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Statement

Printed p. 58 = PDF p. 6, read on the page image. With $f^{(3)}(n;k,s)$ the least number of triples that forces, in every $3$-graph on $n$ vertices, some $k$ vertices spanning at least $s$ triples (the p. 55 definition, recorded on theorem_section_4), the authors close their list of known bounds for $r=3$ with: "Perhaps the most interesting question we were unable to answer is whether $f^{(3)}(n;6,3)=o(n^2)$." Equivalently: does every $3$-graph on $n$ vertices in which no six vertices span at least three triples have $o(n^2)$ triples?

Context in the paper. The list on pp. 57--58, reproduced from Theorem 4 of the authors' earlier paper (their [4]), gives only the lower bound $c_9n^{3/2}<f^{(3)}(n;6,3)$ for this function, and the Theorem of Section 4 with $r=3$, $k=6$, $s=3$ gives the same exponent $3/2$. The paper proves nothing further about the question.

Source. W. G. Brown, P. Erdős and V. T. Sós, Some extremal problems on $r$-graphs, in New Directions in the Theory of Graphs (Proc. Third Ann Arbor Conf., Univ. Michigan, 1971), Academic Press, New York (1973), 53--63, p. 58; the edition is identified in the source digest.

Proof pointer

A question; the paper gives no answer.

Bears on

• Problem 716: the problem's question as worded, with the site's $\mathrm{ex}_3(n,\mathcal F)$ equal to $f^{(3)}(n;6,3)-1$; the answer yes is credited on the problem's claim page to Ruzsa and Szemerédi, not to this paper.
• Problem 1178: for $r=e=3$ the question asks whether six vertices suffice in the definition of $d_3(3)$, that is whether $d_3(3)\le6$; with the Theorem of Section 4 at $k=5$, $s=3$ (exponent $2$), and since three triples on fewer vertices are only harder to find, a yes answer gives $d_3(3)=6$, the conjecture's value at $r=e=3$ (a reading made here).