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%I #20 May 27 2021 09:52:54
%S 0,2,1,0,1,0,2,2,0,1,2,1,0,2,2,1,0,1,0,2,1,0,1,2,2,0,1,0,2,1,0,1,0,2,
%T 2,1,0,1,2,0,1,0,2,1,0,1,0,2,2,0,1,2,1,0,1,0,2,2,1,0,2,2,0,1,2,2,0,1,
%U 0,2,1,0,1,2,0,1,0,1,2,2,0,1,0,2,1,0,1,2,2,0,1,2,1,0,2,2,1,0,1,2
%N Number of 1's separating successive 2's in the Kolakoski sequence A000002.
%C After deleting 0's in this sequence it remains the bisection of Kolakoski sequence A000002(2n+1) n>=1 given by A100428.
%C This is because A100428 gives the lengths of runs of 1's in Kolakoski sequence. - _Jean-Christophe Hervé_, Oct 14 2014
%C The Kolakovski sequence can be obtained back (except the initial 1) by the following substitution rules: insert 2 between two successive nonzero values and 0 -> 22, 1 -> 1, 2 -> 11. - _Jean-Christophe Hervé_, Oct 14 2014
%F a(n) = A078649(n+1)-A078649(n)-2.
%e The Kolakoski sequence begins with 122112122122, thus this one begins 0, 2, 1, 0, 1, 0. - _Jean-Christophe Hervé_, Oct 14 2014
%Y Cf. A000002, A078649, A100428, A248806.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Feb 07 2009
%E Better name from _Jean-Christophe Hervé_, Oct 15 2014
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