gpt pro 5.2 attempt on Math-774

Erdős problem #346

1) FORMAL RESTATEMENT. We consider an increasing sequence $A=\{a_1<a_2<\cdots\}\subseteq\mathbb{N}$ and its subset-sum set \[ P(A)=\left\{\sum_{n\in B} n:\ B\subseteq A\ \text{finite}\right\}. \] A sequence $A$ is complete if every sufficiently large integer lies in $P(A)$. The problem asks whether there exists a sequence $A$ such that:

1. $1$ For every finite subset $B\subseteq A$, the set $A\setminus B$ is complete.
2. $2$ For every infinite subset $B\subseteq A$, the set $A\setminus B$ is not complete.

The statement reports that Graham exhibited an example with these properties. It further asks: if additionally $a_{n+1}/a_n\ge 1+\varepsilon$ for all $n$, must one have \[ \lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\frac{1+\sqrt5}{2}=:\varphi\ ? \] It also notes that if $a_{n+1}/a_n>\varphi$ for all $n$, then property $2$ is "easy".

2) QUICK LITERATURE/CONTEXT CHECK. We do not use Graham's result beyond acknowledging it is asserted in the statement. We focus on proving the "easy" implication about ratios.

3) ATTACK PLAN.

• Prove a standard gap lemma: if a term exceeds $1+$ the sum of all previous terms, completeness fails.
• Use the ratio hypothesis $a_{n+1}\ge c a_n$ with $c>\varphi$ to force such a gap in any subsequence obtained by deleting infinitely many terms.
• Do a small numerical sanity check illustrating the gap phenomenon.

4) WORK.

Lemma 346.1 (gap obstruction to completeness). Let $B=\{b_1<b_2<\cdots\}\subseteq\mathbb{N}$ and let $S_m:=\sum_{i=1}^m b_i$. If for some $m$ one has \[ b_{m+1} > S_m+1, \] then $B$ is not complete. Indeed, every integer in the interval $(S_m,\,b_{m+1})$ fails to lie in $P(B)$.

Proof. Any subset sum of elements among $\{b_1,\dots,b_m\}$ is at most $S_m$. Any subset sum that uses $b_{m+1}$ is at least $b_{m+1}$. Therefore no subset sum can lie strictly between $S_m$ and $b_{m+1}$. In particular, the integers $S_m+1,\dots,b_{m+1}-1$ are missing from $P(B)$, so $B$ cannot be complete. $\square$

Proposition 346.2 (uniform ratio $>\varphi$ forces property $2$). Assume $A=\{a_1<a_2<\cdots\}$ satisfies a uniform ratio bound \[ \frac{a_{n+1}}{a_n} \ge c\quad\text{for all }n, \] for some constant $c>\varphi$. Then for every infinite subset $B\subseteq A$, the set $A\setminus B$ is not complete. Equivalently: deleting infinitely many terms from $A$ always destroys completeness.

Proof. Let $C:=A\setminus B=\{b_1<b_2<\cdots\}$ be the remaining infinite subsequence. Since $C$ is a subsequence of $A$, it also satisfies $b_{i+1}/b_i\ge c$ for all $i$. Hence for each $m$, \[ \sum_{i=1}^m b_i \le b_m\sum_{j=0}^{m-1} c^{-j} \le b_m\cdot \frac{c}{c-1}. \] Now, because $B$ is infinite, there are infinitely many indices $m$ such that at least one term of $A$ lies strictly between $b_m$ and $b_{m+1}$; in particular, for infinitely many $m$ we have $b_{m+1}\ge c^2 b_m$ (skipping at least one ratio-$\ge c$ step multiplies by at least $c^2$). Fix such an $m$. Then \[ b_{m+1} \ge c^2 b_m. \] Since $c>\varphi$ is equivalent to $c^2>\frac{c}{c-1}$ (because $c^2>\frac{c}{c-1}\iff c^2-c-1>0$), we have \[ b_{m+1} - \sum_{i=1}^m b_i \ge c^2 b_m - \frac{c}{c-1}b_m = b_m\left(c^2-\frac{c}{c-1}\right) >0. \] For all sufficiently large such $m$, the right-hand side exceeds $1$, hence $b_{m+1} > \sum_{i=1}^m b_i + 1$. Applying Lemma 346.1 shows that $C$ is not complete. $\square$

FAST REALITY CHECK (illustration with a toy ratio). Take a toy geometric sequence $a_n=\lceil c^n\rceil$ with $c=1.7>\varphi$. If we delete every other term, the remaining subsequence has ratio about $c^2\approx 2.89$, and Lemma 346.1 predicts large gaps in subset sums. A small computation for the first few terms indeed shows missing intervals once a big ratio jump occurs (not included here as a theorem).

5) VERIFICATION. Lemma 346.1 is a standard and fully explicit gap argument. Proposition 346.2 is a direct inequality manipulation that uses the equivalence $c>\varphi\iff c^2>c/(c-1)$.

6) FINAL.

UNRESOLVED

(i) Strongest fully proved partial result obtained here. We proved the "easy" direction stated in the problem: if $a_{n+1}/a_n\ge c>\varphi$ uniformly, then deleting infinitely many terms necessarily destroys completeness (Proposition 346.2).

(ii) Exact first gap. The main open component here is the converse-type question: under the weaker hypothesis $a_{n+1}/a_n\ge 1+\varepsilon$, does the existence of a sequence satisfying $1$--$2$ force $a_{n+1}/a_n\to\varphi$? We have no argument establishing such a limiting ratio.

(iii) Top 3 next moves (concrete targets).

1. Prove a stability version of Proposition 346.2: if deletions always destroy completeness, show that ratios cannot stay bounded away from $\varphi$ on either side.
2. Analyze Graham's claimed example (as given in the statement) and compute its ratio behavior to see whether it converges to $\varphi$; then attempt to extract a general mechanism.
3. Establish necessary conditions for property $1$ (robust completeness under finite deletions), perhaps in terms of lower bounds on local additive coverage or on $\sum 1/a_n$.

(iv) Minimal counterexample structure. A counterexample to "ratio must tend to $\varphi$" would be a sequence satisfying $1$--$2$ and $a_{n+1}/a_n\ge 1+\varepsilon$ but with $\limsup a_{n+1}/a_n>\varphi+\delta$ or $\liminf a_{n+1}/a_n<\varphi-\delta$ for some $\delta>0$. Such a sequence would have to balance being complete after any finite deletion with being fragile under any infinite deletion, yet without the Fibonacci-like ratio.