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On the Size of Finite Sidon Sets

Kevin O’Bryant
City University of New York, The Graduate Center and the College of Staten Island
[email protected]
Abstract

A Sidon set (also called a Golomb Ruler, a B2B_{2} sequence, and a 11-thin set) is a set of integers containing no nontrivial solutions to the equation a+b=c+da+b=c+d. We improve on the lower bound on the diameter of a Sidon set with kk elements: if kk is sufficiently large and π’œ{\cal A} is a Sidon set with kk elements, then diam⁑(π’œ)β‰₯k2βˆ’1.99405​k3/2\operatorname{diam}({\cal A})\geq k^{2}-1.99405k^{3/2}. Alternatively, if nn is sufficiently large, then the largest subset of {1,2,…,n}\{1,2,\dots,n\} that is a Sidon set has cardinality at most n1/2+0.99703​n1/4n^{1/2}+0.99703n^{1/4}. While these are slight numerical improvements on Balogh-FΓΌredi-Roy (arXiv:2103:15850v2), we use a method that is logically simpler than theirs but more involved computationally.

1 Introduction

A Sidon setΒ [1932.Sidon] is a set π’œ{\cal A} of integers with no solutions to a1+a2=a3+a4,aiβˆˆπ’œa_{1}+a_{2}=a_{3}+a_{4},a_{i}\in{\cal A}, aside from the trivial {a1,a2}={a3,a4}\{a_{1},a_{2}\}=\{a_{3},a_{4}\}. Equivalently, π’œ{\cal A} is a Sidon set if the coefficients of (βˆ‘aβˆˆπ’œza)2(\sum_{a\in{\cal A}}z^{a})^{2} are bounded by 2. It is also equivalent, and more useful in this particular work, that π’œ{\cal A} is a Sidon set if there are no solutions to a1βˆ’a2=a3βˆ’a4a_{1}-a_{2}=a_{3}-a_{4} with a1β‰ a2a_{1}\neq a_{2} and a1β‰ a3a_{1}\neq a_{3}. We refer our readers toΒ [mybib] for a survey of Sidon sets and an extensive bibliography. Sidon sets are also known as B2B_{2} sets, Golomb rulers, and 1-thin sets.

Throughout this work, π’œ={a1<a2<β‹―<ak}{\cal A}=\{a_{1}<a_{2}<\dots<a_{k}\} is a finite Sidon set of integers. We seek lower bounds on the diameter diam⁑(π’œ):=akβˆ’a1\operatorname{diam}({\cal A}):=a_{k}-a_{1}. For a given kk, the minimum possible value of diam⁑(π’œ)\operatorname{diam}({\cal A}) is denoted sks_{k}. We let R2​(n)R_{2}(n) be the maximum value of kk with sk≀nβˆ’1s_{k}\leq n-1, i.e., the maximum possible size of a Sidon set contained in {1,2,…,n}\{1,2,\dots,n\}.

In 1938, SingerΒ [1938.Singer] constructed a Sidon set with q+1q+1 elements and diameter q2+q+1q^{2}+q+1, where qq is any prime power, whence sk≀k2s_{k}\leq k^{2} for infinitely many kk. Alternatively, R2​(n)>nR_{2}(n)>\sqrt{n} for infinitely many nn. Using results on the distribution of primes (assuming RH), this construction gives sk≀k2+k3/2+Ο΅,R2​(n)≀n+n1/4+Ο΅s_{k}\leq k^{2}+k^{3/2+\epsilon},R_{2}(n)\leq\sqrt{n}+n^{1/4+\epsilon} for any Ο΅>0\epsilon>0 and sufficiently large k,nk,n. Alternative constructions that have the same asymptotic implications for sk,R2​(n)s_{k},R_{2}(n) have been given by BoseΒ [1942.Bose] and RuzsaΒ [Ruzsa]. Recently, Eberhard and MannersΒ [Eberhard] have given a unified treatment that has these three constructionsβ€”and no others that are asymptotically as goodβ€”as special cases.

In 1941, ErdΕ‘s-TurΓ‘nΒ [ET] use a β€œwindowing” argument to derive the inequality (valid for any positive integers n,Tn,T)

nβˆ’1β‰₯R2​(n)2​TT+R2​(n)βˆ’1βˆ’T,n-1\geq\frac{R_{2}(n)^{2}T}{T+R_{2}(n)-1}-T, (1)

and from this inequality find that R2​(n)=n1/2+O​(n1/4)R_{2}(n)=n^{1/2}+O(n^{1/4}). In subsequent years, analysis of this inequality was improved and finally it was notedΒ [Cilleruelo] that R2​(n)=n1/2+n1/4+12R_{2}(n)=n^{1/2}+n^{1/4}+\frac{1}{2} follows fromΒ (1). In 1969, LindstrΓΆm gave another proof, with a different inequality, which also gives this bound on R2​(n)R_{2}(n). InΒ 2 below, we give an even-more-elementary form of the ErdΕ‘s-TurΓ‘n argument and a β€œbook” analysis ofΒ 1 to give the sharpest expicit bound on sks_{k} that explicitly appears in the literature.

Theorem 1.

For kβ‰₯1k\geq 1, skβ‰₯k2βˆ’2​k3/2+k+kβˆ’1.s_{k}\geq k^{2}-2k^{3/2}+k+\sqrt{k}-1.

In light of the results cited above, the main term in diam⁑(π’œ)\operatorname{diam}({\cal A}) is k2k^{2}, and the secondary term is between k3/2+Ο΅k^{3/2+\epsilon} and βˆ’2​k3/2-2k^{3/2}. Recently, Balogh, FΓΌredi, and RoyΒ [BFR] (hereafter referred to as BFR) improved the bound on the secondary term from βˆ’2​k3/2-2k^{3/2} to βˆ’1.996​k3/2-1.996k^{3/2}. This was the first real improvement on the bound on diam⁑(π’œ)\operatorname{diam}({\cal A}) or R2​(n)R_{2}(n) since 1941. The number (1.9961.996) appearing in their work is not fully optimized, and the important aspect of their work is that the number can be improved to below 22 by combining the ErdΕ‘s-TurΓ‘n and LindstrΓΆm proofs in a nontrivial manner. In this work we further improve the secondary term. The numerical improvement (also not fully optimized) is not the important aspect of this work, but that it is possible to improve the number below 22 by a minor refinement the ErdΕ‘s-TurΓ‘n argument and not using the LindstrΓΆm argument at all. Our method is logically simpler than the BFR method, but computationally more involved.

Theorem 2.

Suppose that π’œ{\cal A} is a finite Sidon set of integers. If kk is sufficiently large, then diam⁑(π’œ)β‰₯k2βˆ’1.99405​k3/2\operatorname{diam}({\cal A})\geq k^{2}-1.99405k^{3/2}. Also, if nn is sufficiently large, then R2​(n)<n1/2+0.99703​n1/4R_{2}(n)<n^{1/2}+0.99703n^{1/4}.

The β€œsufficiently large” phrase is not fundamentally needed. We expect further numerical improvements will appear in the next few years and it is premature to overly optimize at this stage.

We define bkb_{k} by sk=k2βˆ’bk​k3/2s_{k}=k^{2}-b_{k}k^{3/2}, and b∞=lim supkβ†’βˆžbkb_{\infty}=\limsup_{k\to\infty}b_{k}. The ErdΕ‘s-TurΓ‘n work proves that bk<2b_{k}<2, and the Balogh-FΓΌredi-Roy work proves that bβˆžβ‰€1.996b_{\infty}\leq 1.996. Explicit values of bkb_{k} (see FigureΒ 1) are known for 1≀k≀271\leq k\leq 27 thanks to the Distributed.net β€œOptimal Golomb Ruler” projectΒ [Distnet].

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Figure 1: Exact values of bkb_{k} for 1≀k≀271\leq k\leq 27.

ShearerΒ [Shearer] (perhaps around 1986) constructed the Sidon set examples of Singer and Bose for prime powers up to around 150, and thereby gave a reasonably sharp upper bound on sks_{k} for k≀150k\leq 150. More recently, Dogon-RokickiΒ [DR] have taken this computation to its logical extreme. They computed the lower bound on bkb_{k} that arises from looking at all subsets of modular dilations and translations of the sets described by Singer, Bose, and Ruzsa for all prime powers up to 40 200. This reproduces the known optimal sets for all kβ‰₯17k\geq 17 (and most kk below 17, too). Their amazing data, which includes specific candidates for kk-element Sidon sets with minimal diameter for k<40 000k<40\,000, has been instrumental to this work. Their bound for k<40 000k<40\,000 is shown in FigureΒ 2.

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Figure 2: Lower bound on bkb_{k} for 1≀k≀40 0001\leq k\leq 40\,000, as computed by Dogon-Rokicki. The vertical red line separates the known exact values from the mere lower bounds.

2 The ErdΕ‘s-TurΓ‘n Sidon Set Equation

We first give a simplified account of the inequality that ErdΕ‘s-TurΓ‘n derived and used to prove R​(n)<n1/2+O​(n1/4)R(n)<n^{1/2}+O(n^{1/4}), with a sharper derivation of the resulting bound. In fact, the bound given here is sharper than any explicit bound explicitly given in the literature of which this author is aware.

Theorem 3.

For kβ‰₯1k\geq 1, we have skβ‰₯k2βˆ’2​k3/2+k+kβˆ’1s_{k}\geq k^{2}-2k^{3/2}+k+\sqrt{k}-1. For nβ‰₯1n\geq 1, we have R2​(n)<n1/2+n1/4+12R_{2}(n)<n^{1/2}+n^{1/4}+\frac{1}{2}.

Proof.

Let π’œ={0=a1<a2<β‹―<ak}{\cal A}=\{0=a_{1}<a_{2}<\dots<a_{k}\} be a kk-element Sidon set with minimum possible diameter. Let AjA_{j} be the size of the intersection of π’œ{\cal A} with [jβˆ’T,j)[j-T,j).

First, each of the kk elements of π’œ{\cal A} contributes to exactly TT of the AjA_{j}, so that

βˆ‘j=1ak+TAj=k​T.\sum_{j=1}^{a_{k}+T}A_{j}=kT. (2)

By expanding (Ajβˆ’k​Tak+T)2\big{(}A_{j}-\frac{kT}{a_{k}+T}\big{)}^{2} and usingΒ (2), straightforward algebra now confirms that

βˆ‘j=1ak+TAj2=k2​T2ak+T+βˆ‘j=1ak+T(Ajβˆ’k​Tak+T)2β‰₯k2​T2ak+T.\sum_{j=1}^{a_{k}+T}A_{j}^{2}=\frac{k^{2}T^{2}}{a_{k}+T}+\sum_{j=1}^{a_{k}+T}\bigg{(}A_{j}-\frac{kT}{a_{k}+T}\bigg{)}^{2}\geq\frac{k^{2}T^{2}}{a_{k}+T}. (3)

We can combineΒ (2) andΒ (3) to get

βˆ‘j=1ak+T(Aj2)β‰₯12​k2​T2ak+Tβˆ’12​k​T,\sum_{j=1}^{a_{k}+T}\binom{A_{j}}{2}\geq\frac{1}{2}\frac{k^{2}T^{2}}{a_{k}+T}-\frac{1}{2}kT,

and this raises a new combinatorial interpretation. The binomial coefficient (Aj2)\binom{A_{j}}{2} is counting the number of pairs of elements of π’œ{\cal A} in the interval [jβˆ’T,j)[j-T,j). Each pair (ay,ax)(a_{y},a_{x}) of elements of π’œ{\cal A} with ax>aya_{x}>a_{y} contributes 1 to (Aj2)\binom{A_{j}}{2} for j∈[ax+1,ay+T]j\in[a_{x}+1,a_{y}+T] if r=axβˆ’ay≀Tβˆ’1r=a_{x}-a_{y}\leq T-1, and contributes nothing to any (Aj2)\binom{A_{j}}{2} if rβ‰₯Tr\geq T. By the Sidon property, if rβˆˆπ’œβˆ’π’œr\in{\cal A}-{\cal A}, then there is a unique pair (ay,ax)(a_{y},a_{x}) with r=axβˆ’ayr=a_{x}-a_{y}. Ergo, as |[ax+1,ay+T]|=Tβˆ’r|[a_{x}+1,a_{y}+T]|=T-r, we have

2β€‹βˆ‘j=1ak+T(Aj2)=βˆ‘r=1rβˆˆπ’œβˆ’π’œTβˆ’1Tβˆ’r=T​(Tβˆ’1)2βˆ’βˆ‘r=1rβˆ‰π’œβˆ’π’œTβˆ’1Tβˆ’r≀T​(Tβˆ’1)2.2\sum_{j=1}^{a_{k}+T}\binom{A_{j}}{2}=\sum_{\begin{subarray}{c}r=1\\ r\in{\cal A}-{\cal A}\end{subarray}}^{T-1}T-r=\frac{T(T-1)}{2}-\sum_{\begin{subarray}{c}r=1\\ r\not\in{\cal A}-{\cal A}\end{subarray}}^{T-1}T-r\leq\frac{T(T-1)}{2}. (4)

Comparing the upper and lower bounds on βˆ‘(Aj2)\sum\binom{A_{j}}{2}, we now have

12​k2​T2ak+Tβˆ’k​T2β‰₯T​(Tβˆ’1)2.\frac{1}{2}\frac{k^{2}T^{2}}{a_{k}+T}-\frac{kT}{2}\geq\frac{T(T-1)}{2}.

Solving for aka_{k} gives the ErdΕ‘s-TurΓ‘n Sidon Set Inequality: For positive integers TT,

akβ‰₯k2​TT+kβˆ’1βˆ’T.a_{k}\geq\frac{k^{2}T}{T+k-1}-T.

Set T=k3/2βˆ’k+Ο΅T=k^{3/2}-k+\epsilon, with ϡ∈(0,1]\epsilon\in(0,1] chosen to make TT a positive integer. We have the factorization

(k2​TT+kβˆ’1βˆ’T)βˆ’(k2βˆ’2​k3/2+k+kβˆ’1)=(1βˆ’Ο΅)​(k+Ο΅βˆ’1)T+kβˆ’1,\left(\frac{k^{2}T}{T+k-1}-T\right)-\big{(}k^{2}-2k^{3/2}+k+\sqrt{k}-1\big{)}=\frac{(1-\epsilon)\big{(}\sqrt{k}+\epsilon-1\big{)}}{T+k-1},

which is clearly nonnegative for kβ‰₯1k\geq 1. Therefore, sk=akβ‰₯k2βˆ’2​k3/2+k+kβˆ’1s_{k}=a_{k}\geq k^{2}-2k^{3/2}+k+\sqrt{k}-1, as claimed. ∎

We can convert this to a bound on k=R2​(n)k=R_{2}(n) easily. The expression k2βˆ’2​k3/2+k+kβˆ’1k^{2}-2k^{3/2}+k+\sqrt{k}-1 is a polynomial in k\sqrt{k}, and one easily sees that its derivative (with respect to k\sqrt{k}) is positive for kβ‰₯0k\geq 0. Thus, the inequality nβˆ’1β‰₯k2βˆ’2​k3/2+k+kβˆ’1n-1\geq k^{2}-2k^{3/2}+k+\sqrt{k}-1 implies an upper bound on kk. To prove that R2​(n)<n1/2+n1/4+12R_{2}(n)<n^{1/2}+n^{1/4}+\frac{1}{2}, for example, one only needs to derive a contradiction from

nβˆ’1β‰₯(n1/2+n1/4+12)2βˆ’2​(n1/2+n1/4+12)3/2+(n1/2+n1/4+12)+n1/2+n1/4+12βˆ’1.n-1\geq(n^{1/2}+n^{1/4}+\frac{1}{2})^{2}-2(n^{1/2}+n^{1/4}+\tfrac{1}{2})^{3/2}+(n^{1/2}+n^{1/4}+\tfrac{1}{2})+\sqrt{n^{1/2}+n^{1/4}+\tfrac{1}{2}}-1.

While tedious, this purported inequality simplifies algebraically and can be proved impossible by hand.

This proof contains two innovations and a useful consequence. The first innovation is onΒ (3), where we used algebra instead of Cauchy’s Inequality. The algebra of course constitutes a proof of Cauchy’s inequality, but the crux of this article rests on this nontrivial change. The second innovation is the discovery that this bound on sks_{k} produces a pleasantly useful factorization, allowing us to also replace asymptotic analysis entirely with simple algebra.

The consequence is made possible by the first innovation. We can use the equalities inΒ (3) andΒ (4) instead of the inequalities, and we arrive at the following theorem, which drives the rest of this work.

Theorem 4.

Let π’œ{\cal A} be a finite Sidon set of integers, and TT be a positive integer. Then

diam⁑(π’œ)=|π’œ|2​T2T​(T+|π’œ|βˆ’1)βˆ’(2​S​(π’œ,T)+V​(π’œ,T))βˆ’T,\operatorname{diam}({\cal A})=\frac{|{\cal A}|^{2}T^{2}}{T(T+|{\cal A}|-1)-(2S({\cal A},T)+V({\cal A},T))}-T, (5)

where

S​(π’œ,T)\displaystyle S({\cal A},T) :=βˆ‘r=1rβˆ‰π’œβˆ’π’œTβˆ’1(Tβˆ’r),\displaystyle:=\sum_{\begin{subarray}{c}r=1\\ r\not\in{\cal A}-{\cal A}\end{subarray}}^{T-1}(T-r),
V​(π’œ,T)\displaystyle V({\cal A},T) :=βˆ‘i=minβ‘π’œ+1T+maxβ‘π’œ(Aiβˆ’k​TT+maxβ‘π’œ)2,\displaystyle:=\sum_{i=\min{\cal A}+1}^{T+\max{\cal A}}\left(A_{i}-\frac{kT}{T+\max{\cal A}}\right)^{2},

where Ai:=|π’œβˆ©[iβˆ’T,i)|A_{i}:=|{\cal A}\cap[i-T,i)|.

We find it philosophically interesting that TheoremΒ 4 gives an equality. Given |π’œ||{\cal A}|, any lower bound on diam⁑(π’œ)\operatorname{diam}({\cal A}) is equivalent to a lower bound on 2​S​(π’œ,T)+V​(π’œ,T)2S({\cal A},T)+V({\cal A},T), and vice versa. Know any three of |π’œ|,diam⁑(π’œ),V​(π’œ,T),S​(π’œ,T)|{\cal A}|,\operatorname{diam}({\cal A}),V({\cal A},T),S({\cal A},T), and the fourth is uniquely determined.

For i≀Ti\leq T, one has Ai≀R2​(i)A_{i}\leq R_{2}(i), and this gives a β€œtrivial” lower bound on V​(π’œ,T)V({\cal A},T). This allows TheoremΒ 4 to improve the inequality R2​(n)<n1/2+n1/4+12R_{2}(n)<n^{1/2}+n^{1/4}+\frac{1}{2} explicitly by O​(1)O(1). We mention this only because the number 1/21/2 has been called the limit of what can be accomplished with the ErdΕ‘s-TurΓ‘n style argument.

2.1 Exploring the ErdΕ‘s-TurΓ‘n Sidon Set Equation

This subsection contains no logic, but explores the dataset of Dogon-RokickiΒ [DR] to motivate the rest of this work, and perhaps also suggest avenues for further progress.

Asymptotic analysis gives that if 2​S​(π’œ,T)+V​(π’œ,T)β‰₯ν​k​T2S({\cal A},T)+V({\cal A},T)\geq\nu kT (where ν∈(0,1)\nu\in(0,1) is fixed and T=τ​k3/2T=\tau k^{3/2} is allowed to vary), then

diam⁑(π’œ)β‰₯k2βˆ’2​1βˆ’Ξ½β€‹k3/2+O​(k).\operatorname{diam}({\cal A})\geq k^{2}-2\sqrt{1-\nu}k^{3/2}+O(k).

In particular, we need to find a bound on 2​S+V2S+V that is Ω​(k​T)\Omega(kT) to impact the k3/2k^{3/2} coefficient. In FigureΒ 3, we show V​(π’œ,k3/2)/k5/2V({\cal A},k^{3/2})/k^{5/2} for k<4000k<4000.

Refer to caption
Figure 3: V​(π’œ,k3/2)/k5/2V({\cal A},k^{3/2})/k^{5/2} for the Sidon sets π’œ{\cal A} in the Dogon-Rokicki dataset (28≀k<400028\leq k<4000).

In contrast to V​(π’œ,T)V({\cal A},T), which seems to be consistently more than k​TkT in the Dogon-Rokicki dataset, the contribution from S​(π’œ,T)S({\cal A},T) is small and irregular. See FigureΒ 4. The nearly vertical patterns in the data points in FigureΒ 3 andΒ 4 are not visual artifacts, but possibly arise from many of the sets with close kk having small symmetric difference, and so similar values of V,SV,S.

Refer to caption
Figure 4: 2​S​(π’œ,k3/2)/k5/22S({\cal A},k^{3/2})/k^{5/2} for the Sidon sets π’œ{\cal A} in the Dogon-Rokicki dataset (28≀k<400028\leq k<4000).

There is a negative correlation between V​(π’œ,T)V({\cal A},T) and S​(π’œ,T)S({\cal A},T), as can be seen in FiguresΒ 5 andΒ 6.

Refer to caption
Figure 5: (V(π’œ,k3/2)+2S(π’œ,k3/2)/k5/2(V({\cal A},k^{3/2})+2S({\cal A},k^{3/2})/k^{5/2} for the Sidon sets π’œ{\cal A} in the Dogon-Rokicki dataset (28≀k<400028\leq k<4000). Notice that this graph appears smoother than those in FiguresΒ 3 andΒ 4, showing the negative correlation.

[h!]

Refer to caption
Figure 6: The points (V(π’œ,k3/2),2S(π’œ,k3/2)/k5/2(V({\cal A},k^{3/2}),2S({\cal A},k^{3/2})/k^{5/2} for the Sidon sets π’œ{\cal A} in the Dogon-Rokicki dataset (28≀k<400028\leq k<4000). Notice that when VV is particularly large, 2​S2S is particularly small.

From these images, it seems apparent that one should focus more on VV than SS. In FigureΒ 7 we show four plots of Ai:=|π’œβˆ©[iβˆ’k3/2,i)|A_{i}:=|{\cal A}\cap[i-k^{3/2},i)|, with the average value as a red dashed horizontal line, and i=Ti=T,i=aki=a_{k} as vertical black lines. The four, which appear to be typical, show that AiA_{i} is fairly smooth between TT and aka_{k}. In fact, about half of V​(π’œ,T)V({\cal A},T) arises from i≀Ti\leq T and iβ‰₯aki\geq a_{k}. FigureΒ 8 shows the proportion of VV that arises from the edge values of ii.

Refer to caption
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Figure 7: Plot of AiA_{i} as a function of ii for four typical thick Sidon sets. The average value of AiA_{i} is shown as a horizontal red line, and the values i=Ti=T and i=aki=a_{k} are shown as vertical black lines.
Refer to caption
Figure 8: V​(π’œ,k3/2)V({\cal A},k^{3/2}) is defined by a sum with ii running from 1 to ak+k3/2a_{k}+k^{3/2}. This graph shows the proportion of V​(π’œ,k3/2)V({\cal A},k^{3/2}) that comes from i≀Ti\leq T and iβ‰₯aki\geq a_{k}.

3 The BFR Gambit

InΒ [BFR], a small variance (which arose by way of an inequality from coding theory, and a rephrasing of the ErdΕ‘s-TurΓ‘n argument in terms of degree sequences of hypergraphs) was used to imply a lumpiness in the difference set. This lumpiness was then used to supercharge a proof of LindstrΓΆm’s, yielding

R2​(n)<n1/2+0.998​n1/4​ for sufficiently largeΒ n.R_{2}(n)<n^{1/2}+0.998n^{1/4}\text{ for sufficiently large $n$.}

It is, as the proven value was rounded up to 0.9980.998, equivalent to

skβ‰₯k2βˆ’1.996​k3/2​ for sufficiently largeΒ k.s_{k}\geq k^{2}-1.996k^{3/2}\text{ for sufficiently large $k$.}

This work was the first real improvement in the bounds, upper or lower, of R2​(n)R_{2}(n) since 1941. It was the direct inspiration for the current work.

We play the same gambit, but exploit the phenomenon in a different way. Instead of using the LindstrΓΆm proof, we re-use TheoremΒ 4 on a subset of π’œ{\cal A}. That is, either π’œ{\cal A} has a large variance (and so a large diameter), or a large subset β„³βŠ†π’œ{\cal M}\subseteq{\cal A} has a difference set that does not include many small values, and so the small diffs is large, forcing β„³{\cal M} to have a large diameter.

3.1 A Toy Version of our Argument

In this short subsection, we give a simple argument using the ErdΕ‘s-TurΓ‘n Sidon Set Equation to prove that bβˆžβ‰€1.999b_{\infty}\leq 1.999. In the next subsection, we engage in a somewhat harder-to-optimize version of this argument with more parameters that leads to a better bound on b∞b_{\infty} than that given in BFR.

Normalize as above, so that a1=0a_{1}=0, and set T=k3/2T=k^{3/2} (it needs to be an integer, but we will ignore this issue in this subsection, and we shall also be cavalier about inequalities that may only hold for sufficiently large kk). For positive ii, set Ai:=|π’œβˆ©[iβˆ’T,i)|A_{i}:=|{\cal A}\cap[i-T,i)| and Aβˆ’i:=|π’œβˆ©[akβˆ’i+1,ak+T+1βˆ’i)|A_{-i}:=|{\cal A}\cap[a_{k}-i+1,a_{k}+T+1-i)|, and let AΒ―:=k​T/(T+ak)\bar{A}:=kT/(T+a_{k}), the average value of AiA_{i} for 1≀i≀ak+T1\leq i\leq a_{k}+T. If ak≀k2βˆ’Ta_{k}\leq k^{2}-T, then A¯∼k\bar{A}\sim\sqrt{k}. We begin by considering (Ai+Aβˆ’i)/2(A_{i}+A_{-i})/2 as a function of ii; it is monotone increasing for 1≀i≀T1\leq i\leq T. Set

U1\displaystyle U_{1} :={1≀j≀T:Ai+Aβˆ’i2≀89​AΒ―}\displaystyle:=\left\{1\leq j\leq T:\tfrac{A_{i}+A_{-i}}{2}\leq\tfrac{8}{9}\bar{A}\right\}
U2\displaystyle U_{2} :={1≀j≀T:89​AΒ―<Ai+Aβˆ’i2<109​AΒ―}\displaystyle:=\left\{1\leq j\leq T:\tfrac{8}{9}\bar{A}<\tfrac{A_{i}+A_{-i}}{2}<\tfrac{10}{9}\bar{A}\right\}
U3\displaystyle U_{3} :={1≀j≀T:109​A¯≀Ai+Aβˆ’i2}\displaystyle:=\left\{1\leq j\leq T:\tfrac{10}{9}\bar{A}\leq\tfrac{A_{i}+A_{-i}}{2}\right\}

If |U1|+|U3|β‰₯T/16|U_{1}|+|U_{3}|\geq T/16, then

V​(π’œ,T)\displaystyle V({\cal A},T) :=βˆ‘i=1T+ak(Aiβˆ’AΒ―)2\displaystyle:=\sum_{i=1}^{T+a_{k}}(A_{i}-\bar{A})^{2}
β‰₯βˆ‘i∈U1βˆͺU3[(Aiβˆ’AΒ―)2+(Aβˆ’iβˆ’AΒ―)2]\displaystyle\geq\sum_{i\in U_{1}\cup U_{3}}\left[(A_{i}-\bar{A})^{2}+(A_{-i}-\bar{A})^{2}\right]
β‰₯βˆ‘i∈U1βˆͺU32​(Ai+Aβˆ’i2βˆ’AΒ―)2\displaystyle\geq\sum_{i\in U_{1}\cup U_{3}}2(\tfrac{A_{i}+A_{-i}}{2}-\bar{A})^{2}
β‰₯(|U1βˆͺU2|)β‹…2​(19)2​AΒ―2β‰₯1648​k​T.\displaystyle\geq\left(|U_{1}\cup U_{2}|\right)\cdot 2\left(\tfrac{1}{9}\right)^{2}\bar{A}^{2}\geq\frac{1}{648}kT.

We can now appeal to TheoremΒ 4:

diam⁑(π’œ)β‰₯k2​T2T​(T+kβˆ’1)βˆ’116​281​k​Tβˆ’T=k2βˆ’(2βˆ’1648)​k3/2+O​(k).\operatorname{diam}({\cal A})\geq\frac{k^{2}T^{2}}{T(T+k-1)-\frac{1}{16}\frac{2}{81}kT}-T=k^{2}-(2-\frac{1}{648})k^{3/2}+O(k).

Thus, bβˆžβ‰€2βˆ’1648≀1.999b_{\infty}\leq 2-\frac{1}{648}\leq 1.999 if |U1|+|U3|β‰₯T16|U_{1}|+|U_{3}|\geq\frac{T}{16}.

Now suppose that |U1|+|U3|≀T16|U_{1}|+|U_{3}|\leq\frac{T}{16}, so that |U2|β‰₯1516​T|U_{2}|\geq\frac{15}{16}T. Set

β„³:=π’œβˆ©[|U1|+|U2|,akβˆ’|U1|βˆ’|U2|].{{\cal M}}:={\cal A}\cap[|U_{1}|+|U_{2}|,a_{k}-|U_{1}|-|U_{2}|].

Clearly

|β„³|=kβˆ’A|U1|+|U2|βˆ’Aβˆ’(|U1|+|U2|)β‰₯kβˆ’209​AΒ―.|{\cal M}|=k-A_{|U_{1}|+|U_{2}|}-A_{-(|U_{1}|+|U_{2}|)}\geq k-\frac{20}{9}\bar{A}.

Set L=π’œβˆ©[0,|U1|)L={\cal A}\cap[0,|U_{1}|) and R=π’œβˆ©(akβˆ’|U1|,ak]R={\cal A}\cap(a_{k}-|U_{1}|,a_{k}]. By the definition of U1U_{1}, we see that

|LβˆͺR|β‰₯2β‹…89​AΒ―.|L\cup R|\geq 2\cdot\frac{8}{9}\bar{A}.

We also see that LβˆͺRL\cup R has (|L|2)+(|R|2)\binom{|L|}{2}+\binom{|R|}{2} pairs of elements whose difference is at most |U1||U_{1}|. We have

(|L|2)+(|R|2)β‰₯(89​AΒ―)2βˆ’(89​AΒ―)∼6481​k\binom{|L|}{2}+\binom{|R|}{2}\geq(\frac{8}{9}\bar{A})^{2}-(\frac{8}{9}\bar{A})\sim\frac{64}{81}k

differences, each at most T/16T/16. Set T2=38​TT_{2}=\frac{3}{8}T, which is larger than T/16T/16, so that

S​(β„³,T2)\displaystyle S({\cal M},T_{2}) :=βˆ‘r=1rβˆ‰β„³βˆ’β„³T2βˆ’1(T2βˆ’r)\displaystyle:=\sum_{\begin{subarray}{c}r=1\\ r\not\in{\cal M}-{\cal M}\end{subarray}}^{T_{2}-1}(T_{2}-r)
β‰₯βˆ‘r=1r∈(Lβˆ’L)βˆͺ(Rβˆ’R)T2βˆ’1(T2βˆ’r)\displaystyle\geq\sum_{\begin{subarray}{c}r=1\\ r\in(L-L)\cup(R-R)\end{subarray}}^{T_{2}-1}(T_{2}-r)
β‰₯6481​kβ‹…(38​Tβˆ’|U1|)\displaystyle\geq\frac{64}{81}k\cdot\left(\frac{3}{8}T-|U_{1}|\right)

TheoremΒ 4 now gives

diam⁑(π’œ)\displaystyle\operatorname{diam}({\cal A}) =(minβ‘β„³βˆ’a1)+diam⁑(β„³)+(akβˆ’max⁑ℳ)\displaystyle=\left(\min{\cal M}-a_{1}\right)+\operatorname{diam}({\cal M})+\left(a_{k}-\max{\cal M}\right)
β‰₯2​(|U1|+|U2|)+|β„³|2​T22T2​(T2+|β„³|βˆ’1)βˆ’2β‹…(6481​kβ‹…(38​Tβˆ’|U1|))βˆ’T2\displaystyle\geq 2(|U_{1}|+|U_{2}|)+\frac{|{\cal M}|^{2}T_{2}^{2}}{T_{2}(T_{2}+|{\cal M}|-1)-2\cdot\left(\frac{64}{81}k\cdot\left(\frac{3}{8}T-|U_{1}|\right)\right)}-T_{2}
=k2βˆ’(6734729​|U1|Tβˆ’2​|U2|T+63611944)​k3/2\displaystyle=k^{2}-\left(\frac{6734}{729}\frac{|U_{1}|}{T}-2\frac{|U_{2}|}{T}+\frac{6361}{1944}\right)k^{3/2}

The extremal values for |U1|,|U2||U_{1}|,|U_{2}| are 116​T\frac{1}{16}T and 1516​T\frac{15}{16}T, respectively, whence

bβˆžβ‰€115155832<1.999.b_{\infty}\leq\frac{11515}{5832}<1.999.

To get better numbers, we should have more levels than just 8/9,10/98/9,10/9, and we should allow T=τ​k3/2T=\tau k^{3/2} to vary. We pursue this in the next subsection.

3.2 More Parameters Leads to a Better Bound

We begin by fixing real numbers Ο„>0\tau>0, α∈(0,1)\alpha\in(0,1), β∈(0,1)\beta\in(0,1), δ∈(0,1/2)\delta\in(0,1/2), Ο„2>0\tau_{2}>0. Specific values that work out well are:

Ο„\displaystyle\tau =5958β‰ˆ1.017\displaystyle=\frac{59}{58}\approx 1.017 (6)
Ξ±\displaystyle\alpha =80319β‰ˆ0.2508\displaystyle=\frac{80}{319}\approx 0.2508 (7)
Ξ²\displaystyle\beta =195356β‰ˆ0.5478\displaystyle=\frac{195}{356}\approx 0.5478 (8)
Ξ΄\displaystyle\delta =3987737533334382702448810518987915261β‰ˆ0.1628\displaystyle=\frac{398773753333438270}{2448810518987915261}\approx 0.1628 (9)
Ο„2\displaystyle\tau_{2} =51223β‰ˆ0.2287\displaystyle=\frac{51}{223}\approx 0.2287 (10)

The reason for the ridiculous value of Ξ΄\delta is to make two bounds we will encounter exactly equal.

To better wrangle the equations, set minβ‘π’œ=0\min{\cal A}=0, |π’œ|=k|{\cal A}|=k, T=τ​k3/2+Ο΅1T=\tau k^{3/2}+\epsilon_{1} (all of the Ο΅\epsilon variables are chosen in [0,1)[0,1) to make some other variable an integer), and AΒ―=k​Tak+T\bar{A}=\frac{kT}{a_{k}+T}, and assume that sk=ak≀k2βˆ’Ts_{k}=a_{k}\leq k^{2}-T, so that τ​k≀AΒ―<τ​k+Ο΅1/k\tau\sqrt{k}\leq\bar{A}<\tau\sqrt{k}+\epsilon_{1}/k.

For 1≀j≀ak+T1\leq j\leq a_{k}+T, we have Aj:=|π’œβˆ©[jβˆ’T,j)|A_{j}:=|{\cal A}\cap[j-T,j)|. For negative subscripts, we set Aβˆ’j:=Aak+T+1βˆ’jA_{-j}:=A_{a_{k}+T+1-j}. For example, A1=|{0}|=1A_{1}=|\{0\}|=1 and Aβˆ’1=|{ak}|=1A_{-1}=|\{a_{k}\}|=1. As a function of jj, we see that (Aj+Aβˆ’j)/2(A_{j}+A_{-j})/2 is nondecreasing for 1≀j≀T1\leq j\leq T, and increases by at most 1 as jj increases by 1.

Refer to caption
Figure 9: Imagined plot of (Aj+Aβˆ’j)/2(A_{j}+A_{-j})/2 as a function of jj, with the meanings of Ο…1,Ο…2,Ο…3,Ο…4,Ο…5\upsilon_{1},\upsilon_{2},\upsilon_{3},\upsilon_{4},\upsilon_{5} depicted.

We partition the interval [1,T][1,T] into 5 parts:111Setting Ξ²=1\beta=1 forces U2=U4=βˆ…U_{2}=U_{4}=\emptyset, and is thereby substantially simpler, but leads merely bβˆžβ‰€1.999b_{\infty}\leq 1.999.

U1\displaystyle U_{1} :={j∈[1,T]:Aj+Aβˆ’j2≀(1βˆ’Ξ±)​AΒ―},\displaystyle:=\left\{j\in[1,T]:\tfrac{A_{j}+A_{-j}}{2}\leq(1-\alpha)\bar{A}\right\},
U2\displaystyle U_{2} :={j∈[1,T]:(1βˆ’Ξ±)​AΒ―<Aj+Aβˆ’j2≀(1βˆ’Ξ±β€‹Ξ²)​AΒ―},\displaystyle:=\left\{j\in[1,T]:(1-\alpha)\bar{A}<\tfrac{A_{j}+A_{-j}}{2}\leq(1-\alpha\beta)\bar{A}\right\},
U3\displaystyle U_{3} :={j∈[1,T]:(1βˆ’Ξ±β€‹Ξ²)​AΒ―<Aj+Aβˆ’j2≀(1+α​β)​AΒ―},\displaystyle:=\left\{j\in[1,T]:(1-\alpha\beta)\bar{A}<\tfrac{A_{j}+A_{-j}}{2}\leq(1+\alpha\beta)\bar{A}\right\},
U4\displaystyle U_{4} :={j∈[1,T]:(1+α​β)​A¯≀Aj+Aβˆ’j2≀(1+Ξ±)​AΒ―},\displaystyle:=\left\{j\in[1,T]:(1+\alpha\beta)\bar{A}\leq\tfrac{A_{j}+A_{-j}}{2}\leq(1+\alpha)\bar{A}\right\},
U5\displaystyle U_{5} :={j∈[1,T]:(1+Ξ±)​AΒ―<Aj+Aβˆ’j2},\displaystyle:=\left\{j\in[1,T]:(1+\alpha)\bar{A}<\tfrac{A_{j}+A_{-j}}{2}\right\},

We set uiu_{i} by ui​T=|Ui|u_{i}T=|U_{i}|, and set y:=u1+u5,x:=u2+u4y:=u_{1}+u_{5},x:=u_{2}+u_{4}. For example, x+y+u3=1x+y+u_{3}=1 and u1+u2+u3β‰₯1βˆ’xβˆ’yu_{1}+u_{2}+u_{3}\geq 1-x-y.

Claim 1.

If y+Ξ²2​xβ‰₯Ξ²2​δy+\beta^{2}x\geq\beta^{2}\delta, then V​(π’œ,T)β‰₯2​α2​β2​τ2​k​TV({\cal A},T)\geq 2\alpha^{2}\beta^{2}\tau^{2}\,kT.

Proof of Claim:.
V​(π’œ,T)\displaystyle V({\cal A},T) β‰₯βˆ‘j=1T(Ajβˆ’AΒ―)2+(Aβˆ’jβˆ’AΒ―)2\displaystyle\geq\sum_{j=1}^{T}(A_{j}-\bar{A})^{2}+(A_{-j}-\bar{A})^{2}
β‰₯βˆ‘j=1T2​(Aj+Aβˆ’j2βˆ’AΒ―)2\displaystyle\geq\sum_{j=1}^{T}2\left(\tfrac{A_{j}+A_{-j}}{2}-\bar{A}\right)^{2}
=2β€‹βˆ‘i=15βˆ‘j∈Ui(Aj+Aβˆ’j2βˆ’AΒ―)2\displaystyle=2\sum_{i=1}^{5}\sum_{j\in U_{i}}\left(\tfrac{A_{j}+A_{-j}}{2}-\bar{A}\right)^{2}
β‰₯2​(|U1|+|U5|)​α2​AΒ―2+2​(|U2|+|U4|)​(α​β)2​AΒ―2\displaystyle\geq 2(|U_{1}|+|U_{5}|)\alpha^{2}\bar{A}^{2}+2(|U_{2}|+|U_{4}|)(\alpha\beta)^{2}\bar{A}^{2}
β‰₯2​α2​τ2​(y+Ξ²2​x)​k​T.\displaystyle\geq 2\alpha^{2}\tau^{2}(y+\beta^{2}x)\,kT.

By hypothesis of this claim, y+Ξ²2​xβ‰₯Ξ²2​δy+\beta^{2}x\geq\beta^{2}\delta, and so V​(π’œ,T)β‰₯2​α2​β2​τ2​δ​k​TV({\cal A},T)\geq 2\alpha^{2}\beta^{2}\tau^{2}\delta\,kT. ∎

Claim 2.

If y+Ξ²2​xβ‰₯Ξ²2​δy+\beta^{2}x\geq\beta^{2}\delta, then

bβˆžβ‰€Ο„+1Ο„βˆ’2​α2​β2​δ​τ≀38692477564867759220242645451940405707787319054606925942≀1.99405.b_{\infty}\leq\tau+\frac{1}{\tau}-2\alpha^{2}\beta^{2}\delta\tau\leq\frac{3869247756486775922024264545}{1940405707787319054606925942}\leq 1.99405.
Proof of Claim:.

TheoremΒ 4 gives (under the hypothesis that y+Ξ²2​xβ‰₯Ξ²2​δy+\beta^{2}x\geq\beta^{2}\delta)

diam⁑(π’œ)β‰₯k2​T2T​(T+kβˆ’1)βˆ’2​α2​τ2​δ​k​Tβˆ’T=k2βˆ’(Ο„+1Ο„βˆ’2​α2​β2​δ​τ)​k3/2+O​(k).\operatorname{diam}({\cal A})\geq\frac{k^{2}T^{2}}{T(T+k-1)-2\alpha^{2}\tau^{2}\delta kT}-T=k^{2}-\left(\tau+\frac{1}{\tau}-2\alpha^{2}\beta^{2}\delta\tau\right)k^{3/2}+O(k).

With the values of Ο„,Ξ±,Ξ΄\tau,\alpha,\delta given on linesΒ (6),Β (7), andΒ (9) above, this gives the claimed number. ∎

We now work under the hypothesis that y+Ξ²2​x≀β2​δy+\beta^{2}x\leq\beta^{2}\delta. Note that |U1|+|U2|≀y​T+x​T≀(yΞ²2+x)≀δ​T<T|U_{1}|+|U_{2}|\leq yT+xT\leq\left(\frac{y}{\beta^{2}}+x\right)\leq\delta T<T, so that necessarily U3U_{3} is not empty. It can happen that U4,U5U_{4},U_{5} are empty.

Let Xβˆ’XX-X be the set of positive differences of elements of XX. In particular, if XX is a subset of a Sidon set, then |Xβˆ’X|=(|X|2)|X-X|=\binom{|X|}{2}. We set

β„³:=π’œβˆ©(max⁑U3,akβˆ’max⁑U3){\cal M}:={\cal A}\cap(\max U_{3},a_{k}-\max U_{3})

We will find that |β„³|β‰ˆ|π’œ||\cal M|\approx|{\cal A}|, but that if x,yx,y are small then β„³βˆ’β„³{\cal M}-{\cal M} is missing a substantial number of small integers, so that S​(β„³,T2)S({\cal M},T_{2}) is Ω​(k5/2)\Omega(k^{5/2}), which forces the diameter of β„³{\cal M} to be large. This, in turn, forces diam⁑(π’œ)β‰₯2​max⁑U3+diam⁑(β„³)\operatorname{diam}({\cal A})\geq 2\max U_{3}+\operatorname{diam}({\cal M}) to be large, and this gives a bound on b∞b_{\infty}.

Claim 3.

Suppose that y+Ξ²2​x≀β2​δy+\beta^{2}x\leq\beta^{2}\delta. Then |β„³|β‰₯kβˆ’2​(1+α​β)​τ​k+O​(1)|{\cal M}|\geq k-2(1+\alpha\beta)\tau\sqrt{k}+O(1).

Proof of Claim:.

As y+Ξ²2​x≀β2​δy+\beta^{2}x\leq\beta^{2}\delta, we know that U3U_{3} is nonempty. Let z=max⁑U3z=\max U_{3}. By the definition of U3U_{3}, we know that Az+Aβˆ’z2≀(1+α​β)​AΒ―\frac{A_{z}+A_{-z}}{2}\leq(1+\alpha\beta)\bar{A}. Therefore,

|β„³|β‰₯kβˆ’2​(1+α​β)​AΒ―βˆ’2=kβˆ’2​(1+α​β)​τ​k+O​(1).|{\cal M}|\geq k-2(1+\alpha\beta)\bar{A}-2=k-2(1+\alpha\beta)\tau\sqrt{k}+O(1).

∎

We set Li:=π’œβˆ©(Uiβˆ’1)L_{i}:={\cal A}\cap(U_{i}-1) and Ri:=π’œβˆ©(ak+T+1βˆ’Ui)R_{i}:={\cal A}\cap(a_{k}+T+1-U_{i}). Note that L1,L2,R1,R2L_{1},L_{2},R_{1},R_{2} are disjoint, and disjoint from β„³{\cal M}. But they are all subsets of the Sidon set π’œ{\cal A}, so any difference in (L1βˆͺL2)βˆ’(L1βˆͺL2)(L_{1}\cup L_{2})-(L_{1}\cup L_{2}) or (R1βˆͺR2)βˆ’(R1βˆͺR2)(R_{1}\cup R_{2})-(R_{1}\cup R_{2}) is necessarily missing from β„³βˆ’β„³{\cal M}-{\cal M}, and so contributes to S​(β„³,T2)S({\cal M},T_{2}) (provided T2T_{2} is large enough).

Claim 4.

Suppose that y+Ξ²2​x≀β2​δy+\beta^{2}x\leq\beta^{2}\delta.

There are at least (1βˆ’Ξ±)2​τ2​k+O​(k)(1-\alpha)^{2}\tau^{2}k+O(\sqrt{k}) elements of (L1βˆ’L1)βˆͺ(R1βˆ’R1)(L_{1}-L_{1})\cup(R_{1}-R_{1}), each at most y​TyT.

There are at least Ξ±2​(1βˆ’Ξ²)2​τ2​k+O​(k)\alpha^{2}(1-\beta)^{2}\tau^{2}k+O(\sqrt{k}) elements of (L2βˆ’L2)βˆͺ(R2βˆ’R2)(L_{2}-L_{2})\cup(R_{2}-R_{2}), each at most x​TxT.

There are at least (1βˆ’Ξ±β€‹Ξ²)2​τ2​k+O​(k)(1-\alpha\beta)^{2}\tau^{2}k+O(\sqrt{k}) elements of ((L1βˆͺL2)βˆ’(L1βˆͺL2))βˆͺ((R1βˆͺR2)βˆ’(R1βˆͺR2))\big{(}(L_{1}\cup L_{2})-(L_{1}\cup L_{2})\big{)}\cup\big{(}(R_{1}\cup R_{2})-(R_{1}\cup R_{2})\big{)}, each at most (x+y)​T(x+y)T.

Proof of Claim:.

We note that |L1|=Amax⁑U1,|R1|=Aβˆ’max⁑U1,|L_{1}|=A_{\max U_{1}},|R_{1}|=A_{-\max U_{1}}, and the definition of U1U_{1} (together with the nonemptiness of U2U_{2}, which is implied by yΞ²2+x≀δ\frac{y}{\beta^{2}}+x\leq\delta) gives

Amax⁑U1+Aβˆ’max⁑U1≀2​(1βˆ’Ξ±)​AΒ―<A1+max⁑U1+Aβˆ’1βˆ’max⁑U1≀Amax⁑U1+Aβˆ’max⁑U1+2,A_{\max U_{1}}+A_{-\max U_{1}}\leq 2(1-\alpha)\bar{A}<A_{1+\max U_{1}}+A_{-1-\max U_{1}}\leq A_{\max U_{1}}+A_{-\max U_{1}}+2,

and so

2​(1βˆ’Ξ±)​AΒ―βˆ’2≀|L1βˆͺR1|≀2​(1βˆ’Ξ±)​AΒ―.2(1-\alpha)\bar{A}-2\leq|L_{1}\cup R_{1}|\leq 2(1-\alpha)\bar{A}.

Similarly,

|L2βˆͺR2|=2​α​(1βˆ’Ξ²)​AΒ―+O​(1),|L_{2}\cup R_{2}|=2\alpha(1-\beta)\bar{A}+O(1),
|L1βˆͺL2βˆͺR2βˆͺR1|=2​(1βˆ’Ξ±β€‹Ξ²)​AΒ―+O​(1).|L_{1}\cup L_{2}\cup R_{2}\cup R_{1}|=2(1-\alpha\beta)\bar{A}+O(1).

For a Sidon set XX, the size of Xβˆ’XX-X is exactly (|X|2)\binom{|X|}{2}. For example, (L1βˆ’L1)βˆͺ(R1βˆ’R1)(L_{1}-L_{1})\cup(R_{1}-R_{1}) has

(|L1|2)+(|R1|2)β‰₯2​((1βˆ’Ξ±)​AΒ―2)=((1βˆ’Ξ±)​AΒ―)2+O​(k)=(1βˆ’Ξ±)2​τ2​k+O​(k)\binom{|L_{1}|}{2}+\binom{|R_{1}|}{2}\geq 2\binom{(1-\alpha)\bar{A}}{2}=((1-\alpha)\bar{A})^{2}+O(\sqrt{k})=(1-\alpha)^{2}\tau^{2}k+O(\sqrt{k})

elements, each at most diam⁑(U1)=u1​T\operatorname{diam}(U_{1})=u_{1}T, which is at most y​TyT.

Essentially identical computations give the other assertions of this claim. ∎

We set T2:=Ο„2​T+Ο΅2T_{2}:=\tau_{2}T+\epsilon_{2}, with Ο΅2∈[0,1)\epsilon_{2}\in[0,1) chosen to make T2T_{2} an integer.

Claim 5.

Suppose that y+Ξ²2​x≀β2​δy+\beta^{2}x\leq\beta^{2}\delta and Ο„2>x+y\tau_{2}>x+y. Then S​(β„³,T2)S({\cal M},T_{2}) is at least k​Tβ‹…Ο„2β‹…wkT\cdot\tau^{2}\cdot w, where

w\displaystyle w :=Ο„2​(1βˆ’Ξ±β€‹Ξ²)2βˆ’y​(1βˆ’Ξ±)​(1+Ξ±βˆ’2​α​β)βˆ’x​α​(1βˆ’Ξ²)​(2βˆ’Ξ±βˆ’Ξ±β€‹Ξ²)\displaystyle:=\tau_{2}(1-\alpha\beta)^{2}-y(1-\alpha)(1+\alpha-2\alpha\beta)-x\alpha(1-\beta)(2-\alpha-\alpha\beta) (11)
=30590263131179748900463βˆ’66229299056729​yβˆ’508116027794789​x\displaystyle=\frac{30590263131}{179748900463}-\frac{6622929}{9056729}y-\frac{5081160}{27794789}x (12)
Proof of Claim:.

By hypothesis, T2=Ο„2​T>(x+y)​TT_{2}=\tau_{2}T>(x+y)T, so all of the differences described in ClaimΒ 4 contribute to

S​(β„³,T2)\displaystyle S({\cal M},T_{2}) :=βˆ‘r=1rβˆˆβ„³βˆ’β„³T2βˆ’1(T2βˆ’r)\displaystyle:=\sum_{\begin{subarray}{c}r=1\\ r\in{\cal M}-{\cal M}\end{subarray}}^{T_{2}-1}(T_{2}-r)
β‰₯βˆ‘r∈((L1βˆͺL2)βˆ’(L1βˆͺL2))βˆͺ((R1βˆͺR2)βˆ’(R1βˆͺR2))(T2βˆ’r).\displaystyle\geq\sum_{r\in\big{(}(L_{1}\cup L_{2})-(L_{1}\cup L_{2})\big{)}\cup\big{(}(R_{1}\cup R_{2})-(R_{1}\cup R_{2})\big{)}}(T_{2}-r).

For example, the (1βˆ’Ξ±)2​τ2​k(1-\alpha)^{2}\tau^{2}k differences in (L1βˆ’L1)βˆͺ(R1βˆ’R1)(L_{1}-L_{1})\cup(R_{1}-R_{1}) contribute at least

(T2βˆ’y​T)β‹…((1βˆ’Ξ±)2​τ2​k+O​(k))=(Ο„2βˆ’y)​(1βˆ’Ξ±)2​τ2​k​T+O​(k2)(T_{2}-yT)\cdot\big{(}(1-\alpha)^{2}\tau^{2}k+O(\sqrt{k})\big{)}=(\tau_{2}-y)(1-\alpha)^{2}\tau^{2}kT+O(k^{2}) (13)

to S​(β„³,T2)S({\cal M},T_{2}). Similarly, the

Ξ±2​(1βˆ’Ξ²)2​AΒ―2+O​(k)\alpha^{2}(1-\beta)^{2}\bar{A}^{2}+O(\sqrt{k})

differences in (L2βˆ’L2)βˆͺ(R2βˆ’R2)(L_{2}-L_{2})\cup(R_{2}-R_{2}), each at most diam⁑(U2)=u2​T<x​T\operatorname{diam}(U_{2})=u_{2}T<xT, contribute at least

(T2βˆ’diam⁑(U2))β‹…(Ξ±2​(1βˆ’Ξ²)2​AΒ―2+O​(k))=(Ο„2βˆ’x)​α2​(1βˆ’Ξ²)2​τ2​k​T+O​(k2)(T_{2}-\operatorname{diam}(U_{2}))\cdot\big{(}\alpha^{2}(1-\beta)^{2}\bar{A}^{2}+O(\sqrt{k})\big{)}=(\tau_{2}-x)\alpha^{2}(1-\beta)^{2}\tau^{2}kT+O(k^{2}) (14)

to S​(β„³,T2)S({\cal M},T_{2}). Finally, (L1βˆͺL2)βˆ’(L1βˆͺL2)(L_{1}\cup L_{2})-(L_{1}\cup L_{2}) and (R1βˆͺR2)βˆ’(R1βˆͺR2)(R_{1}\cup R_{2})-(R_{1}\cup R_{2}) have many differences that haven’t been counted already. They have

((1βˆ’Ξ±β€‹Ξ²)​AΒ―)2+O​(k)=(1βˆ’Ξ±β€‹Ξ²)2​τ2​k+O​(k)((1-\alpha\beta)\bar{A})^{2}+O(\sqrt{k})=(1-\alpha\beta)^{2}\tau^{2}k+O(\sqrt{k})

elements, but only

(1βˆ’Ξ±β€‹Ξ²)2​τ2​kβˆ’(1βˆ’Ξ±)2​τ2​kβˆ’Ξ±2​(1βˆ’Ξ²)2​τ2​k+O​(k)=2​α​(1βˆ’Ξ±)​(1βˆ’Ξ²)​τ2​k+O​(k),(1-\alpha\beta)^{2}\tau^{2}k-(1-\alpha)^{2}\tau^{2}k-\alpha^{2}(1-\beta)^{2}\tau^{2}k+O(\sqrt{k})\\ =2\alpha(1-\alpha)(1-\beta)\tau^{2}k+O(\sqrt{k}),

each at most diam⁑(U1βˆͺU2)=|U1|+|U2|βˆ’1<(u1+u2)​T≀(y+x)​T\operatorname{diam}(U_{1}\cup U_{2})=|U_{1}|+|U_{2}|-1<(u_{1}+u_{2})T\leq(y+x)T, have not been accounted for already. These contribute an additional

(Ο„2βˆ’(y+x))β‹…2​α​(1βˆ’Ξ±)​(1βˆ’Ξ²)​τ2​k​T+O​(k2)(\tau_{2}-(y+x))\cdot 2\alpha(1-\alpha)(1-\beta)\tau^{2}kT+O(k^{2}) (15)

to S​(β„³,T2)S({\cal M},T_{2}). After some algebra to rearrange terms, this claim is confirmed. ∎

Claim 6.

Suppose that y+Ξ²2​x≀β2​δy+\beta^{2}x\leq\beta^{2}\delta. Then

bβˆžβ‰€Ο„2​(2​τ22​(2​α​β+Ξ²2​δ+1)+2​(Ξ±βˆ’1)​β2​δ​(α​(2β€‹Ξ²βˆ’1)βˆ’1)βˆ’2​τ2​(Ξ±β€‹Ξ²βˆ’1)2+Ο„23)+Ο„2τ​τ22≀38692477564867759220242645451940405707787319054606925942≀1.99405.b_{\infty}\leq\frac{\tau^{2}\left(2\tau_{2}^{2}\left(2\alpha\beta+\beta^{2}\delta+1\right)+2(\alpha-1)\beta^{2}\delta(\alpha(2\beta-1)-1)-2\tau_{2}(\alpha\beta-1)^{2}+\tau_{2}^{3}\right)+\tau_{2}}{\tau\tau_{2}^{2}}\\ \leq\frac{3869247756486775922024264545}{1940405707787319054606925942}\leq 1.99405.
Proof.

We begin with

diam⁑(π’œ)\displaystyle\operatorname{diam}({\cal A}) =min⁑ℳ+diam⁑(β„³)+akβˆ’max⁑ℳ\displaystyle=\min{\cal M}+\operatorname{diam}({\cal M})+a_{k}-\max{\cal M}
β‰₯2​(Tβˆ’x​Tβˆ’y​T)+|β„³|2​T22T2​(T2+|β„³|βˆ’1)βˆ’2​τ2​w​k​Tβˆ’T2\displaystyle\geq 2(T-xT-yT)+\frac{|{\cal M}|^{2}T_{2}^{2}}{T_{2}(T_{2}+|{\cal M}|-1)-2\tau^{2}wkT}-T_{2}

where we have used TheoremΒ 4 and ClaimΒ 5. The first term contributes 2​τ​(1βˆ’xβˆ’y)2\tau(1-x-y) to the k3/2k^{3/2} coefficient, and the last term contributes βˆ’Ο„2-\tau_{2}. The middle term can be converted to a geometric series:

|β„³|2​T22T2​(T2+|β„³|βˆ’1)βˆ’2​τ2​w​k​T\displaystyle\frac{|{\cal M}|^{2}T_{2}^{2}}{T_{2}(T_{2}+|{\cal M}|-1)-2\tau^{2}wkT} =|β„³|21+|β„³|βˆ’1τ​τ2​k3/2βˆ’2​τ​wΟ„22​k1/2\displaystyle=\frac{|{\cal M}|^{2}}{1+\frac{|{\cal M}|-1}{\tau\tau_{2}k^{3/2}}-2\frac{\tau w}{\tau_{2}^{2}k^{1/2}}}
=|β„³|2​(1+(2​τ​wΟ„22​k1/2βˆ’|β„³|βˆ’1τ​τ2​k3/2)+(2​τ​wΟ„22​k1/2βˆ’|β„³|βˆ’1τ​τ2​k3/2)2+β‹―)\displaystyle=|{\cal M}|^{2}\left(1+\left(\frac{2\tau w}{\tau_{2}^{2}k^{1/2}}-\frac{|{\cal M}|-1}{\tau\tau_{2}k^{3/2}}\right)+\left(\frac{2\tau w}{\tau_{2}^{2}k^{1/2}}-\frac{|{\cal M}|-1}{\tau\tau_{2}k^{3/2}}\right)^{2}+\cdots\right)
=|β„³|2​(1+2​τ​wΟ„22​k1/2βˆ’|β„³|βˆ’1τ​τ2​k3/2+O​(1/k))\displaystyle=|{\cal M}|^{2}\left(1+\frac{2\tau w}{\tau_{2}^{2}k^{1/2}}-\frac{|{\cal M}|-1}{\tau\tau_{2}k^{3/2}}+O(1/k)\right)

Defining ΞΌ\mu by |β„³|=kβˆ’ΞΌβ€‹k|{\cal M}|=k-\mu\sqrt{k}, we get

|β„³|2​(1+2​τ​wΟ„22​k1/2βˆ’|β„³|βˆ’1τ​τ2​k3/2+O​(1/k))\displaystyle|{\cal M}|^{2}\left(1+\frac{2\tau w}{\tau_{2}^{2}k^{1/2}}-\frac{|{\cal M}|-1}{\tau\tau_{2}k^{3/2}}+O(1/k)\right) =(kβˆ’ΞΌβ€‹k)2​(1+2​τ​wΟ„22​k1/2βˆ’kβˆ’ΞΌβ€‹kτ​τ2​k3/2)+O​(k)\displaystyle=(k-\mu\sqrt{k})^{2}\left(1+\frac{2\tau w}{\tau_{2}^{2}k^{1/2}}-\frac{k-\mu\sqrt{k}}{\tau\tau_{2}k^{3/2}}\right)+O(k)
=k2+(2​τ​wΟ„22βˆ’2β€‹ΞΌβˆ’1τ​τ2)​k3/2+O​(k).\displaystyle=k^{2}+\left(\frac{2\tau w}{\tau_{2}^{2}}-2\mu-\frac{1}{\tau\tau_{2}}\right)k^{3/2}+O(k).

Thus,

bβˆžβ‰€Ο„β€‹Ο„2βˆ’2​τ​(1βˆ’xβˆ’y)+2​μ+1τ​τ2βˆ’2​τ​wΟ„22.b_{\infty}\leq\tau\tau_{2}-2\tau(1-x-y)+2\mu+\frac{1}{\tau\tau_{2}}-\frac{2\tau w}{\tau_{2}^{2}}. (16)

From ClaimΒ 3, we know that ΞΌ\mu is at most 2​(1+α​β)​τ2(1+\alpha\beta)\tau. The right side ofΒ (16) is a linear expression in xx and yy, which we must maximize in the region xβ‰₯0,yβ‰₯0,y+Ξ²2​x≀β2​δx\geq 0,y\geq 0,y+\beta^{2}x\leq\beta^{2}\delta. The maximum occurs222Ideally, we would set Ο„2=βˆ’2​α2​β3+2​α2​β2βˆ’Ξ±2+2​α​β3βˆ’2​α​β+2β€‹Ξ±βˆ’Ξ²2Ξ²2βˆ’1,\tau_{2}=\sqrt{\frac{-2\alpha^{2}\beta^{3}+2\alpha^{2}\beta^{2}-\alpha^{2}+2\alpha\beta^{3}-2\alpha\beta+2\alpha-\beta^{2}}{\beta^{2}-1}}, and then this expression has a factor of y+Ξ²2​xy+\beta^{2}x, which we could replace with Ξ²2​δ\beta^{2}\delta. The author didn’t realize this until after the optimization work was complete, and including it would make some intermediate optimization steps more delicate. at the vertex (x,y)=(0,Ξ²2​δ)(x,y)=(0,\beta^{2}\delta), and is

bβˆžβ‰€38692477564867759220242645451940405707787319054606925942.b_{\infty}\leq\frac{3869247756486775922024264545}{1940405707787319054606925942}.

∎

ClaimΒ 2 and ClaimΒ 6 complete the proof of TheoremΒ 2. The value of Ξ΄\delta was chosen so as to make the two bounds exactly equal. In general, one sets

Ξ΄=βˆ’Ο„2​(βˆ’2​α2​β2​τ2+4​α​β​τ2​τ2+4​α​β​τ2+Ο„2​τ22+Ο„2​τ2βˆ’2​τ2βˆ’Ο„2+1)2​τ2​(Ξ±2​β2​τ22+Ξ±2​β2βˆ’Ξ±2βˆ’2​α​β+2​α+Ο„22)\delta=-\frac{\tau_{2}\left(-2\alpha^{2}\beta^{2}\tau^{2}+4\alpha\beta\tau^{2}\tau_{2}+4\alpha\beta\tau^{2}+\tau^{2}\tau_{2}^{2}+\tau^{2}\tau_{2}-2\tau^{2}-\tau_{2}+1\right)}{2\tau^{2}\left(\alpha^{2}\beta^{2}\tau_{2}^{2}+\alpha^{2}\beta^{2}-\alpha^{2}-2\alpha\beta+2\alpha+\tau_{2}^{2}\right)}

to accomplish this.

In reality, the bounds were first worked out symbolically using Mathematica, and then optimized using simulated annealing. Having good estimates of where the optimal value occurs then allowed certain streamlining of the argument.

Acknowledgements

I wish to thank Melvyn B. Nathanson and Yin Choi Cheng, who independently showed great patience while I explained early versions of this work.

The author is supported by a PSC–CUNY Award, jointly funded by The Professional Staff Congress and The City University of New York.

References