a note on weak sidon sequences

Discrete Mathematics 299 (2005) 141–144www.elsevier.com/locate/disc

A note on weak Sidon sequences

P. Mark KayllDepartment of Mathematical Sciences, University of Montana, Missoula, MT 59812-0864, USA

Received 15 July 2003; received in revised form 13 May 2004; accepted 14 May 2004Available online 26 July 2005

Dedicated to Professors Brian Alspach and Bernt Lindström on the occasions of their milestone birthdaysin 2003

Abstract

A sequence(ai) of integers isweak Sidonor well-spreadif the sumsai + aj , for i < j , are alldifferent. Letf (N) denote the maximum integern for which there exists a weak Sidon sequence0�a1< · · · < an �N . Using an idea of Lindström [An inequality forB2-sequences, J. Combin.Theory 6 (1969) 211–212], we offer an alternate proof thatf (N) < N1/2 + O(N1/4), an inequalitydue to Ruzsa [Solving a linear equation in a set of integers I, Acta. Arith. 65 (1993) 259–283]. Thepresent proof improves Ruzsa’s bound by decreasing the implicit constant, essentially from 4 to

√3.

© 2005 Elsevier B.V. All rights reserved.

Keywords:Weak Sidon; Well-spread

A sequence(ai) of integers iswell-spread(resp.Sidon) if the sumsai + aj , for i < j

(resp.i�j ), are all different. Such sequences, especially Sidon sequences, have receivedconsiderable attention since Erd˝os and Turán[2] initiated their study in 1941; see, e.g.,[8].Kotzig [5] suggested the term ‘well-spread’—‘weak Sidon’ is a common synonym—butobtaining this reference requires some digging;[7] covers the highlights. For a nonnegativeintegerN, let f (N) denote the maximum integern for which there exists a well-spreadsequence0�a1< · · · < an �N . Our purpose is to present an alternate proof of the followingresult of Ruzsa[9].

Theorem. f (N) < N1/2 +O(N1/4).

E-mail address:mark.kayll@umontana.edu.

0012-365X/$ - see front matter © 2005 Elsevier B.V. All rights reserved.doi:10.1016/j.disc.2004.05.019

142 P.M. Kayll /Discrete Mathematics 299 (2005) 141–144

After the proof, we indicate how our approach improves Ruzsa’s bound. We begin with acruder estimate:

Lemma. If N is sufficiently large, thenf (N) <2.001N1/2.

Proof. Let n := f (N) and 0�a1< · · · < an �N be a well-spread sequence. Since thesumsai +aj , for i < j , are distinct and lie in the set{1,2, . . . ,2N −1}, we have(n

2

)<2N ,

from which the assertion follows easily.�

Proof of Theorem. Let N be large enough to invoke the lemma, setn := f (N), andconsider a well-spread sequence 0�a1< · · · < an �N . The key is to study the positivedifferencesaj − ai . By obtaining both upper and lower bounds for the sum of a certainsubset of these differences—the ‘small’ ones—we shall deduce the desired bound.Following [6], for 1� i < j �n, we callj − i theorderof the differenceaj − ai . Since

the differences of order� >0 can be arranged into sequences of the form

a� − a�, a� − a�, a� − a�, . . . ,

where� − � = � − � = � − � = · · · = �, by ‘telescoping’, we see that the sum of all thesedifferences is at most�N (and less than�N for � >1). Thus, form�2, the sumS of allthe positive differences of order at mostm is less thanm(m + 1)N/2.We callai amean-pointif 2ai = aj + ak for somej, k ∈ {1, . . . , n}; notice that then

ai − ak = aj − ai. Except for the valuesaj − ai , for mean pointsai (or aj ), the differencesak − a�, for 1�� < k�n, are all distinct since(ai) is well-spread. Now the only candidatesfor mean-points area2, . . . , an−1, so we have at mostt := n − 2 differences occurringwith higher multiplicity, and the well-spread property implies that this multiplicity is 2. If1�m < n ands := n− (m+1)/2, then the number of positive differences of order at mostm ismn − m(m + 1)/2= ms. Thus,

S�t∑

i=1

2i +ms−2t∑j=1

(t + j) = ms(ms + 1)

2− t (ms − t).

For 1< m < n, it follows that

ms(ms + 1)

2− t (ms − t) <

m(m + 1)

2N ,

so that

(ms)2

2<

m(m + 1)

2N + mst .

Sinces, t < n, the second term on the right side is less thanmn2, which by the lemma isat most(2.001)2mN <4.5mN . Thus,s2< N(1+ 10/m), and since(1+ x)1/2<1+ x/2

P.M. Kayll /Discrete Mathematics 299 (2005) 141–144 143

for x = 10/m, we have

n = m + 1

2+ s <

m + 1

2+ N1/2

(1+ 5

m

). (1)

With m := N1/4�, this gives the bound in the statement of the theorem.�

Closing remarks. Ourproof uses themain ideaof Lindström[6], asadapted towell-spread,constant-parity sequences in[4]. Ruzsa[9] also based his proof on the idea of studying the‘small’ differencesaj − ai in a “somewhat hidden” fashion (his quote). Here we comparethe resulting implicit constants.To optimize ours, we first perform another iteration of the proof. Instead of applying the

lemma (to boundmn2 from above), we apply the theorem itself. This allows us to replace‘10’ by ‘3 +O(N−1/4)’. To minimize the right side of (the adjusted) inequality (1), we nowchoosem to becN1/4�, for c := √

3. These modifications reduce our upper bound onf (N) to N1/2 + cN1/4 + O(1). Ruzsa’s proof essentially delivers the value 4 in place ofour

√3 : he shows that a weak Sidon sequence contained in the set{1, . . . , N} contains at

mostN1/2 + 4N1/4 + 11 terms. Thus, aside from being more transparent, our proof yieldsan (however slight) improvement to the bound.It should be noted that the theorem compares favourably with the best-known lower

bounds forf (N).UsingSingerdifferencesets (see[10]), it is easy to show thatf (N) > N1/2

for infinitely many integersN; additionally, prime density results (e.g.[1]) imply thatf (N) > N1/2 − N21/80 if N is sufficiently large.Finally,weaddawordonanapplication. In[4], our present theoremwasused todetermine

the growth rate of the maximum label�(n) in a ‘most-efficient’ metric, injective edge-labelling of the complete graphKn for which every Hamilton cycle has identical length.We proved that 2n2−O(n3/2) < �(n) <2n2+O(n61/40), thus settling the main conjecturein [3].

Acknowledgements

Thisnote’s first draftwaswrittenduringmysabbatical leaveat theUniversity of Ljubljana,Slovenia. Thanks to the Department of Mathematics, the Institute of Mathematics, Physics&Mechanics, and especially to BojanMohar for their hospitality. Some of the research tookplace at the Alfréd Rényi Institute of Mathematics of the Hungarian Academy of Sciences.Thanks to Gyula Katona for hosting me, Miki Simonovits for suggesting a related problem,and János Pintz for a helpful discussion. I am grateful to a referee for the reference[9].

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144 P.M. Kayll /Discrete Mathematics 299 (2005) 141–144

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