Thin bases in additive number theory

Discrete Mathematics 312 (2012) 2069–2075

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Thin bases in additive number theory Melvyn B. Nathanson ∗ Department of Mathematics, Lehman College (CUNY), Bronx, NY 10468, United States CUNY Graduate Center, New York, NY 10016, United States

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Article history: Received 1 July 2010 Received in revised form 7 March 2012 Accepted 19 March 2012 Available online 6 April 2012 Keywords: Additive number theory Thin bases Bases of finite order Shatrovski˘i bases

abstract The set A of nonnegative integers is called a basis of order h if every nonnegative integer can be represented as the sum of exactly h not necessarily distinct elements of A. An additive basis A of order h is called thin if there exists c > 0 such that the number of elements of A not exceeding x is less than cx1/h for all x ≥ 1. This paper describes a construction of Shatrovski˘i of thin bases of order h. © 2012 Elsevier B.V. All rights reserved.

1. Additive bases of finite order The central theme in additive number theory is the representation of a nonnegative integer as the sum of a bounded number of elements chosen from a fixed set of integers, such as the squares, the cubes, or the primes. Let N0 and Z denote the sets of nonnegative integers and integers, respectively. For real numbers a and b with a ≤ b, let [a, b] denote the interval of integers between a and b, that is, [a, b] = {n ∈ Z : a ≤ n ≤ b}. Let h ≥ 2, and let (A1 , . . . , Ah ) be an h-tuple of subsets of N0 . We define the sumset

A1 + · · · + Ah = {a1 + · · · + ah : ai ∈ Ai for i = 1, . . . , h}. The h-fold sumset of a set A of nonnegative integers is hA = A + · · · + A .





h terms



The h-tuple (A1 , . . . , Ah ) is called an additive system of order h if A1 + · · · + Ah = N0 . The set A is an additive basis, or, simply, a basis, of order h if hA = N0 . Note that every basis must contain 0 and 1. If (A1 , . . . , Ah ) is an additive system of h order h, then A = i=1 Ai is a basis of order h. The set A is an asymptotic basis of order h if the sumset hA contains all sufficiently large integers. For example, Lagrange’s theorem states that the squares are a basis of order 4, and Wieferich’s theorem states that the nonnegative cubes are a basis of order 9. Let r1 , . . . , rh be relatively prime positive integers, and let

Ai = ri ∗ N0 = {ri vi : vi ∈ N0 } ∗

Correspondence to: Department of Mathematics, Lehman College (CUNY), Bronx, NY 10468, United States. E-mail address: [email protected].

0012-365X/$ – see front matter © 2012 Elsevier B.V. All rights reserved. doi:10.1016/j.disc.2012.03.026

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M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075

for i = 1, . . . , h. There exists an integer C such that every integer n ≥ C can be represented in the form n = r1 v1 + · · · + rh vh with v1 , . . . , vh ∈ N0 [8, Section 1.6], and so the set

A = [0, C − 1] ∪

h 

Ai

i=1

is an additive basis of order h. Let g ≥ 2. We can use the g-adic representation of the nonnegative integers to construct another class of additive bases of order h. Let K1 , . . . , Kh be pairwise disjoint subsets of N0 such that N 0 = K1 ∪ · · · ∪ Kh . For i = 1, . . . , h, let F (Ki ) denote the set of all finite subsets of Ki , and let

Ai =

 

 εk g : F ∈ F (Ki ) and εk ∈ {0, 1, 2, . . . , g − 1} . k

(1)

k∈F

Then (A1 , . . . , Ah ) is an additive system of order h with The counting function of a set A of integers is



A(x) =

h

i =1

Ai = {0}, and A =

h

i=1

Ai is a basis of order h.

1.

a∈A 1≤a≤x

If A is a basis of order h, then for all x ∈ N0 we have

 x+1≤

A(x) + h

 ≤

h

(A(x) + h)h h!

and so A(x)

lim inf

x1/h

x→∞

≥ (h!)1/h .

The basis A of order h is called a thin basis if lim sup x→∞

A(x) x1/h

< ∞.

One of the earliest general problems in additive number theory was that of the existence and construction of thin bases of finite order [11]. In 1937, Raikov [10] and Stöhr [14] independently solved this problem. Using the 2-adic representation and h h the partition N0 = i=1 Ki , where Ki = {k ∈ N0 : k ≡ i − 1(mod h)} for i = 1, . . . , h, they proved that A = i=1 Ai is a thin basis of order h, where Ai is the set defined by (1) with g = 2. Many thin bases have been constructed using variations of the g-adic construction; see, for example, Cassels [2], Grekos [4], Nathanson [9] and Jia and Nathanson [6]. Recently, Blomer [1], Hofmeister [5], and Schmitt [12] constructed other examples of thin bases. In 1940 Shatrovski˘i [13] described a beautiful class of thin bases of order h that depends on the linear diophantine equation n = a1 x1 + · · · + ah xh and does not use g-adic representations. This is an almost forgotten paper.1 It has been mentioned in a few surveys of solved and unsolved problems in additive number theory (e.g. Erdős and Nathanson [3], Nathanson [7] and Stöhr [15]), but MathSciNet lists no publication that cites Shatrovski˘i’s paper. This paper contains an exposition of Shatrovski˘i’s construction of thin bases. 2. The construction Fix an integer h ≥ 2, and define k0 = k0 (h) =



1 21/h − 1



+ 1.

(2)

Then 1+

1 k

< 21/h

for all k ≥ k0 .

1 Shatrovski˘i published his paper in Russian, with a summary in French. The transliteration of the author’s name in the French summary is Chatrovsky, and this is the spelling that appears in Mathematical Reviews. I prefer the AMS and Library of Congress system for transliteration of Cyrillic into English.

M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075

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Let r1 , . . . , rh be a strictly increasing sequence of pairwise relatively prime positive integers, that is, ri < rj and (ri , rj ) = 1 for 1 ≤ i < j ≤ h. Let P be a positive integer such that (i) P ≥ rh − r1 , and (ii) if 1 ≤ i < j ≤ h and if the prime p divides rj − ri , then p divides P. For every positive integer k and for i = 1, . . . , h, we define the integers si,k = ri + kP .

(3)

Then s1,k < s2,k < · · · < sh,k , and s1,k ≥ s1,1 = r1 + P ≥ 2. Let 1 ≤ i < j ≤ h. If a prime number p divides both si,k and sj,k , then p divides sj,k − si,k = rj − ri and so p divides P. Therefore, p divides both ri and rj , which contradicts (ri , rj ) = 1. It follows that s1,k , s2,k , . . . , sh,k is a strictly increasing sequence of pairwise relatively prime positive integers. Moreover, if 1 ≤ i < j ≤ h, then 1 <

s j ,k

rj + kP

=

ri + kP rj − ri

s i ,k

= 1+

r1 + kP rh − r1

≤ 1+

r1 + kP P

≤ 1+

r1 + kP 1

< 1+ . k

If k ≥ k0 , then si,k < sj,k < 21/h si,k . For every positive integer k, we define the integer Sk =

h 

si,k =

i =1

h 

(ri + kP ).

(4)

i=1

Then sh1,k < Sk < shh,k , and shh,k < 2sh1,k < 2Sk for k ≥ k0 . For all k ≥ 1 we have 1 <

=

Sk+1

=

Sk h  

ri + kP

i=1

1+

i =1

<

 h   ri + (k + 1)P

 1+

1



P ri + kP

h

k

.

If k ≥ k0 , then Sk < Sk+1 < 2Sk . For every positive integer k and for i = 1, . . . , h, we define the integers ai , k =

Sk si,k

=

h 

sj,k .

(5)

j=1 j̸=i

Then a1,k > a2,k > · · · > ah,k . For k ≥ k0 , we have Sk ah,k

1/h

= sh,k < 21/h Sk .

(6)

Suppose that there is a prime number p that divides ai,k for all i = 1, . . . , h. Because p divides a1,k , it follows from (5) that p divides sℓ,k for some ℓ ∈ {2, 3, . . . , h}. Because p also divides aℓ,k , it follows from (5) that p divides sm,k for some m ∈ {1, 2, 3, . . . , h} \ {ℓ}. This is impossible because (sℓ,k , sm,k ) = 1. Therefore, a1,k , a2,k , . . . , ah,k is a strictly decreasing sequence of relatively prime positive integers.

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Lemma 1. Let h ≥ 2, and let r1 , . . . , rh be a strictly increasing sequence of pairwise relatively prime positive integers. For every positive integer k and i = 1, . . . , h, let si,k , Sk , and ai,k be the positive integers defined by (3)–(5), respectively. Let k1 ≥ k0 (h). For every positive integer t, there is a unique nonnegative integer ℓ(t ) defined by Sk1 +ℓ(t ) ≤ Skt 1 < Sk1 +ℓ(t )+1 .

(7)

Then ℓ(1) = 0 and the sequence {ℓ(t )}∞ t =1 is strictly increasing. Moreover, 1 Sk1 +ℓ(t )

<

2 Skt 1

.

(8)

Proof. Because the sequence {Sk }∞ k=1 is strictly increasing, for every positive integer t there is a unique nonnegative integer ℓ(t ) that satisfies inequality (7). In particular, Sk1 ≤ Sk11 < Sk1 +1 and so ℓ(1) = 0. If k ≥ k1 ≥ k0 , then Sk+1 < 2Sk and so Skt 1 < Sk1 +ℓ(t )+1 < 2Sk1 +ℓ(t ) ≤ 2Skt 1 < Skt 1+1 < Sk1 +ℓ(t +1)+1 . It follows that ℓ(t ) < ℓ(t + 1). Inequality (8) follows from Skt 1 < Sk1 +ℓ(t )+1 < 2Sk1 +ℓ(t ) . This completes the proof.



Lemma 2. Let h ≥ 2, and let r1 , . . . , rh be a strictly increasing sequence of pairwise relatively prime positive integers. For every positive integer k and i = 1, . . . , h, let si,k , Sk , and ai,k be the positive integers defined by (3)–(5), respectively. For every integer n such that (h − 1)Sk < n ≤ L, there exist positive integers v1 , v2 , . . . , vh such that a1,k v1 + a2,k v2 + · · · + ah,k vh = n and 1 ≤ vi ≤ si,k for i = 1, . . . , h − 1 and 1 ≤ vh < L/ah,k . Proof. By Lemma 1, the integers a1,k , a2,k , . . . , ah,k are relatively prime, and so there exist integers u1 , u2 , . . . , uh , not necessarily positive, such that a1,k u1 + a2,k u2 + · · · + ah,k uh = n. For each i = 1, 2, . . . , h − 1 there is a unique integer vi ∈ {1, 2, . . . , si,k } such that

vi ≡ ui (mod si,k ). Then n =

h 

ai,k ui

i=1

=

=

h−1 

ai,k vi +

i=1

i=1

h−1 

h −1 

ai,k vi +

i=1

=

h −1 

h −1 

ai,k (ui − vi ) + ah,k uh

 ai,k si,k

ai,k vi + Sk

h 

si,k

 h−1   ui − v i si,k

i =1

ai,k vi + ah,k sh,k

i=1

=

ui − v i

i=1

i=1

=

h −1 

i=1

where

i =1

s i ,k

+ ah,k uh

si,k

ai,k vi

 h−1   ui − vi

+ ah , k uh

 h −1   ui − v i

i=1

vh = sh,k



+ uh ∈ Z .

+ ah , k uh

M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075

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The inequality h−1 

ai , k v i ≤

h−1 

i =1

ai,k si,k = (h − 1)Sk

i =1

implies that if n > (h − 1)Sk , then ah , k v h = n −

h−1 

ai,k vi ≥ n − (h − 1)Sk > 0

i =1

and so vh ≥ 1. Moreover, ah,k vh < n ≤ L and so vh < L/ah,k . This completes the proof.



Theorem 1. Let h ≥ 2, and let r1 , . . . , rh be a strictly increasing sequence of pairwise relatively prime positive integers. For every positive integer k and i = 1, . . . , h, let si,k , Sk , and ai,k be the positive integers defined by (3)–(5), respectively. Let k1 ≥ k0 (h). For every positive integer t, let ℓ(t ) be the integer defined by (7), and let

 Vt =



ai,k1 +ℓ(t ) vi : vi ∈ 1, si,k1 +ℓ(t ) for i ∈ [1, h − 1] and vh ∈ 1,





(h − 1)Sk1 +ℓ(t +1) ah,k1 +ℓ(t )



.

Then: (i)

|Vt | <

2(h − 1)Sk1 +ℓ(t +1) ah,k1 +ℓ(t )

.

(ii) hVt ⊇ (h − 1)Sk1 +ℓ(t ) + 1, (h − 1)Sk1 +ℓ(t +1) .





(iii) For every positive integer t1 , the set

At1 =

∞ 

Vt

t = t1

is an asymptotic basis of order h. (iv) The set

A = [0, Sk1 ] ∪

∞ 

Vt

t =1

is a basis of order h. Proof. For i = 1, . . . , h − 1 we have si,k1 +ℓ(t ) =

Sk1 +ℓ(t ) ai,k1 +ℓ(t )

<

Sk1 +ℓ(t +1) ah,k1 +ℓ(t )

and so

|Vt | ≤

h −1 

si,k1 +ℓ(t ) +

(h − 1)Sk1 +ℓ(t +1) ah,k1 +ℓ(t )

i=1

<

2(h − 1)Sk1 +ℓ(t +1) ah,k1 +ℓ(t )

.

This proves (i). Lemma 2 implies statement (ii), and statement (ii) implies that the h-fold sumset hAt1 contains every integer n ≥ (h − 1)Sk1 +ℓ(t1 ) + 1. Thus, At1 is an asymptotic basis of order h. In particular, hA1 contains every integer n ≥ (h − 1)Sk1 + 1. Because

[0, (h − 1)Sk1 ] ⊆ [0, hSk1 ] = h[0, Sk1 ] it follows that A = [0, Sk1 ] ∪ A1 is a basis of order h. This completes the proof. Theorem 2. The set

A = [0, Sk1 ] ∪

∞ 

Vt

t =1

constructed in Theorem 1 is a thin basis of order h.



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M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075

Proof. Let A(x) be the counting function of the set A. For all x ≥ Sk1 , there is a smallest positive integer t ∗ such that Sk1 +ℓ(t ∗ ) ≤ x. Then ∗

Sk1 +ℓ(t ∗ ) ≤ x < Sk1 +ℓ(t ∗ +1) ≤ Skt 1 +1 and 1 ∗

Skt 1 +1

<

1 x

.

(9)

Because Vt (x) ≤ |Vt | for all x, we have A(x) ≤ Sk1 +

∞ 

Vt (x) ≤ Sk1 +

t∗ 

Vt (x).

t =t ∗ +1

t =1

t =1

∞ 

|Vt | +

We shall estimate the two sums separately. From Theorem 1 (i) and inequality (6), we obtain t∗ 

|Vt | ≤

t =1

t∗  2(h − 1)Sk

1 +ℓ(t +1)

ah,k1 +ℓ(t )

t =1

≤ 2(h − 1)

t∗  t =1

Skt 1+1 ah,k1 +ℓ(t ) t∗

= 2(h − 1)Sk1



Skt 1

t =1

ah,k1 +ℓ(t )

t∗

< 2(h − 1)Sk1

 Sk

1 +ℓ(t )+1

t =1

≤ 4(h − 1)Sk1

t∗  Sk1 +ℓ(t ) t =1

2+1/h

<2

ah,k1 +ℓ(t )

ah,k1 +ℓ(t )

(h − 1)Sk1

t∗ 

1/h

Sk +ℓ(t ) 1

t =1 2+1/h

≤2

(h − 1)Sk1

t∗ 

t /h

Sk1

t =1

 <

1/h Sk1

 < <

−1 1+1/h

2+2/h

2





−1 1+1/h

(h − 1)Sk1



1/h

Sk +ℓ(t ∗ ) 1

1/h

1+1/h

22+2/h (h − 1)Sk1 1/h Sk1

1/h

Sk +ℓ(t ∗ )+1 1

Sk1 − 1



t ∗ /h

Sk1

22+1/h (h − 1)Sk1 1/h Sk1



≤

1+1/h

21+1/h h(h − 1)Sk1

 x1/h .

−1

To estimate the second sum, we observe that if ai,k1 +ℓ(t ) vi ∈ Vt and ai,k1 +ℓ(t ) vi ≤ x, then 1 ≤ vi ≤

x ai,k1 +ℓ(t )

≤

x ah,k1 +ℓ(t )

.

Applying inequalities (6), (8) and (9), we obtain ∞  t =t ∗ +1

Vt ( x ) ≤

∞  h 

x

ai,k1 +ℓ(t ) t =t ∗ +1 i =1

< hx

∞ 

1

t =t ∗ +1

ah,k1 +ℓ(t )

M.B. Nathanson / Discrete Mathematics 312 (2012) 2069–2075

< 21/h hx

∞  

∞  

 =

<

1 (1−1/h)t

t =t ∗ +1 Sk1 1−1/h

2hSk1

1−1/h

1−1/h

1−1/h

This completes the proof.

−1

x (1−1/h)(t ∗ +1)

Sk1



x x1−1/h

−1

2hSk1 Sk1



−1

1−1/h 2hSk1

Sk1

 =

∞ 

1−1/h Sk1



1−1/h

2 Sk1 +ℓ(t )+1

t =t ∗ +1

< 2hx

1 Sk1 +ℓ(t )

t =t ∗ +1

< 21/h hx

2075

1−1/h

 x1/h .



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