Representation of Integers by Near Quadratic Sequences

Labib Haddad 120 rue de Charonne 75011 Paris France

                labib.haddad@wanadoo.fr

Charles Helou Department of Mathematics Pennsylvania State University 25 Yearsley Mill Road Media, PA 19063 USA

                     cxh22@psu.edu

Abstract

Following a statement of the well-known Erdős-Turán conjecture, Erdős mentioned the following even stronger conjecture: if the n-th term a_{n} of a sequence A of positive integers is bounded by αn^{2}, for some positive real constant α, then the number of representations of n as a sum of two terms from A is an unbounded function of n. Here we show that if a_{n} differs from αn^{2} (or from a quadratic polynomial with rational coefficients q(n)) by at most o(\sqrt{log n}), then the number of representations function is indeed unbounded.

1 Introduction

In 1941, Erdős and Turán [5] conjectured that if a sequence A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } of positive integers is an asymptotic basis of the set ℕ = {0, 1, 2, … } of natural numbers, i.e., if all large enough integers n are sums of two terms from A, then the number of representations r_{A}(n) = |{(a_{i}, a_{j}) ∈ A × A : a_{i} + a_{j} = n}| of n, as a sum of two terms from A, is unbounded. This is the well-known “Erdős-Turán conjecture”. A few years later (the earliest we are aware of), in 1955 and 1956, Erdős [6], and Erdős and Fuchs [7] asserted that an even stronger conjecture would be that if a_{n} ≤ αn^{2}, for all n, with a real constant α > 0, then lim sup r_{A}(n) = ∞. This came to be known as the “generalized Erdős-Turán conjecture”. It is indeed stronger than the former one, since if A is an asymptotic basis of ℕ, then a_{n} ≪ n^{2} [13, p. 105].

Much work has been done concerning the “Erdős-Turán conjecture”, e.g., [3, 7, 8, 16, 1, 21, 19], including disproofs of analogues of this conjecture in many semigroups other than ℕ, e.g., [20, 16, 17, 11, 12, 2, 14]. In contrast, much less has been done about the “generalized Erdős-Turán conjecture”. In a previous, co-authored, paper [9], we studied the class of sequences that can replace {αn^{2}} in the condition a_{n} ≤ αn^{2} for all n, to imply that r_{A}(n) is unbounded, and we gave several statements equivalent to the “generalized Erdős-Turán conjecture”. In particular, we showed that if the conjecture holds with α = 1, then it holds with any α > 0. Moreover, it is not difficult to see that if a_{n} = o(n^{2}), then the conjecture holds [9, 10]. So we can essentially focus on the case where a_{n} is not too small compared to n^{2}, while bounded by a constant multiple of n^{2}. In particular, we can consider the case where a_{n} is, in a sense, “close” to a constant multiple of n^{2}, or to a quadratic polynomial in n. This is basically the goal of the present paper. We thus show that if |a_{n} −αn^{2}| = o (\sqrt{log n}), with a real constant α > 0, or if |a_{n} − q(n)| = o(\sqrt{log n}), where q(n) is a quadratic polynomial with rational coefficients, then the representation function r_{A}(n) of A is unbounded.

2 Technical tools

Let C = {c_{1} < c_{2} < ⋯ < c_{n} < ⋯ } ⊂ ℝ^{+} be a strictly increasing sequence, in the set ℝ^{+} of real numbers ≥ 0. For any x ∈ ℝ^{+}, let C[x] = C ∩ [0, x] = {c ∈ C : c ≤ x}, and C (x) = |C[x]| the cardinality of C[x]. Note that C(x) is finite for every x ≥ 0 if and only if the sequence C is unbounded. This is in particular true when c_{n+1} − c_{n} ≥ 1 for large enough n, and more particularly if C is a subset of the set ℕ = {0, 1, 2, 3, …} of natural numbers.

The sumset C + C is defined by C + C = {c + d : (c, d) ∈ C × C} .

Now let A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ be a strictly increasing sequence of natural numbers. In addition to the above notions, valid for A as for C, the representation function r_{A} of A is defined by r_{A}(n) = |{(a, b) ∈ A × A : a + b = n}| , for n ∈ ℕ, and we set

s (A) = sup_{n∈ℕ} r_{A} (n) , in \overline{ℕ} = ℕ ∪ \{∞\} .

In the sequel, i, j, k, l, m, n generally denote positive integers, unless it is specified that they lie in ℕ, i.e., that they are integers ≥ 0, while x, y denote real numbers ≥ 0, i.e., they lie in ℝ^{+}.

Note that if A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ^{∗}, where ℕ^{∗} = {1, 2, 3, …} is the set of positive integers, then a_{n} ≥ n for all n ∈ ℕ^{∗}.

For any x ∈ ℝ^{+}, let

U_{A} (x) = |{(a, b) ∈ A × A : a + b ≤ x}| = ∑_{0≤n≤x} r_{A}(n).  (1)

Then

U_{A}(x) = ∑_{n∈(A+A)[x]} r_{A}(n) ≤ ∑_{n∈(A+A)[x]} s(A) = (A + A) (x) · s(A)  (2)

and

A (x)^{2} = |{(a, b) ∈ A × A : a, b ≤ x}| ≤ |{(a, b) ∈ A × A : a + b ≤ 2x}| = U_{A} (2x) ≤
≤ (A + A) (2x) · s(A),  (3)

so that, for all x ∈ ℝ^{+},

\frac{(A + A) (2x)}{A (x)^{2}}s(A) ≥ 1.  (4)

Define

h (A) = lim inf_{x→∞} \frac{(A + A) (2x)}{A (x)^{2}}.  (5)

Lemma 1. If h (A) = 0, then s (A) = ∞.

Proof. This follows immediately from (4).

Corollary 2. If lim inf_{n→∞} \frac{A(x)}{\sqrt{x}} > 0 and lim inf_{n→∞} \frac{(A + A) (x)}{x} = 0, then h (A) = 0, and therefore

s (A) = ∞.

Proof. By assumption, lim sup_{n→∞} \frac{\sqrt{x}}{A(x)} = _{\frac{1}{lim inf_{n→∞} \frac{A(x)}{\sqrt{x}}}} is finite, while lim inf_{n→∞} \frac{(A+A)(2x)}{2x} = 0. So, using

properties of the lower and upper limits, we get

h(A) = lim inf_{x→∞} \frac{(A + A)(2x)}{A(x)^{2}} = 2 lim inf_{x→∞} \frac{(A + A)(2x)}{2x} (\frac{\sqrt{x}}{A(x)})^{2} ≤

≤ 2 (lim inf_{x→∞} \frac{(A + A)(2x)}{2x}) · (lim sup_{x→∞} \frac{\sqrt{x}}{A(x)})^{2} = 0.

The conclusion follows from Lemma 2.1.

Lemma 3. Let A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ^{∗} be a strictly increasing sequence of positive integers, and C = {c_{1} < c_{2} < ⋯ < c_{n} < ⋯ } ⊂ ℝ^{+}. For x ∈ ℝ^{+}, set e(x) = sup_{n≤x} |a_{n} − c_{n}| . We then have, for all x ∈ ℝ^{+},

(A + A) (x) ≤ (4e (x) + 1) · (C + C) (x + 2e (x)) .  (6)

If we further assume that c_{1} ≥ 1 and c_{n+1} − c_{n} ≥ 1 for all n ≥ 1, we then also have, for all x ∈ ℝ^{+},

A(x) ≥ C (x − e(x)) .  (7)

Proof. Note first that the function e(x) is increasing, in the sense that x ≤ y implies e(x) ≤ e(y).

Note also that, since A ⊂ ℕ^{∗}, we have i ≤ a_{i} for all i. So, for n ≤ x, if n = a_{i} + a_{j}, then i ≤ a_{i} ≤ n ≤ x and similarly j ≤ x, and therefore |n − c_{i} − c_{j}| = |a_{i} + a_{j} − c_{i} − c_{j}| ≤ |a_{i} − c_{i}| + |a_{j} − c_{j}| ≤ 2e(x). Hence

(A + A) [x] = {n ≤ x : ∃i, j, n = a_{i} + a_{j}} ⊂ {n ≤ x : ∃i, j, |n − c_{i} − c_{j}| ≤ 2e(x)},

and setting s = c_{i} + c_{j}, we get s ∈ C + C and |n − s| ≤ 2e(x), so that s ≤ n + 2e(x) ≤ x + 2e(x), and therefore

{n ≤ x : ∃i, j, |n − c_{i} − c_{j}| ≤ 2e(x)} ⊂ {n : ∃s ∈ (C + C)[x + 2e(x]), |n − s| ≤ 2e(x)}.

Thus

(A + A) [x] ⊂ ⋃_{s∈(C+C)[x+2e(x)]} ([s − 2e(x), s + 2e(x)] ∩ ℕ) ,

and therefore

(A + A) (x) ≤ ∑_{n∈(C+C)[x+2e(x)]} (4e(x) + 1) = (C + C) (x + 2e(x)) · (4e(x) + 1) .

This proves (6).

Now, if c_{1} ≥ 1 and c_{n+1} − c_{n} ≥ 1 for all n, then c_{n} ≥ n for all n. So if c_{n} ≤ x − e(x), then n ≤ c_{n} ≤ x, so that |a_{n} − c_{n}| ≤ e(x), and therefore a_{n} ≤ c_{n} + e(x) ≤ x.

Hence {n : c_{n} ≤ x − e(x)} ⊂ {n : a_{n} ≤ x}, and thus

C (x − e(x)) = |{n : c_{n} ≤ x − e(x)}| ≤ |{n : a_{n} ≤ x}| = A(x),

which proves (7).

Lemma 4. Let A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ^{∗} and C = {c_{1} < c_{2} < ⋯ < c_{n} < ⋯ } ⊂ ℝ^{+} be two strictly increasing sequences in ℕ^{∗} and in ℝ^{+}, respectively. For x ∈ ℝ^{+}, set e(x) = sup_{n≤x} |a_{n} − c_{n}| . Assume that e(x) is not identically zero, and that c_{1} ≥ 1 and

c_{n+1} − c_{n} ≥ 1 for all n ≥ 1. Then the condition

lim inf_{x→∞} \frac{e (2x) · (C + C) (2x + 2e (2x))}{C(x − e(x))^{2}} = 0  (H)

implies that h (A) = 0, and therefore s (A) = ∞.

Proof. Since e(x) is increasing and not identically zero, there exists a real constant t > 0

such that e(x) ≥ \frac{1}{t} for large enough x. In view of the inequalities (6) and (7) in Lemma 2.3, we have

\frac{(A + A) (2x)}{A (x)^{2}} ≤ \frac{(4e (2x) + 1) · (C + C) (2x + 2e (2x))}{C (x − e(x))^{2}}.

Moreover, for large enough x, we have t·e(2x) ≥ 1, and therefore 4e (2x)+1 ≤ (4 + t)·e (2x) . Thus

\frac{(A + A) (2x)}{A (x)^{2}} ≤ (4 + t) \frac{e (2x) · (C + C) (2x + 2e (2x))}{C(x − e(x))^{2}},

for large enough x, so that the condition (H) implies that lim inf_{x→∞} \frac{(A + A) (2x)}{A (x)^{2}} = 0, i.e.,

h (A) = 0, and therfore, by Lemma 2.1, s (A) = ∞.

Remark 5. The scope of Lemma 2.4 is broader than it seems to be. Indeed, for a subset A of ℕ, modifying, removing or adding finitely many elements does not modify the fact that s(A) is infinite or finite. Thus Lemma 2.4 can be used in more general situations than specified by its assumptions, as shown by the next result.

Fundamental Lemma 6. Let B = {b_{1} < b_{2} < ⋯ < b_{n} < ⋯ } ⊂ ℕ and D = {d_{1} < d_{2} < ⋯ < d_{n} < ⋯ } ⊂ ℝ^{+} be two strictly increasing sequences in ℕ and in ℝ^{+} respectively. Assume that there exists an increasing function f : ℝ^{+} → ℝ^{+} and a positive integer m such

that d_{m} ≥ 1, d_{n+1} − d_{n} ≥ 1 for n ≥ m, and sup_{m≤n≤x} |b_{n} − d_{n}| ≤ f(x) for x ≥ m. Then the

condition

lim inf_{x→∞} \frac{f (2x) · (D + D) (2x + 2f (2x))}{D(x − f(x))^{2}} = 0  (K)

implies that s (B) = ∞.

Proof. For n ∈ ℕ^{∗}, set a_{n} = b_{n+m} and c_{n} = d_{n+m}, and let A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ^{∗} and C = {c_{1} < c_{2} < ⋯ < c_{n} < ⋯ } ⊂ ℝ^{+} be the strictly increasing sequences, in ℕ^{∗} and ℝ^{+}, obtained by deleting the first m terms of B and D respectively. Then c_{1} = d_{m+1} ≥ 2

and c_{n+1} − c_{n} = d_{n+m+1} − d_{n+m} ≥ 1 for n ≥ 1. Moreover, setting e (x) = sup_{n≤x} |a_{n} − c_{n}| , for

x ∈ ℝ^{+}, and using the assumptions on B and D, we have

e(x) = sup_{n≤x} |a_{n} − c_{n}| = sup_{n≤x} |b_{n+m} − d_{n+m}| = sup_{m<i≤x+m} |b_{i} − d_{i}| ≤ f(x + m).

Thus, setting y = x+m, we have e(x) ≤ f(y), and since the functions e and f are increasing,

e (2x) ≤ f (2x + m) ≤ f (2y) .

Also, taking into account that C ⊂ D and C +C ⊂ D+D, so that (C + C) (t) ≤ (D + D) (t) for all t ∈ ℝ^{+}, and that the function t ↦ (C + C)(t) is increasing, we get

(C + C) (2x + 2e (2x)) ≤ (C + C) (2y + 2f (2y)) ≤ (D + D) (2y + 2f (2y)) .

Thus

e (2x) · (C + C) (2x + 2e (2x)) ≤ f (2y) · (D + D) (2y + 2f (2y)) ,  (8)

for x ∈ ℝ^{+}, and y = x + m.

Moreover, for t ≥ m, we have

D(t) − C (t) = |{d_{n} ∈ D : d_{n} ≤ t}| − |{c_{n} ∈ C : c_{n} = d_{n+m} ≤ t}| = m

and

C (t) − C (t − m) = |{c_{n} ∈ C : t − m < c_{n} ≤ t}| ≤ m,

since c_{n+1} − c_{n} ≥ 1 for all n ∈ ℕ^{∗}, so that C (t) ≤ C (t − m) + m and D (t) = C (t) + m ≤ C (t − m) + 2m. Therefore C (t − m) ≥ D (t) − 2m for t ≥ m. Hence, taking into account that the function t ↦ C(t) is increasing and that e(x) ≤ f(y) we get, for large enough x,

C (x − e(x)) ≥ C (x − f (y)) = C (y − m − f(y)) ≥ D (y − f(y)) − 2m.  (9)

It follows from (8) and (9) that, for large enough x and for y = x + m,

\frac{e (2x) · (C + C) (2x + 2e (2x))}{C(x − e(x))^{2}} ≤ \frac{f (2y) · (D + D) (2y + 2f (2y))}{(D (y − f(y)) − 2m)^{2}}.  (10)

Set P (x) = f (2x) · (D + D) (2x + 2f (2x)) and Q (x) = D (x − f(x)), and suppose that the

condition (K) is satisfied, i.e., that lim inf_{x→∞} \frac{P (x)}{Q (x)^{2}} = 0. Then there exists a strictly increasing

sequence (x_{n})_{n≥1} in ℝ^{+}, tending to infinity, such that lim_{n→∞} \frac{P (x_{n})}{Q (x_{n})^{2}} = 0. Since P (x) is an

increasing unbounded function, lim_{n→∞} P (x_{n}) = ∞, and therefore the sequence (Q (x_{n}))_{n≥1} is unbounded. So there exists a subsequence (x_{n_{k}})_{k∈ℕ^{∗}} of (x_{n})_{n≥1} such that lim_{k→∞} Q (x_{n_{k}}) = ∞,

while lim_{k→∞} \frac{P (x_{n_{k}})}{Q (x_{n_{k}})^{2}} = 0. Hence lim_{k→∞} \frac{P (x_{n_{k}})}{(Q (x_{n_{k}}) − 2m)^{2}} = 0, and therefore

lim inf_{y→∞} \frac{f (2y) · (D + D) (2y + 2f (2y))}{(D (y − f(y)) − 2m)^{2}} = lim inf_{x→∞} \frac{P (x)}{(Q (x) − 2m)^{2}} = 0.

It then follows from (10) that lim inf_{x→∞} \frac{e (2x) · (C + C) (2x + 2e (2x))}{C(x − e(x))^{2}} = 0. Thus the condition

(H) of Lemma 2.4 holds, and therefore, in view of this Lemma, s(A) = ∞. As A ⊂ B, it follows that s (B) = ∞ too.

Remark 7. In the statement of Lemma 2.6, we may replace D by D′ = D + γ, i.e., d_{n} by d′_{n} = d_{n} + γ (n ∈ ℕ^{∗}), where γ is any fixed real number, since a translation of the general term of D does not affect the condition (K).

3 Main results

Theorem 8. Let A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ be a strictly increasing sequence of natural numbers, and q(x) = αx^{2} with a real number α > 0. If the function e(x) = sup_{n≤x} |a_{n} − q(n)| ( x ∈ ℝ^{+}) satisfies e (x) = o (\sqrt{log x}) as x → ∞, then s (A) = ∞.

Proof. We apply Lemma 2.6 to B = A and D = {q(1) < q(2) < ⋯ < q(n) < ⋯ }. Indeed, the sequence (q(n))_{n≥1} is strictly increasing and unbounded, with q(n+1)−q(n) = α (2n + 1) unbounded too, so that q(n) ≥ 1 and q(n + 1) − q(n) ≥ 1 for large enough n. There remains to show that the condition (K) holds for f(x) = e(x).

Let S = {n^{2} : n ∈ ℕ^{∗}} . By a classical result of Landau [15], there exists a constant c > 0 such that (S + S) (x) ∼ c\frac{x}{\sqrt{log x}} as x → ∞.

For m, n ∈ ℕ^{∗} and x ∈ ℝ^{+}, as q(m) + q(n) ≤ x is equivalent to m^{2} + n^{2} ≤ \frac{x}{α}, we have

(D + D) (x) = (S + S) (\frac{x}{α}) ∼ \frac{c}{α} \frac{x}{\sqrt{log x}}, so that

(D + D) (x) ≤ c_{1} \frac{x}{\sqrt{log x}},

for large enough x, with a constant c_{1} > \frac{c}{α}.

Moreover, as q(n) ≤ x if and only if n ≤ \sqrt{\frac{x}{α}}, we also have D(x) = [\sqrt{\frac{x}{α}}] > \sqrt{\frac{x}{α}} − 1.

It follows that, for large enough x,

\frac{e (2x) · (D + D) (2x + 2e (2x))}{D(x − e(x))^{2}} ≤\frac{c_{1} · e (2x) · (2x + 2e (2x))}{\sqrt{log (2x + 2e (2x))} (\sqrt{\frac{x−e(x)}{α}} − 1)^{2}} =

= \frac{c_{1}α · e (2x) · (2x + 2e (2x))}{\sqrt{log (2x + 2e (2x))} (\sqrt{x − e(x)} − \sqrt{α})^{2}} .

As e(x) = o (\sqrt{log x}),

\frac{e (2x) · (2x + 2e (2x))}{\sqrt{log (2x + 2e (2x))} (\sqrt{x − e(x)} − \sqrt{α})^{2}} ∼ \frac{2x · e (2x)}{\sqrt{log (2x)} · x} ∼ \frac{2e(2x)}{\sqrt{log (2x)}},

and, since e (x) = o (\sqrt{log x}), we have lim_{x→∞} \frac{2e(2x)}{\sqrt{log (2x)}} = 0. Therefore

lim_{x→∞} \frac{e (2x) · (D + D) (2x + 2e (2x))}{D(x − e(x))^{2}} = 0,

and the condition (K) holds. Thus, by Lemma 2.6, s (B) = ∞, i.e., s (A) = ∞.

Remark 9. In the statement of Theorem 3.1, we may replace q(x) = αx^{2} by q(x) = αx^{2} + γ, where γ is any real constant, in view of Remark 2.7.

Also, if A = {a_{n} = [αn^{2} + γ] : n ∈ ℕ} is the set of the integral parts [αn^{2} + γ] = [q(n)], then s(A) = ∞, since e(x) = sup_{n≤x} |a_{n} − q(n)| ≤ 1 trivially satisfies the condition in Theorem

3.1.

Theorem 10. Let A = {a_{1} < a_{2} < ⋯ < a_{n} < ⋯ } ⊂ ℕ and q(x) be a quadratic polynomial with rational coefficients and positive leading coefficient. If the function e(x) = sup_{n≤x} |a_{n} − q(n)| ( x ∈ ℝ^{+}) satisfies e (x) = o (\sqrt{log x}) as x → ∞, then s (A) = ∞.

Proof. As q(x) has rational coefficients, there exist integers a, b, c, d, with a, d > 0, such that dq(x) = (ax + b)^{2} + c.

Let b_{n} = da_{n} − c and d_{n} = (an + b)^{2}, for n ∈ ℕ^{∗}. Clearly, there exists m ∈ ℕ^{∗} such that b_{m} ≥ 1, d_{m} ≥ 1 and d_{n+1} −d_{n} ≥ 1 for n ≥ m. Set B = {b_{n} : n ≥ m} and D = {d_{n} : n ≥ m}. Then B and D are strictly increasing sequences in ℕ, and, for all n ≥ m,

|d_{n} − b_{n}| = |(an + b)^{2} − da_{n} + c| = d|q(n) − a_{n}|.

For x > m, Let f(x) = sup_{m≤n≤x} |d_{n} − b_{n}|, for x ∈ ℝ^{+}. Then f(x) is an increasing nonnegative

function satisfying f(x) ≤ d · e(x), so that f(x) = o (\sqrt{log x}) (like e(x)). Thus, we may apply Lemma 2.6, provided we show that the condition (K) is satisfied.

Let S = {n^{2} : n ∈ ℕ}. Then D ⊂ S, and therefore D + D ⊂ S + S, so that (D + D)(x) ≤ (S + S)(x), for x ∈ ℝ^{+}.

By Landau’s theorem [15], (S + S) (x) ∼ c_{0} \frac{x}{\sqrt{log x}}, with a constant c_{0} > 0. So there

exists a constant c_{1} > 0 such that (D + D)(x) ≤ (S + S)(x) ≤ c_{1} \frac{x}{\sqrt{log x}}, and therefore

(D + D) (2x + 2f(2x)) ≤ c_{1} \frac{2x + 2f(2x)}{\sqrt{log (2x + 2f(2x))}}.  (11)

Moreover, for x > max(m, b^{2}), if n ≤ \frac{\sqrt{x}−|b|}{a}, then d_{n} = (an + b)^{2} ≤ x. Hence, for large enough x,

D (x) = |\{n ≥ m : d_{n} ≤ x\}| ≥ |\{n ≥ m : n ≤ \frac{\sqrt{x} − |b|}{a}\}|

≥ \frac{\sqrt{x} − |b|}{a} − m ≥ c_{2}\sqrt{x} − c_{3},

with constants c_{2}, c_{3} > 0, and therefore

D (x − f(x)) ≥ c_{2}\sqrt{x − f(x)} − c_{3}.  (12)

It follows from (11) and (12) that, for large enough x,

\frac{f (2x) · (D + D) (2x + 2f (2x))}{D (x − f(x))^{2}} ≤ c_{1} \frac{f (2x) · (2x + 2f (2x))}{\sqrt{log (2x + 2f (2x))} (c_{2}\sqrt{x − f(x)} − c_{3})^{2}} ,

and, since f (x) = o (\sqrt{log x}) , we have

\frac{f (2x) · (2x + 2f (2x))}{\sqrt{log (2x + 2f (2x))} (c_{2}\sqrt{x − f(x)} − c_{3})^{2}} ∼ \frac{2f(2x)}{c_{2}^{2}\sqrt{log x}} = o(1).

Therefore

lim inf_{x→∞} \frac{f (2x) · (D + D) (2x + 2f (2x))}{D (x − f(x))^{2}} = 0.

Thus the condition (K) is satisfied, and by Lemma 2.6, s(B) = ∞. As B is a translate of a homothetic of a subsequence A_{m} = {a_{n} : n ≥ m} of A, namely B = d·A_{m} +|c|, we conclude, e.g., see [9], that s(A_{m}) = s(B) = ∞, and therefore s (A) = ∞.

4 Acknowledgment

We thank an anonymous reader who suggested the use of Landau’s theorem to improve a previous result.

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2010 Mathematics Subject Classification: Primary 11B34; Secondary 11B83. Keywords: sequences, representation functions, quadratic, Erdős-Turán conjecture.

Received July 19 2012; revised version received October 14 2012. Published in Journal of Integer Sequences, October 23 2012.

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