Some remarks on the Erdős Turán conjecture

Transcription

1 ACTA ARITHMETICA LXIII.4 (1993) Some remarks on the Erdős Turán conjecture by Martin Helm (Mainz) Notation. In additive number theory an increasing sequence of natural numbers is called an asymptotic basis of order h of N if every sufficiently large n N can be written as the sum of h elements of A. Let r n (h, A) denote the number of representations of n as n = a a h with a 1,..., a h A and a 1... a h. If A satisfies r n (h, A) g n N, where g is a natural constant, then A is called a B h [g]-sequence (and in the special case g = 1 a B h -sequence). Furthermore, for any given sequence A of natural numbers and any m N we define and for a given N N, δ A (m) := {(a i, a j ) : a i, a j A, m = a j a i }, h A (m) := {(a i, a j ) : a i, a j A [1, N 2 ], m = a j a i }. Introduction. A famous conjecture of Erdős and Turán [2] asserts that there exists no asymptotic basis of order 2 of N that is a B 2 [g]-sequence at the same time. Erdős shows (see [4]) that if A is an arbitrary sequence of natural numbers satisfying A(n)/ > 0 and N is a given natural number then (1) H A (N) := h A (m) N log N, which proves the above hypothesis in the special case g = 1. Almost all known results on B h [g]-sequences are based on considerations concerning the representation of certain natural numbers as a difference of elements of a given sequence A. Therefore for a further proof of the Erdős Turán conjecture it is very interesting to decide whether Erdős estimate (1) is sharp with respect to magnitude. Here we prove by means of an explicit construction that (1) is indeed sharp in the above sense.

2 374 M. Helm Furthermore, (1) unfortunately does not render possible an estimation of h A (m) for any specific m [1, N] but only provides some average information; in particular, it is not possible to decide whether any specific m [1, N] for a given N satisfies e.g. h A (m) > c log N or not, where c is a constant. Here we prove again by means of an explicit construction the existence of two increasing sequences of natural numbers B and M satisfying and M(n) > 0, n log n > 0 δ B (m j ) 1 j j 0. Theorem. There exists an infinite sequence of natural numbers A satisfying (2) and (3) H A (N) = A(n) > 0 h A (m) N log N. P r o o f. We will prove the above theorem by constructing an infinite sequence A of natural numbers as an infinite countable union of finite B 2 - sets (also called Sidon sets). Let µ and ν be arbitrary natural numbers satisfying ν > 2 and µ > 1. Now a sequence (n j ) j N is defined inductively as follows: Therefore We define n 1 := 1, n 2j := µn 2j 1, j N, n 2j+1 := νn 2j, j N. n 2j = µ j ν j 1, j N, n 2j+1 = µ j ν j, j N 0. I k := ]n k 1, n k [ k 2. A well-known result of Erdős and Chowla [1], [2] states that (4) F 2 (n) 1,

3 On the Erdős Turán conjecture 375 where F 2 (n) denotes the maximum number of elements that can be selected from the set 1, 2,..., n to form a B 2 -sequence. Since the B 2 -property of a finite set is invariant under translations, for any j N (4) proves the existence of a Sidon set S 2j I 2j = ]n 2j 1, n 2j [ such that (5) S 2j 2j n 2j 1 = µ 1 2j 1 (j ). We define A := j=1 S 2j and have to show that A satisfies the conditions (2) and (3). P r o o f o f (2). For any m N with m > µ there exists a j 0 N with Therefore and on the other hand, n 2j0 m < n 2j0 +2. m < n2j0 +2 2j0 1, A(m) A(n 2j0 ) A(n 2j0 1) = S 2j0 2j0 1. Thus A(n)/ > 0 and (2) holds. P r o o f o f (3). For a given N N with N > µ there exists a j 1 N such that n 2j1 2 N < n 2j1 and for any m [1, N] we define h 1 A(m) := {(a i, a k ) : a i, a k [1, n 2j1 ] A and m = a k a i }, h 2 A(m) := {(a i, a k ) : a i [1, N 2 ] A, a k ]n 2j1, N 2 ] and m = a k a i }, HA(N) 1 := h 1 A(m), HA(N) 2 := h 2 A(m). Consequently, h A (m) = h 1 A(m) + h 2 A(m), H A (N) = H 1 A(N) + H 2 A(N). E s t i m a t i o n o f HA 1 (N). Obviously HA(N) 1 < (A(n 2j1 )) 2 n 2j1 N. E s t i m a t i o n o f HA 2 (N). Since ν > 2, for any j j 1 the length of the gap between two consecutive Sidon sets S 2j+2 and S 2j is bigger than n 2j1 +1 n 2j1 > n 2j1 > N. Therefore a number m [1, N] can be represented as a difference of two elements a i, a k of A with a k > n 2j1 only if a k and a i are elements of the same Sidon subset of A.

4 376 M. Helm Let Θ N 2 be the number of Sidon subsets S 2j of A satisfying S 2j [1, N 2 ]. Then the B 2 -property of all Sidon subsets of A leads to and consequently, (6) H 2 A(N) = h 2 A(m) Θ N 2 m [1, N] h 2 A(m) NΘ N 2. E s t i m a t i o n o f Θ N 2. For given N N with N > µ, there exists a j 2 N so that and as Θ N 2 j 2 n 2j2 2 N 2 < n 2j2 µ j 2 1 ν j 2 2 N 2 j 2 log N (7) Θ N 2 log N. Thus (6) and (7) imply HA 2 (N) N log N and which completes the proof. H A (N) = H 1 A(N) + H 2 A(N) N log N, Corollary. There exist two infinite increasing sequences B and M of natural numbers satisfying (8) (9) and > 0, M(n) log n > 0 (10) δ B (m) 1 m M. P r o o f. Let A be the infinite sequence of natural numbers generated by the construction of the above theorem in the special case µ = 7/4 and ν = 4 (where the inductive definition of the sequence (n j ) j N is supplemented by the definition n 2 := 2 which does not restrict at all the applicability of the proof). Consequently, in this special case we define Thus n 1 := 1, n 2j+1 := νn 2j, j N, n 2 := 2, n 2j := µn 2j 1, j 2. n 2j = 2 7 j 1, n 2j+1 = 8 7 j 1, j N.

5 For any j N we define On the Erdős Turán conjecture 377 D 2j := {m N : a i, a j S 2j and m = a j a i }. Since S 2j is a Sidon set and lim j S 2j n2j n 2j 1 = 1 there exists a j 0 N so that ( ) S2j D 2j = > n 2 2j 2 j j 0. Since, on the other hand, m D 2j m < n 2j n 2j 1 = 3n 2j 2 j N, for any j j 0 there exists at least one m j D 2j satisfying n 2j 2 < m j < 3n 2j 2. Let M be defined as the sequence m j0, m j0 +1, m j0 +2,... Now we will construct a subsequence B of A by eliminating a negligible number of elements of A so that B will still satisfy and δ B (m j ) = 1 will hold for all j j 0. Since, according to the definition, > 0 n 2j 2 < m j j j 0, we have m j = a k a i a k > n 2j 2. Therefore, since A ]n 2j 2, n 2j 1 [ =, a k satisfies (11) a k > n 2j 1 and a k j 1 h=1 S 2h. On the other hand, for h j the length of the gap between two consecutive Sidon subsets S 2h 2, S 2h is bigger than n 2j 1 n 2j 2 = 3n 2j 2. Therefore as according to the definition m j < 3n 2j 2 for any j j 0, m j can occur as a difference a k a i, a i, a k A, only if both a k and a i are elements of the same Sidon subset S 2h, h j. Therefore the B 2 -property of the sets S 2j implies that D 2j M j j N. Thus j j 0 it is possible to construct a new Sidon set S 2j from S 2j by eliminating less than j elements of S 2j so that D 2j M = {m j },

6 378 M. Helm where We define D 2j := {m N : (a i, a k ) S 2j, m = a k a i }. B := S 2j, where S 2j := S 2j, 1 j < j 0. j=1 Obviously δ B (m j ) = 1 j j 0 and S 2j S 2j > 0. Furthermore, Θ n log n (n ), where Θ n is the number of Sidon subsets S 2j of B satisfying S 2j [1, n]. Consequently which completes the proof. M(n) log n > 0, References [1] S. Chowla, Solution of a problem of Erdős and Turán in additive number theory, Proc. Nat. Acad. Sci. India 14 (1944), 1 2. [2] P. Erdős and P. Turán, On a problem of Sidon in additive number theory and some related problems, J. London Math. Soc. 16 (1941), ; Addendum (by P. Erdős), ibid. 19 (1944), 208. [3] H. Halberstam and K. F. Roth, Sequences, Springer, New York [4] A. Stöhr, Gelöste und ungelöste Fragen über Basen der natürlichen Zahlenreihe. II, J. Reine Angew. Math. 194 (1955), FACHBEREICH MATHEMATIK JOHANNES GUTENBERG-UNIVERSITÄT MAINZ SAARSTR. 21 D-6500 MAINZ, GERMANY Received on and in revised form on (2289)