gpt pro 5.2 attempt on Math-5594

1) FORMAL RESTATEMENT

1.1. Ambiguity in the word “complete”

The problem text itself notes two competing conventions:

* Strong completeness: \[ \Sigma(A):=\left\{\sum_{i\in B} a_i : B\subset \mathbb N\text{ finite}\right\}=\mathbb N\cup\{0\}. \] * Weak completeness: $\Sigma(A)$ contains all but finitely many natural numbers, i.e. \[ \exists N_0\ \forall N\ge N_0:\ N\in\Sigma(A). \]

The statement “Erdős and Graham most likely meant the weaker notion” suggests the intended interpretation is weak completeness.

I will therefore (i) give a precise treatment of both notions and (ii) prove rigorously that the claimed examples (powers of 2; Fibonacci) satisfy the required deletion property in the strong notion (hence also in the weak notion). The general classification for the weak notion remains open (e.g. $m=2,n=3$ is explicitly stated as unknown).

1.2. Conventions

* $\mathbb N:=\{1,2,3,\dots\}$ and $\mathbb N_0:=\{0,1,2,\dots\}$. * A “sequence” $A=\{a_1\le a_2\le\cdots\}$ is an infinite nondecreasing sequence of positive integers. (Duplicates are allowed; distinctness in subset sums is by index.) * For a finite set of indices $D\subset\mathbb N$, $A\setminus D$ denotes the sequence obtained by removing the terms $\{a_i: i\in D\}$ (keeping the remaining terms in nondecreasing order).

1.3. Property to classify

Fix integers $0\le m<n$. We ask for which pairs $(m,n)$ there exists a complete sequence $A$ (say, in the weak sense unless otherwise stated) such that:

1. (Robustness) For every $D\subset\mathbb N$ with $|D|=m$, the sequence $A\setminus D$ is complete. 2. (Fragility) For every $D\subset\mathbb N$ with $|D|=n$, the sequence $A\setminus D$ is not complete.

Edge cases: * $m=0$ means simply “$A$ is complete.” * The “fragility” condition is monotone in $n$ (proved below): if it holds for some $n$, it holds for all $n'\ge n$.

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2) QUICK LITERATURE/CONTEXT CHECK

I do not use any external sources here; I only record what is already in the provided problem text:

* Powers of 2 give an example for $(m,n)=(0,1)$. * Fibonacci $1,1,2,3,5,\dots$ gives an example for $(m,n)=(1,2)$. * $(m,n)=(2,3)$ is stated as not known (for the intended weak completeness). * van Doorn proved that no such sequence exists for $2\le m<n$ under the strong completeness definition $\Sigma(A)=\mathbb N$ (as stated in the problem text).

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3) ATTACK PLAN

Proof track (partial results). 1. Work in the strong completeness notion (it implies the weak notion). 2. Use a sharp necessary-and-sufficient condition for strong completeness of a nondecreasing positive sequence in terms of partial sums. 3. Verify the powers-of-2 and Fibonacci examples with deletion properties by checking this condition after deletions.

Disproof/obstruction track. Try to derive constraints for $m\ge 2$ (for the weak notion) from “gap creation” after deletions. I did not succeed beyond what is already stated ($m=2,n=3$ open).

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4) WORK

Phase 1: FAST REALITY CHECK (computation)

I tested the Fibonacci claims via the strong-completeness partial-sum criterion below. Using the first 60 Fibonacci numbers $1,1,2,3,\dots$:

* After deleting any single term among the first 30, the remaining sequence satisfies the strong-completeness inequalities through the first 50 remaining terms. * After deleting any pair of terms among the first 30, the inequalities fail by (at most) the first 50 remaining terms.

(These are sanity checks only; the proofs below are exact and do not rely on computation.)

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Lemma 4.1 (Strong completeness criterion via partial sums)

Let $A=(a_1\le a_2\le\cdots)$ be a nondecreasing sequence of positive integers, and define $S_k:=\sum_{i=1}^k a_i$.

Claim. The following are equivalent:

(A) $A$ is strongly complete, i.e. every $N\in\mathbb N_0$ is a sum of distinct terms of $A$.

(B) $a_1=1$ and for every $k\ge 1$, \[ a_{k+1}\le S_k+1. \]

Proof.

(B) $\Rightarrow$ (A). We prove by induction on $k$ that the set of subset sums of $\{a_1,\dots,a_k\}$ contains the entire interval $[0,S_k]\cap\mathbb N_0$.

Base $k=1$: since $a_1=1$, subset sums are $\{0,1\}=[0,S_1]$.

Inductive step: assume $[0,S_k]$ is achievable using $a_1,\dots,a_k$. Consider $a_{k+1}$. * Using no $a_{k+1}$ we can make all values in $[0,S_k]$. * Using $a_{k+1}$ plus any subset sum of $a_1,\dots,a_k$, we can make all values in $[a_{k+1}, a_{k+1}+S_k]$.

If $a_{k+1}\le S_k+1$, then these two intervals overlap or touch: $[0,S_k]\cup[a_{k+1},a_{k+1}+S_k]\supseteq [0,S_k+a_{k+1}]=[0,S_{k+1}]$. Hence $[0,S_{k+1}]$ is achievable, completing the induction.

Since $A$ is infinite with positive terms, $S_k\to\infty$, so every $N\in\mathbb N_0$ lies in $[0,S_k]$ for some $k$, hence is representable. Thus $A$ is strongly complete.

(A) $\Rightarrow$ (B). If $a_1\ne 1$, then $1\notin\Sigma(A)$, contradicting strong completeness, so $a_1=1$. Now fix $k\ge 1$. Suppose for contradiction that $a_{k+1}>S_k+1$. Any subset sum using only $a_1,\dots,a_k$ is $\le S_k$, while any subset sum that uses $a_{k+1}$ is $\ge a_{k+1}>S_k+1$. Therefore $S_k+1$ is not representable, contradicting strong completeness. Hence $a_{k+1}\le S_k+1$.

This proves (A)$\Leftrightarrow$(B). $\square$

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Lemma 4.2 (Monotonicity of non-completeness under deletion)

Let $A$ be any sequence of positive integers (duplicates allowed). If $A$ is not complete (strong or weak), then any subsequence obtained by deleting additional terms is also not complete.

Proof. Deleting terms can only remove available subset sums; i.e. if $B\subseteq A$ as multisets (or by indices), then $\Sigma(B)\subseteq\Sigma(A)$. If $\Sigma(A)$ misses infinitely many integers (weak incompleteness) or misses some integer (strong incompleteness), then $\Sigma(B)$ also misses those integers. $\square$

Consequence. If a sequence satisfies the “fragility” condition for some $n$, then it also satisfies it for every $n'\ge n$.

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Proposition 4.3 (Powers of 2 witness $m=0,n=1$ in the strong sense)

Let $A=(1,2,4,8,\dots)$ where $a_i=2^{i-1}$.

1. $A$ is strongly complete. 2. For every single deletion $D=\{t\}$ (remove $2^{t-1}$), the sequence $A\setminus D$ is not complete (hence the pair $(m,n)=(0,1)$ is achievable).

Proof.

1. (Strong completeness.) Here $S_k=1+2+\cdots+2^{k-1}=2^k-1$, so $a_{k+1}=2^k=S_k+1$. By Lemma 4.1, $A$ is strongly complete.

2. (Deleting one term breaks completeness.) Fix $t\ge 1$ and delete $2^{t-1}$. Consider the number $N:=2^{t-1}$. Any subset sum of the remaining powers of 2 is either * $<2^{t-1}$ if it uses only smaller powers $1,2,\dots,2^{t-2}$, since their sum is $2^{t-1}-1$, or * $\ge 2^t$ if it uses any larger power (since the smallest remaining larger power is $2^t$).

Thus $N$ cannot be represented, so $A\setminus\{t\}$ is not strongly complete.

By Lemma 4.2, deleting any $n\ge 1$ terms also yields a non-complete sequence. $\square$

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Proposition 4.4 (Fibonacci witnesses $m=1,n=2$ in the strong sense)

Let $(F_k)_{k\ge 1}$ be Fibonacci numbers with $F_1=1, F_2=1$, and $F_{k+2}=F_{k+1}+F_k$. Let $A=(F_1,F_2,F_3,\dots)=(1,1,2,3,5,\dots)$.

We prove:

1. $A$ is strongly complete. 2. After deleting any single term, the remaining sequence is still strongly complete. 3. After deleting any two terms, the remaining sequence is not strongly complete.

Together, $2$ and $3$ give the required property for $(m,n)=(1,2)$.

Identity used

For all $k\ge 1$, \[ \sum_{i=1}^k F_i = F_{k+2}-1. \] This is proved by induction: it is true for $k=1$, and $\sum_{i=1}^{k+1}F_i=(F_{k+2}-1)+F_{k+1}=F_{k+3}-1$ using the recurrence.

$1$ Strong completeness of Fibonacci

We verify Lemma 4.1(B). We have $F_1=1$. For $k\ge 1$, using the identity, \[ S_k=\sum_{i=1}^k F_i = F_{k+2}-1\quad\Rightarrow\quad S_k+1=F_{k+2}. \] Since $F_{k+1}<F_{k+2}$, we have $F_{k+1}\le S_k+1$. Thus Lemma 4.1 implies $A$ is strongly complete.

$2$ Deleting any one Fibonacci number preserves strong completeness

Fix $t\ge 1$ and delete the single term $F_t$. Let the remaining sequence be $B=(b_1\le b_2\le\cdots)$; note $b_1=1$ always (because either $F_1$ or $F_2$ remains).

We verify Lemma 4.1(B) for $B$. Consider any stage where the next element of $B$ is $F_{k+1}$. There are two cases.

If $k<t$: then none of $F_1,\dots,F_k$ have been deleted yet (since only $F_t$ was deleted), so the partial sum of previous terms equals $\sum_{i=1}^k F_i = F_{k+2}-1$. Hence $F_{k+1}\le F_{k+2}= (F_{k+2}-1)+1$, so the inequality holds.

If $k\ge t$: then among $F_1,\dots,F_k$ we are missing exactly $F_t$. Thus the partial sum of previous terms equals \[ \sum_{i=1}^k F_i - F_t = (F_{k+2}-1)-F_t. \] Adding 1 gives $F_{k+2}-F_t$. Using $F_{k+2}=F_{k+1}+F_k$, the desired inequality $F_{k+1}\le (\text{previous sum})+1$ becomes \[ F_{k+1}\le F_{k+2}-F_t = F_{k+1}+F_k-F_t. \] This holds because $k\ge t\Rightarrow F_k\ge F_t$ (Fibonacci is nondecreasing). Therefore Lemma 4.1 applies and $B$ is strongly complete.

$3$ Deleting any two Fibonacci numbers destroys strong completeness

Fix distinct indices $1\le a<b$ and delete $F_a$ and $F_b$. Let $C$ be the remaining sequence. Consider the stage just before $F_{b+1}$ in $C$: all Fibonacci numbers $F_1,\dots,F_b$ except $F_a,F_b$ are included among the previous terms.

Hence the partial sum of previous terms is \[ \sum_{i=1}^b F_i - F_a - F_b = (F_{b+2}-1) - F_a - F_b. \] Adding 1 gives \[ (F_{b+2}-1)-F_a-F_b +1 = F_{b+2}-F_a-F_b. \] Using $F_{b+2}=F_{b+1}+F_b$, this equals \[ F_{b+1}+F_b -F_a -F_b = F_{b+1}-F_a. \] Since $F_a\ge 1$, we have $F_{b+1}-F_a < F_{b+1}$, i.e. \[ F_{b+1} > (\text{previous sum})+1. \] Thus the necessary condition in Lemma 4.1(B) fails, so $C$ is not strongly complete.

This proves $1$–$3$. $\square$

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5) VERIFICATION

* Lemma 4.1: the forward direction uses a standard interval-cover induction; the backward direction isolates the specific missing number $S_k+1$ when there is a “gap”. Quantifiers are correct. * Proposition 4.3: after deleting $2^{t-1}$, the target number $2^{t-1}$ cannot be made because all sums of smaller remaining powers are $\le 2^{t-1}-1$, and any use of a larger remaining power is $\ge 2^t$. * Proposition 4.4$3$: the failure occurs exactly at the next Fibonacci number after the largest deleted index, so it does not depend on computational cutoffs. * Edge case: deleting both $F_1$ and $F_2$ leaves $2,3,5,\dots$; the proof still detects failure at $F_3$ since the “previous sum” is 0.

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6) FINAL

UNRESOLVED

(i) Strongest fully proved partial result obtained here.

* In the strong sense, powers of 2 give a complete sequence such that deleting any $\ge 1$ element(s) destroys completeness (Proposition 4.3). In particular $(m,n)=(0,1)$ works (and by monotonicity also $(0,n)$ for any $n\ge 1$). * In the strong sense, Fibonacci $1,1,2,3,5,\dots$ remains complete after deleting any one term but becomes incomplete after deleting any two terms (Proposition 4.4). Thus $(m,n)=(1,2)$ works (and by monotonicity also $(1,n)$ for any $n\ge 2$).

(ii) Exact first gap.

Under the intended weak completeness notion (all but finitely many integers representable), the problem asks for a classification of $(m,n)$. The first explicit open instance stated in the problem text is:

Does there exist a weakly complete sequence that remains weakly complete after deleting any $m=2$ terms, but is not weakly complete after deleting any $n=3$ terms?

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

1. Translate weak completeness into a checkable inequality with “finite slack”: find a necessary and sufficient criterion (or a usable sufficient criterion) analogous to Lemma 4.1 but allowing a bounded number of initial gaps. 2. Attempt a “redundant Fibonacci” construction: build a sequence satisfying a near-Fibonacci growth condition that is stable under 2 deletions but where any 3 deletions force a gap in the partial-sum inequality at some later scale. 3. Computational search for $(m,n)=(2,3)$: restrict to “structured” candidates (linear recurrences or block-repeated greedy sequences) and test the deletion property on large finite prefixes to discover patterns that suggest a proof.

(iv) What a minimal counterexample would likely look like.

If $(m,n)=(2,3)$ is impossible in the weak sense, a minimal obstruction would be a theorem of the form:

Any weakly complete sequence has a set of 3 terms whose removal still leaves infinitely many representable integers.

Equivalently, weak completeness would impose enough redundancy that some triple of deletions cannot kill it. Such a result would need a structural description of weakly complete sequences beyond the strong partial-sum criterion.