Let f be defined as in A159885 and f^k be the k-th iteration of f. Then a(n) is the least k for which either {A000120(f^k(2n+1))<A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<=A006694(n)} or {A000120(f^k(2n+1))<=A000120(2n+1)}&{A006694((f^k(2n+1)-1)/2)<A006694(n)}
Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n.
EXAMPLE
Beginning with n=1, we have f(2n+1)=f(3)=5. Here A000120(3)=A000120(5)=2 and A006694((3-1)/2)= A006694((5-1)/2)=1. None of values did not become less than. Therefore a(1)>1. Since f(5)=1 and A000120(1)=1 and A006694(0)=0, then a(2)=2.
