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, 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

Links

Table of n, a(n) for n=1..89.

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

Adjacent sequences: A082843 A082844 A082845 * A082847 A082848 A082849

Keyword

nonn

Author

Benoit Cloitre, Apr 14 2003

Status

approved