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New York Journal of Mathematics

New York J. Math.27(2021) 1060–1084.

On curves with high multiplicity on ℙ(𝒂, 𝒃, 𝒄) for 𝐦𝐢𝐧(𝒂, 𝒃, 𝒄) ≤ 𝟒

David McKinnon, Rindra Razafy, Matthew Satriano and Yuxuan Sun

Abstract. On a weighted projective surfaceℙ(𝑎, 𝑏, 𝑐)withmin(𝑎, 𝑏, 𝑐) ≤ 4, we compute lower bounds for theeffective thresholdof an ample divisor, in other words, the highest multiplicity a section of the divisor can have at a specified point. We expect that these bounds are close to being sharp. This translates into finding divisor classes on the blowup ofℙ(𝑎, 𝑏, 𝑐)that generate a cone contained in, and probably close to, the effective cone.

Contents

1. Introduction 1060

Acknowledgments 1063

2. Proof of Proposition 1.2 and preliminary reductions 1064 3. Ehrhart quasi-polynomials for𝑏𝐷𝑥and𝑐𝐷𝑥 1068 4. Bounding𝑐0(𝑏𝐷𝑥, 𝑛)and𝑐0(𝑐𝐷𝑥, 𝑛) 1072

5. Proof of Theorem 1.5 1077

References 1083

1. Introduction

Given a projective variety𝑋and a point𝑄 ∈ 𝑋, it is, in general, a notoriously difficult problem to calculate the pseudo-effective cone of the blow-up Bl𝑄(𝑋) in terms of the pseudo-effective cone of𝑋. Even addressing thea priorieas- ier question of when Bl𝑄(𝑋)is a Mori Dream Space, where𝑋 = ℙ(𝑎, 𝑏, 𝑐)is a weighted projective surface and𝑄is the identity of its torus, is already challeng- ing and has a rich history [Hun82, Cut91, Sri91, GNW94, CT15, GK16, He17, GGK20]. To gain information about the pseudo-effective cone of Bl𝑄(𝑋), we consider the following quantity, cf. [Fuj92].

Received November 23, 2020.

2010Mathematics Subject Classification. 14C15, 14F43.

Key words and phrases. weighted projective spaces, curves, Ehrhart polynomial, multiplicity.

The first author and third authors were partially supported by a Discovery Grant from the Na- tional Science and Engineering Board of Canada. The second and fourth authors were supported by an Undergraduate Student Research Award from the National Science and Engineering Board of Canada.

ISSN 1076-9803/2021

1060

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Definition 1.1. Let 𝑋 be a projective variety defined over a field𝑘, 𝐷 a 𝑘- rationalℚ-divisor, and 𝑄 a 𝑘-rational point of𝑋. Let 𝜋 be the blowup of 𝑋 at𝑄and𝐸the exceptional divisor of𝜋. We say theeffective thresholdis

𝛾𝑄(𝐷) ∶= sup{𝛾 > 0 ∣ 𝜋(𝐷) − 𝛾𝐸 is pseudo-effective}.

The quantity𝛾𝑄(𝐷)can be reinterpreted concretely as follows:if there is a divisor in the class of𝐷with multiplicity𝑚at𝑄, then𝛾𝑄(𝐷) ≥ 𝑚. Conversely, if𝛾𝑄(𝐷) = 𝑚, then for all𝜖 > 0, the class𝜋𝐷 − (𝑚 − 𝜖)𝐸is pseudo-effective, so 𝐷contains divisors of multiplicity arbitrarily close to𝑚, at least in aℚ-divisor sense. So, computing𝛾𝑄(𝐷)essentially amounts to computing

sup

𝐶,𝑚

{1

𝑚mult𝑄(𝐶)}

as𝑚varies through positive integers and𝐶varies through divisors in the divisor class𝑚𝐷.

In this paper, we give characteristic-free lower bounds for𝛾𝑄(𝐷)in the case where𝑋is the weighted projective surfaceℙ(𝑎, 𝑏, 𝑐)andmin(𝑎, 𝑏, 𝑐) ≤ 4. In fact, we do more than this: we introduce a combinatorial quantity 𝛾expected which is a lower bound on𝛾, and compute𝛾expectedexactly. It is worth remark- ing here that although the motivation for studying𝛾𝑄 is geometric, our lower bounds on 𝛾𝑄 also have consequences for Diophantine approximation prob- lems related to generalizations of Roth’s famous 1955 theorem [Ro55], see e.g., [MR16, Theorem 3.3] and [MS20, Section 8].

In [GGK20], the authors make a series of detailed calculations closely related to what we compute in this paper. In particular, they search the spaces of global sections of toric surfaces of Picard rank one for irreducible curves whose strict transforms have negative self-intersection upon blowing up a point. If there is such a curve, then the pseudo-effective cone of the blowup will be finitely gen- erated by the exceptional divisor of the blowup and another curve of negative self-intersection.

In this paper, we compute not only curves, but also the corresponding value of the effective threshold. We do not prove that the curves we find are always generators of the pseudo-effective cone, but in most cases the value of𝛾 we compute is expected to be equal or very close to the actual value. As the authors of [GGK20] also point out, our quantity𝛾expectedis expected to be very close to the actual value of𝛾.

Since𝑋 = ℙ(𝑎, 𝑏, 𝑐)is a toric surface, if the point𝑄does not lie in the main torus orbit𝑇, then computing𝛾𝑄 is generally straightforward, so we may as- sume that𝑄lies in𝑇. Furthermore, we can choose𝑎,𝑏, and𝑐to be pairwise coprime, with𝑎 ≤ 𝑏 ≤ 𝑐. These inequalities are always strict unless𝑎 = 𝑏 = 1, in which case𝛾can be computed directly. Thus, we may assume that𝑎 < 𝑏 < 𝑐. Finally, since𝑋has Picard rank1, it suffices to compute𝛾𝑄(𝐻), where𝐻is the generator of the Cartier class group.

Our first result concerns the case𝑎 < 4and serves as a warm-up to our main result.

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Proposition 1.2. Let𝑎 < 𝑏 < 𝑐be pairwise coprime, so we may write𝑐 = 𝑝𝑎+𝑞𝑏 with𝑝, 𝑞 ∈ ℤand0 ≤ 𝑞 < 𝑎. Let𝑄be in the torus ofℙ(𝑎, 𝑏, 𝑐)and𝐻be the generator of the Cartier class group. Then

𝛾𝑄(𝐻) ≥ {(𝑞 + 1)𝑏, 𝑝 ≥ 0

(𝑎 − 1)𝑏, 𝑝 < 0and𝑎 ≤ 3.

We do not claim that this Proposition is new. Indeed, in [HKL18], Hausen, Keicher, and Laface prove a number of results along these lines, and moreover they obtain all the results of Proposition 1.2, as a consequence of their Theorems 1.1 and 1.2. Despite the lack of novelty in Proposition 1.2, we present a proof of it to illustrate our techniques in a simpler setting.

Proposition 1.2 yields lower bounds on𝛾𝑄when𝑎 ≤ 3. Moving from𝑎 ≤ 3to 𝑎 = 4is significantly more involved. In order to state our results, we first discuss our main technique of proof. Note that if𝑚 ∈ ℚ+and𝑚𝐻is a Weil divisor such thatℎ0(𝑚𝐻) >(𝜈+1

2

)

, then there is a global section𝑔of𝑚𝐻that vanishes at𝑄 to order𝜈. Writing𝑚 = 𝑚1

𝑚2 with𝑚1, 𝑚2 ∈ ℤ+, we see 𝑔𝑚2 ∈ 𝐻0(𝑋, 𝑚1𝐻) vanishes to order𝜈𝑚2. By definition, it follows that𝛾𝑄(𝐻) ≥ 𝜈𝑚2

𝑚1 = 𝜈

𝑚. This motivates the following definition.

Definition 1.3. For any Weil divisor𝐷, let

𝜈(𝐷) ∶= max {𝑑 ∈ ℤ+∣ ℎ0(𝐷) >(𝑑 + 1 2

) } .

If𝐻denotes the generator of the Cartier class group ofℙ(𝑎, 𝑏, 𝑐)with𝑎 < 𝑏 < 𝑐, then let

𝛾expected(𝐻) ∶= sup {𝜈(𝑚𝐻)

𝑚 ∣ 𝑚 ∈ 1

𝑏ℤ+∪ 1 𝑐ℤ+} .

Remark1.4. Note that the definition of𝛾expectedconsiders only some of the Weil divisors. In particular, since the Weil class group ofℙ(𝑎, 𝑏, 𝑐)is generated by

1

𝑎𝑏𝑐𝐻, every Weil divisor that appears in the definition of𝛾expectedis a multiple of𝑎in the Weil class group.

We can now state our main result. Recall that Proposition 1.2 already yields lower bounds on𝛾𝑄when𝑝 ≥ 0in general, so we turn to the case𝑝 < 0. Theorem 1.5. With notation and hypotheses as in Proposition1.2, assume𝑎 = 4 and𝑝 < 0. Then

𝛾𝑄(𝐻) ≥ 𝛾expected(𝐻) = 𝜈(𝐷0) 𝑚0 ,

where𝐷0 ∼ 𝑚0𝐻 and𝜈(𝐷0)are computed exactly as follows. Given our con- straints, we have2 < 𝑏

−𝑝 < 16

3. Divide the interval[2,16

3]into a countably infinite sequence of intervals of the form

𝐼𝑘 ∶= [ 16(𝑘 + 1)2

8(𝑘 + 1)2− 4(𝑘 + 1) − 1, 16𝑘2 8𝑘2− 4𝑘 − 1]

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with𝑘 ∈ ℤ+. Then the class of𝐷0is given as follows, depending on the value of

𝑏

−𝑝 ∈ 𝐼𝑘: (1) If 𝑏

−𝑝 ∈ 𝐼𝑘,− ∶= [ 16(𝑘+1)

2

8(𝑘+1)2−4(𝑘+1)−1,2𝑘+1

𝑘 ], then𝐷02𝑘+3

𝑐 𝐻with𝜈(𝐷0) = 4(𝑘 + 1).

(2) If 𝑏

−𝑝 ∈ 𝐼𝑘,+ ∶= [2𝑘+1

𝑘 , 4(2𝑘+1)

2

8𝑘2+4𝑘−1], then𝐷02𝑘+1

𝑏 𝐻with𝜈(𝐷0) = 4(𝑘+1).

(3) If 𝑏

−𝑝 ∈ 𝐼𝑘,−′′ ∶= [4(2𝑘+1)

2

8𝑘2+4𝑘−1, 4𝑘

2𝑘−1], then𝐷0𝑘+1

𝑐 𝐻with𝜈(𝐷0) = 2𝑘 + 1.

(4) If 𝑏

−𝑝 ∈ 𝐼𝑘,+′′ ∶= [ 4𝑘

2𝑘−1, 16𝑘

2

8𝑘2−4𝑘−1], then𝐷0𝑘

𝑏𝐻with𝜈(𝐷0) = 2𝑘 + 1.

Remark1.6. The quantity𝛾expected(𝐻)is the lower bound for𝛾(𝐻)obtained by simple linear algebra: the vanishing to order𝑛of a section of𝐻is equivalent to the vanishing of(𝑛+1

2

)

linear forms on the space of sections of𝐻. We therefore have𝛾(𝐻) ≥ 𝛾expected(𝐻)trivially.

However, one also expects that the two quantities are not so different. First of all, the divisibility restriction in𝛾𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑does not exclude any Cartier divisors, and in the examples that we are aware of, does not change the value of𝛾at all.

More significantly, if 𝛾(𝐻) > 𝛾expected(𝐻) at some point 𝑄 in the main torus orbit, then there is a section𝑠of some multiple𝑚𝐻of𝐻that has an order of vanishing that is greater than𝜈(𝑚𝐻)at𝑄. For any element𝜎of the torus, the section𝜎(𝑠)has unusually high order of vanishing at𝜎(𝑄), so foreverypoint of the main orbit, there is a section of𝑚𝐻 that has unusually high order of vanishing there. This is unlikely – though not downright impossible – and so one expects the two quantities to be close.

Nevertheless, there are examples where𝛾and𝛾expecteddo not agree. For ex- ample, if(𝑎, 𝑏, 𝑐) = (5, 33, 49)or(8, 15, 43), Kurano and Matsuoka ([KM09]) showed that𝛾 and 𝛾expected are not the same. Several other authors, includ- ing Gonzalez Anaya, Gonzalez, and Karu, have obtained other very interest- ing results along these lines, of which an excellent summary can be found in [GAGK21].

The rest of the paper is organized as follows. Section 2 proves Proposition 1.2 and describes some preliminary reductions for Theorem 1.5. Section 3 com- putes the main terms in the count of global sections of multiples of𝐻. Section 4 then begins the process of bounding the error terms, and Section 5 finishes the proof of Theorem 1.5.

Acknowledgments

We are grateful to Kalle Karu for many enlightening email exchanges. We thank the anonymous referee for helpful suggestions. This paper is the outcome of an NSERC-USRA project; we thank NSERC for their support.

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2. Proof of Proposition 1.2 and preliminary reductions

Throughout this paper, we let𝑥, 𝑦, and𝑧 be the weighted projective coor- dinates onℙ(𝑎, 𝑏, 𝑐)with weights𝑎,𝑏, and𝑐, respectively. We let𝐷𝑥,𝐷𝑦, and 𝐷𝑧 denote the Weil divisors defined by the vanishing of 𝑥, 𝑦, and𝑧, respec- tively. We let 𝐻 denote the generator of the Cartier class group, so we have 𝐻 ∼ 𝑏𝑐𝐷𝑥 ∼ 𝑎𝑐𝐷𝑦 ∼ 𝑎𝑏𝐷𝑧. Given any Weil divisor𝐷, we let𝑃𝐷 ⊂ ℝ2be the associated polytope with the property thatℎ0(𝐷) = |𝑃𝐷 ∩ ℤ2|. We sometimes abusively denote|𝑃𝐷 ∩ ℤ2|by|𝑃𝐷|.

It will frequently be useful to write our divisors as integer multiples of𝐷𝑥. Note that if𝛿 ∈ {𝑏, 𝑐}and𝑚 ∈ 1

𝛿+, then𝑚𝐻 ∼ 𝑛𝛿𝐷𝑥where𝑛 = 𝑏𝑐𝑚

𝛿 ∈ ℤ+. After a preliminary lemma, we prove Proposition 2.2 which is a slightly more general version of Proposition 1.2.

Lemma 2.1. With notation and hypotheses as in Proposition1.2, if𝑝 < 0, then 𝑞 ≠ 1. In particular, if𝑝 < 0and𝑎 ≤ 3, then𝑎 = 3and𝑞 = 2.

Proof. If𝑞 = 1, then𝑐 = 𝑝𝑎 + 𝑏 < 𝑏which is a contradiction. If𝑝 < 0and 𝑎 ≤ 3, then𝑞 = 0or𝑞 ≥ 2. The former case cannot occur as it implies𝑐 = 𝑝𝑎 and hence𝑝 > 0. The latter case implies2 ≤ 𝑞 ≤ 𝑎 − 1so𝑞 = 2and𝑎 = 3. Proposition 2.2. With notation and hypotheses as in Proposition1.2, we have

𝛾𝑄(𝐻) ≥ {(𝑞 + 1)𝑏, 𝑝 ≥ 0

(𝑎 − 1)𝑏, 𝑝 < 0, 𝑞 = 𝑎 − 1, and −𝑝𝑎

𝑏 ≤ 1.

Furthermore, if𝑎 ≤ 3and𝑝 < 0, then𝑞 = 𝑎 − 1and −𝑝𝑎

𝑏 ≤ 1automatically hold.

Proof. Note that since𝑎,𝑏, and𝑐are pairwise coprime,𝑝 ≠ 0. First suppose 𝑝 > 0. Then the polytope𝑃𝑎𝐷𝑧 is the convex hull of0,(𝑞, −𝑎), and(−𝑝𝑎

𝑏 , −𝑎), so it contains the triangle𝑇with vertices0,(𝑞, −𝑎),(0, −𝑎). By Pick’s Theorem, 1 +(𝑞+2

2

) ≤ |𝑇 ∩ ℤ2| ≤ |𝑃𝑎𝐷𝑧 ∩ ℤ2|, which implies𝜈(𝑎𝐷𝑧) ≥ 𝑞 + 1. Since 𝑎𝐷𝑧1

𝑏𝐻, we find𝛾𝑄(𝐻) ≥ (𝑞 + 1)𝑏. Next, suppose𝑝 < 0,𝑞 = 𝑎 − 1, and −𝑝𝑎

𝑏 ≤ 1. Notice that the polytope𝑃𝑎𝐷𝑧 is given by the vertices as above, and it contains the triangle 𝑇with vertices 0, (𝑞, −𝑎), and(1, −𝑎)by 𝑝 < 0and −𝑝𝑎

𝑏 ≤ 1. From 𝑞 = 𝑎 − 1 and Pick’s Theorem, we have 1 +(𝑎

2

) = |𝑇 ∩ ℤ2|, which, as in the previous paragraph, implies𝛾𝑄(𝐻) ≥ (𝑎 − 1)𝑏.

Finally, we note that if𝑎 ≤ 3and𝑝 < 0, then Lemma 2.1 tells us𝑎 = 3and 𝑞 = 2 = 𝑎 − 1. Then,𝑏 < 𝑐 = 𝑝𝑎 + 2𝑏implies−𝑝𝑎

𝑏 < 1.

The rest of the paper is concerned with the proof of Theorem 1.5. By Lemma 2.1, since𝑝 < 0and𝑎, 𝑏, 𝑐are pairwise coprime, we must have

𝑞 = 3 = 𝑎 − 1.

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We begin by analyzing𝜈(𝐷0).

Proposition 2.3. With notation and hypotheses as in Theorem1.5, if 𝑏

−𝑝 ∈ 𝐼

𝑘 ∶=

𝐼

𝑘,+∪ 𝐼

𝑘,−, resp.𝐼′′

𝑘 ∶= 𝐼′′

𝑘,+∪ 𝐼′′

𝑘,−, and𝐷0is as in the conclusion of the theorem, then𝜈(𝐷0) ≥ 4(𝑘 + 1), resp.2𝑘 + 1.

Proof. Let𝑛0 ∈ ℤbe such that𝐷0𝑛0

𝑏𝐻 ∼ 𝑛0𝑐𝐷𝑥or𝐷0𝑛0

𝑐 𝐻 ∼ 𝑛0𝑏𝐷𝑥. In the former (respectively latter) case,ℎ0(𝐷0)is given by the number of integer lattice points lying in the polytope

𝑃𝑛

0𝑐𝐷𝑥 = Conv ((0, 0), (−𝑛0𝑐

𝑏, 0) , (−3𝑛0, 4𝑛0)) respectively

𝑃𝑛

0𝑏𝐷𝑥 = Conv ((0, 0), (−𝑛0, 0), (−3𝑛0𝑏 𝑐, 4𝑛0𝑏

𝑐)) . First consider the case −𝑝𝑏 ∈ 𝐼𝑘. As in Theorem 1.5,𝐼𝑘,+ ∶= [2𝑘+1

𝑘 , 4(2𝑘+1)

2

8𝑘2+4𝑘−1] and𝐼𝑘,− ∶= [ 16(𝑘+1)

2

8(𝑘+1)2−4(𝑘+1)−1,2𝑘+1

𝑘 ]. If 𝑏

−𝑝 ∈ 𝐼𝑘,+ , then𝐷0 ∼ 𝑛0𝑐𝐷𝑥with𝑛0 = 2𝑘 + 1and if 𝑏

−𝑝 ∈ 𝐼

𝑘,−, then𝐷0∼ 𝑛0𝑏𝐷𝑥with𝑛0= 2𝑘 + 3. Let 𝑃= Conv((0, 0), (−(2𝑘 + 3), 0), (−3(2𝑘 + 1), 4(2𝑘 + 1))).

We then have𝑃 ⊂ 𝑃. Indeed, if 𝑏

−𝑝 ∈ 𝐼

𝑘,+, then the inclusion follows from

−(2𝑘 + 1)𝑐

𝑏 = −(2𝑘 + 1)(4𝑝

𝑏 + 3) ≤ −(2𝑘 + 1)(4 −𝑘

2𝑘+1 + 3) = −(2𝑘 + 3). If

𝑏

−𝑝 ∈ 𝐼𝑘,−, then the inclusion follows from−3(2𝑘 + 3)𝑏

𝑐 ≤ −3(2𝑘 + 1)and 4(2𝑘 + 3)𝑏

𝑐 ≥ 4(2𝑘 + 1). So, in either case, we have

|𝑃∩ ℤ2| ≤ ℎ0(𝐷0).

Note that the area of𝑃is given by 𝐴(𝑃) = 1

2(4(2𝑘 + 1)(2𝑘 + 3)) = 2(2𝑘 + 1)(2𝑘 + 3) and the number of lattice points on its boundary is given by

𝐵(𝑃) = (2𝑘 + 3) + (2𝑘 + 1) + 4 = 4𝑘 + 8.

Since𝑃is a lattice polygon, applying Pick’s Theorem, we have

|𝑃∩ ℤ2| = 𝐴(𝑃) + 1

2𝐵(𝑃) + 1 = 8𝑘2+ 18𝑘 + 11 =(4(𝑘 + 1) + 1 2

) + 1, which shows𝜈(𝐷0) ≥ 𝜈(𝑃) = 4(𝑘 + 1).

For 𝑏

−𝑝 ∈ 𝐼′′𝑘, the same proof works when using

𝑃′′ = Conv((0, 0), (−(𝑘 + 1), 0), (−3𝑘, 4𝑘))

in place of𝑃.

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Proposition 2.3 therefore gives the lower bound 𝛾expected(𝐻) ≥ 𝜈(𝐷0)

𝑚0 , (2.4)

where𝐷0= 𝑚0𝐻is the divisor class described in Theorem 1.5. To obtain upper bounds, we introduce the following quantities and make use of the subsequent lemma. Let

𝛾expected,𝑏(𝐻) ∶= sup {1

𝑚𝜈(𝑚𝐻) ∣ 𝑚 ∈ 1 𝑐ℤ+} 𝛾expected,𝑐(𝐻) ∶= sup {1

𝑚𝜈(𝑚𝐻) ∣ 𝑚 ∈ 1 𝑏ℤ+} .

Then, we may bound 𝛾expected(𝐻)from above by bounding𝛾expected,𝑏(𝐻) and 𝛾expected,𝑐(𝐻)from above, given that

𝛾expected(𝐻) = max{

𝛾expected,𝑏(𝐻), 𝛾expected,𝑐(𝐻)}

. (2.5)

Lemma 2.6. Supposeℙ(4, 𝑏0, 𝑐0),ℙ(4, 𝑏𝐿, 𝑐𝐿), andℙ(4, 𝑏𝑈, 𝑐𝑈)satisfy the hy- potheses of Theorem1.5, except we need not assume𝑝 < 0. Suppose 𝑏𝐿

−𝑝𝐿 < 𝑏0

−𝑝0 <

𝑏𝑈

−𝑝𝑈 and let𝐻0,𝐻𝐿, and𝐻𝑈 denote the generators of the respective Cartier class groups. Then

1

𝑐0𝛾expected,𝑏0(𝐻0) ≤ 1

𝑐𝐿𝛾expected,𝑏𝐿(𝐻𝐿)

and 1

𝑐0𝛾expected,𝑐0(𝐻0) ≤ 1

𝑐𝑈𝛾expected,𝑐𝑈(𝐻𝑈).

Proof. Since𝑐 = 𝑝𝑎+𝑞𝑏 = 4𝑝+3𝑏, we see𝑏

𝑐 = 1

4𝑝

𝑏+3. As a result,𝑏𝑈

𝑐𝑈 < 𝑏0

𝑐0 < 𝑏𝐿

𝑐𝐿. It follows that

𝑃0∶= Conv ((0, 0), (−1, 0), (−3𝑏0 𝑐0, 4𝑏0

𝑐0))

⊂ Conv ((0, 0), (−1, 0), (−3𝑏𝐿 𝑐𝐿, 4𝑏𝐿

𝑐𝐿))

=∶ 𝑃𝐿.

Since𝑃0, resp.𝑃𝐿, is the polytope of 𝑏𝐷𝑥 onℙ(4, 𝑏0, 𝑐0), resp.ℙ(4, 𝑏𝐿, 𝑐𝐿), we have𝜈(𝑛𝑃0) ≤ 𝜈(𝑛𝑃𝐿)for all𝑛 ≥ 1, and so

(1∕𝑐0)𝛾expected,𝑏(𝐻0) ≤ (1∕𝑐𝐿)𝛾expected,𝑏𝐿(𝐻𝐿).

We obtain the inequality(1∕𝑐0)𝛾expected,𝑐(𝐻0) ≤ (1∕𝑐𝑈)𝛾expected,𝑐𝑈(𝐻𝑈)in a similar manner from the inclusion

Conv ((0, 0), (−𝑐0

𝑏0, 0) , (−3, 4)) ⊂ Conv ((0, 0), (−𝑐𝑈

𝑏𝑈, 0) , (−3, 4)) , the left-hand, resp. right-hand, side being the polytope of𝑐𝐷𝑥 onℙ(4, 𝑏0, 𝑐0),

resp.ℙ(4, 𝑏𝑈, 𝑐𝑈).

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Remark 2.7. Lemma 2.6 may be used to reduce the proof of Theorem 1.5 to special classes of weighted projective spaces with desirable arithmetic proper- ties. The key idea is to compare the values of𝛾expected,𝑏(𝐻)and𝛾expected,𝑐(𝐻)on different weighted projective spaces to generate upper bounds.

Fix a weighted projective spaceℙ(4, 𝑏, 𝑐)satisfying the hypotheses of Theo- rem 1.5 and let𝐷0∼ 𝑚0𝐻and𝜈0 ∶= 𝜈(𝐷0)be as predicted by theorem. We will suppose that𝑚0= 𝑛0

𝑐 with𝑛0∈ ℤ+(the case where𝑚01

𝑏+is handled sim- ilarly). We must prove𝛾expected(𝐻) = 𝜈0

𝑚0 = 𝑐𝜈0

𝑛0. Let𝐼 = [𝛽1

𝛼1,𝛽2

𝛼2]be an interval of the form𝐼𝑘,± or𝐼𝑘,±′′ as in Proposition 2.3, where 𝑏

−𝑝 ∈ 𝐼. Assume 𝑏

−𝑝is in the interior of𝐼. We must show𝛾expected(𝐻) = 𝑐𝜈0

𝑛0. By (2.5), this is equivalent to proving

𝛾expected,𝑏(𝐻) ≤ 𝑐𝜈0

𝑛0 and 𝛾expected,𝑐(𝐻) ≤ 𝑐𝜈0 𝑛0.

This may be done as follows: fix increasing sequences of positive integers{𝑏𝑖}𝑖 and {−𝑝𝑖}𝑖 for which 𝛼1𝑏𝑖 − 𝛽1(−𝑝𝑖) = 1 and𝑐𝑖 ∶= 4𝑝𝑖 + 3𝑏𝑖 > 𝑏𝑖 is such that 4,𝑏𝑖,𝑐𝑖 are pairwise coprime. We may always find such sequences, since 𝛼1, 𝛽1 > 0are coprime in all cases listed in Theorem 1.5. Let𝐻𝑖 denote the generator of the Cartier class group ofℙ(4, 𝑏𝑖, 𝑐𝑖). Then for𝑖sufficiently large,

𝑏𝑖

−𝑝𝑖 ∈ 𝐼is monotonically decreasing with 𝑏𝑖

−𝑝𝑖𝛽1

𝛼1. Given 𝑏

−𝑝 ∈ 𝐼, there exists an𝑁large enough such that 𝑏𝑁

−𝑝𝑁 < 𝑏

−𝑝, so by Lemma 2.6, 1

𝑐𝛾expected,𝑏(𝐻) ≤ 1

𝑐𝑁𝛾expected,𝑏

𝑁(𝐻𝑁) ≤ 1

𝑐𝑁+1𝛾expected,𝑏

𝑁+1(𝐻𝑁+1) ≤ … (2.8) Similarly, fix increasing sequences of positive integers {𝑏𝑖}𝑖 and {−𝑝𝑖}𝑖 for which𝛼2𝑏𝑖 − 𝛽2(−𝑝𝑖) = −1and 𝑐𝑖 ∶= 4𝑝𝑖 + 3𝑏𝑖 > 𝑏𝑖 is such that 4,𝑏𝑖, 𝑐𝑖 are pairwise coprime. As above, such sequences always exist. Let𝐻𝑖 denote the generator of the Cartier class group of ℙ(4, 𝑏𝑖, 𝑐𝑖). Then for 𝑖 sufficiently large, 𝑏

𝑖

−𝑝𝑖 ∈ 𝐼is monotonically increasing with 𝑏

𝑖

−𝑝𝑖𝛽2

𝛼2. Choosing an𝑁large enough such that 𝑏

−𝑝 < 𝑏

𝑛

−𝑝𝑛, we have by Lemma 2.6 1

𝑐𝛾expected,𝑐(𝐻) ≤ 1

𝑐𝑁 𝛾expected,𝑐

𝑁(𝐻𝑁 ) ≤ 1

𝑐𝑁+1𝛾expected,𝑐

𝑁+1(𝐻𝑁+1 ) ≤ … (2.9) We claim that to prove Theorem 1.5 forℙ(4, 𝑏, 𝑐), it is enough to show

𝜈0 𝑛0 ≥ 1

𝑛𝜈 (𝑛

𝑐𝑖𝐻𝑖) (2.10)

and

𝑐𝑖𝜈0 𝑛0 > 𝑏𝑖

𝑛𝜈 (𝑛

𝑏𝑖𝐻𝑖) (2.11)

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for all𝑖sufficiently large and all𝑛 ∈ ℤ+. Indeed, (2.10) implies 𝑐𝑖𝜈0

𝑛0 ≥ 𝑐𝑖sup {1 𝑛𝜈 (𝑛

𝑐𝑖𝐻𝑖) ∣ 𝑛 ∈ ℤ+}

= sup {1

𝑚𝜈 (𝑚𝐻𝑖) ∣ 𝑚 ∈ 1 𝑐𝑖+}

= 𝛾expected,𝑏

𝑖(𝐻𝑖)

for all𝑖sufficiently large, which when combined with (2.8) shows 𝛾expected,𝑏(𝐻) ≤ 𝑐𝜈0

𝑛0. Similarly, (2.11) implies

𝑐𝑖𝜈0

𝑛0 > 𝛾expected,𝑐

𝑖(𝐻𝑖), which when combined with (2.9) shows

𝛾expected,𝑐(𝐻) < 𝑐𝜈0 𝑛0 .

which is exactly our goal. Further, the strict inequality in this latter case will allow us to conclude that no𝐷of the form 𝑛𝑐𝐻(other than the case where𝑛is divisible by𝑛0) gives the required upper bound for𝛾expected(𝐻). Together with the lower bound (2.4), this computes𝛾expected(𝐻).

Note that the above computes𝛾expected(𝐻)assuming 𝑏

−𝑝is in the interior of𝐼. If 𝑏

−𝑝 is an endpoint of𝐼, then it is straightforward to check that𝑏 = 2𝑘 + 1and

−𝑝 = 𝑘, in which case𝑐 = 2𝑘 + 3. In that case, one may verify Theorem 1.5 directly using that the Erhart quasi-polynomial computed in Section 3 is an actual polynomial.

3. Ehrhart quasi-polynomials for𝒃𝑫𝒙 and𝒄𝑫𝒙

Our first goal in this section is to give an expression for the number of lattice points in the polytopes𝑃𝑛𝑏𝐷

𝑥 and𝑃𝑛𝑐𝐷

𝑥.

Proposition 3.1. Keep the notation and hypotheses of Theorem1.5, let𝛿 ∈ {𝑏, 𝑐}, and set𝑠 = 𝑏

𝑐. Then

|𝑃𝑛𝛿𝐷

𝑥 ∩ ℤ2| = 𝑐2(𝛿𝐷𝑥, 𝑛)𝑛2+ 𝑐1(𝛿𝐷𝑥, 𝑛)𝑛 + 𝑐0(𝛿𝐷𝑥, 𝑛), where the𝑐𝑖’s are given as follows.

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(1) For𝛿 = 𝑏, we have𝑐2(𝑏𝐷𝑥, 𝑛) = 2𝑠,𝑐1(𝑏𝐷𝑥, 𝑛) = 1

2(1 + 𝑠 + 4

𝑐), and

𝑐0(𝑏𝐷𝑥, 𝑛) = 1 − 1 8𝑠

(

{4𝑠𝑛}2− {4𝑠𝑛}

)

−5 2{𝑠𝑛} +

4 {⌊4𝑠𝑛⌋

4

}

𝑗=0

{3 4𝑗}

+𝑏 − 1 2 {4𝑛

𝑐 } −

𝑏 {⌊4𝑠𝑛⌋

𝑏

}

𝑗=0

{−𝑝 𝑏 𝑗} .

(2) For𝛿 = 𝑐, we have𝑐2(𝑐𝐷𝑥, 𝑛) = 2

𝑠,𝑐1(𝑐𝐷𝑥, 𝑛) =1

2(1 + 1

𝑠+ 4

𝑏), and

𝑐0(𝑐𝐷𝑥, 𝑛) = 1 − {4𝑛

𝑏 } −𝑏 − 1 2 {4𝑛

𝑏 } +

𝑏{4𝑛

𝑏}

𝑗=0

{−𝑝 𝑏 𝑗} . Proof. First consider𝛿 = 𝑏.

|𝑃𝑛𝑏𝐷

𝑥 ∩ ℤ2| ∶= |||Conv(𝐴, 𝐵, 𝐶)||| ∶=

||||

|||

Conv ((0, 0), (−𝑛, 0), (−3𝑛𝑏 𝑐, 4𝑛𝑏

𝑐))||||||| where𝐵𝐶 is given by𝑦 = 𝑏

𝑝𝑥 + 𝑏

𝑝𝑛 and𝐴𝐶 is given by𝑦 = −4

3𝑥. We will compute|𝑃𝑛𝑏𝐷

𝑥 ∩ ℤ2|by counting the number of lattice points lying on each line segment𝐴𝑗𝐵𝑗, where𝐴𝑗 = (−3

4𝑗, 𝑗)lies on𝐴𝐶and𝐵𝑗 = (𝑝

𝑏𝑗 − 𝑛, 𝑗)lies on𝐵𝐶, for𝑗 = 0, 1, … , ⌊4𝑠𝑛⌋. Here, our approach is similar to that of [L11, Theorem 3.1]. Denote𝑀 = ⌊4𝑠𝑛⌋ = 4𝑠𝑛 − {4𝑠𝑛}. Then,

|𝑃𝑛𝑏𝐷𝑥∩ ℤ2|

=

𝑀 𝑗=0

(

−3 4𝑗

−⌈ 𝑝 𝑏𝑗 − 𝑛

⌉ + 1)

= (𝑛 + 1)(𝑀 + 1) +

𝑀 𝑗=0

(−

⌈3 4𝑗

+⌊ −𝑝 𝑏 𝑗

⌋ )

= (𝑛 + 1)(𝑀 + 1) +

𝑀 𝑗=0

(− (3

4𝑗 + 1 − {3

4𝑗}) +−𝑝

𝑏 𝑗 − {−𝑝 𝑏 𝑗}) +

⌊𝑀 4

⌋ + 1

= 𝑛(𝑀 + 1) +𝑀 4 − {𝑀

4 } + 1 +

𝑀 𝑗=0

(−𝑝 𝑏 −3

4) 𝑗 +

𝑀 𝑗=0

{3 4𝑗} −

𝑀 𝑗=0

{−𝑝 𝑏 𝑗} .

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Rewrite the sums involving fractional parts as sums of a linear term in𝑛and a 𝑐−periodic term in𝑛:

𝑀 𝑗=0

{3 4𝑗} =

⌊𝑀 4

⌋∑3

𝑗=0

{3 4𝑗} +

4 {𝑀

4

}

𝑗=0

{3 4𝑗} = 3

2(𝑠𝑛 − {𝑠𝑛}) +

4 {⌊4𝑠𝑛⌋

4

}

𝑗=0

{3 4𝑗} ,

𝑀 𝑗=0

{−𝑝 𝑏 𝑗} =

⌊𝑀 𝑏

𝑏−1

𝑗=0

{−𝑝 𝑏 𝑗} +

𝑏 {𝑀

𝑏

}

𝑗=0

{−𝑝

𝑏 𝑗} = 𝑏 − 1 2 (4𝑛

𝑐 − {4𝑛 𝑐 }) +

𝑏 {⌊4𝑠𝑛⌋

𝑏

}

𝑗=0

{−𝑝 𝑏 𝑗} ,

where we have used the identity

𝑏−1

𝑗=0

{−𝑝

𝑏 𝑗} = 𝑏 − 1 2

given that𝑏and𝑝are coprime. Moreover, by a direct computation,

𝑀 𝑗=0

(−𝑝 𝑏 −3

4) 𝑗 = −(𝑀 + 1 2

) 𝑐

4𝑏 = −2𝑠𝑛2+({4𝑠𝑛} −1 2) 𝑛−1

8𝑠 (

{4𝑠𝑛}2− {4𝑠𝑛}

) .

Thus, we can write |𝑃𝑛𝑏𝐷𝑥 ∩ ℤ2| = 𝑐2(𝑏𝐷𝑥, 𝑛)𝑛2 + 𝑐1(𝑏𝐷𝑥, 𝑛)𝑛 + 𝑐0(𝑏𝐷𝑥, 𝑛), where

𝑐2(𝑏𝐷𝑥, 𝑛) = 2𝑠, 𝑐1(𝑏𝐷𝑥, 𝑛) =1

2(1 + 𝑠 + 4 𝑐) , 𝑐0(𝑏𝐷𝑥, 𝑛) = 1 − 1

8𝑠 (

{4𝑠𝑛}2− {4𝑠𝑛}

)

−5 2{𝑠𝑛}

+

4 {⌊4𝑠𝑛⌋

4

}

𝑗=0

{3

4𝑗} + 𝑏 − 1 2 {4𝑛

𝑐 } −

𝑏 {⌊4𝑠𝑛⌋

𝑏

}

𝑗=0

{−𝑝 𝑏 𝑗} .

Likewise, we may find the Ehrhart quasi-polynomial for|𝑃𝑛𝑐𝐷𝑥 ∩ ℤ2|. To simplify our calculations, we may consider𝑛𝑎𝐷𝑧 = 4𝑛𝐷𝑧 ∼ 𝑛𝑐𝐷𝑥. By linear equivalence,|𝑃𝑛𝑎𝐷

𝑧 ∩ ℤ2| = |𝑃𝑛𝑐𝐷

𝑥∩ ℤ2|. The polytope of𝑛𝑎𝐷𝑧is given by 𝑃𝑛𝑎𝐷𝑧 = Conv(𝐴, 𝐵, 𝐶) ∶= Conv ((0, 0), (−4𝑝

𝑏 𝑛, −4𝑛) , (3𝑛, −4𝑛)) ,

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with𝐴𝐶contained in the line𝑦 = −4

3𝑥and𝐴𝐵contained in the line𝑦 = 𝑏

𝑝𝑥. Similarly as before:

|𝑃𝑛𝑎𝐷

𝑧 ∩ ℤ2| =

4𝑛 𝑗=0

(

⌊3 4𝑗

−⌈ −𝑝 𝑏 𝑗

⌉ + 1)

=

4𝑛 𝑗=0

(3 4𝑗 − {3

4𝑗}) −

4𝑛 𝑗=0

(−𝑝

𝑏 𝑗 + 1 − {−𝑝 𝑏 𝑗}) +

⌊4𝑛 𝑏

+ 1 + 4𝑛 + 1

= 𝑐 4𝑏

(4𝑛 + 1 2

)

−3

2𝑛 + 2(𝑏 − 1)

𝑏 𝑛 − 𝑏 − 1 2 {4𝑛

𝑏 } +

𝑏{4𝑛

𝑏}

𝑗=0

{−𝑝

𝑏 𝑗} + 4𝑛

𝑏 + 1 − {4𝑛 𝑏 } ,

using expressions that we obtained previously. Thus, we can write

|𝑃𝑛𝑐𝐷𝑥 ∩ ℤ2| = |𝑃𝑛𝑎𝐷𝑧 ∩ ℤ2| = 𝑐2(𝑐𝐷𝑥, 𝑛)𝑛2+ 𝑐1(𝑐𝐷𝑥, 𝑛)𝑛 + 𝑐0(𝑐𝐷𝑥, 𝑛), where

𝑐2(𝑐𝐷𝑥, 𝑛) = 2 𝑠, 𝑐1(𝑐𝐷𝑥, 𝑛) = 1

2(1 +1 𝑠 + 4

𝑏) 𝑛, 𝑐0(𝑐𝐷𝑥, 𝑛) = 1 − {4𝑛

𝑏 } −𝑏 − 1 2 {4𝑛

𝑏 } +

𝑏{4𝑛

𝑏}

𝑗=0

{−𝑝

𝑏 𝑗} .

Our next goal is to give upper bounds on the constant terms of the Ehrhart quasi-polynomials of|𝑃𝑛𝛿𝐷𝑥∩ ℤ2|,𝛿 = 𝑏, 𝑐. In Proposition 3.1, notice that the expressions of the last two terms of𝑐0(𝑏𝐷𝑥, 𝑛) and𝑐0(𝑐𝐷𝑥, 𝑛)are of the same form, which we will analyze in depth in Section 4. In the following, we give a uniform upper bound on𝑐0(𝑏𝐷𝑥, 𝑛)minus its last two terms.

Lemma 3.2. In the expression of𝑐0(𝑏𝐷𝑥, 𝑛), we have

−5 2{𝑠𝑛} +

4 {⌊4𝑠𝑛⌋

4

}

𝑗=0

{3 4𝑗} ≤ 1

8 for all𝑛 ≥ 0, where𝑠 = 𝑏

𝑐. Furthermore:

(1) The above expression is positive if and only if 1

4 < {𝑠𝑛} < 3

10. (2) The above expression is greater than −1

32𝑠 only if{𝑠𝑛} < 1

2+ 1

80𝑠. Proof. Let𝑏𝑛 = 𝑚𝑐 + 𝑟with0 ≤ 𝑟 ≤ 𝑐 − 1, so that{𝑠𝑛} = 𝑟

𝑐. Let𝓁 = 0, 1, 2, 3be the integer such that𝓁𝑐

4 ≤ 𝑟 < (𝓁+1)𝑐

4 . Then⌊4𝑠𝑛⌋ = 4𝑚 + 𝓁and so4 {⌊4𝑠𝑛⌋

4

}

= 𝓁.

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Now, we will bound the given expression from the above for each𝓁 = 0, 1, 2, 3. If𝓁 = 0, then the given expression in the lemma is−5𝑟

2𝑐 ≤ 0. If𝓁 = 1, we have

5𝑟

2𝑐 +3

4 ≤ −5

21

4 +3

4 = 1

8. If𝓁 = 2, we have−5𝑟

2𝑐 +3

4 + 2

4 ≤ −5

21

2 + 5

4 = 0. Lastly, if𝓁 = 3, we have−5𝑟

2𝑐+ 3

4+2

4+ 1

4 ≤ −5

23

4+ 3

2 = −3

8.

For the final statements of the lemma, we see the expression is non-positive if 𝓁 ≠ 1, and so we must have1

4 < {𝑠𝑛}. When𝓁 = 1, we computed the expression is equal to−5𝑟

2𝑐+3

4, which is positive if and only if{𝑠𝑛} = 𝑟

𝑐 < 3

10.

Similarly, the expression could be greater than32𝑠−1 in cases𝓁 = 0, 1, 2. (Note that𝑠 < 1because𝑏 < 𝑐.) Working case by case with the expressions obtained, we obtain that{𝑠𝑛} = 𝑟

𝑐 < 1

2+ 1

80𝑠 in order for the expression to be greater than

−1

32𝑠.

Note that−1

8𝑠({4𝑠𝑛}2− {4𝑠𝑛}) ≤ 1

32𝑠since the function𝑥 − 𝑥2is maximized at 𝑥 = 1

2. Combining this observation with Lemma 3.2, we obtain the following corollary.

Corollary 3.3. We have

𝑐0(𝑏𝐷𝑥, 𝑛) ≤ 9 8 + 1

32𝑠 + 𝑏 − 1 2 {4𝑛

𝑐 } −

𝑏 {⌊4𝑠𝑛⌋

𝑏

}

𝑗=0

{−𝑝 𝑏 𝑗} . Moreover, if{𝑠𝑛} ≥ 1

2+ 1

80𝑠, we may improve the above bound as follows:

𝑐0(𝑏𝐷𝑥, 𝑛) ≤ 1 + 𝑏 − 1 2 {4𝑛

𝑐 } −

𝑏 {⌊4𝑠𝑛⌋

𝑏

}

𝑗=0

{−𝑝 𝑏 𝑗} .

4. Bounding𝒄𝟎(𝒃𝑫𝒙, 𝒏)and𝒄𝟎(𝒄𝑫𝒙, 𝒏)

In this section, we prove the key results needed to bound 𝑐0(𝑏𝐷𝑥, 𝑛) and 𝑐0(𝑐𝐷𝑥, 𝑛). This amounts to obtaining bounds for the expression 𝑏−1

2

{4𝑛 𝑐

}

𝑟 𝑗=0{−𝑝

𝑏 𝑗}, where𝑟 = 𝑏{⌊4𝑠𝑛⌋

𝑏 }. We begin by recording the following lemma.

Lemma 4.1. Let𝑛, 𝑏, 𝑐 ∈ ℤ+with4 < 𝑏 < 𝑐andgcd(4, 𝑏, 𝑐) = 1. Let𝑝 < 0be an integer satisfying4𝑝 + 3𝑏 = 𝑐and𝑠 = 𝑏

𝑐. If𝑟 = 𝑏 {⌊4𝑠𝑛⌋

𝑏

} , then 𝑏 − 1

2 {4𝑛 𝑐 } −

𝑟 𝑗=0

{−𝑝

𝑏 𝑗} ≤ (𝑏 − 1)(𝑟 + 1)

2𝑏 −

𝑟 𝑗=0

{−𝑝 𝑏 𝑗} .

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Proof. Since𝑟 is the reminder when𝑏 is divided into ⌊4𝑠𝑛⌋, we have4𝑠𝑛 =

⌊4𝑠𝑛⌋ + {4𝑠𝑛} = 𝑏⌊⌊4𝑠𝑛⌋

𝑏 ⌋ + 𝑟 + {4𝑠𝑛}. So, {4𝑛

𝑐 } = {4𝑠𝑛 𝑏 } = {⎢

⌊4𝑠𝑛⌋

𝑏

⎦+ 𝑟

𝑏 + {4𝑠𝑛}

𝑏 } = 𝑟 + {4𝑠𝑛}

𝑏 ,

where the last equality uses0 ≤ 𝑟

𝑏𝑏−1

𝑏 and0 ≤ {4𝑠𝑛}

𝑏 < 1

𝑏.

By the above lemma, it suffices to bound the expression on the righthand side. In §4.1, we give a general algorithm to obtain bounds on expressions of the form (𝛽−1)(𝑢+1)

2𝛽 −∑𝑢

𝑗=0{𝛼

𝛽𝑗} when𝛼and𝛽 satisfy particular Diophantine equations, see Corollary 4.3.

4.1. An algorithm to bound expressions of the form(𝜷−𝟏)(𝒖+𝟏)

𝟐𝜷 −∑𝒖 𝒋=𝟎{𝜶

𝜷𝒋}.

Our goal in this subsection is to prove:

Proposition 4.2. Suppose that𝛼0> 𝛼1,𝛽0> 𝛽1, and 𝛼1𝛽0− 𝛽1𝛼0= 𝜎 = ±1

where𝛼𝑖, 𝛽𝑖 ∈ ℤ+. Let𝑢0 = 𝛽1𝑡1+ 𝑢1, where𝑢0, 𝑢1, 𝑡1 ∈ ℤ≥0and0 ≤ 𝑢𝑖 < 𝛽𝑖 for all𝑖. Then

(𝑢0+ 1)(𝛽0− 1)

2𝛽0

𝑢0

𝑗=0

{𝛼0

𝛽0𝑗} = (𝑢1+ 1)(𝛽1− 1)

2𝛽1

𝑢1

𝑗=0

{𝛼1𝑗

𝛽1 } + 𝜖(𝜎, 𝑡1, 𝑢1, 𝛽0, 𝛽1), where

𝜖(𝜎, 𝑡, 𝑢, 𝛽, 𝛽) = (𝑢 + 1)(𝜎𝑢 + 𝛽− 𝛽)

2𝛽𝛽 +𝜎𝑡(𝛽(𝑡 − 𝜎) + 2𝑢 + 1 − 𝛽) 2𝛽

+ ∆(𝜎, 𝑡, 𝑢, 𝛽, 𝛽)

and∆(𝜎, 𝑡, 𝑢, 𝛽, 𝛽) = 1if𝛽− 𝛽 ≤ 𝛽𝑡 + 𝑢and𝜎 = −1, and∆(𝜎, 𝑢, 𝛽, 𝛽) = 0 otherwise.

When applied iteratively, we arrive at the following algorithm.

Corollary 4.3. Suppose we have sequences of positive integers𝛼0 > 𝛼1 > ⋯ >

𝛼𝑁and𝛽0> 𝛽1 > ⋯ > 𝛽𝑁such that for all𝑖,

𝛼𝑖𝛽𝑖−1− 𝛽𝑖𝛼𝑖−1= 𝜎𝑖 = ±1.

Let𝑢0, … , 𝑢𝑁and𝑡1, … , 𝑡𝑁be non-negative integers satisfying𝑢𝑖−1 = 𝛽𝑖𝑡𝑖+𝑢𝑖and 0 ≤ 𝑢𝑖 < 𝛽𝑖for all𝑖. Then

(𝑢0+ 1)(𝛽0− 1)

2𝛽0

𝑢0

𝑗=0

{𝛼0

𝛽0𝑗} = (𝑢𝑁 + 1)(𝛽𝑁− 1) 2𝛽𝑁

𝑢𝑁

𝑗=0

{𝛼𝑁𝑗 𝛽𝑁 } +

𝑁 𝑖=1

𝜖(𝜎𝑖, 𝑡𝑖, 𝑢𝑖, 𝛽𝑖−1, 𝛽𝑖), where𝜖is as in Proposition4.2.

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We begin with the following preliminary lemmas.

Lemma 4.4. Let𝛼, 𝛽 ∈ ℤ+be relatively prime. Then

𝛽−1

𝑘=0

⎢⎢

⎣ 𝛼 𝛽𝑘

⎥⎥

= 1

2(𝛼 − 1)(𝛽 − 1) and

𝛽−1

𝑘=0

⎢⎢ 𝛼 𝛽𝑘⎤

⎥⎥

= 1

2(𝛼 + 1)(𝛽 − 1).

Proof. Notice that ∑𝛽 𝑘=0𝛼

𝛽𝑘⌋ + 𝛽 + 1is the number of lattice points in the triangle with vertices0,(𝛽, 0), and(𝛽, 𝛼). So, by Pick’s Theorem,

𝛽 𝑘=0

⎢⎢

⎣ 𝛼 𝛽𝑘⎥

+ 𝛽 + 1 = 1

2(𝛼𝛽 + 𝛼 + 𝛽 + 1) + 1.

Since∑𝛽 𝑘=0𝛼

𝛽𝑘⌋ = ∑𝛽−1 𝑘=0𝛼

𝛽𝑘⌋ + 𝛼, the first result follows. The second result follows from the first and the fact that∑𝛽−1

𝑘=0𝛼

𝛽𝑘⌉ = (𝛽 − 1) +∑𝛽−1 𝑘=0𝛼

𝛽𝑘⌋.

Lemma 4.5. Suppose that𝛼0 > 𝛼1,𝛽0 > 𝛽1, and 𝛼1𝛽0− 𝛽1𝛼0= 𝜎 = ±1 where𝛼𝑖, 𝛽𝑖 ∈ ℤ+. Then

(1) {𝛼0𝑗

𝛽0} = {𝛼1𝑗

𝛽1 } − 𝜎 𝑗

𝛽1𝛽0 for all0 ≤ 𝑗 < 𝛽1, and (2) for any integer𝑢satisfying0 ≤ 𝑢 < 𝛽1,

𝑢 𝑗=0

{𝛼0𝑗 𝛽0 } =

𝑢 𝑗=0

{𝛼1𝑗 𝛽1 } − 𝜎

𝛽0𝛽1 (𝑢 + 1

2 )

(3)

𝛽1

𝑗=0

{𝛼0𝑗

𝛽0 } = 1 − 𝜎 + (𝛽1+ 𝜎)(𝛽0− 𝜎)

2𝛽0 .

Proof. We begin with the proof of (1). The case𝑗 = 0is clear, so we assume 1 ≤ 𝑗 ≤ 𝛽1− 1. Since𝛼0𝑗

𝛽0 = 𝛼1𝑗

𝛽1 − 𝜎 𝑗

𝛽1𝛽0, it suffices to show0 ≤ {𝛼1𝑗

𝛽1 } − 𝜎 𝑗

𝛽1𝛽0 < 1. Since𝛼1and𝛽1are relatively prime, we see 1

𝛽1 ≤ {𝛼1𝑗

𝛽1 } ≤ 1 − 1

𝛽1. The result then follows from the fact that|𝜎 𝑗

𝛽1𝛽0| < 1

𝛽0 < 1

𝛽1. Part (2) follows from (1) by noting

𝑢 𝑗=0

{𝛼0𝑗 𝛽0 } =

𝑢 𝑗=0

({𝛼1𝑗

𝛽1 } − 𝜎 𝑗 𝛽0𝛽1) =

𝑢 𝑗=0

{𝛼1𝑗 𝛽1 } − 𝜎

𝛽0𝛽1 (𝑢 + 1

2 )

.

To prove (3), we use∑𝛽1 𝑗=0{𝛼0𝑗

𝛽0 } = 𝜎+1

2𝜎

𝛽0 +∑𝛽1−1 𝑗=0 {𝛼0𝑗

𝛽0 }and part (2) to see

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