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SOLVED - $10000 For any$C>0$are there infinitely many$n$such that $p_{n+1}-p_n> C\frac{\log\log n\log\log\log\log n}{(\log\log \log n)^2}\log n?$ The peculiar quantitative form of Erdős' question was motivated by an old result of Rankin [Ra38], who proved there exists some constant$C>0$such that the claim holds. Solved by Maynard [Ma16] and Ford, Green, Konyagin, and Tao [FGKT16]. The best bound available, due to all five authors [FGKMT18], is that there are infinitely many$n$such that $p_{n+1}-p_n\gg \frac{\log\log n\log\log\log\log n}{\log\log \log n}\log n.$ The likely truth is a lower bound like$\gg(\log n)^2$. In [Er97c] Erdős revised the value of this problem to \$5000 and reserved the \$10000 for a lower bound of$>(\log n)^{1+c}$for some$c>0$. See also [687]. OPEN -$10000
Let $r_k(N)$ be the largest possible size of a subset of $\{1,\ldots,N\}$ that does not contain any non-trivial $k$-term arithmetic progression. Prove an asymptotic formula for $r_k(N)$.
Erdős remarked this is 'probably unattackable at present'. In [Er97c] Erdős offered \$1000, but given that he elsewhere offered \$5000 just for (essentially) showing that $r_k(N)=o_k(N/\log N)$, that value seems odd. In [Er81] he offers \$10000, stating it is 'probably enormously difficult'. The best known upper bounds for$r_k(N)$are due to Kelley and Meka [KeMe23] for$k=3$, Green and Tao [GrTa17] for$k=4$, and Leng, Sah, and Sawhney [LSS24] for$k\geq 5$. An asymptotic formula is still far out of reach, even for$k=3\$.