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.