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Claim. The answer to Problem 867 is no. The claimed result is Theorem 2.1 of D. Coppersmith and S. Phillips, On a question of Erdős on subsequence sums: for every $N$ there is a set $A\subseteq\{1,\ldots,N\}$ with

$$\lvert A\rvert\ge\frac{13}{24}N-O(1)$$

in which no sum of two or more consecutive members is a member, so $\lvert A\rvert-N/2$ grows like $N/24$ and the bound $N/2+O(1)$ fails. The proof opens from Freud's four-block construction and improves its density $\tfrac{19}{36}$ to $\tfrac{13}{24}$ with the paper's own Table 1; the $O(1)$ is the paper's, which prints no count of the removed boundary elements. The same paper's Theorem 3.7 bounds every such set by $\tfrac23N-\lfloor N/512\rfloor+3\log_4N-\tfrac12$ members, so the maximal density lies in $[\tfrac{13}{24},\tfrac23-\tfrac1{512}]$; its exact value is the paper's Open Question 1 and is not the site's question. Freud's note reports the paper's upper bound as $\tfrac23-\tfrac1{3584}$, a figure the published paper does not print. Both theorems are stated as the source card records them; the proof of Theorem 2.1 is followed but not independently checked.

Depends on. Nothing in this wiki: the proof restates Freud's blocks in its own terms as its starting point, and Table 1 and its verification are the paper's own.

Acceptance. Refereed publication: SIAM J. Discrete Math. 9 (1996), no. 2, 173--177, doi:10.1137/S0895480193244139, received February 1993 and accepted in revised form April 1995, in the May 1996 issue (Crossref record, 2026-10-07; the day is filled to the first of the month). Reviewed: the site's curator, Thomas Bloom, credits [CoPh96] with the best known bounds on the maximal size of such a set, and the thread comment of 2025-09-02 that led to the disproved label cited the paper, by its DOI, for the better lower bound $\tfrac{13}{24}$ and the upper bound $\tfrac23-\tfrac1{512}$. The site's label and its Lean qualifier rest on Freud's construction, recorded on its own claim page.