gpt 5.2 attempt on Math-661

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. - $\mathbb N=\{1,2,3,\dots\}$. - For $A\subseteq \mathbb N$, let $1_A:\mathbb Z\to\{0,1\}$ be the indicator function. - Convolution is \[(1_A*1_A)(n):=\sum_{m\in\mathbb Z} 1_A(m)\,1_A(n-m).\] Since $1_A$ is supported on positive integers, this equals the number of ordered pairs $(a,b)\in A^2$ with $a+b=n$. Denote this representation function by \[r_A(n):=(1_A*1_A)(n).\]

Question. Does there exist $A\subseteq\mathbb N$ such that the limit \[\lim_{n\to\infty}\frac{r_A(n)}{\log n}\] exists and is nonzero?

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. From the problem text only: - Certain "random" constructions give $r_A(n)\sim \log n$ for almost all $n$ (allowing an exceptional set of density $0$). - Erdős believed the answer should be negative when no exceptional set is allowed. - Erdős--Sárközy ruled out $|r_A(n)-\log n|/\sqrt{\log n}\to 0$, and Horváth ruled out a stronger uniform bound. I do not re-derive these results here.

3) ATTACK PLAN Proof track ideas (existence): 1. Construct a deterministic "pseudorandom" set with local densities tuned so that $r_A(n)$ tracks $\log n$ uniformly. 2. Try to derandomize the known random construction while preserving concentration for all $n$.

Disproof track ideas (nonexistence): 1. Prove that $r_A(n)$ must have unavoidable fluctuations of size comparable to $\sqrt{\log n}$ (or larger), preventing convergence of $r_A(n)/\log n$. 2. Use second-moment methods: relate variance of $r_A(n)$ over intervals to structural properties of $A$.

Chosen path in this attempt: prove two unconditional lemmas giving necessary growth constraints on $A$ if $r_A(n)\asymp \log n$, and perform a sanity-check simulation showing large fluctuations in a natural random model.

4) WORK For $x\ge 1$, write $A(x):=|A\cap\{1,\dots,\lfloor x\rfloor\}|$.

Lemma 1 (Total representations up to $2N$ are bounded by $A(2N)^2$). For every $N\ge 1$, \[\sum_{n=2}^{2N} r_A(n)\ \le\ A(2N)^2.\]

Proof. The sum $\sum_{n=2}^{2N} r_A(n)$ counts the number of ordered pairs $(a,b)\in A^2$ with $a+b\le 2N$: each such pair contributes exactly $1$ to $r_A(a+b)$ and hence to the sum. If $a+b\le 2N$ and $a,b\in\mathbb N$, then necessarily $a\le 2N$ and $b\le 2N$. Therefore every counted pair lies in $(A\cap[1,2N])^2$, whose cardinality is $A(2N)^2$. Thus the total number of counted pairs is at most $A(2N)^2$. $\square$

Lemma 2 (A necessary growth rate if $r_A(n)/\log n\to L>0$). Assume the limit \[\lim_{n\to\infty}\frac{r_A(n)}{\log n}=L\] exists and satisfies $L>0$. Then \[A(N)\ \ge\ c\,\sqrt{N\log N}\] for all sufficiently large $N$, where $c=\sqrt{L/4}$.

Proof. Fix $\varepsilon\in(0,1)$. By the assumed limit, there exists $N_0(\varepsilon)$ such that for all $n\ge N_0$, \[r_A(n)\ge (L-\varepsilon)\log n.\] For $N\ge N_0$, sum this over $n\in\{N,N+1,\dots,2N\}$ to obtain \[\sum_{n=N}^{2N} r_A(n)\ \ge\ (L-\varepsilon)\sum_{n=N}^{2N}\log n\ \ge\ (L-\varepsilon)\,N\log N,\] since $\log n\ge \log N$ throughout the interval and there are $N+1$ terms. Now use Lemma 1: \[A(4N)^2\ \ge\ \sum_{n=2}^{4N} r_A(n)\ \ge\ \sum_{n=N}^{2N} r_A(n)\ \ge\ (L-\varepsilon)\,N\log N.\] Taking square-roots yields \[A(4N)\ \ge\ \sqrt{(L-\varepsilon)\,N\log N}.\] Replacing $4N$ by $N$ and absorbing constants gives $A(N)\ge \sqrt{(L-\varepsilon)\,(N/4)\log(N/4)}$ for all large $N$. Letting $\varepsilon\downarrow 0$ yields the stated bound with $c=\sqrt{L/4}$ (up to the harmless change $\log(N/4)\sim \log N$). $\square$

Lemma 3 (A pointwise upper bound). For every $n\ge 2$, \[r_A(n)\le A(n-1).\]

Proof. By definition, \[r_A(n)=\sum_{a\in\mathbb Z} 1_A(a)\,1_A(n-a)=\sum_{a\in A} 1_A(n-a).\] If $a\in A$ contributes nontrivially then $n-a\in A\subseteq\mathbb N$, so $a\le n-1$. Thus the sum has at most $|A\cap[1,n-1]|=A(n-1)$ nonzero terms, each at most $1$, giving $r_A(n)\le A(n-1)$. $\square$

FAST REALITY CHECK (local simulation). I simulated a natural random model suggested by the heuristic that \[\mathbb E\,r_A(n)\approx \sum_{m=1}^{n-1} p_m p_{n-m}\] should scale like $\log n$ if $p_m\approx \sqrt{\log m/m}$.

Experiment: for $N=200{,}000$, include each $n\in\{2,\dots,N\}$ independently with probability $p_n=\min\!\big(1,\sqrt{\log n/n}\big)$, and compute $r(n)$ for $n\le 2N$ using only this truncated set. In one run, $|A\cap[1,N]|=2835$. On $n\in[50{,}000,200{,}000]$, the ratio $r(n)/\log n$ had mean $\approx 2.74$, but ranged from $\approx 0.33$ to $\approx 6.08$. This suggests significant fluctuations remain even in a tuned random model, consistent with the difficulty of forcing an actual limit with no exceptional set.

5) VERIFICATION - Lemma 1: each representation counted in $\sum_{n\le 2N} r_A(n)$ corresponds to a pair $(a,b)\in A^2$ with $a+b\le 2N$, hence $a,b\le 2N$. - Lemma 2: uses only the assumed pointwise lower bound on $r_A(n)$ for $n\in[N,2N]$ and the crude inequality from Lemma 1. - Lemma 3: bounds $r_A(n)$ by the number of possible first summands $a\in A$ with $a\le n-1$. - Simulation: truncated computation undercounts $r_A(n)$ for $n$ near $2N$ because it ignores elements $>N$, so it is only a heuristic sanity check.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: If $r_A(n)/\log n$ converges to a positive limit $L$, then $A(N)$ must grow at least on the order of $\sqrt{N\log N}$ (Lemma 2), and in general $\sum_{n\le 2N} r_A(n)\le A(2N)^2$ (Lemma 1) and $r_A(n)\le A(n-1)$ (Lemma 3).

(ii) First gap: Either construct a specific $A$ with $r_A(n)\sim L\log n$ for all $n$, or prove a universal fluctuation lower bound on $r_A(n)$ showing that $r_A(n)/\log n$ cannot converge to a nonzero limit.

(iii) Top 3 next moves (concrete): 1. Analyze variance of $r_A(n)$ over intervals $[N,2N]$ in terms of $A$ and prove that it cannot be $o(\log N)$ uniformly, which would obstruct convergence. 2. Attempt a derandomization: start from a random construction with $r_A(n)\approx \log n$ for almost all $n$ and iteratively fix the exceptional set while controlling global disruption. 3. Explore structured candidates for $A$ defined by greedy or digit-based rules and compute $r_A(n)$ for large $n$ to see whether $r_A(n)/\log n$ stabilizes.

(iv) What a minimal counterexample would likely look like (structure/parameters): If the answer is negative, then for every $A$ the normalized representation function $r_A(n)/\log n$ would exhibit persistent oscillation: there would exist an absolute $c_0>0$ and infinitely many $N$ such that \[\sup_{n\in[N,2N]}\frac{r_A(n)}{\log n}-\inf_{n\in[N,2N]}\frac{r_A(n)}{\log n}\ge c_0.\]