Frobenius problem for semigroups S(d1,d2,d3) - Springer Link

Funct. Anal. Other Math. (2006) 1: 119–157 DOI 10.1007/s11853-007-0009-5

FAOM

Frobenius problem for semigroups S(d1 , d2 , d3 ) Leonid G. Fel

Received: 9 July 2006 / Revised: 22 October 2006 / Accepted: 3 November 2006 / Published online: 20 November 2007 © PHASIS GmbH and Springer-Verlag 2007

Abstract We find the matrix representation of the set Δ(d3 ), where d3 = (d1 , d2 , d3 ), of integers that are unrepresentable by d1 , d2 , d3 and develop a diagrammatic procedure for calculating the generating function Φ(d3 ; z) for the set Δ(d3 ). We find the Frobenius number F (d3 ), the genus G(d3 ), and the Hilbert series H (d3 ; z) of a graded subring for nonsymmetric and symmetric semigroups S(d3 ) and enhance the lower bounds of F (d3 ) for symmetric and nonsymmetric semigroups. Keywords Restricted partitions · Frobenius problem · Nonsymmetric and symmetric semigroups · Hilbert series of a graded subring Mathematics Subject Classification (2000) Primary 11P8 · Secondary 11N56 · 20F55

1 Introduction Let S(d1 , . . . , dm ) ⊂ N be the numerical subsemigroup generated by a set of integers {d1 , . . . , dm } such that 1 < d 1 < · · · < dm ,

gcd(d1 , . . . , dm ) = 1.

(1)

The set {d1 , . . . , dm } is said to be minimal if there are no nonnegative integers bi,j for which the linear dependence di =

m

bi,j dj ,

bi,j ∈ {0, 1, . . . } for any i ≤ m,

(2)

j =i

holds. It is classically known that d1 ≥ m [1]. For brevity, we let dm denote the tuple (d1 , . . . , dm ), where m is the dimension of dm . Henceforth, dm is a minimal generating set L. G. Fel () Department of Civil Engineering, Technion University, Haifa 3200, Israel e-mail: [email protected]

120

Leonid G. Fel

of S(dm ). The conductor c(dm ) of S(dm ) is defined as c(dm ) := min{s ∈ S(dm )|s + N ∪ {0} ⊂ S(dm )}. The genus G(dm ) of S(dm ) is defined as the cardinality (#) of its complement Δ in N, i.e., Δ(dm ) = N \ S(dm ) and G(dm ) := #Δ(dm ).

(3)

For the reason explained in Sect. 5, it is worthwhile to introduce the generating function Φ(dm ; z) for the set Δ(dm ) of unrepresentable integers in accordance with [2] Φ(dm ; z) = zs , (4) s∈Δ(dm )

Let dm and dn be two tuples of different dimensions, m > n. We define a relation dn ≺ dm if dn is an initial segment of dm as a word in d1 , . . . , dm . If the tuple dn is (d1 , . . . , dn ) for any 2 < n < m, then d2 ≺ · · · ≺ dn ≺ · · · ≺ dm . This implies an ordering in the three sequences Δ(dm ) ⊂ · · · ⊂ Δ(d2 ), G(dm ) < · · · < G(d2 ),

(5)

c(dm ) ≤ · · · ≤ c(d2 ). The semigroup ring k[X1 , . . . , Xm ] over a field k of characteristic 0 associated with S(dm ) is a polynomial subring graded by deg Xi = di , 1 ≤ i ≤ m, and generated by all monomials zdi . The Hilbert series H (dm ; z) of a graded subring k[zd1 , . . . , zdm ] is defined as [2] H (dm ; z) =

Q(dm ; z) zs = m , dj j =1 (1 − z ) s∈S(dm )

(6)

where Q(dm ; z) is a polynomial in z. The number F (dm ) := −1 + c(dm )

(7)

is called the Frobenius number in honor of G. Frobenius, who in his lectures (according to [3]) repeatedly raised the question of determining (or bounding) F (dm ). In fact, all three entities, F (dm ), G(dm ), and H (dm ; z) originate from the same semigroup S(dm ) and are strongly related algebraically (see Sect. 5). For this reason, we call determining F (dm ), G(dm ), and H (dm ; z) the m-dimensional (mD) Frobenius problem. Finding F (dm ) in number theory is also known as the “postage stamp problem” or the “coin exchange problem.” Let R = k[X1 , . . . , Xm ] be the ring of polynomials over a field k and π : R −→ k[S(dm )] be the projection induced by π(Xi ) = zdi . Then k[S(dm )] has a representation k[X1 , . . . , Xm ]/Im , where Im is the kernel of the map π . The semigroup S(dm ) is said to be symmetric iff F (dm ) − s ∈ S(dm ) holds for all s ∈ S(dm ). This kind of semigroup is very important because of Kunz’s theorem [4], which asserts that k[S(dm )] is a Gorenstein ring iff S(dm ) is symmetric. It is classically known that the situation is even simpler in small dimensions m = 2, 3. For every d2 , S(d2 ) is a symmetric semigroup, and k[S(d2 )] is a complete intersection [2]. For m = 3, Herzog [5] proved that k[S(d3 )] is a complete intersection iff S(d3 ) is symmetric.

Frobenius problem for semigroups S(d1 , d2 , d3 )

121

For larger m, the generic semigroup S(dm ) is mostly nonsymmetric, for example, S(d3 ) minimally generated by three pairwise relatively prime elements di is such a semigroup [6]. Concerning the Frobenius numbers, the theorem of Curtis [7] asserts that for m = 3, there is no nonzero polynomial P ∈ C(Y1 , Y2 , Y3 , Z) such that P (d1 , d2 , d3 , F (d3 )) = 0 for all minimal sets (d1 , d2 , d3 ), where d1 and d2 are primes not dividing d3 . In other words, F (d3 ) cannot be determined for all minimal sets (d1 , d2 , d3 ) by any set of closed formulas that could be reduced to a finite set of polynomials.1 Concerning the Hilbert series, there is no way to write the rational function H (dm ; z) for any m ≥ 4 such that its polynomial numerator Q(dm ; z) has a bounded number of nonzero terms for all choices of d1 , . . . , dm [10]. The semigroup S(d3 ) is the first nontrivial and the most elaborated case. Our main results are the expressions for the Frobenius number F (d3 ), the genus G(d3 ), and the numerator Q(d3 ; z) of the Hilbert series for both symmetric and nonsymmetric semigroups S(d3 ). To present them, we introduce auxiliary notions. Following Johnson [11], we define the first minimal relation R1 (d3 ) for a given d3 = (d1 , d2 , d3 ) as R1 (d3 ) :

a11 d1 = a12 d2 + a13 d3 ,

a22 d2 = a21 d1 + a23 d3 ,

a33 d3 = a31 d1 + a32 d2 , where

(8)

a11 = min v11 | v11 ≥ 2, v11 d1 = v12 d2 + v13 d3 , v12 , v13 ∈ N ∪ {0} , a22 = min v22 | v22 ≥ 2, v22 d2 = v21 d1 + v23 d3 , v21 , v23 ∈ N ∪ {0} , a33 = min v33 | v33 ≥ 2, v33 d3 = v31 d1 + v32 d2 , v31 , v32 ∈ N ∪ {0} .

(9)

The uniquely defined values of vij , i = j , that give aii are denoted by aij , i = j . We note that because the set (d1 , d2 , d3 ) is minimal, the elements aij , i, j ≤ 3, satisfy gcd(a11 , a12 , a13 ) = 1, gcd(a21 , a22 , a23 ) = 1,

(10)

gcd(a31 , a32 , a33 ) = 1. The procedure defined in (9) completely determines the elements aij , i, j ≤ 3, of R1 (d3 ) as the functions aij = aij (d1 , d2 , d3 ). For F (d3 ) , G(d3 ), and Q(d3 ; z), we obtain the formulas 1 F (d3 ) = [a, d + J (d3 )] − di , 2 i=1 3

G(d3 ) =

a, d =

3 3 1 aii − di , 1 + a, d − 2 i=1 i=1

3

aii di ,

(11)

i=1

(12)

1 The words “all” and “polynomial” are essential here because there exist infinitely many triples d , d , d of 1 2 3

primes in an arithmetic progression [8] constituting a minimal set and, according to Roberts [9], the Frobenius number associated with them can be represented in the closed, but not polynomial, formula F (d, d + p, d + 2p) = d (d − 2)/2 + (d − 1)p, gcd(d, p) = 1. We use the standard notation a for the integer part of a real number a.

122

Leonid G. Fel

Q(d3 ; z) = 1 −

3

zaii di + z1/2[a,d −J (d

3 )]

3

+ z1/2[a,d +J (d )] ,

(13)

i=1

where J (d3 ) is the positive integer

3

3 aii ajj di dj + 4d1 d2 d3 . J (d ) = a, d 2 − 4

(14)

i>j

Formula (13) for the numerator Q(d3 ; z) of the Hilbert series of the nonsymmetric semigroup S(d3 ) is new and was not obtained earlier. Formula (12) fully agrees with the genus of generic monomial space curves found by Kraft [12], and formula (11) is reducible to the known expressions for symmetric and nonsymmetric semigroups S(d3 ) obtained by Herzog [5] and Fröberg [13].

2 The 3D Frobenius problem: A brief review We start with the 2D Frobenius problem for which F (d2 ), G(d2 ), and Q(d2 ; z) were already known to Sylvester [14]: F (d2 ) = d1 d2 − d1 − d2 , 1 G(d2 ) = (d1 − 1)(d2 − 1), 2

(15)

Q(d2 ; z) = 1 − zd1 d2 . We recall some basic results on the Frobenius problem for the semigroup S(d3 ) following [5], [12], [13], and [15]. Let S(d3 ) be a nonsymmetric semigroup with the first minimal relation R1 (d3 ) defined by (8) and (9). Such relations always exist because the Frobenius number F (d2 ) is finite. We represent (8) as a matrix equation ⎛ ⎞ ⎛ ⎞ d1 0 ⎝ ⎠ ⎝ 0⎠ , d = A 2 3 d3 0

⎞ a11 −a12 −a13 ⎝ a22 −a23 ⎠ , A 3 = −a21 −a31 −a32 a33 ⎛

(16)

where gcd(a11 , a12 , a13 ) = 1,

gcd(a21 , a22 , a23 ) = 1,

gcd(a31 , a32 , a33 ) = 1,

and establish the standard forms of the matrix A 3 satisfying (9) and (16). 2.1 The 3D nonsymmetric semigroups First, let all off-diagonal entries of A 3 be negative integers, aij ∈ {1, 2, . . . }, i = j , i.e., omitting zero. Then, as Johnson showed [11], this necessarily leads to


123

a11 = a21 + a31 , a22 = a12 + a32 , a33 = a13 + a23 , a11 −a13 a22 −a23 det det = d1 , = d2 , −a32 a33 −a31 a33 a11 −a12 det = d3 . −a21 a22

(17)

(18)

Ordering (1) of the integers, d1 < d2 < d3 , imposes additional constraints on the elements aij , a11 > a12 + a13 ,

a22 > a23 ,

a33 < a31 + a32 .

(19)

denote A We let 3 satisfying (17) and (18) and call it the standard form for the nonsymmetric semigroup S(d3 ). Formulas (8), (17), and (18) allow showing that at least one of the aii exceeds 2. The proof can be obtained by contradiction. Let all aii = 2. By (17), we then have aij = 1, i = j , or in accordance with (8), (n) A 3

2d1 = d2 + d3 ,

2d2 = d3 + d1 ,

2d3 = d1 + d2

−→

d1 = d2 = d3 ,

which violates the minimality of (d1 , d2 , d3 ). This implies the inequality a11 a22 a33 ≥ 12. The Frobenius number F (d3 ) for the nonsymmetric semigroup was first found in [5] (also see [13]) calculating only the largest degree of H (dm ; z) (without calculating the Hilbert series itself): F (d3 ) +

3

di = max{a11 d1 + a32 d2 ; a22 d2 + a31 d1 }

i=1

= max{a22 d2 + a13 d3 ; a33 d3 + a12 d2 } = max{a33 d3 + a21 d1 ; a11 d1 + a23 d3 }.

(20)

The genus G(d3 ) of the nonsymmetric semigroup was calculated in algebraic geometry [12]. Dealing with the singularity degrees of the monomial space curve whose corresponding semigroup is S(d3 ), Kraft [12] calculated G(d3 ) given by (12). Concerning the Hilbert series, partial progress was achieved by Székely and Wormald [10], who proved that Q(d3 ; z) consists of a limited number of terms (at most twelve) independent of the values of d1 , d2 , and d3 . This result was essentially refined recently by Denham [15], who gave an algorithm for computing the Hilbert series H (d3 ; z) of a graded of these subring k[S(d3 )] for nonsymmetric semigroups and established a universal property series: Q(d3 ; z) has exactly six terms with the first four of them being 1 − 3i=1 zaii di . 2.2 The 3D symmetric semigroups The number of independent entries aij in (16) can be reduced if at least one off-diagonal element of A 3 vanishes, for example, if a13 = 0 and therefore a11 d1 = a12 d2 . Because the last equality is minimal, the relations [5] follow from (8), a11 = a21 = ⎛

(s) A 3

lcm(d1 , d2 ) , d1

a11 −a22 = ⎝−a11 a22 −a31 −a32

⎞ 0 0 ⎠. a33

a12 = a22 =

lcm(d1 , d2 ) , d2

a23 = 0, (21)

124

Leonid G. Fel

(s) We call A the standard form for the symmetric semigroup S(d3 ). The Frobenius number 3 3 F (d ) is [5]

F (d3 ) = a11 d1 + a33 d3 −

3

di

i=1

= a22 d2 + a33 d3 −

3

di

i=1

= lcm(d1 , d2 ) + a33 d3 −

3

di .

(22)

i=1

The corresponding genus G(d3 ) can also be simplified (see formula (137) in Sect. 5). If S(d3 ) is a symmetric semigroup, then k[S(d3 )] is a complete intersection [5], and the Hilbert series Hs (d3 ; z) is [16] Hs (d3 ; z) =

(1 − za22 d2 )(1 − za33 d3 ) . (1 − zd1 )(1 − zd2 )(1 − zd3 )

(23)

3 Matrix representation of the set Δ(d2 ) In this section, we construct the matrix representation of the set Δ(d2 ) of integers t that are unrepresentable by d1 , d2 . We start with an important statement about matrix representation that dates back to Brauer [3] and results partly from his discussions with I. Schur.2 Lemma 1 [3] Let d1 and d2 be relatively prime positive integers. Then every positive integer s not divisible by d1 or by d2 is representable either in the form s = xd1 + yd2 , x > 0, y > 0, or in the form s = d1 d2 − pd1 − qd2 , p > 0, q > 0, p, q ∈ N. Definition 1 Let integers 2 < d1 < d2 be given. The function σ (p, q) is defined as σ (p, q) := d1 d2 − pd1 − qd2 .

(24)

The next lemma specifies the bounds of the values of p and q introduced in Lemma 1. Lemma 2 Let t be an integer and d2 > d1 , gcd(d1 , d2 ) = 1. Then t ∈ Δ(d2 ) iff t is uniquely representable as t = σ (p, q),

(25)

2 We quote from [3]: “The Theorems in §3–5 result partly from discussions of Schur and the author. It was

formerly intended to publish these results in a joint paper. I conform with Schur’s wishes that the publishing be not longer postponed and that I publish the paper alone.” The paper was submitted for publication on 25 November 1940, less than two months before Schur’s death, and was published two years later.


where

125

d2 1 ≤ p ≤ d2 − , d1 d2 d1 − 1 ≤ d2 − . d1

1 ≤ q ≤ d1 − 1, (26)

Proof In accordance with Lemma 1, every integer t that is unrepresentable by d1 , d2 , gcd(d1 , d2 ) = 1, is representable by (24) and (25). Therefore, the Frobenius number is F (d2 ) = σ (1, 1). Restrictions (26) follow from simple considerations, d2 σ (p, 1) > 0 → p ≤ d2 − , d1 d1 σ (1, q) > 0 → q ≤ d1 − = d1 − 1. d2 The representation of σ (p, q) by (24) is unique. A standard proof by contradiction that (24) is unique is to suppose that there are two such representations σ (p1 , q1 ) and σ (p2 , q2 ). We then have d2 (p1 − p2 )d1 = (q2 − q1 )d2 , 1 ≤ p1 , p 2 ≤ d 2 − , d1 1 ≤ q1 , q2 ≤ d 1 − 1

−→

|p1 − p2 | < d2 ,

|q2 − q1 | < d1 .

But this is impossible because d1 and d2 have no common factors. Finally, we prove the last inequality in (26). In the case d2 ≥ d1 + 2, we obtain d2 −

d2 d2 − 1 − (d1 − 1) − (d1 − 1) ≥ d2 − d1 d1 =

(d2 − d1 )(d1 − 1) −1 d1

≥2

2 d1 − 1 −1=1− > 0. d1 d1

In the case d2 = d1 + 1, we obtain d2 −

d2 d1 + 1 − (d1 − 1) = d1 + 1 − − (d1 − 1) d1 d1 1 = d1 − − (d1 − 1) = 0. d1

Combining both cases, we obtain the last inequality in (26), thus completing the proof of Lemma 2. We show that the integers σ (p, q) given by (25) and (26) exhaust all integers unrepresentable by d1 and d2 or, in other words, they give the genus G(d2 ) obtained by

126

Leonid G. Fel

Sylvester [14] and given in (15). Indeed, counting the number of integers σ (p, q) with properties (25) and (26) successively over the index set q = 1, . . . , d1 − 1, we obtain d1 −1

G(d2 ) =

d2 − q

q=1

d2 d1

1 d2 = q d1 q=d −1 1

=

=

1 2

d 1 −1 q=1

q

d 1 −1 d2 d2 + (d1 − q) d1 d1 q=1

d1 −1 d2 1 d2 (d2 − 1)(d1 − 1) , q + d2 − q = 2 q=1 d1 d1 2

which follows from the equalities d2 d2 d2 q = q + q , d1 d1 d1 d2 d2 d2 d2 = d2 − q − q = d2 − 1 − q , d2 − q d1 d1 d1 d1

(27) q < d1 .

In (27), we let {b} denote the fractional part of a real number b. Representation (25) of all integers σ (p, q) ∈ Δ(d1 , d2 ) is called the matrix representation of the set Δ(d1 , d2 ) and is denoted by M{Δ(d2 )} (see Fig. 1), (28) σ M{Δ(d2 )} = Δ(d2 ). The element σ (p, q) is the integer in the pth row and qth column of M{Δ(d2 )}, for example, σ (1, 1) = d1 d2 − d1 − d2 . Based on M{Δ(d2 )}, we introduce two sets that are important in the following sections. We call the totality of the lowest cells in each column of M{Δ(d2 )} the bottom layer of M{Δ(d2 )}, denoted by BLM {Δ(d2 )}. We also call the totality of the highest cells in each column of M{Δ(d2 )} the top layer of M{Δ(d2 )}, denoted by TLM {Δ(d2 )}. We establish the structure of the sets BLM {Δ(d2 )} and TLM {Δ(d2 )}. Lemma 3 For each number k, 1 ≤ k ≤ d1 − 1, there exists (p, q) ∈ BLM {Δ(d2 )} such that σ (p, q) = k. Proof We define a new function pb (q) as pb (q) = max{1 ≤ p | σ (p, q) > 0},

1 ≤ q ≤ d1 − 1,

(29)

where the subscript b indicates “bottom.” Then (pb (q), q) ∈ BLM {Δ(d2 )}. We establish another representation for the function pb (q). It follows from Definition 1 that d2 d2 σ (p, q) > 0 −→ p < d2 − q −→ pb (q) = d2 − q . d1 d1


127

Fig. 1 Typical matrix representation M{Δ(d2 )} of the set Δ(d1 , d2 ) for the case d1 < d2 < 3/2d1 . The bottom layer BLM {Δ(d2 )} is colored gray. The top layer TLM {Δ(d2 )} coincides with the highest row of the diagram.

Hence, according to (27), we obtain pb (q) = d2 − 1 − qd2 /d1 and further d2 d2 σ (pb (q), q) = d1 − qd2 + d1 q = d1 − d1 q . d1 d1

(30)

For 1 ≤ q ≤ d1 − 1, we have 1/d1 ≤ {qd2 /d1 } ≤ (d1 − 1)/d1 . Combining these bounds with representation (30), we can find the bounds for σ (pb (q), q), 1 ≤ σ (pb , q) ≤ d1 − 1.

(31)

Because representation (24) of σ (p, q) is unique, bounds (31) lead to the conclusion that BLM {Δ(d2 )} is occupied exclusively by the integers 1, . . . , d1 − 1 not necessarily in consec-

utive order. This proves Lemma 3.

The top layer TLM {Δ(d2 )} coincides with the highest row in M{Δ(d2 )}. Hence, in accordance with (28), we can finally write σ BLM {Δ(d2 )} = {1, . . . , d1 − 1}, (32) σ TLM {Δ(d2 )} = {σ (1, 1), . . . , σ (1, d1 − 1)}, where σ (1, 1) = F (d2 ) and σ (1, d1 − 1) = d2 − d1 . 4 Matrix representation of the set Δ(d3 ) In this section, we construct the set Δ(d3 ) from the set Δ(d2 ). We introduce new objects, the associated sets Ωdk3 (d2 ), and use them to define a matrix representation of the set Δ(d3 ).

128

Leonid G. Fel

Here k is an integer variable (see the discussion below). This construction paves the way for solving the 3D Frobenius problem for the nonsymmetric semigroup S(d3 ). According to (5) and (7), the Frobenius numbers F (d1 , d2 ) and F (d1 , d2 , d3 ) satisfy F (d1 , d2 , d3 ) ≤ F (d1 , d2 ). In the next lemma, we show that the equality does not occur. Lemma 4 Let d2 = (d1 , d2 ) be given and gcd(d1 , d2 ) = 1. Each integer d3 ∈ Δ(d2 ) yields a minimal generating set {d1 , d2 , d3 }, which generates the semigroup S(d3 ) such that F (d3 ) < F (d2 ).

(33)

Proof We choose an integer d3 that is unrepresentable by d1 and d2 , d3 ∈ Δ(d2 ). By Lemma 2, there exist pd3 and qd3 satisfying (26) such that d 3 = d1 d 2 − pd 3 d 1 − qd 3 d 2 .

(34)

The triple {d1 , d2 , d3 } represents the minimal set generating S(d3 ) in accordance with (2). We define the set Ωd13 (d2 ) of integers A1 in Δ(d2 ) representable by d1 , d2 , d3 as Ωd13 (d2 ) = {A1 | A1 = u1 d1 + u2 d2 + d3 , 0 ≤ u1 ≤ pd3 − 1, 0 ≤ u2 ≤ qd3 − 1}.

(35)

Because A1 depends on u1 and u2 , we write A1 = A1 (u1 , u2 ). It is clear that the set Ωd13 (d2 ) ∩ Δ(d3 ) = ∅

(36)

is empty because Δ(d3 ) consists of the integers unrepresentable by d1 , d2 , d3 . It follows from (24) that A1 (u1 , u2 ) = σ (pd3 − u1 , qd3 − u2 ). By expressions (15) and (34), we have A1 (pd3 − 1, qd3 − 1) = F (d2 ). In particular, F (d2 ) ∈ Ωd13 (d2 ).

(37)

def

Using F (d2 ) = max{t ∈ Δ(d2 )} by definition (7) and Ωd13 (d2 ) ⊂ Δ(d2 ) by definition (35) and F (d2 ) ∈ Ωd13 (d2 ) by (37), we can obtain the identity max{t ∈ Ωd13 (d2 )} = max{t ∈ Δ(d2 )}. By (5), we have Δ(d3 ) ⊂ Δ(d2 ), which results in the inequality max{t ∈ Δ(d3 )} ≤ max{t ∈ Δ(d2 )}. Combining the last inequality with (36), we obtain max{t ∈ Δ(d3 )} < max{t ∈ Δ(d2 )}, which proves Lemma 4.

4.1 The associated sets Ωdk3 (d2 ) To account for all integers that contribute to the construction of the set Δ(d3 ), we must extend the set Ωd13 (d2 ). First, following (8) and (9), we recall one of the first minimal relations R1 (d3 ) for a given d3 : a33 d3 = a31 d1 + a32 d2 , where a33 ≥ 2.


129

Fig. 2 Typical matrix representation M{Δ(d2 )} of the set Δ(d2 ). Matrix representations M{Ωdk (d2 )} of the 3

kd3 -associated sets Ωd1 (d2 ) and Ωd2 (d2 ) are shown by dashed lines. Their intersection (in gray) contains 3

3

the Frobenius number F (d2 ), which is marked by a black oval. The integers kd3 ∈ Δ(d3 ), 1 ≤ k < a33 , are shown by black boxes.

Definition 2 Let d3 ∈ Δ(d2 ) with the representation d3 = d1 d2 − pd3 d1 − qd3 d2 , where pd3 and qd3 satisfy (26). Let k be a positive integer, 1 ≤ k < a33 . The set Ωdk3 (d2 ) of integers Ak in Δ(d2 ) defined as Ωdk3 (d2 ) = {Ak | Ak = u1 d1 + u2 d2 + kd3 , 0 ≤ u1 ≤ pkd3 − 1, 0 ≤ u2 ≤ qkd3 − 1}

(38)

is called the kd3 -associated set. Because the integer Ak depends on u1 and u2 , we write Ak = Ak (u1 , u2 ). Taking u1 = pkd3 − 1 and u2 = qkd3 − 1, we obtain F (d2 ) = Ak (pkd3 − 1, qkd3 − 1) ∈ Ωdk3 (d2 ),

1 ≤ k < a33 .

(39)

j

It follows from (39) that the intersection of any two associated sets Ωd3 (d2 ) and Ωdk3 (d2 ), where 1 ≤ j, k < a33 , is a nonempty set. It follows from (38) that the matrix representation M{Ωdk3 (d2 )} of Ωdk3 (d2 ) is assigned by the rectangle [1, pkd3 ] × [1, qkd3 ] inside M{Δ(d2 )} with the cardinality #Ωdk3 (d2 ) = pkd3 qkd3 (see Fig. 2). Theorem 1 Let {d1 , d2 , d3 } be a minimal generating set of the semigroup S(d3 ) and σ be an integer σ ∈ Δ(d2 ) representable by d1 , d2 , d3 . Then there exists at least one k, 1 ≤ k < a33 , such that σ ∈ Ωdk3 (d2 ). Before proving Theorem 1, we present an auxiliary lemma based on the theory of the restricted partition function W (σ , dm ). We recall the main recursion relation [17] for

130

Leonid G. Fel

W (σ , dm ), which gives the number of partitions of an integer σ into positive integers d1 , . . . , dm each not greater than σ . Then W (σ , dm ) − W (σ − dm , dm ) = W (σ , dm−1 ),

dm−1 = {d1 , . . . , dm−1 }.

(40)

Lemma 5 Let {d1 , . . . , dm } be a minimal generating set of S(dm ) and Δ(dm ) be the corresponding set of unrepresentable integers. If σ ∈ Δ(dm ) and σ − dk > 0, 1 ≤ k ≤ m, then σ − dk ∈ Δ(dm ). Proof First, we suppose that k = m. If σ ∈ Δ(dm ), then W (σ , dm ) = W (σ , dm−1 ) = 0, and consequently W (σ − dm , dm ) = 0 because of (40). The latter implies σ − dm ∈ Δ(dm ). Now let k be arbitrary, 1 ≤ k < m. The validity of relation (40) is independent of the position of dk in the tuple dm . Therefore, resorting the tuple dm such that dk becomes the last in the list d1 , . . . , dm and repeating the above consideration, we prove Lemma 5. We note that Lemma 5 states a necessary but not sufficient condition for σ , i.e., the opposite implication σ − dk ∈ Δ(dm ) → σ ∈ Δ(dm ) is not true. We now prove Theorem 1. Proof of Theorem 1 Let R1 (d3 ) be the first minimal relation defined in (9). Then kd3 ∈ Δ(d2 ), 1 ≤ k < a33 . We consider an integer σ ∈ Δ(d2 ) that is representable by d1 , d2 , d3 , σ = α1 d1 + α2 d2 + α3 d3 ,

α1 , α2 , α3 ∈ N ∪ {0}.

(41)

It follows from (9) that α3 is not divisible by a33 ; otherwise, α3 σ = α1 d1 + α2 d2 + a33 d3 a33 α3 α3 = α1 + a31 d1 + α2 + a32 d2 ∈ Δ(d2 ), a33 a33 which contradicts our assumption σ ∈ Δ(d2 ). α We prove that σ ∈ Ωd33 (d2 ). For this, we must show that 0 ≤ α1 ≤ pα3 d3 − 1,

0 ≤ α2 ≤ qα3 d3 − 1.

(42)

We apply Lemma 5 with m = 3 α1 times with k = 1 and α2 times with k = 2 and obtain α3 d3 ∈ Δ(d2 ). We consider two cases here. First, letting 1 ≤ α3 < a33 and then substituting (34) in (41), we obtain σ = α1 d1 + α2 d2 + d1 d2 − pα3 d3 d1 − qα3 d3 d2 = d1 d2 − (pα3 d3 − α1 )d1 − (qα3 d3 − α2 )d2 . We apply Lemma 2 to the last representation of σ and obtain pα3 d3 − α1 ≥ 1,

qα3 d3 − α2 ≥ 1.

(43)

Combining (43) with α1 , α2 ∈ N ∪ {0} in (41), we conclude that (42) does hold. This leads α to σ ∈ Ωd33 (d2 ) in accordance with Definition 2.


131

We next consider α3 > a33 and represent α3 d3 as α3 α3 α3 d3 = a33 d3 + a33 d3 . a33 a33

(44)

Substituting a33 d3 from (8) in (44), we obtain α3 α3 α3 d3 = (a31 d1 + a32 d2 ) + a33 d3 . a33 a33 Further, substituting this result in (41), we obtain σ = ξ1 d1 + ξ2 d2 + ξ3 d3 , where

ξ1 = α1 + a31 ξ3 = a33

α3 a33

α3 , a33 < a33 ,

(45)

ξ2 = α2 + a32

α3 , a33

(46)

ξ1 , ξ2 , ξ3 ∈ N ∪ {0}.

Comparing (45) and (46) with (41), we conclude that the second case (α3 > a33 , ξ3 < a33 ) ξ reduces to the first case (α3 < a33 ) and σ ∈ Ωd33 (d2 ) therefore holds with ξ3 instead of α3 . This completes the proof of Theorem 1. Finally, we are ready to prove the main theorem in this section. Theorem 2 Let d3 = (d1 , d2 , d3 ) and the first minimal relation R1 (d3 ) as defined by (8) be given. The set Δ(d3 ) coincides with the complement of the union of all associated sets Ωdk3 (d2 ) in the set of unrepresentable integers Δ(d2 ), where k = 1, . . . , a33 − 1: a 33 −1 Δ(d3 ) = Δ(d2 ) Ωdk3 (d2 ) .

(47)

k=1

Proof First, we show that a 33 −1 Ωdk3 (d2 ) ⊆ Δ(d3 ). Δ(d2 )

(48)

k=1

a33 −1 k 2 Ωd3 (d )} and suppose that σ ∈ Δ(d3 ). Then by the definition We let σ ∈ Δ(d2 ) \ { k=1 3 of Δ(d ), an integer σ is representable by d1 , d2 , d3 . Hence, by Theorem 1, we have σ ∈ a33 −1 k 2 k=1 Ωd3 (d ), which contradicts our assumption on σ . Consequently, (48) holds. Finally, we show that a 33 −1 Ωdk3 (d2 ) ⊇ Δ(d3 ). (49) Δ(d2 ) k=1

Let σ ∈ Δ(d ). Then σ ∈ Δ(d ) by (5). We suppose that 3

2

a 33 −1 σ ∈ Δ(d2 ) Ωdk3 (d2 ) . k=1

132

Leonid G. Fel

Fig. 3 Typical matrix representation M{Δ(d3 )} of the set Δ(d3 ) (striped area) inside M{Δ(d2 )}. The top layer TLM {Δ(d3 )} has a33 convex corners (gray boxes). The integers kd3 ∈ Δ(d3 ), 1 ≤ k < a33 , occupy the a33 −1 k 2 Ωd (d ), which are adjacent to the concave corners of concave corners (black boxes) of the union k=1 3

TLM {Δ(d3 )}.

a33 −1 k 2 Then σ ∈ k=1 Ωd3 (d ). But then σ is representable by d1 , d2 , d3 by Definition 2, which again contradicts our assumption on σ . Hence, (49) holds, and Theorem 2 is proved. Generalizing BLM {Δ(d2 )} and TLM {Δ(d2 )} to m = 3, we call the totalities of the lowest and top elements in each column of M{Δ(d3 )} the respective bottom and top layers of M{Δ(d3 )} with the corresponding notation BLM {Δ(d3 )} and TLM {Δ(d3 )}, BLM {Δ(d3 )} ⊂ M{Δ(d3 )},

TLM {Δ(d3 )} ⊂ M{Δ(d3 )}.

(50)

In Fig. 3, we show a typical matrix representation M{Δ(d3 )} of the set Δ(d3 ) inside M{Δ(d2 )}. The bottom layer BLM {Δ(d3 )} of this diagram coincides with the bottom layer BLM {Δ(d2 )} shown in Fig. 1 (see formula (63) in Sect. 5). The top layer TLM {Δ(d3 )} of this diagram is much more intricate than TLM {Δ(d2 )} given in (32), for example, TLM {Δ(d3 )} has a33 convex corners.

5 Diagrammatic calculation on the set Δ(d3 ) A straightforward reconstruction of the Hilbert series H (d3 ; z) of a graded ring k[zd1 , zd2 , zd3 ] from the set Δ(d3 ) is a difficult problem. To overcome this difficulty, we here develop a diagrammatic calculation procedure on the set Δ(d3 ). We use this procedure to calculate Q(d3 ; z) in Sect. 6 and to solve the 3D Frobenius problem completely. The algebraic approach to the Frobenius problem is based on a strong relation between the Hilbert series H (dm ; z) of a graded ring k[zd1 , . . . , zdm ] over a field of characteristic 0


133

and the generating function Φ(dm ; z) for the set Δ(dm ) [2], which is H (dm ; z) + Φ(dm ; z) =

1 , 1−z

(51)

where Φ(dm ; z) and H (dm ; z) are respectively defined in (4) and (6). Evaluated at a special value of z, the function Φ(dm ; z) gives the Frobenius number and the genus of semigroups in any dimension m. Indeed, according to definitions (3) and (7), we have F (dm ) = deg Φ(dm ; z),

G(dm ) =

1s = Φ(dm ; 1).

(52)

s∈Δ(dm )

Using (6) and (51), we can represent formulas (52) more analytically, F (dm ) = deg Q(dm ; z) −

m

(53)

dj ,

j =1

G(dm ) =

m (−1)m+1 m ∂zm+1 (1 − zdj ) − (1 − z)Q(dm ; z) , (m + 1)! j =1 dj |z=1 j =1

(54)

where ∂zn = d n /dzn denotes the usual nth-order derivative. As can be seen from (53) and (54), the Frobenius problem reduces to finding the numerator Q(dm ; z) of the Hilbert series, which follows if (6) is substituted in (51), Q(dm ; z) =

d m 1 −1 (1 − zdj ) zk − (1 − zd1 )Φ(dm ; z) . j =2

(55)

k=0

Reconstructing the numerator Q(dm ; z) directly from the set Δ(dm ) is a very difficult problem. The main difficulty arises when we deal with the term d1 −1

zk − (1 − zd1 )Φ(dm ; z).

(56)

k=0

But it appears that in the dimension m = 3, an effective procedure can be elaborated for calculating Q(d3 ; z) via geometric transformations (shifts) of a diagram of the matrix representation M{Δ(d3 )}. We call such a procedure diagrammatic calculation. It turns out that diagrammatic calculation in the dimension m = 3 reduces determining Q(d3 ; z) to calculating TLM {Δ(d3 )} and BLM {Δ(d3 )} but not the entire matrix M{Δ(d3 )}. we introduce two functions, τ and its inverse τ −1 , where τ maps each polynomial First, ck zk ∈ N[z] with ck ∈ {0, 1} to the set of degrees {k ∈ N | ck = 0}. In particular, it follows from (4) that τ [Φ(d3 ; z)] = Δ(d3 ),

τ −1 [Δ(d3 )] = Φ(d3 ; z).

(57)

We note that because all the coefficients of the polynomial Φ(d3 ; z) are 1 or 0, we can uniquely reconstruct Φ(d3 ; z) from the set Δ(d3 ) and vice versa. In this sense, τ is an isomorphic map. The map τ is also linear in the following sense. Let d3 be given and a set Δ(d3 ) of all unrepresentable integers be related to its generating function Φ(d3 ; z) by the isomorphic map τ defined in (57). Let two sets Δ1 (d3 ) and Δ2 (d3 )

134

Leonid G. Fel

be given such that Δ1 (d3 ), Δ2 (d3 ) ⊂ Δ(d3 ),

Δ1 (d3 ) ∩ Δ2 (d3 ) = ∅.

(58)

τ −1 [Δ1 (d3 ) ∪ Δ2 (d3 )] = τ −1 [Δ1 (d3 )] + τ −1 [Δ2 (d3 )].

(59)

Then we have

Recalling (28) and (57), we present the relations between the three main entities M{Δ(d3 )}, Δ(d3 ), and Φ(d3 ; z) concerned with the integers that are unrepresentable by d1 , d2 , d3 . These relations involve the two maps σ and τ : σ

τ

M{Δ(d3 )} −→ Δ(d3 ) ←− Φ(d3 ; z). 5.1 Construction of the set τ

d1 −1 k=0

(60)

zk − (1 − zd1 )Φ(d3 ; z)

U1 , which shifts the diagram of the matrix representation We introduce the up-shift operator M{Δ(d3 )} one step up. We define

U1 σ (p, q) := σ (p − 1, q).

(61)

U1 Δ(d3 ) denote the set of It hence follows from (24) that σ (p − 1, q) = σ (p, q) + d1 . We let 3 U1 σ (p, q) such that σ (p, q) ∈ Δ(d ) and define a set Δ (d3 ) = U1 Δ(d3 ). Then all integers Δ (p, q) = Δ(p − 1, q) and

Δ (d3 ) = U1 Δ(d3 ) =

U1 σ (p, q).

(62)

(p,q)∈M{Δ(d3 )}

To calculate term (56) diagrammatically, we need the following results. Let {d1 , d2 , d3 } be the minimal generating set of S(d3 ) and Φ(d3 ; z) be the generating function for the set τ [Φ(d3 ; z)] of unrepresentable integers. For our purpose here, it is important that the construction of M{Δ(d3 )} via Theorem 2 does not affect BLM {Δ(d2 )} (the gray cells in Fig. 1). Indeed, it is clear that the integers (1, . . . , d1 − 1) are unrepresentable by d1 , d2 , d3 . Because of the first equality in (32), this leads to an important result for the bottom layer BLM {Δ(d3 )}, σ BLM {τ [Φ(d3 )]} = σ BLM {τ [Φ(d2 )]} = {1, 2, . . . , d1 − 1}.

(63)

We consider the top layer TLM {τ [Φ(d3 )]}. It is given by σ TLM {τ [Φ(d3 )]} = {σ (pt3 (q), q)},

q = 1, . . . , d1 − 1,

(64)

where the subscript t3 denotes the top of M{Δ(d3 )} and pt3 (q) is defined as pt3 (q) = min{1 ≤ p | σ (p, q) ∈ Δ(d3 )}. Lemma 6 We have σ TLM {τ [zd1 Φ(d3 )]} = {σ (pt3 (q) − 1, q)},

q = 1, . . . , d1 − 1.

(65)


135

Proof We consider the polynomial zd1 Φ(d3 ; z). By (4), (61), and (62), we obtain zs+d1 = zs . zd1 Φ(d3 ; z) = s∈Δ(d3 )

(66)

s∈ U1 Δ(d3 )

Acting on the left- and the right-hand sides of (66) by the map τ , we obtain τ [zd1 Φ(d3 )] = U1 Δ(d3 ) = U1 τ [Φ(d3 )].

(67)

By (61), (62), and (64), this leads to (65).

Corollary 1 Let d3 = (d1 , d2 , d3 ) be given with the first minimal relation defined by (8). Then kd3 ∈ σ TLM {τ [zd1 Φ(d3 )]} , k = 1, . . . , a33 − 1. (68) Proof Let R1 (d3 ) be the first minimal relation for the given d3 . Then kd3 ∈ Δ(d3 ), 1 ≤ k < a33 . First, we consider one such integer kd3 and show that kd3 − d1 ∈ Δ(d3 ). We suppose that kd3 − d1 ∈ Δ(d3 ). Then there exist integers ρ1 , ρ2 , ρ3 ∈ N ∪ {0} such that kd3 − d1 = ρ1 d1 + ρ2 d2 + ρ3 d3 −→

(k − ρ3 )d3 = (ρ1 + 1)d1 + ρ2 d2 ,

1 ≤ k ≤ a33 − 1.

The last equality violates the minimality of the relation R1 (d3 ) given by (8). For every k = 1, . . . , a33 − 1, we now have kd3 ∈ Δ(d3 ),

kd3 − d1 ∈ Δ(d3 ).

(69)

Comparing (69) with (64) and (65), we conclude that the integers kd3 − d1 , 1 ≤ k < a33 , occupy the top layer TLM {τ [Φ(d3 )]} and the integers kd3 , 1 ≤ k < a33 , occupy TLM {τ [zd1 Φ(d3 )]}. This proves Corollary 1. In Fig. 4, we show the matrix representations of the two sets τ [zd1 Φ(d3 ; z)] and τ [Φ(d3 ; z)] with their intersection Π1 (d3 ) := τ [zd1 Φ(d3 ; z)] ∩ τ [Φ(d3 ; z)].

(70)

The matrix M{τ [zd1 Φ(d3 ; z)]} has been shifted one step up with respect to the matrix M{τ [Φ(d3 ; z)]}. It follows from this that τ [Φ(d3 ; z)] = Π1 (d3 ) ∪ σ BLM {τ [Φ(d3 )]} , (71) τ [zd1 Φ(d3 ; z)] = Π1 (d3 ) ∪ σ TLM {τ [zd1 Φ(d3 )]} , (72) and

Π1 (d3 ) ∩ σ BLM {τ [Φ(d3 )]} = ∅, Π1 (d3 ) ∩ σ TLM {τ [zd1 Φ(d3 )]} = ∅.

(73)

We let λq denote the integers occupying the top layer TLM {τ [zd1 Φ(d3 )]}. We hence have σ TLM {τ [zd1 Φ(d3 )]} = {λq | λq = σ (pt3 (q) − 1, q), 1 ≤ q ≤ d1 − 1}. (74) We are now ready to prove the main theorem in this section.

136

Leonid G. Fel

Fig. 4 Matrix representation of two sets: τ [zd1 Φ(d3 ; z)] (inside the plain frame) and τ [Φ(d3 ; z)] (inside the dashed frame). Their intersection Π1 (d3 ) is shown in light gray. The top and bottom layers TLM {τ [zd1 Φ(d3 )]} and BLM {τ [Φ(d3 )]} are respectively shown in dark gray and white. The integers kd3 ∈ σ TLM {τ [zd1 Φ(d3 )]} , 1 ≤ k < a33 , are shown by black boxes.

Theorem 3 We have d1 −1

(1 − zd1 )Φ(d3 ; z) =

q=1

d1 −1

zq −

z λq .

(75)

q=1

Proof We consider the two polynomials Φ(d3 ; z) = τ −1 [Δ(d3 )],

zd1 Φ(d3 ) = τ −1 [ U1 Δ(d3 )],

(76)

and construct their difference K1 = (1 − zd1 )Φ(d3 ; z) acting on the left- and right-hand sides of (71) and (72) by τ −1 , K1 = τ −1 Π1 (d3 ) ∪ σ BLM {τ [Φ(d3 )]} − τ −1 Π1 (d3 ) ∪ σ TLM {τ [zd1 Φ(d3 )]} . Using (58), (59), and (73), we obtain K1 = τ −1 σ BLM {τ [Φ(d3 )]} − τ −1 σ TLM {τ [zd1 Φ(d3 )]} .

(77)

Substituting (63) and (74) in (77), we obtain (75), which completes the proof of Theorem 3. In Fig. 5, we show the matrix representation of the set τ [(1 − zd1 )Φ(d3 ; z)]. Finally, we come to term (56), which is calculated in the next theorem.


137

d1 3 Fig. 5 Matrix representation of the set τ [(1d − z 3)Φ(d ; z)].qThe minus and plus signs in the cells mark λ q 1 the respective terms −z , λq ∈ σ TLM {τ [z Φ(d )]} , and z , q = 1, . . . , d1 − 1, in polynomial (75). The integers kd3 ∈ Δ(d3 ), 1 ≤ k < a33 , are shown by black boxes.

Theorem 4 We have d1 −1

d1 −1

zq − (1 − zd1 )Φ(d3 ; z) =

q=0

z λq ,

λ0 = 0.

(78)

q=0

Proof The proof follows immediately from Theorem 3. We introduce the notation Λ(d3 ) := σ TLM {τ [zd1 Φ(d3 )]} ∪ {0}

(79)

for use in the next section. The basic properties of the set Λ(d3 ) follow from (74) and (79): Λ(d3 ) ⊂ Δ(d3 ),

#Λ(d3 ) = d1 ,

Λ(d3 ) = τ

d 1 −1

z λq .

(80)

q=0

The structure of Λ(d3 ) is very intricate. The set Λ(d3 ) includes the integers λq that do not even belong to the set Δ(d2 ). These are λq = qd2 ,

1 ≤ q ≤ a22 − 1.

(81)

Indeed, it follows from (32) and Lemma 6 that λq ∈ Λ(d3 ). On the other hand, equality (81) means that λq are representable by d2 and therefore λq ∈ Δ(d2 ).

138

Leonid G. Fel

d1 −1 k d1 3 5.2 The polynomial (1 − zd2 ) k=0 z − (1 − z )Φ(d ; z) In the preceding section, we found the set Λ(d3 ) of integers λq that contribute to the polynod1 −1 k mial k=0 z − (1 − zd1 )Φ(d3 ; z). Here, we continue to construct the numerator Q(d3 ; z) according to (55). This is done by further successively applying the diagrammatic calculation procedure to Λ(d3 ). It is not convenient to deal with the matrix representation of the set Λ(d3 ) (see Fig. 5). We start with a representation of the integers λq ∈ Λ(d3 ) that essentially simplifies the calculations. Definition 3 Let integers 2 < d1 < d2 < d3 be given. The function λ(v2 , v3 ) is defined as λ(v2 , v3 ) := v2 d2 + v3 d3 ,

v2 , v3 ∈ N ∪ {0}.

(82)

The next lemma specifies the restrictions on the domain of (v2 , v3 ) introduced in Definition 3. This is the hardest part of the paper. Lemma 7 Let d3 = (d1 , d2 , d3 ) be given with the first minimal relation R1 (d3 ) defined by (8). Let r be an integer. Then r ∈ Λ(d3 ) iff r is uniquely representable as r = λ(v2 , v3 ),

(83)

(v2 , v3 ) ∈ ([0, a22 − 1] × [0, a13 − 1]) ∪ ([0, a12 − 1] × [a13 , a33 − 1])x.

(84)

where

Proof We note that it respectively follows from (81) and Corollary 1 in Sect. 5.1 that v2 d2 ∈ Λ(d3 ),

0 ≤ v2 < a22 ,

v3 d3 ∈ Λ(d3 ),

0 ≤ v3 < a33 .

(85)

We fix v3 such that 0 ≤ v3 < a33 and consider the sequence of integers v

v3 d3 , d2 + v3 d3 , 2d2 + v3 d3 , . . . , v2 d2 + v3 d3 ∈ Ωd33 (d2 ),

(86)

where the maximal element v2 d2 + v3 d3 of sequence (86) is defined by v2 d2 + v3 d3 = max{v2 d2 + v3 d3 | v2 d2 + v3 d3 ∈ Λ(d3 ), 0 ≤ v2 < a22 }.

(87)

We note that the integers of sequence (86) continuously occupy all cells of the corresponding v3 th row in the diagram in Fig. 5 (from right to left without jumps). To calculate v2 d2 + v3 d3 , we should formulate the requirements that must be satisfied. They are based on two facts that follow from (87). First, according to definition (87), the element v2 d2 + v3 d3 is contained in Λ(d3 ). Second, v2 d2 + v3 d3 + d2 belongs neither to Λ(d3 ) (because the element v2 d2 + v3 d3 is maximal in sequence (86)) nor to Δ(d3 ) (because v2 d2 + v3 d3 + d2 is representable by d2 , d3 ). Recalling definition (79) of the set Λ(d3 ) and Lemma 6, we summarize the requirements as v2 d2 + v3 d3 ∈ Λ(d3 )

−→

v2 d2 + v3 d3 − d1 ∈ Δ(d3 ),

(88)


(v2 + 1)d2 + v3 d3 ∈ Λ(d3 ) (v2 + 1)d2 + v3 d3 ∈ Δ(d3 )

139

−→

(v2 + 1)d2 + v3 d3 − d1 ∈ Δ(d3 ).

(89)

Requirement (89) provides the representation (v2 + 1)d2 + v3 d3 − d1 = γ1 d1 + γ2 d2 + γ3 d3 for some γ1 , γ2 , γ3 ∈ N ∪ {0}.

(90)

On the other hand, the integer v2 d2 + v3 d3 − d1 is not representable by d1 , d2 , d3 because of (88). This can happen only if γ2 = 0 because v2 d2 + v3 d3 − d1 is otherwise always representable because of (90), v2 d2 + v3 d3 − d1 = γ1 d1 + (γ2 − 1)d2 + γ3 d3 ,

γ2 ≥ 1.

We return to (90) and consider its solution, substituting γ2 = 0. First, we consider v3 in the interval 0 ≤ v3 < a13 and rewrite (90) in the form (v2 + 1)d2 = (γ1 + 1)d1 + (γ3 − v3 )d3 .

(91)

Comparing (91) with the first minimal relation a22 d2 = a21 d1 + a23 d3 , by the uniqueness (see (9)), we obtain the maximal value of v2 : v2 = a22 − 1,

γ1 = a21 − 1,

γ3 = v3 + a23 .

(92)

Hence, the first kind of the integers λ(v2 , v3 ) ∈ Λ(d3 ) has a representation λ(v2 , v3 ) = v2 d2 + v3 d3 ,

where (v2 , v3 ) ∈ ([0, a22 − 1] × [0, a13 − 1]).

(93)

We next consider the solution of (90) for v3 in the interval a13 ≤ v3 < a33 . Summation of (90) and the first minimal relation a11 d1 − a12 d2 − a13 d3 = 0 leads to the identity a11 d1 + (v2 + 1 − a12 )d2 + (v3 − a13 )d3 = (γ1 + 1)d1 + γ3 d3 , which by the uniqueness gives the maximal value v2 : v2 = a12 − 1,

γ1 = a11 − 1,

γ3 = v3 − a13 .

(94)

Hence, the second kind of the integers λ(v2 , v3 ) ∈ Λ(d3 ) has a representation λ(v2 , v3 ) = v2 d2 + v3 d3 ,

where (v2 , v3 ) ∈ ([0, a12 − 1] × [a13 , a33 − 1]).

(95)

Combining (93) and (95), we obtain (84). Finally, it remains to prove that (83) is unique. A standard proof of uniqueness by contradiction is to suppose that there are two such representations λ(v2a , v3a ), and λ(v2b , v3b ), v2a = v2b , v3a = v3b , and consequently (v3a − v3b )d3 = (v2b − v2a )d2 ,

v3a > v3b ,

v2b > v2a .

On the other hand, using (93) and (95), we obtain |v2b − v2a | ≤ a22 − 1,

|v3a − v3b | ≤ a33 − 1.

(96)

140

Leonid G. Fel

Fig. 6 Typical matrix representation M{Λ(d3 )} of the set Λ(d3 ). All cells in the diagram marked in white are occupied by integers λ(v2 , v3 ) ∈ Λ(d3 ). The integers aii di ∈ Λ(d3 ), i = 1, 2, 3, occupy three cells marked in gray.

Comparing (96) with the third equality, a33 d3 = a31 d1 + a32 d2 , of first minimal relation (8), we obtain a contradiction because the latter equality is unique for the given minimal set d1 , d2 , d3 . This completes the proof of Lemma 7. We note that because matrix representation (82) of all integers λ(v2 , v3 ) ∈ Λ(d3 ) is unique, the following inequalities hold for nonsymmetric semigroups: aii di = ajj dj

for i = j,

1 ≤ i, j ≤ 3.

(97)

The case of the symmetric semigroups admits only one equality in (97) (see Sect. 6.2 for details). Representation (83) of all integers λ(v2 , v3 ) ∈ Λ(d3 ) is called the matrix representation of the set Λ(d3 ) and is denoted by M{Λ(d3 )} (see Fig. 6), λ{M{Λ(d3 )}} = Λ(d3 ).

(98)

The number λ(v2 , v3 ) is the integer in the v2 th row and v3 th column of M{Λ(d3 )}. In Fig. 6, we show the matrix representation M{Λ(d3 )} of the set Λ(d3 ) for the nonsymmetric semigroup S(d3 ). This diagram first appeared in [18] for the algorithmic calculation of F (d3 ). It was later also used in [19], [20], [21], [6], [22], and [23] for the same purpose. Lemma 7 has also two interesting corollaries related to the diagram in Figs. 5 and 6. Corollary 2 The length of the rows in the matrix representation TLM {τ [zd1 Φ(d3 )]} in Fig. 5 that are covered continuously by integers λ(v2 , v3 ) is either a22 −1, a22 , or a12 (defined in unit cells). Proof We consider the upper row of the matrix representation TLM {τ [zd1 Φ(d3 )]} in Fig. 5. According to (81), this is the unique row of length a22 −1 (in unit cells) covered continuously by the integers λq = qd2 , 1 ≤ q < a22 , that does not belong to Δ(d2 ). All the other rows are contained in Δ(d2 ), and in accordance with Corollary 1, their rightmost cells are occupied by one of the integers kd3 , 1 ≤ k < a33 (see Fig. 5). We note that all these rows are mapped one-to-one into vertical columns in the diagram of the matrix representation M{Λ(d3 )} (see Fig. 6). We hence conclude in accordance with Lemma 7 that their length is either a22 or a12 (defined in unit cells).


141

Corollary 3 Let {d1 , d2 , d3 } be a minimal generating set of a nonsymmetric semigroup S(d3 ), and let the first minimal relation be defined by (8). Then a22 + a33 ≤ d1 + 1 ≤ a22 a33 ,

a33 + a11 ≤ d2 + 1 ≤ a33 a11 ,

a11 + a22 ≤ d3 + 1 ≤ a11 a22 .

(99)

Proof The right-hand sides of (99) follow from (18): a22 a33 = d1 + a23 a32 ≥ d1 + 1,

a33 a11 = d2 + a31 a13 ≥ d2 + 1,

a11 a22 = d3 + a12 a21 ≥ d3 + 1. The proof of the left-hand sides of (99) follows from (17) and (18), for example, d1 + 1 − (a22 + a33 ) = a22 a33 − a23 a32 + 1 − (a22 + a33 ) = a22 a33 − (a22 − a12 )(a33 − a13 ) + 1 − (a22 + a33 ) = 1 + a22 (a13 − 1) + a33 (a12 − 1) − a12 a13 ≥ 1 + (a12 + 1)(a13 − 1) + (a13 + 1)(a12 − 1) − a12 a13 = a12 a13 − 1 ≥ 0.

Corollary 3 is thus proved. We now calculate the polynomial d 1 −1 k d1 3 z − (1 − z )Φ(d ; z) (1 − z ) d2

k=0

and apply the diagrammatic calculation procedure the same as in Sect. 5.1. For this, we call the totality of the lowest and top cells in each column of M{Λ(d3 )} the respective bottom and top layers of M{Λ(d3 )}, which are denoted by BLM {Λ(d3 )} and TLM {Λ(d3 )}. As can be seen from Fig. 6, λ{BLM {Λ(d3 )}} = {0, d3 , . . . , (a33 − 1)d3 },

(100)

λ{TLM {Λ(d3 )}} = {λ(a22 − 1, v3 ), 0 ≤ v3 < a13 } ∪ {λ(a12 − 1, v3 ), a13 ≤ v3 < a33 }.

(101)

U2 , which shifts the diagram of the matrix representation We introduce the up-shift operator M{Λ(d3 )} one step up. We define

U2 λ(v2 , v3 ) = λ(v2 + 1, v3 ).

(102)

U2 Λ(d3 ) denote the set of Hence, by (82), λ(v2 + 1, v3 ) = λ(v2 , v3 ) + d2 , and if we let 3 U2 λ(v2 , v3 ) such that λ(v2 , v3 ) ∈ Λ(d ) and define Λ (d3 ) = U2 Λ(d3 ), then all integers Λ (v2 , v3 ) = Λ(v2 + 1, v3 ) and U2 Λ(d3 ) = U2 λ(v2 , v3 ). (103) Λ (d3 ) = (v2 ,v3 )∈M{Λ(d3 )}

142

Leonid G. Fel

Let {d1 , d2 , d3 } be a minimal generating set of S(d3 ), and let Φ(d3 ; z) be a generating function for the set τ [Φ(d3 ; z)] of unrepresentable integers. By Theorem 4, this implies that d1 −1 q d1 3 3 q=0 z − (1 − z )Φ(d ; z) is a generating function for the set Λ(d ). Lemma 8 We have

d 1 −1 d2 k d1 3 λ TLM τ z z − (1 − z )Φ(d ; z) = Γ1 (d3 ) ∪ Γ2 (d3 ),

(104)

k=0

where Γ1 (d3 ) = {λ(a22 , v3 ), 0 ≤ v3 < a13 },

(105)

Γ2 (d3 ) = {λ(a12 , v3 ), a13 ≤ v3 < a33 }. Proof We consider the polynomial zd2 and (103), we obtain z

d2

d 1 −1

d1 −1 k=0

zk − (1 − zd1 )Φ(d3 ; z) . By (78), (102),

d 1 −1 z − (1 − z )Φ(d ; z) = zλq +d2 k

3

d1

k=0

q=0

=

zλ+d2

λ∈Λ(d3 )

=

zλ .

(106)

λ∈ U2 Λ(d3 )

Acting on the left- and right-hand sides of (106) by the map τ , we obtain τ z

d2

d 1 −1

z − (1 − z )Φ(d ; z) = U2 Λ(d3 ). k

d1

3

(107)

k=0

Proving (104) thus reduces to finding the set of the integers U2 Λ(d3 )} λ TLM { occupying the top layer TLM { U2 Λ(d3 )} of the matrix representation. Successively using (102), (103), and (101), we obtain λ TLM { U2 Λ(d3 )} = {λ(a22 , v3 ), 0 ≤ v3 < a13 } ∪ {λ(a12 , v3 ), a13 ≤ v3 < a33 }.

(108)

Introducing the two nonintersecting sets Γ1 (d3 ) and Γ2 (d3 ), Γ1 (d3 ) ∩ Γ2 (d3 ) = ∅, in accordance with (105), we obtain the proof of Lemma 8. We note that according to (100) and (108), λ BLM {Λ(d3 )} ∩ λ TLM { U2 Λ(d3 )} = ∅.

(109)


143

Let Π2 (d3 ) denote the intersection of the sets Λ(d3 ) and U2 Λ(d3 ), U2 Λ(d3 ). Π2 (d3 ) := Λ(d3 ) ∩

(110)

We have the representation Λ(d3 ) = Π2 (d3 ) ∪ λ BLM {Λ(d3 )} , U2 Λ(d3 ) = Π2 (d3 ) ∪ λ TLM { U2 Λ(d3 )} , where

(111) (112)

Π2 (d3 ) ∩ λ BLM {Λ(d3 )} = ∅, Π2 (d3 ) ∩ λ TLM { U2 Λ(d3 )} = ∅.

(113)

We show that (110) necessarily follows from (111), (112), and (109). Indeed, a straightforward calculation gives Λ(d3 ) ∩ U2 Λ(d3 ) = Π2 (d3 ) ∪ λ BLM {Λ(d3 )} ∩ Π2 (d3 ) ∪ λ TLM { U2 Λ(d3 )} = Π2 (d3 ) ∪ λ BLM {Λ(d3 )} ∩ λ TLM { U2 Λ(d3 )} = Π2 (d3 ). We prove an important theorem. Theorem 5 We have d 1 −1 zk − (1 − zd1 )Φ(d3 ; z) (1 − zd2 ) k=0

=

zλ −

zλ .

(114)

λ∈λ TLM { U2 Λ(d3 )}

λ∈λ BLM {Λ(d3 )}

Proof The proof is similar to that of Theorem 3. We consider the polynomials d1 −1

zk − (1 − zd1 )Φ(d3 ; z) = τ −1 [Λ(d3 )],

k=0

z

d2

d 1 −1

z − (1 − z )Φ(d ; z) = τ −1 [ U2 Λ(d3 )] k

d1

3

k=0

d1 −1 k z − (1 − zd1 )Φ(d3 ; z) acting on the and construct their difference K2 = (1 − zd2 ) k=0 −1 left- and right-hand sides of (111) and (112) by τ , K2 = τ −1 Π2 (d3 ) ∪ λ BLM {Λ(d3 )} − τ −1 Π2 (d3 ) ∪ λ TLM { U2 Λ(d3 )} .

144

Leonid G. Fel

d1 −1 k d1 3 Fig. 7 Matrix representation of the set τ (1 − zd2 ) k=0 z − (1 − z )Φ(d ; z) for the semigroup S(d3 ). Positive and negative contributions to polynomial (114) of the terms zλq with λq occupying the cells are respectively marked in gray and white.

Using (113) and (58), (59), we obtain U2 Λ(d3 )} , K2 = τ −1 λ BLM {Λ(d3 )} − τ −1 λ TLM { which leads to (114) in accordance with definition (57) of the inverse map τ −1 .

(115)

The result of the diagrammatic calculation procedure is shown in Fig. 7. d1 −1 k d1 3 5.3 The polynomial(1 − zd3 )(1 − zd2 ) k=0 z − (1 − z )Φ(d ; z) Here, we finish calculating d 1 −1 (1 − zd3 )(1 − zd2 ) zk − (1 − zd1 )Φ(d3 ; z) , k=0

relying on the results obtained in Sect. 5.2. Let {d1 , d2 , d3 } be a minimal generating set of S(d3 ), and let Φ(d3 ; z) be the generating function for the set τ [Φ(d3 ; z)] of unrepresentable integers. Theorem 6 We have d 1 −1 Q(d3 ; z) = (1 − zd3 )(1 − zd2 ) zk − (1 − zd1 )Φ(d3 ; z) k=0

=1−

3

zaii di + zL1 + zL2 ,

(116)

i=1

where L1 = a12 d2 + a33 d3 ,

L2 = a22 d2 + a13 d3 .

(117)

Proof The first equality in (116) follows from (55). Using Theorem 5 and Lemma 8, we obtain d 1 −1 zk − (1 − zd1 )Φ(d3 ; z) = T1 − T2 − T3 , (118) (1 − zd3 )(1 − zd2 ) k=0


145

d1 −1 k d1 3 Fig. 8 Matrix representation of the set τ (1 − zd3 )(1 − zd2 ) k=0 z − (1 − z )Φ(d ; z) for the nonsymmetric semigroup S(d3 ). Positive and negative contributions of the six terms zλ (λ = 0), a11 d1 , a22 d2 , a33 d3 , L1 , and L2 to polynomial (116) are respectively marked in gray and white.

where T1 = (1 − zd3 )

zλ ,

T2 = (1 − zd3 )

λ∈λ BLM {Λ(d3 )}

T3 = (1 − zd3 )

zλ ,

λ∈Γ1 (d3 )

zλ ,

λ∈Γ2 (d3 )

and the sets λ BLM {Λ(d3 )} and Γ1 (d3 ), Γ2 (d3 ) are respectively given in (100) and (105). Calculating the terms T1 , T2 , and T3 separately, we obtain a33 −1

T1 = (1 − zd3 )

zkd3 = 1 − za33 d3 ,

k=0 a13 −1

T2 = (1 − zd3 )

za22 d2 +kd3 = za22 d2 − za22 d2 +a13 d3 ,

(119)

k=0 a33 −1

T3 = (1 − zd3 )

za12 d2 +kd3 = za11 d1 − za12 d2 +a33 d3 .

k=a13

Substituting (119) in (118), we obtain (116).

The matrix representation of the set d 1 −1 τ (1 − zd3 )(1 − zd2 ) zk − (1 − zd1 )Φ(d3 ; z) k=0

for the nonsymmetric semigroup S(d3 ) is shown in Fig. 8. We note that the number of terms contributing to (116) coincides with the number of corners of the polygon that assigned the matrix representation of the set Λ(d3 ) (see Fig. 6). Below, we obtain two important results for the integers L1 and L2 defined in (117). Let d3 = (d1 , d2 , d3 ) be given with the first minimal relation for the semigroup S(d3 ) defined by (8). We show that L1 = L2 .

(120)

146

Leonid G. Fel

To obtain a contradiction, we suppose that the opposite is true, L1 = L2 . By (117), we then have a12 d2 + a33 d3 = a22 d2 + a13 d3 or (a22 − a12 )d2 = (a33 − a13 )d3 . By (17), a22 − a12 = a32 and a33 − a13 = a23 . Hence, a32 d2 = a23 d3 . Also by (17), a23 < a33 and a32 < a22 . But then a32 d2 and a23 d3 are in Λ(d3 ). Hence, because the representation of elements of Λ(d3 ) is unique (see Lemma 7), we have a32 = a23 = 0. We consider the nonsymmetric semigroup S(d3 ) with a23 = a32 = 0. The matrix A 3 of the first minimal relation necessarily has a13 = a12 = 0, which leads to a11 = 0 and contradicts (9). We note that L1 = L2 holds for both nonsymmetric and symmetric semigroups. The next lemma is related only to nonsymmetric semigroups S(d3 ). Lemma 9 We have L1 ≥ a11 d1 + d3 ,

L1 ≥ a33 d3 + d2 ,

L1 ≥ a22 d2 + d1 ,

L2 ≥ a11 d1 + d2 ,

L2 ≥ a33 d3 + d1 ,

L2 ≥ a22 d2 + d3 .

(121)

Proof It clearly follows from the diagram in Fig. 8 that L1 ≥ a11 d1 + d3 ,

L1 ≥ a33 d3 + d2 ,

L2 ≥ a11 d1 + d2 ,

L2 ≥ a22 d2 + d3 .

It can be shown that the other two inequalities, L1 ≥ a22 d2 + d1 and L2 ≥ a33 d3 + d1 , also hold. Indeed, in accordance with (117), we have L1 − a22 d2 = a12 d2 + a33 d3 − a22 d2 = a33 d3 − a32 d2 = a31 d1 ≥ d1 . We use (17) in the last two equalities. The last inequality, L2 ≥ a33 d3 + d1 , can be proved similarly. Both results, (120) and Lemma 9, are used later in Sect. 6.

6 Hilbert series, Frobenius number, and genus In this section, we completely solve the 3D Frobenius problem, i.e., we calculate the numerator Q(d3 ; z) of the Hilbert series for both nonsymmetric and symmetric semigroups S(d3 ) and determine the Frobenius number and the genus based on it. 6.1 The Frobenius problem for the nonsymmetric semigroup S(d3 ) The Hilbert series H (d3 ; z), the Frobenius number F (d3 ), and the genus G(d3 ) of the semigroup S(d3 ) are invariants under permutations of elements di in the generating set {d1 , d2 , d3 }. Therefore, we need to find a more symmetric representation for the integers L1 and L2 defined in (117). This leads to the main theorem in this section. Theorem 7 Let d3 = (d1 , d2 , d3 ) be given with the first minimal relation for the nonsymmetric semigroup S(d3 ) defined by (8). Then the numerator Q(d3 ; z) of the Hilbert series is Q(d3 ; z) = 1 −

3 i=1

zaii di + z1/2[a,d −J (d

3 )]

3

+ z1/2[a,d +J (d )] ,

(122)


J 2 (d3 ) = a, d 2 − 4

3

147

aii ajj di dj + 4d1 d2 d3 ,

a, d =

3

i>j

aii di .

(123)

i=1

(n) of the first minimal relation for the nonsymProof Using (17) and (18) for the matrix A 3 3 metric semigroup S(d ), we note that the integers L1 and L2 defined in (117) satisfy

L1 L2 =

3

aii ajj di dj − d1 d2 d3 ,

L1 + L2 = a, d ,

(124)

i>j

a, d =

3

aii di .

i=1

In other words, L1 and L2 are solutions of the quadratic equation L21,2 − a, d L1,2 +

3

aii ajj di dj − d1 d2 d3 = 0,

(125)

i>j

i.e., 1 L1,2 = [a, d ± J (d3 )], 2 J 2 (d3 ) = a, d 2 − 4

3

(126)

aii ajj di dj + 4d1 d2 d3 ,

i>j

which implies a, d − J (d3 ) = min{2L1 , 2L2 }, a, d + J (d3 ) = max{2L1 , 2L2 }.

(127)

Recalling expression (55) for the numerator Q(dm ; z) of the Hilbert series and substituting (124) in (116), we obtain (122), which proves the theorem. Theorem 8 The Frobenius number and the genus of the nonsymmetric semigroup S(d3 ) are given by 1 F (d3 ) = [a, d + J (d3 )] − di , 2 i=1 3

3 3 1 3 di − aii . G(d ) = 1 + a, d − 2 i=1 i=1

(128)

Proof Implementating formulas (53) and (54) for Q(d3 ; z) given by (122) leads to formulas (128). In Theorem 7, the number J (d3 ) was regarded as a positive integer such that the degrees L1 and L2 of two last terms in (122) are positive integers. In other words, it was also

148

Leonid G. Fel

presumed that the numbers a, d ± J (d3 ) are even positive integers. Here, we prove these statements in the case of the nonsymmetric semigroup S(d3 ). Lemma 10 Let d3 = (d1 , d2 , d3 ) be given with the first minimal relation R1 (d3 ) for the nonsymmetric semigroup S(d3 ) defined by (8). Then the numbers J (d3 ) and 1/2[a, d ± J (d3 )] are respectively nonnegative and positive integers: J (d3 ) = |a12 a23 a31 − a13 a32 a21 |,

(129)

1 1 [a, d ± J (d3 )] = a11 a22 a33 + (a12 a23 a31 2 2 + a13 a32 a21 ± |a12 a23 a31 − a13 a32 a21 |).

(130)

Proof Substituting relations (17) and (18) in expressions (123) for J 2 (d3 ) and a, d and (n) = 0, which leads to using the equality det A 3 a11 a22 a33 − (a11 a23 a32 + a22 a13 a31 + a33 a12 a21 ) = a12 a23 a31 + a13 a32 a21 , we obtain J 2 (d3 ) = (a12 a23 a31 − a13 a32 a21 )2 ,

(131)

a, d = 2a11 a22 a33 + a13 a32 a21 + a12 a23 a31 .

The last relations lead to (129) and (130).

Corollary 4 Let {d1 , d2 , d3 } be a minimal generating set of a nonsymmetric semigroup. Then J (d3 ) ≥ 1.

(132)

Proof According to Lemma 10, the number J (d3 ) is a nonnegative integer. On the other hand, by (120) and (125), J (d3 ) does not vanish, which leads to (132). The unity in (132) is the best possible, as the following example shows. Example 1 The triple {3, 4, 5} generates a nonsymmetric semigroup (with the minimal possible di ). ⎛ ⎞ 3 −1 −1 ⎝ 2 −1⎠ , A a, d = 27, J (d3 ) = 1, 3 = −1 −2 −1 2 Q(d3 ; z) = 1 − z8 − z9 − z10 + z13 + z14 ,

G(d3 ) = 2,

F (d3 ) = 2.

Theorem 8 and Lemma 10 allow expressing F (d3 ) + d1 + d2 + d3 and 2G(d3 ) + d1 + (n) only for a nonsymmetric semigroup d2 + d3 in terms of the elements of the matrix A 3 3 S(d ): F (d3 ) + d1 + d2 + d3 = a11 a22 a33 + max{a12 a23 a31 , a13 a32 a21 },

(133)

2G(d3 ) + d1 + d2 + d3 = 1 + a11 a22 a33 + a13 a32 a21 + a12 a23 a31 .

(134)


149

Formula (133) fully agrees with formula (20) for the Frobenius number obtained in [5], [13]. This can be seen if relations (17) and (18) are substituted in (20). Another important inequality for nonsymmetric semigroups can be proved by considering (133) and (134) and taking their difference, 1 G(d3 ) ≥ 1 + F (d3 ). 2

(135)

We note that (135) is slightly stronger than a similar inequality obtained by Nijenhuis and Wilf [24] for the mD Frobenius problem. Below, we illustrate formulas (122) and (128) obtained for the 3D Frobenius problem for the three triples (23, 29, 44), (137, 251, 256), and (1563, 2275, 2503), which were respectively considered numerically in [25], [11], and [26]. Example 2 ⎛ ⎞ ⎛ ⎞ 23 d1 ⎝d2 ⎠ = ⎝29⎠ , 44 d3 J (d3 ) = 86, F (d3 ) = 239, ⎞ ⎛ ⎞ ⎛ 137 d1 ⎝d2 ⎠ = ⎝251⎠ , 256 d3

⎛

⎞ 7 −1 −3 ⎝ 7 −2⎠ , A 3 = −5 −2 −6 5

a, d = 584,

Q(d3 ; z) = 1 − z161 − z203 − z220 + z249 + z335 , G(d3 ) = 122. ⎛

⎞ 24 −8 −5 ⎝ −7 13 −9 ⎠ , A 3= −17 −5 14

a, d = 10135,

J (d3 ) = 1049,

Q(d3 ; z) = 1 − z3263 − z3288 − z3584 + z4543 + z5592 ,

F (d3 ) = 4948,

G(d3 ) = 2562.

⎞ ⎛ ⎞ ⎛ 1563 d1 ⎝d2 ⎠ = ⎝2275⎠ , 2503 d3

⎛

⎞ 23 −7 −8 ⎝ −17 114 −93 ⎠ , A 3= −6 −107 101

a, d = 548102,

J (d3 ) = 10646, Q(d3 ; z) = 1 − z35949 − z252803 − z259350 + z268728 + z279374 , F (d3 ) = 273033,

G(d3 ) = 138470.

In the next example, we present a special kind of nonsymmetric semigroup, the so-called Pythagorean semigroups [12]. Their generators d1 , d2 , d3 satisfy d12 + d22 = d32 . Example 3 We take 1 ≤ k2 < k1 such that gcd(k1 , k2 ) = 1. Then ⎛ ⎞ ⎛ 2 ⎞ ⎞ ⎛ d1 k1 − k22 k1 + k2 −k1 + k2 −k1 + k2 Pth ⎝d2 ⎠ = ⎝ 2k1 k2 ⎠ , k1 −k2 ⎠ , A = ⎝ −k2 3 d3 k12 + k22 −k1 −k2 k1 a, d = (2k1 − k2 )(k1 + k2 )2 ,

J Pth (d3 ) = k2 (k1 − k2 )2 ,

150

Leonid G. Fel

d1 −1 k Fig. 9 Matrix representation of the set τ (1 − zd3 )(1 − zd2 ) k=0 z − (1 − zd1 )Φ(d3 ; z) for the 3 λ symmetric semigroup S(d ). Positive and negative contributions of the four terms z (λ = 0), aii di , and L = a11 d1 + a33 d3 to the numerator Q(d3 ; z) are respectively marked by gray and white. 2

2

2

2

2

QPth (d3 ; z) = 1 − z(k1 +k2 )(k1 −k2 ) − z2k1 k2 − zk1 (k1 +k2 ) 2 −k (k 2 +k 2 ) 2 1 2

+ zk1 (k1 +k2 )

2

2

+ zk1 (k1 +2k1 k2 −k2 ) ,

F Pth (d3 ) = k1 [k12 − k22 + 2(k1 k2 − k1 − k2 )], GPth (d3 ) =

1 + k13 − k23 + k1 (k1 k2 − k1 − k2 ). 2

6.2 The Frobenius problem for the symmetric semigroup S(d3 ) Being a special type of a nonsymmetric semigroup S(d3 ), the case of a symmetric semigroup essentially simplifies formulas (122) and (128) for Q(d3 ; z), F (d3 ), and G(d3 ). First, the matrix representation of the set d 1 −1 τ (1 − zd3 )(1 − zd2 ) zk − (1 − zd1 )Φ(d3 ; z) k=0

seems much simpler (see Fig. 9) and leads to the known Hilbert series (23) with four nonzero terms in the numerator Q(d3 ; z). We let Fs (d3 ) and Gs (d3 ) respectively denote the Frobenius number and the genus for symmetric semigroup S(d3 ) and derive their expressions. The first minimal relation (21) for the given symmetric semigroup together with (8) and (18) yield a11 d1 = a22 d2 , aii ≥ 2,

a22 a33 = d1 ,

a11 a33 = d2 ,

i = 1, 2, 3.

(136)

Substituting relations (136) in (126) and (128), we obtain Js (d3 ) = a33 d3 ,

1 L1,2 = [2a11 d1 + a33 d3 ± a33 d3 ], 2

Fs (d3 ) = a11 d1 + a33 d3 − (d1 + d2 + d3 ),

(137)

1 Gs (d3 ) = [1 + Fs (d3 )]. 2 The last formula in (137) has the following corollary. Corollary 5 Let {d1 , d2 , d3 } be the minimal generating set for the symmetric semigroup S(d3 ). Then the Frobenius number Fs (d3 ) is always an odd integer.


151

We finish this section with an interesting observation. We recall that according to [1], all elements di of the minimal generating set {d1 , d2 , d3 } for a nonsymmetric semigroup S(d3 ) exceed 2. It appears that this restriction becomes even stronger for a symmetric semigroup. Lemma 11 Let {d1 , d2 , d3 } be the minimal generating set for the symmetric semigroup S(d3 ) with the first minimal relation be defined by (21). Then all elements di of the min-

imal set exceed 3. Proof Let S(d3 ) be a symmetric semigroup with the first minimal relation defined by (21). According to (136), we then have d1 , d2 ≥ 4 because the generating set d1 , d2 , d3 would otherwise not be minimal. Substituting expressions (136) for d1 and d2 in the first minimal relation a33 d3 = a31 d1 + a32 d2 and keeping in mind that a31 , a32 ≥ 1, we obtain a33 d3 = a31 a22 a33 + a32 a11 a33

−→

d3 = a31 a22 + a32 a11 ≥ 4.

Combining all restrictions di ≥ 4, we prove Lemma 11.

We note that coincidence relation (136) aii di = ajj dj , i = j , is not necessarily valid for the indices corresponding to the least and next-to-least generators d1 and d2 . The following example corresponds to the case where a11 d1 = a33 d3 . Example 4 The triple {4, 5, 6} generates a symmetric semigroup (with minimal possible di ). ⎛ ⎞ 3 0 −2 (s) ⎝ 2 −1⎠ , Δ(4, 5, 6) = {1, 2, 3, 7}, A 3 = −1 −3 0 2 H (d3 ; z) =

(1 − z10 )(1 − z12 ) , (1 − z4 )(1 − z5 )(1 − z6 )

G(d3 ) = 4,

F (d3 ) = 7.

7 Lower and upper bounds on the Frobenius number F (d3 ) The history of bounds for the Frobenius number F (d3 ) dates back to I. Schur (see Theorem 1 in [3]) and has been the subject of intensive study for the last 40 years (see [18], [27], [19], [28], [26], [29] and the references therein). The subject is still a very active research area (see [30], [31], [32], [33], and Lecture 4 in [34]). In this section, we enhance the lower bound [28], [26] of F (d3 ) that had been considered the best for the past decade. We also refute two recently advanced conjectures on the upper bound of F (d3 ) [30]. 7.1 Lower bound of F (d3 ) As shown by Rödseth in 1990 [28] and independently by Davison in 1994 [26], the inequality √ ! (138) F (d3 ) ≥ 3 d1 d2 d3 − (d1 + d2 + d3 ) √ holds, where “the constant 3 cannot be replaced by a larger value with the inequality remaining true for all d1 , d2 , d3 ” (Theorem 2.3 in [26]). Obtained combinatorially, it does not distinguish between the triples generating the nonsymmetric and symmetric semigroups. In fact, the lower bound of F (d3 ) for the set {d1 , d2 , d3 } generating symmetric semigroups

152

Leonid G. Fel

is stronger than (138). Moreover, it appears that the case of nonsymmetric semigroups also allows slightly enhancing lower bound (138). To show this, we use the results in Sects. 6.1 and 6.2 and start with the lower bound for the nonsymmetric semigroups S(d3 ). Lemma 12 Let {d1 , d2 , d3 } be the minimal generating set for a nonsymmetric semigroup. Then √ ! (139) F (d3 ) ≥ 3 d1 d2 d3 + 1 − (d1 + d2 + d3 ). Proof We first find the lower bound for a, d . We start with inequalities that follow from (18), a11 a22 > d3 , −→

a22 a33 > d1 ,

a33 a11 > d2

2 2 2 a11 a22 a33 > d1 d2 d3 .

According to (140) and the inequality for symmetric polynomials [35], we obtain ! a, d ≥ 3 (aii di )1/3 > 3 d1 d2 d3 .

(140)

(141)

i=1

Using (97), we can write (a11 d1 − a22 d2 )2 + (a11 d1 − a33 d3 )2 + (a22 d2 − a33 d3 )2 ≥ 12 + 12 + 22 = 6 or, in other words, 1 a11 a22 d1 d2 + a22 a33 d2 d3 + a33 a11 d3 d1 ≤ a, d 2 − 1. 3

(142)

We consider the lower bound of F (d3 ) in two regions for a, d : √ √ 1. a, d > 2 3 d1 d2 d3 + 1 and √ √ 2. a, d ≤ 2 3 d1 d2 d3 + 1. In the first region, we immediately obtain (139) according to expression (128) for F (d3 ). We consider the second region and observe that by (142), J 2 (d3 ) = 4d1 d2 d3 + a, d 2 − 4

3 i>j

1 aii ajj di dj ≥ 4d1 d2 d3 + 4 − a, d 2 ≥ 0. 3

We thus obtain F (d ) + 3

3 i

1 di ≥ (a, d + 2

"

1 4d1 d2 d3 + 4 − a, d 2 ). 3

(143)

√ We let x = a, d and c = d1 d2 d3 + 1 and consider the function f (x) = 1/2 x + ! √ √ 4c2 − x 2 /3 in the interval 3 c2 − 1 < x ≤ 2 3 c. It is easy to find its minimum: √ √ min f (x) = 3 c when x = 2 3 c. Comparing this with (143), we obtain (139) in the second region. Combining the bounds in the two regions, we complete the proof of the lemma. In the next lemma, we find the lower bound of Fs (d3 ) for a symmetric semigroup S(d3 ).


153

Lemma 13 Let {d1 , d2 , d3 } be the minimal generating set for the symmetric semigroup S(d3 ). Then ! (144) Fs (d3 ) ≥ 2 d1 d2 d3 − (d1 + d2 + d3 ). Proof Substituting a11 a33 = d2 from (137) in expression (136) for Fs (d3 ), we obtain Fs (d3 ) + (d1 + d2 + d3 ) = a11 d1 + a33 d3 =

! d1 d2 + a33 d3 ≥ 2 d1 d2 d3 , a33

which proves the lemma.

Lower bound (144) is stronger than lower bound (139) for F (d3 ) in the generic case of a nonsymmetric semigroup S(d3 ). 7.2 Upper bound of F (d3 ) A recent paper [30] stated two conjectures based on numerical calculations for more than √ ten thousand4 randomly chosen admissible triples (d1 , d2 , d3 ) satisfying the condition d1 d2 d3 < 2 · 10 (in fact, the typical values of di did not exceed 750 according to a private communication from M. Beck). We quote from [30]: “For all admissible triples (d1 , d2 , d3 ) the Frobenius number F (d3 ) can be bounded from above, + (d3 ), F (d3 ) ≤ FC;ν

+ FC;ν (d3 ) = C(d1 d2 d3 )ν − (d1 + d2 + d3 ),

(145)

where C = const and ν < 2/3.” Further, a stronger conjecture was posed: “In fact, our data suggests, more precisely, that for all admissible triples (d1 , d2 , d3 ), + (d3 ), F (d3 ) ≤ F1;5/8

i.e.,

+ F1;5/8 (d3 ) = (d1 d2 d3 )5/8 − (d1 + d2 + d3 ).”

(146)

Considering the lower and upper bounds of F (d3 ) for specific triples (d1 , d2 , d3 ), Schlage-Puchta [36] proved that for each integer d1 , there are φ(d1 )/2 pairs (d2 , d3 ) such that d1 < d2 < d3 < 2d1 ,

gcd(d1 , d2 ) = gcd(d1 , d3 ) = 1,

d12 /2 ≤ F (d3 ) ≤ 2d12 ,

(147)

where φ(d1 ) is the Euler function. The further comparison of lower bound (147) with conjectured value (146) led him to claim a disproof of this conjecture (which can be shown for increasing d1 ). In fact, this would be correct if we knew that there exists at least one admissible triple (d1 , d2 , d3 ) satisfying (147). But [36] does not consider the admissibility of the triples where conjecture (146) fails. Hence, strictly speaking, [36] does not refute conjecture (146). In this section, we refute the first conjecture, (145). This necessarily makes the second conjecture, (146), false. Following [30], we recall the terms needed for discussing these conjectures. First, we say that the triple (d1 , d2 , d3 ) constitutes an almost arithmetic sequence if there exist integers a and b such that d2 = ad1 + b,

d3 = ad1 + 2b,

a ≥ 1,

b ≥ 1,

gcd(d1 , b) = 1.

(148)

154

Leonid G. Fel

Next, we say that the triple (d1 , d2 , d3 ) is excluded if at least one of the following holds: 1. One of the elements di is representable by the other two.

(149)

2. One element di divides the sum of the other two.

(150)

3. The elements di represent an almost arithmetic sequence.

(151)

We say that the triple (d1 , d2 , d3 ) is admissible if it is not excluded and consists of pairwise coprime integers. We consider a triple (d1 , d2 , d3 ) such that its elements di have the representation d1 = 2l + 1,

d2 = d1 + 2 = 2l + 3,

d3 = 2d1 + 1 = 4l + 3,

l 1.

(152)

We note that triple (152) differs from those considered in [36]. The minimal set {2l + 1, 2l + 3, 4l + 3} generates a nonsymmetric semigroup S(d3 ) with a matrix A 3 of the first minimal relation and the corresponding Frobenius number (see (133)) ⎛ ⎞ l + 3 −l −1 ⎝ −l l + 1 −1⎠ , A 3= −3 −1 2

F (2l + 1, 2l + 3, 4l + 3) = 2l 2 + 3l − 1.

(153)

We can always choose d1 , d1 + 2, and 2d1 + 1 as pairwise coprime integers (see Example 5). Lemma 14 Triple (152) is admissible. Proof First, (149) is not satisfied, because the set {2l +1, 2l +3, 4l +3} is minimal according to (153). Next, (150) is not satisfied, because 4l + 4 ∈ N, 4l + 3

6l + 4 ∈ N, 2l + 3

6l + 6 ∈ N. 2l + 1

(154)

Finally, calculating a = (2d2 − d3 )/d1 , which satisfies (148), we obtain a = 3/(2l + 1) and conclude that (151) is also not satisfied. Hence, triple (152) is admissible. Lemma 15 Let d3 be an admissible triple given by (152). The Frobenius number F (d3 ) + (d3 ) given by cannot be bounded from above by FC;ν + (d3 ) = C(d1 d2 d3 )ν − (d1 + d2 + d3 ), FC;ν

ν < 2/3, (155)

for any C = const.

Proof We consider triple (152), which is admissible according to Lemma 14, and let δC;ν (l) denote the ratio δC;ν (l) =

+ FC;ν (2l + 1, 2l + 3, 4l + 3)

F (2l + 1, 2l + 3, 4l + 3)

.

(156)

To satisfy conjecture (145), we must find C = const and ν < 2/3 such that δC;ν (l) ≥ 1

(157)


155

for all l > 1. But this is not true. Indeed, we find that the leading term of the asymptotic expression for δC;ν (l) as l → ∞ is δC;ν (l) C24ν−1 l 3ν−2 .

(158)

We note that its decrease as l → ∞ suffices to violate (157) when l exceeds the critical value lcr , l > lcr ,

log2 lcr =

4ν − 1 log2 C + 2 − 3ν 2 − 3ν

for all ν < 2/3. This is true for arbitrary large finite C.

(159)

Therefore, conjecture (145) is refuted. For conjecture (146), the critical value lcr = 212 for ν = 5/8 and C = 1 follows from (159). This explains the results in [30] where all randomly chosen integers di did not exceed 213 . Example 5 shows that the situation changes immediately if di > 213 . Example 5

29 < d1 < 210 ,

⎧ ⎪ ⎨d1 = 1001 = 7 · 11 · 13, d2 = 1003 = 17 · 59, ⎪ ⎩ d3 = 2003 = 1 · 2003,

F (1001, 1003, 2003) = 501499, + (1001, 1003, 2003) = 648574.82, F1;5/8 ⎧ ⎪ ⎨d1 = 10001 = 73 · 137, 13 14 2 < d1 < 2 , d2 = 10003 = 7 · 1429, ⎪ ⎩ d3 = 20003 = 83 · 241, F (10001, 10003, 20003) = 50014999, + (10001, 10003, 20003) = 48745746.76, F1;5/8 ⎧ ⎪ ⎨d1 = 100001 = 11 · 9091, 16 17 2 < d1 < 2 , d2 = 100003 = 1 · 100003, ⎪ ⎩ d3 = 200003 = 1 · 200003, F (100001, 100003, 200003) = 5000149999, + (100001, 100003, 200003) = 3656883908.53. F1;5/8

8 Genera of higher orders The generating function of unrepresentable integers Φ(dm ; z) is a source of more information about the set Δ(dm ). We show how Φ(dm ; z) can be used by computing the power series sn, g0 (dm ) = G(dm ). (160) gn (dm ) = s∈Δ(dm )

156

Leonid G. Fel

The simplest series g1 (d2 ) was calculated by Brown and Shiue [37] and was later generalized by Rödseth for arbitrary n ≥ 0 [38]. In this section, we give a regular approach to that problem and compute some of the g1 for 3D nonsymmetric semigroups based on the results obtained in Sect. 6. Writing the derivative d n /dzn = ∂zn , we find ∂zn Φ(dm ; 1) = gn (dm ) − I1 gn−1 (dm ) + I2 gn−2 (dm ) − · · · ± In−1 g1 (dm ),

(161)

where the coefficients Ik appear as symmetric invariants of the set of integers {1, 2, . . . , n − 1}, I1 =

n−1

j=

j =1

In−2 = In−1

n(n − 1) , 2

n−1 1 j =1

j

j1 j2 ,

...,

j1 >j2 =1

In−1 =

,

n−1

I2 = n−1

j = (n − 1)! ,

j =1

and ∂zn Φ(dm ; 1) denotes ∂zn Φ(dm ; z)|z=1 . Successive calculation of the first three terms gives g1 (dm ) = ∂z Φ(dm ; 1), g2 (dm ) = (∂z2 + ∂z )Φ(dm ; 1), g3 (dm ) = (∂z3 + 3∂z2 + ∂z )Φ(dm ; 1). The first genus g1 (d3 ) for the nonsymmetric semigroup S(d3 ) is g1 (d3 ) =

3 3 1 di + Ai di2 −1+ 12 i=1 i=1 +

3 i>j

Bij di dj −

3 i=1

aii

3

C j dj ,

(162)

i=1

where Ai = (aii − 1)(2aii − 1),

Bij = 3(aii − 1)(ajj − 1) − aii ajj ,

Cj = 2ajj − 3.

We will return to the higher genera gn (d3 ), n ≥ 2, elsewhere. In Example 6, we calculate g1 (d3 ) for the triples presented in Example 2. Example 6 g1 (23, 29, 44) = 9526,

g1 (137, 251, 256) = 2380976,

g1 (1563, 2275, 2503) = 12178811815. Acknowledgements The useful discussions with A. Juhász and his help in preparing the paper are deeply appreciated. The author thanks V.I. Arnold, M. Morales, N. Wormald, and J. Ramirez Alfonsin for providing him with their papers and M. Beck for his communication. The author also thanks the referee for drawing his attention to [23], [28], and [36]. This research was supported by a Kamea Fellowship. Part of the paper was written during the author’s stay at the Isaac Newton Institute for Mathematical Sciences, and its hospitality is deeply appreciated.


157

References 1. Abhyankar S (1967) Local rings of high embedding dimension. Am J Math 89:1073–1077 2. Stanley RP (1996) Combinatorics and commutative algebra, 2nd edn. Progr Math, vol 41. Birkhäuser, Boston 3. Brauer A (1942) On a problem of partitions. Am J Math 64:299–312 4. Kunz E (1970) The value-semigroup of a one-dimensional Gorenstein ring. Proc Am Math Soc 25:748– 751 5. Herzog J (1970) Generators and relations of Abelian semigroups and semigroup rings. Manuscripta Math 3:175–193 6. Fröberg R, Gottlieb C, Häggkvist R (1987) On numerical semigroups. Semigroup Forum 35(1):63–83 7. Curtis F (1990) On formulas for the Frobenius number of a numerical semigroup. Math Scand 67(2):190– 192 8. Van der Corput JG (1938) Über Summen von Primzahlen und Primzahlquadraten. Math Ann 116:1–50 9. Roberts JB (1956) Note on linear forms. Proc Am Math Soc 7:465–469 10. Székely LA, Wormald NC (1986) Generating functions for the Frobenius problem with 2 and 3 generators. Math Chronicle 15:49–57 11. Johnson SM (1960) A linear Diophantine problem. Can J Math 12:390–398 12. Kraft J (1985) Singularity of monomial curves in A3 and Gorenstein monomial curves in A4 . Can J Math 37(5):872–892 13. Fröberg R (1994) The Frobenius number of some semigroups. Commun Algebra 22(14):6021–6024 14. Sylvester JJ et al (1884) Problems from the theory of numbers, with solutions. Educational Times 40–41 15. Denham G (2003) Short generating functions for some semigroup algebras. Electron J Comb 10:R36 16. Stanley RP (1979) Invariants of finite groups and their applications to combinatorics. Bull Am Math Soc NS 1(3):475–511 17. Fel LG, Rubinstein BY (2002) Sylvester waves in the Coxeter groups. Ramanujan J 6(3):307–329 18. Brauer A, Schockley JE (1962) On a problem of Frobenius. J Reine Angew Math 211:215–220 19. Selmer ES (1977) On the linear Diophantine problem of Frobenius. J Reine Angew Math 293/294:1–17 20. Selmer ES, Beyer Ö (1978) On the linear Diophantine problem of Frobenius in three variables. J Reine Angew Math 301:161–170 21. Rødseth, ØJ (1978/1979) On a linear Diophantine problem of Frobenius I, II. J Reine Angew Math 301:171–178, 307/308:431–440 22. Greenberg H (1988) Solution to a linear Diophantine equation for nonnegative integers. J Algorithms 9(3):343–353 23. Arnold VI (to appear) Geometry of continued fractions associated with Frobenius numbers. Funct Anal Other Math 24. Nijenhuis A, Wilf HS (1972) Representations of integers by linear forms in nonnegative integers. J Number Theory 4:98–106 25. Barvinok A, Woods K (2003) Short rational generating functions for lattice point problems. J Am Math Soc 16(4):957–979 26. Davison JL (1994) On the linear Diophantine problem of Frobenius. J Number Theory 48(3):353–363 27. Erd˝os P, Graham RL (1972) On the linear Diophantine problem of Frobenius. Acta Arithm 21:399–408 28. Rødseth ØJ (1990) An upper bound for the h-range of the postage stamp problem. Acta Arithm 54(4):301–306 29. Hamidoune YO (1998) On the Diophantine Frobenius problem. Portugal Math 55(4):425–449 30. Beck M, Einstein D, Zacks S (2003) Some experimental results on the Frobenius problem. Exp Math 12(3):263–269 31. Arnold VI (1999) Weak asymptotics for the numbers of solutions of Diophantine problems. Funct Anal Appl 33(4):292–293 32. Arnold VI (2006) Arithmetical turbulence of selfsimilar fluctuations statistics of large Frobenius numbers of additive semigroups of integers. Preprint IC/2006/037. Abdus Salam Intl Centre Theor Phys, http://www.ictp.it/~pub_off. Cited 22 Oct 2006 33. Arnold VI (2006) Geometry and growth rate of Frobenius numbers of additive semigroups. Math Phys Anal Geom 9(2):95–108 34. Arnold VI (2006) Experimental’noe nablyudenie matematicheskikh faktov (Experimental discovery of mathematical facts). Moscow Center for Continuous Mathematical Education, Moscow 35. Hardy GH, Littlewood JE, Pólya G (1952) Inequalities, 2nd edn. Cambridge Univ Press, Cambridge 36. Schlage-Puchta J-C (2005) An estimate for Frobenius’s Diophantine problem in three dimensions. J Integer Seq 8(1):05.1.7 37. Brown TC, Shiue PJ-S (1993) A remark related to the Frobenius problem. Fibonacci Q 31(1):32–36 38. Rødseth ØJ (1994) A note on T.C. Brown and P.J.-S. Shiue’s paper “A remark related to the Frobenius problem.” Fibonacci Q 32(5):407–408