• Nie Znaleziono Wyników

# n − 1} by f colors so that there is no monochromatic subset the sum of whose elements is n

N/A
N/A
Protected

Share "n − 1} by f colors so that there is no monochromatic subset the sum of whose elements is n"

Copied!
4
0
0

Pełen tekst

(1)

ACTA ARITHMETICA LXXIV.3 (1996)

Sure monochromatic subset sums

by

Noga Alon (Tel Aviv) and Paul Erd˝os (Budapest)

1. Introduction. For an integer n > 1 let f (n) denote the smallest integer f such that one can color the integers {1, . . . , n − 1} by f colors so that there is no monochromatic subset the sum of whose elements is n. Paul Erd˝os [2] asked if for every positive ε, f (n) > n1/3−ε for all n > n0(ε). In this note we prove that this is indeed the case, in the following more precise form.

Theorem 1.1. There exist positive constants c1, c2 so that c1 n1/3

log4/3n ≤ f (n) ≤ c2n1/3(log log n)1/3 log1/3n for all n > 1.

We suspect that the upper bound is closer to the actual value of f (n) than the lower bound but this remains open. The (simple) proof of the upper bound is described in Section 2. The lower bound is established in Section 3.

To simplify the presentation, we omit all floor and ceiling signs, when- ever these are not essential. We make no attempt to optimize the absolute constants throughout the paper. For a set of integers A, let A denote the set of all sums of subsets of A.

2. The upper bound. Given n, we prove that f (n)  n1/3(log log n)1/3

log1/3n

by exhibiting an explicit family of subsets of N = {1, . . . , n−1} whose union covers N , so that n 6∈ A for each subset A in the family. Define

s = n1/3(log log n)1/3 log1/3n .

For each integer k satisfying 1 ≤ k ≤ s, let Ak = {i ∈ N : n/(k + 1) ≤ i <

n/k}. Note that n 6∈ Ak, since the sum of any set of at most k members

[269]

(2)

270 N. Alon and P. Erd˝os

of Ak is less than n whereas the sum of any set of at least k + 1 members of Ak exceeds n. For each prime p ≤ s that does not divide n define Bp = {i ∈ N : p | i}. Since all members of Bp are divisible by p it follows that n 6∈ Bp. It is well known (see, e.g., [4]) that Brun’s sieve method gives that for any set P of primes which are all at most m, the number of integers between 1 and m which are not divisible by any member of P does not exceed O(mQ

p∈P(1 − 1/p)). It follows that there is an absolute constant c so that the number, call it S, of integers in N not covered by the union of all sets Ak and Bp above satisfies

S ≤ c n

s log nQ

p|n, p≤s(1 − 1/p).

(Note that all these integers are smaller than n/s.) However, it is easy to check that

Y

p|n, p≤s

(1 − 1/p)  1 log log n, showing that

S  n2/3(log log n)2/3 log2/3n .

We can now split the set of these remaining integers arbitrarily into dS/se sets Cj of size at most s each. Since each member of Cj is at most n/s, n 6∈ Cj for any Cj. The sets Ak, Bp and Cj together cover N , and their total number is at most

O

n1/3(log log n)1/3 log1/3n

 ,

completing the proof of the upper bound in Theorem 1.1.

3. The lower bound. The proof of the lower bound is based on the following result of S´ark¨ozy [5] (see also [3] and [1] for similar results).

Theorem 3.1 ([5], Theorem 4). Let m > 2500 be an integer , and let A be a subset of {1, . . . , m} of cardinality |A| = 1000(m log m)1/2. Then there are integers d, y, z such that 1 ≤ d ≤ 10m1/2/ log1/2m, z > 10m log m, and y < z/(10 log m), such that {yd, (y + 1)d, (y + 2)d, . . . , zd} ⊂ A.

We also need the following simple lemma.

Lemma 3.2. Let d be a positive integer , and let B be a set of d−1 positive integers, all relatively prime to d. Then for any integer x, B contains a member congruent to x modulo d.

P r o o f. Let B = {b1, . . . , bd−1} and define b0i = bi (mod d), Bi = {b01, . . . , b0i}. Then Bi is a subset of the cyclic group Zd. Let Bi denote

(3)

Monochromatic subset sums 271

the set of all sums of subsets of Bi, computed in Zd. Our objective is to prove that Bd−1= Zd. Note that B1= {0, b01} and Bi = Bi−1 ∪ (Bi−1 + b0i), where the sum is computed in Zd. If for some i, |Bi| = |Bi−1 |, then for every b ∈ Bi−1 , b + b0i is also in Bi−1 , and since 0 ∈ Bi−1 and b0i generates Zd, Bi−1 = Zd, as needed. Otherwise, |Bi| > |Bi−1 | for all i, and hence Bd−1 = Zd, completing the proof.

Corollary 3.3. Let C ⊂ {1, . . . , m} be a set of primes of cardinality

|C| = 1000(m log m)1/2+ 20 m1/2 log1/2m + k,

where m > 2500. Let S denote the sum of the largest k members of C. Then any integer t satisfying 200m3/2/ log1/2m ≤ t ≤ S lies in C.

P r o o f. Let A denote the set of the 1000(m log m)1/2 smallest members of C. By Theorem 3.1 there are d, y, z as in the theorem, so that yd, (y + 1)d, . . . , zd are all in A. Thus, in particular,

(1) zd ≤ m|A| ≤ 1000m3/2log1/2m.

Let B be the set of the 20m1/2/ log1/2m smallest members of C − A.

Claim. Every integer x satisfying yd + md ≤ x ≤ zd lies in (A ∪ B). P r o o f. B contains at least d − 1 elements larger than d, and all of them are relatively prime to d. Therefore, by Lemma 3.2, there is a number x0 which is the sum of at most d−1 members of B and x0≡ x (mod d). Clearly x0≤ md and thus zd ≥ x ≥ x − x0≥ yd. Since x − x0 is divisible by d it lies in A, implying that x ∈ B+ A = (A ∪ B), as needed.

Returning to the proof of the corollary let I denote the interval of all integers between yd + md and zd, and let x1, . . . , xk be all elements in C − (A ∪ B). Then the length of I is at least zd/2 ≥ 5m log m > m and all the k + 1 intervals I, I + x1, I + (x1+ x2), . . . , I + (x1+ . . . + xk) lie in C. The union of these intervals contains all the integers t satisfying yd + md ≤ t ≤ S + zd, and the desired result follows from (1), since

yd ≤ zd

10 log m ≤ 100 m3/2 log1/2m, and md ≤ 10m3/2/ log1/2m.

Corollary 3.4. For all sufficiently large n, and for any set C of at least 200n1/3log2/3n primes between n2/3log1/3n/200 and n2/3log1/3n/100, the number n lies in C.

(4)

272 N. Alon and P. Erd˝os

P r o o f. Apply the previous corollary with m = n2/3log1/3n/100. Here k > 50n1/3log2/3n, 200m3/2/ log1/2m < n and S > kn2/3log1/3n

200 > n, implying that indeed n ∈ C.

P r o o f o f T h e o r e m 1.1 (lower bound). Clearly we may assume that n is sufficiently large, by an appropriate choice of c1. Given a large n, and a coloring of {1, . . . , n−1} by f = f (n) colors without a monochromatic subset whose sum is n, there is, by the prime number theorem, a monochromatic set containing at least

(1 + o(1)) 3n2/3 2f · 200 log2/3n

primes between n2/3log1/3n/200 and n2/3log1/3n/100. By the last corol- lary, this number cannot exceed

200n1/3(log n)2/3, implying the assertion of the theorem.

References

[1] N. A l o n and G. F r e i m a n, On sums of subsets of a set of integers, Combinatorica 8 (1988), 297–306.

[2] P. E r d ˝o s, Some of my recent problems in Combinatorial Number Theory, Geometry and Combinatorics, in: Graph Theory, Combinatorics and Applications, Proceedings of the Seventh Quadrennial International Conference on the Theory and Application of Graphs, Y. Alavi and A. Schwenk (eds.), Wiley Interscience, New York, 1995, 335–350.

[3] G. F r e i m a n, New analytical results in subset-sum problem, Discrete Math. 114 (1993), 205–218; Erratum, ibid. 126 (1994), 447.

[4] H. H a l b e r s t a m and H. E. R i c h e r t, Sieve Methods, Academic Press, 1974.

[5] A. S ´a r k ¨o z y, Finite addition theorems, II , J. Number Theory 48 (1994), 197–218.

DEPARTMENT OF MATHEMATICS MATHEMATICAL INSTITUTE OF THE RAYMOND AND BEVERLY SACKLER FACULTY HUNGARIAN ACADEMY OF SCIENCES

OF EXACT SCIENCES P.O.B. 127

TEL AVIV UNIVERSITY H-1364 BUDAPEST

TEL AVIV, ISRAEL HUNGARY

E-mail: NOGA@MATH.TAU.AC.IL

Cytaty

Powiązane dokumenty

Extending this idea we will introduce Hadamard matrices: such a matrix (of order q) gives sequences which can be generated by finite automata and which satisfy (2) where M 2 is

* Przy zachowaniu odpowiednich środków ostrożności, typu: jednorazowe adresy kryptowalut, używanie anonimowych adresów email koniecznych do weryfikacji, logowanie się do tych

In other words, the problem is to decide whether for every ε &gt; 0, there is a 2-partition {A 1 , A 2 } of the set of positive integers such that the lower asymptotic density of

Let us now recall the notion of α-proper forcing for a countable ordinal α saying that, given an ∈-chain of length α of countable elementary sum- bodels of some large enough structure

In the following we will focus on the system of linear differential equations (0.1) in conjunction with material relations of the type (0.2) in the case that the medium described

Tak, pod kryształow ą tarczą czystego przy­ wiązania, Ludm iła wśród zgrai aw anturników jechała spokojnie, może nie domyślając się na­. w et jakie

The rotating stirrer, for Instance, has to overcome a certain counterpressure, because liquid of lower density has to be pushed into liquid of higher density, thus reducing

The influence of the variability of wave impact force time-history characteristics to the structural response of Wolf Rock lighthouse is presented with the use of

To calculate the probability in question imagine the triangle rotated so a side is perpendicular to the radius.. The chord is longer than a side of the triangle if the chosen

and v can not be together the extremal points. classical families of univalent functions, Trans.. Stankiewicz, J., Neighbourhoods of meromorpnic functions and Hadamard products,

Looking at the darts at dartboard the professor makes the following deduction: the dart which was thrown to a point with a lower number has to be the rst one.. Indeed, assume

In each cell there are infinitely many boxes (enumerated by natural numbers) each containing one real number.. In every cell the boxes, enumeration and the real numbers are

Show that this is equivalent to the assertion that A has strong finite intersection property (sfip), i.e.. every intersection of finitely many elements of A

3 Let A be a Boolean algebra generated by an uncountable almost disjoint family A and the family of finite sets?. How does it Stone space

8 Consider the following version of Martin’s Axiom: for every P - countable partially ordered set, for every family of κ many dense sets in P, there is a filter inter- secting all

(This is Hales-Jewett theorem. See the Blass’ paper linked on the

Can you find a partial order of size at most ω 1 which is not Tukey equivalent to any of them.. (P is Tukey equivalent to Q iff P is Tukey below Q and Q is Tukey

Show that those spaces are Hausdorff but not

1 Show that the existence of countable π-base implies separability which itself implies Knaster property and that Knaster property implies

1 Show that the summable ideal and the ideal of density 0 sets are both dense.. Show that they

Applications of infinitary combinatorics 11

[36] —, —, Pseudo-euclidean Hurwitz pair and generalized Fueter equations, in: Clifford Al- gebras and Their Applications in Mathematical Physics, Proceedings, Canterbury 1985,

“Customer capital is the wealth contained in an organization’s relations with its customers and its suppliers” [9, p. It involves having own brand of the company, customer accounts,