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⇱ Erdős Problem #271


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OPEN This is open, and cannot be resolved with a finite computation.
Let $A(n)=\{a_0<a_1<\cdots\}$ be the sequence defined by $a_0=0$ and $a_1=n$, and for $k\geq 1$ define $a_{k+1}$ as the least positive integer such that there is no three-term arithmetic progression in $\{a_0,\ldots,a_{k+1}\}$.

Can the $a_k$ be explicitly determined? How fast do they grow?
#271: [ErGr80,p.22]
additive combinatorics | arithmetic progressions
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It is easy to see that $A(1)$ is the set of integers which have no 2 in their base 3 expansion. Odlyzko and Stanley [OdSt78] have found similar characterisations are known for $A(3^k)$ and $A(2\cdot 3^k)$ for any $k\geq 0$ and conjectured in general that such a sequence always eventually either satisfies\[a_k\asymp k^{\log_23}\]or\[a_k \asymp \frac{k^2}{\log k}.\]There is no known sequence which satisfies the second growth rate, but Lindhurst [Li90] gives data which suggests that $A(4)$ has such growth ($A(4)$ is given as A005487 in the OEIS).

Moy [Mo11] has proved that, for all such sequences, for all $\epsilon>0$, $a_k\leq (\frac{1}{2}+\epsilon)k^2$ for all sufficiently large $k$. van Doorn and Sothanaphan have noted in the comment section that Moy's proof can be upgraded to give a fully explicit result of\[a_k\leq \frac{(k-1)(k+2)}{2}+n\]for all $k\geq 0$.

In general, sequences which begin with some initial segment and thereafter are continued in a greedy fashion to avoid three-term arithmetic progressions are known as Stanley sequences.

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This page was last edited 20 January 2026. View history

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Formalised statement? No (Create a formalisation here)
Related OEIS sequences: A005487
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Additional thanks to: Ralf Stephan, Nat Sothanaphan, and Wouter van Doorn

When referring to this problem, please use the original sources of Erdős. If you wish to acknowledge this website, the recommended citation format is:

T. F. Bloom, Erdős Problem #271, https://www.erdosproblems.com/271, accessed 2026-04-11
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