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a(n) = (1/2)*(sum of elements of n-th run) using 1 and 2 starting with 1,1.
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%I #27 Jan 03 2025 05:22:46

%S 1,1,2,1,1,1,1,2,1,1,2,1,1,2,2,1,1,2,1,1,1,1,2,1,1,2,2,1,1,1,1,2,1,1,

%T 2,2,1,1,2,1,1,2,1,1,1,1,2,1,1,2,2,1,1,1,1,2,1,1,2,1,1,2,2,1,1,2,1,1,

%U 1,1,2,2,1,1,2,1,1,1,1,2,1,1,2,2,1,1,2,1,1,2,1,1,1,1,2,1,1,2,2,1,1,1,1,2,1

%N a(n) = (1/2)*(sum of elements of n-th run) using 1 and 2 starting with 1,1.

%C We conjecture that the density of 1's in the sequence approaches 2/3 as n -> infinity. This conjecture is proved in the paper of Shallit.

%H Daniel Hoyt, <a href="/A157196/a157196.png">Visual representation of the sequence in the style of the Kolakoski sequence</a>.

%H Daniel Hoyt, <a href="/A157196/a157196.pdf">Representing the sequence visually</a>.

%H Jeffrey Shallit, <a href="https://arxiv.org/abs/2501.00784">Cloitre's Self-Generating Sequence</a>, arXiv:2501.00784 [math.CO], 2025.

%e Write the sums of elements in each run, you obtain: 2,2,4,2,2,2,2,4,2,2,4,2,2,4,4,... dividing by 2 you get: 1,1,2,1,1,1,1,2,1,1,2,1,1,2,2,... the sequence itself.

%p mx:= 1000: l:= [1$2]: a:= n-> l[n]:

%p for h from 2 while nops(l)<mx do

%p t:= 2-irem(h, 2); l:= [l[], t$(l[h]*2/t)]

%p od:

%p seq (a(n), n=1..120); # _Alois P. Heinz_, May 31 2012

%Y Cf. A000002, A157129, A336810.

%K nonn

%O 1,3

%A _Benoit Cloitre_, Feb 24 2009

%E More terms from _Alvin Hoover Belt_, May 31 2012