well distribution of sidon sets in residue classes

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Journal of Number Theory 69, 197200 (1998) Well Distribution of Sidon Sets in Residue Classes Bernt Lindstrom Department of Mathematics, Royal Institute of Technology, S-100 44 Stockholm, Sweden Communicated by V. T. Sos Received April 11, 1997; revised May 26, 1997 Dedicated to Anders Bjorner on his 50th birthday A set A of non-negative integers is a Sidon set if the sums a +b ( a, b # A, a b) are distinct. Assume that a [1, n ] and that | A|=(1+o(1)) n 12 . Let m2 be an integer. In Theorem 1 I prove that asymptotically 1m of all elements in A fall into each residue class modulo m. When m=2 I prove a sharper result in Theorem 2. Assume that | A|n 12 . Then the difference between the number of odd and the number of even elements in A is O( n 38 ). If the interval [1, n ] is divided into m equal parts and the number of elements from A in each part is counted, then similar results hold for these counts. 1998 Academic Press 1. INTRODUCTION A set A of non-negative integers is called a Sidon set if the sums a +b ( a, b # A, a b ) are distinct. Equivalently, A is a Sidon set if the differences a &b ( a, b # A, a {b ) are different. An ordered Sidon set is called a Sidon sequence or B 2 -sequence. If F 2 ( n ) is the maximum size of a Sidon set of integers bounded by n then n 12 &O( n 516 )<F 2 ( n )<n 12 +O( n 14 ) by ([4], Theorem 7 on p. 88). The lower bound is due to Bose and Chowla; the upper bound to Erdo s and Turan. Erdo s and Freud observed in [2] that a Sidon set of size (1+o(1)) n 12 is well-distributed in intervals. I will prove a similar result on well-distribu- tion in residue classes. Erdo s, Sarkozy and So s ([3], p. 345) suggested that the sum-set A+A can be well-distributed in residue classes. This follows easily from my result. Article No. NT972217 197 0022-314X98 25.00 Copyright 1998 by Academic Press All rights of reproduction in any form reserved.

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Page 1: Well Distribution of Sidon Sets in Residue Classes

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Journal of Number Theory�NT2217

Journal of Number Theory 69, 197�200 (1998)

Well Distribution of Sidon Sets in Residue Classes

Bernt Lindstro� m

Department of Mathematics, Royal Institute of Technology,S-100 44 Stockholm, Sweden

Communicated by V. T. So� s

Received April 11, 1997; revised May 26, 1997

Dedicated to Anders Bjo� rner on his 50th birthday

A set A of non-negative integers is a Sidon set if the sums a+b (a, b # A, a�b)are distinct. Assume that a�[1, n] and that |A|=(1+o(1)) n1�2. Let m�2 be aninteger. In Theorem 1 I prove that asymptotically 1�m of all elements in A fall intoeach residue class modulo m. When m=2 I prove a sharper result in Theorem 2.Assume that |A|�n1�2. Then the difference between the number of odd and thenumber of even elements in A is O(n3�8). If the interval [1, n] is divided into mequal parts and the number of elements from A in each part is counted, then similarresults hold for these counts. � 1998 Academic Press

1. INTRODUCTION

A set A of non-negative integers is called a Sidon set if the sums a+b(a, b # A, a�b) are distinct. Equivalently, A is a Sidon set if the differencesa&b (a, b # A, a{b) are different. An ordered Sidon set is called a Sidonsequence or B2-sequence. If F2(n) is the maximum size of a Sidon set ofintegers bounded by n then

n1�2&O(n5�16)<F2(n)<n1�2+O(n1�4)

by ([4], Theorem 7 on p. 88). The lower bound is due to Bose andChowla; the upper bound to Erdo� s and Tura� n.

Erdo� s and Freud observed in [2] that a Sidon set of size (1+o(1)) n1�2

is well-distributed in intervals. I will prove a similar result on well-distribu-tion in residue classes. Erdo� s, Sa� rko� zy and So� s ([3], p. 345) suggested thatthe sum-set A+A can be well-distributed in residue classes. This followseasily from my result.

Article No. NT972217

1970022-314X�98 �25.00

Copyright � 1998 by Academic PressAll rights of reproduction in any form reserved.

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2. WELL-DISTRIBUTION MODULO m

I will prove the following general result. For method cf. [5]!

Theorem 1. Let A�[1, n] be a Sidon set with r=|A|=(1+o(1)) n1�2.For a fixed integer m�2 let Ai=[a # A : a#i (mod m)] and ri=|Ai |,0�i<m. Then ri �- n � 1�m when n � �.

Proof. Let Bi=[(a&i)�m : a # Ai], 0�i<m. It is not hard to see

Bi is a Sidon set in [0, n�m], 0�i<m (2.1)

(Bi&Bi) & (Bj&Bj)=[0], 0�i< j<m (2.2)

Let Bi=[bi, 1<bi, 2< } } } <bi, ri], where ri=|Bi |=|Ai |.

The difference bi, j+v&bi, j has the order v, 0<v<ri .(Suppose that we divide the interval [1, n] into m equal parts and let Ai

be the elements of A in part i, 0�i�m&1. Then let Bi result from trans-lating Ai into [0, n�m]. Also in this case will (2.1) and (2.2) hold.)

We choose now the integers si for 0�i<m such that

si (si+1)�ri<(si+1)(si+2)<(si+3�2)2 (2.3)

The number of differences of order v from Bi is ri&v. The number ofdifferences of order v�si from Bi will therefore be ri si&si (si+1)�2�ri (si&1�2)>ri (r1�2

i &2) by (2.3). Then, by (2.1) and (2.2), we have at least� ri (r1�2

i &2) distinct positive integers. Let S be the sum of these differences.It follows

2S>\ :m&1

0

ri (r1�2i &2)+

2

(2.4)

The sum of all differences of order v�si from Bi equals the sum of alldifferences of order v�ri&si from Bi by [1], Prop. 1.1 of Bermond�Kotzig�Turgeon. The number of these differences is si (si+1)�2�ri �2. Thetotal number is therefore at most r�2. Since these differences are boundedby n�m, we conclude that

2S<rn�m (2.5)

By (2.4) and (2.5) we now have the inequality

m \ :m&1

0

ri (r1�2i &2)+

2

<rn (2.6)

198 BERNT LINDSTRO� M

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Since r=(1+o(1)) n1�2 we find after division by n3�2

m \ :m&1

0

(ri �- n)3�2&(2r�n3�4)+2

<1+o(1) (2.7)

Since r�- n<2 there is, first for fixed i, a subsequence of n � � such thatri �- n � \i (a limit point), then for all i<m. Since r�n3�4 � 0 when n � �,we have

m \ :m&1

0

\3�2i +

2

�1 (2.8)

If we apply Ho� lder's inequality with p=3�2 and q=3 we find

1= :m&1

0

\i�\ :m&1

0

\3�2i +

2�3

m1�3 (2.9)

with equality if and only if all \i are equal. Hence, by (2.8) and (2.9),\i=1�m for 0�i<m, and the theorem follows.

3. EVEN AND ODD NUMBERS

We will assume that r�n1�2. This is satisfied by all Sidon sets of Singerand Bose ([4] Theorems 1$ and 2$ on pp. 80�81). For example, [1, 5, 11,24, 25, 27] is a Singer difference set; [1, 2, 4, 8, 16, 21, 32, 42] is a Sidonset of the Bose type. Note the imbalance between even and odd numbers!We claim that the balance is better when r is larger. More precisely, wehave the following result

Theorem 2. Assume that A�[1, n] is a Sidon set of size r�n1�2. Thenfor the number r0 of even elements and the number r1 , of odd elements

|r0&r1 |<4(r3�20 +r3�2

1 )1�2=O(n3�8).

Proof. Since n�r2, we have by (2.6)

2(r3�20 +r3�2

1 &2r)2<r3.

After moving r3 from right to left

2(r3�20 +r3�2

1 )2&r3+8r2<8((r3�20 +r3�2

1 ) (3.1)

The proof of the following identity is left for the reader.

199ON SIDON SETS

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Lemma. 2(X 3+Y 3)2&(X 2+Y 2)3=[(X 2+Y 2)2+2XY(X 2&XY+Y 2)](X&Y )2.

Put X=r1�20 , Y=r1�2

1 in the Lemma. Then we have, since X 2&XY+Y 2

>0 and X 2+Y 2=r

(r1�20 &r1�2

1 )2 r2<8r(r3�20 +r3�2

1 )&8r2 (3.2)

By Cauchy's inequality is (r1�20 +r1�2

1 )2�2r. If we multiply (3.2) by this anddivide by r2, we find

(r0&r1)2<16(r3�20 +r3�2

1 &r) (3.3)

and the theorem follows readily.

REFERENCES

1. J. C. Bermond, A. Kotzig, and J. Turgeon, On a combinatorial problem of antennas inradioastronomy, in ``Proc. 5th Hungarian Combinatorial Colloquium'' (Hajnal and So� s,Eds.), pp. 135�149, North-Holland, 1976.

2. P. Erdo� s and R. Freud, On sums of a Sidon sequence, J. Number Theory 38 (1991),196�205.

3. P. Erdo� s, A. Sa� rko� zy, and T. So� s, On sums sets of a Sidon set I, J. Number Theory 47(1994), 329�347.

4. H. Halberstam and K. F. Roth, ``Sequences,'' Springer-Verlag, 1983.5. B. Lindstro� m, An inequality for B2-sequences, J. Combinatorial Theory 6 (1969), 211�212.

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200 BERNT LINDSTRO� M