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%I #7 Jun 13 2017 21:44:52
%S 2,3,1,2,3,1,1,2,1,2,3,1,1,1,2,1,1,2,1,2,3,1,1,1,1,2,1,1,1,2,1,1,2,1,
%T 2,3,1,1,1,1,1,2,1,1,1,1,2,1,1,1,2,1,1,2,1,2,3,1,1,1,1,1,1,2,1,1,1,1,
%U 1,2,1,1,1,1,2,1,1,1,2,1,1,2,1,2,3,1,1,1,1,1,1,1,2,1,1,1,1,1,1,2,1,1,1,1,1
%N a(1)=2 and, for all n>=1, a(n) is the length of the n-th run of increasing consecutive integers with each run after the first starting with 1.
%C Unlike the Kolakoski sequence A000002 which is also based on run-lengths and has an unpredictable, complex dynamic behavior, this sequence appears to be completely described by an easily evaluated formula.
%C Removing the initial 2 it remains the fixed point of the morphism: 3-->123, 2-->12, 1->1. Thus the given formulas are exact. Moreover the sequence of length of runs of 1s is given by A004736. - _Benoit Cloitre_, Feb 18 2009
%F Conjecture: Let P(k)=1 + k/3 + k^2/2 + k^3/6. Then a(n)=3 if n=P(k) for some k, a(n)=2 if P(k-1)<n<P(k) for some k and P(k)-n=m(m+1)/2 for some m, else a(n)=1.
%e a(1)=2 requires a(2)=3 to complete the first run of length 2; a(2)=3 then requires a(3)=1, a(4)=2 and a(5)=3 to complete the second run of length 3; etc. (From Labos E.)
%o (PARI) v=[2,3];for(n=2,200,for(i=1,v[n],v=concat(v,i));v);a(n)=v[n]; \\ _Benoit Cloitre_, Feb 18 2009
%K nonn
%O 1,1
%A _John W. Layman_, Aug 29 2002
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