gpt pro 5.2 attempt on Math-6166

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 $A$.