cp's OEIS Frontend

a(n)>=a((f(2n+1)-1)/2), where f is defined as in A159885. E.g., for n=4, we have a(4)=3>a(((9*3+1)/4-1)/2)=a(3)=2.

Let f be defined as in A159885. Then a(n) is the least k such that either f^k(2n+1))<2n+1 or A000120(f^k(2n+1)) < A000120(2n+1) or A006694((f^k(2n+1)-1)/2) < A006694(n).

In connection with A160198, A160267, A160322 we pose a new (3x+1)-problem: does there exist a finite number of sequences A_i(n), i=1,...,T, such that: 1) A_i(0)=0 and A_i(n)>0 for n>=1; 2) if B_i(n) denotes the least k for which A_i(n)>A_i((f^k(2n+1)-1)/2), then B(n)=min_{i=1,...,T}B_i(n)=1 for every n>=1? Note that this problem is weaker than (3x+1)-Collatz problem. Indeed, if the Collatz conjecture is true, then there exist nonnegative sequences A(n) for which A(0)=0 and A(n)>A((f(2n+1)-1)/2) for every n>=1 (see A160348). - Vladimir Shevelev, May 15 2009