Budapest University of Technology and EconomicsInstitute of Mathematics

Department of Stochastics

Some problems related to the additiverepresentation functions

PhD Thesis

Eszter Rozgonyi

Supervisor: Dr. Csaba Sandor

2015

Contents

Introduction 1

1 Preliminaries 51.1 The Probabilistic method . . . . . . . . . . . . . . . . . . . . 51.2 Algebraic tools . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 A converse to an extension of a Theorem of Erdos and Fuchs 102.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2 Probabilistic construction of A . . . . . . . . . . . . . . . . . . 122.3 Proof of Theorem 2.1.2 . . . . . . . . . . . . . . . . . . . . . . 13

3 Generalization of a Theorem of Nathanson and related prob-lems 253.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2 Proof of Theorem 3.1.3, Proposition 3.1.6 and 3.1.8 . . . . . . 313.3 Proof of Theorem 3.1.10 . . . . . . . . . . . . . . . . . . . . . 413.4 Proof of Theorem 3.1.11 . . . . . . . . . . . . . . . . . . . . . 423.5 Proof of Theorem 3.1.12 . . . . . . . . . . . . . . . . . . . . . 46

4 Additive complement sets 484.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.2 Proof of Theorem 4.1.1 . . . . . . . . . . . . . . . . . . . . . . 504.3 Proof of Theorem 4.1.2 . . . . . . . . . . . . . . . . . . . . . . 54

5 Representation functions on groups 555.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 555.2 Proof of Theorem 5.1.3 . . . . . . . . . . . . . . . . . . . . . . 585.3 Proof of Theorem 5.1.6 . . . . . . . . . . . . . . . . . . . . . . 595.4 Proof of Theorem 5.1.7 . . . . . . . . . . . . . . . . . . . . . 605.5 Proof of Theorem 5.1.8 . . . . . . . . . . . . . . . . . . . . . 615.6 Proof of Theorem 5.1.9 . . . . . . . . . . . . . . . . . . . . . . 62

i

6 On Sidon sets which are asymptotic bases 646.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 646.2 Proof of Theorem 6.1.1 . . . . . . . . . . . . . . . . . . . . . . 65

7 Summary 82

Bibliography 83

ii

Acknowledgements

First and foremost I am greatly indebted to my supervisor, Dr. CsabaSandor. I am thankful for drawing my attention to the additive represen-tation functions. I have learned a lot from our consultations. Without hissupport I would never have been able to write my papers and this thesis.

I thank to my coauthor, Sandor Z. Kiss for our common work. I am alsograteful for Professor Imre Z. Ruzsa for the helpful and valuable discussions.

Last but not least I would like to thank the support and the devotion ofmy family and my friends.

iii

Introduction

In this thesis we prove some results which are related to the additive rep-resentation functions and its properties. Let N denote the set of nonneg-ative integers, and let h ≥ 2 be a fixed integer. Let A = a1, a2, . . . ,(0 ≤ a1 < a2 < . . . ) be an infinite sequence of nonnegative integers. Then

for n = 0, 1, 2, . . . the representation functions R(1)h,A(n), R

(2)h,A(n) and R

(3)h,A(n)

denote the number of solutions of the equations:

ai1 + · · ·+ aih = n, ai1 , . . . , aih ∈ A,

ai1 + · · ·+ aih = n, ai1 , . . . , aih ∈ A, ai1 ≤ · · · ≤ aih ,

ai1 + · · ·+ aih = n, ai1 , . . . , aih ∈ A, ai1 < · · · < aih

respectively. (In the first case the order of the summands counts, and theycould be equal. In the second case the order of the elements does not countand they could be equal. In the last case the order of the elements doesnot count and they could not be equal.) We note that there is connectionbetween these three functions. For example (h = 2):

R(2)2,A(n) =

R

(3)2,A(n) + 1 if n is even and n

2∈ A

R(3)2,A(n) otherwise

In most cases we shall express each result in terms of whichever representationfunction is the most appropriate or the most convenient.

The additive representation functions were studied by many aspects. Thefirst question which raised in this topic that R

(i)2,A(n), i = 1, 2, 3 can be

constant or not. In other words is it possible to design a nontrivial set A, sothat, for some misbehaviour at the beginning, R

(i)2,A(n), i = 1, 2, 3 constant?

The answer is NO in every cases. Then how nearly constant can R(1)2,A(n), be

on average? The famous Erdos–Fuchs theorem [ErFu] involves this question.We can say also that this theorem means the beginning of the research ofthe additive representation functions. This result has been generalized andextended by many people.

1

There are many questions and results in the literature which deal withthe properties of these functions (i.e. [ErRe], [HalRo]). For example Erdos

showed that there exists a sequence A ⊂ N for which every n R(1)2,A(n) =

Θ(log n). In few papers Erdos, Sarkozy and T. Sos studied the regularity

property and the monotonicity of the function R(1)2,A(n) ([ErSa85], [ErSa86],

[ErSaTSos87] and [ErSaTSos85]).One can ask inverse questions also. Namely what could we say about the

sets A and B if we have information about their representation functions.Nathanson [Nat78] was the first who studied related questions. In this thesiswe extend and generalize some results of Erdos and Nathanson by usinggenerating function method and probabilistic method. We also prove sometheorems about the additive complement sets, Sidon basis and representationfunctions on groups.

In Chapter 1 we give a short survey about the probabilistic method andabout algebraic tools we are working with. In Section 1.1 we present theErdos–Renyi probabilistic method based on the definitions and notations ofthe Halberstam–Roth book [HalRo]. This method plays a crucial role inthis thesis. In Section 1.2 we present the algebraic tools which we workwith, especially the properties of the cyclotomic polynomials. The next fivechapters of the thesis consist our papers. We divide all of these chapters intotwo parts. The first is an introduction section about the definitions, earlierresults in the appropriate topic and usually the proofs are in the second part.

In Chapter 2 we prove a theorem about a converse of the generalization ofErdos and Fuchs’s result for R

(1)h,A, h > 2. This theorem gives an upper bound

about the deviation from the average behaviour of the R(1)h,A(n) representation

function [RoSa13]. Precisely we show that for every h ≥ 2 there exists asequence A = a1 < a2 < . . . of nonnegatve integers, such that

N∑n=0

R(1)h,A(n) = N +O

(N1− 1.5

h logN).

In Chapter 3 we study inverse problems. In the case h = 2 Nathansondescribed the structure of the sets A and B if their representation functionsare equal from a certain point on ( R2,A(n) = R2,B(n), n ≥ N0). We provesimilar result for the h = ps case, where p is prime and we discuss alsosome related problems in [KiRoSa14a], [RoSa14] using the generating func-tion method and cyclotomic polynomials. The main result of the chapter isthe following. Let h = ps and let the set C ⊆ Z, where C contains incon-

gruent integers modulo p. Let the series A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn,

2

an, bn ∈ C. Then A(z)h−B(z)h = P (z), where P (z) is polynomial if and only

if there exist positive integers N0, m and polynomials FA(z) =

mN0−1∑n=0

dnzn,

FB(z) =

mN0−1∑n=0

enzn, dn, en ∈ C and T (z) =

m−1∑n=0

tnzn, tn ∈ C such that

A(z) = FA(z) +T (z)zmN0

1− zm, B(z) = FB(z) +

T (z)zmN0

1− zmand

(1− zm)h−1|T (z)h−1 (FA(z)− FB(z))

holds.

In Chapter 4 we study additive complement sets. Additive complementis an important concept in additive number theory. In the past few decadesit was studied by many authors. There are several results about the densityof the infinite additive complements A and B. But the case when A or B isfinite is different. Chen and Fang proved that if A has a special form thenthere exists an additive complement B of A such that A(x)B(x)− x = O(1)[CheFa13]. (Here A(x) denotes the number of the elements of A which areless then x.) In [CheFa13] Chen and Fang posed the following conjecture. IfA is not in a special form then for any additive complement B of A we havelimx→∞

(A(x)B(x)− x) = +∞. We prove this conjecture in the prime case and

disprove in the composite case in [KiRoSa14b] by using generating functionmethod and cyclotomic polynomials. Precisely let A be a set of nonnegativeintegers with |A| = n. If A is not of the form A = a + ins + kin

s+1 : i =0, . . . , n− 1 where a > 0, s ≥ 0, ki are integers and n is prime then for anyadditive complement B of A we have lim

x→∞(A(x)B(x)− x) = +∞. If n is a

composite number then there exists an additive complement set B of A suchthat A(x)B(x)− x = O(1).

In Chapter 5 we study the representation functions on groups. Let Gbe a finite (abelian) group. We determine all subsets A ⊂ G such that thenumber of solutions of g = x+ y, x, y ∈ A equals to the number of solutionsof g = x + y, x, y ∈ G \ A. We discuss some related problems [KiRoSa14c]by using generating function method.

Finally in Chapter 6 we study the connection between the asymptoticbases and Sidon sets. For h ≥ 2 integer we say a set A of positive integersis an asymptotic basis of order h if every large enough positive integer canbe represented as the sum of h terms from A. A set of positive integers Ais called Sidon set if all the sums a + b with a, b ∈ A, a ≤ b are distinct.In 1994 P. Erdos, A. Sarkozy, V. T. Sos asked if there exists a Sidon set

3

which is asymptotic bases of order 3. In Chapter 6 we prove the existenceof Sidon sets which are asymptotic bases of order 4 by using probabilisticmethod [KiRoSa14d], especially the Kim–Vu Theorem. In the chapters thedefinitions sometimes repeated, which help the reader to understand thisthesis better.

4

Chapter 1

Preliminaries

1.1 The Probabilistic method

An important problem in additive number theory is to prove that a sequencewith certain properties exists. One of the essential ways to obtain an affirma-tive answer for such a problem is actually to construct a sequence with the re-quired properties. But even when this direct approach proves impracticable,it may still be possible to establish the existence of such a sequence. Manysuch existence theorems have been obtained by Erdos and Renyi. Thesetheorems are the basis of the so-called probabilistic method. There is anexcellent summary of this method in the Halberstam–Roth book [HalRo]. Inthis thesis we use almost the same definitions, notations and terminology likethis book.

The general steps of Erdos and Renyi’s method are the following. First onecan define a probability space and measure. Then to show that a sequencewith a property P exists it is enough to prove that a properly defined randomsequence of nonnegative integers satisfies P with positive probability. In thenext step one can use the Borel–Cantelli lemma, so one need to show thatP(n) fails with sufficiently small probability. According to the lemma this willmean that P(n) holds for all sufficiently large n with probability 1 (furtherdetails see also [Vu00b]).

Now we give a survey of the probabilistic tools and notations which weuse. Let Ω denote the set of strictly increasing sequences of positive inte-gers. In this thesis we denote the probability of an event E by P(E) and theexpectation of a random variable ξ by E(ξ). The following Lemma plays animportant role in our proofs.

Lemma 1.1.1. Let θ1, θ2, θ3, . . . be real numbers satisfying

0 ≤ θn ≤ 1 (n = 1, 2, . . .).

5

Then there exists a probability space (Ω, X, P) with the following two prop-erties:

(i) For every n ∈ N, the event E (n) = A : A ∈ Ω, n ∈ A is measurable,and P(E (n)) = θn.

(ii) The events E (1), E (2), ... are independent.

See Theorem 13. in [HalRo], p. 142. We denote the characteristic functionof the event E (n) by t(A,n) (or tn in a shorter way, if the set A is obvious). Inother way we can say the boolean variable t(A,n) (tn) means that:

t(A,n) = tn =

1 if n ∈ A0 if n /∈ A.

Furthermore, for some A = a1, a2, . . . ∈ Ω we denote the number of solu-tions of ai1 + ai2 + . . . + aih = n with ai1 , . . . , aih ∈ A, 1 ≤ ai1 < ai2 . . . <

aih < n by rh,A(n) (this is the R(3)h,A(n) representation function). Let

rh,A(n) =∑

(ai1 ,ai2 ,...,aih )1≤ai1<...<aih<nai1+ai2+...+aih=n

t(A,ai1 )t(A,ai2 ) . . . t(A,aih ). (1.1.1)

Let r∗h,A(n) denote the number of those representations of n in the form(1.1.1) in which there are at least two equal terms. Thus we have

R(1)h,A(n) = h! rh,A(n) + r∗h,A(n).

It is easy to see from (1.1.1) that rh,A(n) is the sum of random variables.However, for h > 2 these variables are not independent because the samet(A,ai) may appear in many terms.

There are some probabilistic results which can help us to overcome thistrouble. First we present the method of J. H. Kim and V. H. Vu. We givea short survey of this method. Interested reader can find more details in[KimVu], [TaoVu], [Vu00a], [Vu00b].

Assume that t1, t2, . . . , tn are independent binary (i.e., all ti’s are in 0, 1)random variables. (We can regard these ti-s as characteristic functions.)Consider a polynomial Y = Y (t1, . . . tn) in t1, t2, . . . , tn with degree k (wherethe degree of this polynomial equals to the maximum of the sum of theexponents of the monomials). We say a polynomial Y is totally positive if itcan be written in the form Y =

∑i eiΓi, where the ei’s are positive and Γi is

a product of some tj’s. Furthermore, Y is regular if all of its coefficients are

6

between zero and one. We also say Y is simplified, if all of its monomials aresquare-free (i.e. do not contain any factor of t2i ), and homogeneous if all themonomials have the same degree. Thus for instance a boolean polynomial isautomatically regular and simplified, though not necessarily homogeneous.Given any multi-index α = (α1, . . . , αn) ∈ Nn, we define the partial derivative∂α(Y ) of Y as

∂α(Y ) =

(∂

∂t1

)α1

· · ·(∂

∂tn

)αnY (t1, . . . tn).

For instance, if Y = t1t22 and α1 = (1, 1) and α2 = (1, 2), then ∂α1 = 2t2

and ∂α2 = 2. If α equals to zero vector, then ∂α(Y ) = Y . Let denotethe order of α as |α| = α1 + · · · + αn. For any order d ≥ 0, we denoteEd(Y ) = maxα:|α|=dE (∂αY ). Thus for instance E0(Y ) = E(Y ) and Ed(Y ) = 0if d exceeds the degree of Y . We also define E≥d(Y ) = maxd′≥dEd′(Y ). Thefollowing result is due to Kim and Vu.

Theorem 1.1.2. (J. H. Kim and V. H. Vu, 2000)Let k ≥ 1 and Y = Y (t1, . . . , tn) be a totally positive polynomial of n in-dependent boolean variables t1, . . . , tn. Then there exists a constant Ck > 0depending only on k (which is the degree of the polynomial) such that

P(|Y − E(Y )| ≥ Ckλ

k− 12

√E≥0(Y )E≥1(Y )

)= Ok

(e−

λ4+(k−1) logn

)for all λ > 0.

See [KimVu] for the proof. Informally this theorem asserts that when thederivatives of Y are smaller on average than Y itself, and the degree of Y issmall, then Y is concentrated around its mean.

We also need the Borel–Cantelli lemma:

Lemma 1.1.3. (Borel–Cantelli)Let E1, E2, . . . be a sequence of events in a probability space. If

+∞∑j=1

P(Ej) <∞,

then with probability 1, at most a finite number of the events Ej can occur.

See in [HalRo], p. 135.

In this thesis we use another kind of probabilistic method also. There is agroup of theorems in probability theory collectively known as large deviationtheorems. A description of the general nature of these theorems is to be found

7

in [Var84]. The general strategy of this method is to show that the randomvariables are bounded and there is another type of bound which holds onthe sum square of this previous bounds. So finally we have an estimationfor the deviation of the sum of the random variables and its mean. In theproofs we will use the following variant of the Hoeffding-inequality (It is aBernstein-type inequality.):

Theorem 1.1.4. (Hoeffding-inequality, 1963)Let X1, X2, . . . Xn be independent bounded real random variables, ui ≤ Xi ≤ vi.Write Sn =

∑ni=1Xi. Then for every t > 0 we have

P(Sn − E(Sn) ≥ t) ≤ exp

−2n2t2∑n

i=1(ui − vi)2

.

See [Hoe], Theorem 2.To make our proof easier we will use the following modified version of the

original theorem.

Theorem 1.1.5. (Hoeffding-inequality modified)Let X1, X2, . . . Xn be independent bounded real random variables, ui ≤ Xi ≤ viand there is a positive constant D such that

n∑i=1

(ui − vi)2 ≤ D2.

Write Sn =∑n

i=1Xi. For every y > 0 we have

P(Sn − E(Sn) ≥ yD) ≤ exp−2y2.

1.2 Algebraic tools

In this section we give a short survey about the algebraic concept which playa crucial role in the proof of Theorem 4.1.1. In this proof we are workingwith cyclotomic polynomials. Both the definition and the most importantproperties of these polynomials are well-known, so we write only about thoseproperties which we will use in the proof. Interested reader can find these in[LiNie], p. 63-66.

We denote the degree of a polynomial f by deg f . Let θ be an algebraicnumber. We say the monic polynomial f is the minimal polynomial of θ iff is the least degree such that f(θ) = 0. It is well-known that if f is theminimal polynomial of θ, and g is a polynomial such that g(θ) = 0, then f | g.

8

A complex number µ is called primitive nth root of unity if µ is the root ofthe polynomial xn − 1 but not of xm − 1 for any m < n. The cyclotomicpolynomial of order n is defined by

Φn(z) =∏ζ

(z − ζ),

where ζ runs over all the primitive nth roots of unity. This is a monicirreducible polynomial with degree ϕ(n), and Φn(z) has integer coefficients.It is well-known that Φn(z) is the minimal polynomial of ζ and

1 + z + z2 + . . . + zn−1 =∏l|nl>1

Φl(z). (1.2.1)

It is easy to see that

Φps+1(z) = 1 + zps

+ z2ps

+ . . . + z(p−1)ps

. (1.2.2)

9

Chapter 2

A converse to an extension of aTheorem of Erdos and Fuchs

In this chapter we generalize a result which is connected to the famous Erdos-Fuchs theorem. In the first section we give a short summary about the resultsin this topic, and after we present the generalization. The chapter is mostlybased on the article [RoSa13].

2.1 Introduction

For r > 0 let N(r) count the number of lattice points inside the boundaryof a circle with center at the origin and radius r. The famous Gauss-circleconjecture says that

N(r) = r2π +O(r1/2+ε).

Nowadays the best result due to Huxley [Hux] is that here O(r1/2+ε) canbe replaced by O(r131/208). On the other hand, using the techniques of theFourier analysis, Hardy [Har] proved that

N(r) = r2π +O(r1/2(log r)1/4)

can’t hold for all sufficiently large r.Using the definiton of representation functions we can write that

N(r) =∑n≤r2

R(1)2,A(n)

provided A = 0, 1, 1, 4, 4, 9, 9, 16, 16, . . . .

10

In [ErFu] (1956), Erdos and Fuchs proved that for any sequence A andconstant c > 0,

N∑n=0

R(1)2,A(n) = cN + o(N

14 (logN)−

12 )

can’t hold for all sufficiently large n. This theorem asserts that R(1)2,A(n)

cannot behave very regularly. Although here o(N14 (logN)−

12 ) is slightly

weaker than Hardy’s bound O(r1/2(log r)1/4), the Erdos–Fuchs theorem isvalid for any sequence of A of nonnegative integers, rather than only forA = 0, 1, 1, 4, 4, 9, 9, 16, 16, . . . . Subsequently Jurkat (seemingly unpub-lished), and later in 1990 Montgomery and Vaughan [MoVa] improved thistheorem to the following. The formula

N∑n=0

R(1)2,A(n) = cN + o(N

14 )

cannot hold for any sequence A and constant c > 0 for all sufficiently largen. Nowadays, there are several different generalizations of the Erdos-Fuchstheorem [CheTa11], [Hor01], [Hor02], [Hor04], [Ta09]. For example Tang

[Ta09] showed in 2009 that R(1)h,A(n) cannot behave very regularly. Namely

for h ≥ 2 for any sequence A

N∑n=0

R(1)h,A(n) = cN + o(N

14 )

with c > 0 is impossible.In the opposite direction Vaughan asked whether a further improvement

is possible, or in other words, whether there is a sequence A and constantc > 0 such that

N∑n=0

R(1)2,A(n) = cN +O(N

14+ε).

With help of a probabilistic discussion, Ruzsa [Ru97] gave an affirmativeanswer to this question in 1997. In fact, he proved the existence of a sequenceA of integers satisfying

N∑n=0

R(1)2,A(n) = cN +O(N

14 logN)

for all N ≥ 2. It is natural to ask whether Ruzsa’s result can be generalized.In 2012, Dai and Pan [DaPa] extended Ruzsa’s theorem:

11

Theorem 2.1.1. (Dai, Pan, 2012)Suppose that h ≥ 2 is an integer and β < h is a positive real number. Thenthere exists a sequence A = a1 < a2 < a3 < . . . of positive integers,satisfying

N∑n=0

R(1)h,A(n)− CNβ =

O(Nβ−β(h+β)/h2

√logN

)if h > 2β,

O(Nβ−3β/(2h)

√logN

)if h < 2β,

O(Nβ−3/4 logN

)if h = 2β,

where C is constant.

In this section we prove a better error term for the case β = 1.

Theorem 2.1.2. (Rozgonyi, Sandor, 2013)For every h ≥ 2 there exists a sequence A = a1 < a2 < a3 < . . . ofnonnegative integers such that

N∑n=0

R(1)h,A(n) = N +O

(N1− 1.5

h logN). (2.1.1)

2.2 Probabilistic construction of AIn the proof we shall follow the way of Ruzsa in [Ru97]. So at first we define aninfinite increasing sequence S = s0, s1, s2, . . . of nonnegative integers. Wewill choose the nth element of A from sn, sn + 1, . . . , sn+1 by probabilisticmethod.

Denote by En the expectation number that how many times occurs thenumber n in the sequence A. We choose these values En such that

E(R(1)h,A(n)) = 1 ∀n ∈ N

hold.If we don’t care with the independence, then we get

E(R(1)h,A(n)) ≈

∑(i1,...,ih)

i1+···+ih=n

Ei1Ei2 . . . Eih .

So we have to solve the following system of equations:∑(i1,...,ih)

i1+···+ih=n

Ei1Ei2 . . . Eih = 1, ∀n ∈ N.

12

Then( ∞∑n=0

Enzn

)h=∞∑n=0

( ∑(i1,...,ih)

i1+···+ih=n

Ei1Ei2 · · ·Eih)zn =

∞∑n=0

zn =1

1− z.

So∞∑n=0

Enzn = (1− z)−

1h =

∞∑n=0

(− 1h

n

)(−z)n.

Hence a natural choice of En is the following

En = (−1)n(− 1h

n

)=

1

h

h+ 1

2h

2h+ 1

3h· · · (n− 1)h+ 1

nh.

Now we can define the sequence S. Denote by sn the following

sn = min

m :

m∑k=0

Ek > n

.

From this we get the random sequence A in the following way:

P(an = a) =

sn∑k=0

Ek − n if a = sn,

Ea if sn < a < sn+1,

(n+ 1)−sn+1−1∑k=0

Ek if a = sn+1,

0 otherwise.

We will show that with probability 1 this sequence satisfies Theorem 2.1.2.

2.3 Proof of Theorem 2.1.2

The following lemma gives the precise estimation of the integer sn.

Lemma 2.3.1. There exist real numbers c1 = c1(h) > 0 and c2 = c2(h) suchthat

sn = c1(n− c2)h +O(1). (2.3.1)

13

Proof of Lemma 2.3.1. Clearly, E0 = 1 and for k ≥ 1 we have

Ek =1

h

h+ 1

2h

2h+ 1

3h· · · (k − 1)h+ 1

kh= exp

k∑i=1

log

(1− h− 1

ih

)=

= exp

−

k∑i=1

h− 1

ih− 1

2

k∑i=1

(h− 1)2

i2h2− 1

3

k∑i=1

(h− 1)3

i3h3− · · ·

=

= exp

h− 1

h

(− log k + C1 +O

(1

k

))= C2k

−h−1h +O

(k−2+

1h

).

That means

Ek = C2k−h−1

h + ek, where ek = O(k−2+

1h

). (2.3.2)

Using equation (2.3.2) we obtain

m∑k=0

Ek = 1+m∑k=1

(C2k

−h−1h + ek

)= C2

m∑k=1

k−h−1h +C3+O

(m−

h−1h

), (2.3.3)

where C3 = 1 +∞∑k=1

ek.

Using integral approximation we get

m∑k=1

1

kh−1h

=

∫ m+1

1

1

bxch−1h

dx =

=

∫ m+1

1

1

xh−1h

dx+

∫ m+1

1

1

bxch−1h

− 1

xh−1h

dx =

=[hx

1h

]m+1

1+

∫ ∞1

xh−1h − bxch−1

h

xh−1h bxch−1

h

dx−∫ ∞m+1

xh−1h − bxch−1

h

xh−1h bxch−1

h

dx =

= hm1h + C4 +O

(m−

h−1h

).

(2.3.4)

Hence from (2.3.3) and (2.3.4)

m∑k=0

Ek = C2hm1h + C5 +O(m−

h−1h ).

By the definition of the integer sn we have

n <sn∑k=0

Esn ≤ n+ Esn ,

14

thereforeC2hs

1hn = n+O(1).

It means that sn nh, therefore we get

sn∑k=0

Ek = C2hs1hn + C5 +O

(s−h−1

hn

)= C2hs

1hn + C5 +O(n−(h−1)).

On the other hand, using again the definition of the integer sn and equa-tion (2.3.2) we have

n <sn∑k=0

Ek ≤ n+ Esn = n+O(n−(h−1)

),

that issn∑k=0

Ek = n+O(n−(h−1)

).

Now our statement follows from the equality

C2hs1hn + C5 +O

(n−(h−1)

)= n+O

(n−(h−1)

).

In the next step we will use Theorem 1.1.5, the modified Hoeffding inequal-ity. We will choose ai, how we write it in Section 2.2. From the definition ofthe sequence A we easily get the following properties for an:

∞∑l=0

P(an = l) = 1, ∀n ∈ N (2.3.5)

and∞∑n=0

P(an = l) = El, ∀l ∈ N. (2.3.6)

Fix an integer N and define δi1,...,ih and di1,...,ih in the following way:

δi1,...,ih =

1 if ai1 + ai2 + . . .+ aih ≤ N,

0 if ai1 + ai2 + . . .+ aih > N,

(clearly δi1,...,ih depends on N) and

di1,...,ih =∞∑j1=0

. . .

N−j1−...−jh−1∑jh=0

P(ai1 = j1)P(ai2 = j2) . . .P(aih = jh).

15

Then∑(i1,...,ih)

δi1,...,ih = #(i1, . . . , ih) : ai1 + ai2 + . . .+ aih ≤ N =N∑n=0

R(1)h,A(n),

and by (2.3.5) and (2.3.6)

∑(i1,...,ih)

di1,...,ih =∞∑i1=0

. . .

∞∑ih=0

∞∑j1=0

. . .

N−j1−...−jh−1∑jh=0

P(ai1 = j1) . . .P(aih = jh) =

=∞∑j1=0

. . .

N−j1−...−jh−1∑jh=0

∞∑i1=0

. . .

∞∑ih=0

P(ai1 = j1) . . .P(aih = jh) =

=∞∑j1=0

. . .

N−j1−...−jh−1∑jh=0

∞∑i1=0

P(ai1 = j1) . . .∞∑ih=0

P(aih = jh) =

=∞∑j1=0

. . .

N−j1−...−jh−1∑jh=0

Ej1 . . . Ejh =

=∑

(j1,...,jh)j1+...+jh≤N

Ej1 . . . Ejh = N + 1.

Further note that

δi1,...,ih = di1,...,ih =

1 if si1+1 + si2+1 + . . .+ sih+1 ≤ N,

0 if si1 + si2 + . . .+ sih > N.(2.3.7)

So we get the following

N∑n=0

R(1)h,A(n)− (N + 1) =

∑(i1,...,ih)

(δi1,...,ih − di1,...,ih) =

=∑

(i1,...,ih)si1+1+...+sih+1≤N

(δi1,...,ih − di1,...,ih)+

+∑

(i1,...,ih)si1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih − di1,...,ih)+

+∑

(i1,...,ih)si1+...+sih>N

(δi1,...,ih − di1,...,ih).

16

Because of (2.3.7) the first and the third sums are equal to 0. So we onlyneed to consider those h-tuples (i1, . . . , ih) for which si1 + . . . + sih ≤ N <si1+1 + . . .+ sih+1.

We will show that∑(i1,...,ih)

si1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih − di1,...,ih) =

= h!∑

(i1,...,ih)0≤i1<...<ih

si1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih − di1,...,ih) +O(N1− 2

h

). (2.3.8)

In order to prove (2.3.8) we have to calculate the number of h-tuples(i1, . . . , ih), where is = it for some 1 ≤ s < t ≤ h. We may suppose withoutloss of generality that s = h − 1 and t = h. By Lemma 2.3.1 for someconstants C6 and C7 we have

#(i1, . . . , ih) : si1 + . . .+ sih ≤ N < si1+1 + . . .+ sih+1, sih−1= sih =

= #(i1, . . . , ih−1) : si1+. . .+sih−2+2sih−1

≤ N < si1+1+. . .+sih−2+1+2sih−1+1 ≤

≤∑

(i1,...,ih−2)

#

ih−1 :

h−2∑j=1

c1(ij − c2)h + 2c1(ih−1 − c2)h − C6 ≤ N ≤

h−2∑j=1

c1(ij + 1− c2)h + 2c1(ih−1 + 1− c2)h + C7

≤

≤∑

(i1,...,ih−2)

#

ih−1 : h

√√√√N − C7

2c1− 1

2

h−2∑j=1

(ij + 1− c2)h + c2 − 1 ≤ ih−1 ≤

h

√√√√N + C6

2c1− 1

2

h−2∑j=1

(ij − c2)h + c2

≤

≤∑

(i1,...,ih−2)

(h

√√√√ N

2c1− 1

2

h−2∑j=1

(ij − c2)h− h

√√√√ N

2c1− 1

2

h−2∑j=1

(ij + 1− c2)h+O(1)

)=

=∑

(i1,...,ih−2)

(h

√√√√ N

2c1− 1

2

h−2∑j=1

(ij − c2)h− h

√√√√ N

2c1− 1

2

h−2∑j=1

(ij + 1− c2)h)

+O(Nh−2h ).

It is easy to see that the last sum contains telescopic sums. The numberof these telescopic sums is O

(N

h−3h

)and the order of one telescopic sum is

O(N

1h

), which proves formula (2.3.8).

17

By the Borel–Cantelli lemma, it is enough to show that for some con-stant C8

P

( ∣∣∣∣∣ ∑(i1,...,ih)0≤i1...<ih

si1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih − di1,...,ih)

∣∣∣∣∣ ≥ C8N1− 1.5

h logN

)=

= O(N−

h+2h

). (2.3.9)

Unfortunately, the variables δi1,...,ih are not independent. So our main diffi-culty is to decompose the sum in (2.3.9) to a sum of independent variables.

Let (i1, . . . , ih−2), 0 ≤ i1 < i2 < . . . < ih−2 be an (h− 2)-tuple and let

Ii1,...,ih−2,ih := ih−1 : 0 ≤ i1 < i2 < . . . < ih−1 < ih,h∑j=1

sij ≤ N ≤h∑j=1

sij+1.

These Ii1,...,ih−2,ih ’s are intervals. Now we give a suitable partition of theintervals Ii1,...,ih−2,ih in which our variables are independent. We will showthere exists an integer K so that if ih ≡ i,h (mod K) and ih < i,h thenIi1,...,ih−2,ih ∩ Ii1,...,ih−2,i

,h

= ∅. In order to prove it let K be an integer and letus suppose that for some (h−2)-tuple (i1, . . . , ih−2) there exist integers ih andi,h so that ih ≡ i,h (mod K), where ih < i,h and Ii1,...,ih−2,ih ∩ Ii1,...,ih−2,i

,h6= ∅.

Using the definition of Ii1,...,ih−2,ih and Ii1,...,ih−2,i,h

we get

h−1∑j=1

sij + sih ≤ N ≤h−1∑j=1

sij+1 + sih+1

andh−1∑j=1

sij + si,h ≤ N ≤h−1∑j=1

sij+1 + si,h+1.

But sih+K ≤ si,h therefore

si1 + . . .+ sih−1+ sih+K ≤ N ≤ si1+1 + . . .+ sih−1+1 + sih+1.

By Lemma 2.3.1 we get

h−1∑j=1

c1(ij − c2)h + c1(ih +K − c2)h − C9 ≤

≤h−1∑j=1

c1(ij + 1− c2)h + c1(ih + 1− c2)h + C10. (2.3.10)

18

To rearrange (2.3.10) and using the differential mean value theorem we get

c1h(K − 1)(ih + 1− c2)h−1 ≤ c1(ih +K − c2)h − c1(ih + 1− c2)h ≤

≤h−1∑j=1

(c1(ij + 1− c2)h − c1(ij − c2)h

)+C11 ≤ hc1(ih−1 + 1− c2)h−1 +C12.

But it is a contradiction, if K is large enough.

For a (h− 2)-tuple (i1, . . . , ih−2) let

D2i1,...,ih−2

=∑ih

∣∣Ii1,...,ih−2,ih

∣∣2 .It is enough to show that∑

(i1,...,ih−2)

D2i1,...,ih−2

= O(N

h−1h logN

), (2.3.11)

since by Cauchy-Schwarz inequality we have

D =∑

(i1,...,ih−2)

Di1,...,ih−2≤√ ∑

(i1,...,ih−2)

D2i1,...,ih−2

#(i1, . . . , ih−2) =

=

√O(N

h−1h logN

)O(N

h−2h

)= O

(N1− 1.5

h

√logN

).

Therefore by Theorem 1.1.5 (modified Hoeffding’s inequality) we get

P

( ∣∣∣∣∣ ∑(i1,...,ih)

0≤i1<...<ihsi1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih − di1,...,ih)

∣∣∣∣∣ ≥ DK√

2 logN

)=

= P

( ∣∣∣∣∣ ∑(i1,...ih−2)

0≤i1<...<ih−2

∑(ih−1,ih)

ih−2<ih−1<ihsi1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih − di1,...,ih)

∣∣∣∣∣ ≥∑

(i1,...,ih−2)

Di1,...,ih−2K√

2 logN

)≤

≤∑

(i1,...,ih−2)0≤i1<...<ih−2

si1+...+sih−2<N

K−1∑k=0

P

( ∣∣∣∣∣ ∑(ih−1,ih)

ih−2<ih−1<ihih≡k (mod K)

si1+...+sih≤N<si1+1+...+sih+1

(δi1,...,ih−di1,...,ih)

∣∣∣∣∣ ≥ Di1,...,ih−2

√2 logN

)

19

≤∑

(i1,...,ih−2)0≤i1<...<ih−2

si1+...+sih−2<N

K−1∑k=0

exp−2 logN = O(N−1−

2h

)

and we may use the Borel–Cantelli Lemma.

In order to prove (2.3.11) let (i1, . . . , ih−2), 0 ≤ i1 < . . . < ih−2 be a fixed(h− 2)-tuple. Recall the definition of the intervals Ii1,...,ih−2,ih

Ii1,...,ih−2,ih = ih−1 : si1 + . . .+ sih ≤ N < si1+1 + . . .+ sih+1.Denote by k′ the maximum value of ih. Let

H = Hi1,...,ih =N

c1−

h−2∑j=1

(ij − c2)h.

It is easy to see thatH N, (2.3.12)

and H > 0 if N is large enough.To prove Theorem 2.1.2 it is enough to verify the following lemma.

Lemma 2.3.2. 1. k′ = h√H − (ih−2 − c2)h +O(1)

2. |Ii1,...,ih−2,ih | = O(Nh−1

h2 ) for every (h− 2)-tuple (i1, . . . , ih−2).

3. |Ii1,...,ih−2,ih | = O

(Nh−1h

(H−(ih−c2)h)h−1h

), if H − (ih − c2)h) ≥ 1.

First, we finish the proof of Theorem 2.1.2 from Lemma 2.3.2. Since if Cis a fixed constant and N is large enough then∑

(i1,...,ih−2)

D2i1,...,ih−2

=∑

(i1,...,ih−2)

ih−2≤CNh−1h2

D2i1,...,ih−2

+∑

(i1,...,ih−2)

ih−2>CNh−1h2

D2i1,...,ih−2

.

Here for ih−2 ≤ CNh−1

h2 by (2.3.12) we have the following estimation for theD2-bound in Theorem 1.1.5

D2i1,...,ih−2

=∑ih

|Ii1,...,ih−2,ih |2 =

=∑

ih>h√H+c2−1

|Ii1,...,ih−2,ih |2 +∑

ih≤ h√H+c2−1

|Ii1,...,ih−2,ih|2 =

= O(N2(h−1)

h2 ) +∑r≥1

|Ii1,...,ih−2,

h√H+c2−r|2 =

20

=O(N2(h−1)

h2 ) +∑r≥1

(O(N

2(h−1)h )

(( h√H)h − ( h

√H + c2 − r − c2)h)

2(h−1)h

)=

=O(N

2(h−1)

h2

)+O

(N

2(h−1)h H−

2(h−1)2

h2

)∑r

1

r2(h−1)h

= O(N

2(h−1)

h2

).

(2.3.13)

Thus ∑(i1,...,ih−2)

ih−2≤CNh−1h2

D2i1,...,ih−2

= O(N

(h−2)(h−1)

h2

)O(N

2(h−1)

h2

)= O

(N

h−1h

).

Since

c1

h∑j=1

(ij − c2)h − C13 ≤ N,

if ih−2 > CNh−1

h2 , where C is a large fixed constant and N is large enough,then

H − (ih − c2)h ≥ 1.

That means

1 ≤ (ih−2 − c2)h −C13

c1≤ (ih−1 − c2)h −

C13

c1≤ H − (ih − c2)h.

Thus we have

k′ = h√H − (ih−2 − c2)h +O(1) =

h√H

h

√1− (ih−2 − c2)h

H+O(1) ≤

≤ h√H

(1− (ih−2 − c2)h

hH

)+O(1) ≤ h

√H − (ih−2 − c2)h

hHh−1h

+O(1) ≤

≤ h√H + c2 −

ihh−2

2hHh−1h

.

Hence using the above estimations we get∑(i1,...,ih−2)

0≤i1<...<ih−2

ih−2>CNh−1h2

D2i1,...,ih−2

=∑ih−2

ih−2>CNh−1h2

∑(i1,...,ih−3)

0≤i1<...<ih−3ih−3<ih−2

D2i1,...,ih−2

=

21

=∑ih−2

ih−2>CNh−1h2

∑(i1,...,ih−3)

0≤i1<...<ih−3ih−3<ih−2

∑ih

∣∣Ii1,...,ih−2,ih

∣∣2 =

=∑ih−2

ih−2>CNh−1h2

∑(i1,...,ih−3)

0≤i1<...<ih−3ih−3<ih−2

∑ih

O

(N

2(h−1)h

(H − (ih − c2)h)2(h−1)h

)≤

≤∑ih−2

ih−2>CNh−1h2

∑(i1,...,ih−3)

0≤i1<...<ih−3ih−3<ih−2

∑r≥

ihh−2

2hHh−1h

O

(N

2(h−1)h

(H − ( h√H + c2 − r − c2)h)

2(h−1)h

)≤

≤∑ih−2

ih−2>CNh−1h2

∑(i1,...,ih−3)

0≤i1<...<ih−3ih−3<ih−2

N2(h−1)h H−

2(h−1)2

h2

∑r≥

ihh−2

2hHh−1h

O(r−

2(h−1)h

),

where ∑r≥

ihh−2

2hHh−1h

r−2(h−1)h = O

(∫ ∞ihh−2

2hHh−1h

x−2(h−1)h dx

)= O

(i2−hh−2

N(h−1)(h−2)

h2

).

Therefore∑(i1,...,ih−2)

0≤i1<...<ih−2

ih−2>CNh−1h2

D2i1,...,ih−2

= O

( ∑ih−2

ih−2>CNh−1h2

∑(i1,...,ih−3)

0≤i1<...<ih−3ih−3<ih−2

N2(h−1)h N−

2(h−1)2

h2 i2−hh−2N(h−1)(h−2)

h2

).

The number of (h − 3)-tuples (i1, . . . , ih−3) with condition 0 ≤ i1 < · · · <ih−3 < ih−2 is O(ih−3h−2), therefore

∑(i1,...,ih)

0≤i1<...<ih−2

ih−2>CNh−1h2

D2i1,...,ih−2

= O

( ∑ih−2

ih−2>CNh−1h2

Nh−1h i−1h−2

)= O

(N

h−1h logN

),

and this completes the proof of the Theorem 2.1.2.

Proof of Lemma 2.3.2 (1): It is easy to see that the maximum value is at

k′ = maxl : si1+. . .+sih−2+sih−2+1+sl ≤ N ≤ si1+1+. . .+sih−2+1+sih−2+2+sl+1,

22

and using Lemma 2.3.1 a routine calculation gives the statement.

(2), (3): By Lemma 2.3.1 there exist constants C14 and C15 such that

|Ii1,...,ih−2,ih| = #ih−1 : si1 + . . .+sih ≤ N ≤ si1+1 + . . .+sih+1 ≤ #ih−1 :

c1(ih−1 + 1− c2)h ≥ N −h−2∑j=1

c1(ij + 1− c2)h − c1(ih + 1− c2)h − C14;

c1(ih−1 − c2)h ≤ N −h−2∑j=1

c1(ij − c2)h − c1(ih − c2)h + C15,

therefore for some constant C16

H − (ih − c2)h − C16Nh−1h ≤ (ih−1 − c2)h ≤ H − (ih − c2)h +

C15

c1, (2.3.14)

which implies part (2).To prove part (3) we distinguish two cases: If

N

c1−

h−2∑j=1

(ij + 1− c2)h − (ih + 1− c2)h < 0,

thenH − (ih − c2)h = O(N

h−1h ),

thus by (2.3.14)

|Ii1,...,ih−2,ih| = O( h√H − (ih − c2)h) = O

(N

h−1h

(H − (ih − c2)h)h−1h

).

IfN

c1−

h−2∑j=1

(ij + 1− c2)h − (ih + 1− c2)h ≥ 0,

then|Ii1,...,ih−2,ih| ≤

≤ h√H − (ih − c2)h − h

√√√√N

c1−

h−2∑j=1

(ij + 1− c2)h − c1(ih + 1− c2)h +O(1) ≤

≤H − (ih − c2)h − (N

c1−∑h−2

j=1 (ij + 1− c2)h − (ih + 1− c2)h)

(H − (ih − c2)h)h−1h

+O(1) ≤

23

≤∑h−2

j=1

((ij + 1− c2)h − (ij − c2)h

)+ (ih + 1− c2)h − (ih − c2)h

(H − (ih − c2)h)h−1h

+O(1) =

= O

(N

h−1h

(H − (ih − c2)h)h−1h

),

which completes the proof.

24

Chapter 3

Generalization of a Theorem ofNathanson and relatedproblems

In this chapter we study inverse problems about different additive repre-sentation functions. The main result is the generalizations of a theorem ofNathanson for h-term representation functions. The chapter is mainly basedon papers [KiRoSa14a], [RoSa14].

3.1 Introduction

As we mentioned in the Introduction representation functions have been ex-tensively studied by many authors and are still a fruitful area of researchin additive number theory. In this chapter we study inverse questions. Itmeans that we want to say something about the structure of the sets, if wehave information about their representation function. The first results inthis topic are due to Nathanson. Using generating functions, he proved thefollowing result [Nat78].

Let FA, FB and T be finite sets of integers. If each residue class modulo mcontains exactly the same number of elements of FA as elements of FB, thenwe write FA ≡ FB (mod m). If the number of solutions of the congruencea+ t ≡ n (mod m) with a ∈ FA, t ∈ T equals to the number of solutions ofthe congruence b+ t ≡ n (mod m) with b ∈ FB, t ∈ T for each residue classn modulo m then we write FA + T ≡ FB + T (mod m).

Theorem 3.1.1. (Nathanson, 1978)

Let A and B be infinite sets of nonnegative integers, A 6= B. Then R(1)2,A(n) =

25

R(1)2,B(n) from a certain point on if and only if there exist positive integers

N , m and finite sets FA, FB, T with FA ∪ FB ⊂ 0, 1, . . . , N and T ⊂0, 1, . . . ,m− 1 such that FA+T ≡ FB+T (mod m), and A = FA ∪C andB = FB ∪ C, where C = c > N : c ≡ t (mod m) for some t ∈ T.

It is clear that R(2)2,A(n) =

⌈R

(1)2,A(n)

2

⌉and R

(3)2,A(n) =

⌊R

(1)2,A(n)

2

⌋, so for the

sets A, B in Theorem 3.1.1 we have R(2)2,A(n) = R

(2)2,B(n) and R

(3)2,A(n) = R

(3)2,B(n)

also. It is easy to see that the symmetric difference of the sets A and B inTheorem 3.1.1 is finite. A. Sarkozy asked whether there exist two infinitesets of nonnegative integers A and B with infinite symmetric difference, i.e.

|(A ∪B) \ (A ∩B)| =∞

andR

(i)2,A(n) = R

(i)2,B(n)

if n ≥ n0, for i = 1, 2, 3. For i = 1 the answer is negative (see in [Do]). Fori = 2 G. Dombi [Do] and for i = 3 Y. G. Chen and B. Wang [CheWa] provedthat the set of nonnegative integers can be partitioned into two subsets Aand B such that R

(i)2,A(n) = R

(i)2,B(n) for all n ≥ n0. In [Lev] Lev gave a

common proof to the above mentioned results of Dombi [Do] and Chen andWang [CheWa]. Using generating functions Cs. Sandor [Sa] determined thesets A ⊂ N for which either

R(2)2,A(n) = R

(2)2,N\A(n) for all n ≥ n0

orR

(3)2,A(n) = R

(3)2,N\A(n) for all n ≥ n0.

In [Ta08] M. Tang gave an elementary proof of Cs. Sandor’s results. Y. G.Chen and M. Tang studied related questions in [CheTa09].

Let C be a finite set of integers. Let FA(z), FB(z), T (z) denote polynomialsand A(z), B(z) denote power series having coefficients from the set C (i.e.

A(z) =∞∑n=0

anzn, where an ∈ C and z is a complex number, z = r · e2πiθ).

These series converge in the open unit disc. If C = 0, 1 then the generatingfunctions of the sets A, B, FA, FB and T ⊆ N are special kind of thesepolynomials and power series.

To make the proofs easier we will use the following notation: A(z) ∼ B(z)means that A(z) − B(z) = P (z), where P (z) is a polynomial. Using thesewe can rewrite Nathanson’s Theorem in equivalent form:

26

Theorem 3.1.2. (Nathanson, 1978-equivalent form)

Let A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn, an, bn ∈ 0, 1. Then A(z)2 ∼ B(z)2

if and only if there exist positive integers N0, m and polynomials

FA(z) =

mN0−1∑n=0

dnzn, FB(z) =

mN0−1∑n=0

enzn, dn, en ∈ 0, 1 and

T (z) =m−1∑n=0

tnzn, tn ∈ 0, 1 such that

A(z) = FA(z) +T (z)zmN0

1− zm,

B(z) = FB(z) +T (z)zmN0

1− zm

and1− zm

∣∣ T (z) (FA(z)− FB(z))

holds.

S. Z. Kiss, R. Rozgonyi and Cs. Sandor [KiRoSa14a] conjectured thatNathanson’s theorem can be generalized as follows.

Conjecture 3.1.1. (Z. Kiss, Rozgonyi, Sandor, 2014)Let h ≥ 2, A and B be infinite sets of nonnegative integers, A 6= B. ThenR

(1)h,A(n) = R

(1)h,B(n) from a certain point on if and only if there exist positive

integers N0, m and sets FA, FB and T such that FA∪FB ⊂ 0, 1, . . . ,mN0−1, T ⊂ 0, 1. . . . ,m− 1,

A = FA ∪ km+ t : k ≥ N0, t ∈ T,B = FB ∪ km+ t : k ≥ N0, t ∈ T,

and(1− zm)h−1

∣∣ T (z)h−1 (FA(z)− FB(z)) ,

where FA(z), FB(z) and T (z) denote the generating functions of the sets FA,FB and T .

Using generating functions in [KiRoSa14a] they proved the sufficiency partof Conjecture 3.1.1, and they also proved the Conjecture for the case h = 3.

Using power series, we can rewrite Conjecture 3.1.1 in equivalent form.

27

Conjecture 3.1.2. (Equivalent form of Conjecture 3.1.1)

Let A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn, an, bn ∈ 0, 1. Then A(z)h ∼ B(z)h

if and only if there exist positive integers N0, m and polynomials

FA(z) =

mN0−1∑n=0

dnzn, FB(z) =

mN0−1∑n=0

enzn, dn, en ∈ 0, 1 and T (z) =

m−1∑n=0

tnzn,

tn ∈ 0, 1 such that

A(z) = FA(z) +T (z)zmN0

1− zm,

B(z) = FB(z) +T (z)zmN0

1− zm

and(1− zm)h−1

∣∣ T (z)h−1 (FA(z)− FB(z))

holds.

We can prove Conjecture 3.1.2 in the case h = ps, where p is prime.

Theorem 3.1.3. (Rozgonyi, Sandor, 2014)Let h = ps and let the set C ⊆ Z, where C contains incongruent integers

modulo p. Let the series A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn, an, bn ∈ C. Then

A(z)h ∼ B(z)h if and only if there exist positive integers N0, m and polynomi-

als FA(z) =

mN0−1∑n=0

dnzn, FB(z) =

mN0−1∑n=0

enzn, dn, en ∈ C and T (z) =

m−1∑n=0

tnzn,

tn ∈ C such that

A(z) = FA(z) +T (z)zmN0

1− zm,

B(z) = FB(z) +T (z)zmN0

1− zm

and(1− zm)h−1

∣∣ T (z)h−1 (FA(z)− FB(z))

holds.

Corollary 3.1.4. The set C = 0, 1 implies that Conjecture 3.1.1 is truefor the case h = ps.

28

In order to prove Theorem 3.1.3 we verify the following three lemmas.

Lemma 3.1.5. Let C be a set of integers. Suppose that there exist positive

integers N0, m and polynomials FA(z) =

mN0−1∑n=0

dnzn, FB(z) =

mN0−1∑n=0

enzn,

dn, en ∈ C and T (z) =m−1∑n=0

tnzn, tn ∈ C such that

A(z) = FA(z) +T (z)zmN0

1− zm,

B(z) = FB(z) +T (z)zmN0

1− zm

and(1− zm)h−1

∣∣ T (z)h−1 (FA(z)− FB(z))

holds. Then A(z)h ∼ B(z)h.

The following example shows that for any C ⊆ Z, |C| ≥ 2 and h ≥ 2 thereexist different power series A(z), B(z) having their coefficients from C withthe property A(z)h ∼ B(z)h.

Proposition 3.1.6. Let C ⊆ Z, |C| ≥ 2. Then there exist series

A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn, an, bn ∈ C such that A(z) 6= B(z) and

A(z)h ∼ B(z)h.

In the proof of Theorem 3.1.3 we only use the fact that h is a power ofprime in the next Lemma.

Lemma 3.1.7. Let h = ps and let the set C ⊆ Z, where no element of C are

congruent modulo p. Let A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn, an, bn ∈ C. The

condition A(z)h ∼ B(z)h implies that A(z) ∼ B(z).

The condition of Lemma 3.1.7 that C contains incongruent integers modulop is important, because Imre Z. Ruzsa gave the identity(

−1 +2z4

1− z2

)2

−(

2z5

1− z2

)2

= 1− 4z4 − 4z6.

29

It means that there exist power series A(z) and B(z) having coefficients fromthe set C = −1, 0, 2 such that A(z)2 ∼ B(z)2, but A(z) 6∼ B(z), because

−1 +2z4

1− z2− 2z5

1− z2= −1 + 2

∞∑n=4

(−1)nzn.

We can generalize Imre Z. Ruzsa’s construction in the following way:

Proposition 3.1.8. Let h be a prime number. Then there exist a set Ch =c1, c2, . . . , ch+1, c1, . . . , ch+1 ∈ Z such that c1, . . . ch form a complete set ofresidues modulo h and power series A(z) =

∑∞n=0 anz

n, B(z) =∑∞

n=0 bnzn ,

an, bn ∈ Ch such that A(z)h ∼ B(z)h but A(z) 6∼ B(z).

Lemma 3.1.9. Let C be a finite set of integers, A(z) =∞∑n=0

anzn,

B(z) =∞∑n=0

bnzn, an, bn ∈ C. If A(z)h ∼ B(z)h and A(z) ∼ B(z) holds,

then there exist positive integers N0, m and polynomials FA(z) =

mN0−1∑n=0

dnzn,

FB(z) =

mN0−1∑n=0

enzn, dn, en ∈ C and T (z) =

m−1∑n=0

tnzn, tn ∈ C such that

A(z) = FA(z) +T (z)zmN0

1− zm,

B(z) = FB(z) +T (z)zmN0

1− zm

and(1− zm)h−1

∣∣ T (z)h−1 (FA(z)− FB(z)) . (3.1.1)

In 2011, Yang [Ya11] gave another proof of Theorem 3.1.1 without usinggenerating functions. In his paper he posed the following problem.

Problem. (Yang, 2011)If p ≥ 3 is a prime and A is an infinite set of nonnegative integers, thendoes there exist an infinite set of nonnegative integers B with A 6= B suchthat R

(1)p,A(n) = R

(1)p,B(n) for all sufficiently large n?

We show that the answer of Yang’s question is negative.

30

Theorem 3.1.10. (Kiss, Sandor, Rozgonyi, 2014)For every positive p ≥ 2 prime there exists an infinite set of nonnegativeintegers A such that for any infinite set of integers B, A 6= B, we haveR

(1)p,A(n) 6= R

(1)p,B(n) for infinitely many positive integer n.

We studied some similar problems and get the following results.

Theorem 3.1.11. (Kiss, Sandor, Rozgonyi, 2014)For every positive integer H ≥ 2 there exist infinite sets of nonnegative in-tegers A, B, A 6= B such that R

(l)h,A(n) = R

(l)h,B(n), for every l = 1, 2, 3 and

2 ≤ h ≤ H from a certain point on.

In the special case l = 1, Theorem 3.1.11 cannot be extended for infinitelymany h.

Theorem 3.1.12. (Kiss, Sandor, Rozgonyi, 2014)If for some infinite sets of nonnegative integers A and B the representationfunction R

(1)h,A(n) = R

(1)h,B(n), for n ≥ n0(h), for infinitely many positive

integer h ≥ 2, then A = B.

3.2 Proof of Theorem 3.1.3, Proposition 3.1.6

and 3.1.8

As we mentioned in Section 3.1 in order to show Theorem 3.1.3 we will proveLemma 3.1.5, Lemma 3.1.7 and 3.1.9.

Proof of Lemma 3.1.5 We have to show, that A(z)h ∼ B(z)h, that isA(z)h − B(z)h = P (z), where P (z) is a polynomial. Using the binomialtheorem and the assumptions of Lemma 3.1.5 we get that

A(z)h −B(z)h =

(FA(z) +

T (z)zmN0

1− zm

)h−(FB(z) +

T (z)zmN0

1− zm

)h=

=h∑i=1

(h

i

)(FA(z)i − FB(z)i

) T (z)h−i · z(h−i)mN0

(1− zm)h−i.

SinceFA(z)− FB(z)

∣∣ FA(z)i − FB(z)i,

so

T (z)h−i (FA(z)− FB(z))∣∣ (FA(z)i − FB(z)i

)T (z)h−i · z(h−i)mN0 .

31

Therefore it is enough to show that for every 1 ≤ i ≤ h− 1 we have

(1− zm)h−i∣∣ T (z)h−i (FA(z)− FB(z)) . (3.2.1)

From the assumptions we know that (1 − zm)h−1∣∣ T h−1(z) (FA(z)− FB(z))

holds.For a given integer d let denote denote by Φd(z) the dth cyclotomic poly-

nomial. It remains to prove that

Φd(z)h−i∣∣ T (z)h−i (FA(z)− FB(z)) . (3.2.2)

Let T (z) = Φd(z)k1u(z) and FA(z) − FB(z) = Φd(z)k2v(z), where u(z) andv(z) are polynomials with the property Φd(z) - u(z)v(z). By assumptions ofLemma 3.1.5 we know that (h− 1)k1 + k2 ≥ h− 1. Thus either k1 = 0, thenk2 ≥ h− 1, therefore

Φd(z)h−i∣∣ FA(z)− FB(z)

or k1 ≥ 1, therefore Φd(z) | T (z) so

Φd(z)h−i∣∣ T (z)h−i.

This completes the proof of Lemma 3.1.5.

Proof of Proposition 3.1.6: Let Bin(i) denote the parity of the numberof 1’s in the binary form of i. (i.e. Bin(i) = 1, if the number of the 1’s inthe binary form of i is even and Bin(i) = −1, if the number of the 1’s in thebinary form of i is odd.)

It is easy to see, thath−2∏i=0

(1− z2i) =2h−1−1∑i=0

Bin(i)zi.

Suppose, that c1, c2 ∈ C. Let A(z) =∞∑n=0

anzn, where for n ≤ 2h−1 − 1

an =

c1 if Bin(n) = 1

c2 if Bin(n) = −1,

and an = c1, for n ≥ 2h−1.

Let B(z) =∞∑n=0

bnzn, where for n ≤ 2h−1 − 1

bn =

c2 if Bin(n) = 1

c1 if Bin(n) = −1,

32

and bn = c1, for n ≥ 2h−1.

Then

A(z) = FA(z) +c1z

2h−1

1− z,

and

B(z) = FB(z) +c1z

2h−1

1− z.

We may apply Lemma 3.1.5 with FA(z)− FB(z) = (c1 − c2)h−2∏i=0

(1− z2i)

and m = 1 and T (z) = c1 because

(1− zm)h−1∣∣ T (z)h−1(FA(z)− FB(z))

is equivalent with

(1− z)h−1∣∣ ch−11 (c1 − c2)

h−2∏i=0

(1− z2i),

which is obviously true.

Proof of Lemma 3.1.7 It is clear that

A(z)h =

( ∞∑n=0

anzn

)h=∞∑n=0

gnzn,

where

gn =∑

(i1,i2,...,ih)i1+i2+···+ih=n

ai1ai2 . . . aih =

=h∑t=1

∑(j1,...,jt)ji∈N

0≤j1<j2<···<jt

∑(m1,...,mt),mi∈Z+

m1+···+mt=hm1j1+···+mtjt=n

h!

m1!m2! . . .mt!am1j1am2j2. . . amtjt .

(3.2.3)

Using (3.2.3), the formula of Legendre and Fermat’s Theorem we get thatmodulo p we have

h!

m1!m2! . . .mt!am1j1am2j2. . . amtjt =

ps!

m1!m2! . . .mt!am1j1am2j2. . . amtjt ≡

33

≡

0 for t ≥ 2

aps

i1= ap

s

nps≡ a n

psfor t = 1.

(3.2.4)

According to the above calculations we get that gn ≡ 0 (mod p) if ps - n andgn ≡ a n

ps(mod p), if ps | n.

Similarly for B(z)h we get that

B(z)h =

( ∞∑n=0

bnzn

)h=∑

hnzn,

where hn ≡ 0 (mod p) if ps - n and hn ≡ b nps

(mod p), if ps | n.

We know that gn ≡ hn (mod p) for n ≥ n0. For nps ≥ n0 the congruencegnps ≡ hnps (mod p) implies that an ≡ bn (mod p). Using the condition thatthe set C contains incongruent integers modulo p we get that an = bn forn ≥ n0/p

s. And this completes the proof of Lemma 3.1.7.

Proof of Lemma 3.1.9 Since A(z) ∼ B(z) and A(z)h ∼ B(z)h we can writethat A(z) = B(z) + P (z) and A(z)h = B(z)h + Q(z), where P (z) and Q(z)are polynomials. Thus

Q(z) = A(z)h −B(z)h = (B(z) + P (z))h −B(z)h =h∑i=1

(h

i

)P (z)iB(z)h−i

Multiply both side by P (z)h−2, we get that

Q(z)P (z)h−2 = (P (z)B(z))h−1(h+

(h2

)P (z)

B(z)+

(h3

)P (z)2

B(z)2+ . . .

). (3.2.5)

We show that P (z)B(z) is bounded in the open unit disc (z : |z| < 1, z ∈C). Assume that it is not true. Then there is an infinite sequence z1, z2, . . . , zn, . . . ,|zn| < 1 for which |P (zn)B(zn)| → ∞ (so |B(zn)| → ∞), and thus∣∣∣∣∣ (P (zn)B(zn))h−1

(h+

(h2

)P (zn)

B(zn)+

(h3

)P (zn)2

B(zn)2+ . . .

) ∣∣∣∣∣→∞,while the left-hand side is bounded. So P (z)B(z) is bounded in the openunit disc.

34

Next step we show, that P (z)B(z) should be a polynomial. Suppose

indirectly, that P (z)B(z) =∞∑n=0

knzn, kn ∈ Z and kn 6= 0. Let z = re2πiϕ.

For r < 1 then using the Parseval-formula gives∫ 1

0

∣∣P (re2πiϕ)B(re2πiϕ)∣∣2 dϕ =

∞∑n=0

k2nr2n (3.2.6)

Since P (z)B(z) is bounded, if r → 1− then the left-hand side of (3.2.6) isbounded, but the right-hand side of (3.2.6) tends to infinity. So P (z)B(z) =R(z), where R(z) is a polynomial.

Write P (z) in the form P (z) =m∑i=0

λizi (λ0 6= 0). Then bn fulfils the following

linear recursion when n ≥ n1

bn = − 1

λ0(λ1bn−1 + λ2bn−2 + · · ·+ λmbn−m) .

This means that bn is determined by bn−1, bn−2, . . . , bn−m. Since we can choosebn in finitely many way, we conclude that bn is periodic. Therefore there exist

positive integers N0, m and series FA(z) =

mN0−1∑n=0

dnzn,

FB(z) =

mN0−1∑n=0

enzn, dn, en ∈ C for every n = 0, 1, . . . ,mN0 − 1 and

T (z) =m−1∑n=0

tnzn, tn ∈ C for every n = 0, 1, . . . ,m− 1 such that

A(z) = FA(z) +T (z)zmN0

1− zm,

B(z) = FB(z) +T (z)zmN0

1− zm.

In the last step we have to show (3.1.1). We show by induction on k thatfor every 1 ≤ k ≤ h− 1

(1− zm)k∣∣ T (z)k (FA(z)− FB(z)) (3.2.7)

holds. It is clear that

A(z)h −B(z)h =

(FA(z) +

T (z)zmN0

1− zm

)h−(FB(z) +

T (z)zmN0

1− zm

)h= Q(z),

35

where Q(z) is a polynomial. Using the binomial theorem we get

(1−zm)h∣∣ (FA(z)(1− zm) + T (z)zmN0

)h−(FB(z)(1− zm) + T (z)zmN0)h

=

=h∑i=0

(h

i

)T i(z)zimN0

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i . (3.2.8)

If in 3.2.8 i = 0 then (1− zm)h∣∣ (h

0

)T (z)0z0·mN0

(FA(z)h − FB(z)h

)(1− zm)h

and also if in 3.2.8 i = h then (1−zm)h∣∣ (h

h

)T h(z)zhmN0 ·(FA(z)0 − FB(z)0) (1− zm)0

holds. So (3.2.8) is equivalent to

(1− zm)h∣∣∣ h−1∑i=1

(h

i

)T (z)izimN0

(FA(z)h−i − F h−i

B (z))

(1− zm)h−i . (3.2.9)

At first let k = 1. In (3.2.9) if i ≤ h− 2 then

(1− zm)2∣∣∣ (h

i

)zimN0 T (z)i

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i . (3.2.10)

So in (3.2.9) if i = h− 1 then

(1− zm)2∣∣∣ ( h

h− 1

)z(h−1)mN0 T (z)h−1 (FA(z)− FB(z)) (1− zm) (3.2.11)

should be true. Since (z,1− zm) = 1 from (3.2.11) we have

1− zm∣∣ T (z)h−1 (FA(z)− FB(z)) . (3.2.12)

We know, that 1− zm = −∏d|m

Φd(z), where Φd(z) denotes the dth cyclotomic

polynomial. 1− zm has no multiple root, which implies

1− zm∣∣ T (z) (FA(z)− FB(z)) . (3.2.13)

Now assume that for 1 ≤ l ≤ k − 1, k ≤ h− 1 we have

(1− zm)l∣∣ T (z)l (FA(z)− FB(z)) .

We have to prove that

(1− zm)k∣∣ T (z)k (FA(z)− FB(z)) .

36

In (3.2.9)

T (z)i (FA(z)− FB(z)) (1− zm)h−i∣∣∣(

h

i

)zimN0T (z)i

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i . (3.2.14)

Using the assumption of the induction we get

(1− zm)h−i+mink−1,i∣∣∣ T (z)i (FA(z)− FB(z)) (1− zm)h−i . (3.2.15)

From (3.2.14) and (3.2.15) we obtain that

(1− zm)h−i+mink−1,i∣∣∣ (h

i

)zimN0T (z)i

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i .

(3.2.16)So from (3.2.9) we know

(1− zm)k+1∣∣∣ (1− zm)h

∣∣∣h−1∑i=1

(h

i

)T (z)izimN0

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i .

(3.2.17)

If mink− 1, i = i, which means i ≤ k− 1 then h− i+ mink− 1, i = h ≥k + 1. Thus

(1− zm)k+1∣∣∣ (h

i

)T (z)izimN0

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i (3.2.18)

holds. If mink − 1, i = k − 1, then k − 1 ≤ i ≤ h− 1. If k − 1 ≤ i ≤ h− 2then h− i+ mink − 1, i ≥ 2 + k − 1 = k + 1. Thus

(1− zm)k+1∣∣∣ (h

i

)T (z)izimN0

(FA(z)h−i − FB(z)h−i

)(1− zm)h−i . (3.2.19)

So (3.2.9), (3.2.18) and (3.2.19) imply that for i = h− 1 we have

(1− zm)k+1∣∣∣ ( h

h− 1

)T (z)h−1z(h−1)mN0 (FA(z)− FB(z)) (1− zm) , (3.2.20)

which means that

(1− zm)k∣∣ T (z)h−1 (FA(z)− FB(z)) . (3.2.21)

37

We use, that 1− zm = −∏d|m

Φd(z), where Φd(z) denotes the dth cyclotomic

polynomial. This means that for every d, d | m, Φd(z)k | T (z)h−1 (FA(z)− FB(z)).If Φd(z) | T (z) then Φd(z)k | T (z)k is also true. If Φd(z) - T (z) thenΦd(z)k | (FA(z)− FB(z)). So for every d | m, Φd(z)k | T (z)k (FA(z)− FB(z))which means that

(1− zm)k∣∣ T (z)k (FA(z)− FB(z))

also true. And this ended the proof of the induction and the proof of Lemma3.1.9.

Proof of Proposition 3.1.8 Let

a′i =

∏i−1j=0(h

2 + 1− jh)

i!· (−1)ihh−1−i − (−1)i

(h

i

)hh−1,

for 1 ≤ i ≤ h− 1 and

A(z) =h−1∑i=1

a′i(1− zh)i−1 +hh−1zh

2

1− zh,

B(z) =hh−1zh

2+1

1− zh.

We will show that

1. A(z)h ∼ B(z)h

2. There exists a set of integers Ch = c1, c2, . . . , ch+1, such that c1, . . . chform a complete set of residues modulo h and for the previous power

series A(z) =∞∑n=0

anzn, B(z) =

∞∑n=0

bnzn we have an, bn ∈ Ch.

It is easy to see that A(z) 6∼ B(z).At first we prove Statement 1.

A(z)h−B(z)h =

(h−1∑i=1

a′i(1−zh)i−1+hh−1zh

2

1− zh

)h

−

(hh−1zh

2+1

1− zh

)h

. (3.2.22)

From (3.2.22) we see that to prove A(z)h ∼ B(z)h we need to show

(1− zh)h∣∣∣ ( h−1∑

i=1

a′i(1− zh)i + hh−1zh2

)h−(hh−1zh

2+1)h. (3.2.23)

38

Let x = 1− zh. (3.2.23) is equivalent to the following

xh∣∣∣ ( h−1∑

i=1

a′ixi + hh−1(1− x)h

)h− hh2−h(1− x)h

2+1. (3.2.24)

So it is enough to show (3.2.24). For −1 < x < 1 by the binomial seriesexpansion we get

hh2−h(1− x)h

2+1 =(hh−1(1− x)

h2+1h

)h=

=

(hh−1

∞∑i=0

(h2+1h

i

)(−1)ixi

)h=

=

(hh−1

h−1∑i=0

(h2+1h

i

)(−1)ixi + hh−1

∞∑i=h

(h2+1h

i

)(−1)ixi

)h=

=

( h−1∑i=0

hh−1−i∏i−1

j=0(h2 + 1− jh)

i!(−1)ixi + hh−1

∞∑i=h

(h2+1h

i

)(−1)ixi

)h=

=

( h−1∑i=0

hh−1−i∏i−1

j=0(h2 + 1− jh)

i!(−1)ixi

)h+

+h∑k=1

(h

k

)( h−1∑i=0

hh−1−i∏i−1

j=0(h2 + 1− jh)

i!(−1)ixi

)h−k(hh−1

∞∑i=h

(h2+1h

i

)(−1)ixi

)k

=

=

( h−1∑i=0

hh−1−i∏i−1

j=0(h2 + 1− jh)

i!(−1)ixi

)h+∞∑i=h

dixi.

Thus

hh2−h(1−x)h

2+1−

(h−1∑i=0

hh−1−i∏i−1

j=0(h2 + 1− jh)

i!(−1)ixi

)h

=∞∑i=h

dixi =

h2+1∑i=h

dixi.

(3.2.25)Hence

xh∣∣∣ ( h−1∑

i=0

hh−1−i∏i−1

j=0(h2 + 1− jh)

i!(−1)ixi

)h

− hh2−h(1− x)h2+1. (3.2.26)

39

Since for every 1 ≤ i ≤ h− 1

(−1)ihh−1−i∏i−1

j=0(h2 + 1− jh)

i!= a′i + (−1)ihh−1

(h

i

),

from (3.2.26) we get that

xh∣∣∣ ( h−1∑

i=1

a′ixi+hh−1 +hh−1

h−1∑i=1

(−1)i(h

i

)xi

)h

−hh2−h(1−x)h2+1. (3.2.27)

Thus

xh∣∣∣ ( h−1∑

i=1

a′ixi + hh−1

h∑i=0

(−1)i(h

i

)xi

)h

− hh2−h(1− x)h2+1 (3.2.28)

is also true. Using the binomial theorem we get thath∑i=0

(−1)i(h

i

)xi = (1− x)h,

which proves Statement 1.

Now we prove Statement 2. We will show that

∏i−1j=0(h

2 + 1− jh)

i!∈ Z.

When p is a prime number, and pα < h, then (pα, h) = 1. Therefore among

the numbers h2 + 1, h2 + 1−h, . . . , h2 + 1−(⌊

ipα

⌋pα − 1

)h there are

⌊i

pα

⌋,

which are divisible by pα. So

p∑∞α=1b i

pα c∣∣∣ i−1∏j=1

(h2 + 1− jh).

By the Legendre’s formula, in the prime factorization of i! (i < h) the expo-

nent of p is∞∑α=1

⌊i

pα

⌋.

Let

A(z) =h−1∑i=1

a′i(1− zh)i−1 +hh−1zh

2

1− zh=∞∑n=0

anzn.

Then an = 0 or an = hh−1 holds for n 6= ih, 0 ≤ i ≤ h− 2.

Clearly

B(z) =hh−1zh

2+1

1− zh=∞∑n=0

bnzn.

40

Then bn = 0 or bn = hh−1.It remains to show that the integers a0, ah, a2h, . . . , a(h−2)h form primitive

residue class modulo h. If 0 ≤ k ≤ h− 2 we get that

akh =h−1∑j=k+1

(−1)ka′j

(j − 1

k

).

For 1 ≤ j ≤ h− 2 holds then h|a′j thus using Wilson’s Theorem we get

akh ≡ (−1)ka′h−1

(h− 2

k

)≡ (−1)k

∏h−2j=0 (h2 + 1− jh)

(h− 1)!· (−1)h−1

(h− 2)!

k!(h− 2− k)!≡

≡ (−1)k · 1

−1· (−1)h−1

(h− 1)!

(−1)(k + 1)!(h− 2− k)!· (k + 1)(−1)h−1 ≡

≡ (−1)k+h−1(h− 1)!(−1)k+1

(h− 1)(h− 2) . . . (h− k − 1)(h− k − 2)!· (k + 1) ≡

≡ −(−1)h−1(h− 1)!

(h− 1)!(k + 1) ≡ −(k + 1) (mod h).

This show, that Statement 2. holds. This completes the proof of Proposition3.1.8.

3.3 Proof of Theorem 3.1.10

LetA be a sparse, set which means that α(N) < N1p (here α(N) = |[0, N ] ∩ A|).

Let A = a1, a2, . . . . We prove by contradiction. Assume that A, Bare different sets and R

(1)p,A(n) = R

(1)p,B(n) from a certain point on. Since

α(ak) = k < a1/pk it follows that ak > kp. The generating function of A is

A(r) =∑a∈A

ra =∞∑n=0

tA,nrn =

∞∑n=1

(α(n)− α(n− 1)) rn =

= −1 +∞∑n=1

α(n)(rn − rn+1

)= −1 + (1− r)

∞∑n=0

α(n)rn =

= O(

(1− r) · (1− r)−1p−1)

= O(

(1− r)−1p

),

(3.3.1)

as r → 1−. We remind that tA,n is the characteristic function of the set A(i.e. tA,n = 1, if n ∈ A and tA,n = 0, if n 6∈ A ).

Since R(1)p,A(n) = R

(1)p,B(n) it is clear that there is a polynomial P (r) such that

A(r)p −B(r)p = P (r).

41

It is easy to see that there exists a positive integer N0 such that A ∩[N0,+∞) = B ∩ [N0,+∞), because R

(1)p,A(n) ≡ 0 (mod p) if n

p/∈ A, and

R(1)p,A(n) ≡ 1 (mod p) if n

p∈ A. Similarly R

(1)p,B(n) ≡ 0 (mod p) if n

p/∈ B, and

R(1)p,B(n) ≡ 1 (mod p) if n

p∈ B. Thus A(r) differs from B(r) in a polynomial

which means thatB(r) = O

((1− r)−

1p

), (3.3.2)

as r → 1−, as well. So

(A(r)−B(r))(Ap−1(r) + · · ·+Bp−1(r)

)= P (r). (3.3.3)

Therefore there exist relatively prime polynomials R(r) and S(r) such that

R(r)(Ap−1(r) + · · ·+Bp−1(r)

)= S(r). (3.3.4)

As r → 1− in (3.3.3) we get that S(r) and R(r) are bounded, and(Ap−1(r) + · · ·+Bp−1(r)

)→∞.

Therefore r = 1 must be the root of R(r). Thus

R(r) = (1− r)Q(r).

Now we can write (3.3.4) in the form

(1− r)Q(r)(Ap−1(r) + · · ·+Bp−1(r)

)= S(r), (3.3.5)

Since Q(r) is a polynomial it is bounded. It follows from (3.3.1) and (3.3.2)that

Ap−1(r) + · · ·+Bp−1(r) = O(

(1− r)−p−1p

).

So the order of the left hand side of (3.3.5) is O(

(1− r)1p

), as r → 1−.

This means S(r) tends to zero as r → 1−. So S(r) = (1 − r)T (r), and thiscontradicts to (R(r), S(r)) = 1.

3.4 Proof of Theorem 3.1.11

The construction of the sets A and B are the following. Let n be a positiveinteger. Take the binary representation of n

n =

blog2(n)c∑i=0

βi2i,

42

where βi = 0 or 1. Denote again by Bin(n) =∑blog2(n)c

i=0 βi the number of theones in the binary representation of n. Let

FA := kH! | 0 ≤ k < 2H ,Bin(kH!) ≡ 0 (mod 2),

andFB := kH! | 0 ≤ k < 2H ,Bin(kH!) ≡ 1 (mod 2).

We will show that the sets

A = FA ∪ H!2H , H!2H + 1, . . .

andB = FB ∪ H!2H , H!2H + 1, . . .

are suitable. Let h be a fixed integer, 2 ≤ h ≤ H. Then we have

FA(z)− FB(z) =H−1∏i=0

(1− zH!2i

),

and therefore

(1− zh!) . . . (1− z2h−1h!)∣∣ FA(z)− FB(z). (3.4.1)

Hence(1− z) . . . (1− zh−1)(1− zh)

∣∣ FA(z)− FB(z).

The generating function of R(l)h,A(n), l = 1, 2, 3 can be written using a sieve

formula with suitable real numbers Ck1,...,kh :

∞∑n=0

R(l)h,A(n)zn =

∑(k1,...,kh)

k1+2k2+···+hkh=hki≥0,i=1,...,h

Ck1,...,kh

h∏i=1

A(zi)ki . (3.4.2)

We would like to prove that there is a polynomial P (z) such that

∞∑n=0

R(l)h,A(n)zn −

∞∑n=0

R(l)h,B(n)zn = P (z). (3.4.3)

From (3.4.2) we get that the left hand side of (3.4.3) is equivalent to

∑(k1,...,kh)

k1+2k2+···+hkh=hki≥0,i=1,...,h

Ck1,...,kh

( h∏i=1

A(zi)ki −h∏i=1

B(zi)ki). (3.4.4)

43

In view of

A(z) = FA(z) +zH!2H

1− zand

B(z) = FB(z) +zH!2H

1− zwe get that (3.4.4) is equivalent to

∑(k1,...,kh)

k1+2k2+···+hkh=hki≥0,i=1,...,h

Ck1,...,kh

(h∏i=1

(FA(zi)+

ziH!2H

1− zi

)ki−

h∏i=1

(FB(zi)+

ziH!2H

1− zi

)ki).

(3.4.5)It is enough to show that the difference of the products in (3.4.5) is a

polynomial for every h-tuple (k1, . . . , kh). Let the h-tuple (k1, . . . , kh) befixed. Using the binomial theorem we get for suitable constants Dj1,...,jh thisexpression is equal to(

h∏i=1

ki∑ji=0

(kiji

)(FA(zi)

)ji (ziH!2H

1− zi

)ki−ji)−

−

(h∏i=1

ki∑ji=0

(kiji

)(FB(zi)

)ji (ziH!2H

1− zi

)ki−ji)=

=∑

(j1,...,jh)0≤ji≤ki,i=1,...,h

Dj1,...,jh

(h∏i=1

(ziH!2H

1− zi

)ki−ji)( h∏i=1

(FA(zi)

)ji− h∏i=1

(FB(zi)

)ji ).(3.4.6)

We will show that(h∏i=1

(ziH!2H

1− zi

)ki−ji)( h∏i=1

(FA(zi)

)ji − h∏i=1

(FB(zi)

)ji )

is a polynomial. To show this we will prove that there is a polynomial Q(z)such that

h∏i=1

(ziH!2H

1− zi

)ki−ji

=Q(z)

(1− z) . . . (1− zh−1)(1− zh), (3.4.7)

44

and

(1− z) . . . (1− zh−1)(1− zh)∣∣∣ h∏i=1

(FA(zi)

)ji − h∏i=1

(FB(zi)

)ji . (3.4.8)

To deduce (3.4.7) it is enough to show that

h∏i=1

(1− zi)ki−ji∣∣ (1− z) . . . (1− zh−1)(1− zh).

A root of the producth∏i=1

(1− zi)ki−ji is a primitive ith roots of unity, for

some i ≤ h. Let εi denote a primitive ith root of unity. The multiplicity ofεi in the polynomial (1− z) . . . (1− zh−1)(1− zh) is bh

ic. The multiplicity of

εi in the polynomialh∏i=1

(1− zi)ki−ji is

(ki − ji) + (k2i − j2i) + · · · ≤ ki + k2i + . . . .

We know that k1 + 2k2 + · · ·+ hkh = h. Therefore

iki + ik2i + · · · ≤ iki + 2ik2i + · · · ≤ 1k1 + 2k2 + . . . hkh = h.

This means that

ki + k2i + · · · ≤⌊h

i

⌋,

which proves equation (3.4.7).It remains to prove the following lemma, which verifies (3.4.8).

Lemma 3.4.1. If (1− z) . . . (1− zh−1)(1− zh)∣∣ FA(z)− FB(z) then for all

t-tuples (l1, . . . , lt)

(1− z) . . . (1− zh−1)(1− zh)∣∣∣ t∏i=1

(FA(zi)

)li − t∏i=1

(FB(zi)

)li .Proof. We prove this result by induction on t. If t = 1 then we show that

(1− z) . . . (1− zh−1)(1− zh)∣∣∣ (FA(z))l1 − (FB(z))l1 .

Since

(FA(z))l1 − (FB(z))l1 = (FA(z)− FB(z))(

(FA(z))l1−1 + · · ·+ (FB(z))l1−1),

45

we get that the case t = 1 holds.Now assume that the lemma holds for all t or less. For t + 1 we need to

show that

(1− z) . . . (1− zh−1)(1− zh)∣∣∣ t+1∏i=1

(FA(zi)

)li − t+1∏i=1

(FB(zi)

)li . (3.4.9)

The right hand side of (3.4.9) is equal to

(FA(z))l1 . . .(FA(zt+1)lt+1

)− (FA(z))l1 . . .

(FA(zt)

)lt (FB(zt+1)

)lt+1 +

+ (FA(z))l1 . . .(FA(zt)

)lt (FB(zt+1)

)lt+1 − (FB(z))l1 . . .(FB(zt+1)

)lt+1 =

= (FA(z))l1 . . .(FA(zt)

)lt ((FA(zt+1)

)lt+1 −(FB(zt+1)

)lt+1)−

−(FB(zt+1)

)lt+1(

(FA(z))l1 . . .(FA(zt)lt

)− (FB(z))l1 . . .

(FB(zt)

)lt).

Because of our assumption the second term is divisible by(1− z) . . . (1− zh−1)(1− zh). Since

(1− z) . . . (1− zh−1)(1− zh)∣∣ (1− zt+1) . . . (1− zh(t+1))

and

(1− zt+1) . . . (1− zh(t+1))∣∣∣ ((FA(zt+1)

)lt+1 −(FB(zt+1)

)lt+1)

this completes the induction.

3.5 Proof of Theorem 3.1.12

We prove the theorem by contradiction. Assume that for infinite sets ofnonnegative integers A, B, A 6= B, there is an infinite sequence of integers2 ≤ h1 < h2 < . . . hi < . . . and polynomials Pi(r) such that

A(r)hi −B(r)hi =∞∑n=0

(R

(1)hi,A(n)−R(1)

hi,B(n))rn = Pi(r).

(We note that since A 6= B, so Pi(r) 6≡ 0.)Then

Pi(r) = A(r)hi −B(r)hi =

= (A(r)−B(r))(Ahi−1(r) + Ahi−2(r)B(r) + · · ·+Bhi−1(r)

).

46

As r → 1−,

Pi+1(r)

Pi(r)=

A(r)hi−1 + A(r)hi−2B(r) + · · ·+B(r)hi−1

A(r)hi+1−1 + A(r)hi+1−2B(r) + · · ·+B(r)hi+1−1≤

≤hi ·max

A(r)hi−1, B(r)hi−1

max A(r)hi+1−1, B(r)hi+1−1

→ 0.

Let Pi(r) = (1− r)miQi(r), where mi is a nonnegatvie integer, Qi(r) is apolynomial and Qi(1) 6= 0. Thus

Pi+1(r)

Pi(r)=

(1− r)mi+1Qi+1(r)

(1− r)miQi(r),

and mi+1 < mi. We get that m1 > m2 > . . . , which is absurd.

47

Chapter 4

Additive complement sets

This chapter is based on the paper [KiRoSa14b].

4.1 Introduction

Let A ⊆ N and B ⊆ N be finite or infinite sets. Let RA+B(n) such kind ofrepresentation function which denotes the number of solutions of the equation

a+ b = n, a ∈ A, b ∈ B.We put

A(n) =∑a≤na∈A

1 and B(n) =∑b≤nb∈B

1

respectively. We say a set B ⊆ N is an additive complement of the set A ⊆ Nif every sufficiently large n ∈ N can be represented in the form a + b = n,a ∈ A, b ∈ B, i.e., RA+B(n) ≥ 1 for n ≥ n0. Additive complement is animportant concept in additive number theory, in the past few decades it wasstudied by many authors [Dan], [Nar], [Ru99], [SaSze]. In [SaSze] Sarkozyand Szemeredi proved a conjecture of Danzer [Dan], namely they proved thatfor infinite additive complements A and B if

lim supx→+∞

A(x)B(x)

x≤ 1,

thenlim infx→+∞

(A(x)B(x)− x) = +∞.

In [CheFa10] Chen and Fang improved this result and they proved that forany two infinite additive complements A and B, if

lim supx→+∞

A(x)B(x)

x> 2, or lim sup

x→+∞

A(x)B(x)

x<

5

4,

48

thenlim

x→+∞(A(x)B(x)− x) = +∞.

In the other direction they proved in [CheFa11] that for any integer a ≥ 2,there exist two infinite additive complements A and B such that

lim supx→+∞

A(x)B(x)

x=

2a+ 2

a+ 2,

but there exist infinitely many positive integers x such that A(x)B(x)−x = 1.In [CheFa13] they studied the case when A is a finite set. In this case thesituation is different from the infinite case. Chen and Fang proved that forany two additive complements A and B with |A| < +∞ or |B| < +∞, if

lim supx→+∞

A(x)B(x)

x> 1,

thenlim

x→+∞(A(x)B(x)− x) = +∞.

They also proved that if

A = a+ ims + kims+1 : i = 0, ...,m− 1,

where |A| = m, a, s ≥ 0 and ki are integers, then there exists an additivecomplement B of A such that A(x)B(x) − x = O(1). In the special case|A| = 3 they proved that ifA is not of the form a+ i3s + ki3

s+1 : i = 0, 1, 2,where a, s ≥ 0 and ki are integers, then for any additive complement B ofA,

limx→+∞

(A(x)B(x)− x) = +∞

holds.Chen and Fang posed the following conjecture in [CheFa13].

Conjecture 4.1.1. (Chen, Fang, 2013)If the set of nonnegative integers A is not of the form

A = a+ ims + kims+1 : i = 0, ...,m− 1,

where a,m > 0, s ≥ 0 and ki are integers, then, for any additive complementB of A, we have

limx→+∞

(A(x)B(x)− x) = +∞.

In this chapter we prove this conjecture, when the number of elements ofthe set A is prime:

49

Theorem 4.1.1. (Kiss, Rozgonyi, Sandor, 2014)Let p be a positive prime and A be a set of nonnegative integers with |A| = p.If A is not of the form

A = a+ ips + kips+1 : i = 0, ..., p− 1, (4.1.1)

where a > 0, s ≥ 0 and ki are integers, then, for any additive complement Bof A, we have

limx→+∞

(A(x)B(x)− x) = +∞. (4.1.2)

When the number of elements of A is a composite number, we disproveConjecture 4.1.1:

Theorem 4.1.2. (Kiss, Rozgonyi, Sandor, 2014)For any composite number n > 0, there exists a set A and a set B such that|A| = n, B is an additive complement of A and A is not of the form

A(x)B(x)− x = O(1).

A = a+ ins + kins+1 : i = 0, ..., n− 1, (4.1.3)

where s ≥ 0, a > 0, and ki are integers, and

4.2 Proof of Theorem 4.1.1

We will prove that if there exists an additive complement B of A, |A| = psuch that

lim infx→+∞

(A(x)B(x)− x) < +∞,

then A is the form (4.1.1). First we prove that there exists an integer n1

such that RA+B(n) = 1 for n ≥ n1. We argue as Sarkozy and Szemeredi in[SaSze]. As B is an additive complement of A, it follows that

+∞ > C = lim infx→+∞

(A(x)B(x)− x) = lim infx→+∞

((∑a∈Aa≤x

1)(∑

b∈Bb≤x

1)− x)≥

≥ lim infx→+∞

(( ∑a∈A,b∈Ba+b≤x

1)− x)

= lim infx→+∞

( x∑n=0

RA+B(n)− x)≥

≥ lim infx→+∞

( x∑n=n0+1

RA+B(n)− x)≥ lim inf

x→+∞

([x]− n0 +

∑n0<n≤x

RA+B(n)>1

1− x)≥

50

≥ lim infx→+∞

( ∑n0<n≤x

RA+B(n)>1

1)− (n0 + 1),

thus we havelim infx→+∞

( ∑n0<n≤x

RA+B(n)>1

1)< C + n0 + 1,

where C is a positive constant. As B is an additive complement of A, itfollows that there exists an integer n1 such that

RA+B(n) = 1 for n ≥ n1. (4.2.1)

In the next step we prove that A is of the form (4.1.1). Let z = re2iπα =re(α), where r < 1. Let the generating functions of the sets A and B befA(z) =

∑a∈A z

a and fB(z) =∑

b∈B zb respectively. (By r < 1 these infinite

series and all the other infinite series of the proof are absolutely convergent.)In view of (4.2.1) we have

fA(z)fB(z) =(∑a∈A

za)(∑

b∈B

zb)

=+∞∑n=0

RA+B(n)zn =

=

n1−1∑n=0

RA+B(n)zn ++∞∑n=n1

RA+B(n)zn = P1(z) +zn1+1

1− z,

where P1(z) is a polynomial of z. Thus we have

(1− z)fA(z)fB(z) = (1− z)P1(z) + zn1+1. (4.2.2)

Next we prove that fB(z) can be written in the form

fB(z) = FB(z) +T (z)

1− zM, (4.2.3)

where M is a positive integer, FB(z) and T (z) are polynomials with integralcoefficients. We argue as Nathanson in [Nat78].

Let (1 − z)fA(z) =∑N

n=K anzn, where aN 6= 0 and aK 6= 0, and let

fB(z) =∑∞

n=0 enzn, where en ∈ 0, 1. Then we have

(1− z)fA(z)fB(z) =∞∑n=0

cnzn,

where cn = 0 from a certain point on. It is clear that if n is large enough,then cn = en−KaK + en−K−1aK+1 + . . . + en−NaN = 0. This shows that the

51

coefficients of the power series fB(z) satisfies a linear recurrence relation froma certain point on. These coefficients are either 0 or 1 from a certain pointon. It is easy to see that a sequence defined by a linear recurrence relationon a finite set must be eventually periodic, which proves (4.2.3). It followsfrom (4.2.2) and (4.2.3) that

fA(z)

(FB(z) +

T (z)

1− zM

)= P1(z) +

zn1+1

1− z,

hence

(1−zM)fA(z)FB(z)+fA(z)T (z) = (1−zM)P1(z)+(1+z+z2+. . .+zM−1)zn1+1.(4.2.4)

By putting z = 1, we obtain that

fA(1)T (1) = M. (4.2.5)

As fA(1) = |A| = p, it follows from (4.2.5) that p | M . Define k by pk | Mbut pk+1 -M . It follows from (4.2.4) that

(1 + z + z2 + . . . + zM−1)∣∣ fA(z)T (z).

It follows from (1.2.1) that for any 1 ≤ t ≤ k we have

Φpt(z)∣∣ fA(z)T (z).

Assume that for any 1 ≤ t ≤ k we have Φpt(z)∣∣ T (z). Then

T (z) =k∏t=1

Φpt(z)q(z),

where q(z) is a polynomial with integral coefficients. By putting z = 1we obtain that T (1) = pkq(1), hence M = fA(1)T (1) = pk+1q(1) whichcontradicts the definition of k. It follows that there exists an integer 0 ≤ s ≤k − 1 such that Φps+1(z)

∣∣ fA(z), thus fA(z) = Φps+1(z) a(z), where a(z) is apolynomial. As A = a1, . . . , ap, we have fA(z) =

∑pi=1 z

ai . Let ω be theps+1th root of unity i.e.,

ω = exp

(2π

ps+1i

).

It follows that fA(ω) = 0, thus we have∑p

i=1 ωai = 0. Let ai = lip

s+1 + ri,where 0 ≤ ri < ps+1. Without loss of generality we may assume that

0 ≤ r1 ≤ r2 ≤ . . . ≤ rp < ps+1. (4.2.6)

52

Define rp+1 = ps+1 + r1. Since

p∑i=1

(ri+1 − ri) = rp+1 − r1 = ps+1, it follows

that there exists a j with 1 ≤ j ≤ p such that

rj+1 − rj ≥ ps. (4.2.7)

In the next step we prove that this implies

ai − rj+1 = nips+1 + ti, (4.2.8)

where 1 ≤ i ≤ p and 0 ≤ ti ≤ ps+1 − ps holds. Assume that i ≤ j. By thedefinition of ai we have

ai − rj+1 = lips+1 + ri − rj+1.

It follows from (4.2.6) and (4.2.7) that

rj+1 − ri ≤ rj+1 ≤ ps+1 and − ps+1 ≤ ri − rj+1 ≤ rj − rj+1 ≤ −ps.

Thus we have0 ≤ ps+1 + ri − rj+1 ≤ ps+1 − ps,

which implies (4.2.8). In the second case assume that i ≥ j + 2. It is clearfrom (4.2.6) that ri− rj+1 ≥ 0. By the definition of ai and (4.2.6), (4.2.7) wehave

ai − rj+1 = lips+1 + ri − rj+1 ≤ lip

s+1 + ps+1 − ps,

which implies (4.2.8).It follows that there exists an integer a such that ai = a + nip

s+1 + ti,where ni is an integer and

0 ≤ ti ≤ ps+1 − ps. (4.2.9)

As fA(ω) = 0, and the definition of ω we obtain that

p∑i=1

ωai =

p∑i=1

ωa+nips+1+ti =

p∑i=1

ωa+ti = 0.

Let h(z) =∑p

i=1 zti . Thus we obtain that h(ω) = 0. As Φps+1(z) is a minimal

polynomial of ω we have Φps+1(z)∣∣ h(z). It follows from (4.2.9) that

deg

(p∑i=1

zti

)≤ ps+1 − ps = ϕ(ps+1) = deg (Φps+1(z)) .

53

Therefore, by using (1.2.2) we have

p∑i=1

zti = Φps+1(z) = 1 + zps

+ z2ps

+ . . . + z(p−1)ps

and then we have t1, . . . , tp = 0, ps, 2ps, . . . , (p − 1)ps. It follows thatthere exist integers a > 0 and ki, such that A = a + ips + kip

s+1, asdesired.

4.3 Proof of Theorem 4.1.2

Let n = d1d2, d1, d2 > 1 be arbitrary integers, and consider the followingtwo sets:

A = u+ v · d1d2 : 0 ≤ u ≤ d1 − 1, 0 ≤ v ≤ d2 − 1,

B = kd1d22 + w · d1 : k ∈ N, 0 ≤ w ≤ d2 − 1.

It is easy to see that |A| = d1d2. It is clear that A(x) = d1d2 if x is large

enough and B(x) ≤ x

d1d2+O(1), which implies A(x)B(x) − x = O(1). Let

m be a fixed positive integer. It is clear that any positive integer m can bewritten in the form

m = kd1d22 + ud1 + ld1d2 + v,

where k is a nonnegative integer, 0 ≤ u, l ≤ d2 − 1, 0 ≤ v ≤ d1 − 1. HenceB is an additive complement of A. In the next step we prove that the setA is not of the form (4.1.3). Assume that A is of the form (4.1.3). It isclear that the difference of any two elements from A divisible by ns. As Aalso contains consecutive integers we have ns|1, which implies s = 0. ThusA = a + i + kin : i = 0, . . . , n− 1, that is A is a complete residue systemmodulo n, which is a contradiction.

54

Chapter 5

Representation functions ongroups

We can study the representation functions on groups also and usually wehave some similar result like in the case of natural numbers. The chapter ismostly based on the paper [KiRoSa14c].

5.1 Introduction

Let X be a semigroup, written additively. Let A1, . . . ,Ah be nonemptysubsets of X and let x be an element of X. We denote by |A| the cardinalityof the set A. We define the ordered representation function

RA1+···+Ah(x) = |(a1, . . . , ah) ∈ A1 × · · · × Ah : a1 + · · ·+ ah = x|.

If Ai = A for i = 1, . . . , h, then we write

R(1)h,A(x) = |(a1, . . . , ah) : ai ∈ A, a1 + · · ·+ ah = x|.

Let X be an abelian semigroup, written additively. For A ⊂ X, let Ahdenote the set of all h-tuples of A. Two h-tuples (a1, . . . , ah) ∈ Ah and(a′1, . . . , a

′h) ∈ Ah are equivalent if there is a permutation α : 1, . . . , h →

1, . . . , h such that aα(i) = a′i for i = 1, . . . , h. Two other representationfunctions arise often and naturally in additive number theory as we men-tioned earlier. The unordered representation function R

(2)h,A(x) counts the

number of equivalence classes of h-tuples (a1, . . . , ah) such that a1+· · ·+ah =

x. The unordered restricted representation function R(3)h,A(x) counts the num-

ber of equivalence classes of h-tuples (a1, . . . , ah) of pairwise distinct elementsof A such that a1 + · · ·+ ah = x. It is easy to see that the definitions of the

55

unordered and the unordered restricted representation functions make senseonly in Abelian groups.

Alternative definitions for R(2)2,A(x) and R

(3)2,A(x) are the following. Denote

byDA(x) = #a : a ∈ A, a+ a = x

then

R(2)2,A(x) =

1

2R

(1)2,A(x) +

1

2DA(x) (5.1.1)

and

R(3)2,A(x) =

1

2R

(1)2,A(x)− 1

2DA(x). (5.1.2)

Let X = N . Sarkozy asked if there exist two sets A and B with |A∆B| =∞ such that R

(i)2,A(n) = R

(i)2,B(n), for i = 1, 2, 3 and for all sufficiently large n.

For i = 2 Dombi [Do] proved that the answer is positive and for i = 1 theanswer is negative. For i = 3 Chen and Wang [CheWa] proved that the setof natural numbers can be partitioned into two subsets A and B such thatR

(3)2,A(n) = R

(3)2,B(n) for every n large enough. Lev [Lev] and independently

Sandor [Sa] characterized all subsets A ⊂ N such that R(2)2,A(n) = R

(2)2,N\A(n)

or R(3)2,A(n) = R

(3)2,N\A(n) for big enough n. The precise theorems are the

following.

Theorem 5.1.1. (Lev, Sandor, 2004)

Let X = N. Let N be a positive integer. The equality R(2)2,A(n) = R

(2)2,N\A(n)

holds for n ≥ 2N − 1 if and only if |A ∩ [0, 2N − 1]| = N and 2m ∈ A ⇔m ∈ A, 2m+ 1 ∈ A ⇔ m 6∈ A for m ≥ N .

Theorem 5.1.2. (Lev, Sandor, 2004)

Let X = N. Let N be a positive integer. The equality R(3)2,A(n) = R

(3)2,N\A(n)

holds for n ≥ 2N − 1 if and only if |A ∩ [0, 2N − 1]| = N and 2m ∈ A ⇔m 6∈ A, 2m+ 1 ∈ A ⇔ m ∈ A for m ≥ N .

Tang [Ta08] gave an elementary proofs of Lev and Sandor’s results. In[CheYa12a], [CheYa12b], [CheYa13], [Ya14] Chen and Yang studied relatedproblems about weighted representation functions. Similar statement to theabove theorems can not be formulated for the representation function R

(1)2,A(n)

because R(1)2,A(n) is odd if and only if n

2∈ A, therefore either R

(1)2,A(2m) or

R(1)2,N\A(2m) is odd. A nontrivial result is the following in this direction.

Theorem 5.1.3. (Kiss, Rozgonyi, Sandor, 2014)Let X = N. The equality RA+B(n) = RN\A+N\B(n) holds from a certain pointon if and only if |N \ (A ∪ B)| = |A ∩ B| <∞.

56

The modular questions were solved by Chen and Yang [CheYa12a].

Theorem 5.1.4. (Chen, Yang, 2012)

Let X = Zm. The equality R(1)2,A(n) = R

(1)2,Zm\A(n) holds for all n ∈ Zm if and

only if m is even and |A| = m/2.

Theorem 5.1.5. (Chen, Yang, 2012)

Let X = Zm. For i ∈ 2, 3, the equality R(i)2,A(n) = R

(i)2,Zm\A(n) holds for

all n ∈ Zm if and only if m is even and t ∈ A ⇔ t + m/2 6∈ A for t =0, 1, . . . ,m/2− 1.

We extend this theorem to arbitrary finite groupG and the second theoremto finite Abelian group. (From now we denote the groups, subgroups andsubsets with capital letters.)

Theorem 5.1.6. (Kiss, Rozgonyi, Sandor, 2014)Let X = G be a finite group. Then

(i) If there exists a g ∈ G for which the equality RA+B(g) = RG\A+G\B(g)holds, then |A|+ |B| = |G|.

(ii) If |A|+ |B| = |G|, then the equality RA+B(g) = RG\A+G\B(g) holds forall g ∈ G.

We generalize theorem 5.1.5 in the following way.

Theorem 5.1.7. (Kiss, Rozgonyi, Sandor, 2014)Let X = G be a finite group and h ≥ 2 a fixed integer.

(i) If the equality R(1)h,A(g) = R

(1)h,G\A(g) holds for all g ∈ G, then |G| is even

and |A| = |G|/2.

(ii) If h is even and |A| = |G|/2 then R(1)h,A(g) = R

(1)h,G\A(g) holds for all

g ∈ G.

We note that the case when h is odd is still open. When h is odd we canonly prove the following weaker result.

Theorem 5.1.8. (Kiss, Rozgonyi, Sandor, 2014)Let X = Zm and h > 2 be a fixed odd integer. If A ⊂ Zm such that |A| = m/2

then there exists a g ∈ Zm such that R(1)h,A(g) 6= R

(1)h,Zm\A(g).

It would be interesting to characterize all that partitions of a finite Abeliangroup G such that R

(1)h,Ai

(g) = R(1)h,Aj

(g) for every g ∈ G.For the two other representation functions we have the following result

(only for the h = 2 case).

57

Theorem 5.1.9. (Kiss, Rozgonyi, Sandor, 2014)

Let X = G be a finite Abelian group. For i ∈ 2, 3 the equality R(i)2,A(g) =

R(i)2,G\A(g) holds for every g ∈ G, if and only if DA(g) = DG\A(g) for every

g ∈ G.

5.2 Proof of Theorem 5.1.3

LetA(x) =∑a∈A

xa be the generating function of the setA and letB(x) =∑b∈B

xb

be the generating function of the set B. It is easy to see that

A(x)B(x) =(∑a∈A

xa)(∑

b∈B

xb)

=∞∑n=0

RA+B(n)xn.

Since1

1− x− A(x) =

∞∑n=0

xn −∑a∈A

xa =∑a∈N\A

xa

and similarly

1

1− x−B(x) =

∞∑n=0

xn −∑b∈B

xb =∑b∈N\B

xb,

it follows that(1

1− x− A(x)

)(1

1− x−B(x)

)=∞∑n=0

RN\A+N\B(n)xn. (5.2.1)

Hence the condition RA+B(n) = RN\A+N\B(n) holds from a certain pointon 5.2.1 is equivalent to

A(x)B(x)−(

1

1− x− A(x)

)(1

1− x−B(x)

)= P (x),

where P (x) is a polynomial with integral coefficients. This is equivalent to

A(x) +B(x) =1

1− x+ P (x)(1− x). (5.2.2)

Let1

1− x+ P (x)(1− x) =

∞∑n=0

xn +N∑n=0

dnxn =

∞∑n=0

cnxn,

58

where cn = 0, 1 or 2. AsN∑n=0

dn = 0, it follows that the equation (5.2.2) holds

if and only if cn = 1 except for finitely many integer n and the number ofn for which cn = 0 is equal to the number of n for which cn = 2. This isequivalent to the condition |N \ (A ∪ B)| = |A ∩ B| <∞.

5.3 Proof of Theorem 5.1.6

For a given S ⊂ G we denote by tS,g its characteristic function. That istS,g = 1 if g ∈ S and tS,g = 0 if g 6∈ S for every g ∈ G. It follows thattG\S,g = 1 if g ∈ G \ S, i.e., g 6∈ S and tG\S,g = 0 if g 6∈ G \ S, i.e., g ∈ S.Thus we have

tG\S,g = 1− tS,g. (5.3.1)

It is easy to see that

RA+B(g) =∑c+d=g

c,d∈G

tA,c tB,d =∑c∈G

tA,c tB,−c+g,

RG\A+G\B(g) =∑c+d=g

c,d∈G

tG\A,c tG\B,d =∑c∈G

tG\A,c tG\B,−c+g.

It follows from (5.3.1) that for every g ∈ G we have

RA+B(g)−RG\A+G\B(g) =∑c∈G

tA,ctB,−c+g −∑c∈G

tG\A,ctG\B,−c+g =

∑c∈G

tA,ctB,−c+g −∑c∈G

(1− tA,c)(1− tB,−c+g) =∑c∈G

tA,c +∑c∈G

tB,−c+g −∑c∈G

1 =

= |A|+ |B| − |G|.

Hence if there exists a g ∈ G such that RA+B(g) = RG\A+G\B(g) then|A|+ |B| = |G| and if |A|+ |B| = |G| then RA+B(g) = RG\A+G\B(g) for everyg ∈ G.

59

5.4 Proof of Theorem 5.1.7

Assume thatR

(1)h,A(g) = R

(1)h,G\A(g) (5.4.1)

holds for all g ∈ G. It is clear that∑g∈G

R(1)h,A(g) = |A|h,

and ∑g∈G

R(1)h,G\A(g) = |G \ A|h.

It follows from (5.4.1) that |A|h = |G \A|h which implies |A| = |G|/2, whichproves (i).

In the next step we prove (ii). Assume that h is even. For h = 2 thestatement follows immediately from Theorem 5.1.2. Let h > 2. Then wehave

R(1)h,A(g) = |(a1, a2, . . . , ah) : ai ∈ A, a1 + . . . + ah = g| =

=∑

g1,...,gh/2∈Gg1+...+gh/2=g

|(a1, a2, . . . , ah) : ai ∈ A,

a1 + a2 = g1, . . . , ah−1 + ah = gh/2, a1 + . . . + ah = g| =

=∑

g1,...,gh/2∈Gg1+...+gh/2=g

|(a1, a2) ∈ A : a1+a2 = g1| · · · |(ah−1, ah) ∈ A : ah−1+ah = gh/2| =

=∑

g1,...,gh/2∈Gg1+...+gh/2=g

R(1)2,A(g1) · · ·R(1)

2,A(gh/2) =∑

g1,...,gh/2∈Gg1+...+gh/2=g

R(1)2,G\A(g1) · · ·R(1)

2,G\A(gh/2)

=∑

g1,...,gh/2∈Gg1+...+gh/2=g

|(a1, a2) ∈ G\A : a1+a2 = g1| · · · |(ah−1, ah) ∈ G\A : ah−1+ah = gh/2|

=∑

g1,...,gh/2∈Gg1+...+gh/2=g

|(a1, a2, . . . , ah) : ai ∈ G \ A,

a1 + a2 = g1, . . . , ah−1 + ah = gh/2, a1 + . . . + ah = g|

= |(a1, a2, . . . , ah) : ai ∈ G \ A, a1 + . . . + ah = g| = R(1)h,G\A(g).

60

5.5 Proof of Theorem 5.1.8

We prove by contradiction. Assume that for every g ∈ Zm, R(1)h,A(g) =

R(1)h,Zm\A(g). It is easy to see that

R(1)h,A(g) = |(i1, i2, . . . , ih) : ai1 + . . . + aih ≡ g (mod m)| =

=h−1∑j=0

|(i1, i2, . . . , ih) : ai1 + . . . + aih = g + jm|,

thus we have

m−1∑g=0

R(1)h,A(g)zg =

m−1∑g=0

( h−1∑j=0

|(i1, i2, . . . , ih) : ai1 + . . . + aih = g + jm|)zg.

It is clear that

Ah(z) =h−1∑j=0

m−1∑g=0

|(i1, i2, . . . , ih) : ai1 + . . . + aih = g + jm|zg+jm

=m−1∑g=0

h−1∑j=0

|(i1, i2, . . . , ih) : ai1 + . . . + aih = g + jm|zg+jm.

It follows that there exist P1(z) and P2(z) polynomials with integral co-efficients such that

Ah(z) = (1− zm)P1(z) +m−1∑g=0

R(1)h,A(g)zg,

and similarly

(1 + . . . + zm−1 − A(z))h = (1− zm)P2(z) +m−1∑g=0

R(1)h,Zm\A(g)zg.

Thus we have

(A(z))h −(

1− zm

1− z− A(z)

)h=

= (1− zm)(P1(z)− P2(z)) +m−1∑g=0

(R

(1)h,A(g)−R(1)

h,Zm\A(g))zg.

61

As R(1)h,A(g) = R

(1)h,Zm\A(g) for every g ∈ Zm, we have

1− zm∣∣∣ (A(z))h −

(1− zm

1− z− A(z)

)h.

Since1− zm

1− z

∣∣∣ 1− zm and h is odd, we have

1− zm

1− z

∣∣∣ 2(A(z))h.

As the polynomial1− zm

1− z= 1 + . . . + zm−1 has no multiple roots, we

have1− zm

1− z

∣∣∣ A(z),

which is a contradiction.

5.6 Proof of Theorem 5.1.9

We only prove the case i = 2, because the proof of case i = 3 is very similar.For every fixed g ∈ G we have

|A| = |(a, y) : a ∈ A, y ∈ G, a+ y = g| =

= |(a, y) : a ∈ A, y ∈ A, a+y = g|+|(a, y) : a ∈ A, y ∈ G\A, a+y = g| =

= RA+A(g) +RA+G\A(g) = 2R(2)2,A(g)−DA(g) +RA+G\A(g),

thus

R(2)2,A(g) =

1

2|A|+ 1

2DA(g)− 1

2RA+G\A(g).

Replace A by G\A and using also the fact that RA+G\A(g) = RG\A+G\(G\A)(g)we get

R(2)2,G\A(g) =

1

2|G \ A|+ 1

2DG\A(g)− 1

2RA+G\A(g).

Hence for every g ∈ G we have

R(2)2,A(g)−R(2)

2,G\A(g) =1

2|A|+ 1

2DA(g)−

(1

2|G \ A|+ 1

2DG\A(g)

)=

= |A| − 1

2|G|+ 1

2big(DA(g)−DG\A(g)

).

62

Suppose that R(2)2,A(g) = R

(2)2,G\A(g) for every g ∈ G. Then we have(

|A|+ 1

2

)=∑g∈G

R(2)2,A(g) =

∑g∈G

R(2)2,G\A(g) =

(|G \ A|+ 1

2

),

therefore |A| = |G \ A|, that is |A| = |G| − |A|. Hence we get that DA(g) =DG\A(g) for every g ∈ G.

Finally, suppose that DA(g) = DG\A(g) for every g ∈ G. Then we have

|A| = | ∪g∈G a : a ∈ A, a+ a = g| =∑g∈G

|a : a ∈ A, a+ a = g| =∑g∈G

DA(g) =

=∑g∈G

DG\A(g) = |G \ A|,

therefore |A| = |G| − |A|, thus we get that R(2)2,A(g) = R

(2)2,G\A(g) for every

g ∈ G which completes the proof.

63

Chapter 6

On Sidon sets which areasymptotic bases

This chapter is based on the article [KiRoSa14d].

6.1 Introduction

Let A = a1, a2, . . . , a1 < a2 < . . . be an infinite sequence of positiveintegers. A (finite or infinite) set A of positive integers is said to be a Sidonset if all the sums a + b with a, b ∈ A, a ≤ b are distinct. In other words Ais a Sidon set if for every n positive integer R

(2)2,A(n) ≤ 1.

We say a set A ⊂ N is an asymptotic basis of order h, if every largeenough positive integer n can be represented as the sum of h terms from A,i.e., if there exists a positive integer n0 such that R

(2)h,A(n) > 0 for n > n0.

In [ErSaTSos94a] and [ErSaTSos94b] P. Erdos, A. Sarkozy and V. T. Sosasked if there exists a Sidon set which is an asymptotic basis of order 3.The problem was also appears in [Sar] (with a typo in it: order 2 is writteninstead of order 3). It is easy to see [GrHaHePi] that a Sidon set cannot bean asymptotic basis of order 2. A few years ago J. M. Deshouillers and A.Plagne in [DesPla] constructed a Sidon set which is an asymptotic basis oforder at most 7. In [Kis] S. Kiss proved the existence of a Sidon set whichis an asymptotic basis of order 5. In this chapter we will improve this resultby proving that there exists an asymptotic basis of order 4 which is a Sidonset by using probabilistic methods.

Theorem 6.1.1. There exists an asymptotic basis of order 4 which is a Sidonset.

We note that at the same time Javier Cilleruelo [Cil] has proved a slightlystronger result namely the existence of a Sidon set which is an asymptotic

64

basis of order 3+ε. It means, that for any ε > 0 there exists a Sidon sequenceA of positive integers such that all positive, large enough integer n can bewritten as n = a1 + a2 + a3 + a4, a1, a2, a3, a4 ∈ A, a4 ≤ nε. He obtained hisresult independently from our work by using other probabilistic methods.

We recall the definition of the rh,A(n) representation function from Sec-tion 1.1. For some A = a1, a2, . . . ∈ Ω we denote the number of solutionsof ai1 + ai2 + . . . + aih = n with ai1 , . . . , aih ∈ A, 1 ≤ ai1 < ai2 . . . < aih < n

by rh,A(n) (this is the R(3)h,A(n) representation function). Let

rh,A(n) =∑

(ai1 ,ai2 ,...,aih )1≤ai1<...<aih<nai1+ai2+...+aih=n

t(A,ai1 )t(A,ai2 ) . . . t(A,aih ).

Let r∗h,A(n) denote the number of those representations of n in the form(1.1.1) in which there are at least two equal terms. Thus we have

R(1)h,A(n) = h! rh,A(n) + r∗h,A(n). (6.1.1)

It is easy to see from (6.1.1) that in the proof we can use rh,A(n) instead of

R(1)h,A(n).

6.2 Proof of Theorem 6.1.1

Define the sequence θn in Lemma 1.1.1 by

θn = n−57 , (6.2.1)

that is P (A : A ∈ Ω, n ∈ A) = n−57 , for n ∈ N. For a given set A ∈ Ω let

the set B be the following

B = b : b ∈ A,∃a′, a′′, a′′′ ∈ A : b+ a′ = a′′ + a′′′, a′, a′′, a′′′ < b . (6.2.2)

Thus A \ B is a Sidon set. We will prove that A \ B is an asymptotic basisof order 4 with probability 1. This means that there exists integer N0 suchthat with probability 1, r4,A\B(n) > 0 for n ≥ N0. Since

r4,A\B(n) = r4,A(n)−(r4,A(n)− r4,A\B(n)

),

if we get a lower bound for r4,A(n) and an upper bound for(r4,A(n)− r4,A\B(n)

)then we will have a lower bound for r4,A\B(n). So for-

mally we will show that there are positive constants C1 and N1 such thatwith probability 1,

r4,A(n) > C1n17 , n ≥ N1, (6.2.3)

65

and there are positive constants C2 and N2 such that with probability 1,

r4,A(n)− r4,A\B(n) < C2 (log n)6,5 , n ≥ N2. (6.2.4)

In order to prove (6.2.3) and (6.2.4) we use Theoerem 1.1.2.We also need the following Lemma (see in [Nat96], p. 134., Lemma 5.3).

For the sake of completeness we sketch the proof.

Lemma 6.2.1. Let N ≥ 3, α, β > −1. Then

N−1∑n=1

nα(N − n)β = Θα,β

(Nα+β+1

).

Proof of 6.2.1

N−1∑n=1

nα(N − n)β =∑

1≤n≤N2

nα(N − n)β +∑

N2<n<N

nα(N − n)β =

= Θα,β

(Nβ

∑1≤n≤N

2

nα)

+ Θα,β

(Nα

∑N2<n<N

(N − n)β)

=

= Θα,β

(Nβ

∫ N2

1

xαdx

)+ Θα,β

(Nα

∫ N

N2

xβdx

)= Θα,β

(Nα+β+1

).

In the first step we prove (6.2.3) by using Theorem 1.1.2. To do this, weneed the following Lemma.

Lemma 6.2.2. Assume that all of the variables yi’s are different and thetyi’s are random boolean variables.

1. For every nonzero integer a1 and for every integer m

E

( ∑y1

a1y1=m

ty1

)= Oa1 (1) .

2. For every nonzero integers a1, a2 and for every integer m

E

( ∑(y1,y2)

a1y1+a2y2=m

ty1ty2

)= Oa1,a2 (1) .

66

3. For every nonzero integers a1, a2, a3 and for every integer m

E

( ∑(y1,y2,y3)

a1y1+a2y2+a3y3=m

ty1ty2ty3

)= Oa1,a2,a3 (1) .

Proof of Lemma 6.2.2 (1):

E

( ∑y1

a1y1=m

ty1

)=

(ma1

)− 57

= Oa1 (1) if ma1∈ Z+

0 if ma16∈ Z+

(2): We distinguish two different cases.Case 1. Assume, that a1 > 0, a2 > 0, thus m > 0. (Since y1, y2, a1, a2 are

nonnegative, therefore m can not be negative, at this case.) Thus applyingLemma 6.2.1 we get

E

( ∑(y1,y2)

a1y1+a2y2=m

ty1ty2

)= Oa1,a2

ma1∑y1=1

y− 5

71

(m− a1y1

a2

)− 57

=

= Oa1,a2

( ma1∑y1=1

(a1y1)− 5

7 (m− a1y1)−57

)= Oa1,a2

(m−1∑y=1

y−57 (m− y)−

57

)=

= Oa1,a2

(m−

37

)= Oa1,a2 (1) .

Case 2. Now assume that a1 > 0, a2 < 0 and m ≥ 0. (If m is negative,then consider the equation −a1y1−a2y2 = m.) We apply Lemma 6.2.1 again.

E

( ∑(y1,y2)

a1y1+a2y2=m

ty1ty2

)= Oa1,a2

(∞∑y2=1

y− 5

72

(m− a2y2

a1

)− 57

)=

= Oa1,a2

( ∞∑y=1

y−107

)= Oa1,a2 (1) .

The other cases can be deduced from the aboves. So we leave the details tothe reader. ( The case a1 < 0, a2 > 0, either m ≥ 0 or m < 0 is almost thesame like Case 2., we have to change the role of a1 and a2. The case a1 < 0,a2 < 0, thus m < 0 is almost the same like Case 1., we have to get −a1 and−a2 instead of a1 and a2. )

67

(3): We distinguish three different cases.Case 1. Assume, that a1 > 0, a2 > 0, a3 > 0 thus m > 0. (Since y1, y2,

y3, a1, a2, a3 are nonnegative, therefore m can not be negative, at this case.)Thus applying Lemma 6.2.1 we get

E

( ∑(y1,y2,y3)

a1y1+a2y2+a3y3=m

ty1ty2ty3

)=

= Oa1,a2,a3

( ma1∑y1=1

y− 5

71

m−a1y1a2∑y2=1

y− 5

72

(m− a1y1 − a2y2

a3

)− 57

)=

= Oa1,a2,a3

( ma1∑y1=1

y− 5

71

m−a1y1a2∑y2=1

(a2y2)− 5

7 (m− a1y1 − a2y2)−57

)=

= Oa1,a2,a3

( ma1∑y1=1

y− 5

71 (m− a1y1)−

37

)= Oa1,a2,a3

(m−

17

)= Oa1,a2,a3 (1) .

Case 2. Now assume that a1 > 0, a2 > 0, a3 < 0 and m ≥ 0. Thusapplying Lemma 6.2.1 again we get

E

( ∑(y1,y2,y3)

a1y1+a2y2+a3y3=m

ty1ty2ty3

)=

= Oa1,a2,a3

(∞∑y3=1

y− 5

73

m−a3y3a1∑y1=1

y− 5

71

(m− a3y3 − a1y1

a2

)− 57

)=

= Oa1,a2,a3

(∞∑y3=1

y− 5

73

m−a3y3a1∑y1=1

(a1y1)− 5

7 (m− a3y3 − a1y1)−57

)=

= Oa1,a2,a3

(∞∑y3=1

y− 5

73 (m− a3y3)−

37

)= Oa1,a2,a3

( ∞∑y=1

y−87

)= Oa1,a2,a3 (1) .

Case 3. Now assume that a1 > 0, a2 > 0, a3 < 0 and m < 0. By Lemma6.2.1 we get

E

( ∑(y1,y2,y3)

a1y1+a2y2+a3y3=m

ty1ty2ty3

)=

68

= Oa1,a2,a3

(∞∑

y3=b ma3 c+1

y− 5

73

m−a3y3a1∑y1=1

y− 5

71

(m− a3y3 − a1y1

a2

)− 57

)=

= Oa1,a2,a3

(∞∑

y3=b ma3 c

y− 5

73

m−a3y3a1∑y1=1

(a1y1)− 5

7 (m− a3y3 − a1y1)−57

)=

= Oa1,a2,a3

(∞∑

y3=b ma3 c

y− 5

73 (m− a3y3)−

37

)= Oa1,a2,a3

( ∞∑y=1

y−87

)= Oa1,a2,a3 (1) .

The other cases can be deduced from the aboves again. We leave the detailsto the reader. (The case a1 > 0, a2 < 0, a3 < 0, either m ≥ 0 or m < 0 isalmost the same as Case 2. and as Case 3. The case a1 < 0, a2 < 0, a3 < 0thus m < 0 is almost the same as Case 1. In both we have to get −a1, −a2and −a3 instead of a1, a2 and a2.)

Now we are ready to prove (6.2.3). In view of (1.1.1) define Y by

Y = r4,A(n) =∑

(x1,x2,x3,x4)1≤x1<···<x4

x1+x2+x3+x4=n

tx1tx2tx3tx4 .

We want to use Theorem 1.1.2. To do this we have to estimate the expecta-tion of the variable Y and its partial derivatives. Let α = (α1, . . . , αn) be amulti-index. In the first step we prove that for α = 0

E (∂αY ) = E(Y ) = Θ(n17 ), (6.2.5)

and for α 6= 0E (∂αY ) = O(1). (6.2.6)

Let α = 0. By using Lemma 6.2.1 we have

E (∂αY ) = E(Y ) = E

( ∑(x1,x2,x3,x4)1≤x1<···<x4

x1+x2+x3+x4=n

tx1tx2tx3tx4

)=

= Θ

( n−3∑x1=1

x− 5

71

n−x1−2∑x2=1

x− 5

72

n−x1−x2−1∑x3=1

x− 5

73 (n− x1 − x2 − x3)−

57

)=

= Θ

( n−3∑x1=1

x− 5

71

n−x1−2∑x2=1

x− 5

72 (n−x1−x2)−

37

)= Θ

( n−3∑x1=1

x− 5

71 (n−x1)−

17

)= Θ

(n

17

),

69

which shows (6.2.5).Now assume that α = (α1, . . . , αn) 6= 0. If there exists an index i such

that αi ≥ 2 or |α| =∑n

i=1 αi ≥ 5, then ∂αY = 0. It means that in thiscase E(∂αY ) = 0. So we may assume that for every index i, αi ≤ 1 and|α| ≤ 4. Let l1 < l2 < · · · < lw, for which αl1 = . . . = αlw = 1, 1 ≤ w ≤ 4.If 1 ≤ κ1 < . . . < κ4−w ≤ 4, κj ∈ N, then 1, 2, 3, 4 \ κ1, . . . , κ4−w =r1, . . . , rw, where r1 < · · · < rw and xr1 = l1, . . . , xrw = lw. It meansthat the variables xr1 , . . . , xrw occurs in the partial derivative of Y and thevariables xκ1 , . . . , xκ4−w do not. Thus

E (∂αY ) =∑

(κ1,...,κ4−w)1≤κ1<...<κ4−w≤4

E

( ∑(xκ1 ,...,xκ4−w )∑4−w

j=1 xκj=n−∑wj=1 lj

txκ1 . . . txκ4−w

).

Since w ≥ 1, thus 4− w ≤ 3. Since the number of the tuples (κ1, . . . , κ4−w)is bounded, by Lemma 6.2.2 part (2) we get that the expectation is

E (∂αY ) =∑

(κ1,...,κ4−w)1≤κ1<...<κ4−w≤4

O(1) = O(1),

which proves (6.2.6).From (6.2.5) and (6.2.6) we get that

E≥0(Y ) = maxd′≥0Ed′(Y ) = E(Y ) = Θ(n17 ),

andE≥1(Y ) = maxd′≥1Ed′(Y ) = O(1).

Now apply Theorem 1.1.2 with λ = 20 log n and k = 4. We get that

P(|Y − E(Y )| ≥ C4(20 log n)3,5

√E≥0(Y )E≥1(Y )

)= O

(1

n2

). (6.2.7)

Thus by (6.2.7) and Lemma 1.1.3 we get that if n is large enough, withprobability 1,

Y = E(Y ) +O

((log n)3,5

√E≥0(Y )E≥1(Y )

)=

= E(Y ) +O(n

114 (log n)3,5

)= Θ(n

17 ),

which means, that (6.2.3) holds.

70

In the next step we will prove (6.2.4), which shows us that the number ofthose representations in which there is at least one element from the set Bis not too big. Using the definition of the representation functions and thedefinitions of the sets A and B we get

r4,A(n)− r4,A\B(n) =∑

(ai,aj ,ak,al)ai<aj<ak<al∈Aai+aj+ak+al=n

∃m∈i,j,k,l,∃au,av ,az∈Aau,av ,az<amam+au=av+az

taitaj taktal (6.2.8)

To make the analytic calculations easier we estimate (6.2.8) and we have that

r4,A(n)− r4,A\B(n) =1

24

∑(ai,aj ,ak,al)

ai,aj ,ak,al∈A are distinctai+aj+ak+al=n

∃m∈i,j,k,l,∃au,av ,az∈Aau,av ,az<amam+au=av+az

taitaj taktal =

=4

24

∑(ai,aj ,ak,al)

ai,aj ,ak,al∈A are distinctai+aj+ak+al=n∃au,av ,az∈Aau,av ,az<alal+au=av+az

taitaj taktal ≤

≤ 1

6

∑(ai,aj ,ak,al,au,av ,az)

ai,aj ,ak,al,au,av ,az∈A are distinctai+aj+ak+al=nau,av ,az<alal+au=av+az

taitaj taktaltautavtaz

(6.2.9)

Using the variables xi-s we can write this in the following form

r4,A(n)− r4,A\B(n) ≤∑

(x1,x2,x3,x4,x5,x6,x7)x1+x2+x3+x4=n

x1,x2,x3,x4are distinctx4+x5=x6+x7x5,x6,x7<x4

tx1 . . . tx7 . (6.2.10)

So this last estimation means, that always x4 will be the element in the4-tuple (x1, x2, x3, x4) which hurts the Sidon property. It is easy to see thatin the product tx1 . . . tx7 the txi variables are not necessarily independent. Sowe need to transform (6.2.10). For any 7-tuple (x1, x2, x3, x4, x5, x6, x7) withcondition x5, x6, x7 < x4 let x1, x2, x3 ∩ x5, x6, x7 = xi1 , . . . , xis, where

71

1 ≤ i1 < · · · < is ≤ 3. Let x5, x6, x7\xi1 , . . . , xis = xh1 , . . . , xhu, where5 ≤ h1 < · · · < hu ≤ 7 and u ≤ 3− s. Then for every fix s-tuple (i1, . . . , is)there exist s + u tuple (di1 , . . . , dis , bh1 , . . . , bhu) such that we can write thecondition x4 + x5 − x6 − x7 = 0 in the following form:

x4 + di1xi1 + . . .+ disxis + bh1xh1 + . . .+ bhuxhu = 0, (6.2.11)

where x4, xi1 , . . . , xis , xh1 , . . . , xhu are different. In (6.2.11) dij 6= 0, j =1, . . . , s, bhj 6= 0, j = 1, . . . , u, there is only one positive coefficients, whichis equal to 1 and the sum of the negative coefficients is equal to −2. Sincetkx = tx, if k ≥ 1 then tx1 . . . tx7 = tx1tx2tx3tx4txh1 . . . txhu .Thus (6.2.10) is equal to the following∑

(i1,...,is)1≤i1<...<is≤3

∑(di1 ,...,dis ,bh1 ,...,bhu )

dij 6=0,bj 6=0

only one term is positive and=1the sum of the negative terms is−2∑

(x1,x2,x3,x4,xh1 ,...,xhu )x1+x2+x3+x4=n

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

tx1 . . . tx4txh1 . . . txhu .

Let the inner sum be

Ydi1 ,...,dis ,bh1 ,...,bhu =∑

(x1,x2,x3,x4,xh1 ,...,xhu )x1+x2+x3+x4=n

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

tx1 . . . tx4txh1 . . . txhu .

Since the number of the variables Ydi1 ,...,dis ,bh1 ,...,bhu is bounded, it is enoughto show that for every s+ u tuple (di1 , . . . , dis , bh1 , . . . , bhu) with probability1,

Ydi1 ,...,dis ,bh1 ,...,bhu = O((log n)6,5

). (6.2.12)

Let’s fix an s+u tuple (di1 , . . . , dis , bh1 , . . . , bhu). We will use Lemma 6.2.2. Sowe have to estimate both the expectation of Ydi1 ,...,dis ,bh1 ,...,bhu and its partialderivatives. First we will show that for every α = (α1, . . . , αn),

E(∂αYdi1 ,...,dis ,bh1 ,...,bhu

)= O(1) (6.2.13)

holds.

72

Let α = (α1, . . . , αn). If there exists an index i such that αi ≥ 2 or|α| =

∑ni=1 αi ≥ 5 +u, then ∂αYdi1 ,...,dis ,bh1 ,...,bhu = 0. So we may assume that

for every index i, αi ≤ 1 and |α| ≤ 4 + u. Let’s fix α = (α1, . . . , αn) and letβ = (β1, . . . , βn), γ = (γ1, . . . , γn), where βi ∈ N, γi ∈ N and α = β+γ. Hereβ shows the partial derivatives of the variables x1, x2, x3, x4 and γ showsthe derivatives of the variables xh1 , . . . , xhu . So we can write the partialderivatives of Ydi1 ,...,dis ,bh1 ,...,bhu in the following form

∂αYdi1 ,...,dis ,bh1 ,...,bhu =

=∑(β,γ)β+γ=α

∑(x1,x2,x3,x4)

x1+x2+x3+x4=nx1,x2,x3,x4are distinct

(∂βtx1 . . . tx4

)·

·

( ∑(xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

∂γtxh1 . . . txhu

).

Since the number of pairs (β, γ) is bounded, it is enough to show that for everyfixed pair (β, γ), where 0 ≤ βi ≤ 1, 0 ≤ γi ≤ 1, |β| + |γ| =

∑ni=1(βi + γi) ≤

4 + u,

E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=nx1,x2,x3,x4are distinct

(∂βtx1 . . . tx4

)·

·

( ∑(xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

∂γtxh1 . . . txhu

))= O(1) (6.2.14)

holds. Let’s fix now a pair (β, γ). We will show that for every 4-tuple(x1, x2, x3, x4)

E

( ∑(xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

∂γtxh1 . . . txhu

)= O(1). (6.2.15)

73

Let |γ| = w, γl1 = . . . = γlw = 1, 1 ≤ l1 < . . . < lw ≤ n. Letg1, . . . , gw ⊆ h1, . . . hu and j1, . . . , ju−w = h1, . . . , hu \ g1, . . . , gw.These sets shows us that the variables xg1 , . . . , xgw occurs in the partialderivative of Ydi1 ,...,dis ,bh1 ,...,bhu and the variables xj1 , . . . , xju−w do not. Thuswe have

E

( ∑(xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

∂γtxh1 . . . txhu

)=

=∑

(j1,...,ju−w)j1,...,ju−w⊆h1,...,hu

∑π

(xg1 ,...,xgw )=π(l1,...,lw)

E

( ∑(xj1 ,...,xju−w )∑u−wq=1 bjqxjq=

=−x4−di1xi1−...−disxis−∑wq=1 bgqxgq

txj1 . . . txju−w

),

where π(l1, . . . , lw) denotes the permutations of (l1, . . . , lw).Since the number of the u−w tuples (j1, . . . , ju−w) and the permutations

are bounded and since u−w ≤ 3, thus from 6.2.1 we get that this expectationis bounded i.e.,

E

( ∑(xj1 ,...,xju−w )∑u−w

q=1 bjqxjq=−x4−di1xi1−...disxis−∑wq=1 bgqxgq

txj1 . . . txju−w

)= O(1). (6.2.16)

So from (6.2.16) we get that the left hand side of the equation (6.2.14) isequal to the following∑

(x1,x2,x3,x4)x1+x2+x3+x4=n

x1,x2,x3,x4are distinct

E(∂βtx1 . . . tx4

)·

· E

( ∑(xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

∂γtxh1 . . . txhu

)=

= O

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=nx1,x2,x3,x4are distinct

E(∂βtx1 . . . tx4

)).

74

By (6.2.6) we get that if β 6= 0 then this term is equal to O(1). So we mayassume that β = 0. This means, that we have to prove that

E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=nx1,x2,x3,x4are distinct

tx1 . . . tx4·

·

( ∑(xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

x1,x2,x3,x4,xh1 ,...,xhuare distinct

∂γtxh1 . . . txhu

))= O(1). (6.2.17)

If 1 ≤ i1 < . . . < is ≤ 3 then let 1, 2, 3 \ i1, . . . is = f1, . . . , f3−s. Thus(6.2.17) is equivalent to the following

E

( ∑(xi1 ,...,xis ,x4,xh1 ,...,xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

xi1 ,...,xis ,x4,xh1 ,...,xhuare distinct

∂γtxi1 . . . txis tx4txh1 . . . txhu ·

·

( ∑(xf1 ,...,xf3−s )

xf1+...+xf3−s=n−xi1−...−xis−x4xi1 ,...,xis ,x4,xf1 ,...,xf3−sare distinct

txf1 . . . txf3−s

))=

=∑

(xi1 ,...xis ,x4,xh1 ,...,xhu )x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

xi1 ,...,xis ,x4,xh1 ,...,xhuare distinct

(E(∂γtxi1 . . . txis tx4txh1 . . . txhu

)·

· E

( ∑(xf1 ,...,xf3−s )

xf1+...+xf3−s=n−xi1−...−xis−x4xi1 ,...,xis ,x4,xf1 ,...,xf3−sare distinct

txf1 . . . txf3−s

)). (6.2.18)

By using Lemma 6.2.2 we have

E

( ∑(xf1 ,...,xf3−s )

xf1+...+xf3−s=n−xi1−...−xis−x4xi1 ,...,xis ,x4,xf1 ,...,xf3−sare distinct

txf1 . . . txf3−s

)= O(1).

75

It follows that (6.2.18) is equal to

O

(E

(∂γ

∑(xi1 ,...xis ,x4,xh1 ,...xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

xi1 ,...,xis ,x4,xh1 ,...,xhuare distinct

txi1 . . . txis tx4txh1 . . . txhu

)).

Let γ = (γ1, . . . , γn), |γ| = w, γl1 = . . . = γlw = 1, 1 ≤ l1 < . . . < lw ≤n. If v1, . . . , vw ⊂ h1, . . . , hu, then let e1, . . . , eu−w = h1, . . . , hu \v1, . . . , vw. Thus we have

E

(∂γ

∑(xi1 ,...xis ,x4,xh1 ,...xhu )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

txi1 . . . txis tx4txh1 . . . txhu

)=

=∑

v1,...vw⊂h1,...,hu

∑π

(xv1 ,...,xvw )=π(l1,...,lw)

E

(

∑(xi1 ,...,xis ,x4,xe1 ...xeu−w )

x4+di1xi1+...+disxis+bh1xh1+...+bhuxhu=0

xij<x4,j=1,...,s;xhj′<x4,j′=1,...,u

xi1 ,...,xis ,x4,xh1 ,...,xhuare distinct

txi1 . . . txis tx4txe1 . . . txeu−w

), (6.2.19)

where π(l1, . . . , lw) denotes the permutations of (l1, . . . , lw).Since s + u + 1 is the number of the variables in the equation x4 + di1xi1 +. . .+disxis +bh1xh1 + . . .+bhuxhu = 0, it is clear, that s+u+1 is equal to 3 or4. So s+u+1−w ≤ 3, except if s+u+1 = 4 and w = 0. If s+u+1−w ≤ 3then we may use Lemma 6.2.1 again, which implies (6.2.17). So we can onlyshow the case s+ u+ 1 = 4 and w = 0, which is equivalent to γ = 0. To dothis, it remains to prove the following Lemma.

Lemma 6.2.3. The following expectation are bounded.

1. E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=n

tx1tx2tx3tx4

( ∑(x5,x6,x7)

x4+x5=x6+x7

tx5tx6tx7

))= O(1)

76

2. E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=n

tx1tx2tx3tx4

( ∑(x5,x6)

x4+x3=x5+x6

tx5tx6

))= O(1)

3. E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=n

tx1tx2tx3tx4

( ∑(x5,x6)

x4+x5=x3+x6

tx5tx6

))= O(1)

4. E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=n

tx1tx2tx3tx4

( ∑x5

x4+x3=x2+x5

tx5

))= O(1)

5. E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=n

tx1tx2tx3tx4

( ∑x5

x4+x5=x2+x3

tx5

))= O(1)

6. E

( ∑(x1,x2,x3,x4)

x1+x2+x3+x4=nx4+x3=x2+x1

tx1tx2tx3tx4

)= O(1)

Proof of Lemma 6.2.3 (1): Using the definition of the variables ti’s getthat the order of this expectation is

O

(n∑

x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x2−1∑x3=1

x− 5

73 (n− x1 − x2 − x3)−

57 ·

·x4−1∑x5=1

x− 5

75

x4+x5−1∑x6=1

x− 5

76 (x4 + x5 − x6)−

57

). (6.2.20)

Applying Lemma 6.2.1 to the last sum it follows that (6.2.20) is equivalentwith the following

O

(n∑

x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x2−1∑x3=1

x− 5

73 (n− x1 − x2 − x3)−

57 ·

·x4−1∑x5=1

x− 5

75 (x4 + x5)

− 37

). (6.2.21)

77

Since

x4−1∑x5=1

x− 5

75 (x4 + x5)

− 37 = O

(x− 3

74

x4−1∑x5=1

x− 5

75

)=

= O(x− 1

74

)= O

((n− x1 − x2 − x3)−

17

),

it follows that (6.2.21) is equal to

O

(n∑

x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x2−1∑x3=1

x− 5

73 (n− x1 − x2 − x3)−

67

)=

=O

(n∑

x1=1

x− 5

71

n−x1∑x2=1

x− 5

72 (n− x1 − x2)−

47

)= O

(n∑

x1=1

x− 5

71 (n− x1)−

27

)= O(1).

(6.2.22)

(2): Using again the definition of the variables ti’s this expression is equiva-lent with the following

n∑x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x2−1∑x3=1

x− 5

73 (n−x1−x2−x3)−

57

x4+x3−1∑x5=1

x− 5

75 (x4+x3−x5)−

57 .

(6.2.23)Since

x4+x3−1∑x5=1

x− 5

75 (x4 + x3 − x5)−

57 = O

((x4 + x3)

− 37

)= O

(x− 3

74

)=

= O(x− 1

74

)= O

((n− x1 − x2 − x3)−

17

),

it follows that (6.2.23) is equal to

O

(n∑

x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x2−1∑x3=1

x− 5

73 (n− x1 − x2 − x3)−

67

).

It follows from (6.2.22) that this is equal to O(1).

(3): Since x3 < x4, thus x3 <x3+x4

2= n−x1−x2

2. This expectation is

O

n∑x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x22∑

x3=1

x− 5

73 (n− x1 − x2 − x3)−

57 ·

·x4−1∑x5=1

x− 5

75 (x4 + x5 − x3)−

57

), (6.2.24)

78

where

x4−1∑x5=1

x− 5

75 (x4+x5−x3)−

57 =

x4−x3∑x5=1

x− 5

75 (x4+x5−x3)−

57 +

x4−1∑x5=x4−x3+1

x− 5

75 (x4+x5−x3)−

57 =

= O

((x4 − x3)−

57

x4−x3∑x5=1

x− 5

75

)+O

(∫ ∞x4−x3

x−107 dx

)= O

((x4 − x3)−

37

).

So (6.2.24) is equal to

O

n∑x1=1

x− 5

71

n−x1∑x2=1

x− 5

72

n−x1−x22∑

x3=1

(2x3)− 5

7 (n− x1 − x2 − 2x3)− 6

7

.

It follows from (6.2.22) that this is equal to O(1).

(4): Note, that here n = x1 + 2x2 + x5. This expectation is

O

n∑x1=1

x− 5

71

n−x12∑

x2=1

x− 5

72

n−x1−x22∑

x3=1

x− 5

73 (n− x1 − x2 − x3)−

57 (x4 + x3 − x2)−

57

,

(6.2.25)where

(x4 + x3 − x2)−57 = (n− x1 − x2 − x2)−

57 = (n− x1 − 2x2)

− 57 .

So (6.2.25) is equal to

O

n∑x1=1

x− 5

71

n−x12∑

x2=1

x− 5

72 (n− x1 − 2x2)

− 57

n−x1−x22∑

x3=1

x− 5

73 (n− x1 − x2 − x3)−

57

=

=O

n∑x1=1

x− 5

71

n−x12∑

x2=1

x− 5

72 (n− x1 − 2x2)

− 57 (n− x1 − x2)−

37

=

=O

n∑x1=1

x− 5

71

n−x12∑

x2=1

(2x2)− 5

7 (n− x1 − 2x2)− 5

7

=

=O

(n∑

x1=1

x− 5

71 (n− x1)−

37

)= O

(n−

17

)= O(1).

79

(5): It is clear that the n = x1 + x2 + x3 + x4, x4 + x5 = x2 + x3 equation-system is equivalent with n = x1 + 2x4 + x5, x4 + x5 = x2 + x3. Thus thisexpectation is

O

n∑x1=1

x− 5

71

n−x12∑

x4=1

x− 5

74 (n− x1 − 2x4)

− 57

x4+x5−1∑x3=1

x− 5

73 (x4 + x5 − x3)−

57

,

(6.2.26)and

O

(x4+x5−1∑x3=1

x− 5

73 (x4 + x5 − x3)−

57

)= O

((x4 + x5)

− 37

)= O(1).

So (6.2.26) is equal to

O

n∑x1=1

x− 5

71

n−x12∑

x4=1

(2x4)− 5

7 (n− x1 − 2x4)− 5

7

=

= O

(n∑

x1=1

x− 5

71 (n− x1)−

37

)= O

(n−

17

)= O(1).

(6): It is clear that the n = x1+x2+x3+x4, x1+x2 = x3+x4 equation-systemis equivalent with n

2= x1 + x2 = x3 + x4. Thus this expectation is

O

n2∑

x1=1

x− 5

71

(n2− x1

)− 57

n2∑

x3=1

x− 5

73

(n2− x1

)− 57

=

= O

n2∑

x1=1

x− 5

71

(n2− x1

)− 57n−

37

= O(n−

67

)= O(1). (6.2.27)

And this ended the proof of this Lemma.

So from (6.2.13) we get that

E≥0(Ydi1 ,...,dis ,bh1 ,...,bhu ) = maxd′≥0Ed′(Ydi1 ,...,dis ,bh1 ,...,bhu ) = O(1),

and

E≥1(Ydi1 ,...,dis ,bh1 ,...,bhu ) = maxd′≥1Ed′(Ydi1 ,...,dis ,bh1 ,...,bhu ) = O(1).

In Theorem 1.1.2, we have k ≤ 7 and λ = 32 log n. Thus

80

P(∣∣∣Ydi1 ,...,dis ,bh1 ,...,bhu − E(Ydi1 ,...,dis ,bh1 ,...,bhu )

∣∣∣ ≥≥ Ck(32 log n)k−

12

√E≥0(Ydi1 ,...,dis ,bh1 ,...,bhu )E≥1(Ydi1 ,...,dis ,bh1 ,...,bhu )

)=

= Ok

(e−8 logn+(k−1) logn) = Ok

(1

n2

). (6.2.28)

Thus by (6.2.13) and Theorem 1.1.2 we get that with probability 1,

Ydi1 ,...,dis ,bh1 ,...,bhu = E(Ydi1 ,...,dis ,bh1 ,...,bhu )+

+O(

(log n)k−12

√E≥0(Ydi1 ,...,dis ,bh1 ,...,bhu )E≥1(Ydi1 ,...,dis ,bh1 ,...,bhu )

)=

= O((log n)6,5

),

which shows (6.2.13), which proves (6.2.4). And this completes the proof ofthe theorem.

81

Chapter 7

Summary

This thesis was devoted to the different properties of the additive repre-sentation functions and related problems just like Sidon sequences, additivecomplement sets etc. We can say that the existence of these recent resultsshows that the additive representation functions is a famous field of the ad-ditive number theory researches. For possible directions of future research,there is a wide variety of open questions.

For example it is still an open question that in the converse of the Erdos–Fuchs theorem could we give a better error term or not. The original h = 2case is sharp, and in the lower estimate Tang got the same result for everyh ≥ 2. Nowadays the best upper error term is only O(N

1.5h logN) which is

far from N14 . It seems to be sure that for better error term we need to use

other probabilistic construction.It is also an interesting question whether Conjecture 3.1.1 is true when h

is composite number h 6= ps or not. Recently we can not prove this case, andit can happen that the theorem is not true in that case.

We can ask some similar question about additive complement sets likeChen and Fang did and also we can study the properties of the representationfunctions in (Abelian) groups. For example Theorem 5.1.7 is still open inthe case when h is odd. And it would be interesting to characterize all thatpartitions of a finite Abelian group G such that R

(1)h,Ai

(n) = R(1)h,Aj

(n) for everyg ∈ G.

Of course it is still an open question that there exists a Sidon set whichis asymptotic basis of order 3 or not. J. Cilleruelo showed that the modularversoin of this conjecture true, namely for all large enough N , the cyclicgroup ZN contains a Sidon set S ⊂ ZN which is an asymptotic basis of order3 in ZN . So using better probabilistic tools we will be able to prove for thepositive integers also.

Finally we can conclude that in this thesis we worked on interesting prob-

82

lems and there are several ways to continue it.

83

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88