

A073645


a(1)=2 and, for all n>=1, a(n) is the length of the nth run of increasing consecutive integers with each run after the first starting with 1.


1



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, 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, 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
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OFFSET

1,1


COMMENTS

Unlike the Kolakoski sequence A000002 which is also based on runlengths and has an unpredictable, complex dynamic behavior, this sequence appears to be completely described by an easily evaluated formula.
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


LINKS

Table of n, a(n) for n=1..105.


FORMULA

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(k1)<n<P(k) for some k and P(k)n=m(m+1)/2 for some m, else a(n)=1.


EXAMPLE

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.)


PROG

(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


CROSSREFS

Sequence in context: A159956 A053839 A047896 * A294180 A179542 A082846
Adjacent sequences: A073642 A073643 A073644 * A073646 A073647 A073648


KEYWORD

nonn


AUTHOR

John W. Layman, Aug 29 2002


STATUS

approved



