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Statement

Setting (p. 1). For $A\subseteq\mathbb Z$ and $n\in\mathbb Z$, $R_A^{(2)}(n)$ counts the pairs $(a_1,a_2)\in A\times A$ with $a_1+a_2=n$ and $a_1<a_2$. The paper prints the defining set of pairs and compares these functions as counts. $\mathbb N$ is the set of positive integers.

Theorem 1 (Dombi [D02], p. 2). Define $T:\mathbb N\to\{1,-1\}$ by

$$T(1)=1,\qquad T(2n)=-T(2n-1),\qquad T(2n+1)=T(n+1)\qquad(n\in\mathbb N),$$

and put $A=\{n\in\mathbb N:T(n)=1\}$, $B=\{n\in\mathbb N:T(n)=-1\}$. Then $R_A^{(2)}(n)=R_B^{(2)}(n)$ for all $n\in\mathbb N$.

So $\mathbb N=A\cup B$ is a partition into two sets with the same representation function $R^{(2)}$ at every $n$. The theorem is Dombi's (G. Dombi, Additive properties of certain sets, Acta Arith. 103 (2002), no. 2, 137--146); the paper gives it a new proof common with Theorem 2.

Context (p. 1). The paper records Sárközy's question whether there are $A,B\subseteq\mathbb N$ with infinite symmetric difference and $R_A^{(j)}(n)=R_B^{(j)}(n)$ for all but finitely many $n\in\mathbb N$. For $j=1$ the answer is no, by Dombi's observation that $R_A^{(1)}(n)$ is odd exactly when $n=2a$ with $a\in A$; Theorem 1 answers $j=2$ positively.

Essential uniqueness (p. 3, unnumbered). For a partition $\mathbb N=A\cup B$ with $T(n)=1$ on $A$ and $T(n)=-1$ on $B$, the paper says its proof shows that $R_A^{(2)}(n)=R_B^{(2)}(n)$ for all sufficiently large $n$ if and only if $T(1)+\cdots+T(2n)=0$ and $T(2n)=-T(n)$ for all but finitely many $n\in\mathbb N$. It states, leaving the check to the reader, that this is equivalent to the existence of $n_0\in\mathbb N$ with $T(2n)=-T(2n-1)$ and $T(2n-1)=T(n)$ for $n\ge n_0$, and $T(1)+\cdots+T(2n_0)=0$.

Source. Vsevolod F. Lev, Reconstructing integer sets from their representation functions, Electron. J. Combin. 11 (2004), no. 1, Research Paper 78, 6 pp., doi:10.37236/1831: the statement on p. 2, the common proof of Theorems 1 and 2 and the uniqueness remark on p. 3 (Section 2, pp. 3--4). The edition read is identified on the source card.

Read depth. Claims checked: the definitions and the statement were read clause by clause on the printed pages. The proof was read but not checked step by step. Nothing here is independently reviewed.

Proof pointer

P. 3. With $\alpha(x)$, $\beta(x)$ and $\tau(x)$ the generating series of $A$, $B$ and $T$ on $|x|<1$, one has $\alpha+\beta=x/(1-x)$ and $\alpha-\beta=\tau$ (the paper's (1)), and twice the generating series of $R_A^{(2)}$ is $\alpha(x)^2-\alpha(x^2)$ (its (2) with $j=2$). Subtracting the same identity for $B$ leaves $\tfrac{x}{1-x}\tau(x)-\tau(x^2)$. The partial sums $T(1)+\cdots+T(n-1)$ vanish for odd $n$ and equal $-T(n)$ for even $n$, so the difference reduces to a series in $x^{2n}$ with coefficients $-T(2n)-T(n)$, and these vanish by the recursion.

Dependencies

No other result of the paper.

Bears on

No Erdős problem in this corpus. The theorem answers, for $R^{(2)}$, a question the paper attributes to Sárközy.