Numerical semigroups with large embedding dimension satisfy Wilf's conjecture
aa r X i v : . [ m a t h . A C ] D ec Numerical semigroups with large embeddingdimension satisfy Wilf ’s conjecture
Alessio Sammartano ∗ Abstract
We give an affirmative answer to Wilf’s conjecture for numerical semi-groups satisfying 2 ν ≥ m , where ν and m are respectively the embed-ding dimension and the multiplicity of a semigroup. The conjectureis also proved when m ≤ Introduction
A classical problem in additive number theory is the
Diophantine FrobeniusProblem , also known as money-changing problem: given ν coprime positiveintegers g , . . . , g ν determine the largest integer f which is not representableas a linear combination of g , . . . , g ν with coefficients in N . The problem,introduced by Sylvester in [9] for the case ν = 2, has been widely studied inliterature; the monograph [7] gathers a lot of results on the topic.It is natural to study the problem in the context of numerical semigroups,i.e. submonoids of the additive monoid of the natural numbers. It is indeedpossible to provide formulas linking the Frobenius number of a semigroup ∗ email [email protected] ( S ) to its other invariants. With regards to this problem, in 1978 Wilfposed in [10] the following question: Question 1.
Let S be a numerical semigroup with Frobenius number f ( S ) ,embedding dimension ν ( S ) and let n ( S ) = | S ∩ [0 , f ] | . Is it true that f ( S ) +1 ≤ n ( S ) ν ( S ) ? The conjecture is still open, although an affirmative answer has been givenfor various partial cases (see [1], [4] and [5]). Moreover, some computationsmade in [2] strengthen our convinction in a positive answer to the conjecture.In this paper, after some background on numerical semigroups, we findan equivalent form of the conjecture (cf. Proposition 10, Remark 11). Inparticular, while results known so far rely heavily on the particular hypothe-ses (for example symmetric, almost symmetric, three-generatd or maximalembedding dimension semigroups), we try to develop a more general methodto attack the problem. We show that it is affirmatively answered by semi-groups whose embedding dimension is large with respect to the multiplicity(cf. Theorem 18). Finally, we note that the conjecture is also verified bysemigroups with small multiplicity (cf. Corollary 19) and by those generatedby a generalized arithmetic sequence (cf. Proposition 20).A good reference about numerical semigroups is [8].
Let N denote the set of natural numbers, including 0. A numerical semigroup is a submonoid of ( N , +) with finite complement in it. Given a numericalsemigroup S , we define a partial order setting s (cid:22) t if there exists an element u ∈ S such that s + u = t . Each numerical semigroup has a unique minimalsystem of generators { g < g < . . . < g ν } such that every element s ∈ S isrepresentable as s = λ g + . . . + λ ν g ν , with λ i ∈ N . This set coincides withthe set of minimal elements in S \ { } with respect to the partial order (cid:22) .2here are several invariants associated to a numerical semigroup S . Thelargest integer not belonging to the semigroup is called Frobenius number of S and is denoted by f = f ( S ); the number f ( S )+1 is known as the conductor of S . The multiplicity is defined as m = m ( S ) = min { s ∈ S, s > } ; it isclear that m = g . The number of generators ν = ν ( S ) is called embeddingdimension ; it is not difficult to see that the inequality ν ≤ m holds. Aninteger x ∈ Z \ S is called a pseudo-Frobenius number if x + s ∈ S forevery s ∈ S, s = 0; the cardinality of the set of pseudo-Frobenius numbersis known as the type of the semigroup and is denoted with t = t ( S ). We usethe symbol | X | to denote the cardinality of a set X . The number n ( S ) isdefined as n ( S ) = (cid:12)(cid:12) { s ∈ S, s < f } (cid:12)(cid:12) .Throughout the paper we will make an extensive use of an important toolassociated to a semigroup S , which is known as Ap´ery set of S and is definedas Ap( S ) = { w ∈ S, w − m / ∈ S } . This set consists of the smallest elements in S in each class of congruencemodulo m ([8], Lemma 2.4); it follows that | Ap( S ) | = m and 0 ∈ Ap( S ).We name the elements in increasing order setting Ap( S ) = { w < w <. . . < w m − } ; with this notation, we always have w = 0 , w = g and w m − = f + m ([8], Proposition 2.12). It is useful to consider Ap( S ) \ { } asa partially ordered set, with the partial order (cid:22) induced by S : we can indeedstate some properties of S in terms of this poset, as we see in the next result.In order to do this, we define the two subsetsminAp( S ) = { w ∈ Ap( S ) \ { } , w is minimal wrt (cid:22)} maxAp( S ) = { w ∈ Ap( S ) \ { } , w is maximal wrt (cid:22)} . Proposition 2 ([3], Lemma 3.2) . Let S be a numerical semigroup, then:(i) minAp( S ) = { g , . . . , g ν } ; ii) maxAp( S ) = { w, w − m is a pseudo-Frobenius number of S } . We obtain in particular that | minAp( S ) | = ν − | maxAp( S ) | = t ( S ).The next property is an easy consequence of the definitions of Ap( S ) and (cid:22) : Lemma 3 ([5], Lemma 6) . If w ∈ Ap( S ) and u (cid:22) w , then u ∈ Ap( S ) . The following inequality is useful for some particular cases:
Proposition 4 ([5], Theorem 22) . Let S be a semigroup with notation asabove. Then we have f ( S ) + 1 ≤ n ( S )( t ( S ) + 1) . As a consequence, every semigroup S such that t ( S ) + 1 ≤ ν ( S ) satisfiesthe conjecture. Moreover, the above inequality has been used in the samepaper to prove the next result: Corollary 5 ([5]) . If ν ( S ) ≤ , then S satisfies Wilf ’s conjecture. We are now able to prove two results which will be used afterwards.
Lemma 6. If m ( S ) − ν ( S ) ≤ , then S satisfies Wilf ’s conjecture.Proof. Let us distinguish three cases. • If ν = m , then Ap( S ) \ { } = maxAp( S ) = { g , . . . , g ν } and t = ν − • If ν = m −
1, then Ap( S ) \ { } = { g , . . . , g ν , u } with g i (cid:22) u for at leastone index i ∈ { , . . . , ν } ; it follows that g i / ∈ maxAp( S ) and t ≤ ν − • If ν = m −
2, then Ap( S ) \ { } = { g , . . . , g ν , u, v } with u < v . Since u and v are not generators, we have either u = g h + g i , v = g j + g k forsome indexes such that { h, i } 6 = { j, k } , or v = u + g j . In both caseswe find in Ap( S ) \ { } two elements that are not maximal and hence t ≤ ν − t ( S ) + 1 ≤ ν ( S ), hence we can apply Proposition 4.4he technique used in the last Lemma cannot be generalized to highervalues of m − ν , since the inequality t + 1 ≤ ν does not hold in general, asthe following example shows. Example . Let S = h , , , i . Then m − ν = 7 − S ) = { , , , , , , } and the maximal elements are maxAp( S ) = { , , , } , thus t ( S ) = ν ( S ) = 4 and we cannot apply Proposition 4. Corollary 8. If m ( S ) ≤ , then S satisfies Wilf ’s conjecture.Proof. If m ≤
6, then either ν ≤ m − ν ≤ Our aim is to develop a method based on the idea of counting the elementsof S in some intervals of length m . Given an integer k ≥
0, we define k -th interval the set I k = (cid:2) km, ( k + 1) m − (cid:3) = (cid:8) km, km + 1 , . . . , ( k + 1) m − (cid:9) and let n k = (cid:12)(cid:12) { s ∈ S ∩ I k , s < f } (cid:12)(cid:12) . We express the conductor of the semigroupin the form f ( S ) + 1 = Lm + ρ , where 1 ≤ ρ ≤ m and L = (cid:4) fm (cid:5) = (cid:4) w m − m (cid:5) − L is the index of the last interval I k such that I k S , that isto say the only index such that f ( S ) ∈ I k . The next proposition states basicproperties of the n k ’s whose proofs are immediate: Proposition 9.
We have:(i) ≤ n k ≤ m − for k = 0 , . . . , L ; ii) n k = | S ∩ I k | for k = 0 , . . . , L − ;(iii) n k ≤ n k if ≤ k < k ≤ L − ;(iv) n ( S ) = P Lk =0 n k . Now we express Question 1 in an equivalent form, in terms of the quan-tities introduced so far.
Proposition 10.
A semigroup S satisfies Wilf ’s conjecture if and only if (2.1) L − X k =0 ( n k ν − m ) + ( n L ν − ρ ) ≥ . Proof.
Using Proposition 9 we have the following equivalences: f ( S ) + 1 ≤ n ( S ) ν ( S ) ⇔ Lm + ρ ≤ ν L X k =0 n k (cid:16) = L − X k =0 n k ν + n L ν (cid:17) ⇔ L − X k =0 m + ρ ≤ L − X k =0 n k ν + n L ν ⇔ L − X k =0 ( n k ν − m ) + ( n L ν − ρ ) ≥ . Remark . In order to prove Wilf’s conjecture, by means of Proposition 10,we may compute the number of intervals with a fixed amount of elements of S less than Lm , and estimate thus the first part of the sum in 2.1. Moreprecisely, if we consider the quantities ǫ j = (cid:12)(cid:12)(cid:8) k ∈ N , | I k ∩ S | = j, k = 0 , . . . , L − (cid:9)(cid:12)(cid:12) , with j ∈ { , . . . , m − } then, by expanding the sum and gathering the terms with the same value of n k , we have L − X k =0 ( n k ν − m ) = m − X j =1 ǫ j ( jν − m ) . η j = (cid:12)(cid:12)(cid:8) k ∈ N , | I k ∩ S | = j (cid:9)(cid:12)(cid:12) , with j ∈ { , . . . , m − } η j is the number of intervals I k with exactly j elements of S (not necessarilyless than Lm ). The next lemma shows how the two families are related: Lemma 12.
Under the above notation, we have: • ǫ j = η j for j = | I L ∩ S | ; • ǫ j = η j − for j = | I L ∩ S | .Proof. The thesis is straightforward as the only difference in the two defini-tions is made by the interval I L .The following proposition allows us to express the numbers η j in termsof the Ap´ery set. Proposition 13.
For any j = 1 , . . . , m − , we have η j = (cid:4) w j m (cid:5) − (cid:4) w j − m (cid:5) .Proof. Let us fix an interval I k and j ∈ { , . . . , m − } ; we claim that I k contains at least j elements of S if and only if w j − < ( k + 1) m . Inthis case, the interval I k contains exactly j elements of S if and only if w j − < ( k + 1) m ≤ w j and the thesis follows by definition of η j . Let us provethe claim. Set I k ∩ S = { s , . . . , s p } and define s ′ h = min { x ∈ S, x ≡ s h (mod m ) } for each h = 1 , . . . , p : by the characterization of Ap( S ) it followsthat { s ′ , . . . , s ′ p } ⊆ Ap( S ); moreover s ′ h ≤ s h < ( k + 1) m . Conversely, if w ∈ Ap( S ) and w < ( k + 1) m , then w + λm ∈ S ∩ I k for a suitable λ ∈ N and so w = s ′ h for some h ≤ p . Therefore, { s ′ , . . . , s ′ p } is the subset of Ap( S )consisting of all the elements less than ( k + 1) m . Recalling that the elementsin Ap( S ) are listed in increasing order we may conclude the proof: | I k ∩ S | ≥ j ⇔ p ≥ j ⇔ w j − ∈ { s ′ , . . . , s ′ p } ⇔ w j − < ( k + 1) m. Lemma 14.
Let us suppose m − ν ≥ . Then we have (cid:4) w ν +1 m (cid:5) ≥ (cid:4) w m (cid:5) + (cid:4) w m (cid:5) .Proof. Since m − ν ≥
2, there are two non-zero elements in Ap( S ) that arenot generators, that is, two elements in Ap( S ) \ { , g , . . . , g ν } : let u, v bethe smallest of such elements, with u < v . Since u and v are not minimal in S \ { } with respect to (cid:22) , then u = u + u , v = v + v with u , u , v , v positive elements of S ; by Lemma 3 these elements must belong to the Ap´eryset and hence we can write u = w h + w i , v = w j + w k , with h, i, j, k > u (cid:22) v , i.e. v = u + w j , is not excluded: it may occur,under this notation, that u = w k . For the choice of u, v, we have u ≤ w ν and v ≤ w ν +1 . Finally, by u = w h + w i ≥ w + w and v = w j + w k ≥ w + w weobtain: w ν +1 ≥ v ≥ w + w ⇒ j w ν +1 m k ≥ j w m k + j w m k . Lemma 15.
Let us suppose m − ν ≥ . If (cid:4) w m − m (cid:5) = (cid:4) w m (cid:5) + (cid:4) w m (cid:5) , then n L ≥ .Proof. Set w = q m + r , w = q m + r , where q = (cid:4) w m (cid:5) , q = (cid:4) w m (cid:5) , 1, andin particular r < ρ and r < ρ . It follows that { Lm, w + k m, w + k m } ⊆ [ Lm, f ] ∩ S for some k , k ∈ N , and so n L ≥ Lemma 16. Let us suppose m − ν ≥ , then w < f . roof. If w > f , then w i > f for each i = 2 , . . . , m − 1. We claim that w , . . . , w m − ∈ maxAp( S ). Indeed if w i (cid:22) w j for some j > i ≥ 2, then thereexists s ∈ S, s > , such that w j = w i + s . We have s ≥ m , w i ≥ f + 1, hence w j ≥ f + m + 1 and w j − m ∈ S , in contradiction with w j ∈ Ap( S ). Thusthe elements { w , . . . , w m − } are pairwise incomparable with respect to (cid:22) .It follows that, for each j ≥ w j / ∈ minAp( S ) if and only if w (cid:22) w j . Butthis may occur for at most one index j : if w (cid:22) w j , then w j = w + s , with s ∈ S \ { } . By Lemma 3, s ∈ Ap( S ) \ { } and the only possibility is s = w and hence w j = 2 w . Thus at most one element among { w , . . . , w m − } maynot be minimal. We have proved that m − ≤ ν − m − ν ≤ Lemma 17. Let us suppose m − ν ≥ . If n L = 1 , then there are twopossibilities:(1) n L − ≥ ;(2) n L − = 3 , m − ν = 3 and ρ ≤ m − .Proof. By the previous lemma, we have w < w < f . Since n L = 1 it followsthat { Lm, . . . , f } ∩ S = { Lm } and thus w ∈ I h , w ∈ I h for some indexes h , h < L .Let us suppose n L − = | I L − ∩ S | ≤ 3. We have that ( L − m, w + k m, w + k m ∈ I L − ∩ S for suitable k , k ∈ N and so n L − = 3. Recallingthe argument of the proof of Proposition 13, the fact that n L − = | I L − ∩ S | =3 implies w > Lm ; since { Lm, . . . , f } ∩ S = { Lm } we actually have w > f .We want to show, similarly to what we have done within the proof ofLemma 16, that the only possible non-zero elements in Ap( S ) \ minAp( S )are 2 w , w + w , w : in this case we obtain the second assertion m − ν = 3.If w ∈ Ap( S ) \ { } , w / ∈ minAp( S ), then we have w = w i + w j , with i, j > i > 2, by w i ≥ w ≥ f + 1 and w j > m it follows w ≥ f + m + 19nd w − m ∈ S , in contradiction with w ∈ Ap( S ). Thus i, j ≤ { w , w + w , w } .Finally, since w + ( k + 1) m, w + ( k + 1) m ∈ I L and n L = 1, we musthave f < w + ( k + 1) m < ( L + 1) m and f < w + ( k + 1) m < ( L + 1) m ,hence f < ( L + 1) m − ρ ≤ m − Theorem 18. If ν ( S ) ≥ m ( S ) , then S satisfies Wilf ’s conjecture.Proof. By Lemma 6 we may assume m − ν ≥ 3. We want to proceed assuggested in Remark 11: we will count the intervals with 1 element of S andthose with at least 3 elements. The hypothesis 2 ν ≥ m allows us to leave outthose with 2 elements: their contribution to the sum in 2.1, according to thecontent of Remark 11, is ǫ (2 ν − m ) and hence non-negative by hypothesis.Using Proposition 13 we find: η = j w m k − j w m k = j w m k because w = 0; moreover m − X j =3 η j = m − X j =3 (cid:16)j w j m k − j w j − m k(cid:17) = m − X j =3 j w j m k − m − X j =2 j w j m k = j w m − m k − j w m k . Applying Remark 11, Lemma 12 and the above formulas to inequality 2.110e get: L − X k =0 ( n k ν − m ) + ( n L ν − ρ ) = m − X j =0 ǫ j ( jν − m ) + ( n L ν − ρ ) ≥ η ( ν − m ) + (cid:16) m − X j =3 η j − (cid:17) (3 ν − m ) + ( n L ν − ρ ) = j w m k ( ν − m ) + (cid:16)j w m − m k − j w m k − (cid:17) (3 ν − m ) + ( n L ν − ρ ) = (cid:16)j w m − m k − j w m k − j w m k − (cid:17) (3 ν − m ) + j w m k (4 ν − m ) + ( n L ν − ρ ) . Let us distinguish now three possible cases. • The equality holds in Lemma 14.By Lemma 15 we have n L ≥ 3; futhermore (4 ν − m ) ≥ − (3 ν − m ) + ( n L ν − ρ ) = ( n L − ν + ( m − ρ ) ≥ . • The strict inequality holds in Lemma 14 and n L ≥ (cid:16)j w m − m k − j w m k − j w m k − (cid:17) (3 ν − m )+ j w m k (4 ν − m )+( n L ν − ρ ) ≥ n L ν − ρ ≥ ν − m ≥ • The strict inequality holds in Lemma 14 and n L = 1.By Lemma 17 we have either n L − ≥ n L − = 3 , m − ν = 3 and ρ ≤ m − 2. In the first case, we need to add ν to the sum (correspondingto the interval I L − ) and we obtain ν + j w m k (4 ν − m ) + ( ν − ρ ) ≥ ν − ρ ≥ ν − m ≥ . 11n the second case we may assume m ≥ j w m k (4 ν − m ) + ( ν − ρ ) = j w m k (4( m − − m ) + ( m − − ρ ) ≥ j w m k (2 m − 12) + ( m − − m + 2) ≥ (2 m − − m − > . The inequality is valid in each case and the thesis is thus proved.From the previous theorem we immediately get the following result: Corollary 19. If m ( S ) ≤ , then S satisfies Wilf ’s conjecture.Proof. If m ≤ ν ≤ ν ≥ m ; the thesis follows fromCorollary 5 and Theorem 18.We conclude our paper showing another class of numerical semigroupssatisfying the conjecture, which is actually independent of most results ofthe paper. A semigroup generated by a generalized arithmetic sequence isa semigroup of the kind S = h m, hm + d, hm + 2 d, . . . , hm + ld i ; for ourpurpose, we may assume gcd( m, d ) = 1 , m ≥ , l ≤ m − 2. Such semigroupshave been studied in [6]. Proposition 20. If S is a semigroup generated by a generalized arithmeticsequence, then S satisfies Wilf ’s conjecture.Proof. In ([6], Corollary 3.4) the author proved in particular that t ( S ) = m − (cid:4) m − l (cid:5) l − . By definition of ⌊·⌋ we have: m − l < j m − l k + 1 ⇒ m − < j m − l k l + l ⇒ t ( S ) = m − j m − l k l − < l + 1 = ν ( S )and the thesis follows from Proposition 4. Acknowledgments. 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