) -β i l -i t i + x l l + x l-1 (l -1) + . . . + x 1 = pm. (3.1.9)
By substituting (3.1.9) in (3.1.8), we get

m(β i ) = m(k(x -α i ) + 1 -i + x l + . . . + x 1 + x 0 + p).
Consequently,

β i = k(x -α i ) + 1 -i + p + x l + . . . + x 1 + x 0 . (3.1.10)
From (3.1.9), it follows that

β i l = (x -α i )(kl + r) -i t i + x l l + x l-1 (l -1) + . . . + x 1 -pm. (3.1.11)
By multiplying (3.1.10) by l and using (3.1.11), we get the following :

(x -α i )r = p(m + l) + (1 -i )l + i t i + x l-1 + . . . + x 1 (l -1) + x 0 l.
Since i ∈ {0, 1}, x l-1 , . . . , x 0 ∈ N, p ∈ N * and 0 < r < m + l (Proposition 3.0.3), we get that

x -α i ≥ 2.

From (3.1.2), we have

m(β i + i -1) + β i l + i t i = m (x -α i )k + x l + . . . + x 1 + x 0 + (x -α i )(kl + r) + x l l + x l-1 (l -1) + . . . + x 1 .
Since x -α i ≥ 2 and x l , . . . , x 0 ∈ N, it follows that

m(β i + i -1) + β i l + i t i ≥ 2km + 2(kl + r). (3.1.12)
On the other hand, Since β i l + i t i < kl + r (by definition) and

β i + i -1 ≤ 2k (Proposition 3.0.2), it follows that m(β i + i -1) + β i l + i t i < 2km + kl + r. (3.1.13)
From (3.1.12) and (3.1.13) we get a contradiction. Consequently, λ i -m / ∈ S. Hence, w(i) = λ i = m(kα i + β i + i ) + i. Thus, the proof is complete.

Frobenius number of S

Definition 3.2.1. Let the notations be as above. Let

q = r -1 l .
Thus, ql ≤ r -1 and r -1 = ql + t with q, t ∈ N and t < l. Let t ∈ N defines as in 3.0.

1 t = 1 if t ≥ 1, 0 if t = 0.
Proposition 3.2.2 and 3.2.3 give some properties that will be used in this Chapter using the notations used in Definition 3.2.1.

Proposition 3.2.2. Under the above notations, we have

β i ≤ k + q.
Proof. By definition, we have

β i l + i t i ≤ kl + r -1 = (k + q)l + t with t < l.
Consequently,

β i l + i t i < (k + q + 1)l.
Suppose by the way of contradiction that β i > k + q. Hence, β i l + i t i ≥ (k + q + 1)l, which is impossible. Consequently, β i ≤ k + q. Thus, the proof is complete.

Proposition 3.2.3. Under the above notations, we have

q + t ≤ k + 1.
Proof. By definition, we have r -1 ≤ (k +1)l and r -1 = ql+t with q, t ∈ N and t < l. Thus, ql+t ≤ (k +1)l.

Case 1. If t = 0, then t = 0. Hence, ql ≤ (k + 1)l, which implies that q ≤ k + 1. Thus, q + t ≤ k + 1.

Case 2. If t = 1, then t ≥ 1. Hence, ql + t ≤ (k + 1)l with t ≥ 1, it follows that q ≤ k. Therefore, q + t ≤ k + 1. Thus, the proof is complete.

Next, we shall focus on the determination of the Frobenius number of S. We shall start with the following proposition that will enable us to determine the Frobenius number and will help us later in determining the Pseudo frobenius number of S. For all 1 ≤ i < j ≤ m -1 where i and j are written as in Definition 3.0.1, we have :

• If α i = α j -2, β j = j = 0 and β i + i = 2k + 1, then w(i) -w(j) > 0. • If α i = α j -1, β i + i > k + β j + j and β j + j ≤ k, then w(i) -w(j) > 0.
• Otherwise, w(i) -w(j) < 0.

Proof. Let 1 ≤ i < j ≤ m -1, where i = α i (kl + r) + β i l + i t i and j = α j (kl + r) + β j l + j t j be as defined in Definition 3.0.1. By Theorem 3.1.1, we have

w(i) = m(kα i + β i + i ) + i and w(j) = m(kα j + β j + j ) + j.
We claim that α i ≤ α j . In fact, since i < j, then i kl+r < j kl+r , which implies that i kl+r ≤ j kl+r . Hence, α i ≤ α j .

Case 1.

If α i = α j . We aim to show that β i ≤ β j . Indeed, suppose by the way of contradiction that β i > β j , then i = α i (kl + r) + β i l + i t i = α j (kl + r) + β i l + i t i ≥ α j (kl + r) + β i l ≥ α j (kl + r) + (β j + 1)l. Since j t j < l, we get that i > α j (kl + r) + β j l + j t j = j which is a contradiction with i < j. Hence,

β i ≤ β j . • If β i < β j . Then, w(j) -w(i) = m(kα j + β j + j ) + j -m(kα i + β i + i ) -i = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i = m((β j -β i ) + ( j -i )) + j -i.
Since β i < β j , i < j and i , j ∈ {0, 1}, it follows that w(j) -w(i) > 0.

• If β i = β j . We aim to show that i t i < j t j , in particular i ≤ j . Suppose by the way of contradiction that i t i ≥ j t j , then

i = α i (kl + r) + β i l + i t i = α j (kl + r) + β j l + i t i ≥ α j (kl + r) + β j l + j t j = j,
which is a contradiction with i < j. Hence, i t i < j t j . As i , j ∈ {0, 1}, we get that i ≤ j . Therefore,

w(j) -w(i) = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i = m(( j -i )) + j -i.
Since i ≤ j and i < j, we obtain w(j) -w(i) > 0.

Consequently, if i < j and α i = α j , then w(i) -w(j) < 0.

Case 2. If α i < α j .

• If α i ≤ α j -3. By Proposition 3.0.2, we have i + β i ≤ 2k + 1. Then,

w(j) -w(i) = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i ≥ m(3k + (β j -β i ) + ( j -i )) + j -i ≥ m(3k -2k -1 + β j + j ) + j -i.
Since k ≥ 1 and i < j, it follows that w(j) -w(i) > 0. Consequently, if i < j and α i ≤ α j -3, then w(i) -w(j) < 0.

• If α i = α j -2.

• If β j + j > 0. By Proposition 3.0.2, we have i + β i ≤ 2k + 1. Then,

w(j) -w(i) = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i = m(2k + (β j -β i ) + ( j -i )) + j -i ≥ m(2k + β j + j -2k -1) + j -i ≥ m(2k + 1 -2k -1) + j -i.
Since i < j, we get w(j) -w(i) > 0.

• If β j = j = 0. Then,

w(j) -w(i) = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i = m(2k -β i -i ) + j -i.
By Proposition 3.0.2, we have i + β i ≤ 2k + 1.

• If β i + i ≤ 2k. Since i < j, we obtain w(j) -w(i) > 0. • If β i + i = 2k + 1. Since j ≤ m -1, it follows that w(j) -w(i) < 0.
Consequently, if i < j and α i = α j -2, then w(i) -w(j) < 0 unless in the case where β j = j = 0 and β i + i = 2k + 1.

• If α i = α j -1.

• If β i + i ≤ k + β j + j . Then, w(j) -w(i) = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i = m(k + (β j -β i ) + ( j -i )) + j -i ≥ m(k -k) + j -i.
Since i < j, we get w(j) -w(i) > 0.

• If β i + i > k + β j + j . Then, w(j) -w(i) = m((α j -α i )k + (β j -β i ) + ( j -i )) + j -i = m(k + (β j -β i ) + ( j -i )) + j -i ≤ m(k -k -1) + j -i.
Since j < m, it follows that w(j) -w(i) < 0. Note that if

β i + i > k + β j + j , then β j + j ≤ k it follows from β i + i ≤ 2k + 1 (Proposition 3.0.2).
Consequently, if i < j and α i = α j -1, then w(i) -w(j) < 0 unless in the case where β i + i > k + β j + j and β j + j ≤ k.

In conclusion, if i < j, we have w(i) -w(j) > 0, in the case α i = α j -2, β j = j = 0 and

β i + i = 2k + 1, or α i = α j -1, β i + i > k + β j + j and β j + j ≤ k,
and in the other cases w(i) -w(j) < 0.

Thus, the proof is complete.

The following theorem gives a formula for the Frobenius f (S). The Frobenius number f (S) of S is given by :

f (S) = m(kα m-1 + q + t -1) + α m-1 (kl + r) -1 if S satisfies condition (H), m(kα m-1 + β m-1 + m-1 ) -1 otherwise, where (H) : m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 , with α m-1 ≥ 1, β m-1 + m-1 ≤ k, q + t > β m-1 + m-1 and r > 1.
Proof. By Lemma 1.0.21, we have f (S) = max(Ap(S, m)) -m. We are going to show that max(Ap(S, m)) = w((α m-1 -1)(kl + r) + kl + r -1) if S satisfies condition (H) w(m -1) otherwise.

By applying Proposition 3.2.4, we get

max(Ap(S, m)) = max w (α m-1 -2)(kl + r) + kl + r -1 ,w (α m-1 -1)(kl + r) + kl + r -1 , w(m -1) .
Recall that r -1 = ql + t with t < l, t = 0 if t = 0 and t = 1 if t = 0. We have

w (α m-1 -1)(kl + r) + kl + r -1 -w (α m-1 -2)(kl + r) + kl + r -1 = m k(α m-1 -1) + k + q + t + (α m-1 -1)(kl + r) + kl + r -1 -m k(α m-1 -2) + k + q + t -(α m-1 -2)(kl + r) + kl + r -1 = m(k) + kl + r > 0.
Consequently,

max(Ap(S, m)) = max w (α m-1 -1)(kl + r) + kl + r -1 , w(m -1) . Case 1. If α m-1 = 0, then i = (α m-1 -1)(kl + r) + kl + r -1 < 0. Hence, max(Ap(S, m)) = w(m -1) = m(kα m-1 + β m-1 + m-1 ) + m -1. Case 2. If α m-1 ≥ 1.
• If r = 1. Thus, r -1 = 0 and so are q and t . Therefore,

w m -1 -w (α m-1 -1)(kl + r) + kl + r -1 = w(m -1) -w (α m-1 -1)(kl + 1) + kl = m kα m-1 + β m-1 + m-1 + α m-1 (kl + 1) + β m-1 l + m-1 t m-1 -m k(α m-1 -1) + k -(α m-1 -1)(kl + 1) -kl = m(β m-1 + m-1 ) + β m-1 l + m-1 t m-1 + 1 > 0. Consequently, max(Ap(S, m)) = w(m -1) = m(kα m-1 + β m-1 + m-1 ) + m -1. • If r > 1 (r -1 = ql + t where t < l, t = 1 if t ≥ 1 and t = 0 if t = 0). Then, w(m -1) -w (α m-1 -1)(kl + r) + kl + r -1 = m(kα m-1 + β m-1 + m-1 ) + α m-1 (kl + r) + β m-1 l + m-1 t m-1 -m k(α m-1 -1) + k + q + t -(α m-1 -1)(kl + r) + kl + r -1 = m(β m-1 + m-1 -q -t ) + β m-1 l + m-1 t m-1 + 1. • If q + t > β m-1 + m-1 .
We have β m-1 + m-1 ≤ k in this case this follows from Proposition 3.2.3. Since α m-1 ≥ 1 in this case, we get

β m-1 l + m-1 t m-1 + 1 ≤ α m-1 (kl + r) + β m-1 l + m-1 t m-1 = m -1. Consequently, w(m -1) -w (α m-1 -1)(kl + r) + kl + r -1 = m(β m-1 + m-1 -q -t ) + β m-1 l + m-1 t m-1 + 1 ≤ m(-1) + m -1 < 0. Hence, max(Ap(S, m)) = w (α m-1 -1)(kl + r) + kl + r -1 = m(kα m-1 + q + t ) + α m-1 (kl + r) -1. • If q + t ≤ β m-1 + m-1 . Then, w(m -1) -w (α m-1 -1)(kl + r) + kl + r -1 = m(β m-1 + m-1 -q -t ) + β m-1 l + m-1 t m-1 + 1 > 0.
Therefore,

max(Ap(S, m)) = w(m -1) = m(kα m-1 + β m-1 + m-1 ) + m -1.
Hence, if S satisfies condition (H), we get

max(Ap(S, m)) = w((α m-1 -1)(kl + r) + kl + r -1) = m(kα m-1 + q + t ) + α m-1 (kl + r) -1.
Otherwise, we obain

max(Ap(S, m)) = w(m -1) = m(kα m-1 + β m-1 + m-1 ) + m -1.
By applying Lemma 1.0.21, if S satisfies condition (H), we obtain

f (S) = m(kα m-1 + q + t -1) + α m-1 (kl + r) -1, otherwise, we obtain f (S) = m(kα m-1 + β m-1 + m-1 ) -1.
Thus, the proof is complete.

Example 3.2.6. Consider the following numerical semigroups.

• S =< 19, 20, 21, 22, 52 >. By using GAP [8], we get that f (S) = 89. Note that k = 2, l = 3 and r = 8.

In addition, m -1 = 18 = 1(14

) + 1(3) + 1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 with α m-1 = 1, β m-1 = 1, m-1 = 1, t m-1 = 1 and r -1 = 7 = 2(3) + 1 = ql + t with q = 2, t = 1 , t = 1.
We have S verifies condition (H) and verifies the formula given in Theorem 3.2.5 as 89=19(2(1)+2+1-1)+1( 14)-1.

• S =< 11, 12, 13, 14, 36 >. By using GAP [8], we obtain that f (S) = 43. Note that k = 2, l = 3 and r = 8. In addition, m -1 = 10 = 0(14

) + 3(3) + 1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 with α m-1 = 0, β m-1 = 3, m-1 = 1, t m-1 = 1 and r -1 = 7 = 2(3) + 1 = ql + t with q = 2, t = 1 , t = 1.
We have S does not verify condition (H) as α m-1 = 0 and verifies the formula given in Theorem 3.2.5 as 43 = 11(2(0) + 3 + 1) -1.

Genus of S

The following theorem gives a formula for the genus g(S).

Theorem 3.3.1. Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1. The genus g(S) of S is given by :

g(S) = kα m-1 + β m-1 + 1 l(kα m-1 + β m-1 ) 2 + m-1 t m-1 + kα m-1 (α m-1 + 1)r 2 + (q + 1)(r -1 + t)α m-1 2 .
Proof. By using Lemma 1.0.21 and Theorem 3.1.1, we get

g(S) = 1 m w∈Ap(S,m) w - m -1 2 = 1 m m-1 i=0 (m(kα i + β i + i ) + i) - m -1 2 = m-1 i=0 (kα i + β i + i ) + 1 m m-1 i=0 i - m -1 2 = m-1 i=0 (kα i + β i + i ).
Now, we are going to divide the set {0 ≤ i ≤ m -1} into subsets and calculate the value of kα i + β i + i in each subset. We have

{0 ≤ i ≤ m -1} = {j(kl + r) ≤ i ≤ j(kl + r) + kl + (r -1); 0 ≤ j ≤ α m-1 -1} ∪ {α m-1 (kl + r) ≤ i ≤ α m-1 + β m-1 l + m-1 t m-1 } = ∪ 7 =1 A , where A 1 = {j(kl + r); 0 ≤ j ≤ α m-1 -1}, A 2 = {j(kl + r) + yl + 1 ≤ i ≤ j(kl + r) + (y + 1)l; 0 ≤ j ≤ α m-1 -1 and 0 ≤ y ≤ k -1}, A 3 = {j(kl + r) + (k + y)l + 1 ≤ i ≤ j(kl + r) + (k + y + 1)l; 0 ≤ j ≤ α m-1 -1 and 0 ≤ y ≤ q -1}, A 4 = {j(kl + r) + (k + q)l + 1 ≤ i ≤ j(kl + r) + (k + q)l + t; 0 ≤ j ≤ α m-1 -1}, A 5 = {α m-1 (kl + r)}, A 6 = {α m-1 (kl + r) + yl + 1 ≤ i ≤ α m-1 (kl + r) + (y + 1)l; 0 ≤ y ≤ β m-1 -1}, A 7 = {α m-1 (kl + r) + β m-1 l + 1 ≤ i ≤ α m-1 (kl + r) + β m-1 l + m-1 t m-1 }.
Next, we will calculate the value of kα i + β i + i on each subset A .

If i ∈ A 1 , then kα i + β i + i = kj. If i ∈ A 2 , then kα i + β i + i = kj + y + 1. If i ∈ A 3 , then kα i + β i + i = kj + k + y + 1. If i ∈ A 4 , then kα i + β i + i = kj + k + q + 1. If i ∈ A 5 , then kα i + β i + i = kα m-1 . If i ∈ A 6 , then kα i + β i + i = kα m-1 + y + 1. If i ∈ A 7 , then kα i + β i + i = kα m-1 + β m-1 + 1. Therefore, g(S) = m-1 i=0 (kα i + β i + i ) = 7 =1 i∈A (kα i + β i + i ) = α m-1 -1 j=0 kα j(kl+r) + β j(kl+r) + j(kl+r) + k-1 y=0 j(kl+r)+(y+1)l i=j(kl+r)+yl+1 (kα i + β i + i ) + q-1 y=0 j(kl+r)+(k+y+1)l i=j(kl+r)+(k+y)l+1 (kα i + β i + i ) + j(kl+r)+(k+q)l+t i=j(kl+r)+(k+q)l+1 (kα i + β i + i ) + kα α m-1 (kl+r) + β α m-1 (kl+r) + α m-1 (kl+r) + β m-1 -1 y=0 α m-1 (kl+r)+(y+1)l i=α m-1 (kl+r)+yl+1 (kα i + β i + i ) + α m-1 (kl+r)+β m-1 l+ m-1 t m-1 i=α m-1 (kl+r)+β m-1 l+1 (kα i + β i + i ).
Equivalently,

g(S) = α m-1 -1 j=0 kj + k-1 y=0 j(kl+r)+(y+1)l i=j(kl+r)+yl+1 (kj + y + 1) + q-1 y=0 j(kl+r)+(y+1)l i=j(kl+r)+yl+1 (kj + k + y + 1) + j(kl+r)+(k+q)l+t i=j(kl+r)+(k+q)l+1 (kj + k + q + 1) +kα m-1 + β m-1 -1 y=0 α m-1 (kl+r)+(y+1)l i=α m-1 (kl+r)+yl+1 (kα m-1 + y + 1) + α m-1 (kl+r)+β m-1 l+ m-1 t m-1 i=α m-1 (kl+r)+β m-1 l+1 (kα m-1 + β m-1 + 1) = α m-1 -1 j=0 kj + k-1 y=0 (kj + y + 1)l + q-1 y=0 (kj + k + y + 1)l + (kj + k + q + 1)t +kα m-1 + β m-1 -1 y=0 (kα m-1 + y + 1)l+(kα m-1 + β m-1 + 1) m-1 t m-1 = α m-1 -1 j=0 kj + k(kj + 1)l + k(k -1)l 2 + (kj + k + 1)ql + (q -1)ql 2 + (kj + k + q + 1)t +kα m-1 + (kα m-1 + 1)β m-1 l+ β m-1 (β m-1 -1)l 2 + (kα m-1 + β m-1 + 1) m-1 t m-1 = α m-1 -1 j=0 k(kl + 1 + ql + t)j + (k + 1)(kl + 2ql + 2t) + 2qt + q(q -1)l 2 +kα m-1 + (kα m-1 + 1)β m-1 l + β m-1 (β m-1 -1)l 2 +(kα m-1 + β m-1 + 1) m-1 t m-1 = k(kl + 1 + ql + t)α m-1 (α m-1 -1) 2 + kα m-1 + (kα m-1 + 1)β m-1 l + (k + 1)(kl + 2ql + 2t) + 2qt + q(q -1)l α m-1 2 + β m-1 (β m-1 -1)l 2 +(kα m-1 + β m-1 + 1) m-1 t m-1 = (kα m-1 -k)(klα m-1 ) 2 + kα m-1 (α m-1 -1) 2 + (ql + t)(kα m-1 -k)α m-1 2 + (k + 1)klα m-1 2 + (ql + t)(2k + 2)α m-1 2 + 2qt + q(q -1)l α m-1 2 +kα m-1 + (kα m-1 + 1)β m-1 l 2 + β m-1 (klα m-1 ) 2 + β m-1 (β m-1 l) 2 +(kα m-1 + β m-1 + 1) m-1 t m-1 = (kα m-1 + β m-1 + 1) l(kα m-1 + β m-1 ) 2 + m-1 t m-1 + kα m-1 (α m-1 + 1) 2 + (ql + t)(kα m-1 + k + 2)α m-1 2 + 2qt + q(q -1)l α m-1 2 .
Finally,

g(S) = (kα m-1 + β m-1 + 1) l(kα m-1 + β m-1 ) 2 + m-1 t m-1 + kα m-1 (α m-1 + 1) 2 + (k + kα m-1 + 2)α m-1 (r -1) + 2qt + q(q -1)l α m-1 2 = (kα m-1 + β m-1 + 1) l(kα m-1 + β m-1 ) 2 + m-1 t m-1 + kα m-1 (α m-1 + 1)r 2 + (q + 1)(r -1 + t)α m-1 2 .
Thus, the proof is complete. • S =< 19, 20, 21, 22, 52 >. By using GAP [8], we get that g(S) = 50. Note that k = 2, l = 3 and r = 8.

In addition, m -1 = 18 = 1(14

) + 1(3) + 1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 with α m-1 = 1, β m-1 = 1, m-1 = 1, t m-1 = 1 and r -1 = 7 = 2(3) + 1 = ql + t with q = 2, t = 1. We have 50 = 2(1) + 1 + 1 3(2(1)+1) 2 + 1 + 2(1)(1+1)8 2 + (2+1)(8-1+1)(1)

2

. Hence, S verifies the formula given in Theorem 3.3.1.

• S =< 11, 12, 13, 14, 36 >. By using GAP [8], we get that g(S) = 22. Note that k = 2, l = 3 and r = 8.

In addition, m -1 = 10 = 0(14

) + 3(3) + 1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 with α m-1 = 0, β m-1 = 3, m-1 = 1, t m-1 = 1 and r -1 = 7 = 2(3) + 1 = ql + t with q = 2, t = 1. We have 22 = 2(0) + 3 + 1 3(2(0)+3) 2 + 1 + 2(0)(0+1)8 2 + (2+1)(8-1+1)(0)

2

. Hence, S verifies the formula given in Theorem 3.3.1.

Determination of symmetric and pseudo-symmetric numerical semigroups

Next, we shall focus on the determination of symmetric and pseudo-symmetric numerical semigroups. We shall start with a technical Lemma. Lemma 3.4.1. Let the notation be as defined in Definition 3.0.1 and in Definition 3.2.1, we have the following :

2g(S) -f (S) + 1 = F 1 if S satisfies condition (H), F 2 otherwise,
where

F 1 = α m-1 (1 -t )(kl + r) + kt + (q + 1)(t -1) +β m-1 l kα m-1 + β m-1 + 1 -q + 1 -t + m-1 t m-1 kα m-1 + 2β m-1 + 2 -q + 1 -t + 1 -t -q, F 2 = α m-1 -m-1 (kl + r) + (k + q -β m-1 )r + k(l -1) + r + (q + 1)(t -1) +β m-1 l 1 -m-1 + m-1 t m-1 kα m-1 + β m-1 + 2 -m-1 -m-1 -β m-1 and (H) : m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 , with α m-1 ≥ 1, β m-1 + m-1 ≤ k, q + t > β m-1 + m-1 and r > 1.
Proof. By Theorem 3.2.5, we have f (S) = m(kα m-1 + q + t -1) + α m-1 (kl + r) -1 if S satisfies condition (H) and f (S) = m(kα m-1 + β m-1 + m-1 ) -1 otherwise. Now, we use the formulas in Theorem 3.2.5 and Theorem 3.3.1.

Case 1. If f (S) = m(kα m-1 + q + t -1) + α m-1 (kl + r) -1. Then, 2g(S) -f (S) + 1 = (kα m-1 + β m-1 + 1) l(kα m-1 + β m-1 ) + 2 m-1 t m-1 + k(α m-1 + 1)α m-1 r +(q + 1)(r -1)α m-1 + (q + 1)tα m-1 -m(kα m-1 + q + t -1) -α m-1 (kl + r) = α m-1 k 2 lα m-1 + klβ m-1 + 2k m-1 t m-1 + klβ m-1 + kl + k(α m-1 + 1)r +(q + 1)(r -1) + (q + 1)t -km -kl -r + β m-1 l β m-1 + 1 + m-1 t m-1 2β m-1 + 2 -m(q + t -1) = α m-1 kql + kt + qr + (q + 1)(t -1) + β m-1 l kα m-1 + β m-1 + 1 + m-1 t m-1 kα m-1 + 2β m-1 + 2 -α m-1 (kl + r)(q + t -1) -β m-1 l(q + t -1) -m-1 t m-1 (q + t -1) -(q + t -1) = α m-1 (1 -t )(kl + r) + kt + (q + 1)(t -1) + β m-1 l kα m-1 + β m-1 + 1 -q + 1 -t + m-1 t m-1 kα m-1 + 2β m-1 + 2 -q + 1 -t + 1 -t -q. Case 2. If f (S) = m(kα m-1 + β m-1 + m-1 ) -1. Then, 2g(S) -f (S) + 1 = (kα m-1 + β m-1 + 1) l(kα m-1 + β m-1 ) + 2 m-1 t m-1 + k(α m-1 + 1)α m-1 r +(q + 1)(r -1)α m-1 + (q + 1)tα m-1 -m(kα m-1 + β m-1 + m-1 ) = α m-1 k 2 lα m-1 + klβ m-1 + 2k m-1 t m-1 + klβ m-1 + kl + k(α m-1 + 1)r +(q + 1)(r -1) + (q + 1)t -km + β m-1 l β m-1 + 1 + m-1 t m-1 2β m-1 + 2 -m(β m-1 + m-1 ). Therefore, 2g(S) -f (S) + 1 = α m-1 klβ m-1 + kl + kr + (q + 1)(r -1) + (q + 1)t -k + β m-1 l β m-1 + 1 + m-1 t m-1 kα m-1 + 2β m-1 + 2 -α m-1 (kl + r)(β m-1 + m-1 ) -β m-1 l(β m-1 + m-1 ) -m-1 t m-1 (β m-1 + m-1 ) -(β m-1 + m-1 ) = α m-1 -m-1 (kl + r) + (k + q -β m-1 )r + k(l -1) + r + (q + 1)(t -1) +β m-1 l 1 -m-1 + m-1 t m-1 kα m-1 + β m-1 + 2 -m-1 -m-1 -β m-1 .
Thus, the proof is complete.

Determination of symmetric numerical semigroup

The following theorem gives the set of symmetric numerical semigroups. Then, S is symmetric if and only if it satisfies one of the following :

1. S =< 2k + 3, 2k + 4, k(2k + 4) + k + 2 > .
2. S =< 2kl + 3, 2kl + 4, . . . , (2k + 1)l + 3, k((2k + 1)l + 3) + kl + 2 > with l ≥ 2.

S =< β

m-1 + 1, β m-1 + 2, k(β m-1 + 2) + r > with β m-1 ≥ 1.
4. S =< (α m-1 + 1)(k + q + 1), (α m-1 + 1)(k + q + 1) + 1, k((α m-1 + 1)(k + q + 1) + 1) + q + 1 > .

S =< β

m-1 l + 2, . . . , (β m-1 + 1)l + 2, k((β m-1 + 1)l + 2) + ql + t + 1 > with t ≥ 1 and l ≥ 2. 6. S =< (α m-1 + 1)((k + q)l + 2), . . . , (α m-1 + 1)((k + q)l + 2) + l, k((α m-1 + 1)((k + q)l + 2) + l) + ql + 2 > with l ≥ 2. 7. S =< β m-1 l + 2, . . . , (β m-1 + 1)l + 2, k((β m-1 + 1)l + 2) + ql + 1 > with l ≥ 2. 8. S =< α m-1 (kl + 1) + (k -1)l + 2, . . . , α m-1 (kl + 1) + kl + 2, k(α m-1 (kl + 1) + kl + 2) + 1 > with l ≥ 2.
Proof. By Lemma 1.0.34, we have S is symmetric if and only if 2g(S) -f (S) + 1 = 0.

Case 1. If S satisfies condition (H). By Theorem 3.2.5,

f (S) = m(kα m-1 + q + t -1) + α m-1 (kl + r) -1 with α m-1 > 0, β m-1 + m-1 ≤ k, q + t > β m-1 + m-1 and r > 1.
By using Lemma (3.4.1), S is symmetric if and only if

α m-1 (1 -t )(kl + r) + kt + (q + 1)(t -1) +β m-1 l kα m-1 + β m-1 + 1 -q + 1 -t + m-1 t m-1 kα m-1 + 2β m-1 + 2 -q + 1 -t + 1 -t = q.
• If t = 0 (t = 0 and r = ql + 1 in this case). Then, S is symmetric if and only if

α m-1 kl + q(l -1) + β m-1 l kα m-1 + β m-1 + 2 -q + m-1 t m-1 kα m-1 + 2β m-1 + 3 -q + 1 = q. (3.4.1) Since l ≥ 1, α m-1 ≥ 1 and q ≤ k + 1 (Proposition 3.2.3), it follows that kl + q(l -1) ≥ kl, kα m-1 + β m-1 + 2 -q ≥ 1 and kα m-1 + 2β m-1 + 3 -q ≥ 2. Consequently, (3.4.1) implies that α m-1 kl + β m-1 l + m-1 t m-1 (2) + 1 ≤ q. (3.4.2) As α m-1 ≥ 1, l ≥ 1 and q ≤ k + 1 (Proposition 3.2.3), then (3.4.2) implies that l = 1, α m-1 = 1, β m-1 = 0, m-1 t m-1 = 0 and q = k + 1. As r = ql + 1 in this case (as t = t = 0), we get r = k + 2.
Therefore, S is symmetric in this case if and only if

α m-1 = 1, β m-1 = 0, t m-1 = 0, l = 1, r = k + 2. (3.4.3) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1
. By substituting (3.4.3) in S, we obtain

S =< 2k + 3, 2k + 4, k(2k + 4) + k + 2 > . • If t = 1 (t ≥ 1)
. Then, S is symmetric if and only if

α m-1 kt + (q + 1)(t -1) + β m-1 l kα m-1 + β m-1 + 1 -q + m-1 t m-1 kα m-1 + 2β m-1 + 2 -q = q.
(3.4.4)

Since t ≥ 1, α m-1 ≥ 1 and q ≤ k (by Proposition 3.2.3 as t = 1), it follows that kt

+ (q + 1)(t -1) ≥ kt, kα m-1 + β m-1 + 1 -q ≥ β m-1 + 1 and kα m-1 + 2β m-1 + 2 -q ≥ 2β m-1 + 2. Consequently, (3.4.4) implies that α m-1 (kt) + β m-1 l(β m-1 + 1) + m-1 t m-1 (2β m-1 + 2) ≤ q. (3.4.5)
As α m-1 ≥ 1, t ≥ 1 and q ≤ k (by Proposition 3.2.3 as t = 1), then (3.4.5) implies that α m-1 = 1, t = 1, β m-1 = 0, m-1 t m-1 = 0 and q = k. As r = ql + t + 1, we get r = kl + 2. Since t ≥ 1, then l ≥ 2. Therefore, S is symmetric in this case if and only if

α m-1 = 1, β m-1 = 0, t m-1 = 0, r = kl + 2, l ≥ 2. ( 3.4.6)
We have m -

1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1
. By substituting (3.4.6) in S, we obtain

S =< 2kl + 3, 2kl + 4, . . . , (2k + 1)l + 3, k((2k + 1)l + 3) + kl + 2 > with l ≥ 2.
Case 2. If S does not satisfy condition (H). By Theorem 3.2.5,

f (S) = m(kα m-1 + β m-1 + m-1 ) -1 with α m-1 = 0 or β m-1 + m-1 > k or β m-1 + m-1 ≥ q + t or r = 1. (3.4.7)
By using Lemma 3.4.1, S is symmetric if and only if

α m-1 -m-1 (kl + r) + (k + q -β m-1 )r + k(l -1) + r + (q + 1)(t -1) +β m-1 l 1 -m-1 + m-1 t m-1 kα m-1 + β m-1 + 2 -m-1 = β m-1 + m-1 . • If m-1 = 0 (t m-1 = 0). We have r = ql + t + 1, then S is symmetric if and only if α m-1 (k + q -β m-1 )r + k(l -1) + q(l -1) + (q + 2)t + β m-1 (l -1) = 0. (3.4.8)
• If β m-1 = 0. Since m-1 = 0 and m = 1 (S = N), it follows that α m-1 ≥ 1. Since β m-1 = 0, m-1 = 0, and α m-1 ≥ 1, by using (3.4.7), we get r = 1 in particular q = 0 and t = t = 0. In this case by using (3.4.8), we obtain S is symmetric if and only if α m-1 (k + k(l -1)) = 0. As α m-1 ≥ 1, it follows that kl = 0 which is impossible (as k > 0 and l > 0).

• If β m-1 ≥ 1. Since l ≥ 1 and β m-1 ≤ k + q (Proposition 3.2.
2), it follows that α m-1 (k + qβ m-1 )r + k(l -1) + q(l -1) + (q + 2)t ≥ 0 and β m-1 (l -1) ≥ 0. By using (3.4.8), we get

β m-1 (l -1) = 0 (3.4.9) and α m-1 (k + q -β m-1 )r + k(l -1) + q(l -1) + (q + 2)t = 0. (3.4.10)
Since β m-1 ≥ 1, then (3.4.9) implies that l = 1. By substituting l = 1 in (3.4.10), we get

α m-1 (k + q -β m-1 )r + (q + 2)t = 0. (3.4.11)
Now, (3.4.11) implies that α m-1 = 0 (3.4.12)

or (k + q -β m-1 )r + (q + 2)t = 0. (3.4.13)
Since β m-1 ≤ k + q (Proposition 3.2.2), r > 0 and q + 2 > 0, then (3.4.13) implies that β m-1 = k + q and t = 0. As r = ql + t + 1 (with l = 1 proved above), then r = ql + 1 = q + 1 in this case. Therefore, S is symmetric in this case if and only if

α m-1 = 0, β m-1 ≥ 1, m-1 = 0, l = 1 (3.4.14) or β m-1 = k + q, m-1 = 0, l = 1, r = q + 1. (3.4.15) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 .
S =< β m-1 + 1, β m-1 + 2, k(β m-1 + 2) + r > with β m-1 ≥ 1 or S =< (α m-1 + 1)(k + q + 1), (α m-1 + 1)(k + q + 1) + 1, k((α m-1 + 1)(k + q + 1) + 1) + q + 1 > . • If m-1 = 1 (t m-1 ≥ 1)
. Then, S is symmetric if and only if

α m-1 (k + q -β m-1 )r + (q + 1)(t -1) +(t m-1 -1) kα m-1 + β m-1 + 1 = 0.
(3.4.16)

• If t ≥ 1. Since β m-1 ≤ k + q (Proposition 3.2.2) and t ≥ 1, it follows that α m-1 (k + q -β m-1 )r + (q + 1)(t -1) ≥ 0.
We have kα m-1 + β m-1 + 1 > 0 and t m-1 ≥ 1. By using (3.4.16), we get

t m-1 = 1 (3.4.17) and α m-1 (k + q -β m-1 )r + (q + 1)(t -1) = 0. (3.4.18)
Since

t m-1 = 1, it follows that l ≥ 2 ( m-1 t m-1 < l). Now, (3.4.18) gives α m-1 = 0 or (k + q - β m-1 )r +(q +1)(t-1) = 0. If (k +q -β m-1 )r +(q +1)(t-1) = 0, since β m-1 ≤ k +q (Proposition 3.2.
2), r > 0, q + 1 > 0 and t ≥ 1, it follows that t = 1 and β m-1 = k + q. Consequently, S is symmetric if and only if t m-1 = 1 with α m-1 = 0, l ≥ 2 or t m-1 = 1 with t = 1, β m-1 = k + q, l ≥ 2. Since r = ql + t + 1 and t ≥ 1 in this case, it follows that S is symmetric if and only if

α m-1 = 0, t m-1 = 1, t ≥ 1 with l ≥ 2 (3.4.19) or β m-1 = k + q, t m-1 = 1, r = ql + 2 with l ≥ 2. ( 3.4.20)
We have m -

1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 .
S =< β m-1 l + 2, . . . , (β m-1 + 1)l + 2, k((β m-1 + 1)l + 2) + ql + t + 1 > with t ≥ 1 and l ≥ 2 or S =< (α m-1 + 1)((k + q)l + 2), . . . , (α m-1 + 1)((k + q)l + 2) + l, k((α m-1 + 1)((k + q)l + 2) + l) + ql + 2 > with l ≥ 2.
• If t = 0. We have r = ql + 1, then (3.4.16) implies that S is symmetric if and only if

α m-1 (k + q -1 -β m-1 )r + q(l -1) +(t m-1 -1) kα m-1 + β m-1 + 1 = 0. (3.4.21)
We have r = ql+1 and 1 ≤ m-1 t m-1 ≤ l-1 in this case. On the other hand,

β m-1 l+ m-1 t m-1 ≤ kl + r -1 = kl + ql. Hence, β m-1 ≤ k + q -1 (as m-1 t m-1 ≥ 1). Since l ≥ 1 and β m-1 ≤ k + q -1 in this case, it follows that α m-1 (k + q -1 -β m-1 )r + q(l -1) ≥ 0.
We have t m-1 ≥ 1 and kα m-1 + β m-1 + 1 > 0. By using (3.4.21), we get

t m-1 = 1 (3.4.22) and α m-1 (k + q -1 -β m-1 )r + q(l -1) = 0. (3.4.23) Since t m-1 = 1, it follows that l ≥ 2 ( m-1 t m-1 < l). Now, (3.4.23) gives α m-1 = 0 or (k + q - 1 -β m-1 )r + q(l -1) = 0. If (k + q -1 -β m-1 )r + q(l -1) = 0, since β m-1 ≤ k + q -1
in this case (proved above), r > 0 and l ≥ 1, it follows that β m-1 = k +q -1 and q(l -1) = 0. Since t m-1 = 1, it follows that l ≥ 2, in particular q(l -1) = 0 gives q = 0. Thus, in the second case, we have β m-1 = k + q -1 with q = 0 (r = 1 in this case as t = 0). Consequently, S is symmetric if and only if t m-1 = 1 with α m-1 = 0, l ≥ 2 or t m-1 = 1 with q = 0 (r = 1),

β m-1 = k + q -1 = k -1, l ≥ 2.
As r = ql + 1 in this case, it follows that S is symmetric if and only if

α m-1 = 0, t m-1 = 1, r = ql + 1 with l ≥ 2 (3.4.24) or β m-1 = k -1, t m-1 = 1, r = 1 with l ≥ 2. (3.4.25) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1
. By using (3.4.24) and (3.4.25), we get S is symmetric if and only if

S =< β m-1 l + 2, . . . , (β m-1 + 1)l + 2, k((β m-1 + 1)l + 2) + ql + 1 > with l ≥ 2 or S =< α m-1 (kl + 1) + (k -1)l + 2, . . . , α m-1 (kl + 1) + kl + 2, k(α m-1 (kl + 1) + kl + 2) + 1 > with l ≥ 2.
Thus, the proof is complete.

Example 3.4.3. Consider the following numerical semigroups.

1. S =< 9, 10, 35 >. By using GAP [8], we get that S is symmetric. Note that m = 9 = 2k + 3 with k = 3. In addition, l = 1 and r = 5 = k + 2. Hence, S verifies the formula in Theorem 3.4.2.

2. S =< 15, 16, 17, 18, 44 >. By using GAP [8], we get that S is symmetric. Note that l = 3 and m = 15 = 2kl + 3 with k = 2. Moreover, r = 8 = kl + 2. Hence, S verifies the formula in Theorem 3.4.2.

3. S =< 8, 9, 48 >. By using GAP [8], we get that S is symmetric. Note that l = 1. Moreover, m = 8 = 7(1) + 1 where α m-1 = 0, β m-1 = 7 ≥ 1 and t m-1 = 0. In addition, k = 5 and r = 3. Hence, S verifies the formula in Theorem 3.4.2.

4. S =< 10, 11, 35 >. By using GAP [8], we get that S is symmetric. Note that l = 1. In addition, m = 10 = 1(3 + 1 + 1) + 3 + 1 + 1 = (1 + 1)(3 + 1 + 1) where α m-1 = 1, β m-1 = 3 + 1 = k + q such that k = 3 and q = 1, t m-1 = 0 and r = 2 = q + 1. Hence, S verifies the formula in Theorem 3.4.2.

5. S =< 18, 19, 20, 21, 22, 54 >. By using GAP [8], we get that S is symmetric. Note that l = 4 ≥ 2 and m = 18 = 4(4) + 1 + 1 where α m-1 = 0, β m-1 = 4 and t m-1 = 1. In addition k = 2 and r = 10 = 2(4) + 1 + 1 = ql + t + 1 such that t = 1 ≥ 1. Hence, S verifies the formula in Theorem 3.4.2.

6. S =< 16, 17, 18, 40 >. By using GAP [8], we get that S is symmetric. Note that l = 2 ≥ 2, k = 2, q = 1 and r = 4 = ql + 2. In addition,

m = 16 = 1((2 + 1)2 + 2) + (2 + 1)2 + 2 = (1 + 1)((2 + 1)2 + 2) = α m-1 ((k + q)l + 2) + (k + q)l + 2 where α m-1 = 1, β m-1 = k + q = 3 and t m-1 = 1.
Hence, S verifies the formula in Theorem 3.4.2.

7. S =< 18, 19, 20, 21, 22, 75 >. By using GAP [8], we get that S is symmetric. Note that l = 4 ≥ 2 and m = 18 = 4(4) + 1 + 1 where α m-1 = 0, β m-1 = 4 and t m-1 = 1. In addition k = 3 and r = 9 = 2(4) + 1 = ql + 1. Hence, S verifies the formula in Theorem 3.4.2.

8. S =< 89, 90, 91, 92, 93, 94, 565 >. By using GAP [8], we get that S is symmetric. Note that l = 5 ≥ 2, k = 6 and r = 1. In addition, m = 2(5(6) + 1) + (6 -1)(5) + 2, where α m-1 = 2, β m-1 = 5 = k -1 and t m-1 = 1. Hence, S verifies the formula in Theorem 3.4.2.

Determination of pseudo-symmetric numerical semigroup

Now, we shall characterize the set of pseudo-symmetric numerical semigroups.

Theorem 3.4.4. Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1.

Then, S is pseudo-symmetric if and only if it satisfies one of the following :

1. S =< 9, 10, 13 > .

2. S =< 2k + 2, 2k + 3, k(2k + 3) + k + 1 > . 3. S =< (2k -1)l + 3, . . . , 2kl + 3, k(2kl + 3) + (k -1)l + 2 > with l ≥ 2. 4. S =< 2(2l + 2) + 1, . . . , 2(2l + 2) + 1 + l, (2(2l + 2) + 1 + l) + l + 2 > with l ≥ 2. 5. S =< 2k + 1, 2k + 2, k(2k + 2) + 1 > with k ≥ 2.
6. S =< 3, 4, 5 > .

f (S) = m(kα m-1 + q + t -1) + α m-1 (kl + r) -1 with α m-1 > 0, β m-1 + m-1 ≤ k, q + t > β m-1 + m-1 and r > 1.
By using Lemma 3.4.1, S is pseudo-symmetric if and only if

α m-1 (1 -t )(kl + r) + kt + (q + 1)(t -1) +β m-1 l kα m-1 + β m-1 + 1 -q + 1 -t + m-1 t m-1 kα m-1 + 2β m-1 + 2 -q + 1 -t + 1 -t = q + 1.
• If t = 0 (t = 0, r = ql + 1). Since r = ql + 1, it follows that S is pseudo-symmetric if and only if

α m-1 kl + q(l -1) + β m-1 l kα m-1 + β m-1 + 2 -q + m-1 t m-1 kα m-1 + 2β m-1 + 3 -q = q. (3.4.26) Since α m-1 ≥ 1 and q ≤ k + 1 (Proposition 3.2.3), it follows that kα m-1 + β m-1 + 2 -q ≥ β m-1 + 1 and kα m-1 + 2β m-1 + 3 -q ≥ 2β m-1 + 2. Consequently, (3.4.26) implies that α m-1 (kl + q(l -1)) + β m-1 l(β m-1 + 1) + m-1 t m-1 (2β m-1 + 2) ≤ q. As α m-1 ≥ 1, k ≥ 1, l ≥ 1 and q ≤ k + 1 (Proposition 3.2.3), it follows that l = 1, β m-1 = 0, m-1 t m-1 = 0, α m-1 ∈ {1, 2}. (3.4.27)
• If α m-1 = 2. By substituting (3.4.27) in (3.4.26), it follows that S is pseudo-symmetric if and only if 2k = q. As q ≤ k + 1 (Proposition 3.2.3), we get k = 1 and q = 2k = 2. As r = ql + 1 with l = 1 in this case, we obtain r = 3. Thus, we have • If α m-1 = 1. By substituting (3.4.27) in (3.4.26), it follows that k = q. As r = ql + 1 with l = 1 in this case, we obtain r = k + 1. Thus, we have

α m-1 = 2, β m-1 = 0, t m-1 = 0, l = 1, r = 3, k = 1. ( 3
α m-1 = 1, β m-1 = 0, t m-1 = 0, l = 1, r = k + 1. (3.4.29)
We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 . By substituting (3.4.29) in S, we get

S =< 2k + 2, 2k + 3, k(2k + 3) + k + 1 > . • If t = 1 (t ≥ 1)
. Then, S is pseudo-symmetric if and only if

α m-1 kt + (q + 1)(t -1) + β m-1 l kα m-1 + β m-1 + 1 -q + m-1 t m-1 kα m-1 + 2β m-1 + 2 -q = q + 1.
(3.4.30)

Since α m-1 ≥ 1 and q ≤ k (by Proposition 3.2.3 as t = 1), it follows that kα m-1

+ β m-1 + 1 -q ≥ β m-1 + 1 and kα m-1 + 2β m-1 + 2 -q ≥ 2β m-1 + 2. Consequently, (3.4.30) implies that α m-1 (kt + (q + 1)(t -1)) + β m-1 l(β m-1 + 1) + m-1 t m-1 (2β m-1 + 2) ≤ q + 1. ( 3
α m-1 ∈ {1, 2}, β m-1 = 0, m-1 t m-1 = 0, t = 1, l ≥ 2. ( 3.4.32)
• If α m-1 = 1. By substituting (3.4.32) in (3.4.30), it follows that k = q + 1. As r = ql + t + 1 and t = 1, then r = (k -1)l + 2 in this case. Thus, we have • If α m-1 = 2. By substituting (3.4.32), in (3.4.30), it follows that 2k = q + 1. As q ≤ k (by Proposition 3.2.3 as t = 1) in this case, we obtain k = 1 and q = 1. Since r = ql + t + 1 with t = 1 and q = 1 we get r = l + 2. Therefore, we have

α m-1 = 1, β m-1 = 0, t m-1 = 0, r = (k -1)l + 2, l ≥ 2. ( 3
α m-1 = 2, β m-1 = 0, t m-1 = 0, k = 1, r = l + 2, l ≥ 2. (3.4.34) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1
. By substituting (3.4.34) in S, we obtain

S =< 2(2l + 2) + 1, . . . , 2(2l + 2) + 1 + l, (2(2l + 2) + 1 + l) + l + 2 > with l ≥ 2.
Case 2. If S does not satisfy condition (H). By Theorem 3.2.5,

f (S) = m(kα m-1 + β m-1 + m-1 ) -1 with α m-1 = 0 or β m-1 + m-1 > k or β m-1 + m-1 ≥ q + t or r = 1. ( 3
α m-1 -m-1 (kl + r) + (k + q -β m-1 )r + k(l -1) + r + (q + 1)(t -1) +β m-1 l 1 -m-1 + m-1 t m-1 kα m-1 + β m-1 + 2 -m-1 = β m-1 + m-1 + 1.
• If m-1 = 0. We have r = ql + t + 1, then S is pseudo-symmetric if and only if

α m-1 (k + q -β m-1 )r + k(l -1) + q(l -1) + (q + 2)t + β m-1 (l -1) = 1. (3.4.36) • If β m-1 ≥ 1. Since β m-1 ≤ k + q (Proposition 3.2.
2) and l ≥ 1, it follows that α m-1 (k + qβ m-1 )r + k(l -1) + q(l -1) + (q + 2)t ≥ 0 and β m-1 (l -1) ≥ 0. By using (3.4.36), we get that l ≤ 2.

• If l = 2. From (3.4.36), as β m-1 ≥ 1 and β m-1 ≤ k + q (Proposition 3.2.2), it follows that α m-1 = 0 and β m-1 = 1. In this case as m-1 = 0, we get that m = 3. Since l = 2, then ν = 4 > m which is impossible (ν ≤ m). • If l = 1. By using (3.4.36), we get S is pseudo-symmetric if and only if α m-1 (k + qβ m-1 )r + (q + 2)t = 1. Hence, α m-1 = 1 and (k + q -β m-1 )r + (q + 2)t = 1. Since β m-1 ≤ k + q (Proposition 3.2.2), r > 0 and q + 2 > 1, we get that t = 0, β m-1 = k + q -1 and r = 1 (in particular q = 0). Thus, β m-1 = k -1. We have β m-1 ≥ 1, then k ≥ 2. Thus, we have

α m-1 = 1, β m-1 = k -1, t m-1 = 0, r = 1, l = 1 (3.4.37) with k ≥ 2. We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1
. By substituting (3.4.37) in S, we obtain

S =< 2k + 1, 2k + 2, k(2k + 2) + 1 > with k ≥ 2.
• If β m-1 = 0. Since m-1 = 0 and m = 1 (S = N), it follows that α m-1 ≥ 1. As α m-1 ≥ 1, β m-1 = 0 and m-1 = 0, then (3.4.35) implies that r = 1 in particular q = 0 and t = t = 0. Then, (3.4.36) implies that S is pseudo-symmetric if and only if α m-1 k+k(l-1) = 1. Therefore, α m-1 = 1 and kl = 1. Thus, we have

α m-1 = 1, β m-1 = 0, t m-1 = 0, k = 1, l = 1, r = 1. (3.4.38) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1
. By substituting (3.4.38) in S, we get S =< 3, 4, 5 > .

• If m-1 = 1 (t m-1 ≥ 1)
. Then, S is pseudo-symmetric if and only if α m-1 (k + q -β m-1 )r + (q + 1)(t -1) + (t m-1 -1) kα m-1 + β m-1 + 1 = 1.

(3.4.39)

• If t ≥ 1. Since t m-1 ≥ 1, β m-1 ≤ k + q (Proposition 3.2.
2) and t ≥ 1, it follows that α m-1 (k + q -β m-1 )r + (q + 1)(t -1) ≥ 0 and (t m-1 -1) kα m-1 + β m-1 + 1 ≥ 0. From (3.4.39), it follows that

     α m-1 (k + q -β m-1 )r + (q + 1)(t -1) = 0, (t m-1 -1) kα m-1 + β m-1 + 1 = 1 (3.4.40) or      α m-1 (k + q -β m-1 )r + (q + 1)(t -1) = 1, (t m-1 -1) kα m-1 + β m-1 + 1 = 0.
(3.4.41) From (3.4.40) as (t m-1 -1) kα m-1 + β m-1 + 1 = 1, it follows that α m-1 = 0, β m-1 = 0 and t m-1 = 2. Thus, m = 3 but ν ≤ m = 3, then l = 1 for must. On the other hand, t < l = 1 which implies that t = 0 which is impossible as t ≥ 1 in this case. Thus, we do not have case (3.4.40). Now, consider (3.4.41). As kα m-1 + β m-1 + 1 > 0, then (3.4.41) implies that t m-1 = 1, α m-1 = 1 and (k + q -β m-1 )r + (q + 1)(t -1) = 1.

(3.4.42)

Since t ≥ 1, r > 0, q + 1 > 0 and β m-1 ≤ k + q (Proposition 3.2.2), then (3.4.42) implies that q = 0, t = 2 and β m-1 = k + q = k (the case where β m-1 = k + q -1, r = 1 and t = 1 is impossible as r = 1 implies that t = 0 and we get a contradiction). As r = ql + t + 1, we get r = 3. In addition, as t = 2, it follows that l ≥ 3. Thus, we have • If t = 0. We have r = ql + 1, from (3.4.39), it follows that S is pseudo-symmetric if and only if

α m-1 = 1, β m-1 = k, t m-1 = 1, r = 3, l ≥ 3. ( 3
α m-1 (k + q -1 -β m-1 )r + q(l -1) +(t m-1 -1) kα m-1 + β m-1 + 1 = 1. (3.4.44)
We have r = ql+1 and 1 ≤ m-1 t m-1 ≤ l-1. On the other hand,

β m-1 l+ m-1 t m-1 ≤ kl+r-1 = kl + ql. Hence, β m-1 ≤ k + q -1 (as m-1 t m-1 ≥ 1). Since l ≥ 1, t m-1 ≥ 1 and β m-1 ≤ k + q -1, it follows that α m-1 (k + q -1 -β m-1 )r + q(l -1) ≥ 0 and (t m-1 -1) kα m-1 + β m-1 + 1 ≥ 0.
From (3.4.44), it follows that

     α m-1 (k + q -1 -β m-1 )r + q(l -1) = 0, (t m-1 -1) kα m-1 + β m-1 + 1 = 1, (3.4.45) or      α m-1 (k + q -1 -β m-1 )r + q(l -1) = 1, (t m-1 -1) kα m-1 + β m-1 + 1 = 0.
+ β m-1 + 1 ≥ 1, it follows that t m-1 = 1. Since t m-1 = 1 (t m-1 ≤ l -1)
, we get l ≥ 2. In addition, (3.4.46) implies that α m-1 = 1, and

(k + q -1 -β m-1 )r + q(l -1) = 1. (3.4.47)
Since β m-1 ≤ k + q -1 in this case as stated above, r ≥ 1 and l ≥ 1, then (3.4.47) implies that Thus, the proof is complete.

r = 1 (q = 0), β m-1 = k + q -2 = k -2 (3.4.48) or l = 2, q = 1, β m-1 = k + q -1 = k. ( 3
Example 3.4.5. Consider the following numerical semigroups.

1. S =< 9, 10, 13 > . By using GAP [8], we get that S is pseudo-symmetric. Moreover, S verifies the formula in Theorem 3.4.4.

2. S =< 32, 33, 511 >. By using GAP [8], we get that S is pseudo-symmetric. Note that l = 1, k = 15 and r = 16 = k + 1. In addition, m = 32 = 2(15) + 2. Hence, S verifies the formula in Theorem 3.4.4.

3. S =< 15, 16, 17, 18, 19, 44 >. By using GAP [8], we get that S is pseudo-symmetric. Note that l = 4, k = 2 and r = 6 = (k -1)l + 2. In addition, m = 15 = (2(2) -1)4 + 3 = (2k -1)l + 3. Hence, S verifies the formula in Theorem 3.4.4.

4. S =< 17, 18, 19, 20, 25 >. By using GAP [8], we get that S is pseudo-symmetric. Note that l = 3, k = 1, r = 5 = l + 2. Moreover, m = 17 = 2(2(3) + 2) + 1 = 2(2l + 2) + 1. Hence, S verifies the formula in Theorem 3.4.4.

5. S =< 9, 10, 41 >, By using GAP [8], we get that S is pseudo-symmetric. Note that l = 1, r = 1 and k = 4 ≥ 2. Moreover, m = 9 = 2(4) + 1 = 2k + 1. Hence, S verifies the formula in Theorem 3.4.4.

6. S =< 3, 4, 5 >. By using GAP [8], we get that S is pseudo-symmetric. Moreover, S verifies the formula in Theorem 3.4.4. 9. S =< 9, 10, 11, 14 >. By using GAP [8], we get that S is pseudo-symmetric. Note that l = 2, k = 1 and r = 3. Moreover, m = 9 = 4(1) + 5 = 4k + 5. Hence, S verifies the formula in Theorem 3.4.4.

Pseudo-Frobenius Numbers

The aim of this Section, is to determine the set of pseudo-Frobenius Numbers of S. We are going to introduce some Lemmas that will help us in determining the set of pseudo-Frobenius Numbers of S.

Lemma 3.5.1. (see [START_REF][END_REF]) Let S be a numerical semigroup and n ∈ S * . Let Ap(S, n) = {w(i); w(i) ≡ i mod n, 0 ≤ i ≤ n -1} be the Apéry set of S with respect to n and let P F (S) be the set of pseudo-Frobenius numbers of S. Then, w(x) -n ∈ P F (S) if and only if w(x + y) + n ≤ w(x) + w(y) for all 1 ≤ y ≤ n -1 where x + y = x + y mod n.

Proof. Let w(x)-n ∈ P F (S). By definition of P F (S), we have w(x)-n+S * ⊆ S. Then, w(x)+w(y)-n ∈ S (as w(y) ∈ S * ). On the other hand, both w(x) + w(y) -n and w(x + y) are congruent to x + y mod n, then by definition of the elements of the Apéry set of S, we get

w(x + y) ≤ w(x) + w(y) -n, ∀ 1 ≤ y ≤ n -1.
Conversely, suppose that w(x + y) + n ≤ w(x) + w(y), ∀ 1 ≤ y ≤ n -1. By definition of the elements of the Apéry set of S, we have w(x) -n / ∈ S, then it is left to show that w(x) -n + S * ⊆ S to get that w(x)-n ∈ P F (S). Indeed, we have w(x)-n+w(y) ≥ w(x + y) for all 1 ≤ y ≤ n-1 and both w(x)-n+w(y) and w(x + y) are congruent to x + y mod n. Consequently, from the definition of the elements of the Apéry set of S, it follows thatw(

x) -n + w(y) ∈ S, ∀ 1 ≤ y ≤ n -1 which implies that w(x) -n + S * ⊆ S.
The later follows from the fact that for all s ∈ S, there exists (k, w) ∈ N × Ap(S, n) such that s = kn + w and that n ∈ S (Proposition 1.0.13). Hence, our assertion holds. Thus, the proof is complete.

By applying Lemma 3.5.1 on our numerical semigroup, we get Proposition 3.5.2. Proposition 3.5.2, mainly equation (3.5.1), will be used later in determining P F (S). Proposition 3.5.2. Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1. Then, for all 1 ≤ y ≤ m -1. Thus, the proof is complete.

w(x) -m ∈ P F (S) if and only if ∀ 1 ≤ y ≤ m -1, m(kα x+y + β x+y + x+y + 1) + x + y ≤ m(k(α x + α y ) + β x + β y + x + y ) + x + y. ( 3
In Lemma 3.5.3, we give cases where (3.5.1) does not hold. This will allow us to determine some elements that are not in P F (S).

(k + 1)l + 1. Let x = α x (kl + r) + β x l + x t x , y 1 = kl + r, y 2 = 1 and y 3 = l + 1 -x t x .
We have the following :

1. Suppose that x + y 1 ≤ m -1. For all r ∈ N, x does not satisfy (3.5.1) for y 1 .

2. Suppose that x + y 2 ≤ m -1 and x t x = 0. If one of the following conditions holds :

• r -1 = ql + t with t > 0;

• r -1 = ql with q > 0 and β x = k + q;

• r = 1 and β x = k, then x does not satisfy (3.5.1) for y 2 .

3. Suppose x + y 3 ≤ m -1. If one of the following conditions holds :

• r -1 = ql + t with t > 0 and β x = k + q ;

• r -1 = ql with q > 0, β x = k + q -1 and β x = k + q;

• r = 1, β x = k -1 and β x = k, then x does not satisfy (3.5.1) for y 3 .

Proof.

1. We have m(kα

y 1 + β y 1 + y 1 ) = m(k). Thus, m(k(α x + α y 1 ) + β x + β y 1 + x + y 1 ) + x + y 1 = m k(α x + 1) + β x + x + (α x + 1)(kl + r) + β x l + x t x . (3.5.2) Since x + y 1 ≤ m -1, it follows that x + y 1 = x + y 1 . For all r ∈ N, we have x + y 1 = (α x + 1)(kl + r) + β x l + x t x . Hence, m(kα x+y 1 + β x+y 1 + x+y 1 + 1) + x + y 1 = m k(α x + 1) + β x + x + 1 + (α x + 1)(kl + r) + β x l + x t x .
(3.5.3) By using (3.5.2) and (3.5.3), it follows that x does not satisfy (3.5.1) for y 1 .

2. If x = 0, then x = α x (kl + r) + β x l. We have m(kα

y 2 + β y 2 + y 2 ) = m. Therefore, m(k(α x + α y 2 ) + β x + β y 2 + x + y 2 ) + x + y 2 = m kα x + β x + 1 + α x (kl + r) + β x l + 1. (3.5.4) Since x + y 2 ≤ m -1, it follows that x + y 2 = x + y 2 .
If one of the following conditions holds :

In addition, if x = α m-1 (kl +r)+(β m-1 -1)l, then x = 0 and x+y 2 ≤ m-1. Since x does not satisfy (3.5.1) for y 2 if x + y 2 ≤ m -1 and x = 0, it follows that if x = α m-1 (kl + r) + (β m-1 -1)l, then w(x) -m / ∈ P F (S) (Lemma 3.5.3). By the same argument we have if x = (α m-1 -1)(kl + r) + (k + q)l + x t x such that x = 0, then w(x) -m / ∈ P F (S). By using (3.5.12) we deduce that

P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + (β m-1 -1)l + 1 ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x
with x = 1 and (k + q)l + x t x > β m-1 l}.

(3.5.13)

Case 1.3. If m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 , i.e., ( m-1 = 1). We have (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 + (kl + r) ≤ α m-1 (kl + r) + β m-1 l + m-1 t m-1 and (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 + 1 + (kl + r) > α m-1 (kl + r) + β m-1 l + m-1 t m-1 . Consequently, x + y 1 ≤ m -1 iff x ≤ (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 .
Since x does not satisfy (3.5.1) for

y 1 if x + y 1 ≤ m -1, we deduce that if x ≤ (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 , then w(x) -m / ∈ P F (S) (Lemma 3.5.3). In particular, P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 + 1 ≤ x ≤ m -1}. (3.5.14) Moreover, α m-1 (kl + r) + (β m-1 -1)l + l -1 + l + 1 -(l -1) ≤ α m-1 (kl + r) + β m-1 l + m-1 t m-1 and α m-1 (kl + r) + β m-1 l + l + 1 > α m-1 (kl + r) + β m-1 l + m-1 t m-1 . Consequently, x + y 3 ≤ m -1 iff x ≤ α m-1 (kl + r) + (β m-1 -1)l + l -1.
Since x does not satisfy (3.5.1) for y 3 if x + y 3 ≤ m -1 and β x = k + q (Lemma 3.5.3), by using (3.5.14) we deduce that if

x ≤ α m-1 (kl + r) + (β m-1 -1)l + l -1 and x = (α m-1 -1)(kl + r) + (k + q)l + x t x with (k + q)l + x t x > β m-1 l + m-1 t m-1 , then w(x) -m / ∈ P F (S). In particular, P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + β m-1 l ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x with (k + q)l + x t x > β m-1 l + m-1 t m-1 }. (3.5.15)
In addition, if x = α m-1 (kl + r) + β m-1 l, then x = 0 and x + y 2 ≤ m -1. Since x does not satisfy (3.5.1) for y 2 if x + y 2 ≤ m -1 and x = 0, we deduce that if x = α m-1 (kl + r) + β m-1 l, then w(x) -m / ∈ P F (S) (Lemma 3.5.3). By the same argument we have if x = (α m-1 -1)(kl + r) + (k + q)l + x t x such that x = 0, then w(x) -m / ∈ P F (S). By using (3.5.15), we deduce that

P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + β m-1 l + 1 ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x with x = 1 and (k + q)l + x t x > β m-1 l + m-1 t m-1 }. (3.5.16) Case 2. If r -1 = ql for some q ∈ N * . Case 2.1. If m -1 = α m-1 (kl + r), (i.e., β m-1 = m-1 = 0). We have (α m-1 -1)(kl + r) + (kl + r) ≤ α m-1 (kl + r) and (α m-1 -1)(kl + r) + 1 + (kl + r) > α m-1 (kl + r). Consequently, x + y 1 ≤ m -1 iff x ≤ (α m-1 -1)(kl + r).
Since x does not satisfy (3.5.1) for y 1 if x+y 1 ≤ m-1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 -1)(kl+r), then w(x) -m / ∈ P F (S). In particular,

P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + 1 ≤ x ≤ m -1}. (3.5.17) Moreover, (α m-1 -1)(kl + r) + (k + q -1)l + l -1 + l + 1 -(l -1) ≤ α m-1 (kl + r) and (α m-1 -1)(kl + r) + (k + q)l + l + 1 > α m-1 (kl + r).
Consequently,

x + y 3 ≤ m -1 iff x ≤ (α m-1 -1)(kl + r) + (k + q -1)l + l -1.
Since x does not satisfy (3.5.1) for y 3 in the case x+y 3 ≤ m-1, β x = k +q -1 and β x = k +q (Lemma 3.5.3), by using (3.5.17) we deduce that if x ≤ (α m-1 -1)(kl + r) + (k + q -2)l + l -1, then w(x) -m / ∈ P F (S). In particular,

P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + (k + q -1)l ≤ x ≤ m -1}.
(3.5.18)

In addition, if x = (α m-1 -1)(kl + r) + (k + q -1)l, then x = 0, β x = k + q and x + y 2 ≤ m -1. Since x does not satisfy (3.5.1) for y 2 in the case x + y 2 ≤ m -1, x = 0 and β x = k + q (Lemma 3.5.3), we deduce that if x = (α m-1 -1)(kl + r) + (k + q -1)l, then w(x) -m / ∈ P F (S). By using (3.5.18), we obtain

P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + (k + q -1)l + 1 ≤ x ≤ m -1}. (3.5.19) Case 2.2. If m -1 = α m-1 (kl + r) + β m-1 l (i.e., β m-1 > 0, m-1 = 0). We have (α m-1 -1)(kl + r) + β m-1 l + (kl + r) ≤ α m-1 (kl + r) + β m-1 l and (α m-1 -1)(kl + r) + β m-1 l + 1 + (kl + r) > α m-1 (kl + r) + β m-1 l. Consequently, x + y 1 ≤ m -1 iff x ≤ (α m-1 -1)(kl + r) + β m-1 l.
Since x does not satisfy (3.5.1) for y

1 if x + y 1 ≤ m -1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 - 1)(kl + r) + β m-1 l, then w(x) -m / ∈ P F (S). In particular, P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + β m-1 l + 1 ≤ x ≤ m -1}. (3.5.20) Moreover, α m-1 (kl + r) + (β m-1 -2)l + l -1 + l + 1 -(l -1) ≤ α m-1 (kl + r) + β m-1 l and α m-1 (kl + r) + (β m-1 -1)l + l + 1 > α m-1 (kl + r) + β m-1 l. Consequently, x + y 3 ≤ m -1 iff x ≤ α m-1 (kl + r) + (β m-1 -2)l + l -1.
Since x does not satsify (3.5.1) for y 3 in the case x + y 3 ≤ m -1, β x = k + q -1 and β x = k + q (Lemma 3.5.3), by using (3.5.20) we deduce that if x ≤ α m-1 (kl + r) + (β m-1 -2)l + l -1 such that x = (α m-1 -1)(kl+r)+(k +q -1)l+ x t x with (k +q -1)l+ x t x > β m-1 l and x = (α m-1 -1)(kl+r)+(k +q)l with (k + q)l > β m-1 l, then w(x) -m / ∈ P F (S). In particular,

P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + (β m-1 -1)l ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with (k + q -1)l + x t x > β m-1 l} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l}. (3.5.21) If x = α m-1 (kl + r) + (β m-1 -1)l, then x = 0, β m-1 -1 = k + q (as β m-1 ≤ k + q) and x + y 2 ≤ m -1.
Since x does not satsify (3.5.1) for y 2 in the case x + y 2 ≤ m -1, x = 0 and β x = k + q (Lemma 3.5.3), we deduce that if x = α m-1 (kl + r) + (β m-1 -1)l, then w(x) -m / ∈ P F (S). By the same argument we have if x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x such that x = 0, then w(x) -m / ∈ P F (S). By using (3.5.21), we get P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + (β m-1 -1)l + 1 ≤ x ≤ m -1}

∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with x = 1 and (k + q -1)l + x t x > β m-1 l} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l}. Consequently, x + y 3 ≤ m -1 iff x ≤ α m-1 (kl + r) + (β m-1 -1)l + l -1.

Since x does not satisfy (3.5.1) for y 3 in the case x + y 3 ≤ m -1, β x = k + q -1 and β x = k + q (Lemma 3.5.3), by using (3.5.23) we deduce that if x ≤ α m-1 (kl + r) + (β m-1 -1)l + l -1 such that x = (α m-1 -1)(kl+r)+(k+q)l with (k+q)l > β m-1 l+ m-1 t m-1 and x = (α m-1 -1)(kl+r)+(k+q-1)l+ x t x with (k + q -1)l + x t x > β m-1 l + m-1 t m-1 , then w(x) -m / ∈ P F (S). In particular, P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + β m-1 l ≤ x ≤ m -1}

∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l + m-1 t m-1 } ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with (k + q -1)l + x t x > β m-1 l + m-1 t m-1 }. Proof. Suppose by the way of contradiction that x does not satisfy (3.5.1) for y. Consequently, w(x + y) + m > w(x) + w(y).

We have x + y > m -1 and 1 ≤ x, y ≤ m -1, thus x + y = x + y + m, x + y < x and x + y < y. Since w(x)+w(y) and w(x + y) are both elements in S that are congruent to x+y mod m, then from the definition of the element of the Apéry set of S, it follows that w(x) + w(y) = w(x + y) + x 0 m for some x 0 ∈ N. On the other hand, w(x + y) + m > w(x) + w(y). Thus, x 0 = 0 and w(x + y) = w(x) + w(y).

(3.5.39)

As x ≥ 1 and y ≥ 1 , it follows that w(y) > 0 and w(x) > 0.Then, (3.5.39) implies that w(x + y) > w(x) with x + y < x and w(x + y) > w(y) with x + y < y. By Proposition 3.2.4, we have if i < j, then w(i) > w(j) if and only if it satisfies one of the following :

1. α i = α j -2, β j = j = 0 and β i + i = 2k + 1.

2. α i = α j -1, β i + i > k + β j + j and β j + j ≤ k.

In particular, if i < j such that w(i) > w(j), then α i ≤ α j - ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x with x = 1 and (k + q)l + x t x > β m-1 l}.

In addition, if x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x such that x = 1 and (k + q -1)l + x t x > β m-1 l + m-1 t m-1 or x = (α m-1 -1)(kl + r) + (k + q)l such that (k + q)l > β m-1 l + m-1 t m-1 , then x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. In fact, write y = α y (kl + r) + β y l + y t y . Since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. Since x + y ≤ m -1, it follows that y = β y l + y t y (as (k + q -1)l + x t x > β m-1 l + m-1 t m-1 or (k + q)l > β m-1 l + m-1 t m-1 ). Thus, m(kα y + β y + y ) = m(β y + y ). Since x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with x = 1 or x = (α m-1 -1)(kl + r) + (k + q)l, we get that m(kα x + β x + x ) = m(kα m-1 + q). Consequently, m(k(α x + α y ) + β x + β y + x + y ) = m(kα m-1 + q + β y + y ).

(3.5.70)

• If x = (α m-1 -1)(kl + r) + (k + q)l such that (k + q)l > β m-1 l + m-1 t m-1 . We have x + y = (α m-1 -1)(kl + r) + (k + q)l + 1 + β y l + y t y -1 = α m-1 (kl + r) + β y l + y t y -1 with β y l + y t y -1 ≥ 0 as y = β y l + y t y ≥ 1. Then, • If x = (α m-1 -1)(kl+r)+(k+q-1)l+ x t x such that x = 1 and (k+q-1)l+ x t x > β m-1 l+ m-1 t m-1 .

• If β y ≥ 1. We have

x + y = (α m-1 -1)(kl + r) + (k + q -1)l + x t x + β y l + y t y = α m-1 (kl + r) + (β y -1)l + y t y + x t x -1.

Since 1 ≤ x t x ≤ l -1 and y t y ≤ l -1, it follows that 0 ≤ y t y + x t x -1 ≤ 2l -2. Hence, m(kα x+y + β x+y + x+y ) ≤ m(kα m-1 + β y + y ).

• If β y = 0. Then, y = y t y with y = 1 (as y ≥ 1). Since 1 ≤ x t x ≤ l -1 and 1 ≤ y t y ≤ l -1, it follows that 2 ≤ y t y + x t x ≤ 2l -2. If 2 ≤ y t y + x t x ≤ l, then x + y = (α m-1 -1)(kl + r) + (k + q -1)l + x t x + y t y with x t x + y t y ≤ l and if y t y + x t x ≥ l + 1, then x + y = α m-1 (kl + r) + ( x t x + y t y -(l + 1)) with 0 ≤ x t x + y t y -(l + 1) ≤ l -3. Since q ∈ N * , then m(kα x+y + β x+y + x+y ) ≤ m(kα m-1 + q -1 + 1).

Since q ∈ N * and β y + y ≥ 1 (as y ≥ 1), it follows that m(kα x+y + β x+y + x+y ) ≤ m(kα m-1 + β y + q -1 + y ). Hence, if x = (α m-1 -1)(kl+r)+(k+q-1)l+ x t x such that x = 1 and (k+q-1)l+ x t x > β m-1 l+ m-1 t m-1 , or x = (α m-1 -1)(kl + r) + (k + q)l such that (k + q)l > β m-1 l + m-1 t m-1 , then x satisfies (3.5.1) for 1 ≤ y ≤ m -1. Therefore, {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with x = 1 and (k + q -1)l + x t x > β m-1 l + m-1 t m-1 } ∪{w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l} with (k + q)l > β m-1 l + m-1 t m-1 } ⊆ P F (S). ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with x = 1 and (k + q -1)l + x t x > β m-1 l + m-1 t m-1 } ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l + m-1 t m-1 }.

  wx m (y-x-1)+(y-2)+ wy m (x-1)

Remark 1 . 0 . 2 .Example 1 . 0 . 3 .

 102103 All semigroups considered in this thesis are submonoids of (N, +), hence commutative, that is, a + b = b + a for all a, b ∈ S. Consider the following examples :

η

  j = {k ∈ N; n k = j} . Let 0 ≤ k ≤ q -2. If s ∈ S ∩ I k then s + m ∈ S ∩ I k+1 . This implies that n k ≤ n k+1 .Let for example S = 4, 6, 13 . We have c(S) = c = 16 = 4 • 4, hence I k = [km, (k + 1)m[ for all k ≥ 0. Moreover, n 0 = 1, n 1 = 2, n 2 = 2, n 3 = 3, and n k = 4 for all k ≥ 4. We also have η 1 = 1, η 2 = 2, η 3 = 1.

v)<

  By definition, we have w i < w j for all 0 ≤ i < j ≤ m -1. Thus, w i +ρ m By Lemma 1.0.21, we have f = max(Ap(S, m)) -m = w m-1 -m. Hence, w m-1 +ρ m

νν

  ≥ m and w m-1 -m ≥ w x + w y In this Section, we will show that if S is a numerical Semigroup such that w m-1 -m ≥ w x + w y and 2 + wx m (y -x -1) + (y -2) + wy m (x -1) ≥ m, then S satisfies Wilf's conjecture. Theorem 2.5.1. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Let w 0 = 0 < w 1 < . . . < w m-1 be the elements of Ap(S, m). Suppose that w m-1 -m ≥ w x + w y for some 0 < x < y < m -1. If 2 + wx m (y -x -1) + (y -2) + wy m (x -1) wx m + wy m + 2 ν ≥ m, then S satisfies Wilf's conjecture.

  then (2.5.5) gives w

Case 4 .

 4 If wx+ρ m = wx m and wy+ρ m = wy m , then (2.5.5) gives w

≥

  then S satisfies Wilf's conjecture. (2 + α)ν ≥ m.

Example 2. 5 . 4 . 6 ≥

 546 Consider the following numerical semigroup S =< 19, 21, 23, 25, 27, 28 > . Note that 2ν < m. We have w 4 = 27, w 5 = 28 and w m-1 -m = 64 i.e., w m-1 -m ≥ w 4 + w 5 . Moreover, 2 + wx m (y -x -1) + (y -2) + wy m (x -1) 19 = m.

Theorem 3 . 1 . 1 .

 311 Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1.

.1. 4 )

 4 From (3.1.2) and (3.1.4), it follows that

.1. 6 )

 6 Using (3.1.6) and (3.1.5), it follows that

Example 3 . 1 . 2 .

 312 Consider the following numerical semigroup S =< 19, 20, 21, 22, 52 > . Note that m = 19, l = 3, k = 2 and r = 8. Let Ap(S, m) = {0, w(1), . . . , w(m -1)} be the Apéry basis of S. By using GAP [8], we obtain Ap(S, m) = {0, 20, 21, 22, 42, 43, 44, 64, 65, 66, 86, 87, 88, 108, 52, 72, 73, 74, 94} and they verify the formula given in Theorem 3.1.1.

Proposition 3 . 2 . 4 .

 324 Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1.

Theorem 3 . 2 . 5 .

 325 Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1.

Example 3 . 3 . 2 .

 332 Consider the examples in Example 3.2.6.

Theorem 3 . 4 . 2 .

 342 Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1.

  .4.28) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 . By substituting (3.4.28) in S, we get S =< 9, 10, 13 > .

  .4.33) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 . By substituting (3.4.33) in S, we get S =< (2k -1)l + 3, . . . , 2kl + 3, k(2kl + 3) + (k -1)l + 2 > with l ≥ 2.

  .4.35) By using Lemma 3.4.1, S is pseudo-symmetric if and only if

  .4.43) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 . By substituting (3.4.43) in S, we obtain S =< 2kl + 5, . . . , (2k + 1)l + 5, k((2k + 1)l + 5) + 3 > with l ≥ 3.

( 3 .

 3 4.46) From(3.4.45), as kα m-1 + β m-1 + 1 ≥ 1, it follows that α m-1 = β m-1 = 0 and t m-1 = 2. In this case, we have m = 3. As ν ≤ m = 3, it follows that l = 1 which contradicts t m-1 = 2 ≤ l -1. Thus, we do not have case(3.4.45). Now, from(3.4.46), as kα m-1

  .4.49) Since r = ql + 1 in this case, from α m-1 = 1, t m-1 = 1, l ≥ 2, (3.4.48) and (3.4.49), it follows thatα m-1 = 1, β m-1 = k -2, t m-1 = 1, r = 1, l ≥ 2 (3.4.50) or α m-1 = 1, β m-1 = k, t m-1 = 1, l = 2, r = 3. (3.4.51) We have m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 . By (3.4.50) and (3.4.51), we get S =< (2k -2)l + 3, . . . , (2k -1)l + 3, k((2k -1)l + 3) + 1 > with l ≥ 2 or S =< 4k + 5, 4k + 6, 4k + 7, k(4k + 7) + 3 > .

7 .

 7 S =< 45, 46, 47, 48, 49, 50, 203 >. By using GAP [8], we get that S is pseudo-symmetric. Note that l = 5 ≥ 2, r = 3 and k = 4. Moreover, m = 45 = 2(4(5)) + 5 = 2kl + 5. Hence, S verifies the formula in Theorem 3.4.4. 8. S =< 19, 20, 21, 22, 23, 70 >. By using GAP [8], we get that S is pseudo-symmetric. Note that l = 4 ≥ 2, r = 1 and k = 3. Moreover, m = 19 = (2(3) -2)4 + 3 = (2k -2)l + 3. Hence, S verifies the formula in Theorem 3.4.4.

Lemma 3 . 5 . 3 .

 353 Let S be a numerical semigroup minimally generated by m, . . . , m + l, k(m + l) + r with r ≤

( 3 . 5 . 22 ) 2 . 3 .

 352223 Case If m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 , i.e., ( m-1 = 1). We have(α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 + (kl + r) ≤ α m-1 (kl + r) + β m-1 l + m-1 t m-1 and (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 + 1 + (kl + r) > α m-1 (kl + r) + β m-1 l + m-1 t m-1 . Consequently, x + y 1 ≤ m -1 iff x ≤ (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 .Since x does not satisfy (3.5.1) for y 1 if x + y 1 ≤ m -1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 , then w(x) -m / ∈ P F (S). In particular,P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + β m-1 l + m-1 t m-1 + 1 ≤ x ≤ m -1}. (3.5.23) Moreover, α m-1 (kl + r) + (β m-1 -1)l + l -1 + l + 1 -(l -1) ≤ α m-1 (kl + r) + β m-1 l + m-1 t m-1 and α m-1 (kl + r) + β m-1 l + l + 1 > α m-1 (kl + r) + β m-1 l + m-1 t m-1 .

( 3 .Case 3 . 3 Lemma 3 . 5 . 6 .

 333356 5.24) Hence, x does not satisfy (3.5.1) for y if l ≥ 2. Consequently, if x = (α m-1 -1)(kl + 1) + kl and l ≥ 2, then w(x) -m / ∈ P F (S). By using (3.5.32), we getP F (S) ⊆ {w(x) -m; α m-1 (kl + 1) + (β m-1 -1)l + 1 ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + 1) + (k -1)l + 1 with (k -1)l + 1 > β m-1 l} ∪ {w(x) -m; x = (α m-1 -1)(kl + 1) + kl with l = 1 and kl > β m-1 l}. If m -1 = α m-1 (kl + 1) + β m-1 l + m-1 t m-1 , i.e., ( m-1 = 1). We have (α m-1 -1)(kl + 1) + β m-1 l + m-1 t m-1 + (kl + 1) ≤ α m-1 (kl + 1) + β m-1 l + m-1 t m-1and(α m-1 -1)(kl + 1) + β m-1 l + m-1 t m-1 + 1 + (kl + 1) > α m-1 (kl + 1) + β m-1 l + m-1 t m-1 .Consequently,x + y 1 ≤ m -1 iff x ≤ (α m-1 -1)(kl + 1) + β m-1 l + m-1 t m-1 .Since x does not satisfy (3.5.1) fory 1 if x + y 1 ≤ m -1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 -1)(kl + 1) + β m-1 l + m-1 t m-1 , then w(x) -m / ∈ P F (S). In particular, P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + 1) + β m-1 l + m-1 t m-1 + 1 ≤ x ≤ m -1}. (3.5.34) Moreover, α m-1 (kl + 1) + (β m-1 -1)l + l -1 + l + 1 -(l -1) ≤ α m-1 (kl + 1) + β m-1 l + m-1 t m-1 and α m-1 (kl + 1) + β m-1 l + l + 1 > α m-1 (kl + 1) + β m-1 l + m-1 t m-1 .Consequently,x + y 3 ≤ m -1 iff x ≤ α m-1 (kl + 1) + (β m-1 -1)l + l -1.Since x does not satisfy (3.5.1) for y 3 in the case x+y 3 ≤ m-1, β x = k-1 and β x = k (Lemma 3.5.3), by using (3.5.34) we deduce that if x ≤ α m-1 (kl+1)+(β m-1 -1)l+l-1 such that x = (α m-1 -1)(kl+1)+(k-1)l+ x t x with (k -1)l + x t x > β m-1 l + m-1 t m-1 and x = (α m-1 -1)(kl + 1) + kl with kl > β m-1 l + m-1 t m-1 , then w(x) -m / ∈ P F (S). In particular,P F (S) ⊆ {w(x) -m; α m-1 (kl + 1) + β m-1 l ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + 1) + (k -1)l + x t x with (k -1)l + x t x > β m-1 l + m-1 t m-1 } ∪ {w(x) -m; x = (α m-1 -1)(kl + 1) + kl with kl > β m-1 l + m-1 t m-1 }. (3.5.35) If x = α m-1 (kl + 1) + β m-1 l, then x = 0, x + y 2 ≤ m -1 and β x = β m-1 = k (if β m-1 = k, then m -1 = α m-1 (kl +1)+kl + m-1 t m-1, as m-1 = 1 and r = 1, we get a contradiction). Since x does not satisfy (3.5.1) for y 2 in the case x+y 2 ≤ m-1, x = 0 and β x = k (Lemma 3.5.3), we deduce that if x = α m-1 (kl+1)+β m-1 l, then w(x) -m / ∈ P F (S). By the same argument we have if x = (α m-1 -1)(kl + 1) + (k -1)l + x t x such that x = 0, then w(x) -m / ∈ P F (S). By using (3.5.35) we deduce thatP F (S) ⊆ {w(x) -m; α m-1 (kl + 1) + β m-1 l + 1 ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + 1) + (k -1)l + x t x with x = 1 and (k -1)l + x t x > β m-1 l + m-1 t m-1 } ∪ {w(x) -m; x = (α m-1 -1)(kl + 1) + kl with kl > β m-1 l + m-1 t m-1 }.(3.5.36) Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1. Let 1 ≤ x, y ≤ m -1 such that x + y > m -1 and x = α x (kl + r) + β x l + x t x with β x + x > 0. Then, x satisfies (3.5.1) for y.

1 .Theorem 3 . 5 . 7 .Case 1 . 0 . 1 . 1 .

 13571011 Write x + y = α x+y (kl + r) + β x+y l + x+y t x+y and y = α y (kl + r) + β y l + y t y . Since w(x + y) > w(x) with x + y < x and w(x + y) > w(y) with x + y < y, it follows that α x+y ≤ α x -1 and α x+y ≤ α y -1. By proposition 3.0.2, we have β x+y + x+y ≤ 2k + 1. Hence, w(x + y) = m(kα x+y + β x+y + x+y ) + x + y ≤ m(kα x+y + 2k + 1) + x + y = m(kα x+y + 2k) + x + y. (3.5.40) On the other hand, w(y) = m(kα y + β y + y ) + y ≥ m(k(α x+y + 1)) + y. By using β x + x > 0 (hypothesis) and α x ≥ α x+y + 1, we get w(x) = m(kα x + β x + x ) + x ≥ m(k(α x+y + 1) + 1) + x. Consequently, w(x) + w(y) ≥ m(kα x+y + 2k + 1) + x + y. (3.5.41) But (3.5.41) and (3.5.40) contradicts (3.5.39). Therefore, x satisfies (3.5.1) for y. Now, we are ready to determine the set of pseudo-Frobenius Numbers of S. Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1. For all 1 ≤ x ≤ m -1, write x = α x (kl + r) + β x l + x t x as in Definition 3.0.1 and w(x) = m(kα x + β x + x ) + x as in Theorem 3.1.1. We have the following : If r -1 = ql + t for some q, t ∈ N with t < l and t = Case If m -1 = α m-1 (kl + r) (i.e., β m-1 = m-1 = 0), then P F (S) = {w(x) -m; (α m-1 -1)(kl + r) + (k + q)l + 1 ≤ x ≤ m -1}. Case 1.2. If m -1 = α m-1 (kl + r) + β m-1 l (i.e., β m-1 > 0, m-1 = 0), then P F (S) = {w(x) -m; α m-1 (kl + r) + (β m-1 -1)l + 1 ≤ x ≤ m -1}

  m(kα x+y + β x+y + x+y ) ≤ m(kα m-1 + β y + y ).(3.5.71) By using (3.5.70), (3.5.71) and q ∈ N * , we get that x satisfies (3.5.1) for all 1 ≤ y ≤ m -1.

  (3.5.72) By using (3.5.70) and (3.5.72), we get that x satisfies (3.5.1) for 1 ≤ y ≤ m -1.

( 3 .

 3 5.73) By using (3.5.25), (3.5.73) and (3.5.69), we getP F (S) = {w(x) -m; α m-1 (kl + r) + β m-1 l + 1 ≤ x ≤ m -1}

Table des matières

 des Equivalent form of Wilf's conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

	Remerciements	iii
	Introduction
	1 Basics and notations
	2 Wilf 's conjecture
	2.1 wx m +	wy m +2

2.2 Technical results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Numerical semigroups with w m-1 ≥ w 1 + w α and (2 + α-3 q )ν ≥ m . . . . . . . . . 2.4 Numerical semigroups with w m-1 ≥ w α-1 + w α and ( α+3 3 )ν ≥ m . . . . . . . . . . 2.5 Numerical semigroups with 2 + wx m (y-x-1)+(y-2)+ wy m (x-1)

  by hypothesis).

	• If x ≥ 0. By Remark 2.2.1 (ii), we have w 1 +ρ m	≥ 1. From (2.3.1), it follows that
	m-1
	j=1

  by hypothesis).

	Using Proposition 2.1.3, we get that S satisfies Wilf's conjecture. Thus, the proof is complete.
	Example 2.3.2. Consider the following numerical semigroup
	S = < 19, 21, 23, 25, 27, 28 > .

  ≥ 22 = m, thus the conditions of Theorem 2.4.1 are valid. The following Corollary 2.4.3 is an extension for Corollary 2.3.4 using Theorems 2.3.1 and 2.4.1. Let S be a numerical semigroup with multiplicity m and embedding dimension ν. Suppose that

	( 7+3 3 )7 Corollary 2.4.3.	α+3 3 )ν =

by hypothesis).

Using Proposition 2.1.3, we get that S satisfies Wilf's conjecture. Thus, the proof is complete. Example 2.4.2. Consider the following numerical semigroup S =< 22, 23, 25, 27, 29, 31, 33 > .

Note that 3ν < m. We have w 6 = 33, w 7 = 46 and w m-1 = 87 i.e., w m-1 ≥ w 6 + w 7 . Moreover, (

  By substituting (3.4.14) and (3.4.15) in S, we get that S is symmetric if and only if

  .4.31) Since t ≥ 1, it follows that l ≥ 2. As α m-1 ≥ 1, t ≥ 1, l ≥ 1 and q ≤ k (by Proposition 3.2.3 as t = 1), then (3.4.31) implies that

  By applying Theorem 3.1.1, we get w(x) -m ∈ P F (S) if and only if m(kα x+y + β x+y + x+y + 1) + x + y ≤ m(k(α x + α y ) + β x + β y + x + y ) + x + y

.5.1) Proof. By Lemma 3.5.1, we have w(x) -m ∈ P F (S) if and only if w(x + y) + m ≤ w(x) + w(y), ∀ 1 ≤ y ≤ m -1 where x + y = x + y mod m.

for some 1 < α < m-1. If (2 + α-3q )ν ≥ m, then S satisfies Wilf's conjecture.

, then S satisfies Wilf's conjecture.

• r -1 = ql + t with t > 0;

• r -1 = ql with q > 0 and β x = k + q;

• r = 1 and β x = k, then x + y 2 = α x (kl + r) + β x l + 1. We have m(kα x+y 2 + β x+y 2 + x+y 2 + 1) + x + y 2 = m kα x + β x + 2 + α x (kl + r) + β x l + 1.

(3.5.5) By using (3.5.4) and (3.5.5), it follows that x does not satisfy (3.5.1) for y 2 .

3. We have m(kα y 3 + β y 3 + y 3 ) = m(2) if x = 0 and m(kα y 3 + β y 3 + y 3 ) = m if x = 1. Therefore, m(k(α x + α y 3 ) + β x + β y 3 + x + y 3 ) + x + y 3 = m(kα x + β x + 2) + α x (kl + r) + (β x + 1)l + 1.

(3.5.6) Since x + y 3 ≤ m -1, it follows that x + y 3 = x + y 3 . If one of the following conditions holds :

• r -1 = ql + t with t > 0 and β x = k + q ;

• r -1 = ql with q > 0, β x = k + q -1 and β x = k + q;

• r = 1, β x = k -1 and β x = k, then x + y 3 = α x (kl + r) + (β x + 1)l + 1. We have m(kα x+y 3 + β x+y 3 + x+y 3 + 1) + x + y 3 = m kα x + β x + 3 + α x (kl + r) + (β x + 1)l + 1.

(3.5.7) By using (3.5.6) and (3.5.7), it follows that x does not satisfy (3.5.1) for y 3 .

Thus, the proof is complete. Theorem 3.5.4 will determine the elements that do not belong to P F (S). Theorem 3.5.4. Let S be a numerical semigroup minimally generated by m, m + 1, . . . , m + l, k(m + l) + r with r ≤ (k + 1)l + 1. For all 1 ≤ x ≤ m -1, write x = α x (kl + r) + β x l + x t x as in Definition 3.0.1 and w(x) = m(kα x + β x + x ) + x as in Theorem 3.1.1. We have the following : Case 1. If r -1 = ql + t for some q, t ∈ N with t < l and t = 0.

Case 1.1. If m -1 = α m-1 (kl + r) (i.e., β m-1 = m-1 = 0), then

∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x with x = 1 and (k + q)l + x t x > β m-1 l}.

Case 1.3. If m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 (i.e., m-1 = 1), then P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + β m-1 l + 1 ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x with x = 1 and (k + q)l + x t x > β m-1 l + m-1 t m-1 }.

Case 2. If r -1 = ql for some q ∈ N * .

Case 2.1. If m -1 = α m-1 (kl + r) (i.e., β m-1 = m-1 = 0), then P F (S) ⊆ {w(x) -m; (α m-1 -1)(kl + r) + (k + q -1)l + 1 ≤ x ≤ m -1}.

Case 2.2. If m -1 = α m-1 (kl + r) + β m-1 l (i.e., β m-1 > 0, m-1 = 0), then P F (S) ⊆ {w(x) -m; α m-1 (kl + r) + (β m-1 -1)l + 1 ≤ x ≤ m -1} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with x = 1 and (k + q -1)l + x t x > β m-1 l} ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l}.

Case 2.3. If m -1 = α m-1 (kl + r) + β m-1 l + m-1 t m-1 (i.e., m-1 = 1), then

∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x with x = 1 and (k + q -1)l + x t x > β m-1 l + m-1 t m-1 } ∪ {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l + m-1 t m-1 }.

Proof. Case 1. If r -1 = ql + t for some q, t ∈ N with t < l and t = 0.

Since x does not satisfy (3.5.1) for y 1 if x+y 1 ≤ m-1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 -1)(kl+r), then w(x) -m / ∈ P F (S). In particular,

Consequently,

Since x does not satisfy (3.5.1) for y 3 if x + y 3 ≤ m -1 and β x = k + q (Lemma 3.5.3), by using (3.5.8), we get that if x ≤ (α m-1 -1)(kl + r) + (k + q -1)l + l -1, then w(x) -m / ∈ P F (S). In particular,

In addition, if x = (α m-1 -1)(kl + r) + (k + q)l, then x = 0 and x + y 2 ≤ m -1. Since x does not satisfy (3.5.1) for y 2 if x + y 2 ≤ m -1 and x = 0, we deduce that if x = (α m-1 -1)(kl + r) + (k + q)l, then w(x) -m / ∈ P F (S) (Lemma 3.5.3). By using (3.5.9), we obtain

Since x does not satisfy (3.5.1) for

3), then w(x) -m / ∈ P F (S). In particular,

Since x does not satisfy (3.5.1) for y 3 if x + y 3 ≤ m -1 and β x = k + q (Lemma 3.5.3), by using (3.5.11) we

(3.5.12)

as m-1 = 1 and r = ql + 1, we get a contradiction) and x + y 2 ≤ m -1. Since x does not satisfy (3.5.1) for y 2 in the case x + y 2 ≤ m -1, x = 0 and β x = k + q (Lemma 3.5.3), we deduce that if x = α m-1 (kl + r) + β m-1 l, then w(x) -m / ∈ P F (S). By the same argument we have if x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x such that x = 0, then w(x) -m / ∈ P F (S). By using (3.5.24), we obtain

(3.5.25)

and (α m-1 -1)(kl + 1) + 1 + (kl + 1) > α m-1 (kl + 1).

Consequently,

Since x does not satisfy (3.5.1) for y 1 if x+y 1 ≤ m-1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 -1)(kl+1), then w(x) -m / ∈ P F (S). In particular,

and (α m-1 -1)(kl + 1) + kl + l + 1 > α m-1 (kl + 1).

Consequently,

Since x does not satisfy (3.5.1) for y 3 in the case x + y 3 ≤ m -1, β x = k -1 and β x = k (Lemma 3.5.3), by using (3.5.26) we deduce that if x ≤ (α m-1 -1)(kl + 1) + (k -2)l + l -1, then w(x) -m / ∈ P F (S). In particular,

Since x does not satisfy (3.5.1) for y 2 in the case x + y 2 ≤ m -1, x = 0 and β x = k (Lemma 3.5.3), we deduce that if x = (α m-1 -1)(kl + 1) + (k -1)l, then w(x) -m / ∈ P F (S). Hence,

Since x does not satisfy (3.5.1) for y 1 if x + y 1 ≤ m -1 (Lemma 3.5.3), we deduce that if x ≤ (α m-1 -1)(kl + 1) + β m-1 l, then w(x) -m / ∈ P F (S). In particular,

Since x does not satisfy (3.5.1) for y 3 in the case x+y 3 ≤ m-1, β x = k-1 and β x = k (Lemma 3.5.3), by using (3.5.29) we deduce that if

. By using (3.5.30), we get

If we take y = l, then x does not satisfy (3.5.1) for y . Indeed, we have

. By using (3.5.31), we get

with kl > β m-1 l}.

(3.5.32)

Next, let x = (α m-1 -1)(kl + 1) + kl. Suppose that l ≥ 2. If we take y = 2, then x does not satisfy (3.5.1) for y . Indeed, x + y = α m-1 (kl + 1) + 1. We have x + y ≤ m -1 which gives x + y = x + y . In addition, m(kα

If we take y = l, then x does not satisfy (3.5.1) for y . Indeed, we have x + y = α m-1 (kl + 1) + ( x t x -1) with x t x -1 ≥ 1. We have x + y ≤ m -1 which gives x + y = x + y . In addition,

On the other hand, m(k(α x + α y ) +

. By using (3.5.36), we get

(3.5.37)

Next, let x = (α m-1 -1)(kl + 1) + kl. Suppose that l ≥ 2. If we take y = 2, then x does not satisfy (3.5.1) for y . Indeed, x + y = α m-1 (kl + 1) + 1. We have x + y ≤ m -1 which gives x + y = x + y . In addition,

On the other hand,

Hence, x does not satisfy (3.5.1) for y . Consequently, if x = (α m-1 -1)(kl + 1) + kl and l ≥ 2, then w(x) -m / ∈ P F (S). By using (3.5.37), we get

Thus, the proof is complete.

Lemmas 3.5.5 and 3.5.6 give cases where (3.5.1) holds. This will allow us to determine later some numbers that belong to P F (S).

Lemma 3.5.5. Let S be a numerical semigroup minimally generated by m, m

On the other hand, m(kα

Therefore,

Hence, x = m -1 satisfies (3.5.1) for all 1 ≤ y ≤ m -1. Thus, the proof is complete.

Proof. Case 1. If r -1 = ql + t for some q, t ∈ N with t < l and t = 0.

Case 1.1. If m-1 = α m-1 (kl+r) (i.e., β m-1 = m-1 = 0). We claim that if (α m-1 -1)(kl+r)+(k+q)l+1 ≤ x ≤ m -1, then x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. In fact,

• If x = m -1, then by using Lemma 3.5.5 x satisfies (3.5.1) for all 1 ≤ y ≤ m -1.

then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular (α m-1 -1)(kl + r)

and m(kα

In addition, we have m(kα Therefore,

By using (3.5.10) and (3.5.44), we get

In fact, write y = α y (kl + r) + β y l + y t y . Since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. Since x + y ≤ m -1 and (k + q)l + x t x > β m-1 l, it follows that y = β y l + y t y . Thus, m(kα

.5.48)

We have 1 ≤ x t x ≤ t (as By using (3.5.48) and (3.5.49), we get that x satisfies (3.5.1) for 1 ≤ y ≤ m -1. Therefore, {w(x) -m; x = (α m-1 -1)(kl + r) + (k + q)l + x t x with x = 1 and (k + q)l + x t x > β m-1 l} ⊆ P F (S).

(3.5.50) By using (3.5.13), (3.5.50) and (3.5.47), we obtain

x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. Indeed, since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular,

(3.5.52) By using (3.5.52) and (3.5.51), we get that x satisfies (3.5.1) for all y. Consequently,

Furthermore, if x = (α m-1 -1)(kl+r)+(k+q)l+ x t x such that x = 1 and (k+q)l+ x t x > β m-1 l+ m-1 t m-1 , then x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. In fact, write y = α y (kl + r) + β y l + y t y . Since β x + x > 0, then from Lemma 3.5.6 we may assume that x+y ≤ m-1 and this implies that x + y = x+y. Since x+y ≤ m-1 and (k + q)l + x t x > β m-1 l + m-1 t m-1 , it follows that y = β y l + y t y . Thus, m(kα y + β y + y ) = m(β y + y ). Since x = (α m-1 -1)(kl+r)+(k+q)l+ x t x such that x = 1, we get that m(kα

(3.5.56) By using (3.5.16), (3.5.56) and (3.5.53), we get

In fact,

• If x = m -1, then by using Lemma 3.5.5, it follows that x satisfies (3.5.1) for all 1 ≤ y ≤ m -1.

• If (α m-1 -1)(kl + r) + (k + q -1)l + 1 ≤ x ≤ m -2 = (α m-1 -1)(kl + r) + (k + q)l. Indeed, since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular, Therefore,

) for all 1 ≤ y ≤ m -1. Indeed, since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular,

In fact, write y = α y (kl + r) + β y l + y t y . Since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that

By using (3.5.63), (3.5.64) and q ∈ N * , we get that x satisfies (3.5.1) for all 1 ≤ y ≤ m -1.

• If x = (α m-1 -1)(kl + r) + (k + q -1)l + x t x such that x = 1 and (k + q -1)l + x t x > β m-1 l.

• If β y ≥ 1. We have

• If β y = 0. Then, y = y t y with y = 1 (as y ≥ 1). Since 1

Since q ∈ N * and β y + y ≥ 1 (as y ≥ 1), it follows that Hence

x = (α m-1 -1)(kl + r) + (k + q)l with (k + q)l > β m-1 l} ⊆ P F (S).

(3.5.66) By using (3.5.22), (3.5.66) and (3.5.62), we get

) for all 1 ≤ y ≤ m -1. Indeed, since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular,

By using (3.5.68) and (3.5.67), we get that x satisfies (3.5.1) for all y. Consequently,

x ≤ m -1, then x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. In fact,

• If x = m -1, then by using Lemma 3.5.5 x satisfies (3.5.1) for 1 ≤ y ≤ m -1.

• If (α m-1 -1)(kl + 1) + (k -1)l + 1 ≤ x ≤ m -2 = (α m-1 -1)(kl + 1) + kl. Indeed, since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular, (α m-1 -1)(kl • If x = (α m-1 -1)(kl + 1) + (k -1)l + 1 such that (k -1)l + 1 > β m-1 l.

• If β y ≥ 1. We have • If β y = 0. Then, y = y t y with y = 1 (as y ≥ 1). We have x + y = (α m-1 -1)(kl + 1) + (k -1)l + 1 + y t y with 1 + y t y ≤ l (as y t y < l). Hence,

x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. In fact, since β x + x > 0, then from Lemma 3.5.6 we may assume that x + y ≤ m -1 and this implies that x + y = x + y. In particular,

, then x satisfies (3.5.1) for all 1 ≤ y ≤ m -1. In fact, write y = α y (kl + 1) + β y l + y t y . Since β x + x > 0, then from Lemma 3.5.6 we may assume that x+y ≤ m-1 and this implies that x + y = x+y. Since x+y ≤ m-1, it follows that y = β y l+ y t y (as (k - • Thus, the proof is complete. Note that m = 12, k = 2, l = 3 and r = 5 (α m-1 = 1, β m-1 = 0, m-1 t m-1 = 0, q = 1, t = 1). S verifies the formula in Theorem 3.5.7.

2. S =< 18, 19, 20, 21, 47 >. By using GAP [8], we get that P F (S) = {64, 69, 70, 71} = {w(10) -18, w(15) -18, w(16) -18, w(17) -18}.

Note that m = 18, k = 2, l = 3 and r = 5 (α m-1 = 1, β m-1 = 2, m-1 t m-1 = 0, q = 1, t = 1). S verifies the formula in Theorem 3.5.7.

Thèse de Doctorat

Mariam DHAYNI

Abstract

The thesis is made up of two parts. We study in the first part Wilf's conjecture for numerical semigroups.

We give an equivalent form of Wilf's conjecture in terms of the Apéry set, embedding dimension and multiplicity of a numerical semigroup. We also give an affirmative answer for the conjecture in certain cases.

In the second part, we consider a class of almost arithmetic numerical semigroups and give for this class of semigroups explicit formulas for the Apéry set, the Frobenius number, the genus and the pseudo-Frobenius numbers. We also characterize the symmetric (resp. pseudo-symmetric) numerical semigroups for this class of numerical semigroups.

Mots clés

Semigroupes numériques, conjecture de Wilf, semi-groupes presque arithmétiques, nombre de Frobenius, Problème des pièces de monnaie.

Key Words

Numerical semigroups, Wilf's conjecture, almost arithmetic numerical semigroups Frobenius number, money changing problem.