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Claim. Wouter van Doorn, in a post of 6 September 2025 in the discussion thread of Problem 348 and the write-up it links, claims to settle the problem when a multiset $A$ is called complete only if its finite subset sums $P(A)$ contain every positive integer, the sense in which formal-conjectures then stated it: no such multiset exists for any $m,n$ with $2\le m<n$. If $P(A\setminus\{a_i,a_j\})=\mathbb N$ for all indices $i,j$, the post picks $s_1=a_1$ and, for $j\ge2$, $s_j=a_i$ with $i$ such that $a_{i-1}>s_1+\cdots+s_{j-1}$, and asserts that the infinite set $S=\{s_1,s_2,\ldots\}$ still has $P(A\setminus S)=\mathbb N$, so no deletion of any finite size destroys completeness. With the powers of $2$ ($m=0$) and the Fibonacci sequence ($m=1$), which the erdosproblems.com remarks give, the pairs in that sense are those with $m\le1$. The post's later edit says that the argument does not adapt to the eventual sense, since the sequence $a_n=F_n+(-1)^n$ stays eventually complete after every two-term deletion while no infinite deletion keeps it so.

Submission note. Posted to the erdosproblems.com forum by Wouter van Doorn (account Woett) on 6 September 2025:

If I understand correctly, the way this is formalised in Lean requires the multiset $A$ to have the property that the set

$$> P(A) = \left \{\sum_{n \in B} B \subseteq A \text{ finite} \right\} >$$

contains all positive integers, and a lot of texts define completeness this way. In the book by Graham and Erdős however, it is only required that $P(A)$ contains all sufficiently large integers, which is also often taken as a definition of completeness. For what it is worth, with the former interpretation that is used in the Lean formalisation I can fully settle this problem, and show that no such multiset exists for any $m, n$ with $2 \le m < n$.

Assume $A = \{a_1, a_2 \ldots, \}$ is such that $P(A \setminus {a_i, a_j }) = \mathbb{N}$ holds for all indices $i, j$. Set $s_1 = a_1$ and, for any $j \ge 2$, $s_j = a_i$ where $i$ is such that $a_{i-1} > s_1 + \ldots + s_{j-1}$. With the infinite set $S$ equal to $\{s_1, s_2, \ldots \}$, one can then check that $P(A \setminus S) = \mathbb{N}$ holds as well.

For a complete write-up, see here. Perhaps the argument can be adapted to settle the second (intended) interpretation as well.

EDIT: Come to think of it, I don't think this argument can be easily adapted to the intended interpreation, because the sequence $a_n = F_n + (-1)^n$ with $F_n$ the $n$-th Fibonacci number has the property that $P(A \setminus {a_i, a_j})$ contains all large enough integers for all $i, j$, but no infinite subset $S$ exists for which $P(A \setminus S)$ contains all large enough integers. See #346 and references for more information.

EDIT 25-09-2025: I have left a comment on the Lean forum and the formalisation has now been updated to reflect the intended meaning.

Why it is rejected. It answers a variant, not the Statement. The claim reads completeness in the strong sense, every positive integer a finite subset sum. Problem 348 reads it, as its Formulation states, in erdosproblems.com's eventual sense, all sufficiently large integers, and sets the strong sense aside, as the erdosproblems.com remarks and the post's own edit do. Strong completeness is a different property: a deletion can keep a sequence eventually complete while destroying strong completeness, and the post's edit shows that its key step fails in the eventual sense. So its mathematics answers the variant and does not settle this problem; the value answered is its answer to that variant.

Standing. Rejected, as answering a variant. The erdosproblems.com remarks credit van Doorn with the strong-sense result. This corpus has not checked the write-up.

Depends on. No page of this wiki.