Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).
%I #2 Mar 30 2012 18:52:53
%S 2,1,1,5,3,1,1,2,5,1,2,1,1,1,1,5,2,5,3,33,3,1,1,1,1,1,1,1,1,3,1,5,7,1,
%T 5,10,1,1,2,5,5,1,1
%N 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)}
%C Using induction, one can prove that the Collatz (3x+1)-conjecture follows from the finiteness of a(n) for every n.
%e 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.
%Y A000120 A006694 A160267 A006694 A122458 A160266 A159885 A159945 A160198
%K nonn,uned
%O 1,1
%A _Vladimir Shevelev_, May 11 2009