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Statement

With $X(h)$ as defined on the Théorème 1 page:

Conjecture 2 (p. 1720). For every integer $h\ge2$,

$$X(h)\le\frac{h(h+1)}{2}+1 .$$

The paper's grounds (p. 1720): for $2\le h\le6$ the best known upper bound on $X(h)$, and for $h=2,3$ its value, equals $h(h+1)/2+1$, and this quantity appears in several places of the proof as the limit of the method. It adds that if $X(4)$ is indeed $10$, as Li is reported to have announced without publication (p. 1719), a proof of that would have to break this bound for $h=4$.

Scope

A conjecture the paper records and does not prove or refute. It holds for $h=2,3$ by the known values, for $h=4$ by Théorème 1 ($X(4)\le11$) and for $h=5,6$ by (1.7). The paper says (p. 1763) that its Conjecture 28, on Vosper-type subsets of $\mathbb Z/n\mathbb Z$, would imply it.

Read depth. Claims checked: the statement and the sentences around it were read on the page image of p. 1720.

Source. Alain Plagne, À propos de la fonction X d'Erdős et Graham, Annales de l'Institut Fourier 54 (2004), no. 6, 1717--1767; the edition read is named on the source card.

Bears on

• Problem 336: a conjectured upper bound for the paper's function $X$, of order $h^2/2$; the paper states no relation between $X$ and the problem's $h(r)$, and the conjecture says nothing about the limit the problem asks for.