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A082846
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.
1
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
OFFSET
1,2
FORMULA
It seems that limit n ->infinity (1/n)*sum(k=1, n, a(k))= C >2 (C=2.07...)
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,
CROSSREFS
Cf. A082847.
Sequence in context: A073645 A294180 A179542 * A117373 A132677 A010882
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Apr 14 2003
STATUS
approved