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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. 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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

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. - 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(k-1)<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

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Last modified December 8 17:36 EST 2019. Contains 329865 sequences. (Running on oeis4.)