On some invariants in numerical semigroups and estimations of the order bound

On some in v arian ts in n umerical semigroups a nd estimations of the order b ound. Anna Oneto and Grazia T amone 1 Abstract. Let S = { s i } i ∈ I N ⊆ I N be a numeric al semigroup. F or s i ∈ S , let ν ( s i ) denote the n umber of p airs ( s i − s j , s j ) ∈ S 2 . When S is the W eierstrass semigroup of a family {C i } i ∈ I N of one- point algebraic- geometric codes, a goo d bound for the minimum distance of the co de C i is the F eng and R ao or der b ound d O RD ( C i ) := m in { ν ( s j ) : j ≥ i + 1 } . It is well-kno wn that there exists an integer m such that the sequence { ν ( s i ) } i ∈ I N is non-decreasing for s i > s m , t herefore d O RD ( C i ) = ν ( s i +1 ) for i ≥ m . By wa y of some suitable parameters related to the semigroup S , w e find upp er b ounds for s m , w e ev aluate s m exactly in many cases, further we give a low er b ound for several classes of semigroups. Index Ther ms. Numerical semigroup, W eierstrass semigroup, AG code, order b ound on the minimum distance, Cohen-Macaulay type. 1 In tro duction Let S = { s i } i ∈ I N ⊆ I N b e a numerical semigr oup and let e, c, c ′ , d, d ′ denote r espectively the multi- plicit y , the conductor , the subco nductor, the dominant of the semigroup and the greatest element in S preceding c ′ (if e > 1), as in Setting 2.1. F ur ther let ℓ be the num b er o f gaps of S b etw een d and c , and let e s := max { s ∈ S such that s ≤ d a n d s − ℓ / ∈ S } . When S is the W eierstr a ss semigr oup o f a fa mily {C i } i ∈ I N of one-p oint AG co des (see [3],[2]), a go o d bo und for the minimu m distance of C i is the F eng and Rao or der b ou n d d ORD ( C i ) := min { ν ( s j ) : j ≥ i + 1 } where, for s j ∈ S , ν ( s j ) denotes the num b er of pairs ( s j − s k , s k ) ∈ S 2 (see [2]). It is well-kno wn that there exists an in tege r m such that sequence { ν ( s i ) } i ∈ I N is non-decr easing for i ≥ m + 1 (see[7]) and so d ORD ( C i ) = ν ( s i +1 ) for i ≥ m . F or this reaso n it is imp ortant to find the elemen t s m of S . In our pap ers [5 ] and [6 ], we proved that s m = e s + d if e s ≥ d ′ , further we ev a luated s m for ℓ ≤ 2 , e ≤ 6, C ohen - M acaulay ty pe ≤ 3. In this pa per, by a mor e detailed study of the semigroup w e find interesting relatio ns among the int eg ers defined ab ov e; further by using these relations we deduce the F eng and Rao order bo und in several new situations. Mo reov er in e very considered case we show that s m ≥ c + d − e . In Section 2 , we es ta blish v ar io us for m ulas and inequalities among the integers e , ℓ, d ′ , c ′ , d, c and t := d − e s , see in pa rticular (2.5) and (2.6 ). In Section 3, by using the r esults of Se c tion 2 and some result from [6], we improv e the known facts o n s m recalled above; further we sta te the conjecture that c + d − e ia alwa ys a low er b ound for s m and we prov e it in many ca ses. Finally (Section 4) we deduce some particular case by a pplying the previous results and by some direct trick. In conclus io n by glueing tog heter some facts o f [1], [5], [6] and the results o f the prese nt pap er, we see that the v alue of the or der b ound s m depe nds essentially on the po s ition o f the in teger e s in the semigroup. W e summarize b elow the main results for the convenience of the rea de r . If e s ≥ d ′ + c ′ − d then s m = e s + d if e s = 2 d ′ − d then s m = e s + d = 2 d ′ if e s ≤ d ′ + c ′ − d − 1 then s m ≤ max { e s + d, 2 d ′ } . (3.4) 1 The first author i s with Diptem, Universit` a di Geno v a, P .le Kennedy , Pa d. D - 16129 Geno v a (It aly) ( E-mail : oneto@diptem.unige.it). The second author i s with Dima, Univ ersit` a di Geno v a, Via Do decaneso 35 - 16146 Geno v a (Italy) ( E-mail : tamone@dima.unige.it). 1 If [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S then  e s + d ′ − ℓ + 1 ≤ s m ≤ 2 d ′ if 2 d ′ − d < e s < d ′ + c ′ − d s m = e s + d otherwise . (3.11) If    e s ≤ d ′ − 2  e s + 2 , d ′  ∩ I N ⊆ S 2 d ′ − d < e s < d ′ + c ′ − d then  s m = e s + d ⇐ ⇒ e s + 1 / ∈ S and c ′ = d s m ≤ e s + d − 1 otherwise . (3.8) If  e s ≤ 2 d ′ − d  e s + 2 , d ′  ∩ I N ⊆ S then s m = e s + d. (3.8) F ur ther, if H denotes the subset of g a ps o f S inside the in terv al [ c − e, c ′ − 1] and τ is the Cohen- Macaulay t yp e of S , the exact v alue or go o d e s timations for s m are given in the fo llowing cases. If H = ∅ , then s m = e s + d (4.1) If H is a non e mpty inter va l, then s m =  2 d ′ if e s ≥ 2 d ′ + 1 − d e s + d otherwise (4.1) If S is associated to a S u z uk i cur v e, then s m = e s + d (4.14) . If # H ≤ 2 , see (4.4) . If ℓ ≤ 3 , see (4.5) , (4.7) . If τ ≤ 7 , see (4.10) . If e ≤ 8 , see (4 .11 ) . If S is g ener ated b y a n Al most Ar ithmetic S eq uence and embdim ( S ) ≤ 5 , see (4 .1 2) . 2 Semigroups: in v arian ts and relations. W e b e g in by giving the setting of the pa per. Setting 2.1 In all the article w e s hall use the following notation. Let I N denote the set of all non- negative integers. A numeric al semigr oup is a subset S of I N containing 0 , close d under summation and with finite complement in I N; we shall a lw ays a s sume S 6 = I N. W e denote the elemen ts of S by { s i } , i ∈ I N, with s 0 = 0 < s 1 < ... < s i < s i +1 ... . W e set S (1) := { b ∈ I N | b + ( S \ { 0 } ) ⊆ S } W e list below the inv aria n ts related to a semigroup S ⊂ I N we shall need in the sequel. e := s 1 > 1 , the mul tipl icity o f S. c := min { r ∈ S | r + I N ⊆ S } , t he conductor of S d := the greatest element in S prece ding c , the dominant of S c ′ := ma x { s i ∈ S | s i ≤ d and s i − 1 / ∈ S } , the subcondu ctor of S d ′ := the greatest element in S preceding c ′ , when c ′ > 0 k := d − c ′ q := d − d ′ ℓ := c − 1 − d, the num b er of ga ps o f S greater than d g := #(I N \ S ) , the g enus o f S (= the num b er of gaps of S ) τ := #( S (1) \ S ) , the C ohe n − M acaul ay ty pe of S Since c − e − 1 / ∈ S we hav e c − e ≤ c ′ ; we define the following sets H := [ c − e, c ′ ] ∩ I N \ S ⊆ I N \ S S ′ := { s ∈ S | s ≤ d ′ } ⊆ S. 2 F ur ther we shall describ e any semig r oup S with c ′ > 0 a s fo llows: ℓ gaps S = { 0 , ∗ . . . ∗ , e, . . . , d ′ , ∗ . . . ∗ , c ′ ← → d, ∗ . . . ∗ , c →} = S ′ ∪ { c ′ ← → d, ∗ . . . ∗ , c →} , where “ ∗ ” indicates gaps , “ ∗ . . . ∗ ” int erval of al l gaps , and “ ← → ” intervals without gaps . Moreov er for s i ∈ S we fix the following notation. N ( s i ) := { ( s j , s k ) ∈ S 2 | s i = s j + s k } ; ν ( s i ) := # N ( s i ) , η ( s i ) := ν ( s i + i ) − ν ( s i ) . d ORD ( i ) := min { ν ( s j ) | j > i } , the or der bound. A ( s i ) := { ( x, y ) , ( y , x ) ∈ N ( s i ) | x < c ′ , c ′ ≤ y ≤ d } ; α ( s i ) := # A ( s i +1 ) − # A ( s i ) . B ( s i ) := { ( x, y ) ∈ N ( s i ) | ( x, y ) ∈ [ c ′ , d ] 2 } ; β ( s i ) := # B ( s i +1 ) − # B ( s i ) . C ( s i ) := { ( x, y ) ∈ S ′ 2 ∩ N ( s i ) } ; γ ( s i ) := # C ( s i +1 ) − # C ( s i ) . D ( s i ) := { ( x, y ) , ( y , x ) ∈ N ( s i ) | x ≥ c } ; δ ( s i ) := # D ( s i +1 ) − # D ( s i ) . Now we reca ll some definition and former results for co mpleteness. First, a semigr oup S is called or dinary if S = { 0 } ∪ { n ∈ I N , n ≥ c } acute if either S is ordinar y , or c, d, c ′ , d ′ satisf y c − d ≤ c ′ − d ′ [1 , Def . 5 . 6] . Definition 2. 2 We define the invariants e s , m and t as fol lows. e s := max { s ∈ S su ch that s − ℓ / ∈ S } . t := d − e s . m := min { j ∈ I N such that the se quenc e { ν ( s i ) } i ∈ I N is non-de cr e asing for i > j } ( m > 0 ⇐ ⇒ ν ( s m ) > ν ( s m +1 ) and ν ( s m + k ) ≤ ν ( s m + k +1 ) , for e ach k ≥ 1 ) . Theorem 2. 3 L et S = { s i } i ∈ I N b e as in Setting 2.1. (1) ν ( s i ) = i + 1 − g , fo r every s i ≥ 2 c − 1 . [7, Th. 3.8] (2) ν ( s i +1 ) ≥ ν ( s i ) , fo r every s i ≥ 2 d + 1 . [5, P rop. 3 .9.1] (3) If S is an or dinary semigr oup, then m = 0 . [1, Th. 7.3] (4) If e s ≥ d ′ , then s m = e s + d [6, Th. 4.1, Th.4.2] . In p articular: ( a ) if t ≤ 2 , then s m = e s + d , ( b ) if S is an acute semigr oup, then s m = e s + d , with i. either d − c ′ ≥ ℓ − 1 , s m = c + c ′ − 2 = e s + d, ii. or e s = d ( s m = 2 d ) . [5, P rop. 3.4] . (5) If c ′ ∈ { c − e, c − e + 1 } , then S is acute. [6, L emma 5.1 ] . Remark 2. 4 (1) B y the definitio n of e s it is clear that: s − ℓ ∈ S f or each s ∈ S such that e s < s ≤ d. (2). Theor em 2.3 implies that 0 < s m ≤ 2 d for every non-o r dinary semigro up. (3) The condition ( a ) o f (2.3 .4) do es not imply S acute: see (2.5.2). Analogously there exist non-ac ute semigroups satisfying the co nditions (4 .b , i − ii ), as shown in E xample 2.9.2. W e complete this section with some genera l relation amo ng the inv ariants defined a bov e. 3 Prop osition 2. 5 [6, Prop. 2.5] L et c ′ = c − e + q , with q ≥ 0 . Then (1) e ≤ 2 ℓ + t + q . (2) The fol lowing c onditions ( a ) d − c ′ ≥ ℓ − 1 ( i.e. c + c ′ − 2 ≤ 2 d ) . ( b ) e s − ℓ = c ′ − 1 . ( c ) c + c ′ − 2 = e s + d . ( d ) e = 2 ℓ + t + q ar e e quivalent and imply ( i ) c ′ ≤ e s ≤ d ( = ⇒ s m = e s + d ) . ( ii ) S is acute ⇐ ⇒ d − d ′ ≥ 2 ℓ + t . Pro of. (1) By (2.2.1) we ha ve e s − ℓ ≤ c ′ − 1 = c − e + q − 1 , then e s − ℓ ≤ d + ℓ − e + q and so e ≤ 2 ℓ + t + q . (2) The equiv a lences (2 .a ) ⇐ ⇒ (2 .b ) ⇐ ⇒ (2 .c ) are pr o ved in [6, Prop. 2.5]. Clearly the equalit y e = 2 ℓ + t + q holds if and only if e s − ℓ = d − t − ℓ = c ′ − 1. F urther: ( i ) is obvious by (2. b ). ( ii ) If (2. b ) holds, then d − d ′ − (2 ℓ + t ) = e s − ℓ − d ′ − ℓ = ( c ′ − d ′ ) − ( ℓ + 1) = ( c ′ − d ′ ) − ( c − d ) . Then S is a cute ⇐ ⇒ d − d ′ ≥ 2 ℓ + t . Prop osition 2. 6 The fol lowing facts hold. (1) ( a ) If 0 ≤ h < e and d − h ∈ S , then e ≥ h + ℓ + 1 . ( b ) If s , s ′ ∈ S, s ≥ c − e, s − ℓ ≤ s ′ < s , then s ′ ≥ c − e . (2) e s ≥ c − e ( e quivalently, e ≥ t + ℓ + 1) . (3) L et t > 0 and let s ′ := min { s ∈ S | s > e s } . Then e ≥ 2 ℓ + 1 + d − s ′ ≥ 2 ℓ + 1 ( e quivalently, s ′ ≥ c − e + ℓ ) . In p articular, e s + 1 ∈ S = ⇒ e ≥ 2 ℓ + t . (4) One of t he fol lowing c onditions hold ( a ) e s − ℓ > c − e − 1 ( eq uiv al entl y e > 2 ℓ + t, eq uiv al ently e s − ℓ ∈ H ) ( b ) e s − ℓ = c − e − 1 ( eq uiv al entl y e = 2 ℓ + t ) ( c ) c − e − ℓ ≤ e s − ℓ < c − e − 1 ( eq uav ale ntly e < 2 ℓ + t ) (5) Assume e < 2 ℓ + t , then : ( a ) either e s ≤ d ′ , or t = 0 . ( b ) in c ase e s ≤ d ′ we have: [ e s + 1 , c − e + ℓ − 1] ∩ S = ∅ , # H ≥ 2 ℓ + t − e > 0 . Pro of. (1. a ) W e hav e d < d − h + e ∈ S . Hence d − h + e ≥ c = d + ℓ + 1 . (1. b ) If s ≥ c we g et s ′ ≥ c . If s ≤ d, let s ′ = d − k , s = d − h ≥ c − e (hence h ≤ e − ℓ − 1) , then d − ℓ − h ≤ d − k = ⇒ k ≤ h + ℓ ≤ e − 1. Now apply ( a ). (2) Let d = c − e + hℓ + r , with h ≥ 0 , 0 ≤ r < ℓ (r ecall that c − e ≤ d ). If e s < c − e , first note that b y (2.4.1) we ge t d − hℓ ∈ S, d − ( h + 1 ) ℓ ∈ S ; further we get c − e − ℓ ≤ d − ( h + 1) ℓ < c − e , a contradiction b ecause [ c − ℓ − e , c − e − 1] ∩ S = ∅ for every semigr oup. (3) By (2.4.1), e s < s ′ ≤ d = ⇒ s ′ − ℓ ∈ S and so s ′ − ℓ + e ∈ S . Since c − e ≤ e s < s ′ , we get s ′ − ℓ + e > c − ℓ = d + 1; it follows that s ′ − ℓ + e ≥ c . (4) Since c − e − 1 / ∈ S , the statements are a lmost immediate by (2). (5) In cas e e < 2 ℓ + t , by (3) w e have e s + 1 / ∈ S , therefor e we hav e e s / ∈ [ c ′ , d ), hence (5 .a ) holds. (5 .b ) Since e s ≥ c − e (2), we get [ e s − ℓ + 1 , c − e − 1 ] ∩ I N ⊆ [ c − ℓ − e + 1 , c − e − 1] ∩ I N ⊆ I N \ S . W e deduce [ e s + 1 , c − e + ℓ − 1] ∩ I N ⊆ H by (2.4.1). ⋄ 4 Corollary 2. 7 Assume e s < d ( i.e. t > 0) . Then (1) If e s ≤ d ′ , then d − c ′ ≤ ℓ − 2 , d − d ′ ≤ ℓ , c ′ ≥ c − e + 2 . (2) L et c ′ = c − e + q , with q ∈ { 0 , 1 } ; t hen d − c ′ ≥ ℓ − 1 and e = 2 ℓ + t + q . (3) If d − c ′ ≤ ℓ − 2 , then ( a ) d − c ′ + 2 ≤ d − d ′ ≤ ℓ ≤ e − 3 − ( d − c ′ ) . ( b ) If e s ≥ 2 d ′ − d , then t ≤ 2 ℓ . (4) If e s < d ′ and e ≤ 2 ℓ + t , t hen # H ≤ ℓ + t − 2 ( d − c ′ ) − 4 . Pro of. (1) B y (2.5.2) we see that e s ≤ d ′ = ⇒ d − c ′ ≤ ℓ − 2 a nd als o d − d ′ ≤ ℓ . F urther c ′ ≥ c − e + 2 bec ause c − e ≤ e s (2.6.2) and e s ≤ d ′ ≤ c ′ − 2. (2) By the as sumptions and by (5), (4) of Theo rem 2.3, w e hav e d − c ′ ≥ ℓ − 1. Then the o ther statement follows by (2.5.2). (3. a ) Since d ′ ≤ c ′ − 2 the first ine q ualit y holds for any semigr o up. W e hav e d − c ′ ≤ ℓ − 2, by assumption, and d − ℓ ∈ S , by (2.4.1). Hence d ′ ≥ d − ℓ . F or the last inequality see [6, Pro p. 5 .2]]. (3 .b ) follows by (3 .a ) b ecause the a ssumption mea ns t ≤ 2 d − 2 d ′ . (4) By (2.5.2) we ha ve d − c ′ ≤ ℓ − 2, then by )2.6.1-2) and (3 a ), we deduce that { c ′ − ℓ, ...d − ℓ, e s, d ′ } ⊂ S ∩ [ c − e, d ′ ]. Hence # H ≤ c ′ − ( c − e ) − 2 − ( d − c ′ + 1) = 2 c ′ − 2 d − ℓ − 1 − 3 + e ≤ 2 c ′ − 2 d − ℓ − 1 − 3 + 2 ℓ + t = ℓ + t − 2( d − c ′ ) − 4. ⋄ Corollary 2. 8 (1) If c ′ ≤ e s < d , then e ≥ 2 ℓ + t . (2) If e s ≤ d ′ , we have: ( a ) If e ≤ 2 ℓ + t , t hen d + 2 ℓ − e ∈ S ⇐ ⇒ e = 2 ℓ + t . ( b ) If H ⊆ [ d ′ − t + 1 , c ′ − 1] , then e ≤ 2 ℓ + t . ( c ) If H = [ d ′ + 1 , c ′ − 1] ∩ I N and e < 2 ℓ + t , then e s = d ′ . Pro of. (1) is immediate by (2.6 . 3) b ecause e s + 1 ∈ S . (2. a ) Clearly e = 2 ℓ + t = ⇒ d + 2 ℓ − e = e s ∈ S . The conv erse follo ws b y the assumption and by(2.6.5 b ): c − e + ℓ − 1 = d + 2 ℓ − e ∈ S = ⇒ e ≥ 2 ℓ + t ; then e = 2 ℓ + t . (2. b ) e s ≤ d ′ = ⇒ d − ℓ ∈ S = ⇒ d − ℓ ≤ d ′ by (2.7.1). Hence e s − ℓ ≤ d ′ − t : no w the ass umption o n H implies e s − ℓ / ∈ H , i.e., e ≤ 2 ℓ + t (2.64). (2. c ) If e < 2 ℓ + t , we hav e e s + 1 / ∈ S (2.6.3); since c − e < e s + 1 we get e s + 1 ∈ H , and so e s = d ′ . ⋄ Example 2.9 (1 ) If t > 0, for each s i such that e s < s i ≤ d , w e hav e that s i − ℓ ∈ S , hence s i − s i − 1 ≤ ℓ , but it is no t true that for ea c h s i ∈ S such that c − e ≤ s i < d , we hav e s i +1 − s i ≤ ℓ : for instance let S = { 0 , 5 ℓ e , 7 ℓ e s = d − ℓ = d ′ , 8 ℓ d , 9 ℓ + 1 c →} . (2) When t = 0 the inequality e ≥ 2 ℓ + 1 (proved in (2.6.3) for t > 0) in gener al is not true, even for acute semigr oups: S 1 = { 0 , 10 e = d ′ , 17 c ′ , 18 , 1 9 , 2 0 d , 27 c →} : ℓ = 6 , t = 0 , S 1 is acute with d − c ′ ≤ ℓ − 2 , e < 2 ℓ . S 2 = { 0 , 8 e , 12 d ′ , 14 c ′ , 15 , 16 d , 20 c →} : ℓ = 3 , t = 0 , S 2 is non-acute with d − c ′ = ℓ − 1 , e > 2 ℓ . S 3 = { 0 , 7 e , 12 d ′ , 14 c ′ = d , 19 c →} : ℓ = 4 , t = 0 , S 3 is non-acute with d − c ′ ≤ ℓ − 2 , e < 2 ℓ . S 4 = { 0 , 10 e = d ′ , 14 d , 20 c →} ℓ = 5 , t = 0 , S 4 is non-a c ute with d − c ′ ≤ ℓ − 2 , e = 2 ℓ. 5 (3) When e s ≤ d ′ we can have every case ( a ) , ( b ) , ( c ) of (2.6.4 ): S 5 = { 0 , 13 e , 15 d ′ , 20 d, 2 6 c →} : ℓ = t = 5 e < 2 ℓ + t = 15 ; S 6 = { 0 , 15 e , 19 d ′ = e s , 24 d, 3 0 c →} : ℓ = t = 5 e = 2 ℓ + t = 15 ; S 7 = { 0 , 26 e , 28 , 31 d ′ , 33 d , 39 c →} : ℓ = t = 5 e > 2 ℓ + t = 1 5. 3 General results on s m . W e saw in [6], that s m = e s + d , when e s ≥ d ′ . T o giv e estimations of s m in the r emaining cas es we shall use the same to ols as in [6]: w e recall them for the c on venience of the rea der and we add some improv ement, as the ge ner al inequalities (3.1.3) on the difference ν ( s + 1) − ν ( s ). Therefor e great part of the following (3.1),(3.3),(3.4) is alrea dy proved in [6, 4.1, 4.2, 4.3 ]. Prop osition 3. 1 L et S ′ = { s ∈ S, | s ≤ d ′ } . F or s i ∈ S , let η ( s i ) , α ( s i ) , β ( s i ) , γ ( s i ) , δ ( s i ) b e as in (2.1 ) . Then: (1) If e s < d ′ , we have: s i +1 = s i + 1 for s i ≥ e s + d ′ − ℓ , in p articular for s i ≥ 2 d ′ . (2) F or e ach s i ∈ S : η ( s i ) = α ( s i ) + β ( s i ) + γ ( s i ) + δ ( s i ) . F urther α ( s i ) =   − 2 if ( s i +1 − c ′ / ∈ S ′ and s i − d ∈ S ′ ) 0 if ( s i +1 − c ′ ∈ S ′ ⇐ ⇒ s i − d ∈ S ′ ) 2 if ( s i +1 − c ′ ∈ S ′ and s i − d / ∈ S ′ ) . β ( s i ) =   0 if s i ≤ 2 c ′ − 2 or s i > 2 d 1 if 2 c ′ − 1 ≤ s i ≤ c ′ + d − 1 − 1 if c ′ + d ≤ s i ≤ 2 d. γ ( s i ) =   0 if s i ≥ 2 d ′ + 1 − 1 if s i = 2 d ′ − 1 if s i < 2 d ′ and [ s i − d ′ , d ′ ] ∩ I N ⊆ S . δ ( s i ) =   0 if s i +1 − c / ∈ S, s i ≤ 2 c − 1 2 if s i +1 − c ∈ S, s i ≤ 2 c − 1 1 if s i ≥ 2 c . (3) L et s = 2 d − k < 2 d and s + 1 ∈ S , then : ( a ) − h k 2 i − 1 ≤ ν ( s + 1) − ν ( s ) ≤ h k + 5 2 i . ( b ) If s = 2 d ′ − h < 2 d ′ , then − h h 2 i − 1 ≤ γ ( s ) ≤ h h + 1 2 i . Pro of. (1) By ass umption and by (2.6.2) we hav e c − e ≤ e s < d ′ and so d ′ − ℓ ∈ S . It follows d ′ − ℓ ≥ e bec ause d ′ ≥ e ≥ ℓ + t + 1 (2.6.2). Hence s ≥ e s + d ′ − ℓ = ⇒ s ≥ c . (2) By [6 , (3.3)...(3.7)] we have only to prove the last tw o statements for γ . Let s = 2 d ′ − h ∈ S, h ∈ I N , s + 1 = 2 d ′ − h + 1 and as sume [ d ′ − h, d ′ ] = [ s i − d ′ , d ′ ] ∩ I N ⊆ S . Then: C ( s i ) = { ( d ′ − h, d ′ ) , ( d ′ − h + 1 , d ′ − 1) , ( d ′ − h + 2 , d ′ − 2) , ..., ( d ′ − 1 , d ′ − h + 1) , ( d ′ , d ′ − h ) } C ( s i +1 ) = { ( d ′ − h + 1 , d ′ ) , ( d ′ − h + 2 , d ′ − 1) , .., ( d ′ − 1 , d ′ − h + 2) , ( d ′ , d ′ − h + 1) } it follows that γ ( s i ) = # C ( s i +1 ) − # C ( s i ) = h − ( h + 1) = − 1. (3. a ) T o prov e the inequalities for s = 2 d − k , divide the in terv al [ d − [ k / 2 ] , d ] ∩ I N in subsets Λ j := H j ∪ S j , j = 1 , ..., j ( s ), with H j ⊆ I N \ S, S j ⊆ S, S j = [ a j , b j ] ∩ I N interv al such that b j + 1 / ∈ S and H j 6 = ∅ , if j > 1 (i.e. a j − 1 / ∈ S for j > 1, H 1 = ∅ ⇐ ⇒ a 1 = d − [ k / 2] ∈ S ). Hence Λ j = [ ∗ ∗ ∗ a j ← → b j ] . 6 Let N ( s ) j := N ( s ) ∩ { ( x, y ) , ( y , x ) | y ∈ S j } : we ha ve N ( s ) = S j N ( s ) j ∪ D ( s ). Hence: ν ( s + 1) − ν ( s ) = ( P j n j ) + δ ( s ) , where n j = # N ( s + 1) j − # N ( s ) j . F ur ther: − 2 ≤ n j ≤ 2. This fact follows by the same arg umen t use d t o pro ve the formulas for α ( s i ) , β ( s i ) recalled in statement (2) ab ov e. Since 0 ≤ δ ( s ) ≤ 2 (see (2) a b ov e) we conc lude that ( ∗ ) − 2 j ( s ) ≤ ν ( s + 1 ) − ν ( s ) ≤ 2 j ( s ) + 2. More precisely , to ev alua te the larg est and low est p ossible v alues o f ν ( s + 1 ) − ν ( s ), with s = 2 d − k , we consider separa tely four ca ses:     ( A ) k = 4 p ( B ) k = 4 p + 1 ( C ) k = 4 p + 2 ( D ) k = 4 p + 3 . In each ca s e we ca n see that j ( s ) ≤ p + 1 = h k 4 i + 1 . Firs t note tha t d ∈ S , hence j ( s ) is maximal when # H j = # S j = 1, i.e. [ d −  k 2  , d ] = [ ... ∗ × ∗ × ... ∗ d ] (wher e × mea ns element ∈ S ). In ea c h of the ab o ve cases we shall find integers x 1 , x 2 , y 1 , y 2 such that  x 1 ≤ ν ( s + 1 ) ≤ x 2 y 1 ≤ ν ( s ) ≤ y 2 , then the statement will follow b y the obvious inequality x 1 − y 2 ≤ ν ( s + 1) − ν ( s ) ≤ x 2 − y 1 . - If either k = 4 p , or k = 4 p + 1, then j ( s ) is max imal if and only if [ d −  k 2  , d ] = [ d − 2 p ∗ ... ∗ d − 2 ∗ d ], with j ( s ) = p + 1 . Note that when j ( s ) = p + 1 , then 1 ≤ #  N ( s ) \ D ( s )  ≤ 2 p + 1 beca use ( d − 2 p, d − 2 p ) ∈ N ( s ); further we hav e p ≤ j ( s + 1 ) ≤ p + 1 a nd so 0 ≤ # N ( s + 1) ≤ 2 p + 4. If k = 4 p , we hav e 1 ≤ #  N ( s ) \ D ( s )  ≤ 2 p + 1 , since ( d − 2 p , d − 2 p ) ∈ N ( s ), further j ( s + 1) = p , hence 0 ≤ #  N ( s + 1) \ D ( s + 1)  ≤ 2 p . −  k 2  − 1 = − 2 p − 1 ≤ ν ( s + 1) − ν ( s ) ≤ 2 p + 2 − 1 <  k + 5 2  . If k = 4 p + 1, we hav e 0 ≤ #  N ( s ) \ D ( s )  ≤ 2 p + 2 , further s + 1 = 2 d − 4 p , therefore 1 ≤ #  N ( s + 1) \ D ( s + 1)  ≤ 2 p + 1. W e obta in: −  k 2  − 1 = − 2 p − 1 ≤ ν ( s + 1) − ν ( s ) ≤ 2 p + 3 =  k + 5 2  . - If either k = 4 p + 2, o r k = 4 p + 3 , then  k 2  = 2 p + 1, analogo usly w e ge t j ( s ) = p + 1 max imal if  d −  k 2  , d  =  [ ∗ d − 2 p ∗ ... ∗ d − 2 ∗ d ] or [ d − 2 p − 1 ... ∗ × × ∗ ...d ] ( with o ne and only one j 0 such that # S j 0 = 2) . If k = 4 p +2 , in the first sub case we get 0 ≤  N ( s ) \ D ( s )  ≤ 2 p + 2, and 0 ≤ #  N ( s +1) \ D ( s +1)  ≤ 2 p bec ause ( d − 2 p − 1 , d − 2 p ) / ∈ N ( s + 1). Hence −  k 2  − 1 = − 2 p − 2 ≤ ν ( s + 1) − ν ( s ) ≤ 2 p + 2 <  k + 5 2  . In the seco nd sub case we get 1 ≤ #  N ( s ) \ D ( s )  ≤ 2 p + 2, beca use ( d − 2 p − 1 , d − 2 p − 1) ∈ N ( s ) and 0 ≤  N ( s + 1) \ D ( s + 1)  ≤ 2 p sinc e ( d − 2 p − 1 , d − 2 p ) / ∈ N ( s + 1). Hence −  k 2  − 1 = − 2 p − 2 ≤ ν ( s + 1) − ν ( s ) ≤ 2 p + 1 <  k + 5 2  . If k = 4 p + 3, in the first sub c ase we get 0 ≤  N ( s ) \ D ( s )  ≤ 2 p + 2, a nd 0 ≤ #  N ( s + 1) \ D ( s + 1)  ≤ 2 p + 2 b ecause ( d − 2 p − 1 , d − 2 p ) / ∈ N ( s + 1). Hence −  k 2  − 1 = − 2 p − 2 ≤ ν ( s + 1) − ν ( s ) ≤ 2 p + 4 =  k + 5 2  . 7 In the se c o nd sub case we get 0 ≤ #  N ( s ) \ D ( s )  ≤ 2 p + 2 and 0 ≤ #  N ( s + 1) \ D ( s )  ≤ 2 p + 1 bec ause ( d − 2 p − 1 , d − 2 p − 1) ∈ N ( s + 1). Hence −  k 2  − 1 = − 2 p − 2 ≤ ν ( s + 1) − ν ( s ) ≤ 2 p + 3 <  k + 5 2  . (3. b ) The pro of is quite similar to the ab ove o ne: since γ ( s ) = # C ( s + 1 ) − # C ( s ), we do not need to add δ ( s ) and so for m ula ( ∗ ) b ecomes − 2 j ′ ( s ) ≤ γ ( s ) ≤ 2 j ′ ( s ), where j ′ ( s ) is the num b er of subset Λ j as in (3. a ) co n tained in the interv a l [ d ′ −  h 2  , d ′ ] ∩ I N. Then it suffices to pro ceed as ab ov e. ⋄ Example 3.2 The b ounds found in (3.1.3 a ) ar e b oth sharp. T o s ee this fa ct, consider S = { 0 , 10 e , 20 d ′ , 30 d , 40 c →} an the elements s = 2 d − 1 = 59 , s + 1 = 2 d = 60. By a dir e c t computation we easily g et: ν ( s + 1) − ν ( s ) = 3 = h k + 1 2 i + 2 (with k = 1), and ν ( s + 2) − ν ( s + 1 ) = − h k 2 i − 1 (with k = 0 ). Prop osition 3. 3 L et  me an / ∈ S ′ and × me an ∈ S ′ ( r e c al l that for s ≤ d ′ , we have s ∈ S ⇐ ⇒ s ∈ S ′ ) . The fol lowing t ables describ e the differ enc e η ( s i ) = ν ( s i +1 ) − ν ( s i ) for s i ∈ S, s i < 2 c in funct ion of α, β , γ , δ . ( a ) If s i < 2 c :               s i +1 − c s i − d s i +1 − c ′ α δ η ( s i ) / ∈ S ×  − 2 0 β + γ − 2 / ∈ S   0 0 β + γ / ∈ S × × 0 0 β + γ / ∈ S  × 2 0 β + γ + 2 ∈ S ×  − 2 2 β + γ ∈ S   0 2 β + γ + 2 ∈ S × × 0 2 β + γ + 2 ∈ S  × 2 2 β + γ + 4               . Mor e pr e cisely we have the fol lowing sub c ases. ( b ) If s i ≤ 2 d ′ − 1 , then β = 0 :               s i +1 − c s i − d s i +1 − c ′ α β δ η ( s i )  ×  − 2 0 0 γ − 2  × × 0 0 0 γ    0 0 0 γ × ×  − 2 0 2 γ   × 2 0 0 γ + 2 ×   0 0 2 γ + 2 × × × 0 0 2 γ + 2 ×  × 2 0 2 γ + 4               . ( c ) If s i = 2 d ′ , then β = 0 , γ = − 1 : 8               s i +1 − c s i − d s i +1 − c ′ α β δ η ( s i )  ×  − 2 0 0 − 3  × × 0 0 0 − 1    0 0 0 − 1 × ×  − 2 0 2 − 1   × 2 0 0 1 ×   0 0 2 1 × × × 0 0 2 1 ×  × 2 0 2 3               . ( d ) If s i ∈ [2 d ′ + 1 , c ′ + d − 1] , then β ∈ { 0 , 1 } , γ = 0 : ν ( s i +1 ) < ν ( s i ) if and only if the fol lowing r ow is satisfie d  s i +1 − c s i − d s i +1 − c ′  ×   . ( e ) If s i ∈ [ c ′ + d, 2 d ] , t hen β = − 1 , γ = 0 , s i − d ∈ S \ S ′ , s i +1 − c ′ / ∈ S ′ , then ν ( s i +1 ) < ν ( s i ) ⇐ ⇒ s i − ℓ − d / ∈ S . The next theorem collects with some upgrades the r esults [6, Th. 4.1, Th.4.2 , Th. 4.4]: statement (1) improv es [6 , Th.4.2], the las t par t of (5 ) is new. Theorem 3. 4 With Setting 2.1, the fol lowing ine qualities hold. (0) If e s ≥ 2 d ′ − d, then s m ≤ e s + d ; if e s < 2 d ′ − d, then s m ≤ 2 d ′ . Mor e pr e cisely (1) If e s ≥ d ′ + c ′ − d, then s m = e s + d . (2) If e s = d ′ + c ′ − d − 1 , then s m ≤ e s + d − 1 . (3) If 2 d ′ − d < e s < d ′ + c ′ − d − 1 , let U := { σ ∈ [2 d ′ + 1 − d, e s ] ∩ S | σ − ℓ / ∈ S, σ + d + 1 − c ′ / ∈ S } : ( a ) if U 6 = ∅ , then s m = d + max U , in p articular s m = e s + d ⇐ ⇒ e s + d + 1 − c ′ / ∈ S , ( b ) if U = ∅ , then s m ≤ 2 d ′ . (4) If e s = 2 d ′ − d , then s m = e s + d . (5) If e s < 2 d ′ − d , then s m ≤ 2 d ′ , mor e pr e cisely: s m = 2 d ′ ⇐ ⇒ 2 d ′ satisfies either r ow 3 or r ow 4 of T able 3.3 ( c ) . In the c ase e s + d + 1 − c ′ / ∈ S : ( a ) if 2 d ′ − d − 2 ≤ e s ≤ 2 d ′ − d − 1 , then e s + d ≤ s m ≤ 2 d ′ ( b ) if e s = 2 d ′ − d − j, j = 3 , 4 and { d ′ − j, ..., d ′ − 1 } ∩ S 6 = { d ′ − j + 1 } then s m ≥ e s + d . Pro of. (0) is prov ed in [6, (4.4 .1),(4.4.3)]. Now recall that e s ≥ c − e (2.6.2), hence e s + d + 1 ≥ c +1 ∈ S ; further in cases (1 ) and (2) e s + d + 1 − c ′ ≥ d ′ , hence (1) and (2) follow b y (0) a nd by T able 3 .3 ( d ). The cases (3) and (4) follow easily by T ables 3.3 ( d ) and ( c ). (5) W e have s m ≤ 2 d ′ by (0); further 2 d ′ cannot satisfy the first t wo rows of 3.3 ( c ) since e s < 2 d ′ − d . By a direct computation w e can see that w e always hav e γ (2 d ′ − j ) ≤ 1 , for j ≤ 2, while for 9 j = 3 , 4 γ (2 d ′ − j ) ≤ 1 ⇐ ⇒ { d ′ − j, ..., d ′ − 1 } ∩ S 6 = { d ′ − j + 1 } . Now ( a ) a nd ( b ) follow b e- cause ν ( e s + d ) > ν ( e s + d + 1) by T ables 3.3.( b ) and ( c ). ⋄ The follo wing conjecture gives a lo wer b ound for s m , it is justified b y calculations in very many examples. W e are able to pr o ve that it ho lds in many cases. Conjecture 3.5 F or every semigr oup the ine quality s m ≥ c + d − e holds. First we note that (3.5) holds in the following general cases: Prop osition 3. 6 Assu me  either s m ≥ e s + d or s m ≥ 2 d ′ and e s < d ′ then s m ≥ c + d − e . In p articular if either e s + d ≥ c ′ + d ′ or e s + d = 2 d ′ , then s m ≥ c + d − e . Pro of. The first part follows b y (2.6.2 ) and (2.6.3). In fact we have (1) e s ≥ c − e ; (2) in case e s < d ′ , s m ≥ 2 d ′ , we have d ′ ≥ c − e + ℓ (2.6.1 b ) and d − d ′ ≤ ℓ ( 2 .7.1). Hence s m ≥ 2 d ′ ≥ d ′ + c − e + ℓ ≥ c + d − e . Now the par ticular cas e s follow by (3 .4 . 1...4). ⋄ Corollary 3. 7 (1) If s m > 2 d ′ , then s m − d ∈ S . (2) If e s = d ′ − 1 , t hen s m = e s + d ⇐ ⇒ c ′ 6 = d . Pro of.(1) follows by (3.4.1) and by T able 3.3.( d ). (2) If c ′ = d , then s m 6 = e s + d , by (3.4.2). If c ′ 6 = d and e s = d ′ − 1, we have e s ≥ d ′ + c ′ − d then apply (3.4.1). ⋄ Prop osition 3. 8 Assu me e s ≤ d ′ − 2 and [ e s + 2 , d ′ ] ∩ I N ⊆ S . Then s m ≤ e s + d : (1) if 2 d ′ − d < e s < d ′ + c ′ − d , then s m  = e s + d ⇐ ⇒ e s + 1 / ∈ S and c ′ = d ≤ e s + d − 1 , o the rw i se. (2) if e s ≤ 2 d ′ − d , then s m = e s + d . Pro of. In case (1), by applying Theor e m 3.4 we see that s m ≤ e s + d ; further s m = e s + d ⇐ ⇒ e s + d + 1 − c ′ / ∈ S . Since e s + 1 ≤ e s + d + 1 − c ′ ≤ d ′ by the assumptions, we see that s m = e s + d ⇐ ⇒ c ′ = d and e s + 1 / ∈ S . In case (2), by Theor em 3.4 w e have s m ≤ 2 d ′ . Now let e s + d + 1 ≤ s ≤ 2 d ′ . Then by the as sumptions we get        e s + 2 ≤ s + 1 − c ′ ≤ s − d ′ − 1 < d ′ s + 1 ∈ S and s + 1 − c ′ ∈ S ′ { s − d ′ , ..., d ′ } ⊆ S (hence γ ( s ) = − 1 (3.3 . 2)) s − ℓ − d ∈ S (b y (2.4 . 1)) . F r om T ables 3.3 ( b ) − ( c ) we co nc lude that s m < s and also that s m = e s + d ; in fact e s ∈ S, e s + d + 1 − c ′ ∈ S ′ , e s − ℓ / ∈ S, further e s + d − d ′ ≥ e s + 2 , b e cause d − d ′ ≥ 2, therefore γ ( e s + d ) = − 1 by the assumptions and (3.3.2). ⋄ Remark 3. 9 (1) Both situations of (3.8.1) above can happ en, even for ℓ = 3 (see the following (4.7)) : (A) If ℓ = 3 , t = 5, d ′ = d − 3 , c ′ = d − 1, (4.7.case A) we hav e s m < e s + d . (B) If ℓ = 3 , t = 5 , d ′ = d − 3 , d − 4 / ∈ S , c ′ = d , (4 .7.ca se B) w e have s m = e s + d . (2) Assume 2 d ′ − d < e s ≤ d ′ + c ′ − d − 1 and [ d ′ − ℓ + 2 , d ′ ] ∩ I N ⊆ S ; then the set U of (3.4.3) is empt y . I n fact for ea c h s ∈ S, s uch that 2 d ′ + 1 ≤ s ≤ e s + d , we hav e s + 1 ∈ S , and by (2.7.3( a )), d ′ − ℓ + 2 ≤ 2 d ′ + 2 − d ≤ s + 1 − c ′ ≤ d ′ , therefore s + 1 − c ′ ∈ S ′ . (3) If s m < 2 d ′ ≤ e s + d , then    ( a ) e s + d + 1 − c ′ ∈ S ( b ) e s + d + 1 − c ′ − ℓ ∈ S ( c ) { 2 d ′ − d − ℓ, 2 d ′ + 1 − c ′ } ∩ S 6 = ∅ . 10 (3. a ) holds by (3.4 .3); in fact the assumptions imply U = ∅ b ecause e s − ℓ / ∈ S . (3 .b ) is clear by (3 .a ) and by (2.4.1 ) , since e s < e s + d + 1 − c ′ < d . (3 .c ) follows by T a ble 3.3 ( c )). (4) The assumption s m > 2 d ′ in (3.7.1) is necessa ry: for instance if S = { 0 , 2 0 e , 21 , 26 , 27 d ′ , 32 d , 39 c →} we have s m = 2 d ′ , with 2 d ′ − d / ∈ S (w e deduce s m = 2 d ′ by T able 3.3 ( c )). Prop osition 3. 10 If e s < d ′ and [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S , let h = d − c ′ , q = d − d ′ b e as in (2.1) and let σ := max { s ∈ S, s < e s − ℓ } . Then (1) [ e s − ℓ + 1 , d ′ ] ∩ I N ⊆ S and e ≥ 2 ℓ + t . (2) If 2 d ′ − d < e s ≤ d ′ + c ′ − d − 1 , we have ( a ) q + h + 1 ≤ t < 2 q ( ≤ 2 ℓ ) , ( b ) F or s ∈ [ e s + d ′ − ℓ + 1 , 2 d ′ ] ∩ S , we have γ ( s ) = − 1 . ( c ) If 2 d ′ − ℓ − d ∈ S , then σ ≥ 2 d ′ − ℓ − d and γ ( σ + d ) = − 1 . ( d ) We have s m ≤ 2 d ′ . ( e ) L et W := [ e s − 2 ℓ + 1 , 2 d ′ − ℓ − d ] ∩ I N \ S . If W 6 = ∅ , let h 0 := max W , then s m ≥ h 0 + ℓ + d . ( f ) s m < e s + d − ℓ + 1 ⇐ ⇒ [ e s − 2 ℓ + 1 , 2 d ′ − d − ℓ ] ∩ I N ⊆ S , s m < e s + d − ℓ + 1 = ⇒ e ≥ 3 ℓ + t. ( g ) If [ e s − 2 ℓ + 1 , 2 d ′ − d − ℓ ] ∩ I N ⊆ S, then s m ≥ e s + d ′ − ℓ + 1 . Pro of. (1) By the ass umptions and by (2.4.1) we have [ e s − ℓ + 1 , d ′ ] ∩ I N ⊆ S ; the inequa lity e ≥ 2 ℓ + t follows by (2.6.3) (2) Statement ( a ) is immediate by the assumption 2 d ′ − d < e s ≤ d ′ + c ′ − d − 1. ( b ) follows by (1) and by (3.3.2). ( c ) By assumption 2 d ′ − ℓ − d ∈ S , further 2 d ′ − d < e s ; then 2 d ′ − ℓ − d < e s − ℓ , hence σ ≥ 2 d ′ − ℓ − d . Moreov er e s − ℓ + 1 ≤ d ′ − ℓ = 2 d ′ − ℓ − d ′ ≤ σ + d − d ′ < ( d − ℓ ) + e s − d ′ < 2 d ′ − d ′ = d ′ ; then apply ( b ) . ( d ) B y (3.4.3) we know that s m ≤ e s + d . F or each 2 d ′ < s ≤ e s + d we have d ′ − ℓ < 2 d ′ + 1 − c ′ ≤ s + 1 − c ′ ≤ e s + d + 1 − c ′ ≤ d ′ + c ′ − 1 + 1 − c ′ = d ′ . Therefore s + 1 − c ′ ∈ S and s m ≤ 2 d ′ by (3.4.3 b ). ( e ) a nd ( f ). Note that s ∈ [ e s + d − ℓ + 1 , 2 d ′ ] ∩ S = ⇒ e s − ℓ + 1 ≤ s − d ≤ s + 1 − c ′ ≤ s − d ′ ≤ d ′ , hence { s − d, s + 1 − c ′ } ⊆ S ′ and γ ( s ) = − 1, b y ( b ) and the assumptions . By T able 3.3 ( b ) we get ν ( s ) > ν ( s + 1) ⇐ ⇒ s + 1 − c / ∈ S. Then ( e ) follows and the equiv alence ( f ) b ecomes immediate by ( d ), ( e ), recalling that s + 1 − c = s − ℓ − d . W e get e ≥ 3 ℓ + t by (2.6.1 − 2 ), since d − (2 ℓ + t − 1) ∈ S and 2 ℓ + t − 1 < e by ( a ). ( g ) F or s ∈ [ e s + d ′ − ℓ + 1 , 2 d ′ − ℓ ] ∩ I N, w e have γ ( s ) = − 1 (see ( b )). If there ex is ts s ∈ [ e s + d ′ − ℓ + 1 , 2 d ′ − ℓ ] ∩ I N , s + 1 − c ′ / ∈ S , we hav e s − d ∈ S ′ and we have s m ≥ s by T able 3.3 ( b ) (in fact s − d ∈ S ′ by the assumptions ); the claim follows. Assume on the contrary that [ e s + d ′ − ℓ + 2 − c ′ , 2 d ′ − ℓ + 1 − c ′ ] ∩ I N ⊆ S : then [ e s − 2 ℓ + 1 , 2 d ′ − ℓ + 1 − c ′ ] ∩ I N ⊆ S . In fact q + h + 1 ≤ t (3.10.1) = ⇒ e s + d ′ − ℓ + 2 − c ′ = d − q − ℓ − t + 2 + h ≤ d − 2 q − ℓ + 1 = 2 d ′ − ℓ + 1 − d . W e can iter ate the alg orithm lo oking for o ne element s ∈ [2 d ′ − ℓ + 1 , 2 d ′ − ℓ + d + 1 − c ′ ] ∩ I N such that s + 1 − c ′ / ∈ S . If needed we r epeat the arg umen t till we find s ′ such that s ′ + 1 − c ′ / ∈ S : s ′ surely exists since e s − ℓ / ∈ S . ⋄ The previo us results ca n b e summarize d in the following theor em. Theorem 3. 11 A ssume [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S . Then s m ≥ c + d − e . In p articular: (1) if 2 d ′ − d < e s < d ′ + c ′ − d , we have c + d − e ≤ e s + d ′ − ℓ + 1 ≤ s m ≤ 2 d ′ , 11 (2) s m = e s + d in the r emaining c ases. Pro of. (1) Since e s < d ′ , we have [ e s − ℓ + 1 , d ′ ] ∩ I N ⊆ S , e ≥ 2 ℓ + t b y (3.10.1). It fo llows that e s − ℓ + 1 ≥ c − e b ecause [ c − ℓ − e , c − e − 1] ∩ S = ∅ and e s ≥ c − e by (2.6.2). The inequalities follow by items ( d ) , ( e ) , ( f ) , ( g ) of (3.10): if the set W of (3.1 0.2 e ) is no t empty then we see that s m ≥ e s + d − ℓ + 1 ≥ e s + d ′ − ℓ + 3 by (3.10.2 e ), recalling that d ′ ≤ d − 2. If W = ∅ , b y (3.10.2 g ) we get s m ≥ e s + d ′ − ℓ + 1. (2) follows by (3.4 .1) and by (3.8.2). In this case s m ≥ c + d − e by (3.6). T o prove s m ≥ c + d − e in ca se (1), first as sume that W = ∅ : since d ′ ≥ d − ℓ (2.8.1) and e ≥ 3 ℓ + t (3.10.6), w e get s m ≥ e s − ℓ + d ′ + 1 ≥ e s − ℓ + d − ℓ + 1 = c + d − 3 ℓ − t ≥ c + d − e. If W 6 = ∅ , since e ≥ 2 ℓ + t we get s m ≥ e s + d − ℓ + 1 = c + d − 2 ℓ − t ≥ c + d − e , further d > d ′ + 1 = ⇒ e s + d − ℓ + 1 > e s + d ′ − ℓ + 1. ⋄ In Case ( g ) of (3.1 0) we ca n give more precise ev aluations of s m . Corollary 3. 12 L et [ e s − 2 ℓ + 1 , 2 d ′ − d − ℓ ] ∪ [ d ′ − ℓ , d ′ ] ∩ I N ⊆ S , let 2 d ′ − d < e s < d ′ + c ′ − d and let σ b e as in (3.10) . (1) Assume   either c ′ − ℓ − t − 1 ∈ S or  c ′ − ℓ − t − 1 / ∈ S c ′ − 2 ℓ − t − 1 / ∈ S , then s m ≥ e s + c ′ − ℓ − 1 (2) Assume  c ′ − ℓ − t − 1 / ∈ S c ′ − 2 ℓ − t − 1 ∈ S , then : ( a ) If σ ≤ c ′ − ℓ − t − 2 , then s m ≥ σ + d . ( b ) If  σ ≥ c ′ − ℓ − t σ − ℓ / ∈ S , then s m = σ + d . ( c ) If    σ ≥ c ′ − ℓ − t, σ − ℓ ∈ S and either 2 d ′ − 2 ℓ − d / ∈ S or 2 d ′ − ℓ + 1 − c ′ / ∈ S , then s m ≥ 2 d ′ − ℓ. Pro of. (1) Let s = e s + c ′ − ℓ − 1. Then e s + d ′ − ℓ + 1 ≤ s ≤ d ′ + e s − 1 < 2 d ′ and so γ ( s ) = − 1 (3.10.2 b ). F urther s + 1 − c ′ / ∈ S and, by the as sumptions, either s − d ∈ S ′ or ( s − d / ∈ S ′ and s + 1 − c / ∈ S ′ ). The cla im follows b y T able 3 .3 ( b ). (2 .a ) F rom T able 3.3 ( b ) we get ν ( σ + d ) > ν ( σ + d + 1). In fact we hav e σ < σ + d + 1 − c ′ ≤ e s + c ′ − ℓ − 2 + 1 − c ′ = e s − ℓ − 1. Hence σ + d + 1 − c ′ / ∈ S . F urther γ ( σ + d ) = − 1 by the as s umption 2 d ′ − d − ℓ ∈ S and by (3.10.2 c ). (2 .b ) In this case for each s such that σ + d ≤ s ≤ e s + d − ℓ we ha ve s + 1 − c ′ ∈ S : in fact e s − ℓ + 1 ≤ σ + d + 1 − c ′ ≤ s + 1 − c ′ < 2 d ′ − c ′ < d ′ (3.10.2 a ). Moreover we g e t γ ( s ) = − 1 by (3.10.2 b ). If s > σ + d , then s − d / ∈ S hence by T able 3.3 ( b ): ν ( s ) < ν ( s + 1 ) therefore s m ≤ σ + d and ν ( σ + d ) > ν ( σ + d + 1 ) ⇐ ⇒ σ − ℓ / ∈ S . (2 .c ). Since d ′ > e s we hav e γ (2 d ′ − ℓ ) = − 1; in fact 2 d ′ − ℓ ≥ e s + d ′ − ℓ + 1 (3.10.2 b ). Then the claim follows by using the assumption 2 d ′ − ℓ − d ∈ S and T able 3.3 ( b ). ⋄ 4 Some particular case. In this sectio n w e shall estimate or giv e exactly the v alue of s m in some particular case. Since s m = e s + d for each semigro up S satisfying e s ≥ c ′ + d ′ − d , in this s ection we shall often assume e s < c ′ + d ′ − d . 12 4.1 Relations b et w een the order b ound and the holes set H . Let H := [ c − e, c ′ ] ∩ I N \ S b e as in (2.1): when H is an int er v al we deduce the v alue of s m , if # H ≤ 2 and in some other situatio n we give a low er bo und for s m . Prop osition 4. 1 (1) If H = ∅ , then c ′ = c − e and S is acute with s m = e s + d . (2) Assume t hat e s < d ′ . Then the c onditions ( a ) [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S and e = 2 ℓ + t . ( b ) H = [ d ′ + 1 , c ′ − 1] ∩ I N . ar e e quivalent and imply: s m =  2 d ′ if 2 d ′ − d + 1 ≤ e s ≤ d ′ + c ′ − d − 1 e s + d in the remaini ng cases. Pro of. (1) H = ∅ ⇐ ⇒ c ′ = c − e ; then apply (2.3. 5 and 4). (2 ), ( a ) ⇐ ⇒ ( b ). ( a ) implies that [ e s − ℓ + 1 , d ′ ] ∩ I N ⊆ S , and e s − ℓ = c − e − 1 (3.10.1) and (2.6.4). Hence (( b )) holds. On the contrary , ( b ) implies [ c − e, d ′ ] ∩ I N ⊆ S . Since c − e ≤ d ′ − ℓ by (2.6.3), we get [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S , further e ≤ 2 ℓ + t by (2.8.2 b ). B y assumption we als o hav e e s + 1 ∈ S , hence e ≥ 2 ℓ + t by (2.6.3); then ( a ) follows. When 2 d ′ − d < e s < d ′ + c ′ − d , by (2.6.1) c − e ≤ d ′ − ℓ , and so (2.7.1) we hav e c − e − ℓ ≤ d ′ − ℓ − ( d − d ′ ) = 2 d ′ − ℓ − d < e s − ℓ ≤ c − e − 1. W e obtain s m ≥ 2 d ′ by (3.10.2 e ) b ecause the s e t W as in (3.1 0.2 e ) has 2 d ′ − ℓ − d = max W . Now s m = 2 d ′ follows by (3.1 0.2 d ). F or the statement in the remaining case s s ee (3.1 1.2). ⋄ Example 4.2 When e s > d ′ the implica tion ( b ) = ⇒ ( a ) in (4.1 .2) is no t true in general: in fact S := { 0 , 8 e , 12 c − e = d ′ , 14 c ′ , 15 , 16 d ∗ ∗ ∗ 20 c →} has H = { c ′ − 1 } = { 1 3 } , t = 0 , ℓ = 3, e 6 = 2 ℓ + t . Prop osition 4. 3 Assu me e s < c ′ + d ′ − d . L et k := min { n ∈ I N | d ′ − n / ∈ S } , h := d − c ′ , s := d ′ + c ′ − k − 1 . We have (1) s ≤ 2 d ′ ⇐ ⇒ c ′ − d ′ ≤ k + 1 ⇐ ⇒ d − d ′ ≤ k + h + 1 . (2) If e s < d ′ − k and ℓ ≤ k + h + 1 , then s m ≥ s ≥ c + d − e. (3) If 1 ≤ k < ℓ , c ′ − d ′ ≤ k + 1 and { d ′ − ℓ, ..., d ′ } \ { d ′ − k } ⊆ S , then c + d − e ≤ s ≤ s m ≤ 2 d ′ . Pro of. (1) is obvious by the assumptions. (2) W e ha ve [ d ′ − k − ℓ + 1 , d ′ − ℓ ] ∩ I N ⊆ S (2.4.1). F urther by (3 .1.2) we hav e γ ( s ) = − 1; in fact the assumption ℓ ≤ k + h + 1 implies c ′ − d ′ = d − d ′ − h ≤ ℓ − h ≤ k + 1, and so s ≤ 2 d ′ ; since [ s − d ′ , d ′ ] ∩ I N ⊆ S, then γ ( s ) = − 1 (3.1.2). Since e s < d ′ , then d − c ′ ≤ ℓ − 2 , further ℓ ≤ k + h + 1; therefore d ′ − k − ℓ + 1 ≤ s − d = d ′ + c ′ − k − 1 − d ≤ d ′ − ℓ. Hence s − d ∈ S . Moreover s + 1 − c ′ = d ′ − k / ∈ S . Then s m ≥ s by T able 3 .3 ( b ). T o prov e that s ≥ c + d − e , recall that c ′ − d ′ ≤ k + 1. Then by ass umption we hav e e s < d ′ − k < c ′ − k − 1 ≤ d ′ . Then c ′ − k − 1 ∈ S a nd by (2.6.3) we get c ′ − k − 1 ≥ c − e + ℓ . Hence s = d ′ + c ′ − k − 1 ≥ d − ℓ + c ′ − k − 1 ≥ c + d − e (2 .7.3 a ) (3) By a ssumption e s < d ′ − h ≤ d ′ , and so [ d ′ − h − ℓ, d ′ − ℓ ] ∩ I N \ { d ′ − k − ℓ } ⊆ S . Hence [ d ′ − h − ℓ, d ′ − k − 1] ∩ I N \ { d ′ − k − ℓ } ⊆ S . Now recalling that k < ℓ we get: d ′ − h − ℓ ≤ s − d < d ′ − k − h ≤ d ′ − k . Hence s − d ∈ S : in fact s − d 6 = d ′ − k − ℓ b ecause s − d ≥ d ′ − k − ℓ + 1 . F urther s + 1 − c ′ / ∈ S and γ ( s ) = − 1 b y (3.1.2) since s ≤ 2 d ′ and { s − d ′ , ..., d ′ } ⊆ S . Then s ≤ s m by T able 3 .3 ( b ). The inequalit y s ≥ c + d − e can be proved as in (2 ). F or each element 2 d ′ < u < c ′ + d ′ we hav e d ′ ≥ u + 1 − c ′ > s + 1 − c ′ = d ′ − k ; he nc e u + 1 − c ′ ∈ S ′ then ν ( u + 1) ≥ ν ( u ) by T able 3.3 ( d ). 13 Corollary 4. 4 Supp ose e s < c ′ + d ′ − d and # H ≤ 2 . Then s m ≥ c + d − e . Pro of. If # H = 0, we hav e s m = e s + d and we ar e do ne by (4.1.1) and (3.6). If ( # H = 2 and H = { d ′ + 1 , c ′ − 1 } ), o r # H = 1 , then either s m = e s + d , or s m = 2 d ′ (4.1.2); now see (3.6). Finally assume that H = { d ′ − k } ∪ { d ′ + 1 } , with k ≥ 1. In this case we hav e c ′ − d ′ = 2 ≤ k + 1, since k ≥ 1 . Hence the claim s m ≥ c + d − e follows by (3.11), if k > ℓ , and by (4.3.3 ) if k < ℓ . ⋄ 4.2 Case ℓ = 2 . If ℓ = 2 , the conjecture (3.5) is true, more pr ecisely by [6, Thm 5.5] we hav e: Prop osition 4. 5 Assu me ℓ = 2 , then s m ≥ c + d − e and (1) s m = e s + d if   t ≤ 2 , t = 4 t ≥ 5 and d − 3 ∈ S. (2) s m = 2 d − 4 if  either t = 3 and d − 6 / ∈ S or t ≥ 5 and d − 3 / ∈ S. (3) s m = 2 d − 6 if t = 3 and d − 6 ∈ S ( al l t he r emaining c ases ) . Pro of. The v alue of s m is known by [6, Thm. 5.5 ]. Another pro of can b e easily deduced by (2.4.1), ( 3.4.2), (4.3.3), (3.8.2), T a ble 3 .3 ( d ), (3 .10. e, g ). The inequalities s m ≥ c + d − e now follow resp ectiv ely by (3.6) and by (3.11.3). ⋄ 4.3 Case ℓ = 3 . If ℓ = 3, we compute explicitly the p ossible v alues of s m and we show that the conjecture (3 .5) holds. Notation 4 .6 (1) If s i = 2 d − k ∈ S , k ∈ I N, let M ( s i ) := { ( s h , s j ) ∈ S 2 | s i = s h + s j , s h ≤ d, s j ≤ d } . Note that M ( s i ) = { ( d − x, d − y ) ∈ S 2 , | 0 ≤ x, y ≤ k , x + y = k } and that for s i ≥ c , we hav e s i +1 − c = d − ℓ − k ; for short it will b e conv enient to use the following notation. ( ∗ ) ( Σ := { z ∈ I N , | z ≤ d, d − z ∈ S } ( c, h ) ∈ S × Σ , h = d + c − s i instead of the pa ir ( c, s i − c ) ∈ N ( s i ) . If ℓ = 3, then e ≥ t + ℓ + 1 = t + 4 ( 2 .6 . 2). T o calcula te the v alue of s m , we shall assume e s < c ′ + d ′ − d , otherwise s m = e s + d by (3.4.1). Then w e hav e t ≥ 3 , d − 3 ∈ S and d − 3 ≤ d ′ , b y (2.5.2) and by (2.7.3 a ). Three ca ses ar e p ossible: C ase A : S = { 0 , e, ..., d − 3 , ∗ , d − 1 , d, ∗ ∗ ∗ , c = d + 4 →} d ′ = d − 3 C ase B : S = { 0 , e, ..., d − 3 , ∗ ∗ , d, ∗ ∗ ∗ , c = d + 4 →} d ′ = d − 3 C ase C : S = { 0 , e, ..., d − 3 , d − 2 , ∗ , d, ∗ ∗ ∗ , c = d + 4 →} d ′ = d − 2 . T o desc r ibe M (2 d − k ) we shall use the no tations ( ∗ ) fixed in (4.6) a nd fo rwhen neces s ary for an element 2 d − k w e shall list all the pair s ( x, y ) ∈ M ′ (2 d − k ) and the pair ( c, ℓ + k + 1 ) ∈ S × Σ (the pairs underlined ( , ) not necessa rily b elong to Σ 2 ). Prop osition 4. 7 Assu me ℓ = 3 . Then s m ≥ c + d − e . Mor e pr e cisely the values of s m c an b e c omput e d as fol lows. 14 Case A. W e hav e: s m =           e s + d if eithe r t ∈ [0 , 7] \ { 5 } or ( t ≥ 8 , d − 5 ∈ S ) 2 d − 7 if t ≥ 8 , d − 5 / ∈ S if t = 5 : 2 d − 6 if d − 9 / ∈ S 2 d − 7 if d − 9 ∈ S, d − 10 / ∈ S 2 d − 9 if { d − 9 , d − 10 } ⊆ S, d − 12 / ∈ S 2 d − 10 i f { d − 9 , d − 10 , d − 12 } ⊆ S Pro of. S = { 0 , e, ..., d − 3 , ∗ , d − 1 , d, ∗ ∗ ∗ , c = d + 4 →} , with e ≥ ℓ + t + 1 = t + 4 (2.6 . 1) . First we hav e s m = e s + d if t ≤ 2 d − c ′ − d ′ = 4 and s m < e s + d , if t = 5 by (3 .4.1 and 2). Hence we can assume t ≥ 5, so that d − 4 = d − 1 − ℓ ∈ S, d − 3 − ℓ = d − 6 ∈ S , i.e., { 0 , 1 , 3 , 4 , 6 } ⊆ Σ. If t = 5 w e ha ve [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S and 2 d ′ = 2 d − 6 < e s + d = d ′ + c ′ − 1. W e obtain that 2 d − 10 ≤ s m ≤ 2 d ′ , e ≥ 2 ℓ + t and s m ≥ c + d − e by (3.11). Mo r e pr e cisely we can verify that: s m =       2 d − 6 if 9 / ∈ Σ s m = 2 d ′ ≥ c + d − e 2 d − 7 if 9 ∈ Σ and 1 0 / ∈ Σ s m = c + d − 1 1 ≥ c + d − e 2 d − 9 if 9 ∈ Σ , 10 ∈ Σ , 12 / ∈ Σ s m = c + d − 1 3 ≥ c + d − e in f act 10 ∈ Σ = ⇒ e ≥ 14 2 d − 10 i f { 9 , 10 , 12 } ⊆ Σ s m ≥ c + d − e . Note tha t in this ca se we hav e d + d ′ − ℓ − t + 1 = 2 d − 2 ℓ − t + 1 = 2 d − 10 a nd this bo und is achiev ed if { 9 , 10 , 12 } ⊆ Σ (with e ≥ 16). See, e.g. S = { 0 , 16 , 34 , 36 , 37 , 39 , 4 0 e s , 41 , 42 , 43 d ′ , 45 , 46 d , 50 c →} . If t ≥ 6 we hav e e s ≤ 2 d ′ − d a nd we consider the following s ubcases . If t = 6, then e s = 2 d ′ − d = s m by (3.4.4). If t ≥ 7 and 5 ∈ Σ, one has [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S and e s < 2 d ′ − d , hence s m = e s + d , by (3.1 1 ). If t ≥ 7 and 5 / ∈ Σ, we know that s m ≤ 2 d ′ by (3.4.5); one can compute directly that ν (2 d ′ ) < ν (2 d ′ + 1) (see T able 3.3 ( c )) and that ν (2 d − 7) > ν (2 d − 6), hence s m = 2 d − 7 = 2 d ′ − 1 . Since d − 6 = d ′ − ℓ ∈ S, w e get e ≥ 2 ℓ + 1 + 6 = 13 (2.6.3 ), and so s m ≥ c + d − e + 2 . Case B. W e hav e:                   C ase t ≤ 3 : s m = e s + d C ase t ≥ 4 , d − 5 / ∈ S : s m = 2 d − 6 C ase t ≥ 4 , d − 5 ∈ S, d − 4 ∈ S : if t ∈ { 4 , 5 } , s m ∈ [2 d − 9 , 2 d − 6] if t ≥ 6 , s m = e s + d C ase t ≥ 5 , d − 5 ∈ S, d − 4 / ∈ S : if t ∈ { 5 , 6 , 8 } , s m = e s + d if t = 7 , s m ∈ { 2 d − 8 , 2 d − 11 } if t ≥ 9 , d − 7 / ∈ S, s m = 2 d − 8 if t = 9 , d − 7 ∈ S, s m ∈ { 2 d − 10 , 2 d − 11 , 2 d − 13 } if t ≥ 10 , d − 7 ∈ S, s m = e s + d. Pro of. S = { 0 , e, ..., d − 3 , ∗ ∗ , d, ∗ ∗ ∗ , c = d + 4 →} , e ≥ t + 4 . As in case A we can see that s m = e s + d if t ≤ 3 and s m < e s + d for t = 4. Suppos e t ≥ 4. Then { 0 , 3 , 6 } ⊆ Σ. W e deduce the statement by means of the following table:     2 d − 3 (0 , 3 ) 2 d − 4 (0 , 4 ) ( c, 8) 2 d − 5 (0 , 5 ) ( c, 9) 2 d − 6 (0 , 6 )(3 , 3) ( c, 10) . If 4 ∈ Σ , 5 ∈ Σ, w e have [ d ′ − ℓ , d ′ ] ∩ I N ⊆ S and by applying (3.1 1 ) we get s m ≥ c + d − e . More precisely , o ne can e asily verify that for t ∈ { 4 , 5 } we hav e 2 d − 9 ≤ s m ≤ 2 d − 6, for t ≥ 6 we hav e e s ≤ 2 d ′ − d , then by (2.4.1) and b y (3.8.2) we get s m = e s + d . If 5 / ∈ Σ, w e hav e s m = 2 d − 6 . 15 If 4 / ∈ Σ , 5 ∈ Σ: we hav e s m =  2 d − 5 ⇐ ⇒ t = 5 2 d − 6 ⇐ ⇒ t = 6 (8 ∈ Σ , 9 / ∈ Σ); the remaining cases to cons ider sa tisfy { 0 , 3 , 5 , 6 , 8 , 9 } ⊆ Σ , 4 / ∈ Σ , with t ≥ 7, s m < 2 d − 6 :             2 d − 7 (0 , 7) ( c, 11 ) 2 d − 8 (0 , 8)(3 , 5) ( c, 12 ) s m = 2 d − 8 ⇐ ⇒  7 / ∈ Σ or 7 ∈ Σ , 11 / ∈ Σ ( = ⇒ 7 ≤ t ≤ 8) otherw ise 7 , 11 ∈ Σ : { 0 , 3 , 5 , 6 , 7 , 8 , 9 , 1 1 } ⊆ Σ , 4 / ∈ Σ 2 d − 9 (0 , 9)(3 , 6) ( c, 13 ) 2 d − 10 (0 , 10)(3 , 7)(5 , 5) ( c, 14) s m = 2 d − 1 0 ⇐ ⇒ 1 0 ∈ Σ , 13 / ∈ Σ( = ⇒ 9 ≤ t ≤ 10) otherw ise  either ( α ) 10 , 13 ∈ Σ or ( β ) 10 / ∈ Σ ( t = 7) Case ( α ): { 0 , 3 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 13 } ⊆ Σ , 4 / ∈ Σ ( = ⇒ t ≥ 9):                     2 d − 10 (0 , 10)(3 , 7)(5 , 5 ) ( c, 14 ) 2 d − 11 (0 , 11)(3 , 8)(5 , 6 ) ( c, 15 ) s m = 2 d − 1 1 ⇐ ⇒ 14 / ∈ Σ ( = ⇒ t = 9 if 12 / ∈ Σ , t = 11 i f 12 ∈ Σ) otherw ise 1 4 ∈ Σ : { 0 , 3 , 5 ← → 11 , 13 } ⊆ Σ , 4 / ∈ Σ 2 d − 12 (0 , 12) (3 , 9)(5 , 7)(6 , 6) ( c, 16 ) s m = 2 d − 1 2 ⇐ ⇒ t = 12 otherw ise  either ( α 1) 12 / ∈ Σ( ⇐ ⇒ t = 9) or ( α 2) 12 ∈ Σ , 1 5 ∈ Σ 2 d − 13 (0 , 13)(3 , 10)(5 , 8)(6 , 7) ( c, 17) s m = 2 d − 1 3  in case ( α 1) in case ( α 2) ⇐ ⇒ t = 13 otherw ise 1 6 ∈ Σ { 0 , 3 , 5 ← → 16 } ⊆ Σ 4 / ∈ Σ : 2 d − 14 (0 , 14)(3 , 11)(5 , 9)(6 , 8)(7 , 7 ) ( c, 18) s m = 2 d − 1 4 ⇐ ⇒ t = 14 otherw ise 1 7 ∈ Σ , { 0 , 3 , 5 ← → 1 5 , 16 , 17 } ⊆ Σ , 4 / ∈ Σ ... Clearly in cases ( α 2), for each t ≥ 13 we get s m = e s + d. Case ( β ): { 0 , 3 , 5 , 6 , 7 , 8 , 9 , 1 1 , } ⊆ Σ , 4 , 10 / ∈ Σ ( t = 7):  2 d − 11 (0 , 11 )(3 , 8)(5 , 6) ( c, 15) s m = 2 d − 1 1 . Case C . W e hav e: s m =           if t = 3 : 2 d − 4 if d − 4 ∈ S, d − 7 / ∈ S 2 d − 5 if ( { d − 4 , d − 7 } ⊆ S, d − 8 / ∈ S ) or ( d − 4 / ∈ S ) 2 d − 7 if { d − 4 , d − 7 , d − 8 } ⊆ S if t ≥ 4 : e s + d if d − 4 ∈ S 2 d − 5 if d − 4 / ∈ S. Pro of. S = { 0 , e, ..., d − 3 , d − 2 , ∗ , d, ∗ ∗ ∗ , c = d + 4 →} , d ′ = d − 2. As above we se e that t ≥ 3, d − 3 , d − 5 ∈ S (i.e. { 0 , 2 , 3 , 5 } ⊆ Σ), e ≥ 7. Co nsider the table:             2 d − 2 (0 , 2 ) 2 d − 3 (0 , 3 ) ( c, 7 ) 2 d − 4 (0 , 4 )(2 , 2) ( c, 8) s m = 2 d − 4 if 4 ∈ Σ , 7 / ∈ Σ 2 d − 5 (0 , 5 )(2 , 3) ( c, 9) s m = 2 d − 5 if  4 / ∈ Σ or 4 , 7 ∈ Σ , 8 / ∈ Σ . if { 4 , 7 , 8 } ⊆ Σ then e ≥ 1 2 ( d − 8 + e ≥ c ) 2 d − 6 (0 , 6 ) (2 , 4)(3 , 3) ( c, 10) 2 d − 7 (0 , 7 )(2 , 5)(3 , 4) ( c, 11) If t = 3, then 6 / ∈ Σ: we get s m = 2 d − 7. 16 If t ≥ 4 and 4 ∈ Σ, we hav e [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S and e s ≤ 2 d ′ − d . Then s m = e s + d ≥ c + d − e by (3.11). If 4 / ∈ S , by the ab o ve table w e deduce that s m = 2 d − 5 . ⋄ 4.4 Semigroups with CM type τ ≤ 7 . As a consequence of the ab ov e results we obtain low er b ounds o r the exa ct v alue of s m for semig roups with s ma ll Cohen-Maculay type. First, in the next le mma we collect well-kno wn o r easy relations among the CM type τ of S and the other inv aria n ts. Lemma 4. 8 L et τ b e the CM-typ e of the semigr oup S as in (2.1) . (1) # H + ℓ ≤ τ ≤ e − 1 (2) Assume τ = ℓ , t hen H = ∅ and t he fol lowing c onditions ar e e quivalent: ( a ) ℓ = e − 1 ( b ) τ = ℓ, c ′ = d. ( c ) d = c − e . ( d ) S = { 0 , e, 2 e, ..., k e →} . (3) If c ′ > c − e , then τ ≥ ℓ + 1 and τ = ℓ + 1 = ⇒ H = { c ′ − 1 } . (4) Assume e s ≤ d ′ and τ = ℓ + 1 . Then  e ∈ { 2 ℓ + t − 1 , 2 ℓ + t } , if e s = d ′ e = 2 ℓ + t, if e s < d ′ . Pro of. (1) Clearly every gap h ≥ c − e belongs to S (1) \ S , in particular { d + 1 , ..., d + ℓ } ∪ H ⊆ S (1) \ S . The inequality τ ≤ e − 1 . is well-kno wn. (2) (3) are almos t immediate. (4). W e have # H ≤ 1 by (1), e s − ℓ < e s ≤ d ′ < c ′ − 1. If # H = 0, then c ′ = c − e (4.1) and so e = 2 ℓ + t (2.7.2). If # H = 1 = ⇒ e s − ℓ / ∈ H and d ′ = c ′ − 2: it follows e ≤ 2 ℓ + t by (2.6.4). No w apply (2 .6 .3) a nd (2.8.2 c ): if e s < d ′ , then e s + 1 ∈ S , hence e ≥ 2 ℓ + t and so e = 2 ℓ + t . F urther e s = d ′ = ⇒ c ′ ≥ c − e + ℓ = ⇒ e ≥ c − c ′ + ℓ = d + 2 ℓ − d ′ − 1 = 2 ℓ + t − 1. Example 4.9 (1 ) W e rec all that in g eneral c ′ = c − e do es no t imply τ = ℓ . F or instance, let S = { 0 , 1 0 e = d ′ , 16 c − e , 17 , 18 , 19 , 20 , 21 , 22 , 23 , 24 d , 26 c →} . Then τ = 5 6 = ℓ . (2) Analogously H = { c ′ − 1 } do es not imply τ = ℓ + 1: S = { 0 , 1 0 e , 16 c − e , 17 , 18 , 19 , 20 d ′ , 22 c ′ = d , 26 c →} has ℓ = 3 , τ = 5. (3) In (4.8 .4) the conditions e = 2 ℓ + t and e s < d ′ are not equiv alent, further when e s = d ′ bo th the cases with τ = ℓ + 1 , e s = d ′ are po ssible. F or instance { 0 , 9 e = c − e , 10 , 11 d ′ , 13 c ′ , 14 d , 18 c →} has t = ℓ = 3 , e = 2 ℓ + t, e s = d ′ , τ = ℓ + 1; { 0 , 8 e , 9 d ′ , 11 c ′ , 12 d , 16 c →} has t = ℓ = 3 , e = 2 ℓ + t − 1 , e s = d ′ , τ = ℓ + 1. (4) There exist semigro ups with H = { c ′ − 1 } , e s < d ′ , τ = ℓ + 1 as in (4.8 .4): S = { 0 ∗ ... ∗ 11 e ∗ ∗ ∗ 15 d − e ∗ ∗ ∗ 19 , 20 , 21 , 22 , 23 d ′ ∗ 25 c ′ 26 d ∗ ∗ ∗ 30 c →} , has ℓ = 3 , t = 5 , e = 2 ℓ + t, τ = 4. Now we deduce b ounds for s m when τ ≤ 7 . Prop osition 4. 10 F or e ach τ ≤ 7 we have s m ≥ c + d − e . Mor e pr e cisely when e s < c ′ + d ′ − d we have the fol lowing r esults. (1) τ ≤ 3. W e hav e: s m =  2 d − 4 if S non − acute , τ = t = 3 ( ℓ = 2) e s + d in the other ca ses [6, 5.9] [5, 4 .13] 17 (2) τ = 4. W e hav e ℓ ≤ 4 a nd the following s ubcases. If ℓ = 4(= τ ), then H = ∅ (4.8.1 ), therefor e S is acute with s m = e s + d (2.3.4). If ℓ ≤ 3 we are done by the previous (4.5), (4.7), (2.3.4) and (4.1) (rec a ll ℓ = 1 = ⇒ S is acute). More precisely we get: s m =         2 d − 4 if  ℓ = 2 an d e ither ( t = 3 , d − 6 / ∈ S ) or ( t ≥ 5 , d − 3 / ∈ S ) ℓ = t = 3 , e = 9 , c ′ = d, d ′ = c ′ − 2 , d − 4 ∈ S 2 d − 6 if  ℓ = 2 , t = 3 , d − 6 ∈ S ℓ = 3 , t = 5 , e = 11 , c ′ = d − 1 , d ′ = c ′ − 2 e s + d in the other cas e s. (3) τ = 5. As ab ove we know s m in every case : ( a ) If ℓ ≤ 3 we can deduce s m by (4.5), (4.7 ), ( b ) If 4 ≤ ℓ ≤ 5, then we are done by (4 .1), since # H ≤ 1. (4) τ = 6. W e can calcula te s m as follows: ( a ) If ℓ ≤ 3 as in (3. a ). ( b ) If 5 ≤ ℓ ≤ 6, then we are done by (4 .1), since # H ≤ 1. ( c ) If ℓ = 4 and H ⊆ { c ′ − 2 , c ′ − 1 } , we have the v alue of s m by (4.1) a nd (4.1.2). ( d ) If ℓ = 4 a nd H = { d ′ − k , c ′ − 1 } , with k ≥ 1, then d ′ = c ′ − 2, a nd the b ounds for s m are given in (3.1 1) if k > ℓ , a nd in (4.4 ) if ℓ > k ≥ 1 (in fact c ′ − d ′ = 2 ≤ k + 1). (5) τ = 7. W e hav e ℓ ≤ 7 and the following subc a ses. ( a ) If ℓ ≤ 3 , then s m is known a s in (3. a ). ( b ) If 6 ≤ ℓ ≤ 7 then # H ≤ 1 a nd w e ar e done by (4 .1). ( c ) If ℓ = 5, then # H ≤ 2 and we ar e done by (4.1), (4.4). ( d ) If ℓ = 4, then # H ≤ 3 and we ar e done if # H ≤ 2 by (4.1), (4.4). If ℓ = 4 , # H = 3 , consider the following subc a ses ( i ) H = [ d ′ + 1 , c ′ − 1] ∩ I N: then s m is given in (4.1). ( ii ) H = { d ′ − k , c ′ − 2 , c ′ − 1 } , k ≥ 2. If k < ℓ , then s m is given in (4.3.3). If k ≥ ℓ , a pply (3.11). ( iii ) H = { d ′ − 1 , c ′ − 2 , c ′ − 1 } : this c ase cannot exist. In fact s ince S = { e, ...., d ′ − 2 , ∗ , d ′ , ∗ , ∗ , c ′ , ..., d, c →} and b y the ass umption e s < d ′ , we obta in c ′ − ℓ ∈ S and c ′ − ℓ = c ′ − 4 = d ′ − 1 / ∈ S , imp ossible. ( iv ) H = { d ′ − j, d ′ − k , c ′ − 1 } , j > k ≥ 1, hence S = { e, ..., d ′ − 2 , ..., d ′ , ∗ , c ′ , ..., d ( ≤ c ′ + 2) , ..., d + 5 →} , with 2 = c ′ − d ′ ≤ k + 1. As in the pro of o f (4.3.2) for s := d ′ − k + c ′ − 1 we ha ve γ ( s ) = − 1, s + 1 − c ′ / ∈ S , s − d ∈ S ⇐ ⇒ s − d 6 = d ′ − j . Hence (T able 3 .3 ( b )) s m ≥ s if s − d 6 = d ′ − j , i.e., d ′ − k + c ′ − 1 − d 6 = d ′ − j , i.e., d − c ′ 6 = j − k − 1. F o ur s ub cases:       j = k + 1 = ⇒ s m ≥ s if d 6 = c ′ j = k + 2 = ⇒ s m ≥ s if d 6 = c ′ + 1 j = k + 3 = ⇒ s m ≥ s if d 6 = c ′ + 2 j ≥ k + 4 = ⇒ s m ≥ s (since d − c ′ ≤ ℓ − 2 = 2 , j − k − 1 ≥ 3) . In the r emaining three situations we can see that [ d ′ − ℓ , d ′ ] ∩ I N ⊆ S , therefore s m is given by (3.1 1): - If j = k + 1 a nd d = c ′ , since e s < d ′ , ℓ = 4 we g et { d ′ − 4 , d − 4 = d ′ − 2 , d ′ } ⊆ S . Since there are tw o consecutive ho les, then k ≥ 5. It follows [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S . 18 - If j = k + 2, and d = c ′ + 1 , we hav e c ′ − 4 = d ′ − 2 , d ′ − 1 = d − 4 and e s < d ′ − 1 (4.3.1) . Therefor e S = { e, ..., d ′ − 5 , d ′ − 4 , d ′ − 3 , ∗ , d ′ − 2 , d ′ − 1 , d ′ , ∗ , d ′ + 2 = c ′ , d ′ + 3 = d, d + 5 →} . W e deduce [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S . - If j = k + 3, and d = c ′ + 2, ana logously we deduce [ d ′ − ℓ, d ′ ] ∩ I N ⊆ S . ⋄ 4.5 The v alue of s m for semigroups of m ultipicity e ≤ 8 . Corollary 4. 11 F or e ach semigr oup S of mult iplicity e ≤ 8 we have s m ≥ c + d − e . Pro of. Since τ ≤ e − 1 the result follows by (4.10). ⋄ 4.6 Almost arithmetic sequences and Suzuki curv es. Recall that a semigr oup S is gener ate d by an almost arithmetic se quenc e (shortly AAS) if S = < m 0 , m 1 , ..., m p +1 , n > with m 0 ≥ 2 , m i = m 0 + ρ i, ∀ i = 1 , ..., p + 1 , and GC D ( ρ, m 0 , n ) = 1 . (The embedding dimension of S is embdim S = p + 2). Prop osition 4. 12 L et S b e an AA S semigr oup of emb edding dime n sion µ ; then τ ≤ 2( µ − 2) . Pro of. It is a conse q uence of [8, 3.3 - 4 .6 - 4.7 - 5.6 - 5 .7 - 5 .8 - 5 .9 ] after s uita ble calcula tions. Corollary 4. 13 If S is an AAS semigr oup with emb dim S ≤ 5 t hen s m ≥ c + d − e . Pro of. It is an immediate co nsequence of (4.10) and (4.12). ⋄ As a nother co rollary w e obtain the v alue of s m for the W eierstras s semigro up of a S uz uk i cur v e , that is a plane curve C defined b y the equation y b − y = x a ( x b − x ) , with a = 2 n , b = 2 2 n +1 , n > 0 . Some applications of these cur v es to AG co des ca n b e found for ex ample in [4]. Prop osition 4. 14 If S is the Weierstr ass s emigr oup of a S uzuki Curve, then S is symmetric, ther e- for e s m = e s + d . Pro of. In [4, Lemma 3.1] is prov ed that the W e ierstrass semigroup S at a rational place of the function field of C is gener ated a s fo llows: S = < b, b + a, b + b a , 1 + b + b a > W e hav e b = 2 a 2 , with a = 2 n , hence S = < 2 a 2 , 2 a 2 + a, 2 a 2 + 2 a, 2 a 2 + 2 a + 1 > Then co ns ider the semigro up S = < 2 a 2 , 2 a 2 + a, 2 a 2 + 2 a, 2 a 2 + 2 a + 1 >, a ∈ I N . If a = 1 , then S = < 2 , 3 > . If a > 1 , then S is g enerated b y an almo st ar ithmetic s equence, a nd embdim ( S ) = 4; in fact S = < m 0 , m 1 , m 2 , n >, with m 0 = 2 a 2 , m 1 = m 0 + a, m 2 = m 1 + a, n = m 2 + 1 . Since S is AAS , we shall compute the Ap ery set A by means of the algo rithm descr ibed in [8]: let p = embdim ( S ) − 2 = 2 a nd for ea ch t ∈ I N, let q t , r t be the (uniqe) in tege rs such that t = pq t + r t (= 2 q t + r t ) , q t ∈ Z Z , r t ∈ { 1 , 2 } , let g t = q t m 2 + m r t , i.e., g t =  ( q t + 1) m 2 if r t = 2 q t m 2 + m 1 if r t = 1 (in par ticular g 0 = 0). 19 Then by [8] the Aper y se t A of S is: { g t + hn | 0 ≤ t ≤ 2 a − 1 , 0 ≤ h ≤ a − 1 } : therefore the elements of the Ap ery set ar e the 2 a 2 ent r ies o f the following matrix             0 g 1 g 2 g 3 . . . g 2 a − 1 || || || . . . || m 1 m 2 m 1 + m 2 . . . ( a − 1) m 2 + m 1 n g 1 + n g 2 + n . . . . . 2 n g 1 + 2 n g 2 + 2 n . . . . . . . . . . . . . . . . . . . . . ( a − 1) n g 1 + ( a − 1) n g 2 + ( a − 1) n . . . . g 2 a − 1 + ( a − 1) n             Recall that a semigr oup S of multiplicit y e a nd Ap ery set A is sy mmetr ic ⇐ ⇒ for each s i ∈ A , 0 < s i 6 = s e := max A , ther e exists s j ∈ A such that s i + s j = s e . In our case c this condition is satisfied: in fact s i =  αm 2 + h n, h ≤ a − 1 , α ≥ 1 (1) or α m 2 + m 1 + h n 0 ≤ α, h ≤ a − 1 (2) further s e = ( a − 1)( m 2 + n ) + m 1 , and so s e − s i =  ( a − 1 − α ) m 2 + m 1 + ( a − 1 − h ) n ∈ A (1) or ( a − 1 − α ) m 2 + ( a − 1 − h ) n ∈ A (2) Since a semig roup S is symmetric if and only if its CM-type is one, then s m = e s + d by (4.1 0.1). ⋄ References [1] M. Bras- Amoros, “Acute Semigroups, the Or der Bound on the Minimum Distance, and the F eng- Rao Improv ements”, IEEE T r ansactions on Information The ory , vol. 5 0, no. 6, pp.1282-1 289, (2004). [2] G.L. F eng, T.R.N. 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