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A073645
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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.
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1
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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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| 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.
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]
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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(k-1)<n<P(k) for some k and P(k)-n=m(m+1)/2 for some m, else a(n)=1.
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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.)
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PROG
| (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]
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CROSSREFS
| Sequence in context: A159956 A053839 A047896 * A179542 A082846 A117373
Adjacent sequences: A073642 A073643 A073644 * A073646 A073647 A073648
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KEYWORD
| nonn
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AUTHOR
| John W. Layman (layman(AT)math.vt.edu), Aug 29 2002
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