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\begin{center}
\vskip 1cm{\LARGE\bf 
Representation of Integers by Near Quadratic \\
\vskip .1in
Sequences
}
\vskip 1cm
\large
Labib Haddad \\
120 rue de Charonne \\
75011 Paris \\
France  \\
\href{mailto:labib.haddad@wanadoo.fr}{\tt labib.haddad@wanadoo.fr} \\
\ \\
Charles Helou \\
Department of Mathematics \\
Pennsylvania State University \\
25 Yearsley Mill Road \\
Media, PA 19063 \\
USA \\
\href{mailto:cxh22@psu.edu}{\tt cxh22@psu.edu}\\
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\begin{abstract}
Following a statement of the well-known Erd\H os-Tur\' an conjecture,
Erd\H os mentioned the following even stronger conjecture: if the
$n$-th term $a_n$ of a sequence $A$ of positive integers is bounded by
$\alpha n^2$, for some positive real constant $\alpha$, 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 $\alpha n^2$ (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.
\end{abstract}

\bigskip


\section{Introduction} \label{sec: Intro}


In 1941, Erd\H{o}s and Tur\'{a}n \cite {ET} conjectured that if a sequence $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}$ of positive 
integers is an asymptotic basis of the set $\mathbb N = \{ 0, 1, 2, \dots \}$ of natural numbers, i.e., if all large enough integers $n$ are sums of two terms from $A$, 
then the number of representations $r_A(n) = |\{ (a_i, a_j) \in A \times A : a_i + a_j = n \}|$ of $n$, as a sum of two terms from $A$, is unbounded. This is the well-known 
``Erd\H{o}s-Tur\'{a}n conjecture''.  A few years later (the earliest we are aware of), in 1955 and 1956, Erd\H os \cite {E}, and Erd\H os and Fuchs \cite {EF} asserted that 
an even stronger conjecture would be that if $a_n \leq \alpha n^2$, for all $n$, with a real constant $\alpha >0$, then $\limsup r_A(n) = \infty$. This came to be known as 
the ``generalized Erd\H{o}s-Tur\'{a}n conjecture''. It is indeed stronger than the former one,
since if $A$ is an asymptotic basis of $\mathbb N$, then $a_n \ll n^2$ 
\cite[p.\ 105]{HR}. 

Much work has been done concerning 
the ``Erd\H{o}s-Tur\'{a}n conjecture'', e.g., \cite {Di, EF, GHHP1, N1, BCC, S, NS}, 
including disproofs of analogues of this conjecture in many semigroups other than $\mathbb N$, e.g., \cite {P, N1, N2, HH1, HH2, C, KL}. 
In contrast, much less has been done about the ``generalized Erd\H{o}s-Tur\'{a}n conjecture''. In a previous, co-authored, paper \cite {GHHP2}, we studied the class 
of sequences that can replace $\{ \alpha n^2 \}$ in the condition $a_n \leq \alpha n^2$ for all $n$, to imply that $r_A(n)$ is unbounded, 
and we gave several statements equivalent to the ``generalized Erd\H{o}s-Tur\'{a}n conjecture''. In particular, we showed that if the conjecture holds with $\alpha =1$, 
then it holds with any $\alpha >0$. Moreover, it is not difficult to see that if $a_n = o(n^2)$, then the conjecture holds \cite {GHHP2, GHHP3}. 
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 $|a_{n}-\alpha n^{2}| = o\left( \sqrt{\log n}\right)$, with a real constant $\alpha >0$, 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. 





\section{ Technical tools}\label{Technical}



Let $C=\{c_1 < c_2 < \cdots <c_n < \cdots \}\subset \mathbb R^+$ be a strictly increasing sequence, in the set 
$\mathbb{R}^{+}$ of real numbers $\geq 0.$ For any $x\in \mathbb{R}^{+},$ let  
$C[x] =C \cap [ 0, x] =\left\{ c\in C: c\leq x \right\}$, and $C\left( x\right) =\left\vert C[x]\right\vert $ the
cardinality of $C[x] .$ Note that $C(x)$ is finite for every $x\geq 0$ if and only if the sequence $C$ is unbounded. 
This is in particular true when $c_{n+1}-c_{n}\geq 1$ for large enough $n,$ and more particularly if $C$ is a subset of the set 
$\mathbb{N} =\{0,1,2,3,\ldots \}$ of natural numbers.

The sumset $C+C$ is defined by $C+C=\left\{ c+d:\left( c,d\right) \in C\times C\right\} .$

Now let $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb{N}$ \ 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  $r_{A}(n)=\left\vert \left\{ (a,b)\in A\times A:a+b=n\right\} \right\vert ,$ 
for $n\in \mathbb{N}$, and we set $\ s\left( A\right) =\sup\limits_{n\in \mathbb{N}}r_{A}\left( n\right) ,$ in 
$\overline{\mathbb N} = \mathbb N \cup \left\{ \infty \right\} .$ 

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

Note that if $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb{N}^{\ast },$ where $\mathbb{N}^{\ast }=\{1,2,3,\ldots \}$ 
is the set of positive integers, then \ $a_{n}\geq n$ for all $n\in \mathbb{N}^{\ast }.$


For any $x\in \mathbb{R}^{+},$ let 
\begin {equation}\label{U1} 
U_{A}\left( x\right) =\left\vert \left\{ (a,b)\in A\times A:a+b\leq x\right\} \right\vert =\sum\limits_{0\leq n\leq x}r_{A}(n). 
\end {equation}
Then
\begin {equation}\label{U2} 
U_{A}(x)=\sum_{n\in (A+A)[x]}r_{A}(n)\leq \sum\limits_{n\in (A+A)[x]}s(A)=\left( A+A\right) (x)\cdot s(A)
\end {equation} 
and

\begin{align}\label{A & A+A}
A\left( x\right) ^{2} & =\left\vert \left\{ \left( a,b\right) \in
A\times A:a,b\leq x\right\} \right\vert \leq \left\vert \left\{ (a,b)\in
A\times A:a+b\leq 2x\right\} \right\vert =U_{A}\left( 2x\right) \leq  \notag \\ 
& \leq  \left( A+A\right) (2x)\cdot s(A), 
\end{align}

so that, for all $x\in \mathbb{R}^{+}$, 
\begin {equation} \label{s(A)}
\dfrac{\left( A+A\right) (2x)}{A\left( x\right) ^{2}}s(A)\geq 1.
\end {equation}

Define 
\begin {equation} \label{h(A)}
h\left( A\right) =\,\underset{x\rightarrow \infty }{\lim \inf }\dfrac{\left( A+A\right) (2x)}{A\left( x\right) ^{2}}.
\end {equation}


\begin{lemma} \label{Lem1}
If $h\left( A\right) =0$,  then $s\left( A\right) =\infty$.
\end{lemma}
\begin{proof}
This follows immediately from \eqref{s(A)}.  
\end{proof}


\begin{corollary} \label{CorLem1} 
If $\underset{n\rightarrow\infty }{\liminf }\dfrac{A(x)}{\sqrt{x}}>0$  and $\underset{n\rightarrow\infty }{\lim \inf }\dfrac{\left( A+A\right) (x)}{x}=0$, 
then $h\left( A\right) =0$, and therefore $s\left( A\right) =\infty$.
\end{corollary}
\begin{proof}
By assumption, $\underset{n\rightarrow\infty }{\limsup}  \frac {\sqrt x}{A(x)} = \frac 1 {\underset{n\rightarrow\infty }{\liminf} \frac{A(x)}{\sqrt x}}$ is finite, 
while $\underset{n\rightarrow\infty }{\liminf} \frac {(A+A)(2x)}{2x} =0$. So, using properties of the lower and upper limits, we get 
\begin{align}
h(A) & = \liminf_{x\to \infty} \frac {(A+A)(2x)}{A(x)^2} = 2  \liminf_{x\to \infty}  \frac {(A+A)(2x)}{2x}  \left( \frac {\sqrt x}{A(x)} \right)^2  \leq    \notag \\  
& \leq 2 \left( \liminf_{x\to \infty}  \frac {(A+A)(2x)}{2x} \right) \cdot \left( \limsup_{x\to \infty} \frac {\sqrt{x}}{A(x)} \right)^2 =  0.  \notag 
\end{align}
The conclusion follows from Lemma 2.1. 
\end{proof}

\begin{lemma} \label{Lem2}
Let $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb{N}^{\ast }$ be a strictly increasing sequence of positive integers, and 
$C=\{c_{1}<c_{2}<\cdots <c_{n}<\cdots \}\subset \mathbb{R}^{+}.$ For $x\in \mathbb{R}^{+},$ set $
e(x)=\sup\limits_{n\leq x}\left\vert a_{n}-c_{n}\right\vert .$ We then have, for all $x\in \mathbb{R}^{+}$, 
\begin{equation}\label{eq4}
\left( A+A\right) \left( x\right) \leq \left( 4e\left( x\right) + 1\right) \cdot \left( C+C\right) \left( x+2e\left( x\right) \right). 
\end{equation}

If we further assume that $c_{1}\geq 1$ and $c_{n+1}-c_{n}\geq 1$ for all $n\geq 1$, we then also have, for all $x\in \mathbb{R}^{+}$, 
\begin{equation}\label{eq5}
A(x)\geq C\left( x-e(x)\right).
\end{equation}
\end{lemma}

\begin{proof} 
Note first that the function $e(x)$ is increasing, in the sense that $x\leq y$ implies $e(x)\leq e(y).$

Note also that, since $A\subset \mathbb{N}^{\ast },$ we have $i\leq a_{i}$ for all $i.$ So, for \ $n\leq x,$ 
if $n=a_{i}+a_{j},$ then \ $i\leq a_{i}\leq n\leq x$ and similarly $j\leq x,$ and therefore 
$\left\vert n-c_{i}-c_{j}\right\vert =\left\vert a_{i}+a_{j}-c_{i}-c_{j}\right\vert \leq \left\vert a_{i}-c_{i}\right\vert
+\left\vert a_{j}-c_{j}\right\vert \leq 2e(x)$.  Hence
\begin{equation}
\left( A+A\right) \left[ x\right] =\{n\leq x:\exists i,j,\ n=a_{i}+a_{j}\}\subset \{n\leq x:\exists i,j,\ \left\vert n-c_{i}-c_{j}\right\vert \leq 2e(x)\}, \notag
\end{equation}
and setting $\ s=c_{i}+c_{j},$ we get \ $s\in C+C$ \ and \ $\left\vert n-s\right\vert \leq 2e(x),$ so that \ $s\leq n+2e(x)\leq x+2e(x),$ and therefore 
\begin{equation}
\{n\leq x:\exists i,j, |n-c_{i}-c_{j}| \leq 2e(x)\}  \subset  \{n : \exists s\in (C+C)[x+2e(x]), |n-s| \leq 2e(x)\}. \notag \\
\end{equation}
Thus 
\begin{equation}
\left( A+A\right) \left[ x\right] \subset \bigcup _{s\in \left( C+C\right) \left[ x+2e(x)\right] }\left( \left[ s-2e(x),\,s+2e(x)\right]
\cap \mathbb{N}\right),  \notag
\end{equation}
and  therefore 
\begin{equation}
\left( A+A\right) \left( x\right) \leq \sum\limits_{n\in \left( C+C\right) \left[ x+2e(x)\right] }
\left(4e(x)+1\right) =\left( C+C\right) \left( x+2e(x)\right) \cdot \left( 4e(x)+1\right) . \notag
\end{equation}
This proves \eqref{eq4}.

Now, if $c_{1}\geq 1$ and $c_{n+1}-c_{n}\geq 1$ for all $n,$ then $c_{n}\geq n$ for all $n$. So if $c_{n}\leq x-e(x),$ then $n\leq c_{n}\leq x,$ 
so that $\left\vert a_{n}-c_{n}\right\vert \leq e(x),$ and therefore $a_{n}\leq c_{n}+e(x)\leq x.$ 

Hence $\{n:c_{n}\leq x-e(x)\}\subset \{n:a_{n}\leq x\},$ 
and thus 
\begin{equation}
C\left( x-e(x)\right) =\left\vert \{n:c_{n}\leq x-e(x)\}\right\vert \leq \left\vert \{n:a_{n}\leq x\}\right\vert =A(x), \notag
\end{equation}
which proves \eqref{eq5}. 
\end{proof}



\begin{lemma} \label{Lem3}
Let $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb{N}^{\ast }$ and $C=\{c_{1}<c_{2}<\cdots <c_{n}<\cdots \}\subset \mathbb{R}^{+}$ 
be two strictly increasing sequences in $\mathbb{N}^{\ast }$ and in $\mathbb{R}^{+}$, respectively. For $x\in \mathbb{R}^{+}$, set 
$e(x)=\sup\limits_{n\leq x}\left\vert a_{n}-c_{n}\right\vert .$ Assume that $e(x)$ is not identically zero, and that $c_{1}\geq 1$ 
and $c_{n+1}-c_{n}\geq 1$ for all $n\geq 1.$ Then the condition
\begin{equation}\label{(H)}
\liminf_{x\rightarrow \infty} \dfrac{e\left( 2x\right) \cdot \left( C+C\right) \left( 2x+2e\left( 2x\right) \right)}{C(x-e(x))^{2}}=0 \tag {H}
\end{equation}
implies that $h\left( A\right) =0$, and therefore $s\left( A\right) =\infty .$
\end{lemma}

\begin{proof}
Since $e(x)$ is increasing and not identically zero, there exists a real constant $t>0$ such that $e(x)\geq \dfrac{1}{t}$ for large enough $x.$ 
In view of the inequalities \eqref{eq4} and \eqref{eq5} in Lemma 2.3, we have 
\begin{equation}
\dfrac{\left( A+A\right) \left( 2x\right) }{A\left( x\right) ^{2}}\leq \dfrac{\left( 4e\left( 2x\right) +1\right) \cdot 
\left( C+C\right) \left( 2x+2e\left( 2x\right) \right) }{C\left( x-e(x)\right) ^{2}}. \notag
\end{equation}
Moreover, for large enough $x,$ we have $t\cdot e(2x)\geq 1,$ and therefore $4e\left( 2x\right) +1\leq \left( 4+t\right) \cdot e\left( 2x\right) .$ 
Thus 
\begin{equation}
\dfrac{\left( A+A\right) \left( 2x\right) }{A\left( x\right) ^{2}}\leq \left( 4+t\right) \dfrac{e\left( 2x\right) \cdot 
\left( C+C\right) \left( 2x+2e\left( 2x\right) \right) }{C(x-e(x))^{2}}, \notag 
\end{equation}
for large enough $x,$ so that the condition \eqref{(H)} implies that 
$\underset{x\rightarrow \infty }{\lim \inf }\dfrac{\left( A+A\right) \left( 2x\right) }{A\left( x\right) ^{2}}=0$, 
i.e., $h\left( A\right) =0,$ and therfore, by Lemma 2.1, \ $s\left( A\right) =\infty .$ 
\end{proof}


\begin{remark}\label{Rem1} 
The scope of Lemma 2.4 is broader than it seems to be. Indeed, for a subset $A$ of $\mathbb{N}$, 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.
\end{remark}



\begin{fund lem}\label{fund lem}
Let $B=\{b_{1}<b_{2}<\cdots <b_{n}<\cdots \}\subset \mathbb{N}$ and $D=\{d_{1}<d_{2}<\cdots <d_{n}<\cdots \}\subset \mathbb{R}^{+}$ be two strictly 
increasing sequences in $\mathbb{N}$ and in $\mathbb{R}^{+}$  respectively. Assume that there exists an increasing function
$f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$ and a positive integer $m$ such that $d_{m}\geq 1,$ $\ d_{n+1}-d_{n}\geq 1$ for $n\geq m$, and 
$\sup\limits_{m\leq n\leq x}\left\vert b_{n}-d_{n}\right\vert \leq f(x)$ for $x\geq m.$ Then the condition 

\begin{equation}\label{(K)}
\liminf_{x\rightarrow \infty} \dfrac{f\left( 2x\right) \cdot \left( D+D\right) \left( 2x+2f\left( 2x\right) \right)}{ D(x-f(x))^{2}}=0 \tag {K} 
\end{equation}
implies that $s\left( B\right) =\infty .$
\end{fund lem}

\begin {proof} 
For $n\in \mathbb{N}^{\ast },$ set $a_{n}=b_{n+m}$ and $c_{n}=d_{n+m},$ and let $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb{N}^{\ast }$ 
and $C=\{c_{1}<c_{2}<\cdots <c_{n}<\cdots \}\subset \mathbb{R}^{+}$ be the strictly increasing sequences, in $\mathbb{N}^{\ast }$ and $\mathbb{R}^{+},$ 
obtained by deleting the first $m$ terms of $B$ and $D$ respectively.
Then $c_{1}=d_{m+1}\geq 2$ and $c_{n+1}-c_{n}=d_{n+m+1}-d_{n+m}\geq 1$ for $n\geq 1.$ Moreover, setting 
$e\left( x\right) =\sup\limits_{n\leq x}\left\vert a_{n}-c_{n}\right\vert ,$ for $x\in \mathbb{R}^{+},$ and using the assumptions on $B$ and $D,$ 
we have 
\begin {equation}
e(x)=\sup\limits_{n\leq x}\left\vert a_{n}-c_{n}\right\vert =\sup\limits_{n\leq x}\left\vert b_{n+m}-d_{n+m}\right\vert 
=\sup\limits_{m<i\leq x+m}\left\vert b_{i}-d_{i}\right\vert \leq f(x+m). \notag
\end{equation}
Thus, setting $y=x+m,$ we have $e(x) \leq f(y)$, and since the functions $e$ and $f$ are increasing,  
\begin {equation}
e\left( 2x\right) \leq f\left( 2x+m\right) \leq f\left( 2y\right). \notag 
\end {equation}
Also, taking into account that $C\subset D$ and $C+C\subset D+D, $ so that $\left( C+C\right) \left( t\right) \leq \left( D+D\right) \left( t\right) $ 
for all $t\in \mathbb{R}^{+},$ and that the function $t\mapsto (C+C)(t)$ is increasing, we get 
\begin {equation}
\left( C+C\right) \left( 2x+2e\left( 2x\right) \right) \leq \left( C+C\right) \left( 2y+2f\left( 2y\right) \right) 
\leq \left( D+D\right) \left( 2y+2f\left( 2y\right) \right) . \notag
\end {equation}
Thus
\begin {equation}\label{1}
e\left( 2x\right) \cdot \left( C+C\right) \left( 2x+2e\left( 2x\right) \right) \leq f\left( 2y\right) \cdot \left( D+D\right) \left( 2y+2f\left( 2y\right) \right), 
\end {equation}
for $x\in \mathbb{R}^{+},$ and $y=x+m.$ 

Moreover, for $t\geq m,$ we have 
\begin {equation}
D(t)-C\left( t\right)  =\left\vert \left\{ d_n \in D : d_n \leq t \right\} \right\vert -\left\vert \left\{ c_n \in C : c_n = d_{n+m}\leq t\right\}\right\vert = m \notag 
\end{equation}
and 
\begin {equation}
C\left( t\right) -C\left( t-m\right)  =\left\vert \left\{ c_n \in C : t-m<c_{n}\leq t\right\} \right\vert \leq m, \notag 
\end {equation}
since $c_{n+1}-c_{n}\geq 1$ for all $n\in \mathbb{N}^{\ast },$ so that $C\left( t\right) \leq C\left( t-m\right) +m$ and 
$D\left( t\right) =C\left( t\right) +m\leq C\left( t-m\right) +2m.$ Therefore 
$C\left( t-m\right) \geq D\left( t\right) -2m$ for $t\geq m.$ Hence, taking into account that the function $t\mapsto C(t)$ is increasing and that 
$e(x)\leq f(y)$ we get, for large enough $x$, 
\begin {equation}\label{2}
C\left( x-e(x)\right) \geq C\left( x-f\left( y\right) \right) = C\left( y-m-f(y)\right) \geq D\left( y-f(y)\right) -2m.
\end{equation}
It follows from \eqref{1} and \eqref{2} that, for large enough $x$ and for $y=x+m$,  
\begin {equation}\label{3}
\dfrac{e\left( 2x\right) \cdot \left( C+C\right) \left( 2x+2e\left( 2x\right) \right) }{C(x-e(x))^{2}}\leq \dfrac{f\left( 2y\right)
\cdot \left( D+D\right) \left( 2y+2f\left( 2y\right) \right) }{\left( D\left( y-f(y)\right) -2m\right) ^{2}}.
\end{equation}
Set $P\left( x\right) = f\left( 2x\right) \cdot \left( D+D\right) \left( 2x+2f\left( 2x\right) \right) $ and $Q\left( x\right) =D\left( x-f(x)\right)$, 
and suppose that the condition \eqref{(K)} is satisfied, i.e., that $\underset{x\rightarrow \infty }{\lim \inf }\dfrac{P\left( x\right) }{Q\left( x\right)^{2}}=0.$ 
Then there exists a strictly increasing sequence $\left( x_{n}\right) _{n\geq 1}$ in $\mathbb{R}^{+},$ tending to infinity, such that 
$\lim\limits_{n\rightarrow \infty } \dfrac{P\left( x_{n}\right) }{Q\left( x_{n}\right) ^{2}}=0.$  
Since $P(x)$ is an increasing unbounded function, $\lim\limits_{n\rightarrow \infty} P\left( x_{n}\right) =\infty ,$ and therefore the sequence 
$\left( Q\left( x_{n}\right) \right)_{n\geq 1}$ is unbounded. So there exists a subsequence $\left( x_{n_{k}}\right) _{k\in \mathbb{N}^{\ast }}$ of 
$\left( x_{n}\right) _{n\geq 1}$ such that $\lim\limits_{k\rightarrow \infty }Q\left( x_{n_{k}}\right) =\infty ,$ while 
$\lim\limits_{k\rightarrow \infty }\dfrac{P\left( x_{n_{k}}\right) }{Q\left( x_{n_{k}}\right) ^{2}}=0.$ Hence 
$\lim\limits_{k\rightarrow \infty }\dfrac{ P\left( x_{n_{k}}\right) }{\left( Q\left( x_{n_{k}}\right) -2m\right) ^{2}}=0$, and therefore   
\begin {equation}
\underset{y\rightarrow \infty }{\lim \inf }\dfrac{f\left( 2y\right) \cdot \left( D+D\right) \left( 2y+2f\left( 2y\right) \right) }{\left( D\left( y-f(y)\right) -2m\right) ^{2}}
= \underset{x\rightarrow\infty }{\lim \inf }\dfrac{P\left( x\right) }{\left( Q\left( x\right) -2m\right) ^{2}} = 0. \notag
\end{equation}
It then follows from \eqref{3} that 
$\underset{x\rightarrow \infty }{\lim \inf }\dfrac{e\left( 2x\right) \cdot \left( C+C\right) \left( 2x+2e\left( 2x\right) \right) }{C(x-e(x))^{2}}=0.$ 
Thus the condition \eqref{(H)} of Lemma 2.4 holds, and therefore, in view of this Lemma, $s(A)=\infty .$ 
As $A\subset B,$ it follows that $s\left( B\right) =\infty $ too. 
\end{proof}



\begin{remark}\label{Rem2} In the statement of Lemma 2.6, we may replace $D$ by $D'= D + \gamma$, i.e., $d_n$ by $d'_n = d_n + \gamma \ (n \in \mathbb N^*)$, 
where $\gamma$ is any fixed real number, since a translation of the general term of $D$ does not affect the condition \eqref{(K)}.
\end{remark}




\section {Main results }\label{Main}



\begin{theorem}\label{Th2}
Let $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb{N}$ be a strictly increasing sequence of natural numbers, and $q(x)=\alpha x^{2}$ with a real number $\alpha >0.$ 
If the function $e(x)=\sup\limits_{n\leq x}\left\vert a_{n}-q(n)\right\vert $  ($\ x\in \mathbb{R}^{+}$) satisfies $e\left( x\right) =o\left( \sqrt{\log x}\right) $ 
as $x\rightarrow \infty ,$ then $s\left( A\right) =\infty .$
\end{theorem}
\begin{proof}
We apply Lemma 2.6 to $B=A$ and $D=\left\{ q(1)<q(2)<\cdots <q(n)<\cdots \right\}$. Indeed, the sequence 
$\left( q(n)\right) _{n\geq 1}$ is strictly increasing and unbounded, with $q(n+1)-q(n)=\alpha \left( 2n+1\right)$ unbounded too, so that $q(n)\geq 1$ and 
$q(n+1)-q(n)\geq 1$ for large enough $n.$ There remains to show that the condition \eqref{(K)} holds for \ $f(x)=e(x).$ 

Let \ $S=\{n^{2}:n\in \mathbb{N}^{\ast }\}$ . By a classical result of Landau \cite{L}, there exists a constant $c>0$ such that 
$\left( S+S\right) (x)\sim c\dfrac{x}{\sqrt{\log x}}$ \ as $x\rightarrow \infty .$

For $m,n\in \mathbb{N}^{\ast }$ and $x\in\mathbb{R}^{+},$ as $q(m)+q(n)\leq x$ is equivalent to $m^{2}+n^{2}\leq \dfrac{x}{\alpha },$ we have 
$\left( D+D\right) \left( x\right) =\left( S+S\right)\left( \dfrac{x}{\alpha }\right) \sim \dfrac{c}{\alpha }\dfrac{x}{\sqrt{\log x}},$ so that 
\begin{equation}
\left( D+D\right) \left( x\right) \leq c_{1}\dfrac{x}{\sqrt{\log x}}, \notag
\end{equation}
for large enough $x,$ with a constant $c_{1} > \dfrac{c}{\alpha }.$ 

Moreover, as $q(n)\leq x$ if and only if $n\leq \sqrt{\dfrac{x}{\alpha }},$ we also have $D(x)=\left[ \sqrt{\dfrac{x}{\alpha }}\right] >\sqrt{\dfrac{x}{\alpha }} -1.$ 
It follows that, for large enough $x$, 
\begin{align}
\dfrac{e\left( 2x\right) \cdot \left( D+D\right)\left( 2x+2e\left( 2x\right) \right) }{D(x-e(x))^{2}}\leq &
\dfrac{c_{1}\cdot e\left( 2x\right)\cdot \left( 2x+2e\left( 2x\right)\right)}
{\sqrt{\log \left( 2x+2e\left( 2x\right)\right)}\left( \sqrt{\frac{x-e(x)}{\alpha }}-1\right)^{2}} =  \notag   \\
& = \dfrac{c_{1} \alpha \cdot e\left( 2x\right) \cdot \left( 2x+2e\left( 2x\right) \right) }{\sqrt{\log \left( 2x+2e\left( 2x\right) \right) }
\left( \sqrt{x-e(x)}- \sqrt{\alpha } \right) ^{2}}. \notag
\end{align}

As  $e(x) = o\left( \sqrt{\log x}\right)$,  
\begin{equation}
\dfrac{e\left( 2x\right) \cdot \left( 2x+2e\left( 2x\right) \right) }{\sqrt{\log \left( 2x+2e\left( 2x\right) \right) }
\left( \sqrt{x-e(x)}- \sqrt{\alpha } \right)^{2}}\sim \dfrac{2x\cdot e\left( 2x\right) }{\sqrt{\log \left( 2x\right) }\cdot x}
\sim \dfrac{2e(2x)}{\sqrt{\log \left( 2x\right) }}, \notag
\end{equation}
and, since $e\left( x\right) =o\left( \sqrt{\log x}\right)$, we have $\lim_{x\to \infty} \dfrac{2e(2x)}{\sqrt{\log \left( 2x\right) }} =0$. 
Therefore 
\begin{equation}
\lim\limits_{x\rightarrow \infty }\dfrac{e\left( 2x\right) \cdot \left( D+D\right) \left( 2x+2e\left( 2x\right)\right) }{D(x-e(x))^{2}} 
=0, \notag 
\end{equation} 
and the condition \eqref{(K)} holds. Thus, by Lemma 2.6,  $s\left( B\right) =\infty $, i.e., $s\left( A\right) =\infty .$ 
\end{proof}



\begin{remark}\label{Rem3} 
In the statement of Theorem 3.1, we may replace $q(x)= \alpha x^2$ by $q(x)=\alpha x^{2}+\gamma$, where $\gamma$ is any real constant, in view of Remark 2.7.

Also, if $A = \{ a_n = [\alpha n^2 + \gamma] : n \in \mathbb N \}$ is the set of the integral parts $[\alpha n^2 + \gamma] = [q(n)]$, then $s(A)=\infty$, 
since $e(x)=\sup\limits_{n\leq x}\left\vert a_{n}-q(n)\right\vert \leq 1$ trivially satisfies the condition in Theorem 3.1.   
\end{remark}



\begin{theorem}\label{Th3} 
Let $A=\{a_{1}<a_{2}<\cdots <a_{n}<\cdots \}\subset \mathbb N$ and $q(x)$ be a quadratic polynomial with rational coefficients and positive leading coefficient. 
If the function $e(x)=\sup\limits_{n\leq x}\left\vert a_{n}-q(n)\right\vert $  ($\ x\in \mathbb{R}^{+}$) satisfies $e\left( x\right) =o\left( \sqrt{\log x}\right) $ 
as $x\rightarrow \infty ,$ then $s\left( A\right) =\infty .$
\end{theorem}

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

Let $b_n = da_n -c$ and $d_n = (an+b)^2$, for $n \in \mathbb N^*$. Clearly, there exists $m\in \mathbb{N}^*$ such that $b_m \geq 1$, 
$\ d_m \geq 1$ and $d_{n+1}- d_n \geq 1$ for $n\geq m$. Set $B=\left\{ b_n : n \geq m \right\}$ and $D = \{ d_n : n \geq m \}$.
Then $B$ and $D$ are strictly increasing sequences in $\mathbb N$, and, for all $n \geq m$,  
$$|d_n - b_n| = |(an+b)^2 - da_n +c | = d|q(n)-a_n|.$$ 
For $x >m$, Let $f(x) = \sup\limits_{m\leq n\leq x} |d_n - b_n|$, for $x \in \mathbb{R}^{+}$. Then $f(x)$ is an increasing nonnegative function satisfying 
$f(x) \leq d\cdot e(x)$, so that $f(x)= o\left( \sqrt{\log x}\right)$ (like $e(x)$). 
Thus, we may apply Lemma 2.6, provided we show that the condition \eqref{(K)} is satisfied. 
  
Let $S=\{n^{2}:n\in \mathbb N \}$. Then $D \subset S$, and therefore $D+D \subset S+S$, so that $(D+D)(x) \leq (S+S)(x)$, for $x \in \mathbb{R}^{+}$. 

By Landau's theorem \cite{L}, $\ \ \left( S+S\right) (x)\sim c_{0}\dfrac{x}{\sqrt{\log x}},$ with a constant $c_{0}>0.$ So there 
exists a constant $c_1>0$ such that $(D+D)(x) \leq (S+S)(x) \leq c_1 \dfrac{x}{\sqrt{\log x}}$, and therefore 
\begin{equation}\label{*} 
\left( D+D\right) \left( 2x+2f(2x) \right) \leq c_{1}\frac{2x+2f(2x) }{\sqrt{\log \left( 2x+2f(2x)\right)}}. 
\end{equation} 

Moreover, for $x > \max (m, b^2)$, if $n \leq \frac {\sqrt x -|b|}a$, then $d_n = (an+b)^2 \leq x$. Hence, for large enough $x$, 
\begin{align}
D\left( x\right) & =\left\vert \left\{ n\geq m : d_n \leq x  \right\} \right\vert 
\geq \left\vert \left\{ n\geq m : n\leq \frac {\sqrt x -|b|}a  \right\} \right\vert \notag \\
& \geq \frac {\sqrt x -|b|}a - m \geq c_{2}\sqrt{x}-c_{3}, \notag
\end{align} 
with constants $c_{2}, c_{3}>0$, and therefore 
\begin{equation}\label{**} 
D\left( x-f(x)\right) \geq c_{2}\sqrt{x-f(x)}-c_{3}.
\end{equation} 

It follows from \eqref{*} and \eqref{**} that, for large enough $x,$
\begin{equation}
\frac{f\left( 2x\right) \cdot \left( D+D\right) \left( 2x+2f\left( 2x\right) \right) }{D\left( x-f(x)\right) ^{2}} \leq 
c_{1}\frac{f\left( 2x\right) \cdot \left( 2x+2f\left( 2x\right) \right) }{\sqrt{\log \left( 2x+2f\left( 2x\right) \right) }\left( c_{2}\sqrt{x-f(x)}-c_{3}\right) ^{2}}, \notag 
\end{equation}
and, since $f\left( x\right) =o\left( \sqrt{\log x}\right) $ , we have 
\begin{equation}
\dfrac{ f\left( 2x\right) \cdot \left( 2x+2f\left( 2x\right) \right)}{\sqrt{\log \left( 2x+2f\left( 2x\right)\right)}\left( c_{2}\sqrt{x-f(x)}-c_{3}\right)^{2}}
\sim \dfrac{2f(2x)}{c_{2}^{2}\sqrt{\log x}}=o(1). \notag 
\end{equation}
Therefore
\begin{equation} 
\liminf_{x\to \infty} \dfrac{f\left( 2x\right) \cdot \left( D+D\right) \left( 2x+2f\left( 2x\right) \right) }{D\left( x-f(x)\right) ^{2}}=0. \notag 
\end{equation}
Thus the condition \eqref{(K)} is satisfied, and by Lemma 2.6, $s(B)=\infty$. As $B$ is a translate of a homothetic of a subsequence $A_m = \{ a_n : n\geq m \}$ of $A$, 
namely $B = d \cdot A_m +|c|$, we conclude, e.g., see \cite{GHHP2}, that $s(A_m) = s(B) = \infty$, and therefore $s\left( A\right) =\infty$.  
\end{proof}

\section {Acknowledgment}


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




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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B34; Secondary 11B83.

\noindent \emph{Keywords:} sequences, representation functions,
quadratic, Erd\H os-Tur\'an conjecture.


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\vspace*{+.1in}
\noindent
Received July 19 2012;
revised version received  October 14 2012.
Published in {\it Journal of Integer Sequences}, October 23 2012.

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\noindent
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