Search: id:A073645
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%I A073645
%S A073645 2,3,1,2,3,1,1,2,1,2,3,1,1,1,2,1,1,2,1,2,3,1,1,1,1,2,1,1,1,2,1,1,2,1,2,
%T A073645 3,1,1,1,1,1,2,1,1,1,1,2,1,1,1,2,1,1,2,1,2,3,1,1,1,1,1,1,2,1,1,1,1,1,2,
%U A073645 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 A073645 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 A073645 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 A073645 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. [From Benoit 
               Cloitre (benoit784c(AT)orange.fr), Feb 18 2009]
%F A073645 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 A073645 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 A073645 (PARI) v=[2,3];for(n=2,200,for(i=1,v[n],v=concat(v,i));v);a(n)=v[n]; 
               [From Benoit Cloitre (benoit784c(AT)orange.fr), Feb 18 2009]
%Y A073645 Sequence in context: A159956 A053839 A047896 this_sequence A082846 A117373 
               A132677
%Y A073645 Adjacent sequences: A073642 A073643 A073644 this_sequence A073646 A073647 
               A073648
%K A073645 nonn
%O A073645 1,1
%A A073645 John W. Layman (layman(AT)math.vt.edu), Aug 29 2002

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