OFFSET
1,2
COMMENTS
Although the behavior of the partial sums of the Kolakoski sequence (A054353) is mysterious, this sequence is much easier to handle.
FORMULA
a(n)=(3/2)n+O(1). More precisely, let b(n)=3*n-2*a(n); then b(n) satisfies the following recurrence modulo 12: b(n)=1,2,1,0,1,2,3,4,3,2,1 for n=1,2,3,4,5,6,7,8,9,10,11. Then for k>=1 we have b(12k)=b(4k), b(12k+1)=b(4k+1), b(12k+2)=b(4k+2), b(12k+2)=b(4k+2), b(12k+3)=b(4k+2)-1, b(12k+4)=b(4k+2)-2, b(12k+5)=b(4k+2)-1, b(12k+6)=b(4k+2), b(12k+7)=4-b(4k+3), b(12k+8)=4-b(4k+4), b(12k+9)=4-b(4k+3), b(12k+10)=4-b(4k+2), b(12k+11)=b(4k+3).
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Sep 29 2013
STATUS
approved