OFFSET
1,2
COMMENTS
Let x(1)=1 x(2)=n; x(k)=x(k-1)+x(k-2) if x(k-1) and x(k-2) have opposite parities; x(k)=abs(x(k-1)-x(k-2))/2 otherwise. Conjecture : for any n x(k) reaches a cycle among 2 cycles : (1;1;0) and (1;2;3;5). Sequence gives values of n such that (1;1;0) is reached.
FORMULA
a(n)/n seems greater than 2 for n large enough and if limit a(n)/n exists, this limit should be > 2.5. Does a(n)/n = O(log(log(n)))?
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jan 02 2003
STATUS
approved