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%I #29 Aug 12 2022 09:17:30
%S 1,3,7,39,47,111,959,3319,7407,11967,13007,16239
%N Numbers k such that A282442(k) = ceiling(k/2) + 1.
%C All terms are odd.
%C Proof: if A282442(2m) = m + 1, then the step of length m would have to have concluded on exactly the middle step, but a phase with step-length m cannot end on the middle step because the distance from the middle step to the top/bottom of the staircase is equal to m.
%H Mathematics Stack Exchange user Sheljohn, <a href="http://math.stackexchange.com/questions/2145924">A curious sequence</a>.
%Y Cf. A282442.
%K nonn,more
%O 1,2
%A _Peter Kagey_, Feb 15 2017