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    $\begingroup$ If we modify the definition so that we only exclude weakly increasing arithmetic progressions, i.e., $a(j),a(j+k),a(j+2k)$ are in arithmetic progression (with $k\geq 1$) and $a(j)\leq a(j+k)\leq a(j+2k)$, then it is easy to see that terms equal to 1 are indexed by oeis.org/A236246. This suggests looking some more at this modified sequence. Could it coincide with the original sequence? I haven't tried to check this experimentally. $\endgroup$ Commented Aug 17, 2019 at 23:25
  • $\begingroup$ @RichardStanley Let $a$ be the former sequence, let $b$ be the modified one, and assume that the terms indexed by $1$ are unchanged. Observe that $a(27) = 9$ and $a(59) = 5$, but there is no $i \ge 0$ and $k>0$ such that $a(i)=a(i+k)=1$ and $i+2k = 59+59-27 = 91$, so $b(91) = 1$, but $a(91)=2$, contradiction. $\endgroup$ Commented Aug 18, 2019 at 12:29
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    $\begingroup$ Oops, I foolishly assumed that A236246 was actually oeis.org/A003278 without checking. It is clear that the indices for which $b(n)=1$ coincide with A003278. Perhaps it would be interesting to consider $b(n)$ for its own sake. $\endgroup$ Commented Aug 19, 2019 at 14:36
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    $\begingroup$ I wonder if for the question whether every natural occurs at all in this sequence, even though it seems intuitively clear, it is easier to prove that the answer is positive. $\endgroup$ Commented Aug 20, 2019 at 20:13
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    $\begingroup$ A better visualization (see this question on ask.sagemath.org, especially the last comments) shows that the "empty" regions are not empty, but simply too sparse to get a visible representation on the rendered graph. $\endgroup$ Commented Aug 25, 2019 at 15:59