Reused writeup on Math-6166 (gpt pro 5.2)

Erdős Problem 1112 --- collection record

PROVENANCE AND STATUS

Official problem: https://www.erdosproblems.com/1112

Saved collaborative-database status: open (Lean). Snapshot checked: 2026-09-23T17:22:24Z. Latest status and formalization links: https://teorth.github.io/erdosproblems/

This is an attributed collection of existing writeups. The model-folder names below identify their original attribution; placement in GPT_6_Astra_Ultra does not attribute their mathematical work to that model. All locally available source versions selected for this problem are reproduced below, without editing their text.

Proof claims, computations, literature statements, and completion estimates in the reused text have not been independently verified in this collection pass. A source claiming a solution does not change the saved upstream status.

Primary source (latest numbered version in the preferred model folder): attacks/erdos/gpt_pro_5.2/1112.tex.

REUSED SOURCE 1

Original attribution: gpt_pro_5.2. Original file: attacks/erdos/gpt_pro_5.2/1112.tex.

1) FORMAL RESTATEMENT

Fix integers $1\le d_1<d_2$ and $k\ge 3$. Let $r\ge 2$ be an integer.

A (strictly) increasing sequence of positive integers $B=\{b_1<b_2<\cdots\}$ is called $r$-lacunary if \[ b_{i+1}\ge r\,b_i\quad\text{for all }i\ge 1. \]

A (strictly) increasing sequence of positive integers $A=\{a_1<a_2<\cdots\}$ has $(d_1,d_2)$-bounded gaps if \[ d_1\le a_{i+1}-a_i\le d_2\quad\text{for all }i\ge 1. \]

Define the $k$-fold sumset (with repetition allowed) \[ kA := \underbrace{A+\cdots +A}_{k\text{ times}}=\{a_{i_1}+\cdots+a_{i_k}: i_1,\dots,i_k\ge 1\}. \]

Question (as interpreted in the problem statement): does there exist an integer $r=r_k(d_1,d_2)$ such that for every $r$-lacunary $B$ there exists an infinite $A$ with $(d_1,d_2)$-bounded gaps and \[ (kA)\cap B = \varnothing\ ? \] (Edge case note: if one required the sumset to use distinct elements of $A$, the problem changes; the text "$k$-fold sumset" is standardly interpreted as allowing repetition, and the computations/lemmas below use that.)

2) QUICK LITERATURE/CONTEXT CHECK

I did not use web browsing here; I only record what is explicitly stated in the problem text:

- For $d_1=2,d_2=3$, Erdős--Graham noted the existence of $A$ with gaps in $[2,3]$ such that $(A+A)\cap B=\emptyset$ for any $B$ with $b_{1}\ge 5$ and $b_{i+1}\ge 2b_i$. - Bollobás--Hegyvári--Jin proved sharp results for $k=2$ in that regime and also that $r_3(2,3)$ does not exist (in the sense stated in the problem text). - Chen proved $r_2(a,b)\le 2$ for $b\ne 2a$ and lower bounds for $b=2a$. - For $k\ge 3$ beyond the nonexistence examples, the general existence question remains open, with further technical nonexistence results by Tang--Yang.

3) ATTACK PLAN

Proof strategies (if aiming for existence of $r_k(d_1,d_2)$):

- Try to build $A$ as a "near-periodic" bounded-gap sequence and modify it locally near each forbidden value $b_i$, using the lacunarity to keep modifications far apart. - Try modular obstructions: force $kA$ to land in a controlled set of residues mod $q$, then choose $q$ so that $B$ avoids those residues.

Disproof/construction strategies (to show $r_k(d_1,d_2)$ does not exist):

- Construct a universal $r$-lacunary set $B$ that intersects $kA$ for every $(d_1,d_2)$-bounded-gap $A$. (This is what the cited nonexistence results accomplish in special cases.)

Below I only obtain basic structural lemmas and a few explicit examples; they do not settle the general existence/nonexistence problem.

4) WORK

Lemma 1112.1 (Counting bounds for bounded-gap $A$). Let $A=\{a_1<a_2<\cdots\}$ have $(d_1,d_2)$-bounded gaps. Then for every integer $N\ge a_1$, writing $A(N):=|A\cap [a_1,N]|$, we have \[ 1+\frac{N-a_1}{d_2} \le A(N) \le 1+\frac{N-a_1}{d_1}. \] In particular the lower asymptotic density of $A$ is at least $1/d_2$.

Proof. For the upper bound: if $A(N)=t$, then $a_t\le N$. Summing the lower gap bound, $a_t-a_1=\sum_{i=1}^{t-1}(a_{i+1}-a_i)\ge (t-1)d_1$, hence $N-a_1\ge (t-1)d_1$, i.e. $t\le 1+(N-a_1)/d_1$.

For the lower bound: let $t$ be maximal with $a_t\le N$ (so $A(N)=t$). Then $a_{t+1}>N$. Using the upper gap bound, $a_{t+1}\le a_1+(t)d_2$ because $a_{t+1}-a_1=\sum_{i=1}^{t}(a_{i+1}-a_i)\le t d_2$. Thus $N<a_{t+1}\le a_1+t d_2$, so $t> (N-a_1)/d_2$, i.e. $t\ge 1+(N-a_1)/d_2$ since $t$ is an integer. $\square$

Lemma 1112.2 (A uniform syndeticity bound for $kA$ from the upper gap bound). Let $A=\{a_1<a_2<\cdots\}$ be an increasing sequence of integers with $a_{i+1}-a_i\le d_2$ for all $i$ (no lower gap assumption needed). Fix $k\ge 1$ and set $M:=k(d_2-1)$. Then every integer $n\ge k a_1$ belongs to \[ kA + \{0,1,\dots,M\}. \] Equivalently, for every integer $n\ge k a_1$ there exists $s\in kA$ with $n-M\le s\le n$. In particular, the set $kA$ has gaps at most $M+1$ beyond $k a_1$.

Proof. First observe that the condition $a_{i+1}-a_i\le d_2$ implies \[ \{m\in\mathbb{Z}: m\ge a_1\}\subseteq A+\{0,1,\dots,d_2-1\}. \] Indeed, let $m\ge a_1$, and choose $i$ maximal with $a_i\le m$. Then $m<a_{i+1}$, hence $0\le m-a_i\le a_{i+1}-a_i-1\le d_2-1$, so $m=a_i+t$ with $t\in\{0,\dots,d_2-1\}$.

Now take $n\ge k a_1$ and write $n = x_1+\cdots+x_k$ where $x_1=\cdots=x_{k-1}=a_1$ and $x_k:=n-(k-1)a_1\ge a_1$. Each $x_j$ lies in $A+\{0,\dots,d_2-1\}$ by the previous paragraph. Thus we may write $x_j=a_{i_j}+t_j$ with $a_{i_j}\in A$ and $t_j\in\{0,\dots,d_2-1\}$. Summing, \[ n = (a_{i_1}+\cdots+a_{i_k}) + (t_1+\cdots+t_k), \] where the first parenthesis lies in $kA$ and the second lies in $\{0,1,\dots,k(d_2-1)\}=\{0,1,\dots,M\}$. This proves $n\in kA+\{0,\dots,M\}$.

Finally, if a set $S\subset\mathbb{Z}$ satisfies $[N,\infty)\subseteq S+\{0,\dots,M\}$, then every interval $[t,t+M]$ with $t\ge N$ must intersect $S$; otherwise no element of $[t,t+M]$ could be written as $s+u$ with $s\in S$, $0\le u\le M$. This implies the gaps of $S$ beyond $N$ are at most $M+1$. Apply this with $S=kA$ and $N=k a_1$. $\square$

Lemma 1112.3 (A parity-based explicit avoidance example for odd $k$). Assume $2\in[d_1,d_2]$ and $k$ is odd. Let $A:=\{1,3,5,\dots\}$. Then $A$ has $(d_1,d_2)$-bounded gaps (all gaps equal to $2$), and every element of $kA$ is odd. Consequently, $kA\cap B=\emptyset$ for every $B\subseteq 2\mathbb{Z}$ (in particular for $B=\{2^j:j\ge 1\}$).

Proof. All gaps in $A$ equal $2$, so if $2\in[d_1,d_2]$ then $d_1\le a_{i+1}-a_i\le d_2$ holds. If $k$ is odd, the sum of $k$ odd integers is odd, so $kA\subseteq 2\mathbb{Z}+1$. Hence $kA$ is disjoint from any subset of the even integers. $\square$

FAST REALITY CHECK (one concrete instance). Take $(d_1,d_2,k)=(2,3,3)$ and the lacunary set $B=\{2^j:j\ge 0\}$ (ratio $2$). Lemma 1112.3 gives $A=\{1,3,5,\dots\}$, and then $3A$ consists only of odd integers $\ge 3$, while $B$ consists of $1$ and even integers, so $(3A)\cap B=\emptyset$.

(Computation note: a backtracking script was run for this instance; it immediately found the same solution $A$ consisting of consecutive odd integers, which is consistent with the parity argument above.)

5) VERIFICATION

- Lemma 1112.1: the inequalities are obtained by summing gap bounds; all steps are explicit. - Lemma 1112.2 crucially uses the convention that $kA$ allows repetition (we used $a_1$ multiple times). - Lemma 1112.3: check that the endpoint $1\in A$, and the smallest element of $kA$ is $k$, so $1\notin kA$ when $k\ge 3$, ensuring disjointness from $B$ even if $1\in B$.

6) FINAL

UNRESOLVED

(i) Strongest proved partial result: For any increasing $A$ with gaps $\le d_2$, the sumset $kA$ is syndetic with explicit gap bound $\le k(d_2-1)+1$ beyond $k a_1$ (Lemma 1112.2). Also, for odd $k$ and when $2\in[d_1,d_2]$, there is an explicit $A$ with $(d_1,d_2)$-bounded gaps whose $k$-fold sumset avoids any even lacunary $B$ (Lemma 1112.3).

(ii) First gap (crisp): Given arbitrary integers $1\le d_1<d_2$ and $k\ge 3$, prove or disprove the existence of a universal integer $r$ such that every $r$-lacunary $B$ can be avoided by some $(d_1,d_2)$-bounded-gap $A$ in the sense $(kA)\cap B=\emptyset$.

(iii) Top 3 next moves: 1. Try to build $A$ by a lacunary-stage construction: keep $A$ periodic (fixed step pattern) on long blocks, and at stage $i$ locally modify a bounded number of steps inside a window around $b_i$ to eliminate all representations of $b_i$ as a $k$-sum, using the largeness of $b_{i+1}/b_i$ to keep windows disjoint. 2. For fixed $(d_1,d_2,k)$, do systematic computer search for small $r$ and specific lacunary $B$ (e.g. $b_i=r^i$ or random lacunary) to see whether greedy/local-modification heuristics succeed or repeatedly get trapped. 3. Attempt to extract a structural obstruction from Lemma 1112.2: since $kA$ is syndetic with an explicit gap bound, look for constructions of $B$ that force intersection with every syndetic set of that gap type that is representable as a $k$-fold sumset of a $(d_1,d_2)$-bounded-gap set.

(iv) Minimal counterexample structure (if $r_k(d_1,d_2)$ fails to exist): one expects a lacunary $B$ with extremely fast growth but chosen adversarially so that each $b_i$ is forced to lie in $kA$ for every $(d_1,d_2)$-bounded-gap $A$; such a $B$ would likely be constructed inductively to hit the (apparently large) unavoidable subset of $kA$ common to all admissible $A$.

COMPLETION ESTIMATE

The primary source supplies no extractable numerical completion estimate. Its mathematical progress is not numerically assessed here.

Newly verified progress in this collection pass: $0\%$. This measures new verification work, not the amount of mathematics achieved in the reused text.