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A082846
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a(1) = 1, a(2) = 2, a(3) = 3; then a(a(1)+a(2)+..+a(k)) = a(k) and fill the "holes" using the rule : a(m) = 1 if a(m-1) = 3, a(m) = 2 if a(m-1) = 1, a(m) = 3 if a(m-1) = 2.
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1, 2, 3, 1, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 3, 2, 3, 1, 3, 1, 2, 3, 3, 1, 2, 2, 3, 2, 3, 1, 3, 1, 2, 3, 1, 3, 1, 2, 3, 3, 1, 2, 2, 3, 1, 3, 1, 2, 3, 1, 2, 2, 3, 2, 3, 1, 3, 1, 2, 3, 1, 3, 1, 2, 3, 3, 1, 2, 2, 3, 1, 3, 1, 2, 3, 3, 1, 2, 2, 3, 1, 3, 1, 2, 3, 1, 2, 2
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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FORMULA
| It seems that limit n ->infinity (1/n)*sum(k=1, n, a(k))= C >2 (C=2.07...)
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EXAMPLE
| a(1)+a(2)+a(3)=6 so a(6)=a(3)=3. Since a(3)=3 a(4)=1 therefore a(5)=2 and sequence begins : 1,2,3,1,2,3,
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CROSSREFS
| Cf. A082847.
Sequence in context: A047896 A073645 A179542 * A117373 A132677 A010882
Adjacent sequences: A082843 A082844 A082845 * A082847 A082848 A082849
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KEYWORD
| nonn
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AUTHOR
| Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 14 2003
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